Turbomachines

TURBOMACHINES

TURBOMACHINES B. U. Pai Formerly Special Officer, VTU, Belgaum, Karnataka Formerly Principal, SDM College of Engineering & Technology, Dharwad, Karnataka Formerly Professor and Head, Department of Mechanical Engineering, SDM College of Engineering & Technology, Dharwad, Karnataka

TURBOMACHINES Copyright © 2013 by Wiley India Pvt. Ltd., 4435-36/7, Ansari Road, Daryaganj, New Delhi-110002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particular results, and the advice and strategies contained herein may not be suitable for every individual. Neither Wiley India nor the author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Wiley is not associated with any product or vendor mentioned in this book. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionopolis Walk #07-01 Solaris, South Tower, Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2013 ISBN: 978-81-265-3955-0 ISBN: 978-81-265-8046-0 (ebk) www.wileyindia.com Printed at:

Preface I have taught, over the last 30–35 years, different subjects like Thermodynamics, Heat Transfer, IC Engines, Turbomachines, etc. From the beginning, I have been thoroughly involved with my subjects. When I taught Turbomachines, it was a different challenge, something more exciting than Thermodynamics. I could easily identify the opportunity of presenting to my class the different topics of turbomachines in various ways, going beyond the available texts, adding something here, something there. The present book is the result of the compilation of my lecture notes that were prepared when I taught this subject. Quite a few books were referred to, while preparing them, as also during transforming them into this book. Such reference books have been mentioned in the bibliography. Syllabi of various universities have been considered during the compilation and refining of the lecture notes. The contents of this book meet the requirements of the syllabi of many universities. But more important, the contents of this book have been presented in such a way that they offer materials for postgraduate studies also; a discerning student can look at the topics as a source of optimization problems for his or her own work. I was quite fortunate to have many students who could participate in discussions with me to get more and more information, sometimes beyond the prescribed syllabus. Quite often, the explanations in the classroom went beyond the texts; sketches and diagrams were liberally used to supplement the verbal instructions. This book contains many such illustrations (e.g., Figs. 3.5, 6.14, 6.30, 7.5, etc.), where, the primary motive is the understanding of the working principles. There are a few topics with which the students have reasons to be a bit uncomfortable, such as velocity triangles, impulse and reaction, degree of reaction, dynamic action, and so on. This book is intended to offer a firm foundation for the understanding of those topics and the basic concepts with absolute clarity. A sound vocabulary, which is already present in the literature, is rediscovered (not invented). The explanations have been provided in a simple and clear language.

Organization of the Book The nine chapters in this book have been systematically organized as follows: 1. Chapter 1: The purpose of this chapter is some sort of preparation for the subject matter of this book. It is like getting into the mood for an applied subject, after going through, in earlier semesters, the courses in Thermodynamics and Fluid Mechanics, and then taking off to a level of application of the principles of those subjects. 2. Chapter 2: This chapter is meant to throw more light on Thermodynamics, in an area where the basic course in Thermodynamics would not have reached or the applied course in Thermodynamics would have crossed over. This area is precisely the area where the turbomachines operate. 3. Chapter 3: This chapter is devoted first to the topic, “Velocity triangles,” and then to that one important fundamental equation of the turbomachines, the Euler’s turbine equation. No amount of emphasis is too much for the velocity triangles. Even if some extra time is invested on the velocity triangles, it is worth the time and the trouble. And then, the Euler’s turbine equation is like the pulse of a person, that goes on and on, to each and every topic in the remaining chapters.

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Turbomachines

4. Chapter 4: This chapter takes an insight into some general analysis of the turbomachines. Basically this analysis has been arranged in two main categories: radial flow machines and axial flow machines. The analysis of radial flow machines applies to centrifugal pumps and centrifugal compressors. Moreover, the analysis of axial flow machines applies to the axial flow turbines, compressors, and pumps. Since the treatments are common to such machines in two parts, these are grouped together in this chapter and therefore this gives a very good foundation to the various other turbomachines. In fact, in later chapters, this Chapter 4 is being quoted often. Once the first four chapters are meticulously gone through sequentially, the remaining chapters on Steam Turbines (Chapter 5); Hydraulic Turbines (Chapter 6); Centrifugal Pumps (Chapter 7); Fans, Blowers, and Compressors (Chapter 8); and Power-Transmitting Turbomachines (Chapter 9) can be studied very easily, in any order. The topics are straight forward, unfolding in a smooth and easy flow. Plenty of solved examples have been provided at appropriate places in the chapters, and comments have been added at the end of the solutions. It is felt that these comments go a long way in getting the feel of the subject. Moreover, multiple-choice questions and review questions have been judiciously selected and added at the end of each chapter, to enhance the quality of instructions. The Project-Oriented Questions are meant to initiate the students to explore new dimensions of the subject. When such projects, albeit their sizes, are undertaken by them, whether in groups or individually, the students are bound to get better insight and better overall perception of the subject. Detailed solutions available on http://www.wileyindia.com/turbomachines.html If there are a hundred tittles in the subject of Thermodynamics, the subject matter cannot be different in different books. What can be different is the way the subject is presented, the way the examples are stated, the language used, the graphics, and almost every aspect of the book. A few books (not just one or two) generate a pleasant feeling, create a genuine interest, some delight, some fascination. For different persons, it could be different books. I sincerely feel that this book is one among such books on the subject Turbomachines.

B. U. PAI

Acknowledgments I wish to take this opportunity to acknowledge the strength I have derived from the noble life lead by my parents, Sri Subraya Pai and Smt. Subhadra S. Pai. The totality of my being has been molded by my sisters Kasturi, Revathi, Pushpa, Vasanthi, and Jayanthi and by my brothers Namdev, Gopalkrishna, and Seetharam, with all their affection. Some of them are over there, on the Eastern sky, wishing me well-being and success every morning and leading me to light. I wish to express my gratitude to all of them. During the preparation of the background material of this book, I got all the support from the management of Sri Dharmasthala Manjunatheshwara College of Engineering and Technology, Dharwad. I wish to acknowledge, with respectful regards and gratitude, the encouragement of Dr. D. Veerendra Heggade, President, SDME Society, and the support extended by the management, as also all my colleagues at the college. The support of the authorities and my colleagues at VTU is thankfully acknowledged. The book would not have seen the light of the day, but for the efforts of my son Madhusudan and daughter Parineetha and the forbearance of my wife Vasanthi. Words of gratitude are but a small speck in my feelings. I thank Sri. Janardan Joshi and Smt. Tanuja Heddurshetti, who took lots of pain in typing the manuscript with meticulous care of details. Special thanks to Mr. Praveen Settigere (Sr. Manager Acquisitions, Wiley India) for his unstinted coordination during various stages of the book. The editorial support from Ms. Meenakshi Sehrawat (Executive Editor, Wiley India) and Rupnarayan Das (Associate Editor, Wiley India) needs special mention for taking lots of care to put the manuscript in a correct form and giving valuable suggestions while preparing this book. I’d also like to thank Mr. Rakesh Poddar (Production Editor) for meticulously managing the production-related jobs. Finally, the acknowledgements are due to all my teachers and to all my students, especially those students who reposed unparalleled faith on me, those who engaged me in useful discussions and gave way for the enjoyment of the subject in all its rich flavor of the variety and contents.

B. U. PAI

Nomenclature

A A0 A1 A2 B B0 B1 B2 C cb cp cv oC D D0 D1 D2 Dp d d E E1 E2 e e1 e2 F g gc H HE Hm

(Symbols are also explained, when they are introduced in the text.) Area of flow Area of flow through guide vanes Area of flow at the inlet to rotor Area of flow at the outlet from the rotor Breadth or width of the blades Breadth of guide vanes Breadth at the inlet of the rotor Breadth at the outlet from the rotor Constant of proportionality Coefficient of frictional losses Specific heat of fluid at constant pressure Coefficient of velocity for nozzles Unit of temperature, degree Celsius (dimension) Diameter Diameter of guide-vane ring Diameter of rotor at inlet Diameter of rotor at outlet (D1, D2, D3 … Reference diameters of rotors of different machines in model studies, Dm and Dp are also used for model and prototype in model studies) Diameter of penstock pipes Jet diameter in Pelton turbines Runner hub diameter in Kaplan turbines Energy of fluid Energy of fluid at inlet Energy of fluid at outlet Specific energy Specific energy at inlet Specific energy at outlet Force Acceleration due to gravity, 9.81 m/s2 Constant, y 1(kg – m)/(N – s2), an identity Head Head, as identified by Euler’s equation Manometric head

x

Hn h h0 h01 h02 h1 h2 ha hd hf hfs hfd hs hv J K kg L L0 m m. N N

N1 Ns Nsp Nst n P Pa Pa Pn Pr Ps Pu p po

Turbomachines

(H1, H2, H3. Reference heads of different machines in model studies; Hm and Hp are also used for model and prototype in model studies) Net head Enthalpy of fluids Stagnation enthalpy Stagnation enthalpy at inlet Stagnation enthalpy at outlet Static enthalpy at inlet Static enthalpy at outlet Head equivalent to atmospheric pressure Delivery head Friction head (loss) Friction head on suction side Friction head on delivery side Suction head Vapor pressure head Unit of energy, Joules (equal to m – N) (dimension) Unit of absolute temperature, Kelvin (dimension) Unit of mass, kilogram (dimension) Length of pipe Length of guide-vanes Unit of length, meter (dimension) Mass flow rate, kg/s Unit of force, newton (dimension) Revolutions per minute (rpm), speed of rotors (N1, N2, N3, … rpms of different machines in model studies; Nm and Np are also used for rpm of model and prototype in model studies) Unit speed Specific speed Specific speed of pump Specific speed of turbine Revolutions per second (rps) Power, (watts or kilowatts) Available power Unit of pressure, Pascal (equal to N/m2) (dimension) Net power Rotor power; also, overall pressure ratio in multistage units Shaft power Useful power (P1, P2 ... Powers of different machines in model studies; Pm and Pp are also used for powers of model and prototype in model studies) Pressure of fluid Stagnation pressure

nomenclature

p01 p02 p1 p2 Q Q1 R s T T0 T01 T02 T1 T2 U U1 U2 V V1 V2 Vd Vf1 Vf2 Vs Vr Vr1 Vr2 Vru Vu1 Vu2 v W Wa WE Wi Wst w z z1 z2

xi

Stagnation pressure at inlet Stagnation pressure at outlet Static pressure at inlet Static pressure at outlet Volume flow rate of fluid, m3/s Unit flow (Q1, Q2, Q3 … Flow rates of different machines in model studies; Qm and Qp are also used for flow rates of “model” and “prototype” in model studies) Degree of reaction (Also, variable radius in Kaplan turbine) Unit of time, second (dimension) Temperature of fluid on absolute scale (T is also Thrust, newtons, in Eq. 3.2) Stagnation temperature Stagnation temperature at inlet Stagnation temperature at outlet Static temperature at inlet Static temperature at outlet Peripheral velocity of blades, m/s Blade velocity at inlet Blade velocity at outlet Absolute velocity of fluid, m/s Absolute velocity of fluid at inlet Absolute velocity of fluid at outlet Velocity of fluid in delivery pipe Flow component of fluid velocity at inlet Flow component of fluid velocity at outlet Velocity of fluid in suction pipe Relative velocity of fluid, with respect to blade Relative velocity at inlet Relative velocity at outlet Component of relative velocity along U Component of fluid velocity along U, at inlet Component of fluid velocity along U, at outlet Specific volume, m3/kg Specific work Actual specific work Specific work, as per Euler’s equation Ideal specific work Stage-specific work Specific weight of fluid, N/m3 Height above datum Height of inlet above datum Height of outlet above datum

xii

@ @1 @2 A A1 A2 F $ ‹ E Eo E1 G Gh Gm Gn G0 Gr Gs Gv Gco Gdt Gpc Gpt Gst L O1 O2 O3 Q X Xo X1 S V 7 7p 7t A0 A1

Turbomachines

Angle of absolute velocity of fluid relative to blade velocity (Also used as a temporary group in Section 2.9) Angle of fluid velocity relative to U at inlet Angle of fluid velocity relative to U at outlet Blade angle: Angle of tangent to blade profile with U Blade angle at inlet Blade angle at outlet Ratio of specific heats of fluid Change in … Utilization factor of turbines Speed ratio, U/V1 (also, F is speed coefficient) Speed coefficient for guide blades Speed coefficient for rotor blades at inlet Efficiency Hydraulic efficiency Mechanical efficiency Nozzle efficiency Overall efficiency Rotor efficiency Isentropic efficiency Volumetric efficiency carryover efficiency Draft-tube efficiency Polytropic efficiency of compressor Polytropic efficiency of turbine Stage efficiency Efficiencies are also specified for compressor (suffix c) and turbine (suffix t) with subscripts t for total and s for static, in as many combinations, and explained in the context. Coefficient of slip Flow coefficient (model studies) Head coefficient (model studies) Power coefficient (model studies) Density of fluid, kg/m3 Coefficient of pressure Flow coefficient for guide vanes (Francis, Kaplan, Propeller turbines) Flow coefficient for runner (at inlet) (Francis, Kaplan, Propeller turbines) Torque Angular velocity Shape number or dimensionless specific speed Shape number of pump, Eq. 1.27 Shape number of turbine, Eq. 1.28A Area of flow Area of flow through guide vanes Area of flow at the inlet to rotor

nomenclature

A2 B B0 B1 B2 C cb cp cv oC D D0 D1 D2 Dp d d E E1 E2 e e1 e2 F g gc H HE Hm Hn h h0 h01 h02 h1 h2 ha hd hf

xiii

Area of flow at the outlet from the rotor Breadth or width of the blades Breadth of guide vanes Breadth at the inlet of the rotor Breadth at the outlet from the rotor Constant of proportionality Coefficient of frictional losses Specific heat of fluid at constant pressure Coefficient of velocity for nozzles Unit of temperature, degree Celsius (dimension) Diameter Diameter of guide-vane ring Diameter of rotor at inlet Diameter of rotor at outlet (D1, D2, D3 … Reference diameters of rotors of different machines in model studies, Dm and Dp are also used for model and prototype in model studies) Diameter of penstock pipes Jet diameter in Pelton turbines Runner hub diameter in Kaplan turbines Energy of fluid Energy of fluid at inlet Energy of fluid at outlet Specific energy Specific energy at inlet Specific energy at outlet Force Acceleration due to gravity, 9.81 m/s2 Constant, y 1(kg – m)/(N – s2), an identity Head Head, as identified by Euler’s equation Manometric head (H1, H2, H3. Reference heads of different machines in model studies; Hm and Hp are also used for model and prototype in model studies) Net head Enthalpy of fluids Stagnation enthalpy Stagnation enthalpy at inlet Stagnation enthalpy at outlet Static enthalpy at inlet Static enthalpy at outlet Head equivalent to atmospheric pressure Delivery head Friction head (loss)

xiv

hfs hfd hs hv J K kg L L0 m m. N N

N1 Ns Nsp Nst n P Pa Pa Pn Pr Ps Pu p po p01 p02 p1 p2 Q Q1 R s T T0

Turbomachines

Friction head on suction side Friction head on delivery side Suction head Vapor pressure head Unit of energy, Joules (equal to m – N) (dimension) Unit of absolute temperature, Kelvin (dimension) Unit of mass, kilogram (dimension) Length of pipe Length of guide-vanes Unit of length, meter (dimension) Mass flow rate, kg/s Unit of force, newton (dimension) Revolutions per minute (rpm), speed of rotors (N1, N2, N3, … rpms of different machines in model studies; Nm and Np are also used for rpm of model and prototype in model studies) Unit speed Specific speed Specific speed of pump Specific speed of turbine Revolutions per second (rps) Power, (watts or kilowatts) Available power Unit of pressure, Pascal (equal to N/m2) (dimension) Net power Rotor power; also, overall pressure ratio in multistage units Shaft power Useful power (P1, P2 ... Powers of different machines in model studies; Pm and Pp are also used for powers of model and prototype in model studies) Pressure of fluid Stagnation pressure Stagnation pressure at inlet Stagnation pressure at outlet Static pressure at inlet Static pressure at outlet Volume flow rate of fluid, m3/s Unit flow (Q1, Q2, Q3 … Flow rates of different machines in model studies; Qm and Qp are also used for flow rates of “model” and “prototype” in model studies) Degree of reaction (Also, variable radius in Kaplan turbine) Unit of time, second (dimension) Temperature of fluid on absolute scale (T is also Thrust, newtons, in Eq. 3.2) Stagnation temperature

nomenclature

T01 T02 T1 T2 U U1 U2 V V1 V2 Vd Vf1 Vf2 Vs Vr Vr1 Vr2 Vru Vu1 Vu2 v W Wa WE Wi Wst w z z1 z2 @ @1 @2 A A1 A2 F $ ‹ E Eo E1

xv

Stagnation temperature at inlet Stagnation temperature at outlet Static temperature at inlet Static temperature at outlet Peripheral velocity of blades, m/s Blade velocity at inlet Blade velocity at outlet Absolute velocity of fluid, m/s Absolute velocity of fluid at inlet Absolute velocity of fluid at outlet Velocity of fluid in delivery pipe Flow component of fluid velocity at inlet Flow component of fluid velocity at outlet Velocity of fluid in suction pipe Relative velocity of fluid, with respect to blade Relative velocity at inlet Relative velocity at outlet Component of relative velocity along U Component of fluid velocity along U, at inlet Component of fluid velocity along U, at outlet Specific volume, m3/kg Specific work Actual specific work Specific work, as per Euler’s equation Ideal specific work Stage-specific work Specific weight of fluid, N/m3 Height above datum Height of inlet above datum Height of outlet above datum Angle of absolute velocity of fluid relative to blade velocity (Also used as a temporary group in Section 2.9) Angle of fluid velocity relative to U at inlet Angle of fluid velocity relative to U at outlet Blade angle: Angle of tangent to blade profile with U Blade angle at inlet Blade angle at outlet Ratio of specific heats of fluid Change in … Utilization factor of turbines Speed ratio, U/V1 (also, F is speed coefficient) Speed coefficient for guide blades Speed coefficient for rotor blades at inlet

xvi

G Gh Gm Gn G0 Gr Gs Gv Gco Gdt Gpc Gpt Gst L O1 O2 O3 Q X Xo X1 S V 7 7p 7t

Turbomachines

Efficiency Hydraulic efficiency Mechanical efficiency Nozzle efficiency Overall efficiency Rotor efficiency Isentropic efficiency Volumetric efficiency carryover efficiency Draft-tube efficiency Polytropic efficiency of compressor Polytropic efficiency of turbine Stage efficiency Efficiencies are also specified for compressor (suffix c) and turbine (suffix t) with subscripts t for total and s for static, in as many combinations, and explained in the context. Coefficient of slip Flow coefficient (model studies) Head coefficient (model studies) Power coefficient (model studies) Density of fluid, kg/m3 Coefficient of pressure Flow coefficient for guide vanes (Francis, Kaplan, Propeller turbines) Flow coefficient for runner (at inlet) (Francis, Kaplan, Propeller turbines) Torque Angular velocity Shape number or dimensionless specific speed Shape number of pump, Eq. (1.27) Shape number of turbine, Eq. (1.28)

Contents Preface Acknowledgments Nomenclature

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.8 1.9

1.10 1.11 1.12 1.13 1.14 1.15

Basics of Turbomachines Learning Objectives Introduction Classification of Fluid Machines Turbomachines Turbomachines and Positive Displacement Machines Classification of Turbomachines 1.5.1 Examples Units and Dimensions Energy of Fluids 1.7.1 Pressure Energy 1.7.2 Kinetic Energy 1.7.3 Potential Energy 1.7.4 Thermal Energy or Enthalpy 1.7.5 Head Energy Application of the First Law of Thermodynamics Application of the Second Law of Thermodynamics 1.9.1 Efficiencies of Turbomachines 1.9.2 Power-Flow Diagrams in Head Units Model Studies Dimensional Analysis Unit and Specific Quantities Non-Dimensional Parameters and Their Significance Effect of Reynolds Number Specific Speed Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

v vii ix

1 1 1 1 2 3 4 5 8 8 9 9 9 9 11 12 14 15 18 23 24 29 32 33 33 41 41 42 43 43 44 45 46

xviii

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7

2.8 2.9 2.10

3 3.1 3.2 3.3 3.4

3.5 3.6 3.7

Turbomachines

Thermodynamics of Fluid Flow Learning Objectives Introduction Static and Stagnation States Thermodynamics of Turbomachine Processes Isentropic Compression Process Isothermal Compression Process Isentropic Expansion Process Overall Isentropic Efficiency versus Stage Efficiency 2.7.1 Pre-heat Effect in Multi-stage Compressor 2.7.2 Re-heat Effect in Multi-stage Turbines Infinitesimal-Stage or Small-Stage Efficiency or Polytropic Efficiency Reheat Factor for Expansion Processes Overall Isentropic Efficiency versus Finite-Stage Efficiency:Compression and Expansion Processes 2.10.1 Compression Process 2.10.2 Expansion Process or Turbine Process Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

Energy Exchange in Turbomachines Learning Objectives Introduction Velocity Triangles Basic Equations: Linear Momentum Equation, Impulse Momentum Equation, Moment of Momentum Equation, and Euler Turbine Equation Alternate Form of the Euler Turbine Equation 3.4.1 Components of Energy Transfer 3.4.2 Energy Equation of Relative Velocities Impulse and Reaction Utilization Factor of Turbines Speed Ratio Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

47 47 47 47 50 50 55 56 60 62 63 66 70 73 73 75 78 78 79 80 81 82 82 83

85 85 85 85 97 99 101 101 102 105 107 117 117 118 118 120 120 121 122

Contents

4 4.1 4.2 4.3

4.4

4.5

5 5.1 5.2

5.3

5.4

5.5 5.6

General Analysis of Turbomachines Learning Objectives Introduction General Analysis of Radial Flow Machines Radial Flow Machines (Centrifugal Pumps, Centrifugal Blowers, and Centrifugal Compressors): Velocity Triangles 4.3.1 Effect of Blade Outlet Angle on Energy Transfer 4.3.2 Effect of the Blade Outlet Angle on Reaction 4.3.3 Effect of the Blade Exit Angle on the Performance 4.3.4 Flow Analysis in Impeller Blades: Slip and Slip Factor or the Coefficient of Slip 4.3.5 Losses in Impeller Blade Passages 4.3.6 Characteristic Curves: Head–Capacity Relationship 4.3.7 Effect of Prewhirl Axial Flow Machines 4.4.1 Reaction in Axial Flow Machines 4.4.2 Effect of Blade Angles on the Specific Work and Degree of Reaction: Turbines 4.4.3 Effect of Blade Angles on the Specific Work and Degree of Reaction: Compressors 4.4.4 Flow Analysis in Runner Blades: Slip and Slip Factor or Coefficient of Slip 4.4.5 Losses in Blade Passages Fan Laws Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

Steam Turbines Learning Objectives Introduction Classification of Steam Turbines 5.2.1 Single-Stage Impulse Turbine: De Laval Turbine 5.2.2 Single-Stage Reaction Turbine Compounding of Steam Turbines 5.3.1 Pressure Compounding 5.3.2 Velocity Compounding 5.3.3 Pressure–Velocity Compounding Analysis 5.4.1 Rateau Stages 5.4.2 Parsons Stages 5.4.3 Curtis Stage Mass Flow Rate and Blade Heights Efficiencies

xix

123 123 123 123 124 125 129 135 135 137 139 141 141 143 150 151 155 155 155 156 156 157 157 158 159 160 161

163 163 163 164 164 168 169 170 171 172 172 172 178 184 195 196

xx

5.7 5.8

5.9

6 6.1 6.2 6.3

6.4

6.5 6.6 6.7

6.8

6.9

Turbomachines

5.6.1 Nozzle Efficiency 5.6.2 Carryover Efficiency 5.6.3 Stator Efficiency or Blade Passage Efficiency 5.6.4 Rotor Efficiency or Vane Efficiency 5.6.5 Stage Efficiency Reheat Factor Governing of Steam Turbine 5.8.1 Governing by Nozzle Control 5.8.2 Governing by Throttle Control Performance Characteristics of Steam Turbines Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

Hydraulic Turbines Learning Objectives Introduction Classification of Hydraulic Turbines 6.2.1 Selection of Hydraulic Turbines Pelton Turbine 6.3.1 Constructional Details of Pelton Turbine 6.3.2 Analysis of the Pelton Turbine 6.3.3 Efficiencies of Pelton turbine 6.3.4 Design Parameters of Pelton Turbine Francis Turbine 6.4.1 Constructional Features of Francis Turbine 6.4.2 Analysis of the Francis Turbine 6.4.3 Efficiencies of Francis turbine 6.4.4 Design Parameters of Francis Turbine Draft Tube 6.5.1 Analysis of Draft Tube Cavitation Kaplan Turbine and Propeller Turbine 6.7.1 Constructional Features of Kaplan and Propeller Turbines 6.7.2 Analysis of Kaplan and Propeller Turbines 6.7.3 Efficiencies of Kaplan or Propeller Turbine 6.7.4 Design Parameters of Kaplan and Propeller Turbines Governing of Hydraulic Turbines 6.8.1 Parts of the Governing System 6.8.2 Working of the Governing System Characteristics of the Hydraulic Turbine Keywords

196 196 197 197 197 203 208 208 208 209 212 212 212 213 214 215 216 217

219 219 219 220 220 222 222 224 226 227 234 234 236 237 238 247 249 252 253 254 255 255 256 263 263 264 266 267

Contents

Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

7 7.1 7.2 7.3 7.4 7.5 7.6

7.7 7.8 7.9 7.10 7.11 7.12

8 8.1 8.2 8.3

Centrifugal Pumps Learning Objectives Introduction Centrifugal Pump 7.2.1 Advantages of Centrifugal Pumps over Reciprocating Pumps Construction and Working of a Centrifugal Pump Classification of Centrifugal Pumps Different Heads of Centrifugal Pumps Different Efficiencies of a Centrifugal Pump 7.6.1 Mechanical Efficiency 7.6.2 Volumetric Efficiency 7.6.3 Hydraulic Efficiency or Manometric Efficiency 7.6.4 Overall Efficiency Analysis of a Centrifugal Pump Minimum Starting Speed of a Centrifugal Pump Maximum Suction Lift and Net Positive Suction Head Cavitation Priming Pumps in Series and in Parallel 7.12.1 Pumps-in-Series 7.12.2 Pumps-in-Parallel Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

Fans, Blowers, and Compressors Learning Objectives Introduction Radial Flow Compressors or Centrifugal Compressors 8.2.1 Stage Velocity Triangles, Inlet Flow Conditions, and Pre-Whirl Compression Process 8.3.1 Slip in the Compression Process 8.3.2 Isentropic Compression Efficiency

xxi

267 268 268 269 270 272 272

273 273 273 273 274 274 276 278 279 279 280 280 281 281 290 291 294 295 295 295 295 300 301 301 302 303 304 305 305

307 307 307 307 308 310 310 311

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8.4 8.5

8.6

8.3.3 Pressure Coefficient or Loading Coefficient 8.3.4 Power Input Factor or Work-Done Factor 8.3.5 Analysis of Diffuser Flow: Diffuser Efficiency 8.3.6 Stage Work and Stage Efficiency 8.3.7 Pressure Developed in the Rotor and Compressor 8.3.8 Performance of a Radial Flow Compressor Matching of Compressor and System Characteristics: Surging in Compressors Axial Flow Compressors 8.5.1 Stage Velocity Triangles 8.5.2 Stage Work, Stage Efficiency, and Performance Stalling of the Blades Keywords Summary Important Equations Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

9 Power-Transmitting Turbomachines 9.1 9.2 9.3 9.4

9.5

Learning Objectives Introduction Working Principle of a Power-Transmitting Turbomachine Construction and Working of a Power-Transmitting Turbomachine Analysis of Power-Transmitting Turbomachines 9.4.1 Fluid Coupling 9.4.2 Torque Converter Comparison of Fluid Systems with Mechanical Systems 9.5.1 Applications Summary Multiple-Choice Questions Review Questions Exercises Project-Oriented Questions Answers

311 311 312 314 314 315 321 323 324 325 336 338 338 339 340 340 341 342 342

343 343 343 343 344 345 347 351 360 361 365 366 367 367 368 368

Glossary

369

Bibliography

381

Index

383

1

Basics of Turbomachines

Learning Objectives After completing this chapter, you will be able to: v
Understand and be familiar with the hardware details of turbomachines. v
Understand and be familiar with the terminology used in turbomachine practice. v
Understand and be familiar with the concepts of energy flow, including losses, in turbomachines, with both qualitative and quantitative

approaches, to arrive at the different forms of efficiencies. v
Explore the principles of model studies and the application of the same to design turbomachines. v
Arrive at the design parameters of turbomachines from the data of practical and field conditions.

1.1 Introduction The energy consumption of mankind has been steadily increasing. Both the production of mechanical energy from the basic sources of energy and the consumption of that mechanical energy to run the devices to make life easier require various systems. These systems have different components with varied functions and complexities. Some of these components are those in which the fluids flow and cause the conversion of energy. Various types of turbines that produce mechanical form of energy and various types of pumps and compressors that consume the mechanical energy possess some common features. These components, termed under a common name – turbomachines, have the same basic principles of working. Whether it is a tiny fan measuring just 5 cm in diameter to provide the circulation of cooling air to a laptop or a steam turbine producing hundreds of megawatts of power to run electrical generators, the design principles are based on the same fundamentals. This book is intended to make an effort to understand such fundamentals with reference to turbomachines. Further, in this chapter, the fundamentals of fluid mechanics and thermodynamics which form the basis of turbomachines are recalled for the purpose of consolidation and ready reference. The units, dimensions, and model studies are detailed as the first step in the procedure of designing a turbomachine to suit a given situation of available data and required performance.

1.2 Classification of Fluid Machines In order to correctly locate turbomachines in the vast field of fluid machines, a classification of fluid machines is presented in Fig. 1.1.

2

Turbomachines

Fluid machines Turbomachines or Rotodynamic Machines Energy from fluid to rotor; Turbines Energy from rotor to fluid; Pumps Energy transmitting machines Positive Displacement Machines (PDMs) Fluid energy to mechanical energy; Motors Gear motors Piston motors Vane motors Mechanical energy to fluid energy; Pumps Rotary pumps Gear pumps Piston pumps Vane pumps Reciprocating pumps Diaphragm pumps Piston pumps Miscellaneous devices Airbrakes Brake devices Hydraulic brakes Airlift pumps Pump devices

Jet pumps Hydraulic Ram

Figure 1.1

Classification of fluid machines.

1.3 Turbomachines A turbomachine is a machine in which there is a continuous transfer of energy between a rotor with vanes and a fluid flowing through the spaces between the vanes of that rotor, due to dynamic action, resulting in a change in the pressure and/or the momentum of the fluid.

Basics of Turbomachines

3

Some observations may be made with reference to the above definition of a turbomachine; and they are as follows: 1. 2. 3.

4.

5. 6.

It is a machine. It is an equipment whose hardware details can be shown in a sketch. Also, it can be easily visualized. There is a transfer of energy. The transfer is a process and the process is continuous, not periodic or stepwise. The energy-transfer process is between two entities. These two entities are a rotor with vanes and a flowing fluid. The direction of the flow of energy is not specified (as yet). It could be in either direction: From the rotor to the fluid or from the fluid to the rotor. There is a “dynamic action.” The process of energy flow takes place as the fluid “flows.” (Take a cardboard and whirl it or “fan” it in front of your face. You feel a “draft” of air. The moving cardboard has given energy to the air; the air has gained the kinetic energy now. The process is said to be by “dynamic action.”) The fluid does not get confined in some limited space. The net result of the process is the change in the energy of the fluid, either increase or decrease, in the form of change in its pressure, momentum, or both.

Certain generalities exist in turbomachines such as 1. 2.

They may handle either liquids or gases (hence “fluids”) and They may extract energy from fluids or may impart energy to fluids.

The principles or theories involved are the same for the entire class of turbomachines.

1.4 Turbomachines and Positive Displacement Machines Turbomachines differ from another class of machines, “positive displacement machines.” Positive displacement machines are those which handle fluids but in “confined spaces” as in cylinder–piston arrangements. Both the classes of machines have their own applications, being suitable for the purpose for which they are designed. Table 1.1 shows a cursory comparison between the two classes. Table 1.1

Comparison between turbomachines and positive displacement machines

Distinguishing Features

Turbomachines

Positive Displacement Machines (PDMs)

1. Basic mechanism of energy transfer

Energy transfer is by dynamic Energy transfer is by expansion or action and is continuous. compression in confined space with moving boundaries and is periodic.

2. Features of mechanical construction

Parts are few in number. The assemblies are simple. Rotors are not in contact with stators (except in bearings). Hence, the machines are less costly.

Parts are large in number. These parts are of high precision, as in cylinder–piston assemblies. Hence, the machines are costly. (Continued)

4

Turbomachines

Table 1.1

Continued

Distinguishing Features

Turbomachines

Positive Displacement Machines (PDMs)

3. Features of operation

Rotors are well balanced. Because of purely rotating parts, vibrations are almost nil and higher speeds are possible. Because of higher speeds, the fluid-handling capacities are more. This results in higher power per unit weight. Because of “diffusion” processes, the “hydraulic efficiency” or the “adiabatic efficiency” is rather low. Mechanical efficiency is very high, because the mechanical losses are very low, due to shaft rotating in bearings and no sliding contacts. Volumetric efficiency is very high due to a “free passage” or absence of valves. In addition, the “continuous flow,” along with the inertia associated with it, tends to increase the volumetric efficiency. The overall efficiency of turbomachines is much higher than that of PDMs. Generally, power-generating turbomachines are suitable to situations where high powers are involved, as in captive power plants with very high power outputs. In the case of power-consuming turbomachines, a centrifugal fan (a turbomachine) can be as small as of some milliwatts.

Because of reciprocating parts and consequent vibrations, the speeds are limited. Hence, the fluid-handling capacities are also limited. This results in higher specific weight, that is, lower power per unit weight.

4. Efficiencies • Efficiency of the energy transfer process • Mechanical efficiency

• Volumetric efficiency

• Overall efficiency

5. Applications

Because of direct compression or expansion processes, diffusion is of a lower order; compression or expansion processes are more efficient. Due to sliding motions and friction, mechanical losses are high and mechanical efficiency is low.

The presence of valves and complicated flow passages reduce the volumetric efficiency. In addition, the periodic or cyclic flow tends to reduce the volumetric efficiency. In effect, the volumetric efficiency is low. The overall efficiency of PDMs is low.

The PDMs are suitable for low-power applications such as automotive power plants. Very high pressure ratios are possible only in PDMs.

The comparison between the two classes of machines as presented in the table is only qualitative. It need not be concluded that any one type is superior to the other type. Each type has its own merits and applications.

1.5 Classification of Turbomachines There are different types of turbomachines. In order to study turbomachines in a systematic way, it is essential to develop the vocabulary or terminology with reference to turbomachines and get familiar

Basics of Turbomachines

5

with the different parts of turbomachines of different types, as a preliminary step. The classification of turbomachines based on the various criteria is indicated in Table 1.2. Table 1.2

Classification of turbomachines

Basis of Classification

Different Types

The direction of energy flow in the 1. Energy from the fluid to the rotor: machines t
1PXFS
NBDIJOFT t
5VSCJOFT 2. Energy from the rotor to the fluid: t
8PSLJOH
NBDIJOFT

Examples 1. Steam turbine 2. Water turbine 1. Fan, blower 2. Compressor 3. Centrifugal pump Centrifugal pump Francis turbine Kaplan turbine Pelton turbine

Direction of fluid flow in the machines

1. 2. 3. 4.

Type of fluid being handled

1. Hydro turbomachines

Pumps

Type of the dynamic action of the fluid

2. Thermal turbomachines 1. Impulse machines 2. Reaction machines

Compressors Pelton turbine Kaplan turbine

Radial flow machines Mixed flow machines Axial flow machines Tangential flow machines

Some of the turbomachines given as examples in Table 1.2 are schematically shown in the following subsections. The following features may be used as checkpoints to look into, with respect to each of the turbomachine, and the fluid flow path may be clearly identified 1. 2. 3.

4. 5. 6. 7.

There is basically a shaft mounted on bearings. The “vaned rotor” is keyed onto the shaft. “Rotor” is a general name: t
Rotor is also known as “runner” in the work-producing turbomachines or turbines. t
Rotor is also known as “impeller” in the work-absorbing turbomachines, pumps, or compressors. Shaft is “coupled” to either a generator (to produce power) or a motor (to draw power, i.e., to utilize power). A stator encloses the rotor. The stator guides the flow of fluid. The stator houses components such as guide vanes, nozzles, or diffusers.

1.5.1 Examples In the following subsections, some examples of turbomachines have been provided.

1.5.1.1 Centrifugal Pump A centrifugal pump (Fig. 1.2) is a power-consuming turbomachine. A motor drives a shaft. The shaft runs in its bearings and drives an impeller or a rotor. The vanes of the impeller draw the liquid from the suction flange and impart energy to the liquid. The liquid is collected from the periphery of the impeller by a volute casing that has its area progressively increasing. The water is then delivered from the delivery flange into the delivery pipe.

6

Turbomachines Delivery

Casing

Delivery

Coupling

Impeller

Shaft

Suction

Motor

Base

Bearings

Vanes

Figure 1.2

Schematic layout of a centrifugal pump.

1.5.1.2 Francis Turbine A Francis turbine (Fig. 1.3) gets its water supply from the supply flange. The water gets distributed uniformly from a volute casing onto a series of guide vanes that are arranged around the periphery of the rotor. The guide vanes direct the water to the runner or rotor. The water enters the rotor, radially inward, and after imparting its energy to the vanes of the rotor exits the rotor in the axial direction to the draft tube. The water finally gets discharged to the tailrace. The shaft is connected to the generator through a coupling. Water supply

Scroll casing To generator

Shaft

Runner

Guide vanes

Thrust bearing Runner vanes

Draft tube Water discharge Tailrace level

Figure 1.3

Schematic layout of a Francis turbine.

Basics of Turbomachines

7

1.5.1.3 Pelton Turbine A Pelton turbine (Fig. 1.4) has its water supply through the penstock. A nozzle converts the pressure or head energy into kinetic energy. The water coming out from the nozzle in the form of a high-velocity jet impinges on a series of Pelton double cups fixed to the rotor. The cups are the vanes of the rotor. These cups are driven by the jet of water, and get the energy of the water. Energy then flows to the rotor and shaft; the shaft drives a generator through a coupling. The water gets discharged to the tail race. Runner/rotor Pelton cups Bearings Shaft

Coupling

Water supply To generator

Casing

Water jet Nozzle Casing Control spear

Figure 1.4

Base

Schematic layout of a Pelton turbine.

1.5.1.4 Steam Turbine In a steam turbine (Fig. 1.5), the high-pressure steam from the boiler is supplied to the turbine at the supply flange. The steam flows into a series of nozzles arranged on the nozzle ring. This nozzle ring is attached to the stator or casing. The nozzles convert the energy of steam into velocity. The high-velocity steam drives the vanes or blades of the rotor arranged in multiple rows. Having imparted its energy to the rotor, the steam at low pressure is discharged to the condenser. The rotor or shaft drives the generator, through a coupling. Nozzle ring

Rotor

Steam supply Moving blades rotor blades

Bearings Stator blades (fixed blades)

Casing

Shaft Bearings Coupling

To generator Steam discharge

Figure 1.5

Schematic layout of a steam turbine.

8

Turbomachines

Table 1.2 shows that turbomachines are also classified as “impulse machines” and “reaction machines” depending on the type of dynamic action. This dynamic action can be elaborated only after the energy interactions in turbomachines are understood to some extent.

1.6 Units and Dimensions Turbomachines handle fluids that are at various levels of energies. The energies of the fluids are of different forms: pressure energy, kinetic energy, potential energy, and thermal energy. In order to deal with different forms of energy and bring those forms onto a common unit, a dimensional constant is first identified. This constant is “gc.” From Newton’s second law, it is known that the force acting on a mass is proportional to the product of the mass and the acceleration of that mass. Mathematically, it can be written as F t ma ¥ 1´  F  ¦ µ ma § gc ¶ where (1/g c ) is the constant of proportionality. Therefore, gc 

ma F

In SI units, 1 N is the force that acts on a mass of 1 kg and accelerates that mass at 1 m/s2. Substituting m  1 kg, a  1m/s2 and F  1 N in the above equation, gc  or

gc 

ma (1 kg )(1 m/s2 )  F 1N 1 ( kg m ) ( N s2 )

Also, 1 kg m y 1 N s2 With the units of mass, length, time, and force fixed as above (as in the SI system of units), gc is a constant with its numerical value of unity having the dimensions of (kg m)/(N s2). Hence, the dimensions of any physical entity can be multiplied or divided by gc without any loss of character of that physical entity and without any change in the numerical value of the measure of its quantity. This multiplication or division by gc is done where the dimensions are required to be homogeneous in equations, in all the quantities. The different forms of energies can be easily considered, with common dimensions, as illustrated in Section 1.7.

1.7 Energy of Fluids The fluids have different forms of energy: pressure energy, kinetic energy, potential energy, and thermal energy. The dimensions of the different forms of the energy are now considered as follows:

Basics of Turbomachines

9

1.7.1 Pressure Energy The pressure energy of a fluid at a pressure p ( N/m2) and at a specific volume v ( m3/kg) is given by the product pv. The dimensions are as follows: pv 

N m3 mN J s   2 m kg kg kg

This expression is also written in the form p/Q where Q is the density, kg/m3, whenever required.

1.7.2 Kinetic Energy The kinetic energy of a fluid moving at a velocity of V m/s is given by V2/2. The dimensions are as follows: 1 mN J V 2 m2 m2  2  2 s   2 2 s s ( kg m/Ns ) kg kg Hence, V 2 V 2 m2 J   2  2 2 gc s kg This illustrates the use of “division by gc” to get homogeneous dimensions, without altering the numerical value. The kinetic energy of a sample of fluid moving at 100 m/s is 1002  5000 m 2 / s2  5000 (mN)/kg = 5000 J/kg 2

1.7.3 Potential Energy The potential energy of a fluid at a height z (above a given datum level) is given by zg. The dimensions are as follows: zg  m s Hence zg 

m m 2 mN J  2   2 s s kg kg

zg J  g c kg

Again, division by gc is without any alteration in the numerical value. The potential energy of a fluid at a height of 10 m above a given datum is 10 s g  10 s 9.81 m2/s2  98.1 J/kg

1.7.4 Thermal Energy or Enthalpy It is a usual practice to refer to the energy of a gaseous fluid (combustion gas, steam, or air) in terms of its “enthalpy.” Enthalpy is a combined effect of pressure and temperature of a gas. For compressible fluids, enthalpy can be taken as a function of temperature. Changes in this form of energy, therefore, are calculated as

10

Turbomachines

$h  c p $T 

J J s oC  kgoC kg

The equation $h  c p $T above holds good when the specific heat cp remains constant. When the fluid is air, the specific heat can be taken as constant over the ranges of temperature changes that are usually encountered in turbomachines. When the fluid is steam, the enthalpies have to be obtained from the steam tables. A fluid can have different forms of energy as discussed above and its total energy is the sum of the individual forms of energy. Essentially, all the forms must be in the same dimensions to make it possible to add them together. As seen above, whether the dimensions are m2/s2, mN/kg, or J/kg, these are all identically the same.

EXAMPLE 1.1 Calculate the energy of a stream of water at a pressure of 200 kPa, flowing at 8 m/s at an elevation of 5 m above a given datum. Take the density of water as 1000 kg/m3.

Solution: We have Total energy  Pressure energy  Kinetic energy  Potential energy Total energy  

p V 2 zg   R 2 gc gc 200 s 103 82 5 s 9.81   1000 2 s gc gc

N m3 m 2 1 m 1 s  2 s ms 2 s 2 2 m kg s (kg m) /(N s ) s (kg m)/(N s2 )  200  32  49.05  281.05 J/ kg



EXAMPLE 1.2 The pressure and temperature of combustion gases are 6 atm and 560oC, respectively, at the inlet of a gas turbine. The pressure drops to 1 atm while the gases flow through the turbine. Assuming that the expansion is isentropic, determine the enthalpy drop of the gases in the turbine. cp for gases  1.006 kJ/kg K, F  1.4.

Solution: For the isentropic expansion process, ¥p ´ ¦ 2µ T1 § p1 ¶

T2

(G 1)/G

¥ 1´  T2  T1 s ¦ µ § 6¶

0.4 /1.4

Basics of Turbomachines

11

 (560  273) s 0.59935  499.26 K Isentropic enthalpy drop is given by $h  c p (T1 T2 )  1.006 s (560  273 499.26 )  335.74 kJ/kg

1.7.5 Head Energy When liquids are handled by turbomachines, the energy of liquids is also identified as “head energy,” like “so many meters of liquid.” The following are some examples: 1.

2.

The output of a centrifugal pump is 15 m of water or the input to a Pelton turbine is 250 m of water, and so on. This seemingly length dimension for energy is not a misnomer if and only if the name of liquid is mentioned. Thus, “10 m of energy” is wrong, but “10 m of water” is perfectly acceptable as an energy unit. Further, “1 m of water” and “1 m of mercury” are not the same. The specific weight (w) of the liquid is also to be accounted while considering the energy unit.

The forms of energy considered earlier can be related to this head energy in the following ways: 1.

The pressure energy is now

2.

p N 1  s m _____ meters of _____ w m 2 N/m3 The kinetic energy is now 1 V 2 m2  2 s m _____ meters of _____ 2g s 9.81 m/s2

3.

The potential energy is now zg g = z m _____meters of _____

The thermal energy or enthalpy is not converted into “meters of liquid” because enthalpy, in this context, is referred to only gaseous fluids. The energy unit of “meters of liquid” is of a higher magnitude. Thus, 1 m of water is equal to 9.81 J/kg of water.

EXAMPLE 1.3 In a hydropower development project, water is available at a height of 450 m above the datum level. What is the pressure of water at the datum? If the entire energy of water were to be converted into kinetic energy without any loss, what would be the velocity of water at the datum level?

12

Turbomachines

Solution: The energy of available water is 450 m of water. The specific weight of water is 9810 N/m3. So p  450 w Therefore p  450 s w  450 s 9810 m s

N m3

 4414500 N / m 2  4414.5 kPa The total energy of water, 450 m of water, is converted to kinetic energy with a velocity V, without any loss. Then, we have V2  450 2g  V  2 g s 450 

2

s 9.81 s 450  93.963 m/s

The corresponding kinetic energy is given as V 2 (93.963)2   4414.5 J/kg 2 2 Comment: In this case, the same energy of water is expressed in different units: 450 m of water or 93.963 m/s or 4414.5 J/kg. All these quantities are equivalent and represent the very same amount of energy in different units. The fluid stream interacting with the rotor of a turbomachine can have different forms of energy. While analyzing the process of interaction, it is required to add all the forms of energy together, at any given point. It is necessary and common practice to consider the pressure component of energy for liquid streams, whereas for gaseous streams, the enthalpy component of energy is considered, while adding the different forms of energy.

1.8 Application of the First Law of Thermodynamics The first law of thermodynamics gives rise to the steady flow energy equation (SFEE) with a set of assumptions. These assumptions realistically hold good to a turbomachine because for a turbomachine, the inlet or outlet conditions do not vary over the time and there is no depletion or accumulation of mass in the machine as the process is continuous. The heat transfer from or to a turbomachine is taken as negligible.

Basics of Turbomachines

13

Work Fluid entry

Figure 1.6

Fluid exit

2

1

Schematic block diagram of a turbomachine.

Considering the turbomachine shown in Fig.1.6, one can write Work output of the turbomachine  Total fluid energy at inlet Œ Total fluid energy at outlet 

W  E1 – E2

(1.1)

On the basis of unit mass flow rate, W  e1 – e2

Also

¥ V2 V2 g´ ¥ g´  ¦ h1  1  z1 µ ¦ h2  2  z2 µ 2 gc 2 gc gc ¶ § gc ¶ §

(1.2a)

¥ g´ ¥ g´  ¦ (ho )1  z1 µ ¦ (ho )2  z2 µ gc ¶ § gc ¶ §

(1.2b)

W  (h1 h2 ) 

V12 V22 2 gc

 ( z1 z2 )

g gc

(1.3)

All the units are J/kg in the above equations. The quantity on the left-hand side, W, is the “specific work” of a turbomachine. It is the work interaction of 1 kg of fluid while flowing over the rotor of the turbomachine. When the flow rate is m kg/s, the corresponding power of the machine is given by ¥ J´ ¥ kg ´ ¥ J´ P ¦ µ  W ¦ µ s m ¦ µ § s¶ § s ¶ § kg ¶ where P is in watts. W and P are positive when the fluid has higher energy at the inlet than at the outlet. These are “power machines” producing motive power to run other machines. Mathematically, the equations hold good for work-absorbing machines, such as pumps, fans, compressors, etc. The fluid energy at the outlet of these machines is more than the energy at the inlet and the specific work becomes negative in the equation. This signifies that power is the input to such turbomachines. When the fluid is incompressible (namely, liquids), the enthalpy terms are replaced by more appropriate terms of pressure energy. In such cases, the density Q is employed instead of specific volume. The density Q is invariable (Q1  Q2) over a large range of pressures. The expression for specific work in such cases becomes ¥ p V2 V2 g´ ¥p g´ W  ¦ 1  1  z1 µ ¦ 2  2  z2 µ gc ¶ § R 2 gc gc ¶ § R 2 gc

(1.4)

All the units are J/kg. Also W 

p1 p2

R



V12 V22 2 gc



( z1 z2 ) g gc

(1.5)

14

Turbomachines

W 

RV12 ´ 1 ¥ RV22 ´ g 1¥ p 

p   ( z1 z2 ) ¦ µ ¦ µ 1 2 R§ 2 gc ¶ R § 2 gc ¶ gc

So W  and

g 1 ( p01 p02 )  ( z1 z2 ) R gc

(1.6)

P  W s m

Now, since the fluid is incompressible, the head energy terms can also be used. However, it is suggested that one has to be more cautious in using the expression “head-generated” in place of “power-generated” or power output. The terms are somewhat opposite to each other in their signs. If the “head” is the output, obviously the power or work is the input. This corresponds to the turbomachine as a power-consuming machine, using mechanical power and “lifting” or pumping the liquid through a height or the “head” of that liquid. If the “head” is the input, then power or work is the output. This corresponds to a turbine, worked by the liquid head and giving out mechanical power as output. Using H (for head) in place of W, the signs of the quantities are changed, and Eq. (1.5) becomes H

p2 p1 w



V22 V12 2g

 ( z2 z1 )

(1.7)

All units are meters of liquid. In continuation, P  wQH 

N m3 mN J s sm  W 3 m s s s

(1.8)

where Q is the volume rate of flow of the liquid (m3/s) through the turbomachine. In Eqs. (1.2) to (1.7), subscripts 1 and 2 represent the inlet and outlet points of the turbomachine. Thus, z1 and z2 represent the heights of the inlet and outlet, respectively, above a common datum. Equations (1.2)–(1.7) represent the general form. There are a few instances where z1 and z2 are really different, but in a large number of cases, z1 and z2 are equal.

1.9 Application of the Second Law of Thermodynamics In Section 1.8, the energies of a fluid at the inlet and outlet have been considered. The difference between these energies has been taken as the “specific work” and a “corresponding power” has been identified. However, the application of the first law of thermodynamics is not sufficient to completely explain the interaction between the fluid and the rotor. The process of transfer of energy in either direction has its own mechanism, as analyzed by the second law of thermodynamics. Those that are involved are the process losses. The process losses have several components: The internal friction within the body of the fluid, namely, viscous effects, the turbulence, the eddies, etc. All these factors give rise to non-isentropic flow conditions. The losses due to friction between fluid stream and rotor vanes, that is, the skin friction effects, are also included in the above losses. The exit losses (V22 / 2 g c ) or the kinetic energy losses due to the fluid leaving the rotor are the necessary losses because the fluid has to make an exit from the rotor. Losses also occur due to the friction between the shaft and the bearings. The loss of fluid due to leakage is yet another loss. In short, all these losses are due to the irreversibility, and the values of efficiencies keep reducing.

Basics of Turbomachines

15

1.9.1 Efficiencies of Turbomachines Different efficiencies of the power-generating turbomachines and power-absorbing turbomachines have been discussed in the following subsections.

1.9.1.1 Power-Generating Turbomachines or Turbines Any efficiency, in general, can be written as output divided by input. It can also be written as Efficiency 

Input Losses Input

Depending on the type of losses, any particular efficiency can be identified. As an example, Mechanical efficiency 

Input Mechanical losses Input

In order to present an overall view of the different efficiencies of the power-generating turbomachines, a schematic diagram is given in Fig. 1.7 to show the flow of power in such machines. The width of the strip indicates the magnitude of the power P. The power is the product of its two component factors, specific work W and mass flow rate m.  Note that no scale is implied; the diagram is only qualitative. Pn

P

Pa W+

V22 2gc

m

Leakage losses Δm

W+

V22

2gc m − Δm

Fluid friction fluid-rotor friction losses

− losses W − losses W+ 2gc m − Δm m − Δm

Exit losses V22 2gc

Figure 1.7

Pr

V22

Ps

Mechanical losses in bearings

(m − Δm)

Power flow in power-generating turbomachines.

For a turbine, the input power Pa is the power available in the fluid at the inlet. This available power is the product of two factors: 1. 2.

The mass flow rate in kg/s. The “inlet specific work” (J/kg) that again has two components: t
The specific work as evaluated between the inlet and outlet of rotor, W [Eqs. (1.1) and (1.2)]. t
The exit kinetic energy, V 22 / 2g c .

In the schematic diagram shown in Fig. 1.7, the energy or power is represented as a width of a strip and the available power Pa is at the left side. The flow of power is from left to right. The output power is at the right. The diversions downward are the losses. Both specific work and mass flow rates keep reducing due to the losses. The losses, as shown, are only indicative. They occur neither in steps nor one after another. The losses are continuous and simultaneous.

16

Turbomachines

To understand the Fig.1.7, the quantities mentioned in the figure are explained as follows: m  Mass flow rate of fluid at the inlet, kg/s, W  Specific work, as defined by Eqs. (1.1) and (1.2). Wa  Available energy of the fluid at the inlet, J/kg, W 

V22

2 gc Pa  Power available in the fluid at the inlet, Watts, ¥ V2 ´  ¦W  2 µ m 2 gc ¶ § P  Power that can be extracted from the fluid, after the leakage. Generally, leakages are referred to as volumetric losses. Pn  Net power extracted after the losses due to viscous friction and skin friction. So Pn  P – Friction losses. Pr  Power at the rotor or runner of the turbine. So Pr  Pn – Exit losses. Ps  Power at the output of the machine at its shaft that is equal to the runner power minus bearing losses or mechanical losses at the bearings. So Ps  Pr – Bearing losses. The following efficiencies are defined: The utilization factor is

E

Pr Pn



W V 22 W  2 gc

(1.9)

The hydraulic or adiabatic efficiency is

Hh or Ha 

Pr P



Pr Pn

s

Pn P

(1.10)

The volumetric efficiency is ¥ m $m ´ ¥ Q $Q ´ P  Hv  ¦  § m µ¶ ¦§ Q µ¶ Pa

(1.11)

The mechanical efficiency is

Hm 

Ps Pr

(1.12)

The overall efficiency is given by

Ho 

Ps Pa



Ps Pr

s

Pr P

s

P  Hm s Hh s Hv Pa

(1.13)

Basics of Turbomachines

17

It may be noted that hydraulic efficiency is inclusive of the utilization factor and the irreversible effects of viscous and skin frictions, eddies, turbulence, and other losses.

1.9.1.2 Power-Absorbing Turbomachines Power-absorbing turbomachines are fans, blowers, compressors, and pumps. Just similar to the diagram for turbines in Fig. 1.7, a power-flow diagram for work-absorbing machines is provided in Fig. 1.8, applicable to fans, blowers, compressors, and pumps.. Different efficiencies are then identified.

Ps

Pr

Pn

P

Pu

Wn

Wn

W

W

m 
m

m

m

m

Mechanical Return flow Fluid friction losses losses, leakage fluid-rotor at bearings friction m

Figure 1.8

V22 2gc

Exit losses V22 2gc


m

Power flow in power-absorbing turbomachines.

The quantities mentioned in Fig. 1.8 have the following meanings: m  Mass flow rate at the exit of the turbomachine, W  Specific work, as defined by Eqs. (1.1) and (1.2), Wn  Net specific work (W  fluid friction losses), Ps  Shaft power or input power, Pr  Power at the rotor or impeller of the machine. Now, Pr  Wn (m  $m ) Pr  Ps – Mechanical losses where $ m is the fluid that returns to the inlet end of the impeller from the exit end of the impeller and gets recirculated through the impeller along with m and the leakage losses. Also the net fluid power is Pn  Wn s m  Pr Return flow losses P  Pn – Frictional losses The useful fluid power at the outlet is ¥ V2 ´ Pu  ¦W 2 µ m 2 gc ¶ §  P Exit losses

18

Turbomachines

The following efficiencies are defined: Pr



Gm  Mechanical efficiency 



Gv  Volumetric efficiency 



Gh  Ga  Hydraulic or adiabatic efficiency 



Go  Overall efficiency 

(1.14)

Ps P m  n m  $m Pr

(1.15) P Pn

(1.16)

P P Pn Pr  s s Ps Pn Pr Ps

 Hh s Hv s Hm

(1.17)

The purpose of the two power-flow diagrams, Figs. 1.7 and 1.8, is to visualize the flow of energy supplied either in the fluid at the inlet or in the shaft at the coupling. The diagrams when related to the relevant schematic sketches of the equipment (shown in Figs. 1.2–1.6) give a fairly good idea of the conversion of energy from the inlet point to the final useful point along with different loss components. The situation is similar to drawing up a “heat balance” or “enthalpy balance” in Internal Combustion engines. Some more explanations will be added, as and when required. Frequent reference to these diagrams may be necessary from time to time.

1.9.2 Power-Flow Diagrams in Head Units Earlier, the equations for specific work W were identified with reference to the energy of the fluid [Eqs. (1.1)–(1.6)]. Then the same equations were modified to be in the “head” units, particularly with reference to turbomachines having liquids as operating fluids [Eqs. (1.7) and (1.8)]. Likewise, the power-flow diagrams (Figs. 1.7 and 1.8) are now modified in the same way, and are redrawn as shown in Figs. 1.9 and 1.10. It is suggested that all the powers and efficiencies mentioned earlier be written as an exercise. There are the two diagrams – one for power-generating machines (Fig. 1.9 applicable to turbines) and one for power-absorbing machines (Fig. 1.10, applicable to pumps). Once again, it is required to remember that the flow from left to right is power flow (with three factors, w, Q, and H). Diversions downward are the losses. Pa

P

Pn

Pr

w

w

w

w

Q Q

Q Q

Q H


V22

H


2g

Leakage losses (Q)

V22 2g

Fluid friction fluid-rotor friction losses

H


V22 2g

Q Q losses

V22

Ps

H losses

Exit losses 2gc

Figure 1.9

 Shaft Power

Mechanical losses in bearings

w(Q Q)

Power flow from the fluid to the shaft: The fluid is liquid or water.

Basics of Turbomachines

Pr

Pn

P

Ps

w

w

w

Pu Useful fluid w power

Shaft power

Q 
Q

Q

Q

Q

Hn

H

H

Hn

Mechanical Leakage losses Fluid friction losses fluid-rotor return flow at bearings friction Q

Figure 1.10

19



V22 2g

Exit losses

Power flow from the shaft to the liquid.

The endeavors of the mechanical engineers have been to increase the efficiencies of their devices by reducing the losses. Some of the losses in turbomachines, like the ones discussed in Section 1.9.1, are very easy to understand. Bearing friction is one such item. Leakage losses can also be considered easier to understand. What about hydraulic losses, or the losses occurring during expansion or compression processes? What about “exit losses?” (It has to be noted here that the exit velocity is shown as a loss in Section 1.9.1. But it need not be a loss all the time, if the exit velocity is useful, in a succeeding process of compression or expansion, as in multistage systems.) It is required, therefore, to find the nature of the losses and means of assessing and reducing them.

EXAMPLE 1.4 A flow rate of 3 m3/s of water is available at a height of 110 m at a project site. Due to losses in the supply line, the head available at the inlet to powerhouse is estimated to be only 101 m of water. The leakage losses in the powerhouse are negligible. Mechanical losses account for 150 kW. Frictional losses in the rotor blades may be taken as 250 kW. The exit velocity of water from the turbines is 4.5 m/s. Calculate the hydraulic, mechanical, and overall efficiencies of the plant. The specific weight of water may be taken as 9810 N/m3.

Solution: We have Q  3 m3/s, H  101 m, mechanical losses  150 kW, hydraulic losses  250 kW and w  9810 N/m3. Available power is Pa  wQH  9810 s 3 s

101  2972.43 kW 1000

Net power after the losses, Pn  2972.43 – 250  2722.43 kW. Rotor power is Pr  Net power Exit losses  Pn wQ (V 22 / 2g c )

20

Turbomachines

 2722.43 9810 s 3 s

( 4.5)2 2 s 9.81

 2692.055 kW Shaft power is Ps  Rotor power – Mechanical losses  2692.055 – 150  2542.055 kW

Now

Hydraulic efficiency  Mechanical efficiency  Overall efficiency 

Rotor power Available power Shaft power



Rotor poower Shaft power

(no leakage ) 

26992.055  90.57% 2972.43

2542.055  94.43% 2692.055

Available power



2542.055  85.52% 2972.43

EXAMPLE 1.5 A gas turbine works with a pressure ratio of 6:1. The temperature and velocity of gases at the inlet to the turbine are 900 K and 350 m/s, respectively. The isentropic expansion efficiency is said to be 85%. The velocity of gases at the outlet is 50 m/s. The specific heat of gases, cp, over the working range may be taken as 1.004 kJ/kg K and F  1.4. Determine the power output of the turbine for the unit mass flow rate of gases.

Solution: We have pressure ratio  6:1, T1  900 K, V1  350 m/s, V2  50 m/s, cp  1.004 kJ/kgK,

F  1.4. For gas turbines, it may be assumed that z1  z2. For the isentropic expansion process, 1 2 a, ¥p ´ ¦ 1µ T2a § p2 ¶ T1

(G 1)/G

Therefore 900 0.4 /1.4  6  1.6685 T2a Also T2a 

900  539.4 K 1.6685

Isentropic temperature drop  900 – 539.4  360.6. Therefore, the actual temperature drop  (360.6) s 0.85  306.5.

Basics of Turbomachines

21

Now, the exit temperature is T2  900 – 306.5  593.5 K Specific work output is W 

(h1 h2 )  (V12 V22 ) 2 gc

 1.004 (900 593.5) 

(3502 502 ) 2 s 1000

 367.726 kJ/kg The power output for the unit mass flow rate  W s m  367.726 kW. Comment: Although the power output is as above, part of this power is required to drive a compressor, which is part of the gas turbine power plant.

EXAMPLE 1.6 An axial flow air compressor has a pressure ratio of 1:6. The compressor draws air from atmosphere at a temperature of 300 K. The inlet velocity may be considered negligible. The isentropic compression efficiency of the compressor is 0.88. The velocity of air at the outlet is 40 m/s. The specific heat of air over the working range is 1.004 kJ/kg, F  1.4. Find the power required to drive the compressor to compress air at a nominal rate of 1 kg/s.

Solution: We have pressure ratio  1:6, T1  300 K, V2  40 m/s, cp  1.004 kJ/kgK, F  1.4 Assume that z1  z2 for the compressor. For the isentropic compression process, 1 2 a, ¥p ´ ¦ 2µ T1 § p1 ¶

T2a

(G 1)/G

The outlet temperature at the end of isentropic compression is 0.4 /14

T2a  T1 6

 300 s 1.6685  500.55 K

Now Actual temperature rise 

Isentropic temperature rise

0.88 500.555 300  0.88  227.9

Therefore, the actual exit temperature is T2  300  227.9  527.9 K

22

Turbomachines

Specific work required is W  c p (T2 T1 ) 

(V22 V12 ) 2 gc

 1.004(527.9 300) 

( 40)2 2 s 1000

(& V1  0)

= 229.6 kJ/kg The power required to drive the compressor for a flow rate of 1 kg/s is W sm·  229.6 kW Comment: This compressor is required to be matched with the turbine of the previous example. Consider the steps required to match.

EXAMPLE 1.7 Water is required to be pumped to the overhead tank of a building of 20 storeys through a vertical height of 75 m from the sump level. The total equivalent length of the delivery pipe, including the effects of pipe fittings, is 80 m. The pipe diameter is 50 mm; its friction factor is 0.006. Determine the power of the required motor to pump the water at a rate of 10 lps. The mechanical losses of the pump are equal to 0.2 kW. The hydraulic efficiency of the pump is 93.6%. Assume leakage and return flow losses as 0.2 lps. Also find the mechanical, volumetric, and overall efficiencies of the pump.

Solution: We have vertical height  75 m, equivalent length of pipe  80 m, f  0.006, Q  0.01 m3/s, delivery pipe diameter  50 mm, $Q  0.0002 m3/s, Gh  0.936, mechanical losses  0.2 kW. Now, the velocity of water in the pipe is V 

Q 0.01  5.093 m/s  2 Pd 0.052 ´ ¥ ¦§ P s 4 µ¶ 4

Referring to Fig. 1.9, we have Exit losses 

4 fLV 2 4 s 0.006 s 80 s (5.093)2 V 2 (5.093)2   1.322 m; Pipe losses    50.765 m 2 g 2 s 9.81 2 gd 2 s 9.81 s 0.05 Head required  Vertical height  Friction losses  Exit losses



 75  50.765  1.322  127.1 m Now, net head is given by Hn  Net head 

Output head (H) Hydraulic efficiency 127.1  135.8 m 0.936

Basics of Turbomachines

23

Again, Flow across the pump  Pump delivery required  Leakage losses Therefore, Flow across the pump  0.01  0.0002  0.0102 m3/s We have power output of the pump, P  wQH Power of the pump 

9810 s 0.0102 s 135.8  13.59 kW 1000

Referring to Fig.1.10, we have Power of the motor  Power of the pump  Mechanical losses in the pump Power of the motor  13.59  0.2  13.79 kW Power output of the pump Mechanical efficiency  Power input to the pump (from motor) 13.59  98.55% 13.79 Actual flow rate Volumetric efficiency  (Flow rate  Return flow losses) Q  (Q  $Q ) 10   98% 10.2 

From Eq. (1.17), overall efficiency  Hh s Hv s Hm

 0.9855 s 0.98 s 0.936  90.4%

1.10 Model Studies When turbomachines are designed to suit the field conditions, the assumed values of efficiencies must be very near to the possible values. The prototypes so designed are huge. The costs of manufacture, transport, and installation of such machines are also high. Any errors in the designs can result into huge losses. In order to ensure the performance of the prototype machines, certain model studies are undertaken. First, a prototype is designed. Then a model of the prototype is designed and built up. While designing the model, all the geometric dimensions are scaled down from the corresponding dimensions of the prototype machine, keeping the angles of the blades the same as that of the prototype machine. This ensures the geometric similarity between the model and the prototype machine. Further, the “kinematic similarity” and “dynamic similarity” between the model and prototype are also ensured during the testing of the model, namely, the kinetics (velocities) and the forces are maintained proportional. Being small in size, the model is of a much lower cost. Then tests are conducted on this model and the performance characteristics are obtained. Next, the performance is extrapolated to that of the prototype machine. When the geometric, kinematic, and dynamic similarities are ensured between the model and prototype machine, the extrapolated performance can be taken

24

Turbomachines

as a valid performance of the prototype machine. If this performance is not satisfactory, design modifications can be made and the steps can be repeated. Costly errors, therefore, can be avoided by the model studies. The methods of extrapolation of the performance are subject to certain rules that are formulated by the dimensional analysis.

1.11 Dimensional Analysis The dimensional analysis is a simple approach of extrapolating any performance parameter from one machine to another geometrically similar machine in the most general context. The particular context of extrapolation, at present, is from the model to prototype machine. To start with, it is necessary to identify the performance parameters, so that these parameters are not affected by the size of the machines, when the machines are scaled up (or scaled down). Some of these parameters are arrived at as described below: 1.

The volumetric flow rate of a fluid in a turbomachine is Q m3/s. Therefore, Q  Flow area s Flow velocity (m2 s m/s  m3/s) For a geometrically similar machine, the flow area can be written as C1D2 (C1: constant, D: diameter) and the velocity can be proportional to ODN/60 (N: rpm of the machine). Hence Q t C1D 2 s

P DN 60

 Q t ND 3  Q  P1ND 3 where O1 is the constant of proportionality, inclusive of other constants such as C1 above, O/60, scale factors, etc. This O1 is known as the flow coefficient. Flow coefficients have to be of the same value for all geometrically similar machines at a corresponding operating point. Thus Q P1  ND 3 Q3 Q1 Q2    … (1.18) N1D13 N 2 D23 N 3 D33 2.

The subscripts on the right-hand sides, namely, 1, 2, 3, …, signify different similar machines. The head of turbomachine (input or output) is H m. The head H is proportional to the square of the velocity that is V 2. A representative velocity being ODN/60, it is possible to write ¥ P DN ´ H t¦ § 60 µ¶

2

With reference to all the machines with kinematic similarity, it is also possible to write H t N2D2  P2 

H N 2D2

where O2 is the constant of proportionality, inclusive of other constants such as (O/60)2, scale factors, etc. This O2 is known as the head coefficient. Head coefficients have to be of the same value for all geometrically similar machines with kinematic similarity at corresponding operating points. Thus

Basics of Turbomachines

P2  

3.



H N 2D2 H1 N12 D12



H2 N 22 D22



H3 N 32 D32

…

25

(1.19)

The subscripts on the right-hand sides, namely, 1, 2, 3, …, signify different similar machines. The power of a turbomachine is wQH as has already been seen. Expressing Q and H as above, it is now possible to write P t QH (when the fluid density is the same)  P t (ND3) (N2D2)  P t N3D5 P  P3  3 5 N D where O3 is the constant of proportionality, inclusive of other constants, scale factors, etc. This O3 is known as the power coefficient. Power coefficients have to be of the same value for all geometrically similar machines with kinematic and dynamic similarities at corresponding operating points. Thus P3 

P N 3 D5 P P P  31 5  32 5  33 5 … N1 D1 N 2 D2 N 3 D3

(1.20)

The subscripts on the right-hand sides, namely, 1, 2, 3, …, signify different similar machines. The above three coefficients (namely, O1, O2, and O3) are the constants of proportionality. An example is given below to illustrate how to use these coefficients to extrapolate the performance of one machine to that of another. P Np Pp

y P

Nm Pm

x M

Qm

Figure 1.11

QP

Q

Extrapolation of performance.

Suppose that one of the performance characteristics is a plot of power P as a function of the flow rate Q, as obtained by experiment on a model at a speed Nm, curve M in Fig. 1.11. It is required to project the value of power Pp of prototype, at a delivery Qp of prototype at a speed Np of the prototype. Starting from the values of model at point x, the following can be written:

26

Turbomachines

P1 

Therefore

Qm N m Dm3

Q p  Qm s



Qp N p Dp3

N p ¥ Dp ´ N m ¦§ Dm µ¶

3

Thus, Qp is established when the scale factor (Dp/Dm) is known. Further P3 

Pm N m3 Dm5



Pp N p3 Dp5

Therefore 3

¥ N p ´ ¥ Dp ´ Pp  Pm s ¦ µ ¦ µ § N m ¶ § Dm ¶

3

Thus, the value of Pp is also established, when scale factor is known. The characteristic of the prototype machine has a point y (Qp, Pp) on it, as shown. Picking a series of points on curve M, a whole set of points on curve P can be obtained, resulting in the characteristic of the prototype. It is easy to get a family of curves, taking the speed Np as a variable parameter, Np1, Np2, Np3, …, for the prototype machine. The flow rate Q, the head H, the operating speed N, and the power input or output P are the important parameters to be considered in the design of the prototype machine as well as the model. During the experiments on the model, the same parameters are to be studied. Another important parameter is the efficiency of the machines. The objective of a design engineer is to maximize the efficiency. Hence, the efficiency of the model is also studied along with the above parameters (Q, H, N, P) and the efficiency of the prototype machine is extrapolated. A correlation suggested by Moody et al. to connect the efficiencies of the model and prototype machine is as follows: ¥D ´ ¦ mµ 1 Hm § Dp ¶ 1 Hp

1/5

(1.21)

While conducting the experiments on the model, the values of efficiencies of the model, (Gm) can also be evaluated along with Q, H, N, P at different points of operation. Using the above correlation, the values of the efficiencies of the prototype machine (Gp) can also be calculated as part of extrapolation. It is suggested that some caution has to be exercised in the procedure because of some of the uncertainties: 1. 2. 3.

The ratio of surface roughness to any macrodimension may not be the same between the model and prototype. As a result, the flow characteristics may be different. Even between two macrodimensions, the proportions may vary, such as the ratios of clearances to diameters. If the fluid being handled is totally different (with widely varying viscosities) between the model and the prototype, it is possible that the projection may be wrong. Yet, the extrapolation or prediction may be taken as a fair estimate of the performance of the prototype.

Basics of Turbomachines

27

EXAMPLE 1.8 A centrifugal pump is designed for a town water supply project. Before undertaking the manufacture of the pump, a one-eighth size model of the pump is built and trials are run on the model. The observations are shown in Table 1.3. It is required to extrapolate the performance of the prototype pump at a speed of 600 rpm. Particulars of the model: Impeller outlet diameter, Dm: 0.1 m Speed, Nm: 1500 rpm Particulars of the prototype machine: Impeller outlet diameter, Dp: 0.8 m Speed, Np: 600 rpm Table 1.3

Observations

Model head: meters of water, Hm Model flow rate:

m3/s,

Qm

Model input power: kW, Pm

5.8

5.1

4.1

3.2

2.1

0.011

0.0152

0.019

0.022

0.026

0.802

0.927

0.99

0.958

0.85

Solution: Consider the trial with Hm  4.1 m, Qm  0.019 m3/s, and Pm  0.99 kW. The head of

prototype machine is

2

2

2

2

¥ Np ´ ¥ Dp ´ Hp  Hm s ¦ s¦ µ µ § Nm ¶ § Dm ¶ ¥ 600 ´ ¥ 0.8 ´  4.1 s ¦ s¦ µ § 1500 µ¶ § 0.1¶  41.984 m of water Flow rate of prototype is ¥ N p ´ ¥ Dp ´ Q p  Qm s ¦ µ s¦ µ § N m ¶ § Dm ¶

3

¥ 600 ´ ¥ 0.8 ´  0.019 s ¦ s § 1500 µ¶ ¦§ 0.1µ¶  3.89 m3 / s Output power of the model is Pom  w s Qm s H m  9810 s 0.019 s  0.7642 kW

4.1 1000

3

28

Turbomachines

As per the data in Table 1.3, input power for the model, Pim  0.99 kW. Efficiency of the model is

Hm 

0.7642  0.772 0.99

Therefore, the efficiency of prototype is ¥D ´ Hp  1 (1 Hm ) ¦ m µ § Dp ¶

0.2

 0.8496

Output power of the prototype machine is 9810 s 3.89 s 41.984  1602.15 kW 1000 Therefore, the input power required is Output 1602.15   1885.76 kW Efficiency 0.8496 It is suggested that the other trials may be worked out as exercise. The results are shown in the tabular form for all the trials in Table 1.4. Table 1.4

Model Hm m of water

5.8

5.1

4.1

3.2

2.1

Model Qm, m3/s

0.011

0.0152

0.019

0.022

0.026

Model input power, kW

0.802

0.927

0.99

0.958

0.85

Model output power, kW

0.626

0.76

0.7642

0.691

0.536

Model efficiency, Gm

0.781

0.82

0.772

0.721

0.63

Prototype efficiency, Gp

0.855

0.8812

0.8496

0.8159

0.7558

Prototype Hp m of water

59.39

52.22

41.984

32.768

21.50

2.25

3.113

3.89

4.51

5.32

Prototype output power, kW

1310.88

1594.7

1602.15

1449.8

1122.1

Prototype input power, kW

1533.2

1809.7

1885.76

1776.9

1484.6

Prototype Qp,

m3/s

Thus, the performance of the prototype machine can be mapped, even before the machine is manufactured, by conducting tests on a similar, small model machine and extrapolating the performance. As seen above O1, O2, and O3 are the constants of proportionality. Although they are dimensional constants (because they include factors such as O/60, etc.), their usage is limited to the extrapolation of performance, that is, to write Q1 N1D13



Q2 N 2 D23

or

H1 N1D12



H2 N 2 D22

…

Basics of Turbomachines

29

Hence, their numerical values do not directly characterize a turbomachine. Even if it is possible to nondimensionalize the coefficients, their usage still remains restricted. However, to truly characterize a turbomachine, some non-dimensional groups of variables must be identified. A formal method adopted for such identification of non-dimensional parameters is the dimensional analysis using the Buckingham O-theorem.

1.12 Unit and Specific Quantities There are quite a few entities in connection with a turbomachine. Some of them are as simple as physical dimensions, such as the diameter of the rotor, the thickness of the blades, etc. Some entities are the conditions under which turbomachines work, such as the head (input or output), the speed of the machine, the pressure ratios, etc. Now, the pressure ratio is one of the important parameters, influencing the functioning of turbomachines working with compressible fluids (gas, steam, air). Likewise, the head is the important parameter of turbomachines working with incompressible fluids, that is, water turbines and pumps. Thus, the head of water (either available for a turbine or required to be produced by a pump) has been the most important parameter to be considered in the design of a turbomachine. Presently, the effect of the varying head on other parameters is of interest. Consequently, many of the other parameters are being calculated, for the unit value of head. Such reduced values of speed, flow, power, etc. are known as unit speed, unit flow, unit power, etc. when the unit head (equal to 1 m of water) acts on the machine. The machine is not a hypothetical or “altered” (i.e., scaledup or scaled-down) machine. It is the same machine on which the unit head acts, giving rise to the unit speed, unit flow, etc. Thus, the following unit quantities are defined: 1.

The unit speed, N1, of a given turbomachine is the speed of the same machine when the head of the machine is 1 m of water. When the speed of the rotor is N rpm, the peripheral velocity is U

P DN 60

with D being the diameter. As is known, U t 2 gH Hence N tU t H Now N  C1 H Now N  N1 when H  1, by definition. Therefore, using these values, in this above equation, C1  N1 N (1.22) H The unit flow, Q1, of a given turbomachine is the flow through the same machine when the head of the machine is 1 m of water. When the flow through the machine is Q and the flow velocity is Vf we have  N1 

2.

30

Turbomachines

Q t Vf t 2 gH Hence Q t H or Q  C 2 H Now Q  Q1 when H  1. So C2  Q1 Therefore Q1  3.

Q H

(1.23)

The unit power, P1, of a given turbomachine is the power of the same machine when the head of the machine is 1 m of water. It is known that P t QH t H – H t H3/2 Hence P  C3· H3/2 Now P  P1, when H  1. Therefore C3  P1 P (1.24) H 3/ 2 The unit quantities make it possible to compare the different machines working with different heads. In addition to the unit quantities, another set of quantities are the “specific quantities”. In this case, along with the head of a machine, the size of the machine can also be reduced to the unit size for the sake of comparison. The quantities so obtained are known as specific quantities. Thus, the following specific quantities are defined. The specific flow, Q11, of a given turbomachine is the flow corresponding to a similar machine of some unit dimension and working with unit head. The unit dimension is 1 m diameter of the rotor, or in the case of Pelton turbine, it is the unit diameter of the jet. It can be seen that  P1 

4.

Q t D2 H or Q  C4 D2 H or C4  Q11 Therefore Q11  5.

Q (D 2

H)

(1.25)

The specific power, P11, of a given turbomachine is the power of a similar machine, with some unit dimension and working with unit head. It can be seen that P t QH  P t (D2

H )(H)

Basics of Turbomachines

31

 P  C5 D2 H3/2 Now P  P11, when D  1, H  1. Hence C5  P11 Therefore P11 

P ( D 2 H 3/ 2 )

(1.26)

But in continuation, the specific speed is an exception to the above definition, and the specific speed has been elaborated in Section 1.15.

EXAMPLE 1.9 A water turbine of runner diameter 1.12 m works with a head of 101 m of water, 3 m3/s of flow and produces a power of 2542 kW at a speed of 375 rpm. Determine the (a) unit speed, (b) unit flow, (c) unit power, (d) specific flow, and (e) specific power of the turbine.

Solution: Given D  1.12 m, H  101 m, Q  3 m3/s, N  375 rpm, P  2542 kW. (a) Unit speed is given by N1 

N 375   37.31 rpm H 101

(b) Unit flow is given by Q1 

Q 3   0.2985 m3 /s H 101

(c) Unit power is given by P1 

P 2542   2.5 kW 3 2 / H 1013/2

(d) Specific flow is given by Q11 

Q D2

H



3  0.238 m3 /s (1.12 )2 101



2542  1.996 kW (1.12 )2 (101)3/2

(e) Specific power is given by P11 

Q D 2 H 3/ 2

EXAMPLE 1.10 A one-fourth size model of the turbine in Example 1.9 is required to be tested in a laboratory. The maximum head available in the lab is 10 m of water. Find the flow rate required to be planned, the speed at which the model is to be tested, and the output of the model.

32

Turbomachines

Solution: Given Hp  101 m, Hm  10 m, Dp/Dm  4, Np  375, Qp  3 m3/s, Pp  2542 kW. We have

Hp Hm

2

¥ N p ´ ¥ Dp ´ ¦ µ ¦ µ § N m ¶ § Dm ¶

2

2

2 ¥N ´ ¥ Dp ´ H m 10 m ¦ –  42 s µ ¦ µ 101 § Dm ¶ H p § Np ¶

 N m  N p 42 s (10 / 101)  N m  472 rpm Also 3

¥N ´ ¥D ´ ¦ mµ ¦ mµ Pp § N p ¶ § Dp ¶  Pm  4.95 kW Pm

Again

5

3

3 N ¥D ´ 472 ¥ 1 ´  m –¦ m µ  s¦ µ Qp N p § Dp ¶ 375 § 4 ¶

Qm

 Qm  0.059 m3 /s

1.13 Non-Dimensional Parameters and Their Significance The constants O1, O2, and O3, as used in Section 1.11 above, were identified as the constants of proportionality in the consideration of the simple physics of turbomachines [see Eqs. (1.18)–(1.20)]. The use of these constants was also illustrated in the process of extrapolation of performance of turbomachines. These constants can be non-dimensionalized as shown in Table 1.5. The groups of variables obtained in the right side column are the non-dimensional ones. These are like the non-dimensional numbers in fluid mechanics. Table 1.5

Conversion of O -constants to non-dimensional constants

O1  Q/ND3

Replace N (rpm) by n (rps)

O2  H/N2D2

Replace N (rpm) by n (rps) and multiply by g (constant)

O3  P/N3D5

Replace N (rpm) by n (rps) and divide by Q (density of fluid)

Q nD 3 gH P2  2 2 n D

P1 

P3 

P Rn3 D 5

Basics of Turbomachines

33



The non-dimensional form of O1 ( Q/nD3) when expressed as (Q/D2)/(nD)  Q/D2 H becomes the specific flow. Thus, the flow coefficient O1 signifies the specific flow and represents a similar machine with some unit dimension and unit head [see Eq. (1.25)]. In addition, when expressed as (Q/D2)/(nD), this signifies (the inverse of ) the speed ratio (1/E ). This speed ratio is one very important design parameter that will be discussed later in Section 3.6. Similarly, O2 ( H/n2D2) represents the ratio of (fluid velocity)2 to (blade velocity)2. This is also a square of inverse of the speed ratio (1/E 2). One more significance of O2 can be identified when the head H is expressed as proportional to ($p/Q), or O2  $p/QU2. This expression means that the generation of pressure ($p) is proportional to the density of fluid (Q) and the square of the machine speed. The third coefficient O3 in its non-dimensional form can also be identified as O3 t P/[D2(nD)3] t P/D2H3/2 This signifies the specific power, as seen in Eq. (1.26).

1.14 Effect of Reynolds Number One of the non-dimensional numbers is the Reynolds number as applied to turbomachines and is recognized as QnD2/L. This contains the viscosity term of the fluid being handled by the machine. However, the fluids usually handled by turbomachines are those that have very low values of viscosity, that too very little variation during the process, such as air, steam, water, alcohol, etc. Hence, the effect of variation of Reynolds number is very minimal.

1.15 Specific Speed One of the non-dimensional groups obtained by using the Buckingham O-theorem stands apart from the flow coefficient, head coefficient, etc. This group is known as the “specific speed” that is generally meant for machines handling liquids (that too water), although there is nothing wrong in using it (in suitably modified form) for gas-handling machines. This non-dimensional specific speed 7 is given as 7p  7t 

n Q for pumps ( gH )3/ 4 n P for turbines R ( gH )5/ 4

n Q 1 ¥ m3 ´ m s 7p  ( gH )3/ 4 s ¦§ s µ¶ 7t 

(1.27) (1.28)

0.5

s

1 (m 2 / s2 )0.75

n P 1 ¥ mN ´ m s¦ µ R0.5 ( gH )5/ 4 s § s ¶

0.5

s

: dimensionless

1 1 s : dimensionless ( kg/m3 )0.5 (m 2 /s2 )1.25

The specific speed of the pump is in terms of the output of the pump, that is, the flow rate, Q (m3/s). The specific speed of the turbine is in terms of the output of the turbine, that is, the power in watts (or kilowatts). The specific speed of a given pump is the speed of another geometrically similar hypothetical pump of such a size that it can, at that speed, pump 1 m3/s of water against a head of 1 m of water. The specific speed of a given turbine is the speed of another geometrically similar hypothetical turbine of such a size that it can, at that speed, generate 1 unit of power by using a head of 1 m of water.

34

Turbomachines

Both the expressions for the specific speeds, as given by the Eqs. (1.27) and (1.28) do not have in them any physical dimension of the machine (such as the diameter). Hence, the specific speeds characterize the “shape” of the rotors and not the size of the rotors. These specific speeds, therefore, are also termed “shape numbers.” The following discussions are presented to illustrate the use of the specific speed, in the characterization of the machines. A graph of overall efficiency versus delivery (m3/s) or power (kW) is drawn for a turbomachine as shown in Fig. 1.12. (The generality of the nature of this curve has been discussed in Chapter 4. h A C

B

D

Q or P

Figure 1.12

Determination of specific speed at A, that is, at the point of maximum efficiency.

The point of maximum efficiency, such as point A, is where the losses are minimum. The specific speed of a machine is required to be determined at this point of maximum efficiency. It is quite possible to calculate the specific speed of the same machine at various other points on the curve, such as B, C, D, etc., resulting in a large number of values of the specific speeds for the same machine. But all those values, at off-peak points, do not “designate” the machine. Now, such graphs are drawn for different machines having different types of rotors. Figure 1.13 shows the resulting diagram. It is clear from the diagram that different shapes of the impeller or runner give rise to their own specific speeds at their respective maximum efficiencies. For any given shape of the impeller or runner, if the point of operation shifts from its designated specific speed (higher or lower side of its specific speed), the efficiency falls below its maximum value. Hence, the designated specific speed, which gives the maximum efficiency, characterizes the impeller or runner. h

A C D B

Q or P

Figure 1.13

Characterization of turbomachines by the specific speeds.

Basics of Turbomachines

35

Specific speed is an important parameter to characterize the impeller or runner. Another non-dimensional parameter, to accompany the specific speed, is the ratio of the two diameters of the rotor, do/di, where do is the outer diameter and di is the inner diameter. The ratio do/di is a convenient tool to indicate the shape of the impeller or runner. When do/di is 1 (do  di), the impeller or runner is purely axial flow type. As the ratio increases, the shapes change over to mixed flow and then to radial flow types. Figure 1.14 is an improved version of Fig. 1.13, with the x-axis (Q or P) getting a scale of the specific speed also and having representative values of do/di, along with shapes of impellers or runners. The curves represent the performance of such impeller or runner inclusive of the performance at off-peak points. For example, curve x represents the performance of one of the impellers or runners, not only at its peak efficiency but also at other off-peak points over the stretch of curve x. The figure shows the performance of a finite number of impellers or runners. Connecting the points of maximum efficiency of all such impellers or runners, another enveloping curve can be drawn, like the broken line in the figure. Making a choice of a point on this envelope curve and choosing do/di corresponding to such a point give rise to a shape of impeller or runner that must have the maximum efficiency. What Fig. 1.14 shows is also stated in a tabular form, in Table 1.6.

h

x

Q P
do /di 2.2

do di

Figure 1.14

1.8

do di

1.0

di

do

Performance of different types of impellers or runners.

The information of Fig. 1.14 and Table 1.6 is the result of the experimental work conducted on various types of impellers or runners. The specific speeds, the corresponding ratios do/di, and the types of “suitable” impellers or runners are all given only in terms of “ranges” of the values. Hence, the line of demarcation of limits between the types is rather thin. It is because of this that any design requires a model-testing process for the validation of the design.

36

Turbomachines

Table 1.6

Values of specific speeds

Turbine or Pump

Non-dimensional 7 Q H P 3 m /s m W

Q m3/s

MKS H m

P kW

Q cfs

n  rps

n  rad/s

Single jet

0.03–0.10

0.19–0.66

05–35

2.5–7.5

Multijet

0.10–0.21

0.66–1.32

30–70

7.5–16

Slow

0.18–0.36

1.13–2.27

60–120

13.5–27

Medium

0.36–0.54

2.27–3.40

120–180

27–40

Fast

0.54–0.90

3.40–5.67

180–300

40–65

1.80–3.00

11.34–18.90 600–1000

130–225

Slow

1.50–2.80

0.25–0.45

12–24

30–60

Medium

2.20–6.00

0.35–0.95

18–50

45–125

Fast

6.00–12.50 0.95–2.00

50–105

125–250

Mixed flow pump

11–25

1.80–4.00

100–210

245–512

Axial flow pump

20–35

3.20–5.70

170–300

415–732

Pelton turbine

Francis turbine

Kaplan turbine Centrifugal pump

FPS H ft

P HP

The design of a pump or a turbine starts with their respective basic specifications such as flow rate–head (Q–H) or power–head (P–H) values. With a selected value of the rpm (N), the basic data can be converted into a specific speed. This gives rise to the shape of the impeller or runner in terms of its do/di ratio. It is necessary to mention here that the specific speed has long been in use, but not as a dimensionless number. Earlier, the defining equations were N sp 

N Q for pumps H 3/ 4

(1.29)

N st 

N P for turbine H 5/ 4

(1.30)

Because Eqs. (1.29) and (1.30) were not dimensionless, the same basic data (flow rate and head) gave rise to different numerical values of specific speed in a different system of units such as FPS (British) or MKS (Metric). Further, in the FPS system, the flow rate was considered in two different units: cubic feet per second (cfs) or gallons per minute (gpm). Likewise in the MKS system, the power was considered in either kilowatts or metric horse power (MHP). Again, even in SI units, the speed of the rotor can be either “revolutions per second” or “radians per second” (1 rps  2O rad/s or 6.28 rad/s). Because of the above reasons, any reference to the values of the specific speed in the literature has to be made along with the units of the flow rate, head, power, and speed.

Basics of Turbomachines

Table 1.7

Units

Specific speeds in different systems of units

Basic Data Q, H Assumed N and G Calculated P Flow rate

MKS

37

Q

m3/s

Specific Speeds

1.2

Head

H

m of water 100

Speed

N

rpm

1000

Power

P

MHP

1471

Power

P

kW

1083

N sp 

N Q  34.64 H 3/ 4

N st 

N P  121.4 H 5/ 4

(P in MHP)

N st 

N P  104.0 H 5/ 4

(P in kW)

FPS

SI

Flow rate

Q

ft3/s

42.376

Head

H

ft of water

328

Speed

N

rpm

1000

Power

P

HP

1453

Flow rate

Q

m3/s

1.2

Head

H

m of water 100

Speed

N

rev/s

16.667

Power

P

Watts

1083024

Density

Q

kg/m3

1000

Speed

N

rad/s

104.72

N sp 

N Q  84.45 H 3/ 4

N st 

N P  27.3 H 5/ 4

7p 

n Q  0.104156 ( gH )3/ 4

7t 

n P  0.1 [ R s ( gH )5/ 4 ]

When n is in rad/s, 7p  0.6541 7t  0.6274

In Table 1.7, the different units are listed. For the purpose of illustration, the same data are considered (H  100 m of water  328 ft of water; Q  1.2 m3/s  42.376 cfs) to evaluate the specific speed. For the pump, the flow rate and head are the outputs, whereas the same flow rate and head are considered as input to the turbine, and a value of efficiency of energy conversion is assumed as 0.92 to get the output power. The main purpose of Table 1.7 is to illustrate the use of the different systems of units. A relation can also be formed between the different numerical values of the specific speeds, as obtained. For example, (MKS) Nsp  52.9 7p (n: rad/s)

(1.31)

Nst  193.5 7t (n: rad/s)

(1.32)

Also Nsp (FPS)  2.44 Nsp (MKS)

(1.33)

Nst (FPS)  0.225 Nst (MKS)

(1.34)

38

Turbomachines

The specific speed, as seen above, is a parameter based on the value of which the shape of the rotor can be selected for the maximum possible efficiency. Suppose the field conditions give rise to a “medium” value of specific speed (say Nst  120, 7  0.36). In such a case, a mixed flow type rotor gives the best efficiency. For the same data, an axial flow type rotor, if selected, gives only lesser efficiency. The above steps are only the preliminary steps in a design process. A design process is complete only when more details are worked out, such as the diameters at the inlet and outlet of the rotor, other physical dimensions, angles and profiles of vanes, the number of vanes, guide vanes, etc. One more important consideration is whether to have multiple units with the available total flow rate shared between the units. These details have been provided in chapters 5, 6, 7, and 8.

EXAMPLE 1.11 Trials on the model pump of Example 1.8 and the extrapolated values for the prototype pump are tabulated as the solutions of Example 1.8. Calculate the specific speeds for all the trials. Determine the specific speed that designates these pumps (model and prototype).

Solution: The values of Hm, Qm, Hp, and Qp are available in Table 1.4. Further, Nm  1500 rpm; Np  600 rpm. The specific speeds are calculated and stated in Table 1.8, using Eq. (1.29). Table 1.8

Trial No.

1

2

3

4

5

(Ns)m

42.09

54.49

71.76

93

138

(Ns)p

42.07

54.49

71.75

93

138.6

The specific speed of the pumps is 54.49 or 54.5. This value is chosen because the pumps have the maximum efficiency at the second trial values. A slightly better way of finding the designating specific speed is to plot the efficiencies with flow rate on the x-axis and draw a smooth curve through the points. It may be possible to locate the point of maximum efficiency at a value other than the tabulated values on the curve. If such a point is located, then the specific speed has to be calculated at those values. However, for the trials as detailed, the value obtained is 54.49 or 54.5. These values are in the MKS system of units.

EXAMPLE 1.12 A small turbine runs at 600 rpm using water at 50 lps at a head of 12 m with an overall efficiency of 0.87. This turbine is accepted as a model for developing a prototype turbine to be used in a powerhouse with an available head of 60 m and a flow rate of 6.6 m3/s. The speed selected for the prototype is 300 rpm. Find the scale-up factor and the number of turbines required for the powerhouse. Assume the same efficiency for the prototype.

Solution: We have Model: Nm  600; Qm  50 lps  0.05 m3/s; Hm  12 m; G  0.87. Prototype: Np  300; Hp  60 m; total flow available  6.6 m3/s.

Basics of Turbomachines

39

Equating head coefficients, we get Hp Hm



N p2 Dp2 N m2 Dm2

2 ¥ Dp ´ 60 ¥ 300 ´  ¦ s ¦ µ µ 12 § 600 ¶ § Dm ¶

2

Therefore, 2

¥ Dp ´ ¦ D µ  20 § m¶ So the scale-up factor is Dp Dm

 4.472

The power of the model is (wQHG)m  9810 s 0.05 s 12 s 0.87

 5120.8 W  5.1 kW Equating power coefficients we get 3

Pp

¥ N p ´ ¥ Dp ´ ¦ Pm § N m µ¶ ¦§ Dm µ¶

5

Power of the prototype is 3

¥ 300 ´ Pp  5.12 s ¦ s ( 4.472)5  1144.7 kW § 600 µ¶ Total power available is equal to (wQHG)  9810 s 6.6 s 60 s 0.87  3379.74 kW The number of turbines, each of 1144.7 kW is (total power available / power of each machine), equal to 3379.74  2.9525 or 3 1144.7 Comment: Suppose the first step in the solution is taken to equate flow coefficients. The scale factor Dp/Dm becomes too big because it means that only one unit of prototype( to handle all the flow) is to be used. As seen above, the available flow can be split between a number of units; the head is not split between the numbers. Also, the specific speed of the model and prototype can be checked ( N s  N P / H 5/ 4  60.787, 60.782).

40

Turbomachines

EXAMPLE 1.13 Derive an expression for the specific speed of a turbine in the form N st 

N Pt H 5/4

Solution: It is known that U t H and U  P DN / 60. Therefore H N

Dt Also P t QH

and

Q t D2 H

Substituting D as above we get Pt

H H 3/ 2 N2

Pt

H 5/2 N2

N t

H 5/4 P

Therefore N  Constant N st s

H 5/4 P

Hence N st 

N P H 5/4

EXAMPLE 1.14 Derive an expression for the specific speed of a pump in the form N sp 

N Q H 3/4

Solution: It is known that U t H and U  P DN / 60. Therefore Dt

H N

Basics of Turbomachines

41

Also Q t D2 H H Q t 2 s H N H 3/2 Q t 2 N H 3/4 N t Q Therefore N  Constant N sp s Hence N sp 

H 5/4 Q

N Q H 3/4

K eywords Rotor vanes or rotor blades Fluid Dynamic action Positive displacement machines (PDMs) Mechanical efficiency Volumetric efficiency Hydraulic or adiabatic efficiency Radial flow Axial flow Mixed flow Tangential flow Impeller Runner Stator Guide vanes

Nozzles Diffusers Pressure energy Kinetic energy Potential energy Enthalpy Head Power flow Model studies Flow coefficient Head coefficient Power coefficient Unit and specific quantities Specific speed

S ummary 1. A basic definition of a turbomachine is spelt out with some comments. observations of different features (energy-transfer, rotor-fluid, dynamic action, change of pressure/momentum of fluid).

2. Turbomachines are compared with PDMs (energy-transfer mechanism, construction, operation, efficiencies, application). 3. A classification of turbomachines is considered in order to identify the different types

42

Turbomachines

4.

5.

6.

7.

(with respect to energy-flow, fluid-flow, typeof-fluid, dynamic action). Some turbomachines are illustrated, with a purpose of getting familiar with the terminology of the hardware details. Units and dimensions are recalled from the earlier courses such as mechanics, thermodynamics, and fluid mechanics not only as a revision but also as applied to turbomachines. In continuation of the SFEE from thermodynamics, an expression for “specific work” is identified (change of total energy of fluid). The “power flow” is schematically presented with respect to the power-generating as well as power-absorbing turbomachines to show

an over view of the energy flow, including the losses, in these machines. Some important efficiencies are stated with reference to the powerflow diagrams (Figs. 1.7, 1.8, 1.9, 1.10). 8. Unit and specific quantities are studied, with reference to turbomachines (unit quantities: H  1; Specific quantities: H  1 and D  1). 9. The “model studies” is discussed, but in a cursory way (geometric, kinematic and dynamic similarities; O1, O2, O3, extrapolation of performance). 10. Specific speed, one of the most important parameters, is studied in some detail.

I mportant E quations There are quite a few equations, but the equations stated are more like “defining equations.” The material presented is only to facilitate the understanding of the later chapters. 1. Work output of a turbomachine: ¥ V2 V2 g´ ¥ g´ W  ¦ h1  1  z1 µ ¦ h2  2  z2 µ 2 gc 2 gc gc ¶ § gc ¶ § 2. Specific work: ¥ p V2 V2 g´ g´ ¥p W  ¦ 1  1  z1 µ ¦ 2  2  z2 µ gc ¶ § R 2 gc gc ¶ § R 2 gc All the units are J/kg. 3. Also p p2 V12 V22 ( z1 z2 ) g W  1   R 2 gc gc 4. Head H

p2 p1



V22 V12

 ( z2 z1 ). 2g w All units are meters of liquid. In continuation, P  wQH 5. The efficiencies of work-producing turbomachines:

Utilization factor E 

Pr Pn

W



W 

V22 2 gc

Hydraulic or adiabatic efficiency P P P Gh or Ga  r  r s n P Pn P Volumetric efficiency ¥ m $m ´ ¥ Q $Q ´ P  Gv  ¦  § m µ¶ ¦§ Q µ¶ Pa Mechanical efficiency Hm 

Ps Pr

Overall efficiency P P P P Go  s  s s r s  Hm s Hh s Hv Pa Pr P Pa 6. The efficiencies of work-absorbing turbomachines: P Mechanical efficiency Gm  r Ps Volumetric efficiency Gv 

P m  n m  $m Pr

Hydraulic or adiabatic efficiency Gh or Ga 

P Pn

Basics of Turbomachines

Overall efficiency Go 

P P Pn Pr  s s Ps Pn Pr Ps

P1 

10. The non-dimensional specific speed 7 is given as

 Hh s Hv s Hm

7p 

Q ND 3

7t 

7. Flow coefficient

8. Head coefficient

n Q for pumps ( gH )3/ 4 n P for turbines R ( gH )5/ 4

11. The specific speed defining equations:

H P2  2 2 N D

N sp 

N Q for pumps H 3/ 4

N st 

N P for turbines H 5/ 4

9. Power coefficient

P3 

43

P 3 N D5

M ultiple- C hoice Q uestions 1. The dimensions of gc are (a) N s2/(kg m) (b) kg s2/(N m) 2 (c) kg m/(N s ) (d) m s2/(kg N) 2. In a centrifugal pump, the energy flow is (a) fluid to rotor to shaft (b) rotor to fluid to motor (c) shaft to fluid to runner (d) shaft to rotor to fluid 3. In a steam turbine, the input energy is (a) the chemical energy of coal (b) the mechanical energy of shaft (c) the thermal energy of steam (d) the isothermal energy of water 4. In a gas turbine power plant, (a) the compressor is driven by the turbine (b) the compressor drives the turbine (c) the generator drives the turbine (d) the turbine is driven by the compressor 5. The unit of kinetic energy is (a) m/s2 (b) m/kg (d) m/s (c) m2/s2

6. The unit of pressure energy is (b) J/kg (a) N/m2 (c) kg/m N (d) N s/kg 7. The highest amount of energy, among the listed, is (a) 1 m of water (b) 1 m of mercury (c) 1 m of alcohol (d) 1 m N/kg 8. In a Pelton turbine setup, the highest energy is in (a) shaft (b) jet (c) nozzle (d) rotor 9. In a Francis turbine setup, the least energy is in water at (a) inlet guide vanes (b) rotor blade passages (c) draft tube (d) jet 10. The specific speed of a centrifugal pump is to be calculated (a) when its output head is maximum (b) when its flow rate is maximum (c) when its efficiency is maximum (d) when its input power is maximum

R eview Q uestions 1. Define a turbomachine. Explain the salient features of a turbomachine. (Refer Section 1.2)

2. Compare a turbomachine with a positive-displacement machine. (Refer Table 1.1)

44

Turbomachines

3. Classify turbomachines and give examples. (Refer Table 1.2) 4. List the different losses taking place in a power-generating turbomachine. State the corresponding efficiencies. (Refer Section 1.9.1.1 and Fig. 1.7) 5. List the different losses that take place in a power-absorbing turbomachine. State the corresponding efficiencies. (Refer Section 1.9.1.2 and Fig. 1.8) 6. State the objective of the model studies. Write a note on the model studies. (Refer Section 1.10) 7. What are the limitations of the model studies? (Refer Section 1.11)

8. Explain the significance and use of the flow coefficient, head coefficient, and power coefficient. (Refer Section 1.11) 9. Define and derive the expressions for unit flow, unit speed, unit power, specific power, and specific flow. (Refer Section 1.12) 10. Explain the specific speed of (a) a turbine and (b) a pump. Also explain how it is determined. (Refer Section 1.15 and Fig. 1.12) 11. Explain how the characteristics of one machine can be extrapolated to get the characteristics of another similar machine. (Refer Section 1.11and Fig. 1.11)

E xercises 1. The state of air at the inlet to a compressor is as follows: Pressure  60 kPa, temperature  –35oC and velocity  120 m/s. The outlet pressure and velocity are 350 kPa and 280 m/s, respectively. If the isentropic compression efficiency is 85%, calculate the outlet temperature and the power input to the compressor for the flow rate of 5 kg/s, cp  1.004 kJ/kg K, F  1.4. 2. Air enters a turbine at 350 kPa, 780oC, and at a velocity of 280 m/s. The outlet pressure is 60 kPa at a velocity of 180 m/s. The isentropic expansion efficiency is 88%. Calculate the power output of the turbine for the flow rate of 5 kg/s, cp  1.004 kJ/kg K, F  1.4. 3. The mass flow rate in a turbine is 10 kg/s and the output is 4000 kW. The velocity of gases at the outlet is 150 m/s. The mechanical and adiabatic efficiencies are 0.93 and 0.84, respectively. Calculate the rotor power, power lost in the exit, power lost in the runner, and the utilization factor. Take cp  1.004 kJ/kg K, F  1.4. 4. The output of a motor, driving an air compressor, is 320 kW. The mass flow rate through the compressor is 1.5 kg/s. The velocity of air at the outlet is 180 m/s. The mechanical efficiency of the compressor is 94%. The losses in

the runner amount to 33 kW. The leakage or return flow losses are negligible. Determine the hydraulic efficiency, power lost in the bearings, power lost in the exhaust, and overall efficiency. Take cp  1.004 kJ/kg K, F  1.4. 5. The ratings of an available pump are as follows: Q  50 lps, H  20 m of water, N  1440 rpm. Calculate the capacity and head of the pump if it were to be run at 1200 rpm. Also calculate the power output at rated conditions and at 1200 rpm. State your comments on the efficiency of the machine. 6. The design flow rate of water of an axial flow pump is 2.1 m3/s at a speed of 960 rpm. The runner diameter is calculated as 60 cm. The head to be generated is 15 m of water. The pump is expected to have an overall efficiency of 82%. A one-fourth size model of this pump is required to be tested. If the speed of the model is fixed to 1440 rpm, determine the head and the capacity of the model to be expected while testing. Assume that the efficiency of the model is the same as that of the prototype. Calculate the powers required to drive the model and the prototype. Also determine the specific speed of the model or prototype.

Basics of Turbomachines

7. In a test rig in a laboratory, a pump is used to supply water at a rate of 15 lps, at a head of 60 m of water, to run a Pelton turbine. The pump is driven by a motor that has an output of 10 kW at a particular trial. The mechanical efficiency of the pump is 95%. The Pelton turbine has a hydraulic efficiency of 93% and runs a dynamometer that absorbs 7.35 kW of power. Determine the hydraulic efficiency of the pump and the mechanical efficiency of the turbine. The losses in the transmission pipe (from the pump to the turbine) are estimated to be 0.5 kW at the given flow rate. Draw up an energy balance of the entire setup. The volumetric losses of the pump and the turbine are negligible. 8. The speed of the motor in the above test rig is recorded as 2880 rpm and the speed of the turbine as 600 rpm. Calculate the specific speeds and shape numbers of the pump and the turbine. 9. A turbine is designed to run at a speed of 600 rpm for a site where the head available is 80 m of water and the flow rate available is 1.8 m3/s. It is required to test a one-fourth size model of the turbine. The head available in the laboratory is only 8 m of water. Assume an overall efficiency of 82% for both model and prototypes. Calculate

P roject- O riented Q uestions 1. It is required to make a comparative study of the turbomachines (handling air or gas), which are mentioned in the solved and exercise examples of this chapter with respect to the parameters such as pressure ratios, speed, power per unit flow rate, size, efficiencies, etc. Take up this activity as a small project and evaluate/ assess the performances. Also, make an audit of energy balance. 2. It is required to undertake the assessment of the equipments handling water (pumps

45

(a) The flow rate required in the lab for the test. (b) The speed at which the model should be tested. 10. To deliver 0.3 m3/s of water through a total head of 300 m, it is intended to use several pumps in series, with the flow at the outlet of one pump going to the inlet of the next pump and so on, and the heads are additive. A small pump of diameter 10 cm, having an efficiency of 88%, is considered as a model. This model gives a flow rate of 0.02 m3/s against a head of 10 m while running at 980 rpm. Determine the diameter of a similar prototype pump, running at 1440 rpm, and the number of such pumps to meet the above duty. 11. A pump is designed for a flow rate of 200 lps at a head of 150 m of water while running at 1440 rpm. A model of this pump delivers 15 lps with a power input of 0.5 kW. Calculate the scale ratio of the model and the speed of the model. 12. A turbine of speed 375 rpm is designed for a project site where the available head and flow rate are 35 m of water and 15 m3/s. The model of this turbine runs at 250 rpm under a head of 2 m. Calculate the specific speed, scale ratio, flow rate, and power of the model.

S olutions A vailable and turbines separately) mentioned in the solved and exercise examples of this chapter, with respect to the parameters such as head, flowrate, speed, specific speed, power, efficiencies, etc. Take up this activity as a small project. Prepare a table of the design parameters, evaluate and present your assessment. You may use the data of Table of Specific Speeds, if required, to comment.

46

Turbomachines

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5.

(c) (d) (c) (a) (c)

6. 7. 8. 9. 10.

(b) (b) (c) (c) (c)

Exercises 1. 2. 3. 4. 5.

421.53 K, 1081.32 kW 1950 kW 4301 kW, 112.5 kW, 840.7 kW, 0.9726 89.03%, 19.2 kW, 24.3 kW, 83.7% 41.67 lps, 13.89 m, rated power  9.81 kW, power at 1200, rpm  5.678 kW. At any speed other than design speed, the efficiency gets reduced.

6. 2.11 m, 49.22 lps, 1.24245 kW, 363.55 kW, 182.52 7. 92.94%, 94.9% 8. 16.36, 0.0492, 9.74, 0.00935 9. (a) 35.55 lps and more, (b) 759 rpm 10. 21.7 cm, 3 11. 0.706, 307 rpm 12. 316, 0.35857, 0.461 m3/s, 9.045 kW

2

Thermodynamics of Fluid Flow

Learning Objectives After completing this chapter, you will be able to: v
Understand the basic compression process in its theoretical and actual form and the deviations with both qualitative and quantitative approaches. v
Understand the basic expansion process in its theoretical and actual form and the deviations with both qualitative and quantitative approaches.

v
Study the various forms of efficiencies in the above processes and the applicability of such processes. v
Study the effect of stage-wise processes compared to single-stage processes with respect to both compression and expansion. v
Study the effect of compression and expansion as the basic processes, independent of the parameters such as pressure ratios.

2.1 Introduction The fluids involved in the energy-transfer processes with the rotor blades of turbomachines are either incompressible or compressible. The incompressible fluids are liquids. Turbomachines that handle liquids are the hydraulic turbines working with water and the pumps working with water or various liquid chemicals in the process industries. Different fractions of petroleum crude in refineries, milk in dairy industries, liquid dyes in textile industries, naptha and urea in fertilizer industries, paints in paint industries, liquid pulp in paper industries are only some examples of incompressible fluids. The list of such fluids and the variety of pumps handling them are almost endless. The compressible fluids undergo a wide variation in their densities during their transit in turbomachines. The examples of compressible fluids are as follows: 1. 2. 3.

Air in compressors. Combustion gases in gas turbines. Steam in steam turbines, etc.

The behavior of compressible fluids during the processes in turbomachines can be approximated to that of the perfect gases. It is assumed that the laws of perfect gases hold good for these fluids in the range of their properties under consideration. This chapter is intended for the study of compression and expansion of compressible fluids in turbomachines.

2.2 Static and Stagnation States The static state of a fluid is the state described by its various properties without the effect of its velocity. Its properties such as pressure or temperature can be measured by some static probes. On the

48

Turbomachines

other hand, if the fluid is moving at a velocity, then its properties can be identified in two different types: (a) In one type, the effect of the velocity is neglected (as if the fluid is not having any velocity, as if it is at rest stored in a container); (b) In the second type the effect of velocity is considered. In this type, the properties are in the form of the “stagnation properties.” The recognition of the existence of velocity is best illustrated by the very definition of the “stagnation state.” The stagnation state of a moving fluid stream is the state attained by the fluid when it is suddenly or abruptly (or isentropically) brought to rest without any heat transfer or work done. In the process, the kinetic energy associated with the velocity is converted into the internal energy of the fluid, with consequent increase in the values of its properties such as pressure, temperature, enthalpy, etc. Properties of the fluid at the stagnation state are the stagnation properties, “dynamic properties,” or “total properties.” The symbols of stagnation properties are the same as those of static properties, but with a subscript “0” signifying that the velocity has been brought to zero. Thus, when the fluid moves at a velocity V with its static temperature T, static enthalpy h, static pressure p and density (P), its stagnation temperature To, stagnation enthalpy ho and stagnation pressure po are respectively given by the following expressions: T0  T 

V2 2c p g c

(2.1)

h0  h 

V2 2 gc

(2.2)

p0  p 

RV 2 2 gc

(2.3)

The stagnation properties can be shown on the property diagrams, such as the enthalpy–entropy diagram, temperature–entropy diagram, etc., just as the static properties. In the context of the processes taking place in turbomachines, it is most appropriate to use stagnation states because the velocity of the fluid is an integral part of the energy-transfer process. However, there are some situations where it is sufficient to consider only static states.

EXAMPLE 2.1 A stream of combustion gases at the point of entry to a turbine has a static temperature of 1050 K, static pressure of 600 kPa, and a velocity of 150 m/s. For the gases, cp  1.004 kJ/kg K and F  1.41. Find the total temperature and total pressure of the gases. Also find the difference between their static and total enthalpies.

Solution: We have p  600 kPa, T  1050 K, and V  150 m/s. The total temperature is given by V2 T0  T  2c p g c

Thermodynamics of Fluid Flow

49

(150)2 2 s 1004  1061.2 K  1050 

We know that R

G 1 0.41 s cp  s 1004  291.9 J/ kg K G 1.41

Also,

R

p RT

1000 (291.9 s 1050)  1.96 kg /m3  600 s

The total pressure ( p0 )  p 

RV 2 2 gc

1.96 s (150)2 2 s 1000  622.023 kPa  600 

The difference between static and total enthalpies is h0 – h  cp(T0 – T)  1.004 (1061.2 – 1050)  11.245 kJ/kg

EXAMPLE 2.2 Water at a static temperature of 300 K and static pressure of 200 kPa flows in a conduit at a velocity of 10 m/s. Calculate its total temperature and total pressure. cp for water is 4.18 kJ/kg K.

Solution: The total temperature is given by T0  T 

V2 2c p g c

(10)2 2 s 4180  300.012 K  300 

The total pressure is given by ( p0 )  p 

RV 2 2 gc

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Turbomachines

1000 s (10)2 2 s 1000  250 kPa  200 

2.3 Thermodynamics of Turbomachine Processes Turbomachine processes are compression and expansion processes in all their varieties such as isentropic, polytropic, single-stage and multistage processes. An ideal compression process is one in which a fluid is compressed using the least mechanical work. An ideal expansion process is one in which a fluid expands through a given pressure ratio, generating the maximum possible mechanical work. These processes have to be reversible processes for them to become the “most efficient processes.” An ideal compression process with the least amount of mechanical work is a reversible isothermal process. This reversible isothermal process is taken as the standard process to compare and judge the performance of actual processes. An actual compression process is compared with the reversible isothermal process to evaluate the measure of performance, that is, its isothermal efficiency: Isothermal efficiency 

Ideal isothermal work input Actual work input

However, an isothermal process requires that the fluid be continuously cooled to maintain the temperature at its initial level. At the practical flow rates of fluids in the compressors, it is very difficult to provide such cooling during the process. (“Inter cooling” between the stages is different.) Hence, an alternate but more practical standard of comparison is adopted, that is, an isentropic process. Consequently, the measure of performance, the isentropic efficiency, is taken as Isentropic efficiency 

Isentropic work input Actual work input

An ideal expansion process with the maximum amount of mechanical work output is a reversible adiabatic or isentropic process. Hence, this isentropic process is taken as the standard process to compare and judge the performance of an actual process. Thus, an actual expansion process is compared with an isentropic expansion process between the same pressure limits to evaluate the measure of performance, that is, its isentropic efficiency (or adiabatic efficiency): Isentropic efficiency 

Actual mechanical work output Ideal isentropic work output

The efficiencies as mentioned above have the expressions of the type “ideal isothermal work” or “ideal isentropic work.” Both these expressions require the theoretical calculation of the work and to do so, they require the precise initial and final states of the fluid. These states have to be stagnation states because the velocities of the fluids in turbomachine practice cannot be generally neglected.

2.4 Isentropic Compression Process Figure 2.1 represents an enthalpy–entropy diagram that shows the compression process of a fluid (such as a perfect gas) in a compressor. The pressure limits are p1 and p2. The initial state is A. The isentropic and actual

Thermodynamics of Fluid Flow

51

processes are AB and AC, respectively. Process AC indicates the increased entropy due to the irreversibility during the actual process. h p2 C B

p1 A s

Figure 2.1

Compression process between static states.

The isentropic efficiency, as specified in Section 2.3, in terms of the enthalpies is given by

His   His 

Isentropic enthalpy change Actual enthalpy change hB hA

(2.4)

hC hA

Equation (2.4) is more like a defining equation, without specifically mentioning whether the enthalpies are static or stagnation values. Figure 2.2 shows the more detailed compression processes, taking into account the velocities or the kinetic energies. The “actual” compression process has to be between the stagnation states: from the initial stagnation state 01 to the final stagnation state 02. In Eq. (2.4), therefore, the denominator of the expression becomes h02 – h01. h 02 02


P02 P2

2


01

P01 P1

1

s

Figure 2.2

Compression processes between both static and stagnation states.

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Turbomachines

The numerator of Eq. (2.4) now depends on how the efficiency is defined. Four different forms of isentropic efficiencies can be identified, with the numerator suitably defined: 1.

2.

3.

4.

Total-to-total efficiency, with numerator from state 01 to state 02^: This is applicable when the kinetic energies at the inlet and outlet are significant. At the inlet, the kinetic energy is available and at the outlet, the kinetic energy is useful, such as in an intermediate stage of compression. Total-to-static efficiency, with numerator from state 01 to state 2^: This is applicable when some kinetic energy is available at the inlet and the kinetic energy at the outlet is not useful; that is, when the fluid maybe stored in a tank without any effect of the magnitude of its kinetic energy (e.g., compressed air in an air tank). This situation may correspond to the last stage in a series of stages. Static-to-total efficiency, with numerator from state 1 to state 02^: This is applicable when there is no kinetic energy at the inlet, that is, the suction is from a stagnant reservoir (e.g., air from atmosphere), and the kinetic energy at the outlet is useful (in the next stage). This situation may be the first stage in a series of stages of compression. Static-to-static efficiency, with numerator from state 1 to state 2^: This is applicable when the kinetic energies at both the inlet and the outlet are insignificant or the energies are not available at the inlet or not useful at the outlet.

The above situations can be concisely expressed in the following forms: (Hc )t-t  (Hc )t-s  (Hc )s-t  (Hc )s-s 

h02a h01 h02 h01 h2a h01 h02 h01 h02a h1 h02 h01 h2a h1 h02 h01

(2.5) (2.6) (2.7) (2.8)

In the subscripts, the subscript c inside the brackets indicates that the process is compression and in t-t, t-s, s-t, and s-s, t is for total and s is for static. Wherever these suffixes are not mentioned (as in almost all cases), the efficiencies may be taken as total to total. Because the denominators in Eqs. (2.5)–(2.8) are the same, it is very easy to find which efficiency is maximum or which is minimum. In Eqs. (2.5) and (2.7), where the outlet kinetic energy is considered useful, the so-called exit losses are no more losses; this situation modifies the energy flow diagram (Fig. 1.7) in such cases. In all the expressions for the efficiencies [Eqs. (2.5)to–(2.8)], the specific enthalpies can be replaced by the corresponding temperatures, with the assumption that the specific heat cp remains constant over the ranges of pressures or temperatures under consideration. However, steam is not considered in compression processes, and when it is considered, the enthalpies in the expressions cannot be reduced in terms of temperatures. The enthalpies are to be read from steam tables. Further, the isentropic relationship between the temperature ratio and pressure ratio can always be used, as follows (e.g., between states a and b):

Thermodynamics of Fluid Flow

¥p ´ ¦ aµ Tb § pb ¶ Ta

53

(G 1)/G

(2.9)

To illustrate this, Eq. (2.5) can be written as (Hc )t-t  Therefore, T02 T01 

T02a T01 T02 T01

1 (T T01 ) (Hc )t-t 02a



´ T01 ¥ T02a

1µ ¦ (Hc )t-t § T01 ¶



(G 1)/G ´ T01 ¥ ¥ p02a ´

1 ¦¦ µ (Hc )t-t ¦§ § p01 µ¶ µ¶

(2.10)

The actual work of compression can now be written as Wc  c p (T02 T01 ) (G 1)/G · c pT01 ¨¥ p02a ´ ©¦

1¸ (2.11) µ (Hc )t-t ©§ p01 ¶ ¸ ¹ ª If correlations are available for the efficiency values such as (Hc )t-t , then the work requirement (and power) for the process can be calculated in terms of the initial temperature T01 and the required pressure ratio (p02a/p01). A similar procedure as above can be adopted to calculate the work required, starting from Eqs. (2.6) to (2.8). Thus,

 Wc 

Wc 

(G 1)/G · c pT01 ¨¥ p ´ ©¦ 2a µ

1¸ (Hc )t-s ©§ p01 ¶ ¸ ¹ ª

(2.12)

(G 1)/G · c pT1 ¨¥ p02a ´ ¸ ©¦ Wc 

1 (Hc )s-t ©§ p1 µ¶ ¸ ¹ ª

(2.13)

(G 1)/G · c pT1 ¨¥ p ´ 2a © Wc 

1¸ ¦ µ (Hc )s-s ©§ p1 ¶ ¸ ¹ ª

(2.14)

It can now be appreciated why different efficiencies are identified and how they are used.

EXAMPLE 2.3 Air at 100 kPa and 300 K (static) is compressed in a stage of a compressor through a static pressure ratio of 1.9. The velocities at the inlet and outlet are 15 and 120 m/s, respectively. If the total-to-total efficiency is 89%, calculate the total-to-static, static-to-total, and static-to-static efficiencies. For air, take F  1.4 and cp 1.004 kJ/kg K.

54

Turbomachines

Solution: We know that ¥p ´  ¦ 2a µ T1 § p1 ¶

T2a

(G 1)/G

 (1.9)0.4/1.4  1.2

Therefore, T2a  T1 s 1.2  300 s 1.2  360 K Now, T1  300 K, T2a  360 K. So T01  300 

(15)2  300.112 K 2 s 1004

T02 a  360 

(120)2  367.171 K 2 s 1004

We have h02a – h01  cp(T02a – T01)

 1.004 (367.171 – 300.112)



 67.33 kJ/kg K Hence, h02 – h01  67.33 / 0.89  75.65 kJ/kg h2a – h01  67.33 – 7.171  60.159 kJ/kg h2a – h1  67.33 – 7.171  0.112  60.271 kJ/kg h02a – h1  67.33  0.112  67.442 kJ/kg Therefore,

Ht-s  Hs-t  Hs-s 

h2a h01 h02 h01 h02a h1 h02 h01 h2a h1 h02 h01



60.159  79.52% 75.65



67.442  89.15% 75.65



60.271  79.67% 75.65

EXAMPLE 2.4 From an initial total pressure of 100 kPa and a total temperature of 300 K, two fluids are to be compressed to a final total pressure of 600 kPa. Calculate the work required if the process is isentropic. One fluid is water with the density of 1000 kg/m3 and the other is a perfect gas with cp  1.004 kJ/kg K, F  1.4.

Thermodynamics of Fluid Flow

55

Solution: When the fluid is water, the work required is given by $h 

p02 p01

R



600 100  0.5 kJ/kg 1000

When the fluid is a perfect gas T01 T02

¥p ´  ¦ 01 µ § p02 ¶

(G 1)/G

¥ 100 ´ ¦ § 600 µ¶

0.4 /1.4

 0.6

Therefore, T02 

T01 0.6



300  500 K 0.6

Hence, Work required  cp (T02 – T01)

 1.004 (500 – 300)



 200.8 kJ/kg

2.5 Isothermal Compression Process In Section 2.3 it has been mentioned that the least amount of work required for a given compression process is during the reversible isothermal compression (although the standard reference process for comparison cannot be this reversible isothermal process). The ideal work for this process is given by 2

Wideal  ° pdv 1

¥p ´  RT1 ln ¦ 2 µ § p1 ¶

(2.15)

Wactual  c p (T02 T01 )  Q

(2.16)

Also,

where Q is the heat extracted to cool the fluid during the process. Hence, the isothermal compression efficiency can be written as ¥p ´ RT1 ln ¦ 2 µ § p1 ¶ (Hc )isth  (2.17) c p (T02 T01 )  Q Consequently, the actual work of compression can be written as Wactual 

¥p ´ 1 RT1 ln ¦ 2 µ (Hc )isth § p1 ¶

(2.18)

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Turbomachines

2.6 Isentropic Expansion Process The standard expansion process with which the actual expansion process can be compared, as mentioned in Section 2.3, is the isentropic expansion process between the same pressure limits. This isentropic expansion process is reversible and the most efficient one, with the maximum work output. It is also a practical process. Figure 2.3 shows the detailed process with the static and stagnation states, shown at both the inlet and the outlet points on the enthalpy–entropy diagram. The fluid expands from the higher pressure p1 to the lower pressure p2 in a turbine to produce the shaft work. Generally, the intended output of the expansion process is the shaft work. h p01 01

p1

1

02

p02 p2

02
2


s

Figure 2.3

Expansion process.

A turbine cannot work with a fluid from a tank or reservoir because the source should not get depleted. There must necessarily be a continuous stream of fluid entering the turbine. Therefore, the kinetic energy at the inlet (V12 / 2 g c ) is a necessary component of the total energy to be considered, making it essential to use the stagnation state at the inlet. The actual shaft work Ws is up to the outlet state 02, so that WS  h01 – h02 But at the outlet, whether the state to be considered is static or stagnation on the ideal isentropic process depends on the following situations. 1.

Case 1: The expansion process is one of a sequence of processes, such as the one in any intermediate stage (neither the first, nor the last stage) of a multi-stage turbine. The kinetic energy at the outlet of the present process is a part of the inlet energy of the next-stage process. Therefore, it is not a loss. As a consequence, the “accounting” of V22 / 2 g c can be (and has to be) transferred to the next stage. The exit state for the present process can be taken as its stagnation state, 02. The corresponding isentropic process (for comparison) in this case becomes 01–02a. The isentropic efficiency is given by (Ht )t-t  

Output shaft work Ideal enthalpy drop h01 h02 h01 h02a

(2.19)

Thermodynamics of Fluid Flow

2.

In the subscript part, the prefix t is for the expansion process or the turbine process. Because the ideal expansion is from the total state 01 to the total state 02a, this efficiency is known as the totalto-total efficiency. This expression is not applicable to the last stage of a turbine because the exit kinetic energy of the last stage is to be taken as a loss. But one exception here is a turbojet engine. In a turbojet engine, the turbine produces just enough shaft work to run the compressor and accessories. The enormous amount of energy at the exit of the last stage of the turbine (pressure and kinetic energy together) is meant for the expansion in the exit nozzles, to create the velocity and the thrust for the propulsion of the aircraft. The total-to-total efficiency [Eq. (2.19)] is applicable to the above process. Case 2: The expansion process occurs in the last stage of a series of stages, or in only one stage (all by itself ). In both the types, the kinetic energy at the outlet has no further use in any form. In this case, the exit kinetic energy has to be taken as a loss. This loss could have been used or at least reduced by the expansion process extending further. But the expansion process does not use this kinetic energy and hence it is lost. Thus, the ideal expansion process to be considered for comparison is from the stagnation state 01 at the inlet to the static state 2a at the outlet. The isentropic efficiency is given by (Ht )t-s  

3.

57

Output shaft work Ideal enthalpy drop h01 h02 h01 h2a

(2.20)

Because the ideal expansion is from the total state 01 to the static state 2a, this efficiency is known as the total-to-static efficiency. Case 3: In this situation, the useful output comprises both the shaft work and the kinetic energy at the outlet. But the outlet kinetic energy does not run the next stage. Instead, it is useful in some other way, like in the case of steam turbines, the exhaust steam is utilized as process steam in process industries (e.g., controlled heating, constant temperature baths, etc.). Although it is not directly the kinetic energy, the utility of process steam is one of the important factors.

In this context, the ideal expansion is considered from the stagnation state 01 to the static state 2a. The output can now be considered as the shaft work and exit kinetic energy put together. Consequently, the efficiency is given by (Ht )t-s 

h01 h2 h01 h2a

(2.21)

Equations (2.19)–(2.21) give the expressions for the efficiencies in terms of the different enthalpies. The numerators of all the three expressions signify the intended output in the three cases. In the case of steam turbines, the values of enthalpies in the expressions for the efficiencies are to be taken from the steam tables. But in the case of gas turbines, the enthalpies are the functions of temperatures. Hence, it is quite possible to reduce the three expressions in terms of the corresponding temperatures, and then in terms of the initial temperature and the possible pressure ratios, using the usual isentropic relationships. For example, Eq. (2.19) can be written as

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Turbomachines

(Ht )t-t  



c p (T01 T02 ) c p (T01 T02a ) c p (T01 T02 ) ¨ ¥ p ´ (G 1)/G · ¸ c pT01 ©1 ¦ 02a µ ¸ © § p01 ¶ ¹ ª WS ¨ ¥ p ´ (G 1)/G · ¸ c pT01 ©1 ¦ 02a µ © § p01 ¶ ¸ ¹ ª

The actual shaft work becomes ¨ ¥ p ´ (G 1)/G · ¸ Ws  c p (T01 T02 )  (Ht )t-t c pT01 ©1 ¦ 02a µ ¸ © § p01 ¶ ¹ ª

(2.22)

The second case of the efficiency (Eq. 2.20) is reduced as (Ht )t-s  



c p (T01 T02 ) c p (T01 T2a ) WS ¥ T ´ c pT01 ¦1 2a µ § T01 ¶ WS ¨ ¥ p ´ (G 1)/G · ¸ c pT01 ©1 ¦ 2a µ ¸ © § p01 ¶ ¹ ª

The actual shaft work is ¨ ¥p ´ Ws  (Ht )t-s c pT01 ©1 ¦ 2a µ © § p01 ¶ ª

(G 1)/G

· ¸ ¸ ¹

(2.23)

In the third case, the following result is obtained: ¨ ¥ ( p / p )(G 1)/G 1´ V2 · Useful output  (Ht )t-s c p ©T1 ¦ 1 2 (G 1)/G µ  1 ¸ ¶ 2c p g c ¸¹ ©ª § ( p1 / p2 )

(2.24)

Thermodynamics of Fluid Flow

59

EXAMPLE 2.5 Combustion gases flow into a gas turbine stage at a rate of 1.4 kg/s. At the inlet, the static pressure and temperature are 550 kPa and 950 K, respectively, and the velocity is 180 m/s. At the outlet of the stage, the static pressure is 250 kPa and the velocity is 75 m/s. If the power output of the stage is 250 kW, find the efficiency of the stage when (a) the stage is the first one. (b) the stage is one of the middle stages. (c) the stage is the last one. Take cp  1.004 kJ/kg K, F  1.41.

Solution: The gas constant is given by R

(G 1) c p  291.9 J/ kg K G

The density is p 550 s 1000  RT 291.9 s 950  1.9834 kg /m3

R

The total temperature at inlet is given by T01  950 

(180)2  966.14 K 2 s 1004

The total pressure at inlet is given by p01  550  1.9834 s

(180)2  582.13 kPa 2 s 1000

Now by isentropic relation between p1 and p2a, we have ¥p ´  ¦ 2a µ T1 § p1 ¶

T2a

(G 1)/G

¥ 250 ´ ¦ § 550 µ¶

0.41/1.41

 0.7983

Therefore, the static temperature at the end of isentropic process is given by T2a  950 s 0.7983  758.385 K And the corresponding total temperature is given by T02 a  758.385 

(75)2  761.186 2 s 1004

Actual power developed is cp (T01 – T02) m  1.004 (966.14 – 761.186)1.4  288.1 kW

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Turbomachines

(a) and (b) If the stage is the first one or any one of the middle stages, the efficiency may be taken as total-to-total. Therefore, Efficiency 

250  86.78% 288.1

(c) If the stage is the last stage, the efficiency to be considered is total-to-static. Therefore, Efficiency 

250 c p (T01 T2a )



250 1.004(966.14 758.385)



250 292

 85.62% Comment: In an expansion process or a turbine process, a “static” state at the inlet to the first stage cannot be generally considered because the turbine cannot have a steady work output with the inlet from a stagnant source (atmosphere or a storage tank). Hence, for the expansion process, “static-tostatic” or “static-to-total” efficiency need not be considered, unless it is some sort of “instantaneous” efficiency.

2.7 Overall Isentropic Efficiency versus Stage Efficiency The overall isentropic efficiency in the present context refers to the thermodynamic efficiency of multi-stage turbomachines. (In Chapter 1, the overall efficiency was identified as the net effect of mechanical, hydraulic, and volumetric efficiencies. The contexts were different there.) A multi-stage machine comprises several stages. A typical multi-stage air compressor may have about eight or ten stages. However, a multi-stage steam turbine may have even 18 or more stages. In a given machine, generally, each stage is designed to have the same pressure ratio and the same efficiency. The method of manufacturing the components is not different from stage to stage. Because the surface finish of all the blades and the flow passages are the same, each stage has the same efficiency as the other stages. This stage efficiency has already been studied in Section 2.6 for both compression and expansion processes. The overall efficiencies, with all the stages put together, can be substantially different from the individual stage efficiency due to some inherent thermodynamic mechanisms. These effects can now be considered in detail. To start with, one simple result from thermodynamics has to be recalled: The constant pressure lines for a perfect gas on an enthalpy–entropy chart are divergent with the increasing entropy. Two constant pressure lines at p1 and p2, with p2  p1, are shown on the enthalpy–entropy diagram in Fig. 2.4. Two state points A and B are on the constant pressure line p1, with TB  TA.

Thermodynamics of Fluid Flow

61

h

D

p2

C

p1 B A s

Figure 2.4

Divergent constant pressure lines.

A perfect gas is compressed isentropically from A to C and then again from B to D, with both C and D being on the constant pressure line p2. By the isentropic relations, the following can be written: ¥p ´ ¦ 2µ TA § p1 ¶ TC

¥p ´ ¦ 2µ TB § p1 ¶

TD

(G 1)/G

(G 1)/G

¥T ´  K (say )  TC TA  TA ¦ C 1µ  TA ( K 1) § TA ¶ ¥T ´  K (say)  TD TB  TB ¦ D 1µ  TB ( K 1) § TB ¶

Therefore, TD TB TC TA



TB ( K 1) TA ( K 1)



TB TA

1

Therefore, (TD TB)  (TC TA) This proves that the constant pressure lines are divergent. Now, consider a multi-stage compressor with a compressible fluid flow. The design process starts with any specified constant pressure ratio and a constant isentropic efficiency for each stage. As the fluid moves from stage to stage, the density of the fluid keeps increasing and the specific volume of the fluid keeps decreasing. Because the mass flow rate is constant throughout the successive stages, the volume flow rate keeps decreasing. A good design must conform to these changes in the volume flow rate, keeping the area of flow and the velocity of flow suitably adjusted [volume flow rate (m3/s)  area of flow (m2) s velocity (m/s)]. Thus, when correct areas of flow are provided at each stage (neither more nor less), the design pressure ratios can also be maintained the same across each stage in a compressor. In a multi-stage turbine, the fluid keeps expanding; the density decreases stage by stage; volume flow rate keeps increasing; and the areas have to be adjusted with the flow velocities of the successive stages.

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Turbomachines

With this background, the analyses of the multi-stage compressor and multi-stage turbine can now be considered in detail.

2.7.1 Pre-heat Effect in Multi-stage Compressor An enthalpy–entropy diagram is drawn to show the process of compression in a four-stage compressor (Fig. 2.5). h

p5 5


5

5

p4 p3

4
4

4 3 2


p2

3
3

p1

2

1 s

Figure 2.5

Multistage compression process.

In Fig. 2.5, process 1–2–3–4–5 is the actual process of compression, through the four stages, from the initial pressure p1 to the final pressure p5. The total actual work of compression is h5 – h1, through the stages. Because of the staging, the isentropic processes are made up in the stages, such as Wst1  h2a h1 , Wst 2  h3a h2 , Wst 3  h4a h3 , and Wst4  h5a h4 The sum of these isentropic works of the stages is given by

¤Wsti  Wst1  Wst 2  Wst3  Wst 4 i

 (h2a h1 )  (h3a h2 )  (h4a h3 )  (h5a h4 ) This total work ¤Wsi is more than the isentropic work without staging, that is, the single-stage work Wss  h5aa h1 . This is due to (a) the divergent constant pressure lines and (b) the increased temperature at the outlet of a stage, that is, the inlet of the next stage because of the internal fluid friction, eddies, diffusion, etc. It is as if the inefficiency of the previous stage penalizes the next stage by the increase of the inlet temperature of the next stage. The whole effect, ¤ Wsti  Wss , is known as the “pre-heat effect” in the multistage compression. The actual total work is given by ¤ Wai  Wa1  Wa2  Wa3  Wa4 h h h h2 h4a h3 h5a h4  2 a 1  3a   Hst Hst Hst Hst

Thermodynamics of Fluid Flow

¤Wsti



63

(2.25)

Hst

because all efficiencies in the denominators are the same. The actual total work can also be determined by taking the overall isentropic compression efficiency Gc as ¤ Wai 

h5aa h1

Hc



Wss

(2.26)

Hc

Hence, from Eqs. (2.25) and (2.26), we have ¤ Wsti

Hst



Wss

Hc

Therefore,

Hst ¤Wsti  Hc Wss

(2.27)

Now, since ¤Wsti is more than Wss, Gst must be more than Gc. In other words, the overall isentropic compression efficiency is less than the stage efficiency, Gc  Gst.

2.7.2 Re-heat Effect in Multi-stage Turbines The process of expansion, in stages, can also be discussed in the same way as the compression process above. h

p1 p2 1 2


p3

2

p4

3 3

4

3


4

p5 5

4
5
5

s

Figure 2.6

Multi-stage expansion process.

In Fig. 2.6, process 1–2–3–4–5 is the actual process of expansion, through the four stages, from the initial pressure p1 to the final pressure p5. The total actual work of expansion is h1 – h5, through the stages. Because of the staging, the isentropic expansion processes are made up in the stages, such as Wst1  h1 h2a , Wst2  h2 h3a , Wst3  h3 h4a , and Wst4  h4 h5a

64

Turbomachines

The sum of these isentropic works of the stages is given by ¤ Wsti  Wst1  Wst2  Wst3  Wst4  (h1 h2a )  (h2 h3a )  (h3 h4a )  (h4 h5a ) This total work ¤Wsti is more than the isentropic work without staging, that is, the single-stage work Wss  h1 h5aa . This is due to the following: 1. The divergent constant pressure lines. 2. The increased temperature at the outlet of a stage because of the internal fluid friction, eddies, diffusion, etc. It is as if the inefficiency of the previous stage benefits the next stage. The whole effect, ¤ Wsti  Wss , is known as the “reheat effect” in the multi-stage expansion. The actual total work is given by

¤Wai  Wa1  Wa2  Wa3  Wa4 i

 (h1 h2a )Hst  (h2 h3a )Hst  (h3 h4a )Hst  (h4 h5a )Hst  ¤ Wsti – Hst

(2.28)

i

because all stage efficiencies are the same. The actual total work can also be determined by taking the overall isentropic expansion efficiency Gt as

¤Wai  Wss – Ht

(2.29)

i

Therefore,

¤Wsti – Hst  WssHt i

¤Wsti Ht  i Hst Wss

or Now, since

¤Wsti

(2.30)

is more than Wss, Gt must be more than Gst. In a multi-stage expansion process, there-

i

fore, the overall isentropic expansion efficiency Gt is more than the stage efficiency Gst: Gt  Gst.

EXAMPLE 2.6 In a four-stage air compressor, the pressure ratio is 1.3 in each of the stages. The isentropic efficiency of each stage is 0.9. Calculate the final exit pressure and the overall isentropic efficiency. The initial pressure and temperature at the inlet are 100 kPa and 300 K, respectively. For air, cp  1.004 kJ/kg K and F  1.4. What is the pre-heat effect in the compressor?

Solution: We know that Stage pressure ratio  prs  1.3 No. of stages  4

Thermodynamics of Fluid Flow

65

Inlet pressure  p1  100 kPa Inlet temperature  T1  300 K Isentropic stage efficiency  Gst  0.9 Figure 2.5 may be used for reference nomenclature. Now p1  100 kPa, p2  100 s 1.3  130 kPa, p3  130 s 1.3  169 kPa, p4  219.7 kPa, and p5  p4 s 1.3  285.61 kPa. For isentropic process 1–2a, (G 1)/G

¥p ´ ¦ 2µ  (1.3)0.4/1.4  1.07783831 T1 § p1 ¶  T2 a  300 s 1.07783831  323.352 K T2a

Now T2  T1 

(T2a T1 )

Hs (323.352 300)  300  0.9  325.947 K

Proceeding in the same way at p3, p4, p5 levels, we get T3a 351.32 K, T3  354.14 K T4a 381.7 K, T4  384.765 K T5a 414.71 K, T5  418.04 K In a single-stage compression from 100 to 285.61 kPa (without multi-staging), T5aa T1

 (2.8561)0.4/1.4  1.3496

 T5q  300 s 1.3496  404.889 K The overall isentropic efficiency is T5aa T1 T5 T1



404.889 300 418.04 300

 Hc  88.86% Equation (2.45) (to be obtained shortly) may also be used to obtain the overall isentropic efficiency and the result is the same: Gc  88.86%. The pre-heat effect: The total of stage-wise works is given by

¤Wsti  cp[(T2a T1 )  (T3a T2 )  (T4a T3 )  (T5a T4 )] i

 1.004 [(323.352 – 300)  (351.32 – 325.947)  (381.7 354.14)  (414.71 – 384.765)]

 106.655 kJ/kg

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Turbomachines

Also, the single-stage work (between the same initial and final pressures) is given by Wss cp (T5q– T1)

 1.004 (404.889 – 300)



 105.31 kJ/kg So

¤Wsti  106.655 is more than Wss  105.31 by 1.345 kJ/kg, and this is the pre-heat effect. i

Comment: In this solution, it is possible to find the temperatures at all salient points (such as points 1, 2, 2a, 3, 3a, etc., in reference Fig. 2.5) by the application of isentropic relation at the commencement of compression in each stage. Because the number of stages is only 4, it is not difficult. If the number of stages were 8, 9, etc., the procedure would be more time consuming. Equation (2.45) (to be obtained shortly) can be easily applied when there are more stages. However, the stage-wise summation of isentropic works cannot be done, unless all temperatures are found out.

2.8 Infinitesimal-Stage or Small-Stage Efficiency or Polytropic Efficiency It has been seen, in general, that the expressions for various efficiencies contain the ratios of pressures and the ratios of temperatures. Such expressions give rise to the values of efficiency that depend on the pressure ratios of machines. Thus, the efficiency of the basic mechanism of energy transfer gets overlooked. In order to find the efficiency of the basic thermodynamic process, irrespective of the pressure ratios or sizes of the machines, an elemental or infinitesimal change in pressure is to be considered. The efficiency so obtained is known as the infinitesimal- or small-stage efficiency or polytropic efficiency (It is not essential to physically have a stage with such a small change in pressure.). Consider a compression in a small stage. The pressure increases from p to p  dp. The temperature rises from T to T  dT a for the small-stage isentropic process. In an infinitesimal actual process, the rise of temperature is from T to T  dT. Figure 2.7 describes all these processes together. p2

T

p

dp p

T  dT T  dT

dT dT

T

p1

s

Figure 2.7

Infinitesimal-stage compression process.

Thermodynamics of Fluid Flow

67

The finite pressure limits are p1 and p2. Now, for the small isentropic and actual processes, dT a  Hp dT Hence, dT a  Hp s dT For a small isentropic process, we have T  dT a ¥ p  dp ´ ¦ T § p µ¶ 1

dT a ¥ dp ´  ¦1  µ T p¶ §

(G 1)/G

(G 1)/G

Expanding the RHS and neglecting higher order terms, we get 1

dT a G 1 dp 1 s T p G

Replacing dT a by dT s Hp and simplifying, we get dT 1 G 1 dp  s s Hp G T p

(2.31)

Integrating, we have ¥T ´ 1 G 1 ¥p ´ ln ¦ 2 µ  s s ln ¦ 2 µ G § T1 ¶ Hp § p1 ¶ ¥ (G 1)´ ¥ p2 ´ ¦§ G µ¶ ln ¦ p µ § 1¶  Hp  ¥T ´ ln ¦ 2 µ § p1 ¶

(2.32)

(2.33)

Also, from Eq. (2.32), ¥ T2 ´ ¥ p2 ´ [(G 1)/G ](1/Hp ) ¦T µ  ¦ p µ § 1¶ § 1¶

(2.34)

Again, ¨¥ T ´ · T2 T1  T1 ©¦ 2 µ 1¸ ©ª§ T1 ¶ ¹¸ ¥ ¥ p ´ [(G 1)/G ](1/Hp ) ´  T1 ¦ ¦ 2 µ

1µ ¦§ § p1 ¶ µ¶

(2.35)

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Turbomachines

For the polytropic process (where T2 and T1 are connected by the equation pvn  constant), we have ¨¥ p ´ (n 1)/n · T2 T1  T1 ©¦ 2 µ

1¸ ¸ ©§ p1 ¶ ¹ ª

(2.36)

Comparing Eqs. (2.35) and (2.36), we get n 1 1 G 1  s Hp G n  Hp 

G 1 n s G n 1

(2.37)

This equation for the polytropic efficiency gives the expression without the ratio of pressures or the ratio of temperatures and therefore reflects the process efficiency. The process considered is the compression process and therefore a subscript “c” is added to write the expression as

Hpc 

n G 1 s n 1 G

(2.38)

In a very similar procedure as above, the small-stage efficiency or polytropic efficiency for the expansion process or turbine process, Hpt , can be obtained as n 1 G Hpt  s (2.39) n G 1 At this juncture, expressions for stage efficiencies in terms of polytropic efficiencies can be derived for the processes of both compression and expansion. Consider a compression process in a stage as shown in Fig. 2.8. T p2 2 2
1  2
: Isentropic process

p1

1  2: Polytropic process 1

s

Figure 2.8

Compression in a stage.

Stage efficiency is

Hst, c 

c p (T2a T1 ) c p (T2 T1 )

Thermodynamics of Fluid Flow

69

¨¥ T ´ · T1 ©¦ 2a µ 1¸ ©§ T1 ¶ ¸¹  ª ¨¥ T ´ · T1 ©¦ 2 µ 1¸ ©ª§ T1 ¶ ¸¹ p(G 1)/G 1  rs(n 1)/n prs

1 Substituting for (n – 1)/n from Eq. (2.38),

Hst, c 

prs(G 1)/G 1 ¥ 1 ´ ¥ G 1´ ¦ µ s¦ H § G ¶µ prs§ pc ¶

1

Consider the expansion process in a stage as shown in Fig. 2.9. T p1 1 1  2
: Isentropic process 1  2: Polytropic process p2 2 2
s

Figure 2.9

Expansion in a stage.

Stage efficiency is

Hst t 

c p (T1 T2 ) c p (T1 T2a )

¨ T2 · ©1 ¸ T1 ¹  ª ¨ T2a · ©1 ¸ ª T1 ¹ 

1 prs(n 1)/n 1 prs(G 1)/G

(2.40)

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Turbomachines

Substituting for (n – 1)/n from Eq. (2.39), we get H (G 1)/G

Hst t 

1 prspt

1 prs(G 1)/G

(2.41)

2.9 Reheat Factor for Expansion Processes As seen earlier, in a stage-wise expansion, the total of the individual isentropic expansion works of all stages together, namely, ¤ Wsti , is more than the overall isentropic expansion work Wss if the expansion were to i

take place from the initial pressure p1 to the final pressure p2, at one stretch, as a single stage. The reheat factor is defined as the ratio of the total of individual isentropic works of each stage, ¤ Wsti , i to the single-stage isentropic work Wss between the same pressure limits: ¥ ´ ¦ ¤ Wsti µ § ¶ RF  i Wss

(2.42)

The reheat factor is always greater than 1. The reasons are as follows: 1.

2.

The preceding stages give the output at a higher temperature than the isentropic outlet temperature, so that the succeeding stages have a higher temperature at the inlet, and therefore, the inlet state at higher availability. This is as if “the inefficiency of the preceding stage benefits the succeeding stage.” The constant pressure lines are divergent. The expansion process continuously shifts to the regions of higher enthalpy drops.

The reheat factor has the values of the order of 1.04 and 1.05, for the normal designs like 12–15 stages of expansion. This signifies 4%–5% increase in the output. Although staging is essential in connection with compounding (to reduce the speed of rotor), this reheat (and the additional output) is an additional advantage of staging.

EXAMPLE 2.7 The pressure, temperature and velocity of air at the inlet to a single-stage compressor are respectively 100 kPa, 295 K, and 10 m/s. At the exit, the respective values are 210 kPa, 370 K, and 85 m/s. The exit is 1 m above the inlet. Calculate the (a) actual work, (b) isentropic work, and (c) efficiency of the compressor. For air, take cp  1.004 kJ/kg K, F  1.4.

Solution: We have P1  100 kPa, T1  295 K, V1  10 m/s, P2  210 kPa, T2  370 K, V2  85 m/s, z2 – z1  1 m

Thermodynamics of Fluid Flow

71

So T02 T01  (T2 T1 ) 

V2 2 V12



2 g ccp

 (370 295) 

( z2 z1 ) g g ccp

(85)2 (10)2 1 s 9.81  2 s 1004 1004

 78.56 K (a) Therefore, the Actual work  cp (T02 – T01)  1.004 (78.56)  78.874 kJ/kg Also, T02a T01

¥p ´ ¦ 2µ § p1 ¶

(G 1)/G

¥ 210 ´ ¦ § 100 µ¶

0.2857

 1.236

 T02a  295.05 s 1.236  364.68 K

(b) We have

Isentropic work  cp (T02a – T01)

 1.004 (364.68 – 295.05)



 69.91 kJ/kg (c) We also have Compressor efficiency



69.91  88.64% 78.874

EXAMPLE 2.8 Combustion products enter a gas turbine at p01  600 kPa and T01  1000 K. The turbine has four stages of equal stage efficiency and equal pressure ratio. The stage efficiency is 0.86. The exit pressure is 100 kPa. Calculate the (a) isentropic works of all stages, (b) reheat factor, (c) total actual work, and (d) the overall isentropic efficiency of the turbine. Take, for gases, cp  1.005 kJ/kg K and F  1.4.

Solution: We have p01  600 kPa, T01  1000 K, m  4, Gst  0.86, p05  100 kPa, cp  1.005 kJ/kg K, F  1.4. Figure 2.6 may be referred to for nomenclature. Stage pressure ratio is given by ¥p ´ prs  ¦ 05 µ § p01 ¶

1/m

¥ 1´ ¦ µ § 6¶

1/ 4

 0.63894

p01  600(data) p02  p01 s0.63894  383.366 kPa, p03  p02 s0.63894  244.95 kPa p04  p03 s0.63894 156.5 kPa p05  100 (exit pressure, data)

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Turbomachines

For isentropic expansion in the first stage, we get T02a T01

¥p ´  ¦ 02 µ § p01 ¶

(G 1)/G

 0.879873

 T02a  1000 s 0.879873  879.873

Now

T02  T01 – (T01 – T02a) s Gst

 1000 – (1000 – 879.873) s 0.86



 896.69 K Proceeding on the same lines, we get T02a  879.873,

($T)is  120.127,

T02  896.69

T03a  788.973,

($T)is  107.72,

T03  804.05

T04a  707.465,

($T)is  96.585,

T04  720.99

T05a  634.377,

($T)is  86.613,

T05  646.5

(a) Isentropic work of all stages together is given by:

¤Wsti  cp [120.127  107.72  96.585  86.613] i

 1.004 s 411.045  412.689 kJ/ kg Now ¥ 1´ T05aa  T01 s ¦ µ § 6¶

(G 1)/G

 599.35

Hence, Single-stage work  cp (T01 – T05q)

 1.004 (1000 – 599.35)



 402.25 kJ/kg (b) Reheat factor is given by:

¤Wsti Reheat factor 

i

Wss 412.689   1.026 402.25

(c) total actual work is given by:

Total actual work  cp (T01 – T05)  1.004 (1000 – 646.5)  354.9 kJ/kg

Thermodynamics of Fluid Flow

73

(d) the overall isentropic efficiency is given by: 1000 646.5 1000 599.35 353.5  400.65  88.23%

Overall isentropic efficiency 

Comment: As in Example 2.6, all temperatures are obtained using isentropic relations. These are also required for calculating the reheat factor. For getting the overall isentropic efficiency, Eq. (2.49) can also be used (to be derived in Section 2.10). The result obtained has to be the same.

2.10 Overall Isentropic Efficiency versus Finite-Stage Efficiency: Compression and Expansion Processes It is possible to get a resultant expression for the overall efficiency of a series of finite number of stages of either compression or expansion processes. (This does not hold good for the steam processes because the enthalpy values of the steam, in the expressions, cannot be reduced in terms of the temperatures.)

2.10.1 Compression Process Figure 2.10 shows an enthalpy–entropy diagram with an m-stage compression. The pressure ratio of each stage is the same. The isentropic stage efficiency is also the same for each stage. The actual temperature rise $T in each successive stage can be calculated as follows: First stage: ( $T )1  T2 T1 

(T2a T1 )

Hs

¨¥ T2a ´ · ©¦ µ 1¸ ©§ T1 ¶ ¸¹  T1 ª Hs  AT1 where ¨¥ T2a ´ · ©¦ µ 1¸ ©§ T1 ¶ ¸¹ [( prs )(G 1)/G 1] Aª  Hs Hs Second stage: ($T)2  @T2

 @ [T1  ($T)1]

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Turbomachines



 @ (T1  @T1)



 @T1(1  @)

Third stage: ($T)3  @T3

 @ [T2  ($T)2 ]



 @ [T1  ($T)1  ($T)2 ]



 @ [T1  @T1  @T1 (1  @)]



 @ T1 (1  @)2

Fourth stage: ($T)4  @ T1 (1  @)3 mth stage: ($T)m  @ T1 (1  @)m 1 The total of the actual temperature rises in the m stages is given by m

¤ ( $T )i  ( $T )1  ( $T )2  !  ( $T )m i 1

 AT1  AT1(1  A )  AT1(1  A )2  !  AT1(1  A )m 1  T1[(1  A )m 1] ¨¥ ( p )(G 1)/G 1´ m ·  T1 ©¦1  rs µ 1¸¸ Hs ©§ ¶ ¹ ª

(2.43)

The temperature rise in the case of one-stretch compression from T1 to T(m  1)q (Fig. 2.10) is given as T

( m 1)q

¨¥ T ´ · ( m 1)q

T1  T1 ©¦ µ 1¸ ©ª§ T1 ¶ ¸¹  T1( P (G 1)/G 1)

where P is the overall pressure ratio. So T

( m 1)q



T1  T1 ¨ª( prs m )(G

1)/G

1·¹

 T1( prsm(G 1)/G 1)

Hence, the overall isentropic compression efficiency for the process through m stages is given by

Hc 

T( m 1)aa T1

¤ ( $T )i i

(2.44)

Thermodynamics of Fluid Flow

75

Using Eqs. (2.40) and (2.41),

Hc 

T1[( prs )m(G 1)/G 1] ¨¥ ( p )(G 1)/G 1´ m · T1 ©¦1  rs µ 1¸¸ Hs ©§ ¶ ¹ ª

Hence, the overall isentropic compression efficiency Gc is given by

Hc 

[( prs )m(G 1)/G 1]

(2.45)

¨¥ ( p )(G 1)/G 1´ m · ©¦1  rs µ 1¸¸ Hs ©§ ¶ ¹ ª (m
1)

h

pm
1

(m
1) m

(m
1)

pm m

m

p4 4

4

4 3 3

3

p3 p2 p1

2

2 1

s

Figure 2.10

Compression in finite stages.

2.10.2 Expansion Process or Turbine Process Figure 2.11 shows an enthalpy–entropy diagram with an m-stage expansion. The pressure ratio in each stage, prs, is the same for every stage. The isentropic stage efficiency is also the same for each stage. The actual drop in temperature in each successive stage can be calculated as follows: First stage: ( $T )1  T1 T2  (T1 T2 a ) Hs ¥ T ´  T1Hs ¦1 2a µ T1 ¶ §

76

Turbomachines

¨ ¥ p ´ (G 1)/G · ¸  T1Hs ©1 ¦ 2a µ ¸ © § p1 ¶ ¹ ª  T1B where ¨ ¥p ´ B  Hs ©1 ¦ 2a µ © § p1 ¶ ª

(G 1)/G

· ¸ ¸ ¹

Second stage: ($T)2  T2A

 [T1 – ($T)1]A

Third stage:

 T1A (1 – A ) ($T)3  T3A



 [T1 – ($T )1 ($T )2]A



 [T1 – T1A T1A (1 A )]A



 T1A (1 – A )2

Fourth stage: ($T)4  T1A (1 – A )3 mth stage: ($T)m  T1A (1 – A )m 1 h p1 p2 1 2

p3

2 3

3

3 pm m

m

m pm
1

m
1 (m
1)

(m
1) s

Figure 2.11

Expansion in finite stages.

Thermodynamics of Fluid Flow

77

Therefore, the total actual drop in temperature is given by

¤ ( $T )i  ( $T )1  ( $T )2  !  ( $T )m i

 T1B  T1B(1 B )  T1B(1 B )2  !  T1B(1 B )m 1  T1B [1  (1 B )  (1 B )2  !  (1 B )m 1 ] 

T1B [1 (1 B )m ] 1 (1 B )

 T1[1 (1 B )m ]

(2.46)

Therefore, T1 – Tm1  T1(1 – (1 A )m) Also considering the overall isentropic process T1 to T(m1)q in one stretch, we get T1 – Tm1  (T1 – T(m1)q)Gt Also, (T1 – T(m1)q)Gt  T1[1 – (1 A )m] Hence,

Ht 

 Ht 

T1[1 (1 B )m ] ¥ T ´ T1 ¦1 ( m 1)aa µ T1 ¶ § 1 [1 Hst (1 prs(G 1)/G )] m

(2.47)

1 P (G 1)/G

In the denominator, P is the overall pressure ratio for all the stages together: P  (prs) s m  P(F 1)/F  (prs)m(F 1)/F Hence, the overall isentropic expansion efficiency Gt is given as

Ht 

1 [1 Hst (1 prs(G 1)/G )] m

(2.48)

1 ( prs )m(G 1)/G

Equation (2.46) gives an expression for the total actual temperature drop, work output is

¤ ( $T )i .

Therefore, the actual

i

c p ¤ ( $T )i  c pT1[1 (1 B )m ] i

The total of all stage-wise isentropic heat drops is c p ¤( $T ) Hst in the m stages, reheat factor is

¤Wsti . Therefore, the i

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Turbomachines

c pT1[1 (1 B )m ] RF  

¤ Wsti Wss



Hst c pT1[1 ( prs )m(G 1)/G ]

[1 (1 B )m ] Hst [1 ( prs ) m(G 1)/G ] m



1 [1 Hst (1 prs(G 1)/G )]

H st ¨ª1

prsm(G 1)/G

(2.49)

·¹

While deriving the above relations, the enthalpy values were expressed in terms of temperatures and the process-equations for isentropic processes were used. Because of this, the reheat factor, as obtained here, may be applied only to gas turbines. For steam turbines, proper enthalpy values have to be used to obtain the reheat factors.

K eywords Compressible fluids Compression, expansion Static state Stagnation state Isentropic compression process Actual compression process Isothermal compression process Isentropic expansion process Actual expansion process

Stage efficiency Overall isentropic compression efficiency Overall isentropic expansion efficiency Preheat in multistage compression processes Reheat in multistage expansion processes Infinitesimal compression process Infinitesimal expansion process Reheat factor

S ummary In this chapter, it is assumed that the laws of perfect gases hold good for compressible fluids when they undergo compression or expansion processes in turbomachines, without large changes in pressures. t
First, the static and stagnation states are identified (properties at stagnation state  properties at static state  effect of velocity V 2/ 2 in suitable units). t
Different processes such as adiabatic compression, isothermal compression, and isentropic expansion are studied (Actual processes are compared with these ideal processes).

t
In multi-stage units, the overall isentropic efficiencies are compared with the stage efficiencies in both compression and expansion processes (In compression, Gc  Gst; in expansion, Gt Gst). t
The effect of pre-heat in compressors and the effect of reheat in turbines are studied (Preheat effect: more work is required, compared to sum of isentropic works in compression process; Reheat effect: more work is available, compared to sum of isentropic works in expansion process).

Thermodynamics of Fluid Flow

t
Infinitesimal-stage efficiencies or small-stage efficiencies or polytropic efficiencies are studied in detail.

Hpc 

n G 1 s n 1 G

and

Hpt 

n 1 G s n G 1

t
Both in compression and expansion processes, the overall isentropic efficiencies and finite-stage efficiencies are studied in detail.

Hc 

and

Ht 

79

[( prs )m(G 1)/G 1] ¨¥ ( p )(G 1)/G 1´ m · ©¦1  rs µ 1¸¸ Hs ©§ ¶ ¹ ª 1 [1 Hst (1 prs(G 1)/G )] m 1 ( prs )m(G 1)/G

The above study is made in preparation of the study of energy exchange processes in turbomachines.

I mportant E quations 5. Ideal isothermal compression work:

1. Stagnation properties: V2 T0  T  2c p g c

2 ¥p ´ Wideal  ° pdv  RT1 ln ¦ 2 µ § p1 ¶ 1

h0  h 

V2 2 gc

6. Actual work of compression Wactual  c p (T02 T01 )  Q

RV 2 2 gc

7. Actual compression work

p0  p  2. Efficiencies (Hc )t-t  (Hc )t-s  (Hc )s-t  (Hc )s-s 

h02a h01 h02 h01 h2a h01 h02 h01 h02a h1 h02 h01 h2a h1 h02 h01

(G 1)/G · T01 ¨¥ p02a ´ ¸ ©¦ 3. T02 T01 

1 (Hc )t-t ©§ p01 µ¶ ¸ ¹ ª (G 1)/G · c pT01 ¨¥ p02a ´ ¸ © 4. (Wc )t-t 

1 (Hc )t-t ©¦§ p01 µ¶ ¸ ¹ ª

Wactual 

¥p ´ 1 RT1 ln ¦ 2 µ (Hc )isth § p1 ¶

8. The actual shaft work in expansion (for any intermediate stage): ¨ ¥ p ´ (G 1)/G · ¸ WS  c p (T01 T02 )  (Ht )t-t c pT01 ©1 ¦ 02a µ ¸ © § p01 ¶ ¹ ª 9. The actual shaft work in expansion for the last stage (or only one stage) ¨ ¥ p ´ (G 1)/G · ¸ WS  (Ht )t-s c pT01 ©1 ¦ 2a µ ¸ © § p01 ¶ ¹ ª 10. The useful output is ´ ¥ ¥ ¥ p ´ (G 1)/G ´ µ ¦ ¦¦ 1µ

1µ V12 µ ¦ ¦ § p2 ¶ µ  (Ht )t-s c p ¦T1 ¦ (G 1)/G µ 2c p g c µ µ ¦ ¦ ¥ p1 ´ µ µ ¦ ¦§ ¦§ p µ¶ µ¶ ¶ § 2

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Turbomachines

11. Actual total work for a multistage compression:

17. Reheat factor

¤ Wai  Wa1  Wa2  Wa3  Wa4 ¤ Wsti  Hst 12.

Hst ¤Wsti  Wss Hc

¥ ´ ¦ ¤ Wsti µ § ¶ RF  i Wss 18. The overall isentropic compression efficiency

13. Actual work for multistage expansion:

¤Wai  Wss – Ht i

¤Wsti Ht 14.  i Hst Wss

Hc 

[( prs )m(G 1)/G 1] ¨¥ ( p )(G 1)/G 1´ m · ©¦1  rs µ 1¸¸ Hst ©§ ¶ ¹ ª

19. The overall isentropic expansion efficiency

15. Polytropic compression efficiency:

Hpc 

n G 1 s n 1 G

16. Polytropic expansion efficiency:

Hpt 

n 1 G s n G 1

Ht 

1 [1 Hst (1 prs(G 1)/G ] m 1 ( prs )m(G 1)/G

20. The reheat factor for gas turbines only: m

RF 

[1 Hst (1 prs(G 1)/G )]

H st ¨ª1 prsm(G 1)/G ·¹

M ultiple- C hoice Q uestions 1. Pick the correct equation from the following: V2 (a) h01  h1  1 2c p g c (b) p01  p1  (c) T01  T1  (d) T02  T2 

V22 2 gc

(c) Isothermal efficiency  (d) Actual efficiency 

V12

Isentropic work Actual work Actual work Isothermal work

Actual work

Isentropic work 3. In an expansion process, Actual work (a) Isentropic efficiency  Isentropic work

2c p g c V22

2 gc 2. In a compression process, (a) Isentropic efficiency 

(b) Isentropic efficiency 

(b) Isentropic efficiency  Actual work Isentropic work

(c) Isothermal efficiency 

Isentropic work Actual work Actual work Isothermal work

Thermodynamics of Fluid Flow

(d) Actual efficiency 

Actual work

Isentropic work 4. In a multi-stage compressor, the total-to-total efficiency is applicable (a) to the first stage. (b) to any intermediate stage (c) to the last stage only (d) to the intermediate stage only. 5. In a multi-stage turbine, the total-to-static efficiency is applicable (a) to the first stage (b) to the last stage only (c) to any intermediate stage (d) to the intermediate stage only 6. In all stages of multi-stage compressor, pressure ratios are equal and stage efficiencies are also equal. Then (a) overall efficiency  stage efficiency

81

(b) overall efficiency  stage efficiency (c) overall efficiency  stage efficiency (d) any of the above (a), (b), (c) can be true 7. In all stages of multi-stage turbine, pressure ratios are equal and stage efficiencies are also equal. Then (a) overall efficiency  stage efficiency (b) overall efficiency  stage efficiency (c) overall efficiency  stage efficiency (d) any of the above (a), (b), (c) can be true 8. Pick the wrong statement in the following: (a) The pre-heat factor is applicable to a multistage compressor. (b) The reheat factor is applicable to a multistage turbine. (c) In a multi-stage turbine, the reheat factor is more than 1. (d) In a multi-stage compressor, the pre-heat factor is less than 1.

R eview Q uestions 1. Distinguish between static and stagnation properties. (Section 2.2) 2. Why are stagnation properties preferred to static properties for use in the analysis of turbomachine processes? (Section 2.2) 3. Even if isothermal processes require least work for compression, these processes are not used as ideal or standard processes for comparison with actual processes. Why? (Section 2.3) 4. State the four different forms of isentropic compression efficiencies. Distinguish between them. State the applicability of each efficiency. (Section 2.4) 5. State the different viewpoints by which the different forms of isentropic expansion efficiencies can be explained. (Section 2.6) 6. What is a pre-heat effect in a multi-stage compression? (Section 2.7.1) 7. Prove that the overall isentropic compression efficiency is less than the individual stage

8. 9.

10.

11.

12.

13.

isentropic compression efficiency in the case of multi-stage compression. (Section 2.7.1) What is a reheat effect in multi-stage expansion? (Section 2.7.1) Prove that the overall isentropic expansion efficiency in the multistage expansion is more than the individual stage isentropic expansion efficiency. (Section 2.7.1) Explain the polytropic efficiency in the case of compression process. Derive an expression for the polytropic efficiency. (Section 2.8) Explain the polytropic efficiency in the case of expansion process. Derive an expression for the same. (Section 2.8) Derive an expression for the overall compression efficiency in terms of stage pressure ratio. (Section 2.9) Derive an expression for the overall expansion efficiency in terms of stage pressure ratio. (Section 2.10.2)

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Turbomachines

14. Derive an expression for the reheat factor in terms of stage pressure ratio and stage efficiency. (Section 2.10.2)

15. Write a note on the pre-heat effect. 16. Write a note on the reheat effect and/or reheat factor.

E xercises 1. At a particular section in a multi-stage air compressor, the air is at a total temperature of 330 K and a total pressure of 200 kPa. At the end of three stages after this section, the total pressure is 600 kPa. The pressure ratio across each stage is the same. The total-to-total stage efficiency is 0.8 in each stage. For the three stages, determine the overall compression efficiency and the pre-heat effect. 2. Repeat Problem 1 when the stage efficiency is 0.86. 3. Suppose that there are four stages instead of three stages between the same pressure limits (200 and 600 kPa) as in Problem 1. Also the temperature at the beginning is 330 K, as mentioned. For the four stages, determine the overall compression efficiency and pre-heat effect if the stage efficiency is 0.8. 4. Let the stage efficiency be 0.86 instead of 0.8 in each of the four stages between 200 and 600 kPa of Problem 3. The initial temperature is 330 K. Calculate the overall compression efficiency and the pre-heat effect. 5. From the results of the above four problems (1–4), state your comments on the effect of variation of (a) the number of stages and (b) stage efficiency.

P roject- O riented Q uestions 1. In the Exercise Problem 5, you have already stated your comments on the effects of (a) variation of number of stages and (b) variation of stage efficiency on the performance of a multistage compression process, depending on the results of the Problems 1 to 4. Now, include the solved examples of this chapter, with repeated

6. At the inlet of a gas turbine, the gases are at a total pressure of 700 kPa and total temperature of 1100 K. The gases expand in three stages, down to a total pressure of 175 kPa. In each of the three stages, the pressure ratio is the same, and the stage efficiency is 0.85. For the three stages, calculate the overall expansion efficiency and the reheat factor. 7. Repeat Problem 6 with the stage efficiency equal to 0.8 (instead of 0.85). 8. Let there be four stages instead of three, with the initial temperature 1100 K and pressure limits 700–175 kPa in Problem 7. Calculate the overall expansion efficiency if the stage efficiency is 0.85, and the reheat factor. 9. Problem 8 is modified with Gst  0.8. Calculate the overall efficiency and reheat factor. 10. From the results of Problems 6–9, state your comments on the effect of variation of Gst and the number of stages of expansion. 11. Air expands in a single-stage turbine from 280 kPa, 300oC to 60 kPa, 220oC. Calculate the index of expansion and the polytropic efficiency. 12. Air is compressed in a single-stage compressor from 120 kPa, 60oC to 200 kPa, 140oC. Determine the index of compression and the polytropic efficiency of the process. Solutions Available variation of such parameters with more number of stages and more number of values of the parameters to reinforce and fine-tune your comments. 2. In the Exercise Problem 10, you have already stated your comments on the effects of (a) variation of number of stages and (b) variation of

Thermodynamics of Fluid Flow

stage efficiency on the performance of a multistage expansion process, depending on the results of the Problems 6 to 9. Now, include the solved examples of this chapter, with repeated

variation of such parameters with more number of stages and more number of values of the parameters to reinforce and fine-tune your comments.

A nswers Multiple-Choice Questions 1. 2. 3. 4.

(c) (b) (a) (d)

5. 6. 7. 8.

(b) (b) (a) (d)

Exercises 1. 2. 3. 4. 6.

77.91%, 3.267 kJ/kg extra work 84.7%, 1.8735 kJ/kg extra work 77.65%, 3.692 kJ/kg extra work 84.35%, 2.389 kJ/kg extra work 86.634%, 1.02

83

7. 8. 9. 11. 12.

82.06%, 1.0258 86.85%, 1.0216 82.33%, 1.0289 1.36755, 94.07% 1.7286, 67.79%

3

Energy Exchange in Turbomachines

Learning Objectives After completing this chapter, you will be able to: v
Understand and master the concept of velocity triangles as a powerful tool in the practice of turbomachines of all varieties (power-consuming, power-absorbing, radial flow, mixed flow, and axial flow types). v
Understand the most basic equation of energy transfers in turbomachines, that is, the Euler turbine equation.

v
Understand the mechanism of energy transfer process and the fundamentals of the impulse and reaction in turbomachines. v
Quantify the above reaction in terms of the “degree of reaction,” and relate this degree of reaction as an important parameter to other parameters such as the utilization factor, speed ratio, etc.

3.1 Introduction The process of energy transfer, taking place in a turbomachine, has been discussed in chapters 1 and 2, through the steady-flow energy equation (SFEE) from thermodynamics. By knowing the states of a fluid at the inlet and outlet, and by accounting for some losses, it is possible to find the power of the machine. But these details are insufficient to design the hardware details of the machine. The most important details that are the outcome of the design procedure are the details of the rotor vanes or blades, in the form of their profiles, with the required angles at the inlet and outlet, as well as the radii at the inlet and outlet. This chapter aims at understanding the basic details of the mechanism of energy transfer between the flowing fluid and rotating vanes of a rotor of a turbomachine. One of the most fundamental equations for energy transfer, that is, the Euler turbine equation is discussed in detail. The two types of the processes, namely, impulse and reaction, are then studied. Some of the important parameters such as the degree of reaction, utilization factor, and the speed ratio are discussed with respect to the implication of their values and their mutual relationships or dependencies. The first step in this direction is the understanding of “velocity triangles” with respect to a turbomachine.

3.2 Velocity Triangles A velocity triangle with reference to a turbomachine is the representation of three important velocities as the sides of a triangle. The three velocities are as follows: 1. 2. 3.

The peripheral velocity of the rotor blade, U. The relative velocity of the fluid with respect to the rotor blade Vr. The absolute velocity of the fluid V.

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Turbomachines

These above velocities are subscripted to indicate their positions with respect to the blades: Subscript “1” is for the inlet position and subscript “2” is for the outlet position. The three velocities form a triangle only when they satisfy the condition V  U  Vr. This means that the fluid smoothly glides over the blade surface without impact or separation. Therefore, the relative velocity is tangential to the blade profile at the point considered. The “solution” of such a triangle is nothing but finding the three sides of the triangle and its three angles, either graphically or by a simple trigonometric analysis. Such velocity triangles can be drawn for any number of points on the profile of the blade of a rotor, but in general, solutions at the two points, one at the inlet and the other at the outlet, are sufficient at present. The following points serve as the guidelines in understanding the velocity triangles: 1. 2.

All the velocities are vector quantities. Each velocity has a magnitude and a direction. The blade velocity U is invariably tangential to the circular path of the blade, with positive direction in the direction of rotation. The choice of the point of the tangent can be anywhere on the circular path, but it is taken at the uppermost point to keep U horizontal, as shown in Fig. 3.1. U2

U1

Figure 3.1

Blade velocities.

Stream lines

Figure 3.2

Vane-congruent flow.

Energy Exchange in Turbomachines

3.

The instantaneous relative velocity of the fluid (Vr) with respect to the blades is always tangential to the stream line of the flow of the fluid in the passage between the blades of the rotor. One important assumption here is that the stream line is identical in shape with the vanes. This amounts to saying that the flow is “vane-congruent.” Hence, Vr is tangential to the blade profile also. The vane-congruent flow is schematically shown in Fig. 3.2, taking the radial flow rotor as an example. Some blade profiles are shown in Fig. 3.3 along with the direction of relative velocity. The sense of the direction of Vr (inward or outward) has to be prescribed clearly, depending on the flow of the fluid. Vr2

Vr1

(a)

Figure 3.3

5.

6.

(b)

Vr1

(c)

Relative velocities: (a) Blades are bent backwards. (b) Blades are radial at the outlet. (c) Blades are bent forward.

The absolute velocity V of the fluid is the vectorial sum of U and Vr Thus, V  U  Vr. With this relationship, it can be understood that any two known quantities are sufficient to know the third one. The disposition or orientation of the imaginary plane, on which the velocity triangle is visualized, is described in two separate ways: one for the radial flow rotor and the other for the axial flow rotor. For the radial flow rotor, the plane of velocity triangles is perpendicular to the axis of rotation of the rotor (Fig. 3.4). The rotor blades look as if they are encased between the two circular arcs.

Figure 3.4

7.

Vr2

Vr2

Vr1

4.

87

Radial flow rotor.

For the axial flow rotor, the velocity triangles are drawn on a plane that is tangential to the rotor. The blade profiles are also projected on this plane. It is as if the periphery of the rotor drum is slit open

88

Turbomachines

and the surface of the drum is opened up. In other words, this plane is the “developed” surface of the rotor drum along with the blades. The direction of rotation of the rotor drum is seen as parallel to the two parallel lines encasing the row of blades. This is shown in Fig. 3.5.

Figure 3.5

8. 9. 10.

Axial flow rotor.

The directions of fluid flow are radial, axial, or mixed at the entry or exit. Radial flows are seen on planes perpendicular to the axis and axial flows on planes tangential to the rotor. The fluid angles (inlet @1, outlet @2) and blade angles (inlet A1 and outlet A2) are specified with respect to the blade velocity vector U. Finally, perception of the situation is of utmost importance. The situation in radial flow machines is very easy to perceive. For an axial flow machine, the “physics” of the situation can be further explained as per the following thoughts: t
“General flow” of fluid is from one parallel line to the other [Fig. 3.6(a)]. t
The flow direction is from the “nose” to the “tail” of the air-foil section of the blades [Fig. 3.6(b)]. t
If the fluid flow “strikes” on the concave side of the blades, the fluid “drives the blades,” forcing the movement of the blades in that direction. Hence, this type of blade assembly is meant for a turbine [Fig. 3.6(c)]. t
If the fluid flow “strikes” on the convex side of the blades, this type of blade assembly is meant for a compressor [Fig. 3.6(d)]. In this case, the blades force the fluid and move to “scoop” the fluid. (Imagine a shovel scooping sand or a spoon scooping food from a plate. How does the shovel move? How does the spoon move?) This is illustrated in Fig. 3.6(e). t
General flow is as in Fig. 3.6(a). But the “particular flow direction” (at some angle to the parallel lines) must be decided to conform with the geometry of guide vanes, blade profiles, and the velocity triangles.

The above steps are once again written down, in concise form, so that the step-wise procedure can be easily followed and remembered. For the sake of clarity, the procedures for radial flow machines and axial flow machines are shown in separate boxes. The blades of the rotor in a turbomachine are of a shape of an ‘airfoil’. Air foil shapes provide a smooth stream line flow of fluids. In addition, by the suitable design, the passages between the blades can be shaped for any type of flow: expansion flow for turbines, compression flow for compressors, with sectional area suitable for the required pressure ratio, and so on. Further, in axial flow machines, the individual blades are assembled to make up a full rotor. The blades can have very smooth surface finish for the high-performance

Energy Exchange in Turbomachines

89

rotors. Therefore, the air foil shapes of the blades make a meaningful design. In radial outward flow pumps, the entry of fluids is at the inner radius; already the area of flow is restricted. Hence, for cast rotors, the losses (with non-air foil blades) are tolerated, for the sake of low-cost machines.

(a) Tail

Nose

Nose

Tail (b)

(c)

(d)

(e)

Figure 3.6

Orientation of rotor blades in axial flow rotor, related with fluid flow: (a) General flow of fluid; (b) Fluid flow from nose to tail of blades; (c) Turbine blade assembly: fluid forces the blades; (d) Compressor blade assembly: blades force the fluid; (e) Examples of shovel and spoon.

In an airfoil shape, the nose and tail ends are easily identified as shown in the Figs. 3.5 and 3.6 above. The flow of fluid is from nose-end to the tail-end. Once the flow is configured, the mean line of the blade profile is sufficient for the determination of blade angles.

90

Turbomachines

BOX 3.1 

Velocity Triangles for Radial Flow Machines

7KHSODQHRIYHORFLW\WULDQJOHVLVSHUSHQGLFXODUWRWKHVKDIWD[LVUHIHU)LJ  C

D N

A

Figure 3.7     

Radial flow rotor with rotation and blade velocities.

7KHEODGHVWUDYHUVHWKHFLUFXODUSDWK /HWWKHWLSVRIWKHEODGHVDWWKHLQOHWDQGRXWOHWEHDWWKHKLJKHVWSRLQWVRIWKHLUFLUFXODUSDWKVDW$ DQG&UHVSHFWLYHO\ 7KHWDQJHQWVWRWKHFLUFXODUSDWKVDW$DQG&DUH$%DQG&'%RWKDUHKRUL]RQWDO 7KHVSHHGRIWKHVKDIWLV1USPFORFNZLVH 7KHEODGHYHORFLWLHVDWWKHLQOHWDQGRXWOHW8DQG8UHVSHFWLYHO\DUHDORQJ$%DQG&'KDYLQJ ULJKWZDUGVHQVH8O'$18O'&1UHIHU)LJ  A

Figure 3.8 

B

U1

U2

C

B

D

Blade velocities at inlet and outlet points.

7KHWDQJHQWVWRWKHEODGHSURÀOHDWWKHWLSV$DQG&DUH//DQG007KHVHWDQJHQWVUHSUHVHQWWKH GLUHFWLRQV RI 9U ,W LV DOULJKW WR VKRZ WKH WDQJHQWV RQ WZR VHSDUDWH EODGHV EHFDXVH HYHU\ EODGH RFFXSLHVHYHU\SRVLWLRQGXULQJURWDWLRQ UHIHU)LJ  M L

M

L N

Figure 3.9

Orientation of relative velocities.

Energy Exchange in Turbomachines

BOX 3.1  

91

Velocity Triangles . . . . (Continued)

,QWKHSUHVHQWFDVHWKHÁRZLVRXWZDUG7KHVHQVHRI9UDQG9ULVUDGLDOO\RXWZDUG)RULQZDUGÁRZ WKHVHQVHRIWKHVDPH9UDQG9UFDQEHUDGLDOO\LQZDUG $GGWKHEODGHYHORFLW\8DQGUHODWLYHYHORFLW\9U7KHDGGLWLRQLVYHFWRULDO7KHYHFWRU9UVWDUWVZKHUH WKHYHFWRU8VWRSV989U&RPSOHWHWKHWULDQJOHWRUHDOL]H96RPHPRUHGDWDPD\EHUHTXLUHG VXFKDV@AHWFUHIHU)LJ 

V1

V2

Vr1

Vr2

a1 A

Figure 3.10

U1

B

C

U2

D

Addition of U and Vr to form velocity triangles.



'LIIHUHQWFDVHVFDQEHFRQVLGHUHGIRUJHWWLQJVRPHSUDFWLFH      7KHUH LV DEVROXWHO\ QRWKLQJ ZURQJ LQ ZULWLQJ 9  9U  8 WKDW LV YHFWRU 8 VWDUWV ZKHUH 9U VWRSV $OVR DQWLFORFNZLVHURWDWLRQLVDOULJKWIRUWKHURWRU%XWVXFKYDULHWLHVFDQEHWDNHQXSRQO\DIWHUJHWWLQJVRPH SUDFWLFH

For axial flow machines, the blade velocities at the inlet and outlet, U1 and U2, are equal to each other because both the tips of the blade are at the same diameter. It is quite usual, therefore, to draw the velocity triangles on the same base of U  U1  U2. The advantage of having a common base will become clear in the later sections.

BOX 3.2 

Velocity Triangles for Axial Flow Machines

7KHSODQHRIWKHYHORFLW\WULDQJOHVLVWDQJHQWLDOWRWKHURWRUDQGSDUDOOHOWRWKHVKDIWD[LV7KLVSODQHLV ´GHYHORSPHQWµRIWKHURWRUVXUIDFHUHIHU)LJ 

Figure 3.11    

Blades of axial flow rotor, as seen in developed view.

7KHEODGHVORRNDVLIWKH\DUHDUUDQJHGLQDURZEHWZHHQWZRLPDJLQDU\SDUDOOHOOLQHV $VWKHVKDIWURWDWHVWKHEODGHVPRYHDORQJWKHSDUDOOHOOLQHVDVVKRZQ /HWWKHPRWLRQEHIURPOHIWWRULJKWWKHÁXLGÁRZLVXSZDUGLQWKHSUHVHQWFDVH %RWKWKHWLSVRIWKHEODGHVDUHRQWKHVDPHGLDPHWHUWKHUHIRUH88

92

Turbomachines

BOX 3.2 

Velocity Triangles . . . . (Continued)

&RQVLGHUDVLQJOHOLQHEODGHSURÀOHUHIHU)LJ 

M M M

M

L

  

L

L

a1

Figure 3.12

L

Relative velocities are tangential to the blade profiles (at inlet and outlet).

7KHWDQJHQWVWRWKHSURÀOHDUH//DQG00 //DQG00DUHWKHGLUHFWLRQVRI9UDQG9U $GGWKHEODGHYHORFLW\8DQGUHODWLYHYHORFLW\9UWRREWDLQ9989U9UVWDUWVZKHUH8VWRSVUHIHU )LJ  V1

Vr1

V2

Vr2

Vr1

V2

Vr2

a1 = 90°

a1 U

Figure 3.13

V1

U

U

U

Addition of U and Vr to form velocity triangles, corresponding to Fig. 3.12.

 'LIIHUHQWFDVHVFDQEHFRQVLGHUHGIRUJHWWLQJVRPHSUDFWLFH

While explaining the addition of blade velocity (U) and relative velocity (Vr) to get the fluid velocity (V), it is indicated that the rule is V  U  Vr. But there is absolutely nothing wrong in writing V  Vr  U, that is, vector U starts where vector Vr stops. Also anticlockwise rotation is alright for rotor. But such varieties can be considered only after getting some practice. As per the descriptions given in the boxes and the steps (1–10) with reference to Figs. 3.1–3.6, velocity triangles can be sketched if the minimum data are known. Velocity triangles can be graphically drawn to a convenient scale, or can be analytically solved. In continuation, the absolute velocities of fluid, V1 and V2, can also be resolved into the components parallel and perpendicular to the direction of the blade velocity U. The information so obtained becomes the basis of the understanding of the forces exerted, the torques generated or utilized, the power of the machine, and finally the design and performance of the blades. Note: The purpose of the following examples is only to practice the drawing of velocity triangles. The utility of velocity triangles is yet to be discussed. However, it can be said that the study of turbomachines becomes very simple when velocity triangles are clearly understood.

EXAMPLE 3.1 A rotor is shown in Fig. 3.14. The entry of fluid is radial and the flow is outward. The blades are radial at the outlet. Draw representative velocity triangles.

Energy Exchange in Turbomachines

93

Solution: The velocity triangles are shown alongside. M

L

U2

M

1. Outward flow 2. Radial entry (a1 = 90°) (b2 = 90°)

U1

L N

V1

V2

Vr1

Vr2 b1

a1

U2

U1

Figure 3.14

b2

a2

Rotor of Example 3.1.

EXAMPLE 3.2 A rotor is shown in Fig. 3.15. The entry of fluid is radial and the flow is outward. The blades are bent forward at the outlet. Draw representative velocity triangles.

Solution: The velocity triangles are shown alongside. M U2

M L

1. Outward flow 2. Radial entry (a1 = 90°)

U1

L N

V1

V2

Vr1

Vr2 b1

a1

b2 U2

U1

Figure 3.15

Rotor of Example 3.2.

EXAMPLE 3.3 A rotor is shown in Fig. 3.16. The entry of fluid is at an angle and the flow is inward. The blades are radial at the inner diameter. Draw representative velocity triangles.

Solution: The velocity triangles are shown alongside. L M

N

M L

a1

U1

L U2

Inward flow

U1

U2

V1

Vr1 V2

Figure 3.16

Rotor for Example 3.3.

Vr2

94

Turbomachines

EXAMPLE 3.4 A rotor is shown in Fig. 3.17. The entry of fluid is at an angle and the flow is inward. The blades are such that at the outer diameter they are at 90o to blade velocity and the flow is radial at the inner diameter. Draw representative velocity triangles.

Solution: The velocity triangles are shown alongside. L U1 L M

1. Inward flow 2. Radial exit (a2 = 90°)

Figure 3.17

a2

b1

M U2

N

U2

U1 a1

V1

b2

V2

Vr1

Rotor for Example 3.4.

EXAMPLE 3.5 A rotor is shown in Fig. 3.18. The entry of fluid is at an angle and the flow is inward. The blades are inclined at the outer diameter as shown and the flow is radial at the inner diameter. Draw representative velocity triangles

Solution: The velocity triangles are shown alongside. L

U1

U1

U2 a1

L M N

M

U2

1. Inward flow 2. Radial exit (a2 = 90°)

Vr1 V1

Figure 3.18

V2 Vr2

Rotor for Example 3.5.

EXAMPLE 3.6 An axial flow rotor is shown in Fig. 3.19. State the direction of the flow, whether this is turbine or compressor and the direction of rotation. Sketch the representative velocity triangles.

Energy Exchange in Turbomachines

95

Solution: The velocity triangles are as shown. M

M

V1

Vr2

V2

Vr1

1. Flow is upward 2. This is a turbine blade 3. The blades move toward the right


L

U

L

Figure 3.19

Rotor for Example 3.6.

EXAMPLE 3.7 An axial flow rotor is shown in Fig. 3.20. State the direction of the flow, whether this is turbine or compressor and the direction of rotation. Sketch the representative velocity triangles.

Solution: The velocity triangles are as shown. M

M L

1. Flow is upward 2. This is a compressor blade 3. The blades move toward the right


V1

Vr1

V2

Vr 2

U L

Figure 3.20

Rotor for Example 3.7.

EXAMPLE 3.8 An axial flow rotor is shown in Fig. 3.21. State the direction of the flow, whether this is turbine or compressor and the direction of rotation. Sketch the representative velocity triangles.

96

Turbomachines

Solution: The velocity triangles are as shown. L

U

1. Flow is downward 2. This is a turbine blade 3. The blades move V r1 toward the left


L

M

V2 V r2

V1 M

Figure 3.21

Rotor for Example 3.8.

EXAMPLE 3.9 An axial flow rotor is shown in Fig. 3.22. State the direction of the flow, whether this is turbine or compressor and the direction of rotation. Sketch the representative velocity triangles

Solution: The velocity triangles are as shown. M

V2 M

1. Flow is upward 2. This is a turbine blade

L

3. The blades move toward the right


V1

Vr2

Vr1

U L

Figure 3.22

Rotor for Example 3.9.

EXAMPLE 3.10 An axial flow rotor is shown in Fig. 3.23. State the direction of the flow, whether this is turbine or compressor and the direction of rotation. Sketch the representative velocity triangles.

Solution: The velocity triangles are as shown.

Energy Exchange in Turbomachines

97

Vr2 U V2

1. Flow is toward the right 2. This is a compressor blade 3. The blades move upward

L

Vr1

L M M

Figure 3.23

V1

Rotor for Example 3.10.

3.3 Basic Equations: Linear Momentum Equation, Impulse Momentum Equation, Moment of Momentum Equation, and Euler Turbine Equation The Euler turbine equation is the basic equation on which the study of turbomachines is built up. The assumptions made while deriving the SFEE hold good for the Euler turbine equation also. Thus, the constancy of states of fluid separately at the inlet section and outlet section, non-accumulation of mass and steady invariable states over time intervals are also the assumptions in the case of the Euler turbine equation. Acceleration during starting and deceleration during stopping are excluded from the application of this equation. According to Newton’s second law, we have F  ma, which is now written as dV d or F  (mV ) dt dt This form of equation is known as a linear momentum equation. Again, the same equation when written as F m

Fdt  d (mV ) is known as impulse momentum equation (Fdt is the impulse of the force F). In continuation of the above, the equation applicable to turbomachines is rmdV Fr  dt Here, r is the radial distance of the force, Fr is the moment of the force and is known as the torque of the machine. rmdV/dt or rd(mV)/dt is the moment of the rate of change of momentum or “angular momentum.” In words, the above equation means that “the applied torque is the rate of change of angular momentum.” Thus, the equation d (mr s V ) T dt

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Turbomachines

is known as the angular momentum equation. Vax2 Vrad2 V1

V2

Vr2

Vax1 V1

Vrad1

Vu2

Vu2 U2

Vu1

V1

Vr1

Vu1 U1

Figure 3.24

Velocities in a generalized rotor.

In Fig. 3.24, a generalized rotor of a turbomachine is shown, with a typical flow of fluid, represented by the absolute velocities of fluid: V1 at the inlet and V2 at the outlet. Also shown are all the components of the fluid velocities:Vu1, Vu2, Vax1, Vax2, Vrad1 and Vrad2. Here, Vu is the component of the absolute velocity of fluid, along U; this is the “whirl” component, contributing to the torque. Vax is the component of the absolute velocity of fluid, along the axial direction; this causes the axial thrust. Vrad is the radial component of the absolute velocity of fluid. For a purely axial flow machine, the radial components Vrad1 and Vrad2 are reduced to zero. The axial components Vax1 and Vax2 become the flow components Vf1 and Vf2. Similarly, for a purely radial flow machine, Vrad1 and Vrad2 become the flow components Vf1 and Vf2. In this case, Vax1 and Vax2 are reduced to zero. Now, r can be taken as the generalized radius vector and V as the generalized velocity vector, each having components in axial, radial, and tangential directions. The cross product also gives the three components in the axial, radial, and tangential directions. However, the torque and power developed in a turbomachine are due to the tangential components only, while the axial thrust developed is due to the axial components. The radial components have the effect of producing a couple or moment of forces on the bearings, without any contribution to the torque. Thus

T  m (r s V ) and

P  TW  m W (r s V )

where V is the angular velocity. Considering the net torque from the inlet to the outlet section, the above equation is written as P  m (W1rV 1 u 1 W 2 r2Vu 2 )

Energy Exchange in Turbomachines

or

P  W  U1Vu 1 U 2Vu 21 m

99

(3.1)

where W is the specific work, and U1  V1r1; U2  V2r2. Equation (3.1) for the specific work is the Euler turbine equation. From this point onward, the presence or absence of gc should not be a problem anymore (m2/s2 is the same as J/kg.). Just as the tangential components of fluid velocities give rise to the tangential force (and then the torque, as above), the axial components of fluid velocities give rise to the axial thrust to be taken up by the bearings. Designating the thrust as T, we have T 

d (mVf )  m (Vf1 Vf2 ) dt kg m  s s N s

(3.2)

In Eq. (3.1) for W, the specific work W when obtained as a positive value represents the work output of a turbine. When the value of W obtained is negative, W is the input to the machine. This machine may be a fan, a blower, a compressor, or a pump. When the thrust T obtained is positive, that thrust on the bearings is in the direction of the inlet stream (irrespective of whether the machine is a pump or a turbine) and a negative value of T represents the thrust in the opposite direction of the inlet stream. As seen above, the complete details of the absolute fluid velocities V1 and V2 along with the fluid angles @1 and @2 are sufficient to determine the torque of the shaft, the axial thrust, and the specific work. However, the outlet velocity V2 depends on the inlet velocity V1, the flow characteristics of the fluid from the inlet to outlet through the passage between the blades, the cross-sectional area of the passage and its variation, the losses taking place in the passage, etc. One convenient method to find V2 is to calculate the relative velocity at the inlet, Vr1, and to find its variation (or otherwise) to Vr2 along with the blade angle A2 at the outlet, and then to solve for V2. As will be seen later, there are different methods to determine V2, but in general, all the methods have to be toward only one aim, that is, completing the velocity triangles at both the inlet and outlet, with the available or assumed data. In fact, the flow passage between the blades is to be designed so as to arrive at the required value of Vr2 and consequent V2.

3.4 Alternate Form of the Euler Turbine Equation One more form of the Euler turbine equation in terms of the six velocities (three from each velocity triangle) can be derived. Consider a general velocity triangle ABC, with V  U  Vr, as shown in Fig. 3.25, without subscripts “1” and “2” for the velocities. From triangle ACM CM2  AC 2 AM2  Vf 2  V 2 Vu2

Also, from triangle BCM CM2  BC 2 BM2  Vf2  Vr2 (U Vu )2  Vr2 U 2 Vu2  2UVu

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Turbomachines

C V

Vr

Vf a

b

M

A

U

Vu

B

U

Figure 3.25

General velocity triangle.

From the above two equations, for Vf2 , one can write V 2 Vu2  Vr2 U 2 Vu2  2UVu  2UVu  V 2  U 2 Vr2  UVu 

(V 2  U 2 Vr2 ) 2

Hence U1Vu1  and

U 2Vu2 

(V12  U12 Vr12 ) 2 (V22  U 22 Vr22 ) 2

at the inlet velocity triangle at the outlet velocity triangle

Substituting these values into the Euler turbine Eq. (3.1), we get W  U1Vu1 U 2Vu2 ¥ V 2  U12 Vr12 ´ ¥ V22  U 22 Vr22 ´ ¦ 1 µ¶ ¦§ µ¶ 2 2 § ¥ V 2 V22 ´ ¥ U12 U 22 ´ ¥ Vr12 Vr22 ´ ¦ 1 µ¶  ¦§ µ¶ ¦§ µ¶ 2 2 2 §

(3.3)

The specific work, as given by Eqs. (3.1) and (3.3), is sometimes denoted by WE, subscript E signifying the name Euler. Equation (3.3) is the second form of Euler turbine equation. This expression for the specific work, along with the equation obtained as a consequence of the first law of thermodynamics (and SFEE), gives a lot of information regarding the design of the blade profiles. The expressions for the specific work, in the form of Euler turbine equations, have been obtained starting from the “vane-congruent” flow. The vane-congruent flow is the ideal flow. The actual flow differs from the

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101

ideal flow due to various factors such as fluid viscosity, skin friction, eddies, turbulence, etc., as studied in fluid mechanics. All the variables can be accounted for, but this will be done later.

3.4.1 Components of Energy Transfer The energy transfer between a fluid and a rotor has been expressed in different ways. As seen earlier, the application of the first law of thermodynamics has given rise to two equations [Eqs. (1.2) and (1.3) in Chapter 1] for the specific work. The alternate form of the Euler turbine Eq. (3.1) has again given one more equation Eq. (3.3) for the specific work. One common component in all the equations is the quantity (V12 – V22)/2, which is the kinetic component of the energy transfer. The other components are as follows: 1. 2. 3. 4. 5.

The elevation component, (z1 – z2)g. The pressure-drop component, (p1 – p2)/Q. The enthalpy-drop component, (h1 – h2). The centrifugal energy component, (U12 U 22 ) / 2. The relative velocity component, (Vr22 Vr12 ) / 2.

The kinetic component is considered as the major component. All other components as grouped in Eqs. (1.2), (1.3), (1.4) in Chapter 1 and Eqs. (3.1) and (3.3) in this chapter can be considered as another combined component. The kinetic component can directly interact between the fluid and rotor; whereas the other components, in suitably formed groups, (that is, different groups in different equations) can have the interaction somewhat in an indirect way. The indirect way stated in simple terms is that the components have to simultaneously “transform and transfer.” This process is elaborated in Section 3.5.

3.4.2 Energy Equation of Relative Velocities The relative velocities at the inlet and outlet of the blades form a group, (Vr22 Vr12 )/ 2, as seen in Eq. (3.3). This group can be identified in an interesting way. Suppose that it is possible to consider the reference frame (i.e., the flow channel in the rotor) in isolation, or alternately, the observer also moves along with the frame. It can be realized that the variation of the relative velocity is directly connected to the area of flow by the equation

R1 A1Vr1  R2 A2Vr2 where A1 and A2 are the areas at the inlet and outlet of the blade passage, respectively. If further the density remains constant, as it is in a large number of cases, then A1Vr1  A2Vr2 

Vr 2 Vr1



A1 A2

Hence, Vr1 and Vr2 can be designed to be in a definite required proportion by designing the area ratio accordingly. As a result, the flow can now be like the conduit flow, where Bernoulli’s equation can be applied (as a first approximation), the flow channel acting as a diffuser or nozzle, with the pressure being developed or consumed.

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Turbomachines

However, the development of pressure is not totally due to this variation of Vr1 and Vr2 because the centrifugal component (U 22 U12 ) / 2 also participates in the mechanism by acting like the external force on the fluid. The effects have to be overlapped, as readily seen in the grouping of this component in the energy equation.

3.5 Impulse and Reaction The process of energy transfer in a turbomachine is by the dynamic action. The “dynamic action” is named so because the transfer process is during the flowing of the fluid. In fact, the transfer of energy, whether from or to the fluid, is possible only if that energy is in the kinetic form. What happens when the energy of the fluid is of some other form (like pressure or enthalpy)? 1.

In the case of turbines, the transformation of other forms of energy into kinetic form must take place before the transfer process, as in nozzles, as shown in the schematic view in Fig. 3.26. Turbomachine

Fluid

Fluid

inlet

outlet

Nozzle

Figure 3.26

2.

Rotor

Schematic view of a turbine.

In the case of pumps or compressors, the transformation must take place after the transfer process, as in diffusers, as shown in the schematic view in Fig. 3.27. Turbomachine

Fluid

Fluid

inlet

outlet

Rotor

Figure 3.27

Diffuser

Schematic view of a pump or compressor.

The above two cases are presented as if 1. 2. 3.

The two processes, transfer of energy and transformation of energy, are totally decoupled. The two processes are one after the other. The two processes take place separately in different parts.

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103

In actual practice, any design is one of the following three possible designs. It is possible to design a turbomachine in three different categories in such a way that: 1.

2. 3.

The two processes, transformation or conversion of energy (of the fluid) and transfer of energy (between the rotor and the fluid), are really decoupled, that is, they occur separately in different parts, as depicted in Figs. 3.26 and 3.27. The two processes, the conversion and the transfer, occur together, simultaneously, in the flow passages between the blades of the rotor. The process of transformation of energy or conversion of energy from one form to another form takes place “partly separately” and “partly simultaneously” with the transfer of energy, that is, the process of expansion or compression occurs partly in nozzles or diffusers and partly in the flow passages of the rotor. The process is shared between the parts.

The above three cases are distinctly different from one another. In the light of the above, the equations for the specific work W are recalled for further discussion (completely dropping gc): W  W  W 

V12 V22 2 2 V1 V22 2 V12 V22 2



p1 p2

R

 ( z1 z2 ) g

 (h1 h2 )  ( z1 z2 ) g 

U12 U 22 2

Vr12 Vr22 2

In all the expressions for W, the first component (V12 V22 ) / 2 is the change in the kinetic energy of the fluid between the inlet and outlet of the rotor, and this component gets transferred between the fluid and rotor directly, without requiring the presence of either nozzles or diffusers. The remaining components ( p1 p2 ) ( z1 z2 )  g R or

(h1 h2 )  ( z1 z2 ) g

or

(U12 U 22 ) 2

(Vr12 Vr22 ) 2

(all are equal to one another) are the components that require the transformation in the rotor itself, simultaneously with the process of transfer of energy. These components are termed as “reaction components.” The ratio of the components of energy transferred, due to the change in the pressure of fluid between the inlet and outlet of the rotor, to the total energy transferred is known as the degree of reaction, R. Thus ¨ ( p1 p2 ) · © ¸  ( z1 z2 ) g R ª ¹ R W (h h )  ( z1 z2 ) g  1 2 W

[3.4(a)] [3.4(b)]

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Turbomachines

(U12 U 22 ) (Vr12 Vr22 )

2 2  W 

(U12 U 22 ) (Vr12 Vr22 ) (V12 V22 )  (U12 U 22 ) (Vr12 Vr22 )

[3.4(c)] [3.4(d)]

It is possible to visualize the reaction in different ways. Imagine a skating board in which the wheels are locked, “braked,” or restrained by stoppers. Unless the brakes are released, the skating board cannot move. Let there be a person on the board. If this person tries to jump from the board, he/she can jump over a long distance, the board remaining firmly where it is [Fig. 3.28(a)].

(a)

(b)

Figure 3.28

Reaction in linear motion: (a) Board remains firm and (b) board moved backward.

Suppose the brakes on the wheels are released. The board is now free to move. The person on the board is ready to jump. He/she bends and his/her powerful thigh muscles and calf muscles, having their stored energy, suddenly unwind, releasing the energy. What happens? The person jumps, but now his/her forward movement is much less. Instead, the skating board moves backward [Fig. 3.28(b)]. This backward movement is due to the reaction. The stored energy of the person on the board is released and the muscles expand, but, the effect is to push the board backward in the opposite direction.

Energy Exchange in Turbomachines

105

+

Figure 3.29

Reaction in rotation.

Again, imagine a cylindrical drum, having nozzles in some tangential direction. The drum is supported in bearings. It is now charged with high-pressure air (Fig. 3.29). If the drum is restrained by brakes and air is released through the nozzles, the velocity of air can be very high. But if the drum is not restrained (the brakes are released), for the same velocity of air from the nozzles (with reference to the nozzle), the drum moves backward. This is by the reaction. A water sprinkler in a garden is yet another example of the movement by reaction. All the movements, seen above, can be precisely quantified and related to the energy spent.

3.6 Utilization Factor of Turbines The utilization factor of a turbine is defined as the ratio of the ideal work output of the turbine to the energy available for conversion into work at the inlet stream of the fluid: 

W Wa

(3.5)

Here, W is the ideal specific work, as defined by the Euler turbine equation. The energy available in the fluid at the inlet, Wa, has two parts: its kinetic energy, V12/2, and the energy that is possible to be obtained due to the pressure drop (known as the reaction component). Thus Wa 

V12 2



p1 p2

R

( with z1  z2 )

Also Wa 

V12 2



(U12 U 2 2 ) 2

( Vr12 Vr22 ) 2

W 

V2 2 2

Hence 

(V12 V22 )  (U12 U 22 ) (Vr12 Vr22 ) V12  (U12 U 22 ) (Vr12 Vr22 )

(3.6)

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Turbomachines

Also 

U1Vu1 U 2Vu2 [ V12

 (U12 U 22 ) (Vr12 Vr22 )]/2

(3.7)

The utilization factor is defined only for turbines, which may be of any type, either axial flow or radial flow or mixed flow type. The fluids can be either compressible or incompressible. The utilization factor of a turbine is a performance parameter: It indicates how well the turbine performs. This factor can be maximized, or conditions for the maximum utilization factor can be laid down, while designing the blades. The utilization factor is not defined for work-absorbing type of turbomachines. Now, from Eq. [3.4(d)] we have R

(U12 U 22 ) (Vr12 Vr22 ) (V12 V22 )  (U12 U 22 ) (Vr12 Vr22 )

So V12 V22 1  1 2 R (U1 U 22 ) (Vr12 Vr22 ) 

V12 V22 (U12 U 22 ) (Vr12 Vr22 )



 (U12 U 22 ) (Vr12 Vr22 ) 

1 1 R

1  R R (V12 V22 )R 1 R

(3.8)

Substituting Eq. (3.8) into the expression for the utilization factor [Eq. (3.6)], the utilization factor is obtained as 

V12 V22

(3.9)

V12 RV22

Clearly, the equation does not hold good for the condition R  1, but otherwise, for any value 0 b R  1, it is applicable for any turbine in general. Thus, the utilization factor can be seen as a function of the fluid velocities V1, V2 and the degree of reaction R. As mentioned earlier, the degree of reaction can be taken as a design parameter and therefore can have a prescribed value. The inlet velocity V1 is also a part of the prescribed data. Hence, the utilization factor can be varied by varying the exit velocity V2. By the very definition of the utilization factor, the least value of V2 gives rise to the maximum possible utilization factor, m. The minimum value of V2 occurs when the exit angle @2 is 90o, with V2 being perpendicular to U2. Now, for axial flow machines (U1  U2  U) and for Vf1  Vf2  V2  V1 sin@1, the expression for m readily reduces to m   m 

V12 V22 V12 RV22



cos2 A1 1 R sin 2 A1

V12 V12 sin 2 A1 V12 RV12 sin 2 A1



1 sin 2 A1 1 R sin 2 A1 (3.10)

Energy Exchange in Turbomachines

107

3.7 Speed Ratio The speed ratio (E) is the ratio of the blade peripheral velocity U1 to the absolute fluid velocity V1 at the inlet. In the context of the design of a turbomachine, these two velocities U1 and V1 are two important parameters. V1, the fluid inlet velocity, is guided by the available head of fluid or available pressure drop or enthalpy drop that is accepted as the basic data. U1, the blade velocity, is a function of the diameter and speed of the rotor (U1  ODN/60). The speed, N (rpm), is an assumed value that is selected from the synchronous speeds for turbines or “slip speeds” for pumps or compressors. Any logical conclusion to arrive at a particular value of speed ratio (E), therefore, leads to the determination of a suitable value of the diameter of the rotor and the speed of the machine. The speed ratio E  U1/V1, therefore, is a representation of all the basic and assumed data. In fact, along with the maximization of the utilization factor , the speed ratio is also to be “suitably” arrived at. From Eq. (3.8), we have (U12 U 22 ) (Vr12 Vr22 ) 

(V12 V22 )R 1 R

Hence ¥ V 2 V22 ´ ¥ V12 V22 ´ ¥ R ´ W ¦ 1 µ¶  ¦§ µ¶ ¦§ 1 R µ¶ 2 2 § 

V12 V22 ¥ R ´ ¦§1  µ 2 1 R¶

Therefore ¥ V 2 V22 ´ ¥ 1 ´ W  U1Vu1 U 2Vu 2  ¦ 1 µ¶ ¦§ 1 R µ¶ 2 §

(3.11)

Now, for maximization of the utilization factor,   m, @2  90o, Vu2  0, and V2  Vf1  V1 sin@1. Therefore ¥ V2´ V12 ¦1 22 µ U § V1 ¶ 1 1 F 1  s s 2(1 R ) V1 V1 Vu 1 F  F 

1 sin 2 A1 2(1 R )cos A1

¥ Vu1 ´ ¦ V  cos A1 µ § 1 ¶

cos A1

(3.12)

2(1 R )

Thus, it is seen that when F  cos A1 / 2(1 R ) , we have m  continuation, the maximum possible W is given by W  Wa s m

cos2 A1 1 R sin 2 A1

as seen by Eq. (3.10). Also, in

(3.13)

In Eq. (3.12), the values of E obtained may be taken as if it is a specified condition, and the corresponding value of m may be taken as the derived result. Table 3.1 gives some representative values of such results.

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Turbomachines

Table 3.1

Representative values of derived results

cos2 A1

General expressions

F

cos A 2(1 R )

Impulse turbine R0

F

U cos A1  2 V1

m  cos2 A1

Reaction turbine R  0.5

F

U  cos A1 V1

m z cos2 A1

m 

1 R sin 2 A1

W  Wa s m ¥V 2 ´ W  ¦ 1 µ 4F 2 § 2 ¶

 4F 2

¥ V2´ W  ¦ 2 s 1 µ F2 2 ¶ §

z F2 (for @ b about 25o)

U2

The equations for m and E [Eqs. (3.10) and (3.12)] indicate that for a given inlet fluid velocity V1 and inlet fluid angle @1, the blade speed U for the impulse turbine (R  0) is half that of the 50% reaction turbine (R  0.5). Hence, if both types of the turbines were to run under the same values of V1 and @1, at the same speed, the diameter of the impulse-type machine would be half that of the 50% reaction-type machine. Alternately, if the diameters were to be the same, the speed (rpm) of the impulse-type machine would be half that of the 50% reaction-type machine. To understand the further significance of the above results, first, a representative value of the fluid inlet velocity, V1, is assumed to be 100 m/s. This is taken as the field data. Two cases of the degree of reaction (R  0 and R  0.5) are worked out for various values of the fluid inlet angle (@1  10 − 30o). The results are shown in Tables 3.2 and 3.3. Table 3.2

Result when V1 = 100 m/s and R = 0

@

E

U

m

Wm

10o

0.492

49.2

0.97

4850

15o

0.483

48.3

0.933

4665

20o

0.47

47.0

0.883

4415

25o

0.453

45.3

0.821

4105

30o

0.433

43.3

0.75

3750

Table 3.3

Result when V1 = 100 m/s and R = 0.5

@

E

U

m

Wm

10o

0.985

98.5

0.98

4900

15o

0.966

96.6

0.965

4825

20o

0.94

94.0

0.938

4690

25o

0.906

90.6

0.902

4510

30o

0.866

86.6

0.857

4285

Energy Exchange in Turbomachines

109

The results obtained further indicate that the 50% reaction-type machines give rise to better values of utilization factors than those given by impulse machines. These results are without taking into account the losses, friction, etc. If the effects of such losses are also to be considered, the reaction machines score still better. This is because the fluid velocities in the reaction machines are generally lower than those in the impulse machines. Consequently, the losses in the reaction machines are less. The results also suggest that the fluid inlet angles @1 are to be as low as possible for better utilization.

EXAMPLE 3.11 The velocity triangles at the inlet and outlet are given in Fig. 3.30. State with reasons the following: (a) Whether the machine is radial flow type or axial flow type. (b) Whether the machine is work-producing type or work-absorbing type. (c) Specific work W. (d) Power per unit flow rate. (e) Degree of reaction R. (f ) Axial thrust. (g) Utilization factor, if applicable. V1

V2

Vr1

=Vf1

= Vf2

Vf1 = Vf2

90°

b1 = 70°

90°

U1 = 4 m/s

Figure 3.30

Vr2

U2 = 8 m/s

Velocity triangles for Example 3.11.

Solution: (a) Because U2  U1, this is a radial outward flow machine. (b) Because V2  V1, this is a work-absorbing-type machine, pump or compressor. (c) We know that V1 U1

 tan B1  tan70o

Hence V1  U1tan B1  4tan70o  11 m /s  Vf1  Vf2  Vr2 Now V2  U 22  Vr22  82  112  13.6 m/s and Vu2  U 2  8 m / s So W  U1Vu1 U 2Vu2  0 8 s 8  64 J/ kg This is the work input.

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Turbomachines

(d) We have P  64 J/s per unit mass flow rate  64 W/(kg/s). (e) Kinetic component of specific work is given by V22 V12 2



13.62 112  31.98 m 2 /s2 2

Therefore, the reaction component of the specific work is 64 31.98  32.02 Therefore, the degree of reaction is 32.02 R  0.5 64 (f ) Axial thrust: This is a radial flow machine; therefore, there is no axial thrust due to the flow in the rotor. However, the fluid moves initially in the axial direction, because of which an axial thrust is created. So Axial thrust  U1 s m ¥m k ´  4 s1  4¦ s s µ  N § s s¶ (g) Utilization factor is applicable only to turbines; hence it is not applicable here.

EXAMPLE 3.12 The velocity triangles at the inlet and outlet are given in Fig. 3.31. State with reasons the following: (a) Whether the machine is radial flow type or axial flow type. (b) Whether the machine is work-producing type or work-absorbing type. (c) Specific work W. (d) Power per unit flow rate. (e) Degree of reaction R. (f ) Utilization factor, if applicable. V2

Vr2

V1

Vr1 = Vf1 = Vf2

a1 = 30°

b2 = 50°

b1 = 90°

U1 = U2 = 15 m/s

Figure 3.31

Velocity triangles for Example 3.12.

Solution: (a) Because U1  U2, this is an axial flow machine. (b) Because V2  V1, this is a work-producing machine, that is, turbine.

Energy Exchange in Turbomachines

(c) To calculate Specific work: U  cos A1 V1 Therefore V1 

and

U 15   17.32 m /s cos A1 cos 30o

Vr1  U tan30o  15 tan30o  8.66 m/ s  Vf2  Vf1

Outlet velocity triangle is shown in Fig. 3.32. V2

Vr2

V f2 b2 = 50° V u2

x U2 = 15

Figure 3.32

Outlet velocity triangle for Example 3.12.

Let x  U – Vu2. Now Vf2 x

 tan50o

Therefore Vf2

x

tan 50o



8.66  7.27 m /s tan 50o

Therefore Vu2  U 2 x  7.73 m /s Also W  U (Vu1 Vu2 )  15 (15 7.73)  109.05 m 2 /s 2 or J/kg (d) Power  P  109.05 W per unit flow rate. (e) Degree of reaction R: We have x  cos50o Vr2

Therefore Vr 2 

x 7.27   11.31 m /s cos 50o cos 50o

111

112

Turbomachines

Because U1  U2, we have W 

V12 V22 2



Vr22 Vr12 2

Therefore, ¨ (Vr22 Vr12 ) · ¨ (11.312 8.662 ) · ¸ © © ¸ 2 2 R© ¸© ¸  0.243 W 109.05 ¸ © © ¸ ¸¹ ©ª ©ª ¹¸ (f ) Utilization factor : 

W 109.05   0.62 V22 11.62 109.05  W  2 2

Comment: Problems 3.11 and 3.12 have been worked out using basic trigonometric relations. Graphically, by using triangles drawn accurately to scale, various velocities or components can also be measured on the diagrams. By practice, however, the analytical procedures like the above may look easier.

EXAMPLE 3.13 The velocity triangles at the inlet and outlet are given in Fig. 3.33. State with reasons the following: (a) Whether the machine is radial flow type or axial flow type. (b) Whether the machine is work-producing type or work-absorbing type. (c) Specific work W. (d) Power per unit flow rate. (e) Degree of reaction R. (f ) Axial thrust. (g) Utilization factor, if applicable.

V2

V1 = 210 m/s

Vr2

Vr1 90°

22° U1 = U2 = 100 m/s Vf1 = Vf2

Figure 3.33

Velocity triangles for Example 3.13.

Solution: (a) Because U1  U2, this is an axial flow machine.

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113

(b) Because V2  V1, this is a work-producing machine (i.e., turbine). (c) We have Vu1  V1cosA1  210 cos 22o  194.7 m /s Vf1  V1sinA1  210 sin 22o  78.67 m /s  Vf2  V2 So W  U1Vu1 U 2Vu2 (Vu2  0)  100 s 194.7  19470 J/kg Specific work is W  19.47 kJ/kg. (d) Power  W· m ( m  1 kg/s)  19.47 kW per unit mass flow rate (e) Kinetic component is given by V12 V22 2



2102 78.67 2  18955 J/kg 2

Therefore degree of reaction is R

19470 18955  2.67% 19470

(f ) Axial thrust is zero because Vf1  Vf2. (g) Utilization factor is given by 

W 19470   0.863 V22 78.67 2 19470  W  2 2

Comment: The reaction is only 2.67% or 0.0267. This is almost like zero. The machine may be termed an impulse machine.

EXAMPLE 3.14 The velocity triangles at the inlet and outlet are given in Fig. 3.34. State with reasons the following: (a) Whether the machine is radial flow type or axial flow type. (b) Whether the machine is work-producing type or work-absorbing type. (c) Specific work W. (d) Power per unit flow rate. (e) Degree of reaction R. (f ) Axial thrust. (g) Utilization factor, if applicable.

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Turbomachines

V1

Vr2 Vf2

Vr1

V2

Vf1 35° 30° 20° U1 = U2 = 85 m/s

Figure 3.34

Velocity triangles for Example 3.14 (Vr2 = 0.98 Vr1)

Solution: (a) Because U1  U2, this is an axial flow machine. (b) Because V2  V1, this is a work-producing machine (i.e., turbine). (c) From the inlet velocity triangle, tan 35o 

Vf1 x

Therefore x tan 35o  Vf1 That is, (V1cos20o 85) tan 35o  V1sin20o  (0.9397V1 85) 0.7  V1 s 0.342 Hence V1  188.4 m /s Therefore Vu1  V1cos20o  177 m /s Vf1  V1sin20o  64.44 m /s Now Vr1  [(Vu1 U )2  Vf12 ]0.5 Vr2

 [(177 85)2  64.442 ]0.5  112.32 m/s  0.98 s Vr1  110 m / s

So Vu2  U Vr2 cos30o  85 110 cos30o  10.26 m /s Therefore Vu1 Vu2  177 ( 10.26)  187.26 m/s So specific work is W  U (Vu1 Vu2 )  85 s 187.26  15917 J/kg

Energy Exchange in Turbomachines

(d) Power  P  15917 W per unit flow rate. (e) Reaction component is Vr22 Vr12



2

1102 112.322  258 m 2 /s2 2

Therefore, the degree of reaction is R

Reaction component Specific work

258   0.0162% or 1.62% 15917 (f ) Axial thrust: We have Vf2  Vr2 sin30o  55 m /s So Axial thrust  (Vf1 Vf2 )m  (64.44 55)1  9.44 N in the direction of Vf1 (g) Utilization factor : We have V2  (10.262  552 )0.5  55.95 m /s Therefore 

W 15917   0.836 2 (55.95) (55.95)2 W  15917  2 2

Comments: 1. 2. 3.

The negative reaction is alright. Very small value of R (0.0162). This machine can be said to be an impulse machine. There is an axial thrust.

EXAMPLE 3.15 The velocity triangles at the inlet and outlet are given in Fig. 3.35. State with reasons the following: (a) Whether the machine is radial flow type or axial flow type. (b) Whether the machine is work-producing type or work-absorbing type. (c) Specific work W. (d) Power per unit flow rate. (e) Degree of reaction R. (f ) Utilization factor, if applicable.

115

116

Turbomachines U1 = U2 = 30 m/s a1 = 60°

V1 = 40 m/s a1 = 60°

V2 at 90° to U

Vf1 = Vf1 = V2

Vr1

Figure 3.35

V1

Vr2

V2

Velocity triangles for Example 3.15.

Solution: (a) Since U1  U2  U, this is an axial flow machine. (b) Since V2  V1, this is a turbine. (c) We have Vu2  0; Vu1  40 cos60o  20 m /s So specific work is W  UVu1  30 s 20  600 J/kg (d) Power  P  W m  600 W per unit flow rate. (e) We have V2  Vf2  Vf1  40 sin60o  34.64 m /s The kinetic component of W is V12 V22 2 Therefore



402 34.642  200 J/kg 2

Reaction component  W Kinetic component  600 200  400 J/kg

Therefore, the degree of reaction is R (f ) Utilization factor is 

400  0.67 600

600 W   0.5 2 ¥ V2 ´ 600  600 W ¦ § 2 µ¶

Comment: This problem is yet another one for practice. However, the utilization factor  is too low and is to be improved. How can it be improved?

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117

Suppose it is not possible to change the magnitude and direction of the inlet fluid velocity V1. The other option is to increase the velocity of blades U, that is, to increase the speed of the machine. The increased values of U can be tried, and it is easy to trace the steps in calculations (Table 3.4). (The kinetic component 200 J/kg is not altered.) Table 3.4

Effect of increasing speed (U) in Example 3.15

U

W  UVu1

Reaction Component

R

Š W/(W V22/2)

40

40 s 20  800

600

0.75

800/(800  600)  0.57

50

50 s 20  1000

800

0.80

1000/(1000  600)  0.625

60

60 s 20  1200

1000

0.83

1200/(1200  600)  0.67

It may be observed that such variations can improve the utilization factor. Further, it follows that Vr1 and Vr2 also change their angles (A1 and A2). The blades of the rotor have to be at different angles. Hence, the blade shape and orientation have to be different.

K eywords Absolute fluid velocity Alternate form of Euler turbine equation Angular momentum equation Axial flow machines Blade velocity Degree of reaction Euler turbine equation Impulse machines

Linear momentum equation Radial flow machines Reaction machines Relative velocity Speed ratio Utilization factor Vane-congruent flow Velocity triangles

S ummary The focus of study in this chapter is on the mechanism of the transfer of energy and momentum between the flowing fluid and the blades of the rotor. t
The velocity triangles are studied in detail, with respect to both radial flow machines and axial flow machines. t
V  U  Vr t
The Euler turbine equation is the most basic principle of working of any type of turbomachine. This equation, for specific work, is derived from Newton’s law of motion and its corollary, the impulse momentum equation. The second alternate form of the equation for the specific

work is also derived in terms of the velocity components of the fluid, that is, W  U1Vu1 U 2Vu2 t
The two different components of the energy transferred are identified: The kinetic component and the reaction component. Consequently, a distinction is drawn between the two types of energy transfer processes: The impulse type and the reaction type. t
Kinetic component  (V12 – V22)/2 t
Reaction component (U12 – U22)/2  (Vr22 – Vr12)/2

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Turbomachines

t
With respect to the machine, the degree of reaction R, the utilization factor , and the speed ratio E are identified as important parameters

that facilitate the design of the blades as also the evaluation of the performance of the turbomachine.

I mportant E quations 1. Specific work

Hence

P  W  U1Vu 1 U 2Vu2 m 2. Specific work



(V12 V22 )  (U12 U 22 ) (Vr12 Vr22 ) V12  (U12 U 22 ) (Vr12 Vr22 )

Also

¥ V 2 V22 ´ ¥ U12 U 22 ´ ¥ Vr12 Vr22 ´ W ¦ 1 µ¶  ¦§ µ¶ ¦§ µ¶ 2 2 2 § 3. Degree of reaction ( p1 p2 ) ( z1 z2 )  g R R W (h h )  ( z1 z2 ) g  1 2 W 2 2 (U1 U 2 ) (Vr12 Vr222 )

2 2  W (U12 U 22 ) (Vr12 Vr22 )  2 (V1 V22 )  (U12 U 22 ) (Vr12 Vr22 ) 4. Utilization factor



[ V12

U1Vu1 U 2Vu2  (U12 U 22 ) (Vr12 Vr22 )] 2

5. Utilization factor 

V12 V22 V12 RV22

6. Maximum utilization factor m 

cos2 A1 1 R sin 2 A1

7. Specific work ¥ V 2 V22 ´ ¥ 1 ´ W  U1Vu1 U 2Vu 2  ¦ 1 µ¶ ¦§ 1 R µ¶ 2 § 8. Speed ratio

W  Wa

F

cos A1 2(1 R )

M ultiple- C hoice Q uestions (Note: In all the following questions, U is the blade velocity, V is the absolute fluid velocity, and Vr is the relative velocity of the fluid with reference to the blade. Subscript 1 is for the inlet and 2 is for the outlet.) 1. U is always tangential to (a) the blade profile (b) the stream line of the fluid

(c) the rotor wheel (d) the rotor axis 2. Vr is always tangential to the (a) stream line of fluid (b) rotor axis (c) guide vanes (d) none of these

Energy Exchange in Turbomachines

3. V is the vectorial sum of U and Vr (a) only at the inlet (b) only at the outlet (c) only in the turbines (d) everywhere 4. In a radial flow rotor, the velocity triangle is visualized on a plane that is (a) parallel to the axis (b) perpendicular to the axis (c) tangential to the blade profile (d) none of these 5. In an axial flow rotor, the velocity triangle is visualized on a plane that is (a) parallel to the axis (b) perpendicular to the axis (c) tangential to the blade profile (d) none of these 6. Choose the wrong equation in the following: (a) V  U  Vr (b) U  V Vr (c) Vr  U  V (d) V  Vr  U 7. Velocity triangles are drawn for (a) stator vanes (b) rotor vanes (c) guide vanes (d) none of these 8. For a centrifugal pump, (a) U1  U 2 (b) U1  2U 2 (c) U1  U 2 (d) U1  U 2 9. For an axial flow turbine, the fluid strikes (a) on the convex side of the blade (b) on the concave side of the blade (c) tangential to the blade. (d) perpendicular to the blade 10. For a radial flow machine, the fluid enters at (a) the outer rim (b) the inner rim (c) perpendicular to the rim (d) any of (a), (b), and (c). 11. If there is no change in the fluid pressure between the inlet and outlet of the rotor of an axial flow machine, then (a) R x 0 (b) R  0 (c) R  0 (d) R  0

119

12. If in an axial flow machine, the absolute fluid velocities at the inlet and outlet are equal, then (a) R  0 (b) R  0 (c) R  0 (d) R  1 Use Fig. 3.36 to answer the next three MultipleChoice Questions (13–15).

Figure 3.36

Blade assembly for Multiple-Choice Questions 13, 14, and 15.

13. In the blade assembly shown, the fluid flow is (a) top to bottom (b) bottom to top (c) left to right (d) right to left 14. In the blade assembly shown, the rotation is (a) left to right (b) right to left (c) top to bottom (d) bottom to top 15. The assembly shown is of (a) a turbine (b) a compressor (c) a pump (d) none of these 16. The speed ratio is (a) V1 /V2 (b) U1 /U 2 (c) U1 /V1 (d) U 2 /V2 17. The utilization factor is (a) (W  V12 / 2) / W (b) W / (W  V12 / 2) (c) W / (W V2 2 / 2) (d) W / (W  V2 2 / 2) 18. The Euler turbine equation is an expression for (b) axial thrust (a) power (c) torque (d) specific work

120

Turbomachines

R eview Q uestions 1. What is a vane-congruent flow? By means of neat sketches, show the vane-congruent flow for at least two types of radial flow blades and one type of axial flow blades. (Refer Section 3.2.3) 2. What is the purpose of air foil shapes for blades? (Refer Section 3.2) 3. Derive the Euler turbine equations. Do the equations hold good for pumps/compressors? (Refer Section 3.3)

4. Distinguish between impulse and reaction processes in turbomachines. Give examples. (Refer Section 3.5) 5. Explain the term utilization factor. (Refer Section 3.6) 2 2 V V2 6. Prove that  12 (Refer Section 3.6) V1 RV22 7. Prove that F 

cos A1

(Refer Section 3.7) 2(1 R ) 8. Compare the impulse and reaction (R  0.5) machines. (Refer Section 3.7)

E xercises (For the sake of practice, nonstandard speeds are also used in problems.) 1. The rotor of a radial flow machine has the inlet diameter of 40 cm and outlet diameter of 90 cm. At the outlet, the blades are radial. The speed of the rotor is 8000 rpm. The fluid enters the rotor in a radial direction. Calculate the inlet blade angle, absolute velocity of fluid at the outlet, and the specific work, if the inlet fluid velocity is 150 m/s. The velocity of flow remains constant in the rotor. Also find the angle of outlet velocity. 2. The blade angles at the entry and exit of an axial flow turbine rotor are 25o each. The blades have a mean diameter of 60 cm and the speed is 3000 rpm. The absolute velocity of fluid at the outlet is axial. Calculate the absolute velocities at the inlet and outlet, angle of inlet velocity, specific work, degree of reaction, and utilization factor. The velocity of flow remains constant in the rotor. 3. The rotor of a radial flow machine has its entry and exit diameters as 20 cm and 50 cm, respectively. Its blades are bent backward so that the blade tangent at the outlet makes an angle of 60o with the blade velocity. The fluid velocity

at the inlet is radial, without any whirl component. The flow components remain constant at 10 m/s in the rotor. The speed of the rotor is 800 rpm. Draw the velocity triangles at the inlet and outlet of the rotor. Calculate the blade angle at the inlet and the specific work. Also calculate the specific work when the blade outlet angle is 80o, instead of 60o. 4. In an axial flow machine, the mean diameter of the rotor is 50 cm and the speed of the rotor is 10000 rpm. The fluid enters with a velocity of 380 m/s at 25o to the blade velocity. The blade angle at the outlet is 35o. The flow component remains constant in the rotor. Calculate the specific work and degree of reaction. If the machine is turbine, calculate the utilization factor. 5. The mean diameter of the rotor in an axial flow machine is 0.6 m and the speed is 10000 rpm. The fluid enters the rotor in the axial direction at a velocity of 180 m/s. The fluid leaves the rotor at an angle of 35o with the blade velocity. Calculate the blade angles at the inlet and outlet and the specific work. The flow components do not vary in the rotor. 6. The fluid enters the rotor at a velocity of 60 m/s at a diameter of 90 cm at an angle of 30o

Energy Exchange in Turbomachines

7.

8.

9.

10.

with the tangent to the wheel. The wheel speed is 1200 rpm. The absolute exit velocity of the fluid is radial, at a diameter of 30 cm. The flow components remain constant in the rotor. Calculate the blade angles at the inlet and outlet, the specific work, the degree of reaction, and the utilization factor. An axial flow turbine has a mean rotor diameter of 70 cm and the rotor speed is 1000 rpm. The fluid enters with a velocity of 100 m/s at an angle of 28o with the plane of rotor. The blade outlet angle is 28o. Calculate the blade inlet angle, the specific work, the degree of reaction, and the utilization factor. The blade velocity in an axial flow turbine is 125 m/s. The blade angles at the inlet and outlet are 40o and 25o, respectively. The absolute fluid velocity at the inlet is 220 m/s. The relative velocity at the outlet is 50% more than that at the inlet due to expansion. Calculate the angle at which the fluid enters the rotor, specific work, degree of reaction, utilization factor, and axial thrust. Water enters a radial flow pump without any whirl component at a velocity of 12 m/s. The entry diameter of the rotor is 10 cm and the exit diameter is 20 cm. The speed of the rotor is 2000 rpm. Calculate the blade angle at the inlet. The relative velocity of water at the outlet gets reduced to 75% of that at the inlet, due to an increase of flow areas, and the blade outlet angle is 80o. Calculate the specific work and degree of reaction. Air enters a radial flow compressor with a velocity of 50 m/s, without any whirl component. The diameters of the rotor are 30 cm and 60 cm at the inlet and outlet, respectively.

P roject- O riented Q uestions 1. The solved and unsolved examples of this chapter are meant to illustrate the various forms of velocity triangles and the variety of the turbomachines. In addition, their working

121

The blades are radial at the outlet. Calculate the blade inlet angle, speed of the rotor, and the degree of reaction if the specific work of the rotor is 30 kJ/kg. The flow component of velocity remains constant in the rotor. Repeat the problem, if the outlet blade angle is 75o instead of 90o. 11. A radial flow pump impeller has diameters of 3.5 cm and 8 cm at the inlet and the outlet, respectively. The blades are inclined backward, at 65o to the blade velocity, at the outlet. Water enters the rotor with a velocity of 8 m/s, without any whirl component. The radial components of the velocity remain constant in the rotor. Determine the speed of the rotor to generate a head of 16 m of water. If the speed is 2880 rpm, what is the theoretical head generated? 12. Water enters the rotor of an axial flow pump, without any whirl component. The diameters of the rotor are 45 cm at the blade tip and 10 cm at the hub. The speed of the rotor is 800 rpm. The axial flow component of water velocity is 8 m/s, which may be assumed to remain constant through the entire cross section of flow. Draw the velocity triangles at the inlet of the blade, separately at tip end and hub end, and determine the blade angles at the inlet at both the points. The outlet blade angles are 40o at the tip and 90o at the hub. Calculate the specific works separately at the tip and hub ends of the blade. If the variation of the specific work can be taken as linear between the hub and tip, determine its value at 0.6 times the length of the blade. Taking this value of specific work as an average, uniform value, calculate the theoretical power required to drive the rotor.

S olutions A vailable parameters also present a wide range of values of such parameters. Now, with respect to all the problems, conduct an overall survey of all the parameters, such as (a) speed, (b) specific

122

Turbomachines

work, (c) specific speed, (d) power, (e) degree of reaction, (f ) utilization factor, (g) speed ratio etc. Present your findings in the form of a table. What are the different conclusions you can draw from this table? 2. In Problem 11, an impeller of a radial flow pump with inlet diameter of 3.5 cm and outlet diameter of 8.0 cm has been mentioned. Imagine that you are actually fabricating an impeller of these diameters and that you are equipped with a motor that can be set

to run at three different speeds (750, 1000 and 1500 rpm). Design three blade profiles for the impellers, with a common inlet angle of 70° and outlet angles of 55°, 65° and 75°. Calculate the theoretical flow rates when each impeller is run at the three different speeds. Plot the results in a suitable set of coordinates. Draw up your conclusions. Frame at least one more such possibility of conducting experiment to prove a chosen theoretical result.

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5. 6. 7. 8. 9.

(c) (a) (d) (b) (a) (c) (b) (c) (b)

10. 11. 12. 13. 14. 15. 16. 17. 18.

(d) (b) (d) (d) (d) (a) (c) (d) (d)

Exercises 1. 41.84o, 405.745 m /s, 142.12 kJ/kg, 21.7o 2. 193.556 m /s, 43.95 m /s, 13.124o, 17.766 kJ/kg, 0, 0.948 3. 50o, 317.72 J/kg, 401.7 J/kg

8. 18.6o, 28.9765 kJ/kg, 0.2566, 0.9159, 1.03 N /( kg /s ) 9. 48.89o, 395 J/kg, 0.556 10. 30o, 5513 rpm, 0.5; 29.05o, 5729.6 rpm,

4. 81.67 kJ/kg, 0.28, 0.8588

0.5374 11. 3469.5 rpm, 10.2 m of water 12. 23o, 62.3o, 175.6 J/kg, 17.64 J/kg, 112.42 J/kg, 136 kW

5. 29.81o, 72.4o, 80.76 kJ/kg 6. 81.3o, 57.86o, 2.938 kJ/kg, 0.54, 0.867 7. 42.28o, 5.128 kJ/kg, 0.5, 0.678

4

General Analysis of Turbomachines

Learning Objectives After completing this chapter, you will be able to: v
Study the effects of variation of exit blade angle on the energy transfer, reaction, and machine performance. v
Study the mechanism of slip and its effects on the performance of the machines.

v
Study the overall performance of machines as a relationship between the specific work or head and the capacity flow rate of the fluid. v
Study the reaction and to quantify the same in finite terms.

4.1 Introduction The flow of fluids, both compressible and incompressible, through the blade passages of the rotor was considered in Chapter 3. The overall nature of the losses during the flow and the consequent efficiencies were studied in some details in chapter 1. The study, however, was limited to the “fluid front.” The attention was on the parameters of the fluid, such as the head, pressure, temperature, enthalpy, velocity, flow rate, and so on. The purpose of this chapter is to study the topics with respect to the parameters of the machine. One of the important machine parameters is the shape of the blades of the rotor. The shape of the blade decides not only the magnitude of the energy transferred, but also the “quality” of the energy, that is, whether it is of kinetic form or other forms. The characterization of the blade shape is by its angles at the inlet and outlet, A1 and A2, respectively. Some of the machine parameters are the physical dimensions of the machine such as the inlet or outlet diameters of the rotor, the speed of the rotor, and whether the machine is a radial flow type or axial flow type. The variations of the parameters of the machine affect the performance of the machine in general. The performance parameters are the specific work, the efficiencies, the reaction or the degree of reaction, the utilization factor, and so on. In this chapter, we study the effects of the machine parameters on the performance parameters, first with respect to radial flow machines and then with respect to axial flow machines.

4.2 General Analysis of Radial Flow Machines In a radial flow machine, the two ends of the rotor blade have different linear velocities. The velocity triangles are constructed with these blade velocities as the bases. In general, the velocity triangle at the smaller radius is made up of lower velocities and that at the larger radius is made up of higher velocities. When U is small, V and Vr are also comparatively small; when U is large, V and Vr are also large. This is how the pumps

124

Turbomachines

and compressors are evolved with radially outward flow, with higher energy at the outlet, at the outer radius. For the same reason, the radial flow turbines are inward flow turbines, with discharge velocities of smaller magnitudes and therefore with lower values of exit losses.

4.3 Radial Flow Machines (Centrifugal Pumps, Centrifugal Blowers, and Centrifugal Compressors): Velocity Triangles The velocity triangles for the blade of an impeller of a radially outward flow machine are generally of the form as shown in Fig. 4.1. (The variations from this general form may be considered step-by-step.) The absolute fluid velocity V1 at the inlet is shown at @1  90o to the blade velocity U1. In compressors or pumps of smaller sizes, guide vanes are not present to direct the fluid into the impeller at any particular angle. Hence, the fluid enters the impeller in a radial direction, giving rise to Vu1  0. V1 = Vf1

V2

Vr1

Vr2

Vf2 b2

b1

U2

U1

Figure 4.1

Vu2

x

Velocity triangles for a radial flow machine.

By the Euler turbine equation [Eq. (3.1)] and with Vu1  0, the specific work is given as WE  U 2Vu2

(4.1a)

In terms of the head developed by a pump, this equation can be written as HE 

U 2Vu2 g

(4.1b)

From Fig. 4.1, Vu2  U 2 x  U 2 Vf2 cot B2 Substituting this value of Vu2 into Eq. (4.1a) and dropping the negative sign we get WE  U 2 (U 2 Vf2 cot B2 )  U 22 U 2Vf2 cot B2

(4.2a) (4.2b)

General Analysis of Turbomachines

125

A pump, a blower, or a compressor is usually run by a motor of constant speed N. Hence, U 2 (  P D2 N / 60) is also a constant. Further Vf2, the flow component, can be written as Q/A2, where A2 is the exit area of the impeller and Q is the volume flow rate. This results in WE  C1 C 2Q

(4.3)

where C1  U 22 and C 2  (U 2 cot B2 ) / A2 .

4.3.1 Effect of Blade Outlet Angle on Energy Transfer Equation (4.3) depicts the energy transfer or specific work as a function of the volume flow rate Q. For a given machine, the blade outlet angle A2 is a specified value, and therefore, C 2  (U 2 cot B2 ) / A2 is a constant. The volume flow rate Q is taken as an independent variable, because it can be controlled and set for any required value by operating a simple valve at the outlet of the machine. Equation (4.3) represents a straight line relationship and is shown in Fig. 4.2 for the three representative values of A2: (a) Less than 90o, (b) equal to 90o, and (c) more than 90o. WE

b2 > 90° b2 = 90° b2 < 90°

Q

Figure 4.2

WE–Q relationship for a radial flow machine.

In Fig. 4.2, A2 is taken as a parameter. 1.

2. 3.

When A2 is less than 90o, that is, when the blades are bent backward to the direction of rotation of the rotor, the slope of the line is negative. As the flow rate increases, Vf2 increases, and along with it, Vu2 decreases. Consequently, the specific work (or head) reduces as the flow rate is increased. When A2 is equal to 90o, the variation in the flow rate or the variation in Vf2 does not affect Vu2 . The specific work (or head) remains constant. When A2 is more than 90o, that is, when the blades are bent forward, the slope of the line is positive. As the flow rate is increased, Vu2 also increases. Therefore, the specific work (or head) also increases.

The above ideas of variation of the outlet angle and its effects can also be reinforced by knowing the variation of Vu2 in the three cases, as can be seen with reference to the velocity triangles in Fig. 4.3. The figure shows the blade shapes corresponding to the three types of blade outlet angles. The respective velocity triangles at the outlet are also shown in Fig. 4.3. In all the three cases, the blade inlet angles are kept the same. The broken lines in the triangles represent the effect of increase of Vf2 (namely, flow rate Q) on Vu2 , (namely, specific work WE or Euler Head HE).

126

Turbomachines

Vr2

V2

V2

V2

Vr2

Vr2

Vf2 b2

Vu2 Vu2

Vu2 Vu2

+

Figure 4.3

Vu2 Vu2 b2 > 90°

b2 = 90°

b2 < 90°

Vf2

b2

b2

+

+

Effect of increasing Vf 2 (namely, Q) on Vu2 (namely, WE or HE).

For a given value of the flow rate, there is one more important effect of variation of the blade outlet angle. For any outlet velocity triangle, as the height of the triangle, Vf2 , remains constant, Vu2 keeps on increasing as the blade outlet angle A2 increases. This can easily be seen in Fig. 4.4. This can also be substantiated by Eq. (4.3a), where the magnitude of WE increases as A2 increases. V2

V2

V2

V2

U2

Vu2 Vu2 Vu2 Vu2

Figure 4.4

Effect of A2 on Vu2.

Interestingly, the specific work W reduces to zero when Vf2 cot B2  U 2  cot B2  U 2 /Vf2 This type of impeller is known as the zero-work impeller. It is like the fluid particle moving in the blade passage, neither pushing the blade ahead nor getting pushed by the blade from behind. The absolute velocity of the fluid at the inlet and that at the outlet are equal in magnitude and direction, unaffected by the movement of the blade. This situation of W  0 arises out of a combination of other parameters also, such as (a)

General Analysis of Turbomachines

127

the blade inlet angle A1 and (b) the ratio of diameters D2/D1 of the impeller at the outlet and inlet. This is illustrated by the following cases, along with Fig. 4.5: 1.

B1  45o, D2 /D1  2.0 When A1  45o, U1 Vf1; when D2/D1  2.0, U2  2U1  2Vf1  2Vf2; hence, U2/Vf2  2; therefore ¥U ´ B2  cot 1 ¦ 2 µ  cot 1 (2)  26.5o § Vf2 ¶

2.

A1  45o, D2/D1  2.5, by similar steps as above, U2/Vf2  2.5 and therefore, ¥U ´ B2  cot 1 ¦ 2 µ  cot 1 (2.5)  21.8o § Vf2 ¶

3.

A1  60o, D2/D1  2, by similar steps as above, U2/Vf2  2/ 3 and therefore, ¥U ´ ¥ 2U ´ 1 B2  cot 1 ¦ 2 µ  cot 1 ¦ µ  40.89o V 3 U § § f2 ¶ 1¶ V1 Vr1 =Vf1

V2 =Vf2 =U1

b1 = 45°

=U1 U1

U2 = 2U1

(a) V1 Vr1 =Vf1

b2

V2 =Vf2 =U1

b1 = 45°

b2 U2 = 2.5U1

U1 (b) V1 Vr1 =Vf1

V2 =Vf2 b1 = 60°

b2 U2 = 2U1

U1 (c)

Figure 4.5

Zero-work impellers. (a) Inlet and outlet velocity triangles for A1 = 45o, D2/D1 = 2. (b) Inlet and outlet velocity triangles for A1 = 45o, D2/D1 = 2.5. (c) Inlet and outlet velocity triangles for A1 = 60o, D2/D1 = 2.

Consider any one of the above cases. When the exit blade angle is less than A2 as calculated above (retaining the stated parameters as constant), Vu2 becomes negative and therefore specific work W becomes positive, resulting in a turbine action of the machine (fluid “pushing the blade”). When the exit blade angle A2 is more than A2 as obtained above, Vu2 becomes positive and W becomes negative, signifying the pumping action of the machine with work inflow. When A2 further keeps on increasing, Vu2 also increases; W increases in the negative direction and the pump capacity also increases. Figure 4.6 shows this variation of W on the base of A2.

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b1 = 45° D2 /D1 = 2.5 W = 0 at b2 = 21.8°

Turbine W=0

Pump or compressor

b2

21.8°

Figure 4.6

Effect of A2 on specific work.

EXAMPLE 4.1 In a radial flow pump, the impeller has its smaller diameter of 5 cm and bigger diameter of 12.5 cm. Its speed is 1500 rpm. The inlet blade angle is 50o. The fluid enters the impeller without any whirl component. The flow component remains constant. (a) Find the specific work and degree of reaction at a blade outlet angle of 70o. (b) Also find at what outlet angle the impeller becomes a zero-work impeller.

Solution: Given D1  0.05 m, D2  0.125 m, N  1500 rpm, A1 50o, Vu1  0 (@1  90o), A2  70o. If W  0, A2  ? Now

P D1N  3.93 m /s 60 P D2 N  9.82 m /s Blade velocity at outlet, U 2  60 Velocity triangles at inlet and outlet are as shown in Fig. 4.7. Blade velocity at inlet,

U1 

V2

Vr1

V1 =Vf1 =Vf2

Vr2

Vf2

50°

U1 = 3.93

Figure 4.7

Vu2

x 70°

U2 = 9.82

Inlet and outlet velocity triangles for Example 4.1.

From the inlet velocity triangle, we have: Vf1  U1 tan A1 V1  Vf1  Vf2  3.93 tan 50o  4.68 m/s From outlet velocity triangle, we have

Vf2 x

 tan 70

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Therefore x

Vf2 tan 70



4.68  1.7 m/s tan 70

Also, the whirl component at outlet is given by Vu2  U 2 x  9.82 1.7  8.12 m/s Now, the specific work is given by W  U 2Vu2  9.82 s 8.12

(Vu1  0)

 79.74 J / kg For zero work, V2 is the same as V1 (magnitude and direction); tan B2 

Vf2 U2



4.68  0.47658 9.82

 B2  25.48o

4.3.2 Effect of the Blade Outlet Angle on Reaction Reaction, as defined earlier, is the ratio of the energy transfer due to change in fluid pressure to the total energy transfer. The variation of the blade exit angle affects the degree of reaction also. As in Section 4.3.1, the value of the degree of reaction depends on the inlet blade angle A1 and the ratio of the diameters D2/D1, along with the blade exit angle. Once again, to find the isolated effect of the blade exit angle A2, the other variables are assumed to be constant at some illustrative values. For example, let A1  45o and @1  90o so that V1  Vf2  Vf2  U1; Vu1  0. By definition R 





Reaction component Total energy transfer Total work Kinetic component Total work (V22 V12 ) 2 U 2Vu2

U 2Vu2

U 2Vu2

[(Vf22  Vu22 ) Vf12 ] 2 U 2Vu2

So ¥V ´ R  1 ¦ u2 µ § 2U 2 ¶

(4.4)

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Also R  

(2U 2 Vu2 ) 2U 2 (2U 2 U 2  Vf2 cot B2 ) 2U 2 · 1 ¨ Vf2 cot B2 ¸ ©1  2 ª U2 ¹

(4.5)

This result is applicable for A1  45o and @1  90o so that V1  Vf2  Vf2  U1; Vu1  0. Further substituting D2/D1  2.5 and U2  2.5U1  2.5 Vf2 and therefore Vf2/U2  1/(2.5), in Eq. (4.5), it follows that 1 R  (cot B2  2.5) (4.6) 5 This Eq. (4.6) is applicable for the case illustrated in Fig. 4.5 (b). When A2, in the above conditions, becomes equal to 158.2o, the degree of reaction reduces to zero, the machine becomes impulse type, and the centrifugal head balances the relative velocity head. The variation of degree of reaction is shown in Fig. 4.8.

b1 = 45° D2/D1 = 2.5 R = 0 at b2 = 158.4°

R

R=0

b2

Figure 4.8

158.4°

Effect of A2 on the degree of reaction R.

If the reference values of A1 and D2/D1 were chosen as in Fig. 4.5(a) (i.e., A1  45o, D2/D1  2), the nature of variation of R would be the same, but the values would be different (W  0 at A2  26.5o; R  (2  cot A2)/4; R  0 at A2  153.4o). Equations (4.4), (4.5) and (4.6) are not general results. These results are with some assumptions (@1  90o, A1  45o). The assumption that @1  90o is generally acceptable, but the other assumption A1  45o puts limitation on the use of the three equations.

EXAMPLE 4.2 In a radially inward flow turbine, the diameter of the runner at the inlet is 50 cm and the diameter at the outlet is 15 cm. The speed of the machine is 1500 rpm. The fluid at a velocity of 35 m/s enters the

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runner at 20o to the tangent and leaves the runner without any whirl component. The flow component of the fluid velocity remains constant in the runner. Calculate the blade angles at the inlet and outlet, the specific work, and the reaction.

Solution: We know D1  50 cm, D2  15 cm, N  1500 rpm, V1 35 m/s, @1  20o, Vu2  0, Vf1  Vf2  constant. To calculate: A1, A2, W, R. Now

P D1N P s 0.5 s 1500   39.27 m /s 60 60 P D2 N P s 0.15 s 1500   11.78 m /s Blade velocity at outlet, U2 is given by: U 2  60 60 Blade velocity at inlet, U1 is given by: U1 

The velocity triangles are shown in Fig. 4.9. U1 = 39.27

U2 = 11.78

U1 = 39.27 20°

U2

b1

V1 = 35

b2 V2 Vr2

Figure 4.9

Blade shapes and velocity triangles for Example 4.2.

Now, the whirl component Vu1, of inlet velocity V1, is given by V u1  V1 cos @1 Hence,

Vu1  V1 cos 20  35 cos 20  32.89 m/s

Therefore Vru1  39.27 32.89  6.38 m/s The flow component of V1 is given by Vf1  V1 sin 20o  11.97 m/s From inlet velocity triangle, Fig. 4.9, ¥ 11.97 ´ B1  tan 1 ¦  61.94o § 6.38 µ¶ From the outlet velocity triangle Fig. 4.9, ¥V ´ ¥ 11.97 ´ B2  tan 1 ¦ f2 µ  tan 1 ¦  45.46o § 11.78 µ¶ §U2 ¶ So W  U1Vu1  39.27 s 32.89  1291.6 J / kg Now V1  35, V2  11.97. Therefore, the kinetic component in specific work is V12 V22 2

 540.86

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Hence, the degree of reaction is W Kinetic component W 1291.6 540.86   0.58 1291.6

R=

EXAMPLE 4.3 In a radially outward flow machine, the outer diameter of the rotor is 1.5 times the inner diameter and the blade angles at the inlet and outlet are equal. The drive for the machine is at 3000 rpm. Determine the diameters for a specific work of 2400 J/kg. Also find the degree of reaction. Assume no whirl at the inlet and Vf1  Vf2 .

Solution: Given D2  1.5 D1, A1  A2, N  3000, W  2400 J/kg or m2/s2, Vu1  0, Vf1  Vf2 . To

calculate , D1, D2, R.

C

F Vr1

V1

V2

b1 A

B

U1

Figure 4.10

G

Vr2

b2 D U2 = 1.5 U1

E

Velocity Triangles for Example 4.3.

Consider the velocity triangles, Fig. 4.10, with U2  1.5U1 and A1  A2. We have $ABC  $DEF Because A1  A2, Vf1  Vf2 and CAB  FDE , we have GD  0.5 U1  Vu2 Since Vu1 0, we have W  U 2Vu2  (1.5U1 )(0.5U1 )  2400  0.75U12 2400  U12   3200 0.75  U1  56.57 m / s We know that

P D1N  U1 60 60U1 60 s 56.57  D1    0.36 m (P s 3000) PN

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D2  1.5D1  1.5 s 0.36  0.54 m

and

As seen in velocity triangles, Vr1  Vr2 . Hence, the reaction component in the specific work is (U 22 U12 ) 2



(56.57 s 1.5)2 (56.57 )2  2000 m 2 / s2 2

We know W 

(V22 V12 ) U 22 U12  2 2

Because this is a work-absorbing machine, the degree of reaction is R 

Reaction component Total specific work 2000  0.833 2400

Comments: The reaction component is entirely due to the centrifugal energy, because the relative velocities at the inlet and outlet are equal to one another. It may be concluded that 83.3% of the energy transferred is converted into pressure and only 16.7% is in the kinetic form.

EXAMPLE 4.4 The inner and outer diameters of the rotor of a radial flow machine are 10 cm and 20 cm, respectively. The blade angles at the inlet and outlet are 40o and 70o, respectively. The fluid enters the rotor at 90o to the blade velocity, and the flow velocity remains constant in the rotor. Determine the speed of the machine for a specific work of 1000 J/kg.

Solution: Given D1  10 cm, D2  20 cm, A1  40o, A2  70o, @1  90o, Vf1  Vf2 , W  1000 J/kg. To

calculate N. It is possible to draw the shapes of the velocity triangles as shown in Fig. 4.11. V1 =Vf1 =Vf2

V2

Vr1

x 70°

40°

U2 = 2U1

U1

Figure 4.11

Vr2

Velocity Triangles for Example 4.4.

Now, Vf1  U1 tan B1  Vf2

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In the outlet velocity triangle, x

Vf2 tan B2

 x  U1 s



U1 tan B1 tan B2

tan 40o  0.3054 U1 tan 70o

Therefore Vu2  (U 2 x )  2U1 0.3054U1  1.6946U1 Now W  U 2Vu2  2U1 s 1.6946U1  1000  3.3892U12  U12  295.0549  U1  17.177 m /s Because OD1N/60  U1 we have N 

60U1

P D1



60 s 17.177  3280 rpm P s 0.1

Comments: The speed is not a suitable one because generally the induction motors have to run at a speed slightly less than the synchronous speed, such as 2880, 1440, 980, etc. Suppose the speed of 2880 rpm is selected. The resulting specific work can be calculated with all the other data of the problem. This specific work is surely less than 1000 J/kg. An interesting problem is to determine D2, the outer diameter, keeping all other data intact, including specific work, 1000 J/kg. Starting from the speed, N, equal to 2880 rpm, the blade velocity at inlet, U1, is calculated as U1 

P D1N P s 0.1 s 2880   15 m/s 60 60

Then, the flow velocity, Vf1, is given by, (Refer Fig. 4.11), Vf1  15 tan 40  12.586  Vf2 From the outlet velocity triangle, Fig. 4.11, x

Vf2 tan 70



12.586  4.58 tan 70

W  U 2 (U 2 4.58)  1000  U 22 4.58 U 2  U 22 4.58U 2 1000  0

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Solving for U2  U2  Now

(4.58  4.582  4000 )  34 m/s 2

P D2 N  U2 60 60U 2 60 s 34  D2    0.2255 PN P s 2880  D2  22.55 cm

4.3.3 Effect of the Blade Exit Angle on the Performance The performance of a machine is the totality of the specific work or energy transfer, the reaction, the power consumption, the efficiency, and so on. Equation (4.3) represents the energy transfer W in a radial flow pump or compressor. In a pump, the head developed may be written as W/g. In a compressor, the pressure developed may be written as W s Q. Either way, W is identified as a function of the flow rate Q. The flow rate Q is taken as an independent variable that can be varied by the operation of a valve at the outlet. In practice, the flow rate is as per “demand” or “load.” A plot of W (or H to a different scale) on the base of the flow rate Q, therefore, represents one of the important characteristics of the machine. Figure 4.2 shows the characteristic in the form of three lines with different slopes. The lines represent the cases of different values of the blade exit angle A2. Equation (4.3) and Fig. 4.2 are the outcome of the starting from Euler’s equation. The energy transfer, as discussed, is due to the “vane-congruent flow.” The expression for the actual energy transfer can be obtained when the factors causing the deviation from the vane-congruent flow are considered.

4.3.4 Flow Analysis in Impeller Blades: Slip and Slip Factor or the Coefficient of Slip As the fluid flows through the passage between the rotor blades, the fluid experiences a “push” by the blades. The blades impart energy to the fluid, exerting that push on it. But in the process, the fluid also experiences a slightly reduced pressure at the backside of the blade. This is shown in Fig. 4.12. The net effect of this differential pressure between the two sides of the vanes is that the fluid deviates from its vane-congruent nature, as it flows out of the impeller. It looks as if the blade-exit angle is slightly reduced. − + + − −+ + + −− ++ −− −+ +

Figure 4.12

Fluid deviation due to pressure distribution.

Vane-congruent flow also indicates that the velocity distribution of the fluid in the flow channel is uniform. But actually, the velocity distribution is altered due to the inertia of the fluid particles. To understand this, suppose that there is a particle of fluid, frozen with an arrow mark on it, as shown in Fig. 4.13(a). As the

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Turbomachines

impeller rotates clockwise, this particle, due to its own inertia, tries to maintain its orientation, pointing to the same absolute direction as it started with. With fluid inertia Uniform velocity profile

+ (a)

Figure 4.13

(b)

Fluid deviation due to inertia effects. (a) Retention of particle-orientation due to inertia and (b)Change of velocity profile due to inertia.

When the impeller rotates through 180o, the particle still points to its original direction, but relative to the impeller, it now points to the center of the impeller. This relative rotation (anticlockwise with respect to the impeller) completes 360o when the impeller also completes (clockwise) 360o. The net effect is that there is a relative rotation of the fluid particle with respect to the impeller. When this relative rotation is superimposed on the uniform vane-congruent velocity profile, the resultant velocity profile becomes as shown in Fig. 4.13(b). Again the effect is a slight reduction of the exit angle of fluid. The third alteration of exit velocity of fluid is due to the finite thickness of blades. At the exit of the impeller, the blades do not continue to exist, the area available for the flow is increased. Consequently, the flow velocity is reduced. All these effects are shown in Fig. 4.14. Because of the combined effects, the whirl component of the exit velocity V is reduced. As a result, the magnitude of the energy transfer is reduced. u2

Vr2 V2

Fluid deviations

Vu2

Slip

U2

Vu2

Figure 4.14

Reduction of Vu2 due to pressure distribution, inertia of fluid, and finite thickness of blades.

This overall reduction of the specific work, from the specific work as calculated by the Euler turbine equation, is known as slip. This slip is accounted for by the “slip factor” or “coefficient of slip” L. When the specific work due to Euler’s equation is WE, and after accounting for slip, it becomes the ideal specific work Wi, then Wi  WE s M

(4.7a)

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137

In continuation, the head generated can also be written as Hi  HE s M

(4.7b)

Slip is not a loss. The occurrence of slip just means that a given impeller with all its geometrical features merely behaves as if it has a different geometry. (It is like an electric bulb, designed as 40 W that might act as a 38 W bulb. It gives light and heat equivalent to 38 W, but also draws 38 W from the mains, not 40 W.) The usual values of the coefficient of slip are around 0.97, 0.98, etc., accounting for 2%–3% of deviation. In high-power machines, the slip may be less than this value, with L  0.985 also. When the machine operates at the off-design points (farther away from the maximum-efficiency point), the coefficient of slip may be quoted at a value as low as 0.88, 0.9 etc. Figure 4.2, showing the variation of the specific work on the base of flow rate, can now be thought of as a plot of Wi on the base of Q. The nature of the lines is still the same, with a scale factor. Figure 4.15 is a replica of Fig. 4.2, showing Wi or Hi along with the earlier ordinates. WE Wi W

Q

Figure 4.15

Effect of slip on the specific work.

4.3.5 Losses in Impeller Blade Passages The losses that occur as the fluid flows in the passages between the blades of the rotor are like those that occur as studied in fluid mechanics. These losses are due to wall friction, turbulence, eddies, viscous resistance, etc., which reduce the ideal work Wi of Fig. 4.15. The losses are of two types. 1.

2.

The first type is due to the skin friction because of the viscosity of the fluid. These losses are proportional to the square of the flow velocity, to the surface roughness, and to the length of the passage. The second type of losses is due to the flow deviating from the vane-congruent flow. This occurs when the flow rate varies (either increases or decreases) from the design conditions. Offdesign flow rates give rise to off-design flow velocities or relative velocities. The fluid in such cases tends to either “impinge on” or “separate from” the blade surface, instead of smoothly gliding over the surface. These losses are known as “turning losses” and follow a square law. The minimum losses are at the rated flow and increase in both directions of lower and higher flow rates.

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Turbomachines

Both the types of losses are shown in Fig. 4.16. The total of the two losses is also shown.

Losses Total losses Flow losses Turning losses Flow rate

Figure 4.16

Losses in turbomachines due to fluid flow.

EXAMPLE 4.5 A fluid is directed at a velocity of 60 m/s on the concave side of the blades of a radially inward flow turbine at an angle of 25o with the blade tip velocity. The blade outlet angle is 50o with respect to blade velocities. The outlet diameter of the runner is half of that at the inlet. The speed of the runner is 1000 rpm. The flow velocity remains constant. The discharge of fluid is at 90o to the blade velocity. Determine the (a) blade inlet angle, (b) runner diameter at the inlet, (c) Runner diameter at the outlet, (d) the specific work, (e) the degree of reaction, and (f ) the utilization factor.

Solution: Given V1  60 m/s, @1  25o, A2  50o, D2  0.5 D1, N  1000 rpm, @2  90o, Vu2  0. The approximate shape of the blade and the velocity triangles are as shown in Fig. 4.17. V1 = 60 m/s U1

A

U1

B

a1 = 25°

E

D b1

U2 = U1 /2 50°

U2 +

V1 = 60

Figure 4.17

Vr1 C

V2 G

Vr2

Shape of blades and velocity triangles for Example 4.5.

Now At inlet, the whirl velocity is:Vu1  AD  V1 cos 25o  60 cos 25o  54.38 m/s At inlet, the flow velocity is: Vf1  DC  V1 sin 25o  60 sin 25o  25.357 m/s  Vf2  V2 Therefore U2 

Vf2 tan 50o



25.357  21.277 m/s (tan 50o  1.192) tan 50o

F

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139

We have D1  2D2. Therefore U1  2 U2  42.554 m/s. (a) The blade inlet angle is given by: 23.357 ¨ · B1  tan 1 © ¸  65o ( 54 . 38

42 . 554 ) ª ¹ (b) and (c) The runner diameters are determined as follows: We know that P D2 N  U2 60 60U 2

60 s 21.277  0.4063 m  40.6 cm PN P s 1000 Also, runner inlet diameter, D1  2D2  81.2 cm. (d) Specific work is given by Runner outlet diameter,  D2 



W  U1Vu  42.554 s 54.38  2314 J/kg 1

The kinetic component of W V12 V22 2



(602 25.357 2 )  1478.5m 2 /s2 2

(e) The degree of reaction is (W kinetic component) W 2314 1478.5 R  0.361 2314 R

(f ) The utilization factor 

W 2314  0.878  V22 ¥ (25.357 )2 ´ 2314  W  µ¶ ¦§ 2 2

4.3.6 Characteristic Curves: Head–Capacity Relationship One of the important characteristics of a turbomachine is a plot of its actual specific work or energy transfer on the base of the fluid flow rate. As has been seen, the energy transfer can also be in terms of the “head generated” in the case of centrifugal pumps. This characteristic is basically known as the head–capacity relationship. This head–capacity curve is the resultant of the ideal energy transfer and the losses in the machines. Hence, Figs. 4.15 and 4.16 are combined in Fig. 4.18, with W  Wi – losses. To illustrate, the case of A2 < 90o, that is, the backward bent vanes, is considered for Wi. The ordinate in Fig. 4.18 can also be the head.

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Turbomachines

Wi W = Wi − Total losses

Specific work

W H Total losses

Flow rate

Figure 4.18

Head–capacity relationship.

The head–capacity curve, as in Fig. 4.18, is generated at a constant speed. The variation of the losses gives rise to another important characteristic – the efficiency versus flow rate. The efficiency has to be maximum where the losses are minimum. On either side of the point of maximum efficiency, the curve droops. A plot of the efficiency is also shown in Fig. 4.19, along with the head generated. When the flow rate reduces to zero, the efficiency also has to reduce to zero. The point of maximum efficiency is the “design point” or the “duty point.” The rating of the machine is considered at the point of this maximum efficiency. Figure 4.19 contains one more line in the form of a plot of the power of the machine on the same base flow rate. The power P is proportional to the product of the flow rate and the head (P  wQH). Some portion of the head–capacity curve (resembling a hyperbolic nature) gives rise to a somewhat “flat” portion of the power characteristic. On both sides of this flat portion, the power is also reduced. This feature adds to the stability of the operation of the machine in the region around the design point. Duty point W or Head

W H

h

Efficiency

Power P

flow rate

Figure 4.19

Characteristics of a radial flow machine.

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141

The actual H–Q curve, the efficiency–flow rate curve, and the power–flow rate curve are the results obtained from the ideal characteristic for the case of the backward bent blade with the outlet angle less than 90o. Even if the angle were 90o or more, the trends of variation of all the characteristic lines would be more or less the same. However, there are reasons to maintain A2 less than 90o (backward bent blades) and this will be discussed in connection with the centrifugal pumps.

4.3.7 Effect of Prewhirl The analysis of the radial flow machine is with the assumption that the fluid entry into the impeller is without any pre-whirl, that is, with Vu1  0. This is a valid assumption, because no inlet guide vanes can be correctly provided in the suction pipe of a pump or even a blower. This is more so for the cases where the flow rate is variable over a large range. In the case of the multi-stage units, where the inlet to the later stage is directly from the outlet of the previous stage, a certain amount of pre-whirl or pre-rotation can exist. In such cases, Vu1 cannot be zero. The specific work, as given by the Euler turbine equation [Eq. (3.1)], is made up of two terms having opposite signs. Hence, the contribution of the term U1Vu1 is opposite to that of U2Vu2, giving rise to a “subtractive” effect to the specific work or head generated. Thus, the input work to a pump or compressor can be reduced by an amount equal to U1Vu1 when Vu1 is not zero. WhenVu1 is not zero, the fluid entry is at an angle @1 (@1 x 90o). Because the angles @1 and A1 are constant, with the blade velocity U1 also remaining the same, the flow velocity Vf1 in the inlet velocity triangle has to remain constant. This value of the flow velocity has to correspond to the design value of the flow rate. Once again, a variable flow rate, at values other than that of the design flow rate, gives rise to turning losses. The “subtractive” effect is to slightly reduce the specific work. The net effect of U1Vu1 is shown in Fig. 4.20.

J kg m2 s2

U1Vu1

Flow rate

Figure 4.20

Effect of pre-whirl.

4.4 Axial Flow Machines In axial flow machines, the blade velocities at the inlet and outlet are equal, U1  U2  U. Therefore, the two velocity triangles have equal bases. Both the triangles can be drawn on a common base. Two typical sets of velocity triangles are drawn as in Figs. 4.21 and 4.22. The two sets are almost identical. Triangles ABC (in each set) are the inlet triangles with U, V1 and Vr1 . Triangles ABD are the outlet triangles

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with U, V2, and Vr2 . As can be seen, vertices C and D are exchanged between the two sets. Figures 4.21 and 4.22 also include the blade shapes that are drawn in conformity with the velocity triangles. C V1

Parallel to Vr2 , BD

D Vr1

V2

A

Vr2

U B Parallel to Vr1 , BC

Figure 4.21

Typical velocity triangles, blade shapes (compressors).

D V2

Parallel to Vr2 , BD

C Vr2

A

V1

Vr1

U B Parallel to Vr1 , BC

Figure 4.22

Typical velocity triangles, blade shapes (turbines).

It may be observed that the relative velocities Vr1 at the inlet are tangential to the blade profiles at the inlet in the two diagrams. Similarly, the relative velocities Vr2 at the outlet are tangential to the blade profiles at the outlet. Figure 4.21 represents an increase in the fluid velocity from the inlet to outlet, that is, from V1 to V2, with an increase in the fluid energy. This is the case of a pump or compressor. The blade shape confirms this situation. The movement of the blade is to “scoop” the fluid, “lift” it, and impart energy to it. In a similar way, Fig. 4.22 represents a decrease in the fluid velocity from V1 to V2, a decrease in the fluid energy, and consequently an increase in the shaft energy. This is the case of a turbine. The blade shape confirms this situation. The fluid striking the concave side of the blade forces the blades to move and gives energy to the blade. The velocity triangles and blade shapes of Figs. 4.21 and 4.22 are drawn to indicate how the blade shapes can be derived from the velocity triangles and vice versa. The analysis of the processes of compression and expansion can be almost the same. The magnitude of the energy transfer is the same in the configurations of blades of Figs. 4.21 and 4.22, the sign being negative for the compressor or pump (Fig. 4.21) and positive for the turbine (Fig. 4.22). In the compressor, there is no pre-whirl. V1 is shown at 90o to the blade velocity U and therefore Vu1 is zero. In the turbine, the exit velocity V2 is shown at 90o to the blade velocity U, with Vu2  0. But these

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143

need not be compulsorily so. There can be some variation in the blade shapes and velocity triangles. The fluid velocity at the inlet to the compressor blades (Fig. 4.21) can be at any angle, and so is the fluid velocity at the outlet of the turbine blades (Fig. 4.22). Since axial flow machines are generally multi-stage machines, there can be different values of @1 in compressor blades in later stages after the first stage, or different values of @2 in turbine blades in the stages, except the last stage. Figures 4.21 and 4.22 exhibit one more feature. The case is particularly illustrated to indicate that the analysis of compressors and turbines can be very much parallel to each other, or better still, the analysis can be the same, but with some discretion regarding the signs of the energy flow. This observation is perhaps a little bit complex, but the complexity is only to a certain extent. It can be imagined that in the turbine (Fig. 4.22), the fluid flow and rotation of the blade can be “reversed” by exchanging the nose and tail of the blade, but not angles. The reversed situation clearly falls in line with the compressor blade (Fig. 4.21). Once this is understood in all its details, the generality or universality of the analysis can be appreciated. As a matter of precaution, however, it is necessary to mention one important feature of compressors. The pressure at the exit of the compressor is higher than that at the entry. Naturally, the fluid has to flow from a lower pressure to a higher pressure, as it moves through the rotor, getting the kinetic energy from the blades of the rotor and transforming it into pressure energy. Hence, the design has to be for the higher rates of diffusion, that is, the flow passages must accommodate the higher rates of conversion of velocity into pressure. Avoiding the separation of flow at the blade surfaces is an important requirement and, therefore, it becomes the limiting factor in the design. In this context, the pressure ratio of each stage has to be small, with general values of the degree of reaction about 65%, even at the cost of efficiencies. Hence, although the analyses of compressors and turbines are the same, as mentioned above, the range of values has to be restricted with respect to the design of compressor blades. Clearly, the specific works of the compressor blades are required to be lower compared to the turbine blades.

4.4.1 Reaction in Axial Flow Machines One interesting result to express R is derived with respect to axial flow machines (U1  U2), with an assumption that the flow components of velocities remain constant (Vf1  Vf2 ) . When Vf1  Vf2 , the axial thrust reduces to zero. Hence, this assumption is realistic and holds good in the design process. Then from Eq. [3.4(iii)],

(Vr12 Vr22 ) 2 R for the axial flow machine (U1  U 2 ) W Now, 2 V 2 Vr12  Vru1 f1

Vr2  Vru2  Vf2 2

2

2

W  U (Vu1 Vu2 ) Hence 2 V 2 V 2 V 2 )

(Vru1 f1 ru2 f2 2 R U (Vu1 Vu2 )

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Turbomachines

2 V 2 ) (Vru1 ru2 2  U (Vu1 Vu2 )

(Vru1 Vru2 )(Vru1  Vru2 ) 2  U (Vu1 Vu2 ) Now, because Vru1 – Vru2  Vu1 –Vu2, (Vru1  Vru2 ) 2 U

R

(4.8)

Therefore, R

Vru

mean

,

U

(4.9)

where Vru mean is the projection of Vr mean on U. CM = MD

D V2

A

C

M Vr2

V1 Vr1

NU B Vru mean U

Figure 4.23

Reaction in an axial flow machine, R = −BN/AB.

Figure 4.23 shows the velocity triangles, ABC and ABD, where C and D are vertices of the inlet and outlet velocity triangles, respectively. The mid-point of CD is M. Vru mean is the projection of Vr mean on U, that is, BN is the projection of BM on AB. Thus, R  BN/AB. It may be noted that the negative sign in the expression for R and the reverse direction of BN (opposite to AB, i.e., U) cancel each other and the degree of reaction R becomes positive in the case illustrated in Fig. 4.23. If N lies beyond B on extended AB, R becomes negative. Also, if N lies beyond A on extended BA, R becomes more than 1. If BN  0, that is, if N coincides with B and M is directly above B, Vru mean is zero and R  0. Also Vr1  Vr2 and p2  p1. This is shown in Fig. 4.24. This type of machine is an impulse machine. In impulse machines, therefore, there is no change in the pressure of the fluid between the inlet and outlet of the rotor, the flow passage between the blades is of constant area of cross-section, and the relative velocity remains constant in magnitude as the fluid moves through the rotor.

General Analysis of Turbomachines D

V1

Vr1

U B

A

Figure 4.24

C

M Vr2

V2

145

Impulse machine, R = 0, Vr1 = Vr2, Vru mean  0.

When V1  V2, the absolute velocity of fluid remains constant as the fluid flows through the passages between adjacent blades. The second component, in the expression for the specific work [Eqs. (1.3), (1.5)] or the first component in Eq. (3.3) becomes zero. The result is R  1. Because the degree of reaction is 1, all the energy transfer is simultaneous with transformation or conversion of energy. In order that the energy is “transformed” or “converted” from the pressure form to kinetic form (or vice versa), the flow passages between adjacent blades of the rotor (corresponding to this situation) must act like nozzles (or diffusers). As a result, the pressure of the fluid varies as it flows through such passages. The machine is known as a reaction machine. The corresponding velocity triangles are shown in Fig. 4.25. D

C

M Vr2

V2

V1

Vr1

B

A, N

Vru mean  U ; R  1 Figure 4.25 M

D V2

Vr2

Vr1

M

D

C V1

Velocity triangles for the reaction R = 1.

V2

Vr2

C V1

C

M

D

V1

Vr2

Vr1

Vr1

V2

NU B

A

A

N

U B

BN  0.25AB R  0.25

BN  0.5 AB R  0.5

(a)

(b)

Figure 4.26

A

U B

N

BN  0.75AB R  0.75 (c)

Velocity triangles for constant U, V1, and @1; R = 0.25, 0.5, 0.75.

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Turbomachines

In addition to the above two types (R  0, R  1), the third type is actually a combination of the first two types. The degree of reaction can vary. The fraction of specific work that gets “converted” and “transferred” between the fluid and rotor can be anywhere between 0 and 1. Three values of degrees of reaction (R  0.25, R  0.5, and R  0.75) are illustrated in Fig. 4.26 [(a)–(c)]. All such machines are known as reaction machines, including the type R  1 of Fig. 4.25. The above discussion indicates that the degree of reaction can be pre-determined. With reference to the velocity triangles in Fig. 4.23, if R  0.3, point N on AB can be such that BN  0.3 s BA and M can be located directly above this point. Then C and D can be placed equidistant from M. Theoretically, the degree of reaction can be less than 0 or even more than 1. But practical realization of such values is more difficult than their theoretical realization. Consider U1  U2 for an axial flow machine. Then R 

(Vr22 Vr12 ) [(V12 V22 )  (Vr22 Vr12 )] Y (X Y )

(4.10a) (4.10b)

where X  V12 V22 and Y  Vr22 Vr12 . If R, the degree of reaction, is to be negative, then Y must be negative and |X|  |Y|. This situation is when V1  V2 and Vr2  Vr1 or p2  p1. In other words, a stream of fluid, with very high kinetic energy, while passing through the passage between the rotor blades, should not only transfer part of its kinetic energy to the rotor, but also transform some of the remaining kinetic energy into pressure energy. This is the case of energy output at the shaft and compression of the fluid, with velocity triangles as shown in Fig. 4.27. M

D V2

A

U B

C V1

Vr2

Vr1

N

BN = 0.25 AB R = − 0.25

Figure 4.27

Negative reaction.

To repeat once again, a negative reaction means that a high-velocity fluid stream enters the flow passages between the rotor blades, imparts some energy to the shaft, and gets itself compressed to a higher pressure. If R, the degree of reaction, is to be more than 1, then in Eq. (4.10), X must be negative and Y must be positive. This is the situation when V2  V1 and Vr2  Vr1 . The fluid passage between the rotor blades must be such that it accelerates the flow. The energy input at the shaft contributes to accelerate the flow and not to increase the pressure. The velocity triangles are as shown in Fig. 4.28.

General Analysis of Turbomachines M

D V2

C

Vr2

V1

N

147

Vr1

A

U

B

BN = −1.3 AB R = 1.3

Figure 4.28

Degree of reaction is more than 1.

To repeat, a degree of reaction being more than 1 means that the fluid, at a higher pressure at the inlet to the rotor, expands in the blade passage and gets accelerated when the shaft also is driven by external energy and contributes to accelerate the fluid flow. In all the discussions so far, it is mentioned, sometimes, that V2  V1, p2  p1, Vr2  Vr1 , etc. These are only ideal cases with friction being neglected for the purpose of discussions. Slight reductions such as V2  0.98V1 or Vr2  0.98Vr1 (accounting for 2% losses as example) can always occur, without losing the generality of the discussions above. Further, when the velocity triangles are solved at the inlet and outlet, the degree of reaction can also be determined. But conversely, given a value of the degree of reaction, the velocity triangles can be solved to suit the value of R. Hence, the degree of reaction can be a design parameter, implying that its value can be prescribed or chosen and blade angles can be arrived at.

EXAMPLE 4.6 Prove that R 

Vf ¥ tan B1  tan B2 ´ . 2U ¦§ tan B1 s tan B2 µ¶

Solution: We know, from Section 4.4.1, Eq.(4.8), that (Vru1  Vru2 ) 2 2 R U But Vru1  Vf1 cot B1 and Vru2  Vf2 cot B2 . Substituting these values and with Vf1  Vf2  Vf in Eq. (4.8), Vf

(cot B1  cot B2 ) 2U V ¨ 1 1 ·  f ©  ¸ 2U ª tan B1 tan B2 ¹ V ¨ tan B1  tan B2 ·  f © ¸ 2U ª tan B1 s tan B2 ¹

R

148

Turbomachines

EXAMPLE 4.7 ¨ tan B1 tan B2 · Show that W  UVf © ¸ and state the assumptions. ª tan B1 s tan B2 ¹

Solution: Assumption: For an axial flow machine, U1  U2  U. Further, Vf1  Vf2  Vf . From the Euler turbine equation

W  U [Vu1 Vu2 ] where Vu1  U Vf1 cot B1 Vu2  U Vf2 cot B2 Therefore W  U (U Vf1 cot B1 U  Vf2 cot B2 )  UVf (cot B2 cot B1 ) ¥ 1 1 ´  UVf ¦

µ § tan B2 tan B1 ¶ ¥ tan B1 tan B2 ´  UVf ¦ µ § tan B1 s tan B2 ¶

EXAMPLE 4.8 In an axial flow turbine, the blade velocity is 60 m/s. The fluid enters at 30o to the plane of the wheel at a velocity of 80 m/s. Calculate the blade inlet angle. If the blades are to be designed for (a) R  0.25 and (b) R 0.5, calculate the blade outlet angle and specific work in each case.

Solution: Given U1  U 2  U  60 m /s, A1  30o, V1  80 m /s The inlet velocity triangle is drawn as shown in Fig. 4.29. C V r1

V1 = 80

Vf1

30° A

Figure 4.29

x U = 60

B

Inlet velocity triangle for Example 4.8.

General Analysis of Turbomachines

149

Now Vf1  V1 sinA1  80 sin30o  40 VU1  V1 cos A1  80 cos 30o  69.282 In Fig. 4.29, the segment x is given by x  AC cos30o U Therefore, x  69.282 60  9.282 Blade inlet angle is the angle A1 of relative velocity Vr1 (BC) with blade velocity U (AB), as in Fig. 4.29. Vf1 40 ¥ 40 ´ This angle A1 is given by, tan B1    B1  tan 1 ¦  76.9o. This angle § 9.282 µ¶ (Vu1 U ) 69.282 60 is shown in Fig. 4.30. (a) For the required value of R  0.25, we have from Eq. (4.9), Vru mean  U – R  BN (in Fig. 4.30) Vru mean  U s 0.25  60 s 0.25  15 Mark N on BA such that BN  15, as shown in Fig. 4.30. Referring to Section 4.4.1, the vertex D of outlet velocity triangle ABD must be placed so that the midpoint of CD must be above N. This gives rise to the projection of Vr2 on U, that is, Vru2  BN  (half of CD). Therefore, Vru  BN  (BN  x )  15  15  9.282  39.282 2

D V2

C

M V1

Vr2

Vr1

45.5° A

N

60

Figure 4.30

15

76.9° B

9.282

Inlet and outlet velocity triangles, Example 4.8.

Vru2  BN  (BN  x )  15  15  9.282  39.282

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Turbomachines

Now, outlet blade angle A2 is given by tan B2 

Vf2 Vru2



40 39.82

¥ 40 ´  B2  tan 1 ¦  45.5o § 39.282 µ¶ Now, to calculate specific work, W, $Vu  (BN  x ) s 2  (15  9.282) s 2  24.282 s 2  48.564 And therefore, W  U s $Vu  60 s 48.564  2914 J/kg (b) For the required value of R  0.5, the triangles are required to be symmetrical. Hence

B2  A1  30o So $Vu  U  2x  60  (9.282)2  78.564 Therefore specific work is given by W  U($Vu)  60 s 78.564  4714 J/kg

4.4.2 Effect of Blade Angles on the Specific Work and Degree of Reaction: Turbines In an axial flow turbine, the exit fluid velocity V2 must be minimum in order to obtain the maximum utilization factor, . Therefore, V2 must be axial, that is, V2 must be at 90o to the blade velocity U. This imposes the condition that the exit blade angle A2 must be equal to tan 1 (Vf2 / U ). ΔVu = 2u D

C Vr 2

V2

V1

M

Vr1

b2 A

Figure 4.31

U

b1

b1 = b2

B

Velocity triangles for an impulse machine, R = 0, W = 2U2.

Now, considering the limits of the degree of reaction R as 0 and 1, Fig. 4.31 indicates the following: 1.

At A1  A2, R  0, ($Vu)  2U, we have Wmax  U(2U)  2U2. This is the case of an impulse turbine.

General Analysis of Turbomachines

2. 3.

151

At A1  90o, R  0.5, ($Vu)  U we have W  U2. As A1 tends to reach (180 A2), R tends to be 1, but this is trivial because W tends to be 0.

4.4.3 Effect of Blade Angles on the Specific Work and Degree of Reaction: Compressors In an axial flow compressor, the inlet fluid angle @1 is taken as 90o for axial entry. The inlet blade angle A1 has to be constant at A1  tan 1(Vf1/U). The angle A2 can vary from a minimum value equal to A1 up to (180

A1), for the values of degree of reaction R to be between 0 and 1, respectively, as shown in Fig. 4.32(a). For a compressor, however, the degree of reaction has to be high, somewhere around 0.65 and more, as indicated in Fig. 4.32(b). However, it is not possible to reach R  1 when the specific work tends to be zero. As a limit, R and W are set at Rmin  0.5 and Wmax  U2. C

D V2

Vr1

V1

Vr2

V1

V2

U

B

Vr2

U B

A

(a)

Figure 4.32

D Vr1

b2

b1 A

C

(b)

Velocity triangles: Compressor. (a) R = 0 (not desirable) and (b) R  0.5 (generally acceptable).

EXAMPLE 4.9 The impeller diameter of an axial flow pump is 50 cm. The impeller speed is 750 rpm. The fluid enters the impeller at a velocity of 15 m/s, without any whirl component. (a) Determine the blade inlet angle. Calculate the specific work and reaction when the outlet blade angle is (b) 65o, (c) 80o, and (d) 110o. Assume that the flow component remains constant.

Solution: Given D1  0.5 m  D2, N  750 rpm, V1  15 m/s, Vu1  0, Vf1  Vf2  V1 . The outlet blade angle  A2  (b) 65o, (c) 80o, (d) 110o Blade velocity at inlet and outlet:U1  U 2 

P DN 60

 P s 0.5 s

750  19.635 m / s 60

(a) Blade inlet angle A1 is given by tan A1 

V1 U1

 B1  37.38



15 19.635

152

Turbomachines

The velocity triangles are as shown in Fig. 4.33: V1 = 15 = Vf1 = Vf2 x U1 = U2 = 19.635

Figure 4.33

Velocity triangles for Example 4.9.

In the outlet velocity triangles, the projection of Vr2 on U is designated as x for easy reference. (b) When A2  65o, tan B2  x

Vf2 x

Vf2 tan 65o



15  7.00 m/s tan 65o

Therefore, the whirl component of V2 is given by Vu2  19.635 7.00  12.635 Now, the specific work is given by W  U 2Vu2  19.635 s 12.635  248.1 J/kg Referring to Section 4.4.1, to calculate degree of reaction R: We have (U  x ) (19.635  7) Vru mean  2   13.32 2 2 Therefore using Eq. (4.9), degree of reaction is given by R

13.32  0.68 19.635

Continuing on the same lines as above we have for the cases of (c) A2  80o, (d) A2  110o, we have: A2

X

Vu2

W

R

80

2.645

16.99

333.6

0.567

110

5.46

25.10

492.7

0.361

Comments: 1. 2.

As A2 increases, the specific work also increases. As A2 increases, the degree of reaction decreases.

General Analysis of Turbomachines

153

EXAMPLE 4.10 A fluid stream is available at 250 m/s at the outlet of the nozzles that can be placed at 25o to the plane of the runner wheel in an axial flow turbine. The blade inlet angle is 40o. Calculate the blade outlet angle, the diameter of the runner, and the specific work if the speed of the runner is 1500 rpm. Take (a) R  0.5 and (b) R  1.0.

Solution: Given V1  250 m/s, @1  25o, A1  40o, N  1500, R  0.5 or 1.00. To calculate A2, D, W. Since U(U  U1  U2) is not known, let it be AB. V1 = 250

C

25° U

40° B

Vr1

V2

125.9 A U = 100.66 B

E

(a)

Figure 4.34

C V1

Vr2

Vr1 Vf1

A

453.15

D

E

(b)

Velocity triangles for Example 4.10. (a) Inlet velocity triangle and (b) Inlet and outlet velocity triangles, for R = 1.0.

First, to calculate the diameter of rotor and the blade velocity, the inlet velocity triangle, ABC, in Fig. 4.34(a) is solved. Referring to the triangle ABC, AEC, Vf1  V1 sin25o  250 sin25o  105.655 Vf1  tan 40o BE Therefore, BE 

Vf1 tan 40o



105.655  125.9 (tan 40o  0.839) tan 40o

Similarly AE  250 cos25o  226.577 Also, AB  AE BE  226.577 125.9 AB  100.66  U P DN  U 60 60U 60 s 100.66 D   1.28 m PN P s 1500 (a) Consider the case of R  0.5. When R  0.5, the velocity triangles are symmetrical. So B2  A1  25o

154

Turbomachines

The specific work is given by W  U ( $Vu )  U ( AE  BE )  100.66 s (226.577  125.9)  35480 J/kg (b) Consider the case of R  1.0. When R  1.0, the velocity triangles are as shown in Fig. 4.34(b) with CD  2 –AE (refer Section 4.4.1). So

$Vu  2 – AE  226.577 s 2  453.15

Now, in triangle ABD, tanA2  CE/ (CD – BE) tan B2 

105.655 453.15 125.9

 B2  17.89o Also, the specific work is U ( $Vu )  100.66 s 453.15  45610 J/kg Comments: 1.

2. 3.

The kinetic energy of the fluid is V12 / 2 at the inlet ( 31250 J/kg). This holds good for both cases R  0.5 and R  1.0. The energy output is not only due to this kinetic energy, but also due to the reaction, that is, due to the expansion of fluid. Higher reaction means higher conversion of pressure energy into output. In the velocity triangles, this can be seen as Vr1 “expanding” to Vr2 , in the two cases of the reaction. Velocity triangles can be drawn to scale and quantities such as AB, BE, CD, etc. can be measured. Accuracy, however, is limited to only three digits, which is sometimes acceptable. Because the problem is on turbine, the utilization factor can also be determined as W / (W  V22 / 2 ). V2 is required to be found out for this purpose.

(When R  0.5, V2  Vr1  167.37,   0.72. When R  1.0, V2  V1  250,   0.59.)

4.4.4 Flow Analysis in Runner Blades: Slip and Slip Factor or Coefficient of Slip As in the case of radial flow machines, axial flow machines (compressors and turbines) have the slip due to the pressure distribution on the two sides of the blades, finite thickness of blades, and nonuniform velocity profiles. Figures 4.35 and 4.36 show these effects for the turbines and compressors, respectively. The specific work is reduced in both the cases of turbines and compressors. The coefficient of slip is of the order of 0.98 for smaller machines to 0.99 for larger machines, while operating at rated conditions.

General Analysis of Turbomachines

155

ΔVu D1

D +

+

Fluid

Figure 4.35

− +

−

+

+

−

−

C

− A

U

(a)

B

(b)

Slip in turbines. (a) Pressure distribution around turbine blades and (b) Velocity triangles, with slip (D–D1). Vu








C V1





Figure 4.36

D

Slip

Vr1


A

(a)

D1

Fluid

U

B

(b)

Slip in compressors. (a) Pressure distribution in compressor blades and (b) Velocity triangles with slip (D–D1).

The specific work is reduced in both the cases of turbines and compressors. The coefficient of slip is of the order of 0.98 for smaller machines to 0.99 for larger machines, while operating at rated conditions.

4.4.5 Losses in Blade Passages The losses that occur in axial flow machines are like those in radial flow machines in all their details. There are the losses due to the friction between the fluid and the blade surfaces, proportional to the square of the velocity, but these are comparatively less due to better surface finishes of the blades (made possible by the manufacturing methods). The viscous flow losses depend on the viscosity of the fluids, but these are more in axial flow machines due to the flow through multiple rows of blades. The turning losses at off-design performance (at part loads) take place much like in radial flow machines with a minimum at the design point. The losses that occur in axial flow turbines are to be subtracted from the ideal specific work. Thus, outputs are reduced. In axial flow compressors, for a desired output, the input is required to be increased by the amount of losses.

4.5 Fan Laws Some considerations of the simple physics of turbomachines give rise to the relations between the various parameters such as the speed (N), diameter of rotor (D), head created or consumed (H), power produced or utilized (P), etc., as discussed in Chapter 1. The coefficients in the form O1, O2, and O3 are useful in model testing. The non-dimensional forms of these coefficients are also discussed in Chapter 1 to extend the applicability of the numbers so obtained.

156

Turbomachines

The interdependencies of the parameters in the form of relations between them are known as fan laws. These laws are stated as follows: 1.

First fan law: In an incompressible flow through a rotor of specified diameter, the volume flow rate of fluid is directly proportional to the speed of the rotor, that is, Q tN

2.

Second fan law: In an incompressible flow through a rotor of specified diameter, the stagnation pressure rise is directly proportional to the product of the density of the fluid and the square of the speed, that is, ( $p )o t R N 2

3.

Third fan law: In an incompressible flow through a rotor of specified diameter, the power required to drive the machine is proportional to the product of the density and cube of the speed, that is, P t RN 3

Fan laws are yet another form of stating the similarity laws and model studies. The variation of diameter is of constructional nature, the variation of speed is of operational nature, and the variation of density is indicative of handling different fluids. The Fan laws reinforce the ideas of operating a given machine at various operating parameters and indicating the performance during such variation.

K eywords Machine parameters Blade angles Rotor diameters Radial flow machines Axial flow machines Performance parameters Reaction Utilization factor Specific work–capacity flow rate relationship Zero-work impellers Specific work as a function of A2

Reaction as a function of A2 Slip, coefficient of slip Pressure distribution on blade surfaces Inertia effects Blade thickness effects Slip not a loss Friction losses Turning losses Reaction in an axial flow machine 0 b R b1 Negative R, R  1

S ummary In this chapter, the analysis of turbomachines is considered with respect to the machine parameters. Radial flow machines and axial flow machines are discussed separately. In radial flow machines: t
The effects of the variation of the blade outlet angle A2on the specific work and degree of

reaction are studied (as A2 increases, the magnitude of W increases and R decreases). t
The case of zero-work impeller is discussed (Absolute velocity of fluid, V, does not change in the rotor, due to some combinations of parameters A1, D2/D1 A2). t
The degree of reaction is detailed [R  (W – Kinetic component)/W].

General Analysis of Turbomachines

t
The slip and the coefficient of slip are discussed in detail, with causes and effects (Wi  WE s L, Hi  HE s L). t
The losses in the rotor are discussed in detail. t
The head–capacity relationship is arrived at step by step, and the performance characteristics are studied, finally to locate the duty point (Figs. 4.18, 4.19). In axial flow machines: t
The degree of reaction is studied in detail, as a basic parameter of importance, almost as a design parameter (R  –Vru mean/U).

157

t
The effects of variation of blade angles on the specific work and degree of reaction are studied with respect to both turbines and compressors (As A2 increases, the magnitude of W increases and R decreases). t
The slip and the coefficient of slip are studied with respect to turbines and compressors (Wi  WE s L, Hi  HE s L). t
The losses in axial flow machines are likened to those of radial flow machines. The general approach to the study of turbomachines is presented in order that the study can now be continued with respect to the individual machines in the later chapters.

I mportant E quations 1. Specific work

5. Ideal head WE  U 2Vu2

Hi  HE s M

2. Head developed HE 

6. Degree of reaction U 2Vu2

(Vru1  Vru2 ) 2 R U

g

3. WE  C1 C 2Q where C1  U 22 and C 2 

(U 2 cot B2 ) A2

.

7. Degree of reaction R

4. Ideal work

Vru mean U

Wi  WE s M

M ultiple- C hoice Q uestions 1. When the outlet angle of a blade is 90o in a radially outward flow machine, the theoretical specific work (a) is directly proportional to the flow rate (b) is inversely proportional to the flow rate (c) is not dependent on the flow rate. (d) none of these. 2. In a radial flow machine, as A2 keeps on increasing, the specific work

(a) changes from the positive to the negative value (b) changes from the negative to the positive value (c) remains constant (d) remains positive 3. In a radial flow machine, as A2 keeps on increasing, the degree of reaction

158

4.

5.

6.

7.

Turbomachines

(a) changes from the positive to the negative value (b) changes from the negative to the positive value (c) remains constant (d) remains positive Due to slip, in a radial flow impeller, the specific work (a) increases (b) decreases (c) remains constant (d) none of these As the flow rate increases in a radial flow machine, the losses (a) increase (b) decrease (c) first remain constant and then decrease (d) first decrease and then increase The turning losses in a radial flow machine depend on (a) the specific work (b) the degree of reaction (c) the flow rate (d) none of these As the speed of an axial flow rotor increases, the specific work (a) increases (b) decreases (c) first increases and then decreases (d) does not change

8. The expression R  Vru mean / U holds good (a) only when U1  U2 (b) only when Vf1  Vf2 (c) at all times (d) none of these 9. The expression R  Vru mean / U for an axial flow machine holds good (a) for impulse machines (b) for reaction machines (c) when the reaction is negative (d) all of these 10. In an axial flow impulse machine, (a) Vr1  Vr2 (b) A1  A2 (c) R  0 (d) all of these 11. In an axial flow machine with reaction of 100%, (a) V1  V2 (b) A1  A2 (c) Vr1  Vr2 (d) all of these 12. In a radial flow turbine, the flow is generally (a) inward (b) outward (c) tangential (d) any of these 13. In a radial flow machine, when the machine runs without any flow rate of fluid, the head generated is (a) U12 /2 g (b) U 22 /2 g (c) (U 22 U12 )/2 g (d) U 22 cot B2 /A 14. In a radial flow machine, it is not possible to have (a) R  0 (b) R  1 (c) R  1 (d) none of these

R eview Q uestions 1. Explain why, in general, radial flow turbines are inward flow machines and radial flow pumps or compressors are outward flow machines. (Refer Section 4.1) 2. Explain the effect of increasing the blade outlet angle on the energy transfer in radial flow machines. (Refer Section 4.3) 3. Explain the effect of increasing the blade outlet angle on the reaction in (a) a radial flow machine and (b) an axial flow machine. (Refer Sections 4.3.2 and 4.4.1)

4. What is a zero-work impeller? (Refer Section 4.3.1) 5. Explain the mechanism of slip, as it occurs in (a) radial flow machines and (b) axial flow machines. (Refer Sections 4.3.4 and 4.4.4) 6. Explain the losses taking place in the machines. (Refer Section 4.3.5) 7. Explain the characteristics of a radial flow pump and/or compressor. (Refer Section 4.3.6)

General Analysis of Turbomachines

8. Prove that R  Vru mean / U , and explain the significance of this equation. Also mention the assumptions. (Refer Section 4.4.1) 9. Sketch some general form of velocity triangles and derive an expression for the specific work. (Refer Section 4.3, Fig. 4.1, Eq. 4.1(a)) 10. State the effects of variation of blade angles on head and reaction in the case of centrifugal pump. (Refer Sections 4.3.1–4.3.3)

159

11. Write a note on slip as referred to centrifugal pumps. (Refer Section 4.3.4) 12. What are the losses that take place in centrifugal pumps? What are the causes of such losses, and what are their effects? (Refer Section 4.3.5) 13. Sketch the general characteristics of centrifugal pumps. (Refer Section 4.3.6)

E xercises 1. The impeller of a radial flow pump has the diameters of 4.5 cm and 10 cm at the inlet and outlet. The blades are bent backward, at 65o to the blade velocity at the outlet. Water enters the impeller radially at a velocity of 10 m/s. The flow component of the velocity gets reduced to 60% from the inlet to outlet, in the rotor due to the variation of the area. The speed of the rotor is 2500 rpm and a slip of 3% may be assumed. Draw the velocity triangles and calculate the specific work. Also determine the power required to drive the pump with an overall efficiency of 87.5%, and the degree of reaction. The flow rate is 30 kg/s. 2. The inlet and outlet diameters of the impeller of a radial flow pump are 6.5 and 15 cm, respectively. The speed is 1440 rpm. The blade angle at the inlet is 70o. The velocity of flow gets reduced to 90% in the impeller due to the increase of area. If the outlet blade angle is 85o, calculate the specific work with no slip and with a slip of 5%. Also determine the degree of reaction in the two cases. 3. The runner of an inward flow turbine has a diameter of 75 cm at the inlet. The speed of the rotor is 500 rpm. Water enters the turbine at an angle of 28o. The velocity of flow gets reduced to 75% due to the variation of area in the rotor. The water gets discharged without any whirl component at a rotor diameter of 30 cm. The blade outlet angle is 55o. The losses in

4.

5.

6.

7.

the runner are 15%. Find the velocity of water at the inlet, the inlet blade angle, the specific work output, the degree of reaction, and the utilization factor. In an axial flow turbine with a speed of 1500 rpm, the fluid velocity at inlet is 180 m/s at an angle of 30o to the wheel plane. The blade inlet angle is 60o. Find the rotor diameter. Assuming the flow component to remain the same, calculate the specific work and utilization factor if the degree of reaction is (i) 0.5 and (ii) 0.75. The blade angles at the inlet and outlet are 50o and 25o, respectively, in an axial flow turbine. The inlet fluid velocity is 150 m/s at an angle of 20o to the wheel plane. The rotor diameter is 0.85 m. Calculate the speed of the rotor. If the relative velocity is increased in the rotor by 50%, due to expansion, calculate the specific work, degree of reaction, utilization factor, and axial thrust. The diameter of the rotor of an axial flow impulse turbine is 0.85 m and the speed is 3000 rpm. The fluid velocity at the inlet to the rotor is 320 m/s. Calculate the blade angles at the inlet and outlet, the nozzle angle, and the specific work for a rotor efficiency of 0.86, with flow velocity remaining constant in the rotor and for a utilization factor of 0.7. The rotor of an axial flow blower has a tip diameter of 30 cm and hub diameter of 10 cm. The speed of the rotor is 2880 rpm. The

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blower is designed to draw 1.2 kg/s of air at a pressure of 100 kPa and temperature of 25oC. The blade outlet angle is more than the inlet angle by 20o. Assume that the air enters the rotor without any whirl and that the axial component of velocity remains constant in the rotor. Calculate the volume flow rate, the blade angles at the inlet and outlet, the specific work, the power, and the degree of reaction. Base the calculations on the mean diameter of the rotor. 8. An exhaust fan has a blade tip diameter of 30 cm and hub diameter of 4 cm. It runs at 600 rpm. The blade angles at the tip at the inlet and outlet are 30o and 60o, respectively. Air at 100 kPa and 25oC flows into the rotor with no whirl component, and the axial flow component remains constant. The average specific work over the length of the blade may be taken as 75% of that at the tip of the blade. Calculate the air flow rate, specific work, and power required to drive the fan at an overall efficiency of 60%. If the same design were to be used for an axial flow pump to pump water

P roject- O riented Q uestions 1. The 10 solved examples and 10 exercise problems of this chapter make up a variety of turbomachines. With a broad bifurcation under radial flow and axial flow machines, prepare a table of all the parameters, such as speed, specific speed, specific work, head or pressure, degree of reaction, angles, efficiencies, etc. Find the gaps in the information and obtain those values, if it is possible. For example, if specific speed is neither a part of data nor has been asked as part of answers, then, it may be calculated and provided in the table. Are there any particular observations to be made from the table?

(with stronger blades), what would be the power? 9. An axial flow compressor is required to work with a degree of reaction of 0.6. The speed is 6000 rpm. The mean diameter of the rotor blades is 45 cm, and the height of the blades is 6 cm. The air velocity at the inlet is axial and is equal to 80 m/s. Calculate the blade angles at the inlet and outlet, the mass flow rate, and the power required to drive the compressor. Assume that the air is at 100 kPa and 25oC, and that the flow component of velocity remains constant in the rotor. 10. An inward flow water turbine has its rotor diameters as 75 cm and 40 cm at inlet and outlet, respectively. The speed is 375 rpm. The velocity of flow remains constant in the rotor. The water leaves the rotor without any whirl component. The blade outlet angle is 50o. For a degree of reaction equal to 0.8 and a utilization factor of 0.95, calculate the angle of entry of water to the rotor, blade inlet angle, and specific work. If the flow rate is 1.1 m3/s, determine the theoretical output power of the rotor.

S olutions A vailable 2. In order to study the effect of the outlet blade angle A2 on the performance of a radial flow machine, calculate the specific work W and degree of reaction R for an impeller with the inlet diameter D1  3 cm, outlet diameter D2  6 cm, speed N 1000 rpm, inlet blade angle A1 50°, and a series of values of A2 (15°, 20°, 22°, 24°, 25°, etc.). Neglect the effects of slip and losses. Now plot the values of W and R, so obtained, on a base of A2. If required, extend the values of A2 for this “experiment”. Write your observations from the plot. Can you correlate your answer with any equation in the text?

General Analysis of Turbomachines

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A nswers Multiple-Choice Questions 1. 2. 3. 4. 5. 6. 7.

(c) (a) (a) (b) (d) (c) (a)

8. 9. 10. 11. 12. 13. 14.

(d) (d) (d) (a) (a) (c) (c)

Exercises 1. 130.7 J/kg, 4.48 kW , 0.84 2. 115.93 J/kg, 110.13 J/kg, 0.695, 0.7256 3. 31.856 m /s, 60.4o, 469.43 J/kg, 0.053, 0.882 4. 1.323 m, 21.6 kJ/kg, 0.8, 27 kJ/kg, 0.74 5. 2200 rpm, 13.13 kJ/kg, 0.214, 0.9342, 8.85N /( kg /s)

6. 49.3o, 49.3o, 36.8o, 44.032 kJ/kg 7. 1.0263 m3 /s, 28.44o, 48.44o, 0.4728 kJ/kg, 0.567 kW , 0.74 8. 0.44173 kg /s, 44.416 J/kg, 32. 7 W , 27.97 kW 9. 29.5o, 70.6o, 7.93 kg /s, 126.8 kW 10. 27.157o, 105.2o, 286.46 J/kg, 315 kW

5

Steam Turbines

Learning Objectives After completing this chapter, you will be able to: v
Get familiar with the different types of steam turbines and the various terms associated with such turbines. v
Study the basic impulse and reaction turbines, their operating parameters, and effects of such parameters on their performance. v
Study the necessity of compounding of steam turbines and the methods and arrangements of compounding. v
Analyze the performance of each type of steam turbine, with the effects of various parameters on such performances.

v
Study the reheat factor in multi-stage steam turbines, factors affecting such reheat, and the effects of such reheat. v
Understand the need and mechanism of governing of steam turbines and the implications of different methods. v
Study the performance characteristics of steam turbines.

5.1 Introduction Steam turbines are power-producing turbomachines that use steam as the fluid. The steam absorbs the thermal energy from the combustion zone in boilers or from the coolant circuit of nuclear reactors. The energy is available in the steam at high pressure and temperature. The steam is then taken to the turbine where it expands to lower pressures in nozzles or blade passages and the resulting velocity imparts energy to the rotating shaft of the turbine. The input energy to the turbine is the enthalpy drop of the steam when it expands from its pressure at the entry to the pressure at the exit. The enthalpy drop of the steam in the steam turbine can be found by tracing the process of expansion, knowing the initial state of the steam and the final pressure, and assuming isentropic expansion. Steam pressures at the inlet to the turbines can be very high, such as 35 bar, 40 bar, etc. Exit pressures are of the order of 0.03–0.04 bar. When steam goes through such large pressure drops, the velocity generated in the nozzles can be very high. Such high velocities give rise to large losses in the flow passages. If attempts are made to absorb the energy in a single stage, the exit losses are also of high magnitudes. Hence, different arrangements are made to have maximum utilization of the energy. This chapter is an attempt to study the different types of steam turbines, the variations of pressures and velocities of steam in such turbines, the efficiencies, the methods to improve such efficiencies, and the different arrangements to accomplish the same. Further, the principles of thermodynamic flow of steam, with different effects, as studied in chapters 3 and 4, are recalled and applied to steam turbines.

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5.2 Classification of Steam Turbines In Chapter 4, it was seen that there exists a distinct variation between radial flow machines and axial flow machines. Although radial flow steam turbines are conceptualized, steam turbines are generally axial-flowtype turbines (U1  U2  U). One basis of the classification of steam turbines is with reference to the impulse and reaction types. Steam turbines are, therefore, of two types: Impulse turbines and reaction turbines. Another type of classification is based on “staging” arrangements, that is, turbines are either single-stage or multi-stage. Turbines are also named after the persons who designed such arrangements: (a) (b) (c) (d)

De Laval turbine, named after its designer, Gustafe de Laval; Curtis turbine, named after its designer, Charles Curtis; Parsons turbine, named after its designer, Charles Parson and Rateau turbine, named after its designer, Auguste Rateau.

5.2.1 Single-Stage Impulse Turbine: De Laval Turbine In an impulse turbine (Fig. 5.1), high-pressure steam enters a set of nozzles that are part of the stator or casing, placed in the form of a ring. The steam then expands in the nozzles, generating streams of very high velocity or kinetic energy, and the pressure drops. The high-velocity streams or jets strike the blades of the rotor, giving the impulse to the rotor. The kinetic energy is transferred to the rotor. There is no pressure drop as the steam flows through the passages between the rotor blades and hence the relative velocity remains constant over the blades. The reaction is zero. The variation of the pressure of the steam and the corresponding velocity are shown in the diagram. The figure also shows the combined velocity triangles for the rotor blades for one set of values (A1, A2, @1, and @2). While the blade angles A1 and A2 must be the same for this type of turbine, @1 and @2 can vary over some range of values. The whole arrangement is a basic impulse turbine. This turbine is known as the De Laval turbine, named after its designer, Gustafe de Laval. Stator

Rotor V2

V1

Vr 2

Vr 1

b2 a1 Inlet pressure

V1

b1 U b2

b1

a1 Approach velocity

Exit pressure Exit velocity Axial distance

Figure 5.1

Impulse turbine: De Laval turbine, R  0, Vr2 Vr2, A1  A2.

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165

5.2.1.1 Analysis: Blade Efficiency, Maximum Blade Efficiency The expansion of steam and its pressure drop occur only in the nozzles and there is no expansion of the steam as it flows through the blade passages of the rotor. Therefore, the relative velocity of steam in the blade passages remains constant, Vr1  Vr2 . The rotor blade angles A1 and A2 are equal in the impulse turbine. The specific work is given by W  U (Vu1 Vu2 ) or W 

(V12 V 22 ) 2

The input energy for the blades is only of the kinetic form and is equal to V 12 / 2 A rotor blade efficiency or vane efficiency is defined as

Hr  

Output Input (V1 )2 (V2 )2

1 1 1

(V1 )2 (V2 )2 (V1 )2 (Vf2 )2  (Vu2 )2 (V1 )2 (Vf1 )2 (V1 ) 2

 1 sin 2A1

(Vu2 )2

(&Vf1  Vf2 )

(V1 ) 2 (Vu2 )2 (V1 ) 2

So

Hr  cos2 A1

(Vu2 )2 (V1 )2

The maximum rotor efficiency is obtained when the subtractive term (Vu22 /V 12 ) is reduced to zero that happens when @2  90o and Vu2  0. Hence

Hr max  cos2 A1

(5.1)

When the relative velocity at the outlet of the blades is considered equal to that at the inlet, there are no losses in the blade passages. This is only theoretical situation. When the steam flows through the blade passages, losses take place due to various factors such as friction, turbulence, by-pass at the tip of the blades, and so on. The relative velocity at the outlet, Vr2 , gets reduced to c b s Vr1 , where cb is known as “coefficient of blade friction” that has a value less than 1 and is of the order of 0.96, 0.97, etc. If the surface finish of the blades is not smooth, the value of cb can be still less. During the design of the blades, the aim of a good design is to maximize the specific work (and efficiency), reduce the axial thrust (to zero), etc. It is possible to choose the blade angles, areas of the flow, the

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operating speed, and similar other parameters toward this aim. As seen above, the relative velocities and blade angles need not essentially be equal at the inlet and outlet. Further, for a specified inlet steam velocity, the speed and diameter of the rotor can also be a matter of choice. In this context, a general approach to optimization is now considered. The general velocity triangles shown in Fig. 5.1 can now be considered in more details. From the Euler turbine equation, the expression for specific work is stated as follows: W  U (Vu1 Vu2 )  U (| Vru1|  | Vru2|)  U (Vr1 cos B1  Vr2 cos B2 ) So

¥ V cos B2 ´  U (Vr1 cos B1 ) ¦1  r2 s µ § Vr1 cos B1 ¶ W  UVr1 cos B1(1  c bC )

where cb is the coefficient of blade friction, etc., and C is the ratio of cosines of blade angles at the outlet and inlet. Also Vr1 cos B1  V1 cos A1 U Therefore W  U (V1 cos A1 U )(1  c bC ) The blade efficiency or rotor efficiency is W (V1 )2 / 2 U (V1 cos A1 U )(1  cbC )  (V1 )2 / 2

Hr 

2s

U¥ U´ cos A1 µ (1  c bC ) ¦ V1 § V1 ¶

 Hr  2F(cos A1 F )(1  c bC ) Thus, the rotor efficiency or blade efficiency is identified as a function of speed ratio E, nozzle angle @1, coefficient of blade-friction cb, and the ratio of cosines of angles. In other words, the effects of available enthalpy drop, rotor diameter, rotor speed, nozzle angle, and blade angles are included in the expression for rotor efficiency. An optimum value of the speed ratio E can be obtained by equating dGr/dE to zero, with @1, cb, and C taken as constants. The result is cos A1 cosA1 2F  0  F  2 Substituting this value of E into the expression for Gr, we get

Hr  2 s  Hr 

cos A1 ¥ cos A1 ´ cos A1 (1  cbC ) ¦ § 2 2 µ¶

cos2 A1(1  cbC ) 2

Steam Turbines

167

When cb  1 and C  1 (no losses and equiangular blades), the efficiency becomes maximum, and the above expression reduces to (Hr )max  cos2 A1 This equation is the same as Eq. (5.1). In the case of impulse turbine, the utilization factor is the same as the rotor efficiency. Hence, the maximum utilization factor is given by max  cos2A1

(5.2)

The velocity triangles corresponding to this situation are the particular cases of the general velocity triangles of Fig. 5.1, and this particular case of @2 equal to 90o is shown in Fig. 5.2. ΔVu = 2U D V2

V1

Vr 2

Vr 1

b2 A

Figure 5.2

C

b1 U

B

Velocity triangles of the De Laval turbine [(Gr)max and max].

The corresponding maximum output is Wmax  U ( $Vu )  U –2U  2U 2

(5.3)

5.2.1.2 Stage Efficiency A stage with respect to the De Laval turbine is identified as the set of nozzles in one stator ring and the one ring of the rotor blades. The input energy to this stage is the enthalpy drop of steam in the nozzles. Because there is no enthalpy drop in the rotor blades, the input to the stage is the enthalpy drop only in the nozzles. The nozzles have their own losses due to the fluid friction, turbulence, etc., and therefore a nozzle efficiency is identified as the ratio of the nozzle output to the nozzle input, as given by

Hn 

(V1 )2 /2 ( $h )stage

(5.4)

The usual values of the nozzle efficiency are of the order of 0.96–0.99. When the nozzles are convergent– divergent, as they are in the case of De Laval turbine, the losses are more and the efficiencies are in the lower range of 0.96–0.97. Higher efficiencies are possible when the pressure drops are less (as in compounding situations to be seen later) and when nozzles are convergent ones. A stage efficiency Gstage is given by

Hstage 

Wstage ( $h )stage

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Turbomachines



(V1 )2 / 2 W s (V1 )2 / 2 ( $h )stage

Hstage  Hr s Hn

or

(5.5)

5.2.2 Single-Stage Reaction Turbine A reaction turbine (Fig. 5.3) is the one in which the expansion and pressure drop of steam take place not only in the stationary nozzles but also in the flow passages between the rotor blades. The process of expansion gets shared between the stator and rotor blades. The velocity is generated in the nozzles and in the rotor blades. The kinetic energy so generated, in the nozzles and the rotor blades, gets transferred to the rotor, as the steam flows through the rotor. The reaction generated is due to the part of expansion in the rotor. Depending on the proportions of such expansion, the degree of reaction may vary (0  R  1), but generally the design is for 50% reaction (R  0.5). This means that of the total energy transferred from the steam to the rotor, 50% is of impulse form (kinetic energy form) and 50% is due to the pressure drop in the rotor, that is, of reaction form. This type of turbine is the Parsons turbine, named after its designer, Charles Parson. Figure 5.3 also shows the variation of pressure and velocity, and the combined velocity triangles for the rotor of Parsons turbine. Stator

Rotor

U

V1

Vr 2

V2

Vr1

a2

a1 b2

b1

U b2 Inlet pressure V1 Approach velocity

a1

b1 V1

U

Exit pressure Exit velocity Axial distance

Figure 5.3

Reaction turbine: Parsons turbine. R  0.5, @1  A2, A1  @2.

5.2.2.1 Analysis: Blade Efficiency, Maximum Blade Efficiency In Chapter 3, expressions have been derived for the utilization factor, specific work, speed ratio, and maximum utilization factor for a turbine in terms of the degree of reaction R [Eqs. (3.10)–(3.13)]. These hold good for any value of R, in general, 0 b R b 1. In Chapter 4, the velocity triangles have been discussed for various values of the degree of reaction, R. It has been seen that the velocity triangles are symmetrical when the degree of reaction R is 0.5. Even with R  0.5, there are different possibilities and one such general case has been shown in Fig. 5.3.

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169

In the case of one stage of Parsons turbine, when the condition of maximum utilization factor (i.e., @2  90o) is combined with R  0.5, the resulting velocity triangles are as shown in Fig. 5.4. ΔVu = U D V2

Vr 2

V1

A

Figure 5.4

C Vr 1

U B

Velocity triangles for the Parsons turbine:   max, @2  90°, R  0.5.

From the velocity triangles, it is readily observed that ( $Vu )  U Therefore W  U ( $Vu )  U 2

(5.6)

Also from Eq. (3.10) m 

cos2 A1 1 0.5 sin 2 A1

(5.7)

In a Parsons stage, the flow passages in both the stator and the rotor blades are like the convergent nozzles. Because the pressure drop is less in comparison with that in impulse-type turbines, the nozzle efficiency Gn, applicable to both the stator and the rotor passages, is higher than the nozzle efficiency of impulse turbines and is of the order of 0.97, 0.985, etc. The input to the stage is the enthalpy drop in both the stator and the rotor and therefore the stage efficiency is stated as

Hstage 

W ( $h )stator  ( $h )rotor

(5.8)

5.3 Compounding of Steam Turbines The steam available for turbines is at pressures of the order of 40 bar (or more). The condenser pressure (or “back pressure”) to which it can expand is of the order of about 0.04 bar. Whether the expansion occurs in nozzles, blade passages of the rotor or both, the velocity generated can be very high, of the order of 1400 m/s. Utilization of this kinetic energy in a single ring of rotor blades has numerous disadvantages: 1. 2. 3. 4.

5.

The process of expansion involves convergent–divergent nozzles, with lower expansion efficiencies. The process of energy conversion is not efficient and lot of energy is wasted at the exit. The frictional losses are also high due to the flow of steam at high velocities. The high velocities of the steam impose high speeds of the rotor, such as 30000 rpm, 40000 rpm, etc. These high speeds give rise to very high stresses in the blades, making the mechanical design more complex and resulting in costly materials. The high speeds are unmanageable in the driven side also, unless the speeds are reduced by powerconsuming gear boxes.

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When the velocity of steam at the exit of nozzles is very high, or when the pressure energy itself is very high, the method devised to utilize such energy at a lower speed of the turbine is to absorb the energy in multiple rings of rotor blades. This method of reducing the speed is known as compounding. Compounding is needed to avoid the above disadvantages of high flow-velocity of steam and lower efficiency of expansion in nozzles. Compounding is the method adopted for the transfer of energy from the steam to the rotor in multiple rows of rotor blades, in order to run the turbine at lower speeds. 1. 2.

If the process of expansion of steam is shared in multiple rows of nozzles (or nozzle-shaped flow passages between the blades of stator rings), then the method is known as pressure compounding. If the process of expansion of steam is in one set of stator nozzles and the absorption of the resultant high velocity is in multiple rows of rotor blades, then the process is known as velocity compounding.

Pressure compounding and velocity compounding can be combined together to have pressure–velocity compounding.

5.3.1 Pressure Compounding The method of pressure compounding in a steam turbine is shown in Fig. 5.5. The turbine has alternate rows of stator blades (“fixed” blades) and rotor blades (“moving” blades). The stator blade passages act as nozzles and the steam expands in these passages with continuous drop in pressure and increase in velocity. The blade passages of the rotor may have cross-sections of two types:

Supply pressure

No ΔP

Nozzle ring

No ΔP

Exit pressure

Approach velocity

Figure 5.5

Exit velocity

Stator blades

Moving

Stator blades

Moving

Nozzle ring

Moving

The cross-sectional area of the passages between the rotor blades remains constant, with no expansion of steam in the rotor blades, as shown in Fig. 5.5(a). The turbine is known as the pressure-compounded impulse turbine. This arrangement is known as the Rateau turbine, named after its designer, Auguste Rateau. Moving

1.

Pressure drop in all the rings Supply pressure

Approach velocity

Exit velocity Exit pressure

Pressure compounding: (a) Rateau stages and (b) Parsons stages. In both the parts, the top portions of the figures indicate lay out of the nozzle-rings, the rotor and stator blade-rings; the lower portions indicate the variations of pressure and velocity along the flow path.

Steam Turbines

2.

171

The cross-sectional area of the passages between the rotor blades is such that the passages act as nozzles, with expansion of steam occurring in the rotor blades also, as shown in Fig. 5.5(b). The turbine is known as the pressure-compounded reaction turbine. This arrangement can have any degree of reaction R, but the Parsons turbine has only one value of degree of reaction, R  0.5. These stages are repeated one after the other, each stage comprising one set of stator blades followed by one set of rotor blades. In Fig. 5.5(b), two Parsons stages are shown.

5.3.2 Velocity Compounding

Moving rotor

Stator

Moving rotor

Stator

Stator nozzle ring

Moving rotor

In this type of turbine (Fig. 5.6), a stator ring has the nozzles. The steam, while flowing through these nozzles, gets accelerated to higher velocities as the pressure drops. The velocity (and the kinetic energy) gets transferred to the rotor in multiple rows (or rings) of rotor blades. There are stator blade rings between the rotor blade rings. The function of the stator blades is only to redirect the steam onto the next row of rotor blades. After the first set of nozzles, there is no pressure drop or expansion of steam in either the rotor blades or the stator blades. The whole arrangement is one stage of the turbine, identified by the one pressure drop. The number of rows of rotor blade rings in a stage can vary from 2 to about 6. A velocity-compounded turbine is invariably of impulse-type, because there is no pressure drop in the rotor blades. This type of turbine is known as the Curtis turbine, named after its designer, Charles Curtis. Figure 5.6 shows one Curtis stage with three rows of rotor blade rings.

(a) Supply pressure

Exit pressure Axial distance Approach velocity

Figure 5.6

Exit velocity (b)

Curtis stage with three rotor blade rings. (a) Layout of the nozzle-ring and the rotor/stator rings and (b) Variation of pressure and velocity of steam along its flow-path.

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Turbomachines

5.3.3 Pressure–Velocity Compounding This type of turbine has multiple stages of the velocity-compounded turbine mentioned above. The exit pressure of the last stage is the condenser pressure. It is to be observed here that the Euler turbine equation is for one row or one ring of the rotor blades and the velocity triangles are drawn at the inlet and outlet of the row or the ring of the rotor blades. As seen in Section 5.3, there can be multiple rows of rotor blades, assembled or keyed onto the same rotor shaft. This calls for the drawing of the velocity triangles, in pairs, and application of the Euler turbine equation for each row of the successive rotor blades. The energy transferred in each row can then be summed up to get the total energy transferred. The specific works of successive rotor blade rings may or may not be the same, as the fluid angles and blade angles keep varying, depending on the type of the compounding adopted.

5.4 Analysis The analyses of a single-stage impulse turbine (De Laval turbine) and a single-stage reaction turbine (Parsons turbine) are discussed in the sections 5.2.1.1 and 5.2.2.1. As has been seen above, the compounding of the steam turbines results into the following types of turbines, with multiple stages: 1. 2. 3. 4.

Rateau stages: Pressure-compounded impulse turbine. Parsons stages: Pressure-compounded reaction turbines with 50% reaction. Curtis stages: Velocity-compounded impulse turbines with more than one row of rotor blades for each pressure drop in the nozzles. Combination stages: Repetition of the above stages, either of the same type or of different types.

When there are multiple stages, the effect of a previous stage gets carried on to the next stage and hence the analysis has to be inclusive of this effect. The analyses of the above cases can be considered now in detail.

5.4.1 Rateau Stages A Rateau turbine is like a number of impulse turbines (such as the De Laval turbine) connected in series. The steam flows from the outlet of one impulse stage to the inlet of the next stage. The separation of such stages is by very small gaps and it looks as if the stages are continuous. In the De Laval turbine, the pressure drop of steam from the supply pressure to the final exhaust pressure could be in one stretch in the single row of nozzles, with associated losses, but in Rateau stages, such a pressure drop is shared in many rings of nozzles, or nozzle-shaped stator blade passages. This feature of Rateau turbines makes it possible to keep the flow velocities of steam at lower values, decreasing the losses. In Fig. 5.7 the three successive Rateau stages of a pressure-compounded steam turbine are shown. The possible number of Rateau stages can be decided depending on the boiler pressure and the condenser pressure. The velocity generated in the nozzles of each stator ring (or “diaphragm”) depends on the enthalpy drop across the nozzles, according to the following equation: V  2( $h )  Va2

Stator blades

No ΔP

Stator blades

No ΔP

173

Moving

Nozzle ring

Moving

Moving

Steam Turbines

No ΔP

Supply pressure

Exit pressure Exit velocity Axial distance

Approach velocity

Figure 5.7

Rateau stages (pressure-compounded impulse type). (a) Layout of the nozzle-ring and the rotor/stator rings. (b) Variation of pressure and velocity of steam along its flow-path.

where $h is the enthalpy drop across the nozzles and Va is the inlet or approach velocity of steam at the entry of the nozzles. The velocity triangles for a stage are the same as those in Fig. 5.2 and are drawn as shown in Fig. 5.8. The data required for the complete solution of the velocity triangles are as follows: 1. 2. 3. 4.

The velocity at the inlet to the rotor blades, V1. The angle of V1 with U, namely, the fluid angle @1. The blade velocity U. The blade angles A1 and A2. V1

Vr2

V2

Vr1

b1

b2 A

Figure 5.8

U

B

Velocity triangles for one row of a Rateau stage.

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Since this is the impulse blade, we also have Vr1  Vr2 and A1  A2. The requirement of maximum utilization factor imposes the condition that the absolute exit velocity of steam from the rotor blade must be axial (@2  90o). In the next stage, if the exit velocity from the second set of nozzles is the same as the exit velocity from the first set of nozzles [i.e., if the enthalpy drop $h for the second set of nozzles is the same as that of the first set of nozzles and if all the angles (@1, @2, A1, A2) are also the same], the same set of velocity triangles holds good for the second row of rotor blades. In such a case, the specific work of the second row of rotor blades is the same as that of the first row. This is the situation in the case of Rateau stages, and this may continue for the subsequent stages also. The total output is the sum of the outputs from each row. n

WT  ¤ Wi i 1

 nWi where Wi  [U (Vu1 Vu2 )]i

(5.9)

where n is the number of Rateau stages. Now, this result may indicate that the nozzles in all the rows and the rotor blades of all the rows have the same geometry and design. However, this is only partly true. As the steam expands to lower pressures, the specific volume of the steam increases. Hence, for the same mass flow rate, the volume flow rate increases. This increased volume flow rate has to be accommodated by increasing the area of flow as may be required. Increasing the height of the blades is one method of increasing the area of flow, even when the blade angles are kept the same. But after a few stages, increasing the blade height may not be possible. Then the mean diameter of the blade ring can be increased, causing an increased U. In such cases, the whole design can be changed without requiring the condition of maintaining the enthalpy drops to be equal to the previous nozzles. If the design is such that the enthalpy drop across each set of nozzles is the same and the angle of velocity is also the same, then the rotor blade angles are the same for each row, that is, A2  A1.

5.4.1.1 Losses in Nozzles and Blade Passages The discussions so far have been based on ideal conditions. The actual conditions are different from ideal conditions due to various factors: Wall friction, turbulence, viscous resistance, leakage near the tips, finite gaps between the rows of nozzles and blades, and so on. Although the various factors are not separately accounted for, an approximate method to combine the effects of all the factors together is to introduce a coefficient of blade friction cb such that Vr2  c b s Vr1 . Instead of making Vr2  Vr1 , now Vr2 may be taken as 0.98Vr1 where the losses are taken as 2% (as an example). The nozzle losses can be accounted for by considering the efficiency of the nozzle, Gn , such that the ideal enthalpy drop ($h)i is corrected to the actual enthalpy drop ($h)a, with ( $h )a  Hn s ( $h )i Two to three percent losses in nozzles can be accounted for by having Gn as 0.97–0.98. With these corrections, the velocity triangles can be drawn as shown in Fig. 5.9.

Steam Turbines

V1

Vr2

V2

b2 A

Figure 5.9

175

Vr1

b1 B

U

Velocity triangles of one Rateau stage with losses B  B; Vr 2  cb . Vr1( cb z 0.98 ).

EXAMPLE 5.1 The mean rotor diameter of a De Laval turbine is 0.5 m. Its speed is 3000 rpm. Steam is supplied from nozzles at a velocity of 200 m/s at a nozzle angle of 22o. The relative velocity of steam reduces by 5% in the blades due to friction. Taking the blade angles at the inlet and outlet to be equal, find the angles, the power output, and the axial thrust for a flow rate of 10 kg/s. If the axial thrust has to be reduced to zero, what modifications of the blades are required?

Solution: Given D  0.5 m, N  3000 rpm, V1  200 m/s, @1  22o, Vr2  0.95 Vr1, A1  A2, m  10

kg/s. To calculate (a) A1, P, Axial thrust. (b) Modifications of blades for zero axial thrust We know that blade velocity is given by

P DN 60 3000  P s 0.5 s 60  78.54 m / s

U1  U 2  U 

Also whirl component of inlet velocity is given by Vu1  V1 cos A1  200 s cos 22o  185.44 m / s The velocity triangles are shown in Fig. 5.10, U and V1 are marked; CE  Vf1, DF  Vf2, V2  AD; BC and BD are respectively Vr1 and Vr2. V 1 = 200

D

22° F

A

U = 78.54

Figure 5.10

b1

b2

C

x

B

Velocity triangles for Example 5.1.

E

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In the triangle ABC and AEC together, AB  U  78.54 and AE  Vu1  185.44. Therefore, BE  AE – AB  185.44 – 78.54 BE  x  106.9 m/s, x as shown in triangle BCE. The flow component of inlet velocity, Vf1  CE  V1 sin A1  200 s sin 22o  74.92 m / s (a) Inlet blade angle is given by B1  tan 1

74.92  35o  B2 106.9

Outlet blade angle is A2  A1  35o Relative velocity at inlet,

Relative velocity at outlet, In the triangle Therefore

Vr1  BC 

x 106.9  cosB1 cos35o

Vr1  130.5 m / s Vr2  Vr1 – c b  130.5 s 0.95  124 m / s BDF, BF  Vr2 cos B2  124 cos35o  101.57

Vu2  AF  BF AB  101.57 78.54  23.03 m /s Specific work,

W  U (Vu1 Vu2 )  78.54(185.44  23.03)  16.3756 kJ/kg Power  Wm  163.756 kW Axial thrust  m (Vf1 Vf2 )  10(74.92 Vr2 sin B2 )  37.97 N

(b) To make the axial thrust zero, A2 can be increased, so that Vf1  Vf2 . Then

B2  sin 1

74.92  37.17o 124

EXAMPLE 5.2 An impulse-type steam turbine has a speed of 1500 rpm. The mean rotor diameter is 0.8 m. The steam velocity at the nozzle exit (inlet to the blades) is 200 m/s, at an angle of 20o to the plane of the wheel. (a) Find the blade angle at the inlet. (b) Find also the outlet blade angle for the utilization factor of 0.8. If the mass flow rate of steam is 10 kg/s, determine the power of the turbine and the axial thrust of the rotor on the bearings. Take Vr1  Vr2 neglecting friction.

Solution: Given N  1500 rpm, D  0.8 m, V1  200 m/s, @1  20o,   0.8, m  10 kg/s. To find A1, A2, P, Axial thrust. We know that P DN P s 0.8 s 1500 Blade velocity is given by U   62.83 m /s 60 60 Also Whirl component is Vu1  V1cos A1  200 cos20o  187.9 m /s. Flow component is

Vf1  V1sin A1  200 sin20o  68.4 m /s.

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177

¨ Vf1 · B1  tan 1 © ¸ ªVu1 U ¹ 68.4 ¨ ·  tan 1 © ª187.9 62.83 ¸¹  28.67o

(a) The inlet blade angle is

(b) Since this is an impulse machine, the reaction component is zero. Therefore W 

(V12 V22 ) 2

Now, utilization factor is 

W W  2  0.8 2 V ¥V ´ 1 W ¦ 2 µ 2 2 § ¶

Therefore, the specific work is ¥V 2 ´ W  ¦ 1 µ s 0.8 § 2 ¶ ¥ 2002 ´ ¦ s 0.8  16 kJ/ kg § 2 µ¶ The power is given by P  Wm  16 s 10  160 kW To solve the velocity triangles at the inlet and outlet, we have V22 2



V12 2

W  20000 16000  4000

Therefore V2  4000 s 2  89.44 m/s The specific work is W  16000  U (Vu1 Vu2 )  U ( $Vu ) Therefore $Vu 

16000  254.655 62.83

At this juncture, it is possible to draw the velocity triangles (Fig. 5.11). Take U  62.88 m/s, Vu1  187.9 m/s, and $Vu  254.655 m/s, all on the base line, and complete the inlet velocity triangle. (Graphically, the outlet velocity triangle can be completed, with V2  89.44 and Vr1  Vr2.)

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C V1

D V r2

Vr1

V2 89.44

a2 F

A

20° U = 62.83

28.67 = b1

B

E V u1 = 187.9

V u2 = −66.755 (ΔV u ) = 254.655

Figure 5.11

Velocity triangles for Example 5.2.

Now, whirl component at outlet, Vu2, is BF – BA in the triangle BDF; so, Vu2  254.655 187.9  66.755 ¥ 66.755 ´  41.72o A 2  cos 1 ¦ § 89.44 µ¶ Therefore Vf2  66.75 tan ( 41.72o)  59.53 And hence, the blade outlet angle A2 is, from triangle BDF, angle DBF, that is, 59.53 ¨ ·  24.67o B2  tan 1 © ª 66.75  62.83 ¸¹ The axial thrust is given by Axial thrust  m (Vf1 Vf2 )  10(68.4 59.53)  88.7 N

5.4.2 Parsons Stages Parsons stages of steam turbines have a degree of reaction R equal to 0.5. One stator blade-ring followed by a rotor blade-ring together make up the one stage. The blades are designed so that the passages between the blades act like nozzles in both the stator and rotor. The expansion of steam takes place in both the rings. The areas of flow have to be increased continuously to accommodate the increased volume flow rates. The velocity triangles at the inlet and outlet of every rotor blade ring become symmetrical. The usual practice is to have the same geometry (@1, @2, A1, A2) of the blades with continuously increasing heights. The same set of velocity triangles and analysis hold good for a few of the rotor rings in succession. When the increase in the height of blades becomes limited after a few rings, the mean diameter of rotor rings can be increased so that another set of velocity triangles and analysis can hold good for another series of rotor rings.

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179

5.4.2.1 Losses in the Parsons Stages For the purpose of analysis, one Parsons stage is considered as made up of two distinct parts: Blade passages in the stator and blade passages in the rotor. (It may be noted here that there are many Parsons stages in a turbine and the stage, now under consideration, has a stage previous to it and a stage later). Because the expansion of steam takes place equally in both the stator and rotor passages, it is as if there are two sets of nozzles, first in the stator and second in the rotor. In the first part, that is, the stator passage, the exit velocity is V1 and the inlet velocity is V2 (being the exit velocity of the previous stage). In the second part, it is the relative velocity Vr1 increasing to Vr2 . Considering the stator blade passage as the nozzle with a nozzle efficiency Gn, we have V12 2

HcoV22 2

 Hn ( $h )is 

(5.10)

where ($h)is is the isentropic or ideal enthalpy drop and Gco is the “carryover efficiency.” This carryover efficiency is to account for the discontinuity between the rotor blade ring (moving blades) of the previous stage and the present stator blade ring under consideration, as if V2 from the previous moving blades does not arrive at the present stator blade uninterrupted (that is, due to loss in the gap between the blade rings). The usual losses due to turbulence, viscous friction, etc. are included in the value of Gco. It may be noted that if Gco were equal to 1, Eq. (5.10) would confirm to the steady flow energy equation as applied to a nozzle. Equation (5.10) can be written as ( $h )is, stator 

V12 HcoV22

(5.11)

2Hn

Further, the enthalpy drop in the rotor blade passages ($h)is, rotor in a reaction turbine can be related to ($h)is, stator [refer Eq. (3.8)] as ( $h )is, rotor  ( $h )is, stator s or

( $h )is, rotor 

R 1 R

R (V12 HcoV22 ) 2Hn (1 R )

(5.12)

Therefore, the total enthalpy drop for the stage, inclusive of stator and rotor, is given by ( $h )is, stage  ( $h )is, stator  ( $h )is, rotor 



(V12 HcoV22 ) 2Hn



R (V12 HcoV22 ) 2Hn (1 R )

V12 HcoV22

(5.13)

2(1 R )Hn

Recalling Eq. (3.11) we have W  U (Vu1 Vu2 ) 

V12 V22 2(1 R )

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Turbomachines

U (Vu1 Vu2 ) 

or

Vu12 Vu22

2(1 R )  2U (1 R )  Vu1  Vu2  Vu2  Vu1 2U (1 R )

for Vf1  Vf2

Hence W  U [Vu1  (Vu1 2U (1 R ))]  W  2U [Vu1 U (1 R )] Now, the stage efficiency is

Hst  Hst 

 

W ( $h )is, stage 2U [Vu1 U (1 R )] ¨ (V12 HcoV22 ) · © ¸ ª 2(1 R )Hn ¹ 4U (1 R )Hn [V1 cos A1 U (1 R )] V12 HcoV22 4F(1 R )Hn [cos A1 F(1 R )] ¥V 2 ´ 1 Hco ¦ 22 µ § V1 ¶

(5.14)

where E  U/V1 is the speed ratio. Now, from (V12 V22 )  2(1 R )U (Vu1 Vu 2 )  4(1 R )U [V1 cos A1 U (1 R )] we have V22 V12

 1 4(1 R )F[cos A1 F(1 R )]

Substituting this value of V22 / V12 into Eq. (5.14), we have

Hst 

4FHn (1 R )[cos A1 F(1 R )] 1 Hco [1 4(1 R )F{cos A1 F(1 R )}]

(5.15)

For maximum utilization, @2  90o; from velocity triangles, E  cos@1. Substituting these conditions, namely, @2  90o, and E  cos@1, in Eq. (5.15),

Hst 

4Hn R (1 R )cos2 A1 1 Hco [1 4(1 R )R cos 2 A1 ]

(5.16)

The stage efficiency Gst can be visualized as the product of nozzle efficiency or stator efficiency Gn and the rotor efficiency or vane efficiency, Gr, that is

Hst  Hn s Hr

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181

Therefore

Hr 

4 R (1 R )cos2 A1

(5.17)

1 Hco [1 4 R (1 R )cos2 A1 ]

For the case of R  0.5, that is, Parsons stage, we have

Hr   Hr 

cos2 A1 1 Hco (1 cos2 A1 ) cos2 A1

(5.18)

1 Hco sin 2 A1

EXAMPLE 5.3 In a reaction turbine, the inlet and outlet blade angles are 50o and 20o, respectively. The steam enters at 18o to the plane of the rotor wheel and leaves at 40o. The rotor speed is 260 m/s. Calculate the speed ratio, specific work, degree of reaction, utilization factor, and axial thrust.

Solution: Given A1  50o, A2  20o, @1  18o, @2  40o, U  260 m/s. To calculate E , W, R, .

Since all the angles are given as data, this example can be solved by the graphical method. Figure 5.12 is drawn to scale. The required quantities can be measured. However, the analytic calculations are also given. In triangles ABC and ACE of Fig. 5.12, Vu1  AE  V1 cos 18o  0.951 V1 The blade velocity, U  260 m/s, is shown as AB. Therefore, BE  AE – AB  0.951 V1 – 260 CE  BE tan 50o  (0.951 V1 260)1.19 D

Vr2 V2 V1

C Vr1

40° F

199.2

Figure 5.12

A

18° b2 = 20° U = 260

b1 = 50° B

Velocity triangles for Example 5.3.

Also from triangle ACE Therefore

Vf1  CE  V1sin18o  0.309 V1 (0.951V1 260)1.19  0.309V1  1.13169V1 0.309V1  309.4

E

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Solving for inlet velocity,  V1  376 m/s So Whirl component is Vu1  AE  V1cos18o  357.6. Flow component is Vf1  CE  V1sin18o  116.2. Applying the same logic to triangles ABD and DBF (Vr2 cos20o 260) tan40o  Vr2 sin40o Solving for relative velocity at outlet  Vr2  488.66. So, from the triangle BDF, Vf2  DF  488.66 sin20o  167 Vu2  Vr2 cos20o 260  199.2 V2 

From the triangle ADF,

Vu2 sin 40o

 259.8

The speed ratio is U 260   0.69 V1 376 The specific work is W  U (Vu1 Vu2 )  260 (357.6  199.2)  144768 J/kg  144.8 kJ/kg The kinetic component is V12 V22 2



3762 259.82  36.94 kJ/kg 2

Therefore, the degree of reaction is W Kinetic component W 144.8 36.94   0.745 144.8

R

Utilization factor is W V2 W  2 2 144.8   0.81 (259.8)2 / 2 144.8  1000



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183

The axial thrust is m (Vf1 Vf2 )  1 s (116.2 167 )  50.8 N per unit mass flow rate This means 50.8 N of thrust (force) is in the direction opposite to that of Vf1.

EXAMPLE 5.4 In a Parsons stage, the steam inlet velocity is 200 m/s at an angle of 18o to the plane of the rotor. The rotor diameter is 0.75 m and the speed is 3000 rpm. Calculate the blade angles, specific work, and utilization factor.

Solution: Given V1  200 m/s, @1  18o, N  3000 rpm, D  0.75 m. The Parsons stage is of 50% reaction; therefore, the velocity triangles are symmetric: @1  A2; A1  @2. Blade velocity is U

P DN P s 0.75 s 3000   117.8 m/s 60 60

We have whirl component , Vu1  V1 cos @1  200 cos18o  190.2 m/s and, component, Vf1  V1 sin @1  200 sin18o  61.8 m/s Now ¨ Vf1 · B1  tan 1 © ¸ ªVu1 U ¹

Blade angle at inlet is

61.8 ¨ ·  tan 1 © ¸ ª (190.2 117.8) ¹  40.48o  A 2 Now, velocity triangle at inlet is drawn, Fig. 5.13, (ABC), and the outlet triangle (ABD) can be completed by symmetry. D

Vr2

C

V1 = 200

V2

Vr1

40.48° F

A

Figure 5.13

40.48° 18° 18° B (72.4) U = 117.8 Vu1 = 190.2

Velocity triangles for Example 5.4.

E

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Turbomachines

$Vu  Vu1 Vu2  190.2  (190.2 117.8)  262.6 Therefore specific work is W  U ( $Vu )  117.8 s 262.6  30934 J/kg  30.9 kJ/kg Now, in the triangle BCE, Vr1  BC  CE/sin 40.48o  61.8/sin40.48o; And because of the symmetry in 50% reaction turbines, we have V2  Vr1  61.8/sin 40.48o V2  95.2 m /s

or, The utilization factor is 

W 30934   0.872 V22 (95.2)2 30934  W  2 2

5.4.3 Curtis Stage For the purpose of analysis, a Curtis stage with three rows of moving blades (three rotors) is considered (Fig. 5.6). The pairs of velocity triangles, each pair for one row, can be drawn either separately or on a common base of U as shown in Fig. 5.14. There is no pressure drop in the flow across any row. The relative velocity Vr11 can be shifted geometrically to Vr12, as is done in Fig. 5.14 (BC1 to BD1). Similarly, the exit absolute velocity V12 of the first row can be shifted to the inlet absolute velocity of the second row V21, and the procedure can be continued. The geometrical transfer of velocities is shown by the arcs of circles with the arrows indicating the direction of transfer. For better understanding, the arrows are marked by letters. These letters denote the centers of the corresponding circular arcs. The following points may be noted when drawing the velocity triangles, as in Fig. 5.14: 1.

2. 3.

4. 5.

All the velocities and angles have two subscripts. The first subscript represents the order of the moving blade ring: “1” – the first row, “2” – the second row, and “3” – the third row. The second subscript represents the inlet “1” or outlet “2.” Thus, Vr21 is the inlet relative velocity of steam in the blades of the second row. The blades are all equiangular: A11  A12 or A31  A32. It is not necessary to have A11  A21. The blade geometries can be independent for different rows. It is usual to have (as far as possible) the constant axial components of the fluid velocities, individually for each row. The purpose of this is to have zero axial thrust on the rotor ring (or at least to reduce it as much as possible). It is not absolutely essential to have the same axial component (Vf ) for all rows together. It is possible, however, to have such a design among other designs. The last ring of blades in a stage has the absolute outlet fluid velocity (here V32) in the axial direction, so as to have maximum utilization, but here again this is only desirable, not essential. Except as above, any condition can be laid out regarding the values of any other angles.

To illustrate the above, three types of successive pairs of velocity triangles are drawn in Figs. 5.14–5.16. Each of the figures is an illustration with a different set of conditions; although all of them refer to the Curtis stage with three rings of rotor blades.

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185

In Fig. 5.14, the absolute velocity of steam at the inlet to each of the rotor blade ring is assumed to be at the same angle, that is, A11  A 21  A 31 In Fig. 5.15, the rotor blade angles at the inlet and outlet for all the three rings are assumed to be the same. This means that

B11  B12  B21  B22  B31  B32 In Fig. 5.16, the axial components of absolute steam velocities of all the three rings are assumed to be the same. The way Figs. 5.14–5.16 are explained and drawn indicates that the procedure of drawing velocity triangles is started with reference to the point of inlet of the steam at the first rotor ring, that is, triangle ABC1. For the purpose of illustration, the absolute values of angles and velocities are assumed to be known, but are not specified in the diagram; only the equalities of entities are specified (either Vr11  Vr12 or B11  B12 ) . In such cases, one cannot be sure that the last exit velocity, V32 or AD3, would be exactly at 90o to U or AB. The data of U, V11, @11, etc., being arbitrarily fixed, the final exit velocity (V32 or AD3) can be of any magnitude and at any angle. D3 Rotor C3 B

Stator D2 Rotor

A

B D1

A

F2

D1

C2 B C3

D3

Figure 5.14

Rotor

C1 Nozzles C1 V11 = AC1

D2

F1

C2 Stator

A

U B

E3

E2

E1

Velocity triangles for a three-row Curtis stage (@11  @21  @31). ABC1 and ABD1 are the inlet and outlet triangles for the first rotor ring blades. ABC2 and ABD2 are the inlet and outlet triangles for the second rotor ring blades. ABC3 and ABD3 are the inlet and outlet triangles for the third rotor ring blades.

The procedure, therefore, is a little modified. V32 or AD3, with a definite magnitude (a reasonable percentage of V11) and at 90o to U, is taken as the starting point. Then with the data of angles, one can proceed in the reverse direction (D3 to C3, C3 to D2, and so on). Actual field problems may require some trials to converge to a good solution.

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If, in a general problem, it is not absolutely essential to have the last exit velocity at 90o to U, the procedure can start with the inlet velocity triangle of the first ring, ABC1. Further, a Curtis stage with two rotor rings may be even easier than the illustrated case, but a Curtis stage with four rotor rings should not be more difficult than the three-rotor-ring stage. Table 5.1

Legend for the velocity triangles in Fig. 5.14

Processes 1.

2.

3.

4.

5.

Steam flow in the first rotor ring, from the inlet to outlet: The relative velocity remains the same. The blade angles at the inlet and outlet are the same. The axial components of velocity are the same. Steam flow in the stator blade: The absolute steam velocity remains the same from the outlet of the first ring to the inlet of the second ring. Steam flow in the second rotor ring, from the inlet to outlet: The relative velocity remains the same. The blade angles are the same. The axial components are the same.

Implementation in the Velocity Triangles

$Vu

Vr11  Vr12 , that is, BC1  BD1

E 1F 1

˜

˜

A11  A12, that is, C1 B E1  D1BF1 Vf11  Vf12 , that is, C1E1  D1F1 V12  V21, that is, AD1  AC2

Vr21  Vr22 , that is, BC2  BD2 ˜

˜

E 2F 2

A21  A22, that is, C 2 B E 2  D2 B F2 Vf21  Vf22, that is, C2E2  D2F2

Steam flow in the stator blade: The absolute steam velocity is the same, from the outlet of the second ring to the inlet of the third ring.

V22  V31, that is, AD2  AC3

Steam flow in the third rotor ring: The relative velocity remains the same. The blade angles are the same. The axial components are the same.

Vr31  Vr32 , that is, BC3  BD3 ˜

E3A ˜

A31  A32, that is, C 3 B E3  D3 B F3 Vf31  Vf32 , that is, C3E3  D3A

The total specific work for all the three rings of rotor blades is the sum of the specific works of each of the rings and is given as WT  W1  W2  W3  U (Vu11 Vu12 )  U (Vu21 Vu22 )  U (Vu31 Vu32 )  U [($Vu )1  ( $Vu )2  ( $Vu )3 ]

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187

B

A

B A

D1

C1

D2 B

C2

D3 C3 F1

Figure 5.15

A

F2

U B

E3

E2

E1

Velocity triangles for a three-row Curtis stage. (Blade angles are the same: A11  A12  A21  A22  A31  A32.) (The legend is the same as for Fig 5.14, Table 5.1).

B

A B A D2

D1

C3

C2

M

F1

Figure 5.16

B D3

F2

A

U B

E3

E2

C1

E1

Velocity triangles for a three-row Curtis stage. (Flow components are the same: Vf11  Vf12  Vf21  Vf22  Vf31  Vf32 ). (The legend is the same as for Fig 5.14, Table 5.1).

Some interesting results are obtained in the case of Fig. 5.16, in which the flow components of the absolute steam velocities in all the rotor rings are assumed to be the same. Consider a pair of velocity triangles, ABC3 and ABD3, for the third rotor blade ring. For the impulse wheel, the vertices of the triangles, C3 and D3, are required to be equidistant from point M, which is directly

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above B, refer Fig. 4.19 or Fig. 5.2. Because MD3 is equal to AB (i.e., U), C3D3  2U. Hence, the specific work of the third rotor blade ring is W3  U ( $Vu )3  U (2U )  2U 2 Again, D3 is the center of D2C3 because AD2  AC3. This results in D2D3  2U, and further D2M  3U  MC2, giving rise to C2D2  6U. Hence, the specific work of the second rotor blade ring is W2  U ( $Vu )2  U (6U )  6U 2 Proceeding in the same way, one gets C1D1  10U, and hence, the specific work of the first rotor blade ring is W1  U ( $Vu )1  U (10U )  10U 2 Thus, for a Curtis stage with three rotor wheels, the specific works are 10U2, 6U2, and 2U2, that is, in the ratio 10:6:2 or 5:3:1. If the number of rotor rings are 3, the specific works are (4i – 2)U2, i  1, 2, 3. This can be extrapolated for a four-wheel (or five-wheel) Curtis stage, the corresponding specific works being (4i – 2)U2, i  1, 2, 3, 4, that is, 14U2, 10U2, 6U2, and 2U2, (18U2, 14U2, 10U2, 6U2, and 2U2, when n  5). Once again Fig. 5.16 also shows that AE1  6U, and angle C1AE1 is @11. Hence, cosA11  F 

AE1 AC1



6U  6F V11

cos A11 6

This can also be written in the general form as

F

cos A11 2n

where n is the number of rotor blade rings in a Curtis stage. For maximum utilization, V11, U, and @11 are related through the number of rotor blade rings n in a stage, such that

F

cos A11 U  2n V11

(5.19)

5.4.3.1 Losses in a Curtis Stage The losses that occur during the flow of steam through the various passages of stator and rotor are due to the wall friction, viscous flow, turbulence, eddies, etc. and these losses depend on the surface finish of the blades, length of flow, viscosity of steam, etc. A blade friction factor cb can be used to account for all such losses together. In multi-stage impulse turbines, these blade friction factors can be easily employed while transferring the velocities as in Figs. 5.14–5.16 Figure 5.17 is a typical representation of Fig. 5.14, including the effect of the losses in the form of cb. Consider, for example, the shifting of Vr11 to Vr12. This is to represent the

Steam Turbines

189

flow of steam from the inlet of the first rotor blade ring to its outlet. This is BC1 to BD1. Now, if the losses are 3%, then the starting point of the arc is at 97% of BC1. In a similar way, when the steam flows in the stator blades, the shifting of V12 (AD1) to V21 (AC2) can be done, with an arc starting at 97% of AD1. The three situations of Figs. 5.14–5.16 are now modified, accounting for the losses and are shown in Figs. 5.17–5.19. It is to be noted that losses occur in all the five rows of blades, that is, three rows of the rotor blades and two rows of the stator blades. B

A C1 D1

B A

C2

D2 B

D3

F1 Figure 5.17

F2

A U

B

C3

E2

E3

E1

Velocity triangles for a three-row Curtis stage with losses (@11  @21  @31).

B

A

D1

C1 B

D2 A

C2 B

D3

C3 F1 Figure 5.18

F2

A

U B

E3

E2

E1

Velocity triangles for a three-row Curtis stage with losses (A11  A21  A31, A12  A22  A32).

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Turbomachines

B

A D1

A D2

F2

F1

Figure 5.19

D3

A

B

U

B

C3

E3

C2

B

E2

C1

E1

Velocity triangles for a three-row Curtis stage with losses. (Flow components are all equal.)

One simple observation in Figs. 5.17–5.19 is that the two conditions (a) A1  A2 and (b) zero axial thrust do not go together. When the losses were not considered earlier, these two conditions almost meant one another. The deviation between A1 and A2, to make the flow components equal, is very small. Alternately, when A1  A2, the flow components differ from one another by very little amount, still giving rise to some axial thrust. A good design is to aim at zero axial thrust. Simple trigonometric calculations are required to calculate A1 and A2 to find the complete solutions to the velocity triangles with zero axial thrust.

EXAMPLE 5.5 The velocity of steam at the entry to the first row of rotor blades is 300 m/s, in a two-row Curtis turbine. The first row has the blade angles of 25o each at the inlet and outlet. The second row has the blade angles of 30o each at the inlet and outlet. The blade coefficients of friction are 0.92 in each of the rotor blades and 0.95 in the intermediate stator blades. The exit velocity of steam from the second row is axial. Find the blade velocity, the specific work in each of the two rotor rows, and the axial thrust due to each row. Also determine the rotor efficiency.

Solution: Given V11  300 m/s, A11  A12  25o, A21  A22  30o, cb1  cb2  0.92, (cb)st 0.95, @22  90o. To calculate U, W1, W2, axial thrust, rotor efficiency. This example is solved by the graphical method. As the blade velocity is not known, the base of the velocity triangles may be arbitrarily taken as 2 cm, with the scale of the drawing to be decided later (Fig. 5.20).

Steam Turbines

(Δ Vu)1

X3

(Δ Vu)2

D1 Vf12

X1 D 2

(D1 F1)

Vf22 F1

C1

V11

A

X2 U B

Vf11

Vr11 C2

E2

(D2 A)

(C1 E1)

E1 Vf21 (C2 E2)

AC1 = 8.8 cm; E1F1 = 12 cm; E2A = 4.2 cm; C1E1 = 2.95 cm; D1F1 = 2.7 cm; D2E2 = 1.3 cm; D2A = 1.18 cm;

Figure 5.20

Velocity triangles for Example 5.5.

The stepwise procedure is as given below, starting from the exit velocity triangle of the second row: 1. 2.

Mark AB  2 cm on a horizontal line. This represents blade velocity, U. Draw BD2 at 30o to BA and AD2 at 90o to AB. ABD2 is the exit triangle of the second row. AD2 is V22 ; BD2 is Vr22 .

3.

Extend BD2 to BX1:BD2  BC2 x 0.92 ; BC2  BD2/0.92 BX1 

4. 5.

6. 7.

8. 9. 10.

BD2 0.92

Transfer BX1 to BC2 BC2 is at 30o to extended AB (BD2  BC2 s 0.92). Draw AC2 and extend to X2: AC 2 AX 2  0.95 ABC2 is the inlet triangle of the second row. Transfer AX2 to AD1, where BD1 is at 25o to BA. ABD1 is the exit triangle of the first row. Draw BD1 and extend to X3: BD1 BX 3  0.92 Transfer BX3 to BC1 at 25o to extended AB. Draw AC1. ABC1 is the inlet triangle of the first row. AC1 is the inlet velocity of steam to the first row. Measure AC1: AC1  8.8 cm; this must represent V11, that is V11  300 m/s Therefore, 8.8 cm represents 300 m/s; or, 1 cm represents (300/8.8)  34.1 m/s. Hence the scale of the drawing is given as: 1 cm  34.1 m /s Blade velocity  U  2 cm  68.2 m/s

191

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Turbomachines

Also ( $Vu )1  E1F1  12 cm  409.2 m /s Therefore, the specific work of the first row is U ( $Vu )1  68.2 s 409.2  27.907 kJ/kg Now ( $Vu )2  E 2 A  4.2 cm  143.22 m /s Therefore, the specific work of the second row is U ( $Vu )2  68.2 s 143.22  9.768 kJ/kg Axial thrust due to the first row is m (Vf11 Vf12 )  1(C1E1 D1F1 )  (2.95 2.7 ) s 34.1  8.525 N per (kg /s) Axial thrust due to the second row is m (Vf21 Vf22 )  1(C 2 E 2 D2 A )  (1.3 1.18) s 34.1  4.092 N per (kg /s) The rotor efficiency is Rotor efficiency 

(27.907  9.768)  83.72% 3002 (2 s 1000)

EXAMPLE 5.6 The mean diameter of a two-row rotor of a Curtis stage is 0.9 m and the speed is 1500 rpm. For the first row of the rotor blades, the blade inlet angle (A11) is 30o. For the stator blade, the inlet angle @12 is 40o. For the second row of the rotor, the blade inlet and outlet angles (A21, A22) are 20o each. The final discharge of steam from the second row is axial (@22  90o). For each row of the blades (rotors, stator), the friction coefficient is 0.96. Calculate the following: (a) Nozzle angle @11 of the first row. (b) Steam velocity V11 at the inlet to the first row. (c) Stator blade angle @21 at its outlet (entry to the second row). (d) Outlet blade angle A12 of the first row. (e) Total axial thrust. (f ) Total specific work. (g) Rotor efficiency.

Solution: Given D  0.9 m, N 1500 rpm, A11  30o, @12  40o, A21  A22  20o, @22  90o, cb  0.96.

Because the steam velocity is not known and because the final discharge of steam is required to be axial, the calculation or the construction of the velocity triangles is in the reverse direction, that is, the last triangle first. Both analytic calculations and procedure for drawing the velocity triangles are given in the following table with the left-hand side column showing the calculations and the right-hand side column showing the corresponding steps of drawing. The velocity triangles drawn to scale are shown in Fig. 5.21.

Steam Turbines

1.

2. 3.

4.

5.

6.

P s 0.9 s 1500  70.686 60

Mark AB  70.686 on a base line

Vf22  70.686 tan20o  25.7276

AD2 at 90o, BD2 at 20o with AB

U 

Vr 22 

70.686  75.2225 cos 20o

Vf22  V22  AD2; BD2 is

Vr21 

Vr22  78.357 0.96

Increase BD2 m BD2/0.96

Vf21  78.357 sin20o  26.81 BE2  78.357 cos20o  73.63 ˜

A 21  C 2 A E 2

193

Vr22

BC2 (BD2/0.96), at 20o at BE2 BE2  73.63 Angle C2AE2 by measurement

26.81 (70.686  73.63)  10.52o  tan 1

7.

8.

V21 

26.81  146.84 sin10.52o

AC2  147

V12 

V21  152.9 0.96

Increase AC2 m AC2/0.96  153

9.

Vf12  152.9 sin40o  98.3  D1F1

10.

AF1  152.9 cos40o  117.13 BF1  117.13  70.686  187.816

11.

12.

13.

14.

A12  tan 1

98.3  27.62o 187.816

AD1  153, AD1 at 40o

˜

D1 B F1  27.62o

Vr12 

187.816  211.98 cos 27.62o

BD1  212

Vr11 

Vr12 211.98   220.82 0.96 0.96

Increase BD1 m BD1/0.96

Vf11  220.82 sin30o  110.41

BC1  220.81, at 30o, C1E1  110.41 (Continued)

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(Continued) 15.

BE1  220.82 cos30o  191.23

16.

A11  C1 A E1

˜

˜

Angle C1 A E1  23o

110.41 ¥ ´  tan 1 ¦ § (70.686  191.23)µ¶  22.86o

17.

V11  AC1 

AC1 by measurement  284

110.41 sin 2.86o

 284.24

C1

V11

Vr12

D1

Vr11

V12

Vf11

Vf12 D2 V22

V21 Vr21

F1

40° A

Figure 5.21

(a) (b) (c) (c) (e) (f )

U

20°

B 30° 20°

C2 E2

E1

Velocity triangles for Example 5.6.

Nozzle angle at the first row  @11  22.86o Steam velocity at the inlet to the first row  V11  284.24 m/s Stator blade angle at its outlet or the angle of steam entry to the second row  @21  10.52o Outlet blade angle at the first row  A12  27.62o Total axial thrust  (110.41 – 98.3)  (26.8 – 25.7276)  13.18 N/(kg/s) Total specific work  U[($Vu)1  ($Vu)2]  70.686[(70.686  191.3  117.13) (70.686  73.63)]  37 kJ/kg

(g) Rotor efficiency 

37000  91.6% (284.24)2 / 2

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195

5.5 Mass Flow Rate and Blade Heights It has been mentioned, while explaining the different parameters of design in the successive rotor rings, that the blade heights have to be adjusted as per requirement in the different types of compounding. What is this requirement? Basically, there is the volume flow rate of steam through the turbine, associated with a specified mass flow rate, confirming to the equation Volume flow rate (m3/s)  Mass flow rate (kg/s) s Specific volume (m3/kg) It is known that as the steam expands from the high pressure to the low pressure, the specific volume of steam keeps on increasing. Therefore, the volume flow rate also keeps on increasing for the specified mass flow rate of the steam through the turbine. Now, the volume flow rate is also guided by the equation Volume flow rate (m3/s)  Flow velocity (m/s) s Flow area (m2) If the flow velocity is constant as the steam flows through the turbine passages, the area of flow has to keep on increasing so as to match with the increasing volume flow rate. If the flow velocity varies, even then the variation of area of flow has to keep pace, to correctly adjust with the volume flow rate. In steam turbines, the flow area is of annular shape, and is calculated by ODh, where OD is the circumference at the mean diameter D and h is the height of the blade (subject to the correction of the area of flow due to the blade thickness). It is in this context that the blade heights have to continuously increase, so that at any section in the turbine, the equation mv   V  P DhVf holds. As an example, let the steam be at 2 bar, 200oC at the entry to a stage in a turbine. Let the pressure be 0.75 bar at the exit of the stage. The isentropic expansion (i.e., the entropy is constant) of the steam shows that the exit temperature is 100oC. From the steam tables, at the entry (2 bar, 200oC), the specific volume is 1.08 m3/kg, and at the exit (0.75 bar, 100oC), the specific volume is 2.28 m3/kg. Also, from the known values of enthalpy drop between the two states and approach velocity to stage, the velocity triangles can be constructed, and the velocity of flow can be found. Let this flow velocity be, say, 25 m/s. Also let the mass flow rate be a nominal 5 kg/s and the mean diameter of rotor be 1 m. At the inlet to the stage

P DhVf  mv  mv  h  P DVf 5 s 1.08  = 6.88 cm P s 1 s 25 At the outlet from the stage mv  P DVf 5 s 2.28  = 14.51 cm P s 1 s 25

h

Thus, it can be seen how the expansion of steam to lower pressures and its increase of specific volume finally decide the increase in the blade height (Refer Example 5.9).

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Turbomachines

5.6 Efficiencies Right from Chapter 1, various efficiencies have been defined in different types of machines, in different processes, and with different viewpoints. Even a single process has been seen in different contexts and efficiencies have been defined accordingly. In the sections 5.2 to 5.5, in the analysis of steam turbines, different efficiencies were stated. These efficiencies are now again explained for the purpose of clarity, consolidation, and review.

5.6.1 Nozzle Efficiency Nozzles are ones that are present in the first stationary ring in a steam turbine, attached to the casing or stator. The steam from the steam chamber (or steam chest) first passes through these nozzles before meeting the first ring of rotor blades. Generally, it is assumed that in the steam chamber, steam has only negligible velocity. Hence, the enthalpy of steam at the entry to the nozzles may be considered either at the static state or at the stagnation state. As the steam expands in the nozzles, the pressure and enthalpy drop, and the velocity is generated. The nozzle efficiency (Gn) is stated as

Hn 

h01 h2 h01 h2a

The exit state (2a) is the isentropic exit state and state 2 is the actual exit state. (The initial condition is recognized in the stagnation condition.) Both these states are static states because the velocity is separately accounted as V2. Thus V2  2(h01 h2 )  2(h01 h2a ) s Hn  2( $his )Hn There are two types of nozzles: 1. 2.

Convergent nozzles: They have high efficiencies of the order of even 98%–99%. They are used for lower pressure drops and lower enthalpy drops, in reaction turbines, compounded turbines, etc. Convergent–divergent nozzles: They are employed for the generation of supersonic velocity for impulse stages. These nozzles have lower efficiency compared with convergent nozzles. The lower efficiency of convergent–divergent nozzles may be attributed to (a) higher velocity, hence higher friction losses and (b) the possibility of separation of flow.

5.6.2 Carryover Efficiency When compounding is arranged in steam turbines, the expansion of steam takes place in stator blade passages (in addition to that in the nozzles of the first ring). At the inlet to the stator blade passages, the inlet velocity is the exit velocity of the previous rotor blades, which cannot be neglected. However, due to the gap between the rotor and stator rings and due to the possible eddies, discontinuities, and a sort of “fluidshear” between the rings, the exit velocity of the previous rotor blades, V2 (in velocity triangles), is somewhat reduced when approaching the stator blades for further expansion. This reduction is quantified by employing the “carryover efficiency” (Gco) such that the approaching kinetic energy to the next set of blades is taken as Gco s V22/2. The value of carryover efficiency depends on the gap between the respective rings, the speed of rotation, the diameter, the eddies, etc., and it is of the order of 0.96–0.97.

Steam Turbines

197

5.6.3 Stator Efficiency or Blade Passage Efficiency As the expansion continues in the stator blades, the velocity again rises due to pressure drop and enthalpy drop. Just like the nozzle efficiency, a stator efficiency or blade passage efficiency (Gp) is given by the expression V12 2

 Hp ( $h )is 

HcoV22 2

where GcoV22/2 is the part due to the velocity at the exit of the previous blade ring. It is interesting to note that if the carryover efficiency were equal to unity, the blade passage efficiency Gp would be equivalent to the nozzle efficiency Gn. In the expression of V12/2 as above, the term that includes Gp is an independent term (i.e., with or without Gco). The value of Gp, therefore, is just like that of Gn. Because Gp refers to compounded turbines, the blade passages are only convergent (not convergent–divergent), and Gp has values of the order of 98%–99%.

5.6.4 Rotor Efficiency or Vane Efficiency The rotor efficiency is defined, in the simplest way, as the ratio of work output from the rotor to the energy available for conversion to work. In an impulse turbine, the rotor efficiency is expressed as

Hr 

W (V1 )2 / 2

because the rotor blades have the inlet energy of kinetic type only. In reaction turbines, the energy available for conversion includes two parts: The kinetic part (V12/2) and the part that could arise due to the pressure drop (or enthalpy drop) in the rotor. It is in this context that two separate expressions are there for rotor efficiency: One for an impulse turbine and the other for a reaction turbine.

5.6.5 Stage Efficiency In steam turbines, one stage comprises those components that have one pressure drop and at least one ring of rotor blades absorbing the energy. In impulse compounding, there are more than one rotor ring to absorb the energy from one pressure drop, as seen in Curtis stages. In reaction compounding, one stator ring followed by one rotor ring makes up one stage, as seen in Parsons stages. A stage efficiency is the ratio of the output of the stage to the input of the stage. In the case of impulse stage, the stage efficiency (Gst) is given by Gst  Gn s Gr. In the reaction stage, the expression for the stage efficiency is Gst  Gp s Gr.

EXAMPLE 5.7 Steam supply to the nozzles of an impulse turbine is at 4.0 bar, 200oC. The expansion of steam in the nozzles is down to 2.5 bar with a nozzle efficiency of 86%. The nozzles are inclined to the plane of the rotor at 20o. The mean diameter of the rotor is 1.3 m and the speed is 3000 rpm. (a) Calculate the speed ratio and the blade angle at the inlet. (b) Also determine the outlet blade angle for zero axial thrust when the friction coefficient of blades is 0.96. (c) Find the specific work, the rotor efficiency, and the stage efficiency.

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Turbomachines

Solution: Steam at the inlet: 4 bar, 200oC, outlet 2.5 bar. Isentropic efficiency  86%, @1  20o, D  1.3 m, N  3000 rpm. Enthalpy of steam at the inlet to the nozzle (from tables)  2860.4 kJ/kg. Under constant entropy, enthalpy at the outlet  2760 kJ/kg. Actual enthalpy drop (at 86% efficiency) is (2860.4 2760)0.86  86.34 kJ/kg Velocity triangles are shown in Fig.5.22, along with the analytical steps as follows: D V2

C

V1 = 415.55

Vr2

Vr1

b2

20°

A

F

U = 204.2

Figure 5.22

b1 E

B

Velocity triangles for Example 5.7.

Steam velocityV1, is given by V1  2 s $h  2 s 86.34  415.55 m/s Blade velocity is U

P DN P s 1.3 s 3000   204.2 m/s 60 60

(a) Speed ratio is U 204.2   0.49 V1 415.55 In the triangle ACE, CE  Vf1  AC sin 20  V1 sin 20  415.55 sin 20  142.13 And, AE  Vu1  AC cos 20  V1 cos 20  415.55 cos 20  390.5 Inlet blade angle,

B1  tan 1

Vf1 (Vu1 U )

¨ 142.13 ·  tan 1 © ª 390.5 204.2 ¸¹  37.34o ´ ¥ Vf (b) Relative velocity at outlet, for zero axial thrust, Vr2  0.96 Vr1  0.96 ¦ § sin37.34o µ¶  225 m/s

Steam Turbines

¥V ´ ¥ 142.13 ´ B2  sin 1 ¦ f2 µ  sin 1 ¦ § 225 µ¶ § Vr2 ¶  39.17o

Outlet blade angle,

(c) Whirl velocity at outlet, Specific work

199

Vu2  U Vr2 cos B2  204.2 225 cos39.17o  29.76 W  U (Vu1 Vu2 )  204.2(390.5 29.76)  73.66 kJ/kg

Now Inlet energy to the rotor  Kinetic energy V12 2



( 415.55)2  86.34 kJ / kg 2

Therefore Rotor efficiency 

Output 73.66   85.3% Input 86.34

Stage efficiency is

Hr s Hn  0.8531 s 0.86  73.37%

EXAMPLE 5.8 The nozzles are inclined at 18o to the plane of the rotor in a two-row Curtis stage. The outlet angles of the rotor blades are 25o and 30o in the first and second rows, respectively. The stator blades (between the two rotor blade rings) have the outlet angle of 20o. The friction coefficients of the blades in all the three rows are each 0.92. The rotor speed is 3000 rpm. The mean diameter of the three rings of blades is 70 cm. The final discharge velocity of the steam is axial. Based on a nominal flow rate of steam at 1 kg/s, find the following: (a) Initial velocity of steam from the nozzles. (b) Power output from each row of the rotor. (c) Rotor efficiency. (d) Power lost in friction. (e) Axial thrust.

Solution: Given @11 18o; A12  25o; A22  30o; @21  20o; cb  0.92 for all rows. Also N  3000 rpm,

D  0.7 m; @22  90o, m  1 kg/s. Because the final discharge is required to be at @22  90o with blade velocity, and the initial velocity of steam is required to be found out, the construction of the velocity triangles has to be in the “reverse” direction i.e., the last triangle first).

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Turbomachines

It is also helpful to sketch a rough, free-hand velocity triangle (Fig. 5.23) set initially, so as to trace the triangles in the reverse direction.

D1

Vr12

C1AB

a11 = 18°

C2AB

a21 = 20°

D1BA

b12 = 25°

D2BA

b22 = 30°

V12 Vf12

D2 V22

F1

Vr21

Vf22

A

Figure 5.23

C2

V21

Vr22

C1 V11

U B

Vr11 Vf11

Vf21

E2

E1

Velocity triangles for Example 5.8.

A stepwise procedure for the construction of the velocity triangles is as follows: 1.

We have the blade velocity, U, as

P DN P s 0.7 s 3000   110 m/s. 60 60 Take a base line of enough length and mark AB  U  110 m/s (Adopt a scale). Draw AD2, BD2: AD2 at 90o to AB and BD2 at A22  30o to BA. This is the exit velocity triangle, at the outlet of the second row of rotor. By measurement, AD2  V22  63.5 m/s and BD2  Vr  127 m/s. (For analytical solution, we 22 have V22  AD2  AB tan A22  110 tan 30o  63.5; and Vr22  AB/cos A22  110/cos30o  127) Now, we have Vr22  cb sVr21, where cb is 0.92. Therefore, U

2. 3. 4. 5.

Vr22  Vr21 s 0.92; BD2  Vr22  BC 2 (  Vr21 ) s 0.92.

6.

7.

Draw BC2  (BD2/0.92)  (127/0.92)  138, to intersect a line AC2 at @21  20o with AB. (Arc with center B and radius 138, and line from A at 20o, intersecting at C2) By measurement, AC2  234 m/s  V21. (For analytical solution, we have “cos rule”, applied to triangle ABC2, where we can write: BC22  AB2  AC22 – 2(AB)(AC2)cos @21 , and this results in a quadratic equation in AC2. Solving this, we get AC2  234 m/s.) We have V21  cb V12. And cb  0.92. Therefore, V12 

V21 0.92

That is AD1  V12  8.

AC 2 (V21 ) 0.92



234  254.4 0.92

Draw arc with center A, radius 254.4, and line from B at 25o to BA, to locate D1; draw AD1 and BD1. By measurement, BD1  Vr12  348.8. (Analytical solution is just like step 6 above, triangle is ABD1). BD1  348.8 m/s. Analytic value is quoted here.)

Steam Turbines

9. 10. 11. 12.

Vr12

201

BD1

348.8   379.1  BC1 0.92 0.92 0.92 Draw arc with center B, radius 379.1, and line from A at 18o to AB, to locate C1; draw AC1and BC1. By measurement, AC1  V11  480 m/s (This value is by analytic solution, as in step 6 above.). Draw perpendiculars from C1 to E1, C2 to E2, and D1 to F1 to the base line. By measurement, (Analytical values are: ($Vu)1  Vu11  Vu12  664; ($Vu)2  Vu21  Vu22  220) We have Vr12  cb Vr11; hence, Vr11 



E1F1  ( $Vu )1  664 E 2 A  ( $Vu )2  220 Now, although both the construction of velocity triangles and analytical solutions have been stated above (steps 1 to 12), it is not essential to have both procedures; any one procedure is sufficient. (a) Initial velocity of steam, V11  (AC1)  480 m/s. (b) Power output from the rings: Specific work of the first rotor ring i W1  U ( $Vu )1  110 s 664  73040 J/kg Power output, of the first rotor ring: 73.04 kW/ (kg/s) Specific work of the second rotor ring is W2  U ( $Vu )2  110 s 220  24200 J/kg Power output, of the second rotor ring: 24.2 kW/ (kg/s) Total work is W1  W2  97.24 kJ/kg (c) Rotor efficiency: The input kinetic energy is V112 2



4802  115.2 kJ / kg 2

Therefore the rotor efficiency is 97.24  84.4% 115.2 (d) Power lost in friction: Now power lost in the first rotor ring is U (Vr11 Vr12 )  110(379.1 348.8)  3333 W /( kg / s ) Power lost in the second rotor ring is U (Vr21 Vr22 )  110(138 127 )  1210 W /( kg / s ) Therefore, total power lost in friction is 3333 1210  4543 W/(kg/s) Also (e) Axial thrust  m [(Vf11 Vf12 )  (Vf21 Vf22 )]

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Turbomachines

 1;(148.33 147.4 )  (80.03 63.5)=  17.43 N /( kg /s ) Comments: 1. 2. 3. 4.

Although the graphical method (drawing velocity triangles and taking measurements) takes much less time, the accuracy is a little less, but the results can be accepted. Instead of the initial steam velocity as part of the data, the data can be in the form of enthalpy drop across the nozzles, with a nozzle efficiency [V  2( $h )Hn ]. Data can also be in the form of initial state of the steam at the entry to nozzles and pressure at the exit from nozzles. The enthalpy drop can be taken from the steam tables or Mollier charts. The problem can also be extended to find blade heights.

EXAMPLE 5.9 Steam expands from 2 bar, 200oC to 1.1 bar in the nozzles of efficiency 0.96. These nozzles are inclined at an angle of 18o to the rotor plane of an impulse turbine, running at 3000 rpm. The mean diameter of the blade assembly is 70 cm. The outlet blade angle is 25o and blade friction coefficient is 0.92. Calculate (a) the blade inlet angle, (b) specific work, (c) blade height, (d) axial thrust, and (e) exit velocity of steam. Take the mass flow rate of steam as 10 kg/s.

Solution: Given Initial state of steam: 2 bar, 200oC Pressure at nozzle exit: 1.1 bar Nozzle efficiency: 0.96 N  3000 rpm, @1  18o, D  70 cm, cb  0.92, A2  25o, m  10 kg/s. To calculate A1, blade height, W, axial thrust. At initial state, 2 bar, 200oC, we have Enthalpy of steam (from steam tables)  2870.5 kJ/kg Entropy  7.507 kJ/kg K At the same entropy (7.507), exit pressure 1.1 bar, we have Enthalpy  2750.5 kJ/kg Therefore Isentropic enthalpy drop  ($h)is  2870.5 – 2750.5  120 kJ/kg At 0.96 efficiency, the actual enthalpy drop is ( $h )act  120 s 0.96  115.2 kJ/kg The velocity of steam, at the exit of nozzles or inlet to blades, is V1  2( $h )act  2 s 115.2 s 1000  480 m / s. (The velocity triangles of the first rotor blade ring of the Example 5.8 may be referred.). The whirl velocity, Vu1  V1 cos 18  480 cos 18  456.5 m/s The flow velocity, Vf1  V1 sin 18  480 sin 18  148.33 m/s

Steam Turbines

The blade velocity,

U

203

P DN P s 0.7 s 3000   110 m / s 60 60

¥ Vf1 ´ (a) The blade inlet angle from the inlet triangle, B1  tan 1 ¦ µ § (Vu1 U )¶ ¨ 148.33 ·  23.17o  tan 1 © ª 456.6 110 ¸¹ Relative velocity at inlet from the inlet triangle, Vr1 

Vf1 sin B1



148.33  377 m / s sin 23.17o

Therefore, outlet relative velocity, Vr2  Vr1 s 0.92  347 m/s From the triangle, Vu2  Vr2 cos A2 – U  347 cos25o – 110  204.5 (b) Specific work, W  U(Vu1 Vu2)  110[ 456.5 ( 204.5)]  72710 J/kg  72.71 kJ/kg (c) Blade height: The specific volume of steam at the exit of nozzles  1.72 m3/kg (from tables). For 10 kg/s, the volume flow rate is mv   Area of flow s Flow velocity  17.2 m3 /s  A s Vf 1  Area of flow 

17.2 m 2  1160 cm 2 148.33

This area is PDh (annular area, with mean diameter D and blade height h). Therefore h

Blade height is: (d) Axial thrust: Now, So

1160 1160   5.3 cm P D P s 70 Vf2  Vr2 sin 25o  146.65

Axial thrust  m (Vf1 Vf2 )  10(148.33 146.65)  16.8 N

(e) Exit velocity:

V2 

Vf22  Vu22  146.52  204.52  251.6 m/s.

Comment: It may be noted that this example is one part of the previous example, with only the first rotor blade ring and the data are common for blade angles, rotor geometry, etc. It may also be noted how graphical and analytical solutions differ from each other.

5.7 Reheat Factor The reheat factor was discussed in Chapter 2 for multi-stage expansion processes. The reheat factor is applicable to pressure-compounded turbines with two or more stages of pressure drops.

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Turbomachines

The reheat factor of multi-stage steam turbines is defined as the ratio of the sum of individual isentropic enthalpy drops or isentropic works of each stage ¤ Wsti to the single-stage isentropic enthalpy drop or i

isentropic work Wss, if the expansion of the steam were to take place between the same initial state and final pressure in a single stage. p1

h

p2 1 2

p3

2

p4

3 3

3

4 pm  1

4

4

m

pm

m

m (m
1)

m
1 (m
1)

Figure 5.24

s

Expansion in m stages.

With reference to Fig. 5.24

¤Wsti  Wst1  Wst2  !  Wstm i

 (h1 h2a )  (h2 h3a )  !  (hm h(m 1)a ) Wss  h1 h(m 1)aa The sum of individual stage isentropic enthalpy drops, (h1 h2a ), (h2 h3a ), …, (hm h(m 1)a ), is higher than the isentropic enthalpy drop between the same initial and final pressures, during the one-stretch expansion, due to the fact that the constant pressure lines in an h–s chart are divergent. Also, the increased temperatures at the ends of any individual stage expansion processes, which are at the beginning of the expansion processes of the next stages, indicate the higher availability of steam for the next stages. Thus, the work lost due to fluid friction in one stage is made up in the next stage. This continues throughout the stages. But only in the last stage, the frictional loss is not recovered because of the absence of further stages. Overall, the whole system has an increased output. Thus

¤Wsti RF 

i

Wss

; (RF  1)

(5.20)

The usual values of reheat factor are of the order of 1.03, 1.05, etc., signifying 3–5% increase in output due to the multi-staging. The reheat factor depends on the initial state of the steam (pressure and temperature), the pressure ratio of expansion of each stage, the isentropic efficiency of each stage, and the number of stages in a turbine. Although multi-staging is adopted as a means of reducing the velocity of steam in the turbine passages and the speed of the turbine, the reheat effect is an unintentional advantage of multi-staging.

Steam Turbines

205

EXAMPLE 5.10 In a section of a turbine, there are four stages of the Rateau compounded types. The steam at the inlet to the first stator blades of this section is at 1 bar, 200oC. The pressure of the steam at the exit of the fourth row of the rotor blades is 0.3 bar. The overall isentropic expansion efficiency is 0.85. The actual expansion process may be represented by a straight line on an enthalpy–entropy chart. The steam velocity at the entry to each ring of the rotor blades is 380 m/s at an entry angle of 20o to the rotor plane. The exit blade angle is 23o. The enthalpy drops, work produced, and blade angles are the same in all the rotor rings. The mean diameter of the rotor is 1.2 m and the design speed is 3000 rpm. The coefficient of blade cb is 0.96. The flow rate of steam is 20 kg/s. Draw the velocity triangles and calculate the following: (a) Pressure at the entry to each rotor blade ring. (b) Specific work of a rotor ring. (c) Total power of the four stages. (d) Blade heights of successive rotor rings. (e) Axial thrust. (f ) Reheat Factor.

Solution: Four-stage Rateau turbine: Steam: 1 bar, 200oC to 0.3 bar, Gis  0.85, m  20 kg/s V1  380 m/s, @1 20o, A2  23o, D  1.2 m, N  3000 rpm, cb  0.96. The state of steam at the inlet and exit are located on an h–s chart and the values obtained are recorded. The same is shown in Fig. 5.25 (a). The procedure of locating the points and the details are explained as follows: 1.

2. 3. 4. 6. 7. 8. 9. 10.

On the Mollier chart (h-s chart for steam), locate initial state-point of steam, at pressure of 1 bar, temperature of 200oC, mark it as point 1, in Fig.5.25 (a). Record the enthalpy at this point, 2875 kJ/kg as shown. Locate point, 5q, vertically below point 1, (constant entropy), on the constant pressure line of 0.3 bar. Record the enthalpy at this point, 2655 kJ/kg as shown. Find the enthalpy drop; that is, (2875 – 2655)  220 kJ/kg. This is isentropic drop of enthalpy during expansion. Actual enthalpy drop is given by 220 s Gis ( 0.85)  187 kJ/kg. With this drop, that is, 187 kJ/ kg, locate point 5 on 0.3 bar pressure line to represent the actual exit point. Draw a straight line, joining points 1 and 5. This represents the actual expansion of steam in the stages. The line 1–5 is now divided equally into 4 parts, and the points are marked 1-2-3-4-5 (inclusive of 1 and 5). Values of pressures, enthalpies and specific volumes are marked on the figure as shown. The pressures at the entry points to successive rotors are shown as 1.0, 0.84, 0.67, 0.49 and 0.3 bars, respectively, starting from the first rotor of the segment of the turbine as specified in the example. The actual enthalpy drop in a stage is 46.75 kJ/kg, as shown. The corresponding velocity generated is given by (2 s 46750) which is 305.78 m/s. (This is in order. The inlet velocity to each rotor is given as 380 m/s; it is inclusive of the effect of the exit velocity of the “previous” stage.) With this background, the solution of the example is continued: (a) The pressures at the entry of successive rotors are: 1.0, 0.84, 0.67, 0.49, and 0.3 bars. (b) Specific work of a rotor ring: The whirl component at inlet is Vu1  V1 cos A1  380 cos 20  357 m/s The flow component at inlet is

Vf1  V1 sin A1  380 sin 20  130 m / s

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Turbomachines

P DN P s 1.2 s 3000   188.5 m/s 60 60 In inlet velocity triangle ABC, along with ACE, the segment BE  AE – AB; that is, BE  357 – 188.5  168.5 130 Inlet blade angle is B1  tan 1  37.65o 168.5 168.5 Vr1   212.82 cos 37.65o Vr2  212.82 s 0.96  204.3 V2  Vr2 – sin23  79.8 m /s Vr2 – cos23  188.06 U

The blade velocity is

This indicates that V2 is almost perpendicular to the blade velocity U and that Vu2  0. 2.57 m3/kg

Enthalpy kJ kg 1.0 bar

2875

2

2828.25

3.14

1 2

0.84

3 3

187

4.65

2781.50 0.67

46.75

3.81

5.714 m3/kg

4

220

2734.75 4

0.49 2688

5

0.3 bar 5

2655

5

Entropy

1  5
Single stage isentropic process 1  2, 2  3, 3  4, 4  5; Stagewise isentropic processes 1-2-3-4-5; Actual process Full lines; Constant pressure lines Broken lines; Constant specific volume lines (a) C

V1 = 380 D

Vr1

20°

A

Vf1 = 130

23°

U = 188.5

B

E

Vu1 = 357

(b)

Figure 5.25

(a) Extract of h–s chart. (b) Velocity triangles at inlet and outlet of any one stage for Example 5.10.

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207

Now Vf2  Vr2 sin 23  79.8 m / s Therefore, the specific work is W  U (Vu1 Vu2 )  188.5 s 357  67.295 kJ/kg (c) Total power from all four stages is (67.295 s 20) s 4  5383.6 kW (d) Heights of blades on successive rotor rings: The specific volumes of the steam, after expansion in each of the stator rings, are obtained from the Mollier diagram (h–s chart): Refer Fig. 5.25 (a): Exit of the first rotor: 3.14 m3/kg  v1 Second rotor: 3.81 m3/kg  v2 Third rotor: 4.65 m3/kg  v3 Fourth rotor: 5.714 m3/kg  v4 Let H1, H2, H3 and H4 be the heights of the blades from the first rotor to the last rotor, respectively. Then (P DH1 ) s Vf1  m s v (mv  1)  H1   0.128 m  12.8 cm P DVf1 Similarly H2  H3 

(mv  2)

P Dvf1 (mv  3)

 0.1553 m  15.53 cm

 0.1895 m  18..95 cm P DVf1 (mv  4) H4   0.2328 m  23.29 cm P DVf1 (e) Axial thrust for each row is Axial thrust  m (Vf1 Vf2 )  20(130 79.8)  1004 N per ring  4016 N for four rotor rings (f ) The reheat factor is Reheat factor 

¤( $h )is ( $h )total



55  56  57  58  1.03 220

Comments: 1.

As mentioned in the example itself, this turbine is a part of a bigger turbine (or a bigger “project”). The overall enthalpy drop has to be from a much higher pressure and temperature, expansion taking place down to a much lower pressure. A small segment is chosen as an example to illustrate the design activity (to some extent).

208

2.

3. 4.

5.

6.

Turbomachines

Steam tables and Mollier charts are used to read the values of specific enthalpy and specific volume. The accuracy is subject to verification, especially because of the interpolation required to locate the exact values. However, the results obtained amply indicate the nature of variation of the parameters. In the text, it has been mentioned that the very same velocity triangles hold good for a series of rotor rings, provided the enthalpy drops are maintained the same for the stages. This example supports the same. A look at the values of V1 and V2 throws some more light on the subject: V2 is the absolute velocity of steam as it leaves the rotor blades and enters the next stator blades that act as nozzles. Hence, V2 is the “approach velocity” for these nozzles. With $h as the enthalpy drop, the velocity generated (to enter the next set of rotor blades) is 2( $h )  V22 , including the effect of the approach velocity. In the design process, this exercise involves the balancing of the values. The axial thrust is very much present because Vf1 x Vf2 . In a full size turbine, this can be balanced by splitting the steam flow into two streams, flowing in the opposite direction, with a common shaft. (Even with a single stream, 50% of turbine can be arranged in opposite directions.) The present example is with Rateau compounding stages. A similar procedure can be adopted with Parsons stages with R  0.5.

5.8 Governing of Steam Turbines In thermal power plants, the generators are driven by steam turbines. The generators are required to be run at a constant speed, in order to supply the electrical energy at a constant frequency of operation. Therefore, the turbines also have to run at a constant speed, even with varying loads on the generator. The function of the governor is to maintain the constant speed by controlling the input energy to the turbine so as to match the same with the required output. The steam turbine has its input energy in the form of mass flow rate of steam multiplied by the enthalpy drop of steam per unit mass. To vary the input, therefore, either the mass flow rate or the enthalpy drop can be controlled. The two methods are explained in the following sections.

5.8.1 Governing by Nozzle Control The steam inlet to the rotor blades is through the first series of nozzles that are distributed over the annular space. During the running of the turbine on full load, all the nozzles are open, and the torque generated on the rotor is uniform over the entire periphery. When the load on the generator decreases, if all the nozzles are continued to be open, the input energy exceeds the output energy requirements, and the turbine and generator tend to increase the speed. The flyballs of a governor (connected to the main shaft) get raised due to the higher speed. The upward movement of the sleeve of the governor is converted to affect the closure of some of the nozzles. Generally, the closure is by sets of segmental plates so that pairs of the opposite nozzles are closed. In nozzle control governing, therefore, the flow rate of steam is controlled. The inlet (boiler) pressure and outlet (condenser) pressure are not altered. Enthalpy drop per kg of steam is also not altered. Along with the nozzle control, it is required that the steam generation itself has to be controlled. However, these controls are associated with the boilers.

5.8.2 Governing by Throttle Control In this method of governing the turbines, the steam flow to the turbine is through a throttle valve. The throttle valve is operated by the governor through some links as shown in Fig. 5.26 (a).

Steam Turbines

209

Governor

Levers Drive from main shaft Turbine Steam supply

Drive to generator Throttle valve

Exhaust steam (a)

A

h

B

C

D

ly pp Su ssure pre

e

sur

res

tp aus

Exh

s (b)

Figure 5.26

(a) Governing of steam turbines by throttle control. (b) Drop in enthalpy gets reduced due to throttling process A-B-C-D (Drop in enthalpy at D is less than that at C).

The process of throttling is shown in the enthalpy–entropy diagram in Fig. 5.26 (b). The state of steam at supply is shown as A. During the throttling process, the enthalpy remains constant. The pressure gets reduced to lower values. These states are shown as B, C, D, etc. Because the condenser pressure does not vary, the enthalpy drop keeps on reducing as shown. Hence, when speed of the turbine increases due to drop in load, the throttle control comes into effect, the energy input to the turbine is reduced and the speed is brought back to the rated speed.

5.9 Performance Characteristics of Steam Turbines Generally, the performance characteristic of any machine is a plot of some performance parameter on the base of either percentage load on the machine or any of the various design parameters of the machine. Thus, 1. 2.

The y-axis may represent efficiency, power output, head created, thrust generated, etc. The x-axis may represent percentage load, capacity flow rate, specific speed, etc.

210

Turbomachines

In the case of steam turbines, there is a certain peculiarity. As an example, in IC engines, every kg of fuel spent can represent a definite amount of energy fed to the machine. But in steam turbines, the input energy to the machine, per kg of steam flow, depends on a whole range of pressures and temperatures of that steam, expanding up to the condenser pressure. Considering the schematic enthalpy–entropy chart shown in Fig. 5.27, the steam condition at the inlet can be anywhere in the super-heated region (any pressure, any temperature) and the steam condition at the outlet can be up to the condenser pressure that can also vary. With this background, it is necessary to specify the conditions of steam at the inlet and outlet for the curves of efficiency or steam consumption.

Superheated region

Enthalpy

Condenser pressure Dry saturated line

Entropy

Figure 5.27

Schematic enthalpy–entropy chart.

B A

Efficiency

C

20%

50%

Steam Pressure = PA = Steam Temperature = TA = Condenser Pressure = Speed of the Turbine = (PB < PA < PC)

100% Load

Figure 5.28

Efficiency versus load.

In the light of the above facts, the characteristic of efficiency versus load for a steam turbine can be discussed, as referred to Fig. 5.28. Suppose that the steam supply is at a pressure PA and temperature TA at 100% load. Since this is the design point, all the parameters such as flow areas, flow angles, etc. are as per the design values, the losses are the least and the efficiency is the maximum. If the load varies from the design value to any lower value such as 80%, 60%, etc., the flow quantity gets reduced due to the governing process, but the velocity triangles and angles do not vary. Hence, the efficiency remains almost constant over a large portion of the varying load. However, if the load is too low, the efficiency naturally decreases. This is shown in Fig. 5.28 by a line passing through A. The entire line represents one inlet steam pressure.

Steam Turbines

211

Suppose that the supply pressure is at a lower value than the pressure at A. The enthalpy drop per kg of steam is also less, but the governor adjusts the flow rate to be higher. The net effect is that the losses are comparatively less and hence the efficiency increases marginally. At part-loads also, that is, as the load on the turbine decreases below the rated load, the efficiency is maintained at the same trend. This is shown by the line through B, with PB  PA. Suppose that the supply pressure is increased. The tendency is now in the opposite direction, the decreased efficiency line passing through C and PC  PA. Another representation of the performance is a plot of steam consumption versus load. This is shown in Fig. 5.29. The entire line, showing the consumption, is at constant supply pressure. The trend of variation is almost the same, with the consumption (kg/h) being lower at higher pressures due to the higher enthalpy drops per kg of steam.

Steam Consumption kg/hr

Steam Pressure = PA = Steam Temperature = TA = Condenser Pressure = Speed = PA > PB

PB PA

20%

100%

50% Load

Figure 5.29

Steam consumption versus load.

Steam Pressure = Steam Temperature = Condenser Pressure = Speed =

SSC kg/kWh

20%

50%

100%

Load

Figure 5.30

Specific steam consumption versus load.

Yet another characteristic is the steam consumption in terms of “specific steam consumption (SSC; kg/ kWh),” as shown in Fig. 5.30. This specific steam consumption is the lowest at full load. As in other situation, here also the line shifts to lower levels as the supply pressure increases.

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Turbomachines

K eywords Impulse steam turbine Reaction steam turbine Single-stage turbine Multistage turbine Compounding of steam turbine Pressure compounding Velocity compounding

De Laval turbine Parsons turbine Rateau turbine Stage efficiency Blade efficiency Maximum utilization factor Reheat factor

S ummary Steam turbines are power-producing turbomachines in thermal power plants and nuclear power plants. In this chapter, 1. Steam turbines are classified and identified as impulse turbines or reaction turbines. Impulse turbine: Expansion of steam is only in nozzles, no expansion in rotor blades, Vr1  Vr2, W has only kinetic component, no reaction component. Reaction turbine: Expansion of steam is in both nozzles and rotor blades, W has both kinetic and Reaction components; Degree of reaction is the fraction of reaction component in total W. 2. The designs are also classified as De Laval turbines, Parsons turbines, and Rateau turbines. Turbines are also single-stage or multistage units. 3. De Laval turbine: Impulse machine. One stator ring of nozzles and one rotor ring of impulse blades 4. Parsons turbine: Reaction is 50%; half expansion in stator blades and half expansion in rotor blades. 5. Rateau turbine: Basically, this is impulse turbine, in multiple stages. Similar to De Laval

6.

7.

8.

9. 10.

turbine in-series, expansion of steam in many stages; range of pressure drop in one stage is small, unlike De Laval turbine. The purpose of the compounding of steam turbines is clearly understood as a means of avoiding the disadvantages of single-stage turbines. Disadvantages of single stage turbine: Too high velocities of steam, higher losses, low efficiency, too high rotor speeds, higher stresses, costly materials. All these are avoided by multistaging. After studying the basic physical arrangements, the different types are analyzed with respect to the blade efficiency, stage efficiency, and maximum utilization factors. A reheat factor of pressure-compounded steam turbines is studied in detail. RF is ratio of sum of individual isentropic works to single-stage isentropic work. RF is about 1.03, 1.04, etc. Applicable to only multistage turbines (whether impulse, reaction or combination). Steam turbines are the important prime movers in the power sector.

I mportant E quations 1. Impulse turbine rotor efficiency Hr max  cos2 A1

2. Maximum utilization factor max  cos2A1

Steam Turbines

3. Maximum output 

Wmax  U ( $Vu )  2U 2

(V12 HcoV22 )

V12 / 2 ( $h )stage

5. Stage efficiency



4F(1 R )Hn [cos A1 F(1 R )] ¨ ¥ V22 ´ ·

H 1 © co ¦ 2 µ ¸ § V1 ¶ ¸¹ ©ª

Hstage  Hr s Hn 6. Reaction turbine output 14. Hst 

W  U ( $Vu )  U 2 7. Maximum utilization factor for R  0.5 m 

15. Hst 

cos2 A1 1 0.5 sin 2 A1

8. For reaction turbines

Hstage 

W [( $h )stator  ( $h )rotor ]

9. Effect of carryover efficiency V12 2

 Hn ( $h )is  Hco

10. ( $h )is, stator  11. ( $h )is, rotor 

2(1 R )Hn

13. Stage efficiency W Hst  ( $h )is, stage

4. Nozzle efficiency

Hn 

213

V22 2

(V12 HcoV22 )

16. Hr 

17. Hr 

4FHn (1 R )[cos A1 F(1 R )] 1 Hco [1 4(1 R )F{cos A1 F(1 R )}] 4Hn R (1 R )cos2 A1 1 Hco [1 4(1 R )R cos 2 A1 ] (for  max ) 4R (1 R )cos2 A1 1 Hco [1 4R (1 R )cos2 A1 ] (Rotor efficiency ) cos2 A1 1 Hco sin 2 A1

18. Speed ratio for n rows

F

2Hn R (V12 HcoV22 )

cos A11 U  2n V11

19. Reheat factor

¤Wsti

2Hn (1 R )

12. ( $h )is, stage  ( $h )is, stator  ( $h )is, rotor

(for R  0.5)

RF 

i

Wss

M ultiple- C hoice Q uestions 1. The De Laval turbine is a (a) reaction-type steam turbine (b) multistage impulse steam turbine (c) single-stage impulse steam turbine (d) none of these 2. The Rateau turbine is a (a) reaction steam turbine (b) impulse steam turbine

(c) three-stage steam turbine (d) 50% reaction steam turbine 3. The Parsons steam turbine is a (a) 50% reaction turbine (b) 50% impulse turbine (c) three-stage impulse turbine (d) none of these

214

Turbomachines

4. The Curtis turbine is a (a) 50% reaction turbine (b) 50% impulse turbine (c) impulse-type turbine (d) none of these 5. In a single-stage impulse turbine with no axial thrust and friction, (b) V1  V2 (a) Vr1  Vr2 (c) @1  @2 (d) A2  A1 6. In a single-stage reaction turbine with no axial thrust and friction, (a) Vr1  Vr2 (b) V1  V2 (d) none of these (c) A1  A2 7. In a 50% reaction turbine with no axial thrust and friction, (a) @1  @2 (b) A2  A1 (d) Vr1  Vr2 (c) @1  A2 8. The following statement, where cb is a friction factor, represents accounting for the frictional losses in the rotor blades of a steam turbine. (a) Vr21  c bVr12 (b) Vr22  c bVr21 (c) Vr32  c bVr23 (d) V22  c bV21

9. The degree of reaction in a Rateau stage is given by (a) R  1 (b) R  0.25 (c) R  0.75 (d) none of these 10. The purpose of multi-staging in steam turbines is (a) to run the turbine at a constant speed (b) to run the turbine with a reheat factor more than 1 (c) to run the turbine at a low speed (d) to run the turbine with no axial thrust 11. The reheat factor of a multi-stage steam turbine does not depend on (a) the pressure ratio of each stage (b) the efficiency of each stage (c) the degree of reaction of each stage (d) none of these 12. The rotor efficiency of a stage does not depend on (a) the pressure ratio of that stage (b) the state of the supply steam at that stage (c) the speed of the rotor of that stage (d) the surface finish of the blades in that stage

R eview Q uestions 1. Explain with a neat sketch the working of a De Laval turbine. (Refer Section 5.2.1) 2. Derive an expression for the rotor efficiency of a De Laval turbine. What are the conditions for this efficiency to be maximum? What is the maximum rotor efficiency? (Refer Section 5.2.1) 3. Explain the working of a single-stage, reactiontype steam turbine, with relevant diagrams. (Refer Section 5.2.2) 4. Why is compounding of steam turbine necessary or what are the disadvantages of singlestage steam turbines? How can we overcome such disadvantages? (Refer Section 5.3) 5. Distinguish between pressure compounding and velocity compounding. Add further ideas such as in velocity compounding, initial rotors

6. 7.

8.

9.

10.

have very high steam velocities and therefore higher losses. (Refer Section 5.3.1) Explain with neat sketches the scheme of pressure compounding. (Refer Section 5.3.1) Explain the velocity compounding in steam turbines with relevant sketches. (Refer Section 5.3.2) Explain a Curtis stage with two rotor rings. Make suitable changes in the number of stages and number of pairs of velocity triangles. (Refer Section 5.3.3) Explain the losses in a reaction stage of a steam turbine. Also explain how to account for the same. (Refer Section 5.4.1.1) Explain the losses in a Curtis stage of a steam turbine. Also explain how to account for the same. (Refer Section 5.4.3.1)

Steam Turbines

11. Explain the reheat factor and its significance. (Refer Section 5.5) 12. What is governing of steam turbines? How is governing done in steam turbines? Explain. (Refer Section 5.6)

215

13. Explain nozzle governing in steam turbines. (Refer Section 5.6.1) 14. Explain the throttle control governing in steam turbines. (Refer Section 5.6.2) 15. Write a note on the performance characteristics of steam turbines. (Refer Section 5.7)

E xercises 1. In an impulse turbine with a single row of rotor blades, the absolute exit velocity from the blades is axial. The blade angle at the exit is 22o. The mean diameter of the rotor is 0.8 m and the speed is 1500 rpm. For zero axial thrust, calculate the blade inlet angle, nozzle angle, the steam velocity at the inlet, and the specific work. Assume the coefficient of blade friction, cb as 0.95. 2. In a steam turbine with 50% reaction, the nozzle angle is 18o. The blade velocity is 160 m/s and the steam velocity is 220 m/s. Calculate the blade inlet angle, utilization factor, and power for a steam flow rate of 15 kg/s. 3. A De Laval steam turbine is being designed for zero axial thrust with a mass flow rate of 12 kg/s. The steam velocity at the exit of the nozzles is 300 m/s. The nozzles are inclined to the plane of the wheel at 19o. The mean diameter of the rotor is 1.6 m and the speed is 1500 rpm. The friction factor of the blades can be taken as 0.95. Calculate the blade angles at the inlet and outlet, the specific work, and the utilization factor. If the specific volume of steam during the transit through the blades is 1.6 m3/ kg, calculate the blade height. 4. In a De Laval turbine, the steam supply to the nozzles is at 3 bar, 300oC with negligible approach velocity. The pressure at the exit of the nozzles is 2.5 bar. The nozzles are placed at 23o to the plane of the rotor wheel. The diameter of the rotor measured at the mid-height of the blades is 1.1 m and the speed is 3000 rpm. The nozzle efficiency is 98%. Calculate

the velocity of steam at the exit of the nozzles and the angle of the blade at the inlet. Taking the blade outlet angle as 2o more than the inlet angle and blade friction factor as 0.98, calculate the specific work. If the flow rate of the steam is 10 kg/s, calculate the blade height required, the power, the axial thrust, and the utilization factor. 5. The nozzles are inclined at 20o to the plane of the wheel in a Curtis stage with two rotor wheels. The velocity of steam from the nozzles is 350 m/s. The friction coefficients of the two rotor wheels and the stator wheel between them are 0.95 each. The stator blades again direct the steam at 20o to the plane of the rotor wheel. The outlet blade angles for both the rotor wheels are 25o each. Calculate the blade inlet angles for the first and the second row, specific works of the first and second row, and the total axial thrust. Also calculate the blade heights at the inlet and outlet of the first row and the inlet and outlet of the second row. The specific volume of the steam remains constant at 1.2 m3/kg and the mass flow rate is 10 kg/s. The mean diameter of the rotor is 0.9 m and the speed is 1500 rpm. 6. In a reaction turbine, the steam velocity at the inlet to the rotor wheel is 240 m/s at a nozzle angle of 20o to the plane of the wheel. The mean diameter of the rotor is 1.15 m and the speed is 1500 rpm. The axial flow component remains constant. The degree of reaction is 0.6 and the steam flow rate is 10 kg/s. Calculate the blade angles at the inlet and outlet, specific work,

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power, and blade height. The mean specific volume of steam may be taken as constant at 1.2 m3/kg. 7. The inlet velocity of steam to a stage of Parsons turbine is 180 m/s at an angle of 20o with the plane of the wheel. The speed of the rotor is 1500 rpm and the mean diameter is 75 cm. The steam flow rate is 8 kg/s. Calculate the blade inlet angle and the power developed. If the available enthalpy drop in a power plant is 800 kJ/kg, and each stage of the above type has an efficiency of 0.85, determine the number of stages possible in a turbine. 8. In a Curtis stage with two rows of rotor blades, the final discharge of steam from the second

P roject- O riented Q uestions 1. Make an observation that out of the 10 solved examples and the 8 exercise problems of this chapter, only 4 problems required the reference to the Mollier Diagram (enthalpy–entropy chart for steam). Locate these four problems. Then, take a Mollier diagram and mark all the four isentropic expansion processes (not actual processes) of the problems on the same chart, starting from initial state to the final pressure line. These four lines, representing the isentropic expansion, are for reference only. Now, in the same chart, mark any one of the following isentropic expansion processes: (a) 12 bar, 300°C to 0.05 bar (b) 10 bar, 350°C to 0.1 bar (c) 8 bar, 325°C to 0.05 bar For the chosen process, determine the number of stages that may be required for the expansion process, with every stage having equal enthalpy drop, about 40 kJ/kg. Because of the integer number of stages, this enthalpy drop may have to be corrected. Also, because you are referring to the Mollier chart, the numerical values may not be exact. Refer to the steam tables to refine your values. Switch over to the actual processes with an isentropic

row is axial. The outlet blade angle of the second rotor blade row is 30o. The outlet angle of absolute steam velocity from the first row is 40o. For a design with the steam inlet velocity of 320 m/s and the axial thrust in both rows equal to zero, determine the following: (a) Inlet blade angle of the first row. (b) Outlet blade angle of the first row. (c) Inlet blade angle of the second row. (d) Angle of entry of steam to the first row. (e) Angle of entry of steam to the second row. (f ) Blade velocity (g) Diameter of rotor if the speed is 1500 rpm. (h) Total specific work. (i) Rotor efficiency.

S olutions A vailable efficiency of 0.83, as you proceed with the expansion process, stage-by-stage. Record the values of enthalpy and specific volume at every point, as has been illustrated in the Solved Example 5.10. With this background, now calculate the generation of velocity in each stage. Increase this velocity by about 20%. Let this be the entry velocity of steam in a given stage (Refer to the comment No. 4, at the end of the Example 5.10.). Now, let this turbine be made up of Parsons stages, for which the velocity triangles are symmetrical. Choose a convenient speed, and a diameter of the rotor, so that the velocities of steam and blades are in agreement with a speed ratio, of your choice. Now follow the steps of choosing/designing the other parameters of the turbine, such as inlet angle of steam, blade angles, blade heights(mass flow rate 10 kg/s), etc. Check, whether the exit velocity, together with generated velocity, gives rise to the entry velocity; and if not, by how much it differs. Complete your project by calculating the efficiencies and the reheat factor.

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217

(b) A convergent–divergent nozzle can accommodate only a fixed ratio of pressures; (c) This pressure ratio (downstream to upstream pressures, p2/p1) is about 0.53; (d) If this pressure ratio is different, then, the shockwaves occur in the flow, with losses. As a first approximation, for this problem, assume that shockwaves are absent. An enthalpy drop of about 100 kJ/kg generates higher velocities. Let the exit velocity from each stage be axial. Choose a convenient speed, and a diameter of the rotor, so that the velocities of steam and blades are in agreement with a speed ratio, of your choice. Now follow the steps of choosing/designing the other parameters of the turbine, such as inlet angle of steam, blade angles, blade heights (mass flow rate  10 kg/s). Check, whether the exit velocity, together with genereated velocity, gives rise to the entry velocity; and if not, by how much it differs. Complete your project by calculating the efficiencies and the reheat factor.

2. In the above problem, you have chosen one of the three alternate enthalpy drops. Now, choose one of the remaining two enthalpy drops to design Curtis stages, each stage having two rotors. For the chosen process, determine the number of stages that may be required for the expansion process, with every stage having equal enthalpy drop about 100 kJ/kg. Because of the integer number of stages, this enthalpy drop may have to be corrected. Also, because you are referring to the Mollier chart, the numerical values may not be exact. Refer to the steam tables to refine your values. Switch over to the actual processes with an isentropic efficiency of 0.8 as you proceed with the expansion process, stage-by-stage. Record the values of enthalpy and specific volume at every point, as has been illustrated in the Solved Example 5.10. Recall from your earlier course on Thermodynamics that: (a) Any enthalpy drop that generates a velocity more than sonic velocity requires a convergent–divergent nozzle;

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5. 6.

(c) (b) (a) (c) (a) (b)

7. 8. 9. 10. 11. 12.

(c) (b) (d) (c) (c) (b)

Exercises 1. 2. 3. 4.

20.85o, 11.09o, 132 m/s, 8.136 kJ/kg 54o, 0.922, 620 kW 31.7o, 33.6o, 38.324 kJ/kg, 0.885, 3.91 cm 439.4 m/s, 36.54o, 78.227 kJ/kg, 1.97 cm, 782.27 kW, 43.8 N, 0.8244 5. 24.87o, 30.4o, 35.576 kJ/kg, 16.3 kJ/kg, 194.6 N, 3.54 cm, 3.71 cm, 6.267 cm, 7.9 cm

6. 31.26o, 18.62o, 34.213 kJ/kg, 342.13 kW, 4.05 cm 7. 29.18o, 131.65 kW, 42 8. Hint: For analytical or graphical solution, assume a blade velocity. Finally, scale up or scale down to V11  320 m/s. 26. 5o, 28o, 20.37o, 15.6o, 76.614 m/s, 97.55 cm, 45.3 kJ/kg, 88.48%

6

Hydraulic Turbines

Learning Objectives After completing this chapter, you will be able to: v
Understand the different types of hydraulic turbines, their ranges of the basic parameters, such as head, discharge, specific speeds, and suitability for a site. v
Study the Pelton turbine with its working features, suitability, efficiencies, and design parameters. v
Study the Francis turbine with its working features, suitability, efficiencies, and design parameters. v
Study the Kaplan and propeller turbines with their working features, suitability, efficiencies, and design parameters.

v
Understand the application of a simple draft tube, how it saves the head that would have been lost otherwise, and how it reduces exit losses. v
Understand the mechanism of cavitation, its causes, effects, and methods to prevent its occurrence. v
Study the application of the governor to hydraulic turbines. v
Study the characteristics of hydraulic turbines.

6.1 Introduction Hydraulic turbines are power-producing turbomachines, using water as the fluid. The water has to be available at a reasonable height or head, in fairly large quantities so that some economically feasible power projects may be developed. As was seen in Chapter 1, the power available in the water is proportional to the product of the flow rate and the head (P  wQH). In a project site, the available flow rate depends on the rainfall in the region, the extent of the catchment area, and the possibility of storage of water (natural or built-up). The available head is a characteristic of the topography of the project site. A schematic layout of a typical hydroelectric power project is shown in Fig. 6.1. 1

4

2

5

6

3 Hg

7 This distance depends on the land topography 9

8 10

Figure 6.1

Schematic layout of a typical hydroelectric power project. (1) Dam; (2) reservoir fed from rainfall in catchment; (3) trash gate; (4) forebay (to take care of daily fluctuations); (5) surge tank; (6) valve house; (7) penstocks; (8) powerhouse; (9) turbines; (10) tailrace.

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Turbomachines

The type of turbine to be employed for the power project depends on the head available and the round-the-year uniform flow rate that is possible at the site of the project. This chapter aims at identifying the different types of turbines, based on the characteristics of the modern-day turbines, their suitability for a given project site, the constructional details, and the design of each type of turbine.

6.2 Classification of Hydraulic Turbines Hydraulic turbines are classified based on several criteria. Some prominent classification details are shown in Table 6.1. Table 6.1

Classification of hydraulic turbines

Criteria

Types

Examples

Dynamic action of water on rotor blades

1. Impulse turbines

Pelton turbine

2. Reaction turbines

Kaplan turbine

Direction of flow of water through the runner

1. Radial flow

Slow Francis turbine

2. Mixed flow

Medium Francis turbine

3. Axial flow

Kaplan turbine

4. Tangential flow

Pelton turbine

Ns: 05 35

Single-jet Pelton turbine

Ns: 30 70

Multi-jet Pelton turbine

Ns: 60 120

Slow Francis turbine

Ns: 120 180

Medium Francis turbine

Ns: 180 300

Fast Francis turbine

Ns: 300 1000

Kaplan turbine

Low head (5 75 m of water) Medium head (50 250 m of water) High head (150 m of water and above)

Kaplan turbine

Specific speed, Ns N: rpm P: kW H: meters of water

Head on the turbine

Francis turbine Pelton turbine

Apart from the criteria of classification, it can be seen that basically there are three types of turbines: Pelton, Francis, and Kaplan turbines, named after their designers. 1. 2. 3.

The Pelton turbine is an impulse turbine, with tangential flow, for high-head applications. The Francis turbine is a reaction turbine, with radial or mixed flow, for medium head applications. The Kaplan turbine is a reaction turbine, with axial flow, for low-head applications.

6.2.1 Selection of Hydraulic Turbines The power projects, where hydraulic turbines are to be installed, are generally huge projects, involving very high investments on head works and machinery. Because of the wide variation of the two basic data, namely, the head and the flow rate, each project requires some unique design. Therefore, the selection and design of a particular type of turbine must be undertaken with some discretion, so as to have the highest possible efficiency of the turbine.

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221

There are two approaches to decide the type of turbine suitable for a given project site, with specified head and flow rate: 1.

One criterion is the head (meters of water) available. Shown in Fig. 6.2 is a scale that indicates the head and the corresponding suitable turbine. 05 Head (meters of water)

50

150

75

Kaplan

Pelton Francis

Suitable turbine

Figure 6.2

2.

250

Selection of turbines on the basis of head.

It may be observed that there are some ranges of overlaps, such as the 50 75 m or 150 250 m stretches. In these ranges, the turbine can be selected by the criterion of the specific speed, mentioned in the next paragraph, to include the effect of the available flow rates also. Another criterion is the specific speed corresponding to the site data. The head, H (m of water), and the flow rate, Q (m3/s), are taken as data. An overall efficiency G of the order of 0.85 or 0.88 can be assumed. Then, the power P is given by P  (wQH H / 1000) kW Now, the speed of the turbine N rpm has to be selected that must be one of the synchronous speeds (N  3000/p, p  1, 2, 3, …, to have a frequency of 50 Hz of electrical power supply) because turbines drive the alternators. With this, the specific speed can be calculated by Ns 

N P H 5/ 4

Figure 6.3 indicates a scale of specific speeds and the corresponding suitable turbines. Ns

30

5

Types of turbine

60

35

300 Kaplan

Multi-jet Pelton (D2 / D1)

Figure 6.3

70

Pelton

Francis 0.5

1.0

Selection of turbines on the basis of specific speed.

It may be noted here that even on this scale, there are overlaps. Also, the limits of the specific speed to select any particular type of turbine are, however, not very sharp. There are many other considerations, such as cost factors, control factors, applications, etc., and the calculated specific speed can be logically manipulated. If the calculated specific speed happens to be like 400, then one can assume two equal-sized turbines with the available flow rate equally divided between the two. The new specific speed for each of the turbine becomes

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Turbomachines

Ns 

N P /2 H 5/ 4

This new specific speed is brought in the range 60−300 to select a Francis turbine. This method of crossing over the range (from Kaplan to two or more units of Francis, or from Francis to multiple-jet Pelton) has to be evaluated with alternate plans and other possible cost factors. Selection of any type of turbine for a site is by any of the above two approaches. However, the selection also has to focus on the further steps of the design. As an illustration, on the stretch of the Francis turbine in Fig. 6.3, another scale of the diameter ratio D2/D1 is also given. This indicates that the ratio D2/D1 varies with specific speeds. As the specific speed increases, the Francis turbine tends to be nearer to the axial flow machine. The name “Francis turbine” therefore is not an all-time standard design. Just like the diameter ratios, all the comparative and absolute dimensions have to be determined for a given project. The guidelines start from the values of the specific speed. It has to be mentioned here that Figs. 6.2 and 6.3 are derived from Fig. 1.13, where the different types of turbines are compared for their suitability for the situation dictated by the specific speed. The aim is to maximize the efficiency in a given set of data for a project site.

6.3 Pelton Turbine The Pelton turbine belongs to the range of the low specific speeds (5 to 70) and the range of high heads (150 m of water and above). It is an impulse-type turbine.

6.3.1 Constructional Details of Pelton Turbine A Pelton turbine setup is shown in Fig. 6.4. Some of the terms used are the shaft, the rotor, the nozzle, the jet, etc., which are shown in the figure. The blades or vanes of the rotor in the case of Pelton turbine are the “Pelton double cups” or “buckets,” as shown in Fig. 6.5. These double cups are mounted on the periphery of a circular disk and together they form the rotor of the Pelton turbine. Runner/rotor Pelton cups

Bearings

Shaft

Coupling

Water supply To generator

Water jet Nozzle Control spear

Figure 6.4

Base Casing

Schematic layout of a Pelton turbine.

Hydraulic Turbines

223

A

A Section AA

Figure 6.5

Pelton double cup.

The water, supplied from the head-works to the power house through the penstocks (steel pipes), is led to these buckets in the form of a high-speed jet issued from a nozzle (Fig. 6.6). The kinetic energy of water jet is transferred to the series of the buckets (and to the rotor) that come in succession in the line of the jet as the rotor rotates. The jet gets divided into two equal halves by the jet splitter of the double cup, with each half striking the cups on either side. From penstocks

Water jet Movement controlled by seromotor of governor

Figure 6.6

Spear

Nozzle of Pelton turbine.

The potential energy of the water at the head-works is converted into the kinetic energy of the jet of water in the nozzle, with a nozzle efficiency Gn. This efficiency is of the order of 0.98. A concentric spear inside the nozzle controls the rate of flow of water coming out of the nozzle. The movement of the spear is controlled by a servomotor of the governor that is intended to maintain a uniform speed of the turbine. In multi-jet Pelton turbines, water is led around the rotor into the identical nozzles equally spaced around the periphery of the rotor. The spear assemblies are also identical in all the nozzles. Their movement is controlled by the same source, so that all the jets are equally controlled. Whenever a Pelton turbine is required to be stopped, a shut-off valve in the supply mains has to be closed. But this should not be done suddenly, as otherwise water hammer is likely to occur in the pipes. A deflector can be actuated so that the jet can be deflected from the Pelton cups, in the opposite direction, so that it can act, in the meantime, as a brake jet. This jet is in the opposite direction to the main jet, with water striking the backside of the cups. These details are shown in Fig. 6.7. A separate brake jet can also be arranged in the direction opposite to the main jet.

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Turbomachines

Jet: Braking position From penstocks Open

Close

Open

Close

Jet: Running position Deflector

Figure 6.7

Deflector assembly and brake jet.

Pelton turbines can be designed with horizontal or vertical shaft arrangements. In a horizontal setup, the turbine wheel can be between two journal bearings. The design can also be in the over-hung form, with bearings on one side and its shaft coupled to the generator shaft. In vertical installations, the supports are the thrust bearings, the turbine wheel being at the bottom level.

6.3.2 Analysis of the Pelton Turbine The gross head at the project site, Hg, is the difference of water levels between the reservoir and tailrace. The head available at the power house or nozzles is H. The difference (Hg H ) is due to the topography of the land and the losses in the penstocks, valves, height of nozzles above the tailrace, etc. The velocity of the water jet at the outlet of the nozzle or inlet to the rotor cups is V1  c v 2 gH m/s

(6.1)

where cv is the coefficient of velocity of the nozzle, with a value of the order of 0.96 0.98. The velocity of the Pelton cups is P DN m/s U 60 where D is the pitch diameter (m) and N is the rotational speed (rpm) of the rotor shaft. The centerline of the water jet is tangential to this pitch circle (Fig. 6.8). The jet interacts with the Pelton cups over a stretch of the travel, from position A to position B in Fig. 6.8. The mean position is shown in the cross-section in Fig. 6.9. The outlet is a little divergent so that the water leaves the buckets without splashing over the back of the next bucket.

Jet A

Figure 6.8

B

Jet striking Pelton buckets.

Hydraulic Turbines

225

15°

Jet

Jet splitter

15°

Figure 6.9

Jet is split between the two cups.

The water takes a turn of about 165o from A1  0o to A2  165o. The velocity triangles corresponding to the upper bucket of Fig. 6.9 are shown in Fig. 6.10. It may be noted that the inlet velocity triangle is reduced as a straight line, as shown. When the blade friction factor is cb we have Vr2  c b s Vr1 Vr2 = cb × Vr1 U b1 = 0

Vr1

V2

V1

Figure 6.10

b2 = 165° U

Velocity triangles for the Pelton cup at the inlet and outlet.

With reference to inlet velocity triangle of Fig. 6.10, we have

and

Vr1  V1 U

(6.2a)

Vu1  V1

(6.2b)

Also, with reference to the outlet velocity triangle, we have Vu2  U Vr2 cosB2

(6.3a)

Using Eq. (6.2a) and Eq. (6.2b) in Eq. (6.3a), we get Vu2  U cb s Vr1cosB2  U c b (V1 U )cosB2 The specific work W is given by

(6.3b)

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Turbomachines

W  U (Vu1 Vu2 )  U [V1 (U c b (V1 U )cosB2 )]  U [V1 U  c b (V1 U )cosB2 ]  U (V1 U ) (1  c b cosB2 ) 

U ¥ U´ 1 1  cb cos B2 V12 V1 ¦§ V1 µ¶

(6.4a)

So W  F(1 F ) (1  c b cosB2 )V12

(6.4b)

where E  U/V1 is the speed ratio. For a given installation, cb, A2, and V1 are constants. The specific work W is maximized when dW/dE is taken as zero, resulting in E  0.5. Hence Wmax  0.25(1  c b cosB2 )V12

(6.5)

The hydraulic efficiency Gh  W/(V12/2) works out to a maximum value of

Hhmax  

Wmax (V12 ) / 2 1  c b cos B2 2

(6.6)

6.3.3 Efficiencies of Pelton turbine The hydraulic efficiency is obtained as in Eq. (6.6) in the form

Hh 

1  c b cos B2 2

This expression holds for the defining equation Gh or Ga 

Pr P



Pr Pn

s

Pn P

where the hydraulic efficiency is defined as the ratio of the power of rotor to the power available in the inlet stream of water (Gh = Pr/P), including the effect of exit losses, fluid friction on the blade surface, etc. One can recall the definition of volumetric efficiency Gv as

Hv 

(Q $Q ) Q

(6.7)

The reduction in the volume flow rate in a turbine ($Q) was earlier attributed to the leakage. In the case of Pelton turbine, this loss can also be attributed to the “ineffective” volume flow rate that is in the outer layers of water in the jet, which may not be as effective as the core of the jet in exerting the force on the buckets. The mechanical efficiency (Gm) has also been defined as

Hydraulic Turbines

Hm  

227

Shaft power (Ps ) Rotor power (Pr ) Pr Mechanical losses Pr

(6.8)

The overall efficiency is then given by

Ho  Hh s Hv s Hm

(6.9)

There are quite a few factors that affect the above efficiencies, individually and collectively. The surface finish of the blades, the friction factor cb , the sand content of minute sizes in the water jet, the speed, the smoothness of the bearings, the disk friction of the rotor, and the magnitudes of all these in relation to the total power contribute to the final value of the overall efficiency. It is common to use a value of approximately 0.85−0.88 for the overall efficiency.

6.3.4 Design Parameters of Pelton Turbine Although the speed ratio, E  0.5 is a theoretical value, in practice, E is taken as about 0.45 or 0.46, so that U  (0.45 to 0.46)V1m /s

(6.10)

With this, the values of U and N (as assumed earlier) are known in the equation U  ODN/60, from which we can calculate 60U D m (6.11) PN Now, from the specific speed, as decided by the site data, the number of jets is determined. The total available volume flow rate is equally divided between the jets. The flow rate per jet is Qj 

Q , n  1, 2, 3, n

(6.12)

Now, Volume flow rate in a jet  Area of jet s Velocity of jet ¥ Pd 2 ´  Qj  ¦ V § 4 µ¶ 1 and therefore d

4Q j

PV1

(6.13)

For the Pelton wheel, the jet diameter is an important parameter to decide the geometrical proportions of the Pelton double cups. These proportions are so fixed that almost the entire jet is usefully employed in striking the cups at the middle splitter line to generate the maximum possible torque. The notch at the tip of the bucket helps in this aspect. If the notch were not present, the cups would come in contact with the jet, starting from the tip, where a sizable quantity of the jet would disperse over and around the tip with a lot of losses as shown in Fig. 6.11.

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Turbomachines

Pelton bucket

Jet

Figure 6.11

Losses in the jet when buckets do not have notches.

T

L

d B N

Figure 6.12

Proportions of Pelton cups.

The geometrical features of the Pelton double cup are shown in Fig. 6.12. These features are the length (L), breadth (B), and depth (T ) of the double cup. These parameters are optimized in terms of the jet diameter d, and are specified by the following equations: L  2.3d to 2.8d B  2.8d to 3.2d T  0.6d to 0.9d

(6.14) (6.15) (6.16)

The width of the notch, N, in Fig. 6.12 is approximated as 2 5 mm more than the jet diameter d. The number of buckets is also optimized and is given as Z

0.5D  15 d

(6.17)

EXAMPLE 6.1 Design a Pelton turbine for a project site where the available head is 510 m and the uniform flow rate is 0.03 m3/s. Assume an overall efficiency of 0.867, coefficient of velocity of nozzle as 0.985, a speed ratio of 0.46, and a speed of 1500 rpm.

Solution: Given H  510 m, Q  0.03 m3/s, Go  0.867, cv  0.985, E  0.46, N 1500. Now the power output is

Hydraulic Turbines

P  wQH Ho 

229

9810 s 0.03 s 510 s 0.867  130.13 kW 1000

The specific speed is Ns 

N P 1500 s 130.13   7.06 H 5/ 4 5105/ 4

Select a single-jet Pelton turbine. Velocity of the jet is V1  c v 2 gH  0.985 2 s 9.81 s 510  98.53 m /s Diameter of jet is d

4Q 4 s 0.03   0.0197 m  1.97 cm z 2 cm PV1 (P s 98.53)

Diameter of rotor is D  0.46 V1 s

60 60  0.46 s 98.53 s  0.577 m  57.7 cm PN P s 1500

The jet ratio is D 57.7   28.85 d 2 Therefore, Number of buckets  0.5 s 28.85  15  14.42  15  29.42 That is Z  30. Now Length of bucket  2.3d  2.3 s 2  4.6 cm Breadth of bucket  2.8d  2.8 s 2  5.6 cm Depth of bucket  0.6d  0.6 s 2  1.2 cm Inlet angle  5o Outlet angle  180 165  15o Comment: 1. This is a very small turbine, 130 kW. The head is normal but the flow rate available is only 30 lps. The project might be a “micro hydroelectric plant,” which is possible only if the cost of the head works is very small. 2. Becausee Ns is 7, the lower range is adopted for bucket dimensions. The bucket angles are also adopted by the thumb rule. 3. If not part of the data, Go, cv, and E can be assumed. 4. The example may be reworked with N  3000 rpm. Compare the results.

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Turbomachines

EXAMPLE 6.2 A Pelton turbine running at 600 rpm has a net head of 260 m at its nozzles. It is supplied with water at a rate of 2 m3/s. The speed ratio of the machine is 0.46 and the velocity coefficient of the nozzle is 0.98. Calculate the jet diameter, wheel diameter, and salient dimensions of the Pelton cups, with outlet angle A2  20o. It is estimated that 0.015 m3/s of water is ineffective in the system. Windage and bearing losses are 60 kW. Take an initial approximation of overall efficiency as 91%. Then calculate the actual hydraulic efficiency, volumetric efficiency, mechanical efficiency, and overall efficiency.

Solution: Given N  600 rpm, Hn  260 m, Q  2 m3/s, E  0.46, cv  0.98, A2  20o, $Q  0.015 m3/s, mech. losses  60 kW, initial Go  0.91. To calculate d, D, cup dimensions, various efficiencies. Initially, Power 

wQH H0 1000



9810 s 2 s 260 s 0.91  4642 kW 1000

Therefore, the specific speed is Ns 

N P 600 s 4642   39.15 H 5/ 4 2605/ 4

At this specific speed, two jets have to be chosen and the flow rate of each jet has to be half of the total flow rate. Hence, the flow rate per jet  1 m3/s. The jet velocity is V1  c v 2 gH n  0.98 s 2 s 9.81 s 260  70 m /s We know that

Pd 2 V  Qj 4 1 Therefore d

4Q j

PV1



4 s1  0.1349 m  13.5 cm (P s 70)

Also

F

U  0.46 V1

Therefore U  0.46V1  0.46 s 70  32.2m /s Moreover U

P DN 60

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231

Therefore D

60U 60 s 32.2   1.025 m  102.5 cm PN P s 600

Hydraulic efficiency is

Hh 

1  cos B2

2  Hh  97%



1  cos 20o 2



1  0.93969  0.9698 2

Volumetric efficiency is Q $Q 2 0.015   0.9925 Q 2  Hv  99.25%

Hv 

Rotor power is given by Pr  (wQH ) s Hv s Hh  9810 s 2 s 260 s 0.9925 s 0.97 /1000  4911 kW Mechanical efficiency is

Hm 

Rotor power Mechanical losses Rotor power

4911 60  0.9878 4911  Hm  98.78% 

Actual overall efficiency is

Ho  Hv s Hn s Hm  95.1% Comment: The initial approximate overall efficiency, as per data, is 91%. The final value, after calculations, is Go  95.1%. This is alright. The purpose of the initial approximation is only to decide the specific speed and the consequent number of jets. Once the number of jets is fixed as 2, further calculations do not depend on the initial approximate value, and it should not be carried on to decide other efficiencies.

EXAMPLE 6.3 The gross head available at a project site is 350 m of water. The penstock pipe is estimated to be 600 m long. The pipe friction factor is f  0.007. The total pipe losses have to be limited to 4% of gross head. The expected power from the project is 2600 kW. The turbine speed is 600 rpm. Calculate (a) the required flow rate Q m3/s, (b) the pipe line diameter Dp, (c) the jet diameter d, and (d) the rotor mean diameter D. The speed ratio is 0.46, the nozzle velocity coefficient is 0.985, and the overall efficiency is 0.92.

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Solution: Given Hg  350 m, L  600 m, f  0.007, losses in pipe: 4% of gross head, P  2600 kW, N  600 rpm, E  0.46, cv  0.985, Go  0.92. The net head is H n  H g s 0.96  350 s 0.96  336 m

The power is given as 2600 kW, hence, P

wQH H0 1000

 2600 kW

(a) The required flow rate, therefore, given by Q

2600 s 1000  0.8574 m3 /s 9810 s 336 s 0.92

(b) To calculate the pipe diameter: The velocity of water in the pipeline is Vp 

Q 0.8574  2 P Dp P Dp2

4 4 where Dp is pipe diameter. Loss in the pipe (at 4% of head)  350 s 0.04  14 m. Therefore 4 fLVp2

4 s 0.007 s 600 ¨ 0.8574 s 4 · 14   s© ¸ P Dp2 ¸¹ 2 gDp 2 s 9.81 s Dp ª© 4 s 0.007 s 600 s 0.85742 s 42  Dp5  14 s 2 s 9.81 s P 2  Dp  0.592289 m

2

 Dp  59.23 cm (c) The velocity of the jet is V1  c v 2 gH n  0.985 s 2 s 9.81 s 336  80 m /s Now Flow rate in the jet  Area of jet s Velocity of jet Q 

Pd 2 V 4 1

Therefore the jet diameter is d

4Q 4 s 0.8574   0.1168 m PV1 (P s 80)

 d  11.68 cm (d) To calculate rotor diameter: We know that blade velocity is U

P DN  0.46 2 gH n 60

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Therefore the wheel diameter is 60 s 0.46 s 2 s 9.81 s 336  1.1888 m P s 600  D  1.19 m D

EXAMPLE 6.4 The mean diameter of a Pelton wheel is 2.6 m. The bucket outlet angle is 20o. The net head available at the nozzle is 500 m of water. Calculate (a) the speed of the rotor, (b) the theoretical efficiency, (c) the theoretical power, when the flow available is 3.5 m3/s, and (d) the specific speed.

Solution: Given D  2.6 m, A2  20o, Hn  500 m, . Assume cv  0.985, E  0.46. To calculate N, G, P for Q  3.5 m3/s as also Ns. (a) The jet velocity is V1  0.985 2 gH n  0.985 s 2 s 9.81 s 500  97.56 m /s The blade velocity is given by U  FV1  0.46 s 97.56  44.88 m /s Also U

P DN 60

Therefore the rotor speed is N 

60U 60 s 44.88   330 rpm PD (P s 2.6)

(b) The theoretical efficiency is

H

1  cos B2

2  H  97%



1  cos 20o 1  0.96939   0.969845 2 2

(c) The theoretical power is P

wQH H 9810 s 3.5 s 500 s 0.97   16652 kW 1000 1000

(d) The specific speed is Ns 

N P 330 s 16652   18 H 5/ 4 5005/ 4

Comments: 1. 2.

For Ns  18, the turbine has a single jet. If the jet diameter were to be calculated, it would be d  4Q / PV1  0.2137 m  21.4 cm. Both the diameter of wheel and jet diameter (D and d ) are on the larger side.

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If the wheel diameter D were to be redesigned, it could be adjusted for N  500 or 600 (both are synchronous speeds). The 330 rpm obtained above is non-synchronous. The nearest one is 300 rpm or 375 rpm. For N  500, D  60U/ON  1.7143 m. The corresponding specific speed is N s  N P / H 5/ 4  27.29

4.

This design with D  1.7143 m and d  21.4 cm would have been better. Adopting two jets would have been still better (Why?) Important: The above comments are offered with a view to analyze the design process. However, given data need not be (and should not be) altered or commented upon, unless comments are invited. In fact, a wheel diameter of the order of 2 2.5 m is preferred when huge quantities of water are available and three, four, or five nozzles are required to be placed around one rotor to work on it.

6.4 Francis Turbine The Francis turbine is a reaction turbine suitable for a medium range of specific speeds (60−300) and a medium range of heads (50−250 m). The Francis turbine is designed as a radial flow machine in the range of specific speeds of 60−120. As the specific speed corresponding to the data of the project site keeps on increasing, the design shifts to a mixed flow machine and then to almost an axial flow machine.

6.4.1 Constructional Features of Francis Turbine An installation of the Francis turbine is shown in Fig. 6.13. Water from the penstock pipe enters an outer spiral casing that may be fabricated out of steel plates or cast in concrete with a lining of steel plates. This casing is arranged around a ring of guide vanes, and its area of cross-section diminishes progressively giving a uniform distribution of water at a continuous outlet, inward toward a number of guide vanes, around the rotor. Water supply

Scroll casing

To generator Shaft

Runner

Guide vanes

Thrust bearing Runner vanes

Draft tube Water discharge Tailrace level

Figure 6.13

Schematic layout of a Francis turbine.

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235

The guide vanes have airfoil shapes so that the passages between them act like nozzles that convert a part of the pressure energy of water into kinetic energy. The water coming from the casing is directed on to the rotor vanes. Each guide vane has its own axis about which it can swing, so as to vary the area of flow of water. The swinging of all the guide vanes (about their individual axes) is controlled by a governor-actuated regulator ring, so that the flow of water can be controlled. The details are shown in Fig. 6.14.

Open

Regulator ring (actuated by governor)

Close

Close

Water entry from casing

Guide-vanes

Figure 6.14

Control of guide vanes.

This control is only to the extent of maintaining constant speed over the fluctuation of loads over the turbine. The water from the guide vanes enters the rotor, with both kinetic energy and pressure energy. The rotor vanes absorb these energies and the water is discharged to a component known as a draft tube (Fig. 6.13). As a reaction turbine installation, the casing, guide-vane ring, and rotor of the Francis turbine run full without the water coming in contact with the atmosphere. At the exit of the rotor, that is, in the draft tube also, the water is not open to atmosphere. The draft tube is a slightly divergent tube, connecting the outlet of the runner to the tailrace level. The draft tube also runs full. The water column in the draft tube is under sub-atmospheric pressure and effectively saves the head that otherwise would have been lost, when the turbine is installed at a higher level than the tailrace. Because of the divergent portion, a part of the exit kinetic energy is also recovered. Francis turbine installations can be designed either with horizontal shafts or vertical shafts. With horizontal shafts, the draft tubes have to be provided with a bend that reduces the effectiveness of the draft tube in recovering the lost head. The vertical shaft installations have the turbine rotor at the lowest level so that the axial discharge from the runner becomes vertically downward. This becomes a very good feature of the vertical draft tube, with better efficiencies.

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Francis turbines with huge capacities are usually designed with vertical shafts. Some dams are built primarily for the purpose of irrigation. The water level in the dams may reach such levels that it is possible to use that head while letting the water out to the canals through the turbines in power houses at the bottom of the dams. Generally, Francis turbines become the most suitable choices for such “dam power houses.”

6.4.2 Analysis of the Francis Turbine Velocity triangles were discussed in detail in Chapter 3 with an understanding that the planes of the triangles were either perpendicular to the axis (for radial flow machines) or parallel to the axis (for axial flow machines). In continuation with the same discussion, for the mixed flow pattern of the Francis turbine, the plane of the inlet velocity triangle is perpendicular to the axis (radially inward flow), but the outlet velocity triangle is on a plane that shifts its orientation (from perpendicular to) parallel to the axis (axial discharge flow). This agrees well with the expression for the specific work (W ) because the components of velocities responsible for the energy transfer are the whirl components, Vu1 and Vu2, whether in a radial flow pattern or an axial flow pattern. With this in mind, a typical set of velocity triangles is shown in Fig. 6.15. The orientation of the rotor blades is also shown in Fig. 6.15. The range of specific speeds corresponding to the design of Francis turbine is stated as 60−300, and this is a wide range. The details shown in Fig. 6.15 apply to the lower stretch of the specific speeds 60−120 in this wide range. As the specific speed increases, the shapes of blades of the runner also change. The changing phases of the blades and the applicable velocity triangles are shown in Fig. 6.16, to cover the whole range of the specific speeds. Figure 6.16 may be considered as continuation of Fig. 6.15. The velocity triangles at the outlet of the blades in all the cases of Fig. 6.16 have @2  90o, namely axial discharge, as in Fig. 6.15. In all the cases, Vu2  0 . This results in the specific work W being equal to U1 Vu1. The input to the rotor is in both kinetic and potential forms of energies, adding up to a total of gH. Therefore, the hydraulic efficiency is given by UV Hh  1 u1 (6.18) gH Vu1 B0

1

U1 2 b1

b1

D1 D2 Parallel to Vr1

Vr1 V1

Vu2 = 0

Ns = 60 − 120 a1 = 15° − 25° f1 = 0.62 − 0.68 b1 = 90° − 120°

U2

Vf2 V2

1 2

Figure 6.15

Vf1

Vr2

Parallel to Vr2

Francis runner with velocity triangles.

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237

Vu1

B0

U1 b1 Vf1

Vr1 V1 ° 1 a1 = 25° − 32 2 f1 = 0.68 − 0.72 b1 = 90°

(a) Ns = 120 to 180

B0

U1 Vu1 b1 Parallel to Vr1 V1 ° 1° 1 a1 = 32 2 − 37 2 f1 = 0.72 − 0.76 b1 = 60° − 90°

Vf1 Vr1

(b) Ns = 180 to 300

Figure 6.16

Francis runner with velocity triangles. Runner and velocity triangles for specific speed (a) Ns  120 180 and (b) Ns  180 300.

6.4.3 Efficiencies of Francis turbine The hydraulic efficiency of the Francis turbine is obtained in the form

Hh 

U1Vu1 gH

which is same as Eq. (6.18). The expressions for volumetric efficiency [Eq. (6.7)], mechanical efficiency [Eq. (6.8)] and overall efficiency [Eq. (6.9)], which were detailed for the Pelton turbine in Section 6.3.2, are very much valid for Francis turbine also, because, these expressions are independent of the (a) geometrical features and the (b) mechanism of energy transfer of these turbines. Also, the reasons for the various losses in the turbines are largely of same nature. The loss due to the leakage in the clearances in Francis turbine is one difference that exists between Pelton and Francis turbines. The numerical values of the efficiencies also approximately match between the machines.

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6.4.4 Design Parameters of Francis Turbine The design parameters of Francis turbine are discussed in the following sub-sections.

6.4.4.1 Volute or Spiral Casing The casing is the outer conduit of the Francis turbine assembly. The penstock pipe is connected to the inlet of this casing (through a stop valve). The cross-section at the inlet is circular with the diameter equal to that of the penstock pipe. The cross-sectional area of the casing has to gradually decrease. At the same time, the height of the inlet to the guide vanes, from the inner side of the casing, has to remain invariant. Hence, the crosssection gets into oval shape, gradually reducing the area to almost zero, when it reaches back through 360o. At the inlet, the area A  (Od2)/4. At any angle P (refer Fig. 6.17), we have AQ 

P d 2 (360 Q )o s 4 360o

where P is measured at the central axis, starting from the line of commencing the decrease of area. The overall purpose is to reduce the eddies and to effect a smooth and uniform distribution of water all around the rotor. d B

B q Aq

B

B

Figure 6.17

Spiral casing with decreasing area (B = Bo + 5 mm).

6.4.4.2 Guide Vanes As explained earlier, the guide vanes swing about their own individual axes. In their “full open” position, let their tail ends be on a diameter D0. For the purpose of the design procedure, the following two assumptions are made: 1. 2.

The guide vanes also rotate about the axis of the shaft, like the runner does, at the speed of the runner, N rpm. There is a reference velocity, 2 gH , of the water if the entire head energy were to be converted into kinetic energy without loss.

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239

(Both the above are only assumptions; neither the guide vanes rotate around the runner nor the potential energy of water is totally converted into velocity in the Francis turbine.) With these assumptions, the hypothetical tangential velocity of the guide-vane tip is U0 

P D0 N 60

Also U 0  F0 2 gH

(6.19)

So D0 

60F0 s 2 gH

PN

(6.20)

Here E0 is a hypothetical “speed ratio” or a simple coefficient, varying between 0.7 at Ns  60 and 1.31 at Ns  300. The length of the guide vanes, L0, is taken as L0  0.3D0

(6.21)

The height of the guide vanes, B0, is calculated from the equation of flow rate: Volume flow rate (m3/s )  Flow area around the rotor (m2) s Flow velocity (m/s)  Q  P D0 B0Vfo  P D0 B0Y 0 2 gH So B0 

Q P D0Y 0 2 gH

(6.22)

where X0 is the flow coefficient, varying from 0.15 at Ns  60 to 0.35 at Ns  300. The number of guide vanes is decided as a thumb rule. Too few numbers do not serve the purpose of uniform distribution of water. Too many numbers may give rise to some loss, with added costs. Depending on the diameter D0, as determined above, the number of guide vanes varies from 8 to 24 (8, 10, 12, 14, 16, 20, 24) as D0 varies from 20 cm to 2 m. (Numbers chosen are even to facilitate the shipping of the component parts in segments.)

6.4.4.3 Rotor The rotor diameter at the inlet (namely, the outer diameter) and the tangential velocity of the blades at the inlet are related, as usual, by the relation U1 

P D1N 60

The velocity U1 is taken as U1  F1 2 gH

(6.23)

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Turbomachines

where E1 is the speed ratio with respect to the reference velocity 2 gH . This speed ratio has values ranging from 0.62 to 0.82 over the range of specific speeds 60−300. Equating both the expressions for U1 we get D1 

(60F1 2 gH )

PN

(6.24)

The outlet diameter D2 of the runner (inner diameter) is calculated from the ratio D2/D1, varying from 0.5 to 1, with the specific speed range of 60−300. The height B1 of the blades at the inlet is taken as equal to the height of the guide vanes B0. The number of blades on the runner is generally within p1 of that of the guide vanes. The blade angles of the runner are also chosen over a range of values, depending on the specific speed. These values are indicated in Figs. 6.15 and 6.16. It may be noted that the values of @1 and A1 shown in Figs. 6.15 and 6.16 are the values at full load on the turbine.

EXAMPLE 6.5 A Francis turbine is to be designed for the flow rate of 2 m3/s available at a project site at a net head of 10 m of water. The expected overall efficiency is 80%. The speed coefficient (or speed ratio) and the flow coefficient can be assumed as 0.8 and 0.6, respectively. The hydraulic losses in the turbine are 15% of available energy. Design the turbine rotor, with the salient dimensions and angles, to run at 300 rpm. The water leaves the rotor without any whirl component.

Solution: Given Q  2 m3/s, Hn  10 m, Go  80%, E  0.8, X  0.6, hydraulic losses  15%,

N  300 rpm, @2  90o, Vu2  0. The salient dimensions and angles of the rotor are to be calculated. Blade velocity at inlet is given by

U1  F 2 gH  0.8 2 s 9.81 s 10  11.2 m / s Flow velocity is Vf1  Y 2 gH  0.6 2 s 9.81 s 10  8.4 m /s Power output is P  wQH Ho 

9810 s 2 s 10 s 0.8  156.96 kW 1000

Specific speed is Ns 

N P 300 s 156.96   211.36 H 5/ 4 105/ 4

Now

P D1N  U1  11.2 60 Therefore the wheel diameter is

Hydraulic Turbines U1 = 11.2 a1

241

U2 = 7.84

7.445

b1 Vf1= 8.4

a2 = 90° Vf2 =Vf1 =V2

Vr1

V1

Figure 6.18

b2

Vr2

Inlet and outlet velocity triangles.

D1 

60 s 11.2  0.713 m (P s 300)

Diameter at the inlet  D1  71.3 cm. Now, because hydraulic losses are 15%, the hydraulic efficiency is

Hh  1 0.15  0.85 

U1Vu1 gH

Therefore Vu1 

0.85 s 9.81 s 10  7.445 11.2

With these values of U1, Vu1, and Vf1, it is possible to construct inlet velocity triangle, as shown in Fig. 6.18. The guide-vane angle, as obtained from the triangle, is ¥V ´ ¥ 8.4 ´ A1  tan 1 ¦ f1 µ  tan 1 ¦  48.45o § 7.445 µ¶ § Vu1 ¶ Blade angle at the inlet is ¨ Vf1 · 8.4 ¨ · B1  tan 1 © ¸  tan 1 © ¸¹  65.9o

U V 11 . 2 7 . 445

ª u1 ¹ ª To calculate the width of the runner at inlet, we have Area of flow s Flow velocity  Volume flow rate  P D1B1Vf1  Q1 Therefore wheel width at the inlet is Q 2   0.10629 P D1Vf1 P s 0.713 s 8.4  B1  10.63 cm B1 

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Now, salient features of the runner include the diameter, width, and blade angle at outlet. To calculate these parameters, the ratio D2/D1 can be assumed as 0.7 (at a specific speed of 211) and Vf2  Vf1. The diameter at the outlet is D2  D1 s 0.7  0.713 s 0.7  0.5 m  50 cm Wheel width at the outlet is B2 

Q 2   0.15 m  15 cm P D2Vf1 P s 0.5 s 8.4

Blade angle at the outlet is

B2  tan 1

Vf2 U2

 tan 1

8.4 8.4  tan 1   47o U1 s D2 7.84 D1

The outlet velocity triangle is also drawn in Fig. 6.18. Comments: 1. 2. 3. 4.

With the coefficients such as E1 and X1 as part of the data, the procedure is straightforward. The stator guide vanes are not yet designed. Eo and Xo are required. Coefficients (including that of the rotor also) can be remembered in their ranges, approximately. The dimensions and blade angle at outlet were obtained, after the value of D2/D1 was assumed as 0.7. Therefore, these parts of answers can vary over some range. As an alternate problem, if the speed were not given for the values of the product D1N, the possible values of speed and diameter could be listed. The cost factors can also be considered in choosing the appropriate speed and diameter.

EXAMPLE 6.6 A Francis turbine is supplied with water at a rate of 1.7 m3/s under a net head of 60 m and runs at 500 rpm. The blade angle of the rotor at the inlet is 90o. The ratio D2/D1 is 0.5. The velocity of flow through the rotor is constant and is equal to 8 m/s. The water leaves the rotor at 90o. The thickness of blades accounts for 5% loss of flow area. Assume that the hydraulic losses are negligible. Determine the following: (a) Inlet and outlet diameters of the rotor, D1 and D2. (b) Power. (c) Specific speed. (d) Widths of the rotor at the inlet and outlet, B1 and B2. (e) Guide-vane angle, @1. (f ) Rotor blade angle at the outlet, A2.

Solution: Given Q  1.7 m3/s, H  60 m, N  500 rpm, A1  90o, D2/D1  0.5, Vf1  Vf2  8 m/s, @2  90o, loss of area due to thickness  5%. Due to negligible hydraulic losses, we have U1Vu1  gH Because A1  90o, we can write Vu1  U1, and therefore U12  gH

Hydraulic Turbines

Thus, the blade velocity at inlet is U1  9.81 s 60  588.6  24.26 m /s The blade velocity at outlet is given by U2 

U1 2

 12.13 m /s (because D2 / D1  0.5)

Now it is possible to draw the velocity triangles, as shown in Fig. 6.19. U1 = 24.26 a1 = 18.25°

U2 = 12.13 b1 = 90° 8 Vf1 Vr1 V1

Figure 6.19

a2 = 90° Vf2 8 V2 Vr2

Inlet and outlet velocity triangles.

Also U1 

P D1 N (U s 60)  D1  1 60 (P s N )

(a) Hence, we have Rotor inlet diameter  D1  0.93 m Rotor outlet diameter  D2  D1/2  0.463 m (b) Power is given by P

wQH 9810 s 1.7 s 60   1000.62 kW 1000 1000

(c) Specific speed is given by Ns 

N P 500 s 1000.62   94.7 H 5/ 4 605/ 4

(d) To calculate the widths of the rotor at inlet, we have Flow area s Flow velocity  Volume flow rate  P D1B1Vf s (1 0.05)  Q Blade width at inlet is given by B1 

1.7 Q  P D1Vf s 0.95 P s 0.93 s 8 s 0.95

Inlet width of rotor  B1  0.0766 m  7.66 cm Outlet width of rotor  B2  0.153 m  15.3 cm (e) To calculate guide-vane angle: We know that

b2 = 33.4°

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Turbomachines

tan A1 

Vf

1

U1

So guide-vane angle is

A1  tan 1

8  18.25o 24.26

(f ) Blade angle at the outlet is

B2  tan 1

Vf1

¥ 8 ´  tan 1 ¦  33.4o § 12.13 µ¶ U2

All these values are shown in Fig. 6.19.

EXAMPLE 6.7 A project site has a possibility of 8 m3/s of water flow at a net head of 70 m. The overall efficiency is expected to be 0.85. Table 6.2 may be used in the selection of ratios or coefficients for the Francis turbine. Design all the elements of a Francis turbine for this project. Table 6.2

Ns

60

120

180

@1 E1

15o 0.62

25o 0.68

32½o 37½o 0.72 0.76

A1

120o

90o

90o

60o

0.5

0.6

0.7

0.8

D2/D1

300

Solution: Given Q  8 m3/s, Hn  70 m, Go  0.85. Now Output power  wQH nHo 

9810 s 8 s 70 s 0.85  4670 kW 1000

To start with, let E1  0.7. Inlet blade velocity is U1  0.7 2 gH  0.7 2 s 9.81 s 70  25.94 m /s Also

P D1N 60 Therefore 60U 60 s 25.94 D1N    495.4 P P Prepare a table for D1N  495.4 (Table 6.3). U1 

Table 6.3

N D1

500 0.9908

375 1.321

300 1.651

250 1.98

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Higher speeds involve smaller diameters of the rotors, the torques are lower, and blade thicknesses have to be high. Lower speeds are with higher diameters, high torques (but the same power). More important is the fact that lower speeds are to be coupled to alternators with higher number of pairs of poles. Hence, lower speeds impose increased costs of both turbine and alternator. The cost of the entire project, therefore, increases. It is required that: 1. 2.

Either a few speeds (such as 375, 300, 250 in this case) have to be selected, and after arriving at a complete design, each has to be evaluated with cost factors. Or a possibility of splitting the flow and having multiple units has to be explored.

At present, select N  300. Specific speed is Ns 

N P 300 s 4670   101.25 H 5/ 4 705/ 4

Now, make the choice from the data table (Table 6.3): E1  0.64, A1  100o, D2/D1  0.55. Blade velocity at inlet is U1  F1 2 gH  0.64 2 s 9.81 s 70  23.72 m /s Blade velocity at outlet is U 2  U1 s 0.55  13.05 m /s Also U1 

P D1N 60

Therefore diameter at inlet is D1 

U s 60 23.72 s 60   1.51 m PN P s 300

The diameter at outlet is D2  D1 s 0.55  0.83 m Because @2  90o (assumed), Vu2  0. (This is the usual case.) Therefore W  U1Vu1 J/ kg The power is 4670 kW and flow rate is 8 m3/s or 8000 kg/s. Therefore, from P  W s mass flow rate, we have 4670 s 1000  U1Vu1 s 8000 4670 s 1000  Vu1   24.61 m /s 8000 s U1 It is possible to show the velocity triangles now as in Fig. 6.20.

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Turbomachines Vu1 = 24.61 U1 = 23.72

U2 = 13.05 100° V1

Figure 6.20

b2 = 21.15°

Vf2 = 5.05 V2 Vf1 Vr2

Inlet and outlet velocity triangles.

Now, from inlet triangle, Vf1

 tan80o Vu1 U1  Vf1  (Vu1 U1 ) tan80o  5.05 m /s  Vf2 From the outlet velocity triangle, we have the outlet blade angle as

B2  tan 1

Vf2

¥ 5.05 ´  tan 1 ¦  21.15o § 13.05 µ¶ U2

Guide-vane internal diameter is Do  D1  10 mm  1.61 m Length of guide vanes is Lo  0.3 Do  0.48 m Assume the loss of area of flow due to the blade thickness as 10%. We have

P D1B1Vf s (1 0.1)  8 Therefore rotor width at the inlet is B1 

8  0.371 m  37.1 cm P s 1.51 s 5.05 s 0.9

Rotor width at the outlet is B2 

B1 0.55



0.371  0.675 m  67.5 cm 0.55

Number of guide-vanes on the stator is chosen as a thumb rule, depending on the diameter  Zo  20. Guide-vane height is Bo  B1  5 mm  37.6 cm Number of vanes on rotor is Z1  19 (this is Zo  or 1). Comments: 1.

Some comments have been given in the course of solution. The same is reinforced; a different speed and diameter may be chosen. The complete solutions may be listed as alternate solutions. Evaluation, with cost analysis, can be performed to make a final selection.

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2.

3.

247

Not all the data are required from the data table. Example is @1. In fact, after the design process, it is seen that @1 does not check with value in the table. This is because of the “range of values.” Hence, it is alright. Now, it can be appreciated why “model testing” is required (as mentioned in the first chapter).

6.5 Draft Tube A draft tube is a simple, slightly divergent conduit that is connected to the outlet of the runner of the reaction turbine at its smaller end, as shown in Fig. 6.21. Its larger end is submerged below the tailrace level. A draft tube also runs full, just like the casing, guide vanes, and runner of a reaction turbine. (The Pelton turbine does not have a draft tube. The runner of Pelton turbine does not run full, and the jet is a free jet into the atmosphere.) Runner

Draft tube pa Tailrace

Figure 6.21

Draft tube.

Generally, the turbine runners are installed at a level about 3−4 m above the tailrace level so as to avoid any eventual flooding of the entire turbine floor (in case the tailrace is blocked for the flow of water for any reason). Thus, that much head would be lost for use in the generation of the power output of the turbine. The main purpose of the draft tube is to recover that head. Because of the draft tube, the pressure at the exit of the runner is lower than the atmospheric pressure. The second purpose of the draft tube is accomplished by making the draft tube a little divergent. At the exit of the runner, there is the exit loss, V22 /2g, but due to the divergent shape, this loss is reduced to V32/2g, where V3 is the velocity at the exit of the draft tube and V3 is less than V2. The divergent angle cannot be too large; otherwise there may be separation of flow from the walls. The usual semi-cone angle of divergence is about 4o–5o. Although a very good positioning of the draft tube is vertically downward with the least length, some different orientations have been designed to meet various requirements. Some of the shapes have been discussed as follows: 1. 2.

A simple, straight type of draft tube is shown in Fig. 6.21. Moody’s design of the draft tube is shown in Fig. 6.22. This type of draft tube has a well spreadout shape with a solid cone inside. The semi-cone angle can be more, without the possibility of separation, because of the inside cone. With this, a large exit area is possible, without requiring large lengths, with a consequent lower value of exit velocity and lower exit losses.

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3.

A simple, divergent, elbow-type draft tube is shown in Fig. 6.23. The purpose of the elbow type is to accommodate a longer length and larger exit area, keeping the cone angle low, as also avoiding separation. This is suitable at places where the danger or possibility of flooding of the turbine is very remote, and therefore turbine installation need not be too high above the tailrace level.

Tailrace

Figure 6.22

Moody’s draft tube.

Tailrace

Figure 6.23

4.

Elbow-type draft tube.

A simple, divergent, elbow-type draft tube with a rectangular outlet is shown in Fig. 6.24. This has the same features as the simple elbow type, but is more suitable where excavations are required in order to place the draft tube.

Tailrace

Figure 6.24

Elbow-type draft tube with a rectangular exit.

Hydraulic Turbines

5.

249

Elbow-type draft tubes with split exit areas, shown in Fig. 6.25, are made possible by piers that have the same functions as Moody’s center cone. This also adds the advantages of constructional stability. pa Tailrace

Figure 6.25

Elbow-type draft tube with a split exit.

6.5.1 Analysis of Draft Tube For the purpose of analysis and to indicate its position with respect to the runner, a straight-type draft tube is shown in Fig. 6.26. Consider sections “2” and “3” with reference to Fig. 6.26. Assume that the losses between these points are hf due to friction, inclusive of losses due to bends (if any), etc., without any other energy interactions. Applying Bernoulli’s equation for the two sections, we have p2 w



V22 2g

 z2 hf 

p3 w



V32 2g

 z3

Rotor 2

Draft tube

hs z2

pa h3

3 Reference

Figure 6.26

z3

Draft tube.

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Turbomachines

Therefore p2 w

¥ V 2 V32 ´

( z 2 z3 ) ¦ 2 µ  hf w § 2g ¶ p3



But p3 w



pa w

 h3

where pa is the atmospheric pressure. Therefore p2 w



pa w

( z2 z3 h3 )

V22 V32 2g

 hf

From Fig. 6.26, z2 z3 h3  Hs, where Hs is the height of the rotor outlet above the tailrace. It is usual to take hf as a fraction k of (V22 V32 ) / 2 g . The equation for p2/w reduces to p2 w



pa w

H s (1 k )

V22 V32 2g

(6.25)

This equation indicates that at the outlet of the runner, “2,” the pressure head is less than the atmospheric pressure head by the following two quantities: 1. 2.

Hs, the height of installation of runner above the tailrace level. (1 k )(V22 V32 ) / 2 g , a part of the exit losses.

If the draft tube were not there, the pressure head at the exit of the runner would be the atmospheric head and the exit losses would be V22 /2g. Thus, the draft tube saves the above two heads for power output. The factor (1 k) in Eq. (6.25) is also known as the draft tube efficiency. This draft tube efficiency can be 100% when k reduces to zero, that is, when the losses in the draft tube (friction, bend, etc.) reduce to zero. Even while it is not possible to completely avoid the losses, it is always advisable to reduce k by having the internal surfaces of the draft tube very smooth and avoiding any bends, and making the length of the draft tube as small as possible.

EXAMPLE 6.8 The head available at the inlet to a Francis turbine is 28 m. The output of the turbine is 1000 kW at an overall efficiency of 0.88, when there is no draft tube attached to the exit of the turbine. The level of the turbine is 2.5 m above the tailrace. The flow coefficient of the turbine is 0.3. Assess the effect of installing (a) a straight cylindrical draft tube and (b) a draft tube with the half-cone angle of 5o at the exit of the turbine. The efficiency of the draft tube may be taken as 95% in either case.

Solution: Francis turbine, H  28 m, P  1000 kW, Go  0.88, Hs  2.5 m, X  0.3. Draft tube efficiency  95%. We know that P  wQH Ho

Hydraulic Turbines

Therefore the flow rate is Q

P 1000  4.137 m3 /s  0.88 ´ wH H0 ¥ ¦§ 9810 s 28 s µ 1000 ¶

Flow velocity is Vf  Y 2 gH  0.3 s 2 s 9.81 s 28  7.0315 m / s This is the velocity at the exit of turbine or the inlet to the draft tube. The inlet diameter of the draft tube is calculated from

P d12 V Q 4 f 0.5 4Q ¥ 4 s 4.137 ´  d1  ¦  0.8655 m µ PVf § P s 7.03515 ¶ Draft tube with inlet diameter 0.8655 m and half-cone angle 5o is shown in Fig. 6.27. d1 = 0.8655 m

2.5 m

5° d2 = 1.303 m

Figure 6.27

Draft tube (half-cone angle 5o).

Outlet diameter of the draft tube is d 2  d1  2(2.5 tan 5o)  0.8655  0.4374  1.303 m Now Velocity of water at the outlet 

1.

Q 4.137 s 4   3.103 m /s P d 22 (P s 1.3032 ) 4

Pressure at the exit of runner, when a straight draft tube is used is pa

W

( H s  0)  10 2.5  7.5 m of water

This means that 2.5 m (of water) saved.

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252

2.

Turbomachines

Pressure at the exit when a conical draft tube is used is ¨ ¥ V 2 V32 ´ · 10 © H s  0.95 ¦ 2 µ ¸  10 (2.5  1.928)  5.575 m of water § 2 g ¶ ¸¹ ©ª

This means 4.428 m (of water) saved. The extra power would have been: 1.

¥ 9810 s 4.137 s 2.5 s 0.88 ´ wQH extra Ho  ¦ µ¶  89.28 kW, that is, 8.9% extra power. § 1000

2.

¥ 9810 s 4.137 s 4.428 s 0.88 ´ wQH extra Ho  ¦ µ¶  158.14 kW, that is, 15.8% extra power. § 1000

6.6 Cavitation Cavitation is a phenomenon that starts in low-pressure regions in liquids adjacent or close to solid surfaces. If the low pressure is equal to or less than the vapor pressure of the liquid at the prevailing temperature, the liquid starts vaporizing. The vapor bubbles are formed on the solid surfaces. When, due to eddies and turbulence, the higher pressure liquid comes adjacent to such vapor bubbles, the pressure in the bubbles also increases and the vapor condenses. The bubbles collapse and the surrounding liquid rushes from all the directions to occupy the space vacated by the condensing vapors. This generates very high localized pressures on the surfaces, enough to even chip off the metal particles and cause pits or cavities. The overall process is named pitting or cavitation. Such cavitation occurs over a large number of points on a surface. It is accompanied by the vibrations due to the collapse of vapor bubbles and rush of water from all around. Cavitation, even from its starting, causes the loss of surface finish, and therefore increases losses. This reduces the efficiency of the machine. Continuous cavitation can cause even complete mechanical failure of the parts. Cavitation can occur in low-pressure regions that are at the discharge side of the reaction turbine blades (Francis or Kaplan) when connected to the draft tubes. In Fig. 6.28 the turbine outlet with draft tube is shown. Turbine outlet

2

Draft tube

Hs

pa Tailrace

3

Figure 6.28

Cavitation

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253

In Fig. 6.25, the point of the likely occurrence of cavitation is at a height Hs m above the free surface level of water in the tailrace. The free surface level of water in the tailrace is at the atmospheric pressure pa. Due to the water column in the draft tube, the pressure at the blade surfaces is less by Hs. Also, the pressure is further reduced by Hd (V22 V32 ) / 2 g in the draft tube as seen by Eq. (6.25). Thus p2 w



¥ V 2 V32 ´

H s Hdt ¦ 2 µ w § 2g ¶ pa

(6.26)

When the pressure p2, as indicated by Eq. (6.26), reduces to the level of vapor pressure of the water at the temperature of the surroundings, the cavitation is likely to commence. When p2 reduces further (below the vapor pressure), cavitation is almost certain to occur. Cavitation is not a desirable phenomenon. Efforts to avoid cavitation can be made by preventing the formation of low-pressure regions by the installation of the turbine as near to the tailrace level as possible, or even below that level. Cavitation damage also can be controlled by the use of high-strength materials (e.g., cast iron, cast steel, stainless steel, etc.).

6.7 Kaplan Turbine and Propeller Turbine The Kaplan turbine or propeller turbine is a low head (up to 75 m), large flow rate, high specific speed (300–1000), axial flow, reaction turbine. A schematic view of the Kaplan turbine is shown in Fig. 6.29. Inlet

Spiral casing

Guide vanes

Shaft a b

Runner vanes

Draft tube

B0 Whirl chamber

Hub

Tailrace

Outlet

Figure 6.29

Kaplan turbine with an ellipsoidal whirl chamber.

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Turbomachines

6.7.1 Constructional Features of Kaplan and Propeller Turbines A schematic view of the Kaplan turbine is shown in Fig. 6.29. Water enters a spiral casing and passes on to the ring of guide vanes, just like in the Francis turbine. The guide vanes have their individual axes, about which they can swing, to control the quantity of flow of water. As the water passes through the guide vanes, its head energy partly gets converted into kinetic energy, and it acquires a whirl component of its velocity. The water then enters an ellipsoidal whirl chamber, where it gets into a spiral vortex flow pattern before meeting the leading edge of the runner blades. As the water flows through the passages between the runner blades, the runner blades absorb the energy from the water. The water is then discharged with an axial flow pattern into a draft tube and then to the tailrace. As a reaction turbine, the Kaplan turbine runs full through the casing, the guide vanes, the runner, and the draft tube. One very important feature of the Kaplan turbine is that, unlike the Francis turbine, its runner blades can be swung about their own axes so as to change their blade angles, by a governor-actuated mechanism, to suit the demand of the load even when the turbine is running. The propeller turbine is almost identical to the Kaplan turbine in its features such as the head, specific speed, design, construction, etc. The only difference is that the blades of the runner are integral with the hub. It is not possible to swing the runner blades. Hence, propeller turbines are suitable for constant load (base load) applications, with peak efficiency only at the design load, whereas Kaplan turbines can have the peak efficiency over a large stretch of part-load performance. There are some installations of Kaplan or propeller turbines where the available head may be as low as 6 m to 9 m or 10 m, but with large flow-rates. These are the “run-of-the-river” power plants where the storage reservoir is of minimal capacity, but water flows in huge quantities round the clock. In such installations, there is no need of penstock pipe or the spiral casing. Water enters the guide-vane ring from all around as if directly from the reservoir. After passing the runner blades, the water is carried to the tailrace by a bent draft tube. The turbine installation is at the bottom of the dam that impounds the water. The height of the dam may not be more than some 10 12 m. The generator floor can be above that level also. The schematic diagram of this type of installation is shown in Fig. 6.30. pa Upstream

Power house Trash gate

Shaft

Guide vanes

H

Down stream Tailrace

Turbine

Draft tube

Figure 6.30

Kaplan turbine without spiral casing.

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255

6.7.2 Analysis of Kaplan and Propeller Turbines Because of the large volume flow rates of Kaplan or propeller turbines, the area of flow across the runner blades has to be quite large. Hence, the diameter of the runner (D1) is generally quite big. The length of the runner blade is (D1−d)/2, where d is the hub diameter. The peripheral velocity of the blade at the tip (U1t ) and that at the hub (U1h ) are not the same. This is true also at the outlet end of the blade. As a result, the blade is required to have a “twist” from the hub to the tip, and it is required to have two sets of velocity triangles: one set (inlet and outlet) at the tip and another set (inlet and outlet) at the hub. The two sets are shown in Fig. 6.31 along with the Kaplan blade. U1h Vu1h

U1t Vu1t Vr1h

Vr1t

V1t Inlet

V1h Inlet

U2t

Vr2t

V2t

Outlet (a)

Figure 6.31

U2h

Vr2h

V2h

Outlet

(b)

(c)

Velocity triangles of Kaplan turbine: (a) Kaplan blade, (b) tip edge, and (c) hub edge.

At the outlet, or discharge end, the absolute velocity of water is axial, without any whirl component, at both the tip and hub. The specific works when calculated at the tip and hub are given by Wtip  U1tVu1t and

Whub  U1hVu1h

However, the flow at the inlet to the blade is like a free vortex (VuR  constant). Therefore, it may be easily seen that the specific works at the tip and hub are the same and hence the expression for the overall specific work can be written as W  (U1Vu1 )tip (6.27) The hydraulic efficiency is given by

Hh   Hh 

W gH (U1 Vu1 )tip gH

(6.28)

6.7.3 Efficiencies of Kaplan or Propeller Turbine The comments written in connection with the efficiency of Francis turbine may be read as being referred to Kaplan turbine or propeller turbine.

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Turbomachines

6.7.4 Design Parameters of Kaplan and Propeller Turbines Kaplan or propeller turbine is designed by approximately the same procedure as the Francis turbine. The numerical values of the flow ratios, the speed ratios, etc. are, however, different and these ratios are based on the specific speeds ranging from 300 to 1000. Because Kaplan or propeller turbines are purely axial flow turbines, the outlet diameter D2 is equal to the inlet diameter D1. One new parameter for Kaplan or propeller turbine is the diameter d of the hub. This is calculated from yet another specified ratio d/D1 that depends again on the specific speed. All the parameters are decided by extended experimentation and experiences, as stated with reference to Fig. 1.13. The values quoted are shown in Table 6.4. Table 6.4

Design parameters of Kaplan or propeller turbines

Head (m)

75

50

40

20

5

Specific speed

300

400

600

800

1000

E0

1.3

1.5

1.8

2.1

2.5

E1

1.2

1.4

1.7

2.0

2.4

X0

0.12

0.16

0.24

0.32

0.39

X1

0.35

0.4

0.51

0.62

0.75

d/D1

0.7

0.55

0.5

0.4

0.3

No. of blades on the runner

10

6

5

4

3

As in the Francis turbine calculations, the design parameters can be used in the following equations to design the Kaplan or propeller turbines: 1.

Spiral casing: The inlet diameter D can be calculated for the available flow rate Q with a maximum velocity V of about 8−10 m/s as D

4Q PV

(6.29)

With this diameter, the inlet area of the spiral casing is (OD2)/4. This area can be uniformly reduced, over 360o, around the guide-vane ring, as per the following equation (refer Fig. 6.17): ¨ (P D 2 ) · © 4 ¸ (360 Q ) ¹ AQ  ª 360 2.

The inner diameter D0 of the guide-vane ring is given by D0 

3.

(6.30)

60F0 2 gH

PN

(6.31)

The length of the guide vanes is L0  0.3 D0

(6.32)

Hydraulic Turbines

4.

The height of the guide vanes is B0 

5.

Q P D0Y 0 2 gH

60F0 2 gH

PN

d D1

10.

11.

(6.36)

From Hh  U1Vu1 / gH , we have Vu1 

9.

(6.35)

The flow velocity is Vf  Vf1  Vf2 , Vf  Y 1 2 gH

8.

(6.34)

The runner hub diameter is d1  D1 s

7.

(6.33)

The runner tip diameter is D1 

6.

257

gH Hh U1

(6.37)

where U1  P D1N / 60. From the inlet velocity triangle at the tip, the guide-vane angle is ¥V ´ A1  tan 1 ¦ f µ § Vu1 ¶

(6.38)

¥ Vf ´ B1  tan 1 ¦ µ § U Vu1 ¶

(6.39)

The blade angle at the tip is

At any radius R, the whirl component (Vu 1 )R can be calculated by (Vu1 )R 

Vu1R1 R

(6.40)

where R1  D1 / 2. 12.

Accordingly, the blade inlet angle at any radius R is given by ¨ · Vf (B1 )R  tan 1 © ¸ ªU (Vu1 )R ¹

13.

(6.41)

The blade outlet angle at the tip is ¨V · (B2 )t  tan 1 © f ¸ ªU 2t ¹

(6.42)

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14.

Turbomachines

The blade outlet angle at the hub is ¨V · (B2 )h  tan 1 © f ¸ ªU 2h ¹

15.

The blade outlet angle at any radius R is ¨V · (B2 )R  tan 1 © f ¸ ªU 2R ¹

16.

(6.44)

The semi-major axis a of the ellipsoidal whirl chamber (refer Fig. 6.29) is a  0.16D1 to 0.2D1

17.

(6.43)

(6.45)

The semi-minor axis b of the ellipsoidal whirl chamber (refer Fig. 6.29) is b  0.13D1

(6.46)

EXAMPLE 6.9 The speed ratio and flow ratio of a Kaplan turbine are 2.0 and 0.65, respectively. The hub diameter to tip diameter ratio is 0.3. The hydraulic efficiency and overall efficiency are estimated as 92% and 85%, respectively. The head acting on the turbine is 4 m. One single rotor develops as much as 8000 kW of power. Determine (a) flow rate, (b) the tip and hub diameters of the runner, (c) the speed, (d) the specific speed, (e) and the blade angles of the runner. Assume that the discharge is axial and flow components remain constant through the rotor.

Solution: Given E  2.0, X  0.65, Dh /Dt  0.3, Gh  0.92, Go  0.85, H  4 m, P  8000 kW. Inlet velocity at the tip (U1)t is given by (U1 )t  F 2 gH  0.2 s 2 s 9.81 s 4  17.7 m /s Flow velocity Vf1 is given by Vf1  Y 2 gH  0.65 s 2 s 9.81 s 4  5.76 m /s (a) To find the flow rate, we know that P  wQH nHo Therefore the flow rate is Q

P 8000  239.85 m3 /s  0.85 ´ (w s H n s H0 ) ¥ ¦§ 9810 s 4 s µ 1000 ¶

(b) To find diameters of the runner: Flow area s Flow velocity  Flow rate ¨ P ( Dt2 0.32 Dt2 ) · © ¸ Vf  Q 4 ª ¹

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259

So runner tip diameter is Dt2 

Q s4 239.85 s 4   7.633m [Vf s P s (1 0.09)] (5.76 s P s 0.91)

As per data, Dh /Dt  0.3. Therefore, runner hub diameter is Dh  Dt s 0.3  7.633 s 0.3  2.29 m (c) To find the speed, we have (U1 )t 

P Dt N 60

Therefore, speed of the runner is N 

60 s (U1 )t

P Dt



60 s 17.7  44.29 rpm P s 7.633

(d) Specific speed is Ns 

N P 44.29 s 8000   700.286 H 5/ 4 45/ 4

So N s  700 (e) To find the runner blade angles: There are four angles to be found at inlet and outlet at the tip; and inlet and outlet at the hub. We start with hydraulic efficiency, which is

Hh  

U1Vu1 gH

U1Vu1

 0.92 (as per data) gH  (Vu1 )t  0.92 s 9.81 s 4 / 17.7  2.04 m /s (Vu1)t = 2.04

(Vu1)h = 6.8 (U1)t = 17.7

(U1)h = 5.31 (b1)h = 75.5°

(b1)t = 20.2° (V1)t (Vr1)t

Figure 6.32

(V1)h

(Vr1)h

(Inlet)

Inlet velocity triangles at tip and at hub.

For free vortex, Vu R  constant

260

Turbomachines

Therefore (Vu1 )h  (Vu1 )t s

Rt Rh

 2.04 s

1  6.8 0.3

With this, the inlet velocity triangles can be drawn as shown in Fig. 6.32. At the inlet: The inlet blade angle at the tip is Vf ¨ · 5.76 · ¨ 1 (B1 )t  tan 1 © ¸  tan 1 © ¸  20.2o ( )

( ) ( . . ) U V 17 7 2 04

¹ ª ©ª 1 t u1 t ¸ ¹ The inlet blade angle at hub is ¨ · V f1 5.76 (B1 )h  tan 1 ©  75.5o ¸  tan 1 (5.31 6.8) ª (U1 )h (Vu1 )h ¹ Velocity triangles at the outlet, for both tip and hub, are now drawn in Fig. 6.33. (U2)t = 17.7

(U2)h = 5.31 (b2)h

(b2)t = 18.03°

(Vf2)t

(Vf2)h

= 47.33°

(V2)h

(V2)t (Outlet)

Figure 6.33

Outlet velocity triangles at tip and at hub.

At the outlet: The outlet blade angle at the tip is ¨ 5.76 · (B2 )t  tan 1 © ¸  18.3o ©ª17.7 ¸¹ Outlet blade angle at the hub is ¨ 5.76 · (B2 )h  tan 1 © ¸  47.33o ©ª 5.31 ¸¹ Comments: 1.

2.

This is a typical run-of-the-river power plant. The land topography, in such cases, does not permit a large reservoir. The head of 4 m, mentioned in the example, may not require even a penstock conduit or a spiral casing as shown in Fig. 6.30. Water just arrives, passes through the turbine, and goes off. It is a good experience to visualize the blade of the runner with the angles as calculated.

Hydraulic Turbines

261

EXAMPLE 6.10 A project site has a capacity of providing with 150 m3/s of water flow at a head of 18 m. The overall efficiency can be 0.87 for the development. Select the type of turbine and calculate the salient dimensions and angles of the runner. The speed can be assumed as 375 rpm.

Solution: The power that can be developed is given by P  wQH Ho 

9810 s 150 s 18 s 0.87  23044 kW 1000

Therefore the specific speed is Ns 

N P 375 s 23044   1535.4 H 5/ 4 185/ 4

This specific speed, calculated for a single unit of turbine, indicates that there must be multiple units of turbines. The specific speeds corresponding to the number of units are listed as follows: (Ns can be calculated as 1535.4/ x , where x is the number of units.) 1. 2. 3. 4.

Two units, power per unit 11522 kW, Ns  1086 Three units, power per unit 7681 kW, Ns  886 Four units, power per unit 5761 kW, Ns  768 Five units, power per unit 4609 kW, Ns  687

It is better to select (except Ns  1086) the highest Ns, 886, so that three identical units of Kaplan-type turbines work to produce the total power, each unit having a power of 7681 kW. Flow rate per unit is Qu  150/3  50 m3 /s Assume X1  0.65 (corresponding to Ns  886). Therefore, the flow velocity is Vf  0.65 2 gH  0.65 2 s 9.81 s 18  12.215 m /s Assuming d1 D1, runner tip diameter D1 is given by having Flow area s Flow velocity  Volume flow rate

P (D12 0.32 D12 ) Vf  Q u 4 4Qu  D12 (1 0.09)  PVf 4 s 50  D12  P s 12.215 s 0.91



So Runner tip diameter  D1  2.393m  D2 As assumed above, Hub diameter  Tip diameter s 0.3

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Turbomachines

The runner hub diameter is obtained as d1  2.393 s 0.3 m  0.718m  d 2 Blade velocity at inlet (U1)t and outlet (U2)t at the tip are calculated as (U1 )t  P D1N /60  46.99 m /s  (U 2 )t Similarly, blade velocity at inlet (U1)h and outlet (U2)h at the hub are calculated as (U1 )h  P d1N /60  14.10 m /s  (U 2 )h The power is given by P  Wm where m  50

kg m3 s 1000 3  50000 kg / s s m

Therefore, specific work is W 

P 7681 s 1000  153.62 J / kg  (U1Vu1 )tip m 50000

The whirl component of inlet velocity at the tip is 153.62 (Vu1 )tip   3.27 m/s 46.99 Now, in the whirl chamber of Kaplan turbine, the flow is as per the free vortex theory, that is, Vu R  Constant Therefore, the whirl component of velocity at the hub is D 3.27 (Vu1 )hub  (Vu1 )tip s 1   10.9 m/s d1 0.3 Knowing the values of blade velocities (at tip and hub), the whirl components at the inlet (at tip and hub), and a constant flow component of velocity in the entire flow passage, it is possible to draw the velocity triangles at (a) inlet tip, (b) inlet hub, (c) outlet tip and (d) outlet hub, as shown in Fig. 6.34. (U1)t = 46.99 Vf1 12.215

(b1)t = 15.61°

(V1)t

(Inlet)

(Vr1)t (a)

(U1)h = 14.1 Vf1 (b 1 )h = 75.32° 12.215 (V1)h (b) (U2)h = 14.1

(U2)t = 46.99

(b 2 )h = 40.9°

(b 2 )t = 14.57° (V2)t

(Outlet)

(Vr2)t (c)

Figure 6.34

(Vr1)h

(Vr2)h (d)

Velocity triangles: (a) and (b) Inlet triangles at tip and at hub; (c) and (d) outlet triangles at tip and at hub.

Hydraulic Turbines

263

The blade angles are calculated with reference to the triangles, as follows: 12.215  15.61o 43.72 12.215  tan 1  75.332o 3.20 12.215  tan 1  14.57o 46.99 12.215  40.90o  tan 1 14.10

(B1 )tip  tan 1 (B1 )hub (B2 )tip (B2 )hub

Comment: Kaplan turbines, as have been seen, are meant for very high flow rates at rather low heads. Hence, their working on part loads does not generally arise. As a result, as far as possible, one big unit is preferable to two/three smaller units. Yet, in this example, three units are recommended to maintain the specific speeds within the normally accepted limits, that is up to Ns  1000 (refer Table 6.1).

6.8 Governing of Hydraulic Turbines In hydroelectric power plants, hydraulic turbines are employed to drive the alternators. These alternators are required to run at constant speeds in order to supply the electrical power at constant frequency at all loads. Hence, hydraulic turbines have to run at a constant speed. Under steady-state conditions, the alternator supplies a constant load, that is, the output power of the alternator is constant. The rate of water supply to the turbine also remains constant. Suppose that the load on the alternator decreases. If the water supply is not correspondingly decreased, the turbine picks up the speed because of the less output at constant input (i.e., excess input energy). Similarly, if the load on the alternator increases, without change in the rate of supply of water to the turbine, the speed of the turbine and alternator drops. Thus, in both cases, some unbalance is created between the input and output, and the speed suffers a change. Governing is the process that maintains the speed constant by regulating the supply of water as required by the variation of the load.

6.8.1 Parts of the Governing System A schematic layout of a governing system for hydraulic turbines is shown in Fig. 6.35. The various units that make up the overall governing system and their functions are as follows: 1.

2.

3.

Governor: This is the speed-sensing unit. The input to this unit is the speed of the turbine in the form of a drive from the main shaft of the turbine. The output of the unit is the vertical movement of an end of a lever as shown. As the speed increases, the flyballs get raised and end A of the lever moves up. As the speed decreases, end A of the lever moves down. Actuating unit: A pump creates the pressure in the circulating oil that is supplied to a distributing valve. This distributing valve has its input from the lever, adjusting the position of a double piston, so as to guide the oil flow to the servomotor. Servomotor: This is a cylinder−piston arrangement. When high pressure oil is supplied to one side of the piston, the other side gets connected to the drain (and vice versa) through the distribution

264

Turbomachines

valve. Thus, the movement of the piston is controlled, and the piston is connected further to the inlet side of the turbine. Governor

A

Flyballs Lever

B

Fulcrum Drive from turbine-shaft

Distribution valve F E C Pump

D

G

Oil sump Servomotor

Figure 6.35

4.

Piston

Governing system.

Regulators: This is the final unit of the governing system. Depending on the turbine type, there can be different types of regulators.

In the case of Pelton turbine, the piston of the servomotor is connected to the spear that is placed inside the nozzle. The movements of the piston and the spear are therefore the same, controlling the flow of water from the nozzle (also refer Figs. 6.6 and 6.7). In the case of the Francis or Kaplan turbine, the movement of the piston of the servomotor is integrated to the rotation of a regulating wheel. As this wheel rotates, all the inlet guide vanes swing about their individual axes so as to vary the area of flow for water, thereby controlling the flow of water to the rotor (also refer Fig. 6.14).

6.8.2 Working of the Governing System Referring to the layout diagram (Fig. 6.35), when the speed of the turbine increases, the flyballs of the governor are raised, lifting end A of the lever and lowering end B. In the distribution valve, the high-pressure oil from valve C is connected to valve D, simultaneously connecting valve E to valve F. The left side of the piston of the servomotor is now subjected to the supply of high-pressure oil, and the right side is exposed to the drain, through valves E and F. The spear (of the nozzle assembly) moves to the right, reducing the flow of water and bringing down the speed of the turbine back to the rated speed. When the speed of the turbine is reduced due to higher loads, the flyballs are lowered, end A of the lever is lowered, end B is raised, valves C and E are connected, valves D and G are connected, the piston of servomotor moves to the left, and the spear moves back, allowing more water to flow to meet higher load demand.

Hydraulic Turbines

265

Regulating ring Guide vanes Open

Open Closed

H Servomotor Close

Figure 6.36

Governing of reaction turbines.

In the case of reaction turbine (Francis, Kaplan or propeller), the movement of the piston of the servomotor is connected to another cranking lever, H, which in turn is connected to the regulating wheel. As shown in Fig. 6.36, the rightward movement of the piston of the servomotor rotates the cranking lever clockwise, the regulating wheel also moves clockwise, and the guide vanes close in, lowering the water flow rate. The withdrawal of the piston of the servomotor causes the opening of the guide vanes to allow more water to flow to meet the higher loads on the turbine. Main shaft Pinion Sleeve Helical groove Blades

Gear

Sleeve Hub Shaft

Figure 6.37

Schematic view of rotor blade control in the Kaplan turbine.

For Kaplan turbines, apart from controlling the guide vanes, the governing system also controls the swinging action of the rotor blades, as shown in Fig. 6.37. The entire mechanism is enclosed inside the hub of the main shaft. The complete assembly of this mechanism rotates along with the main shaft. When required to control the swing of the rotor blades, the motions of the parts are relative to the shaft, and therefore the understanding of the movements is better facilitated if the shaft is considered stationary. The up and down movement of the governor sleeve is connected to another sleeve with helical grooves on its surface. A gearwheel, having its stubs engaged in the helical grooves, gets a rotating movement that in turn is used to swing all the blades through a conical pinion on the axes of the swing of the blades. For the sake of clarity, the sleeve, the gearwheel, and the conical pinion are also shown separately and only one direction of movement is marked in Fig. 6.37.

266

Turbomachines

As the speed of the main shaft is increased due to reduced load, the sleeve moves up and the gearwheel rotates in the direction shown, causing the pinions (as many as number of blades) also to rotate and flatten the blades. The flow area is reduced, the flow rate is also reduced, and the speed is brought back to the rated value.

6.9 Characteristics of the Hydraulic Turbine Hydraulic turbines are run to drive alternators or generators in powerhouses. These turbines are required to run at a constant speed with the loads (on the generator) that do not remain constant. Because of the governors, the input (i.e., the flow rate of water) is varied to match the output, keeping the speed constant. The plots of the power output and efficiency on the base of input are known as the operating characteristics of turbines. The speed of the machine, over the stretch of the operations, has to remain constant. The variations of power, efficiency, torque, etc. are different for different types of turbines. Some variations are shown in Fig. 6.38. Pelton 1.0

4.0

Kaplan

0.5

0

p/pmax

h/hmax

Propeller

25

50

Francis

Kaplan

Propeller

Pelton

75

% Rated flow

Figure 6.38

Francis

0.5

100

0

25

50

75

100

% Rated flow

Operating characteristics of hydraulic turbines.

As the flow rate is varied in the Pelton turbine, the jet diameter is varied. But the velocity of the jet and the angle of jet do not vary. As a result, the velocity triangles remain the same. Theoretically, there is no change in the mechanism of energy transfer. However, the disk friction losses, hydraulic losses, etc. continue to be the same. Hence, the efficiency is slightly reduced. On the whole, the efficiency is considered to remain the same over a wide stretch. In the Francis turbine, the situation is slightly different. Because of the governing action, the guide vanes allow the water to flow at different angles. The water strikes the rotor blades at angles other than the design values. The flow velocity is also reduced. The smooth glide of water on the blade surface gets disturbed. The losses increase, and the efficiency is reduced. A similar situation holds good in the case of propeller turbine. In the Kaplan turbine, because of the dual control of the governor, the blade angles are also varied along with the reduction of flow velocity. This gives rise to the smooth flow of water at the altered load also (with new velocity triangles). The losses increase a little, but not too much. The efficiency also remains approximately the same over a wide stretch of loads.

Hydraulic Turbines

267

The characteristics obtained with varying speeds are known as main characteristics. Actually, the hydraulic turbines, as they are employed in power houses, are supposed to run at a rated speed. The speed is one of the main parameters with which the whole procedure of analysis and design is started. The blade angles are designed for the constant speed. Hence, the main characteristics are not of much practical application.

K eywords Hydroelectric power project Head and flow rates at site Impulse and reaction turbines Selection based on the head Selection based on the specific speed Pelton turbine, construction, double cup Nozzle, jet, deflector, brake jet Multi-jet Pelton turbine Velocity triangles, specific work Efficiencies, design parameters

Francis turbine, construction, runner Guide vanes, swing, control Velocity triangles, specific work Efficiencies, design parameters Draft tube, types, efficiency Cavitation, causes, and prevention Kaplan and propeller turbines Swing of runner blades of Kaplan turbine Velocity triangles, specific work Efficiencies, design parameters

S ummary 1. Hydraulic turbines are the prime movers that drive alternators in hydroelectric power stations. The energy flow to the turbines is in the form of flow rate of water (Q m3/s) and head of water (H m of water). 2. Input power, P  wQH. 3. The different types of turbines and the criteria of their selection are: (a) Pelton turbine: High head (150 m and above) and low specific speed (5 70). (b) Francis turbine: Medium head (50 250 m) and medium specific speed (60 300). (c) Kaplan turbine: Low head (5 75 m) and high specific speed (300 and above). (d) Propeller turbine: Same as Kaplan turbine; this is for constant load application. 4. The components of the basic types of turbines (in the order of flow of water) are: (a) Pelton turbine: Nozzle, double-cups, rotor. (b) Francis turbine: Volute casing, guide-vanes, rotor, draft tube. (c) Kaplan turbine: Volute casing, guide-vanes, ellipsoidal whirl chamber, rotor, draft tube.

5. The use of the draft tube is to recover the head lost in the installations of reaction turbines when the turbines are required to be installed at some level above the tailrace level. 6. Cavitation occurs due to low pressures (below the vapor pressure of water), in the discharge ends of rotor-blades of reaction turbines, when draft tubes are installed. Its causes are: (a) Higher heights of installation above the tailrace. (b) Lower atmospheric pressures (higher altitudes). (c) Higher atmospheric temperatures. (d) Higher losses in the draft tubes (bends, length, rough inside finishes). (e) Effects of cavitation are loss of efficiency and (eventual) failure of parts. (f ) Prevention of cavitation is by installation of turbines as near to tailrace as possible. 7. Governing is the process of maintaining constant speed of turbines with varying loads, by adjusting the flow rates of water. 8. Components of governing system are: speed sensing mechanism (fly-balls), actuating

268

Turbomachines

mechanism (pump, actuating valve), servomotor, and regulator. Regulator works on the spear-nozzle assembly of Pelton turbine and on the guide-vane assembly of Francis turbine and propeller turbines. In Kaplan turbines, it works on both guide-vane assembly and rotor blades.

9. The operating characteristics of the turbine are plots of the power output and efficiency on the base of input. Pelton and Kaplan turbines give better performance (less drop in efficiency) during part load working than the Francis and propeller turbines.

I mportant E quations Pelton Turbine: 1. Velocity at the nozzle outlet: V1  c V 2 gH m /s 2. W  E (1 – E) (1  cb cos A2)V12 , where E  U/V1 3. Wmax  0.25(1  cb cos A2) V12 4. Ghmax  (1  cb cos A2)/2 5. U  (0.45 0.46)V1 m/s

4Q j

6. d 

PV1

7. L  2.3d to 2.8d 8. B  2.8d to 3.2d 9. T  0.6d to 0.9d 0.5D 10. Z   15 d

Francis Turbine: 1. Hh  2. D0 

U1Vu1

7.

gH

w

5. D1  6.

p2 w



PN

Hh 

Q P D0Y 0 2 gH

9. Vu1 

60F1 2 gH

w

H s (1 k )

¥ V 2 V32 ´

H s Hdt ¦ 2 µ w § 2g ¶ pa

(U1Vu1 )tip gH gH Hh U1

10. A1  tan 1

PN pa



8. Hydraulic efficiency

60F0 2 gH

3. L0  0.3 D0 4. B0 

p2

V22 V32 2g

, where U1 

P D1N 60

Vf Vu1

¨ Vf · 11. B1  tan 1 © ¸ ª (U Vu1 ) ¹

M ultiple- C hoice Q uestions 1. Hydraulic turbines are not classified based on the (a) specific speed (b) specific work

(c) flow direction (d) dynamic action of fluid 2. The diameter ratio D2/D1 of Francis turbine is selected based on the

Hydraulic Turbines

3.

4.

5.

6.

7.

(a) specific speed (b) specific work (c) degree of reaction (d) head The input energy to the Pelton turbine is of the form (a) pressure energy. (b) head energy (c) kinetic energy (d) none of these To stop the Pelton turbine, the equipment used is the (a) deflector valve (b) distributor valve (c) control valve (d) governor The efficiency of the Pelton turbine does not depend on the (a) speed (b) specific speed (c) specific work (d) surface finish of cups One of the different parts of Pelton turbine setup is a (a) guide vane (b) nozzle (c) rotor vane (d) draft tube The input energy to the Francis turbine rotor is of the form of

269

(a) kinetic energy (b) head energy (c) both kinetic and head energy (d) none of these 8. The velocity components of the fluid, responsible for the energy transfer from the fluid to rotor in the Francis turbine, are (a) flow components (b) tangential components (c) radial components (d) axial components 9. The angle of the exit velocity of fluid in the Francis turbine depends on the (a) specific speed (b) specific work (c) speed (d) none of these 10. The main function of the draft tube is (a) to guide the way-out for the water (b) to recover the exit velocity (c) to recover the head lost due to higher level installation (d) none of these

R eview Q uestions 1. Write a note on the classification of hydraulic turbines. (Refer Section 6.2) 2. Discuss the selection of hydraulic turbines for a project site. (Refer Section 6.2.1) 3. Explain with a neat sketch the single-jet Pelton turbine installation. (Refer Section 6.3.1) 4. What is the function of the braking jet? How does it work? (Section 6.3.1, Explanation with a figure of a nozzle/jet in the opposite direction) 5. When is the multi-jet Pelton turbine to be used? (Refer Section 6.2.1) 6. Explain with a neat sketch the setup of a Francis turbine. (Refer Section 6.4.1) 7. Describe the regulation of the movement of guide vanes and its purpose. (Sections 6.8.1, 6.8.2. Explanation can be with reference to Fig. 6.36)

8. Discuss the twin objectives of the draft tube. Also discuss how these objectives are met with. (Refer Section 6.5) 9. Sketch the different types of draft tubes and mention their advantages. (Refer Section 6.5, Figs. 6.21−6.25) 10. What are the factors affecting the efficiencies of the Francis turbine? (Refer Section 6.4.3) 11. Define the draft tube efficiency. (Refer Section 6.5.1) 12. Explain the phenomenon of cavitation. Where is it likely to occur in turbine installations? How to reduce its effects? (Refer Section 6.6) 13. Explain with a neat sketch the setup of Kaplan turbine. Where is it suited? (Refer Section 6.7)

270

Turbomachines

14. What is the special feature of Kaplan turbine? Discuss the purpose of special feature. (Refer Section 6.8.2) 15. Discuss the governing of Kaplan turbine. (Refer Section 6.8.2) 16. Write a note on the need of the governing process of the hydraulic turbines. (Refer Section 6.8) 17. Discuss with a neat sketch the governing system with respect to hydraulic turbines. (Refer Section 6.8.1)

18. Discuss with a neat sketch the working of a governing system. (Refer Section 6.8.2) 19. Draw the typical operating characteristics of hydraulic turbines. Discuss their features. (Refer Section 6.9) 20. Draw the typical velocity triangles of the different types of hydraulic turbines. (Refer Figs. 6.10, 6.15, 6.31)

E xercises 1. As a small laboratory project, a table-top working model of a Pelton turbine is required to be designed. It may be assumed that water is available at a constant net head of 4 m from the overhead tank. A nozzle of 5 mm diameter is the starting point. Calculate the following parameters of the Pelton turbine with a proposed speed of 600 rpm, assuming a speed ratio of 0.46, nozzle velocity coefficient of 0.98, and an overall efficiency of 83%: (a) power, (b) diameter, (c) specific speed, (d) number of Pelton cups, (e) length of the cup and (f) breadth of the cup. 2. As a small laboratory project, a table-top working model of a Francis turbine is required to be designed. The water is available at a constant head of 4 m from the over-head tank. As a starting point, assume a specific speed of 120, a rotor speed of 800 rpm, and an overall efficiency of 75%. Also in standard notations, assume the following: Eo 0.75, X0  0.36, E1  0.68. Calculate the following: (a) power, (b) flow rate, (c) inner diameter of guide-vane ring, (d) length of guide vanes, (e) breadth or height of guide vanes, (f ) inlet diameter of runner, (g) breadth of runner inlet, (h) number of runner vanes, (i) guide-vane angle, (j) runner blade angle at the inlet, and (k) runner blade angle at outlet. 3. The net head at the nozzle of a Pelton turbine is 260 m of water and the flow rate available is

estimated as 4.8 m3/s. The speed ratio and the coefficient of velocity of nozzle are 0.46 and 0.96, respectively. The overall efficiency may be taken as 85%, as a starting point. The selected speed is 500 rpm. Calculate the (a) specific speed, (b) diameter of rotor, (c) number of jets, and (d) jet diameter. 4. A Pelton turbine is coupled to an alternator, running at 600 rpm. The output of the alternator is 4300 kW. At this output, the alternator is specified to have an overall efficiency of 88%. On the turbine side, the mechanical loss on the shaft is found to be 100 kW. The Pelton cups turn the jet through an angle of 160o ( A2  20o). The relative velocity of water experiences a loss of 10% due to friction, as the water moves in the cups. The head measured at the nozzle is 340 m of water. The nozzle efficiency is 0.975. Calculate the (a) output power of the turbine, (b) mechanical efficiency of the turbine, (c) hydraulic efficiency, (d) water flow rate, (e) specific speed, and (f ) jet diameter. 5. The inlet blade angle of a Francis runner is 55o. The flow rate of water is 2.8 m3/s at a net head of 65 m of water. The discharge of water from the runner is without any whirl component. The overall efficiency of the turbine is 82%. The outlet diameter of runner

Hydraulic Turbines

is 0.6 times the diameter at the inlet. The blade velocity at the inlet is 26 m/s. For the runner wheel, the diameter to width ratio is 8. The flow component of velocity of water remains constant in the runner. Calculate the (a) power of the turbine, (b) specific work, (c) the speed ratio of the runner E1, (d) runner inlet diameter, (e) runner width, (f ) speed, (g) specific speed, (h) guide-vane angle, (i) guide wheel diameter, (j) guide wheel width, and (k) utilization factor. 6. A Francis turbine of a dam power house develops 450 kW of power at a speed of 1000 rpm when the head available is 85 m. An overall efficiency of 85% can be assumed. The runner wheel diameter is 12 times the width of the runner. The guide-vane angle is 20o. The runner diameter at the outlet is half that at the inlet. The flow component remains constant. The discharge of water is without any whirl component. Determine the (a) mass flow rate of water, (b) specific work, (c) diameter of runner at the inlet, (d) diameter of runner at the outlet, (e) blade width, (f ) blade angle at the inlet, and (g) blade angle at the outlet. 7. For a Francis turbine, the net head available is 12.5 m of water, and the available flow rate is 0.35 m3/s. The blades are radial at the inlet (A1  90o). A velocity of flow, equal to 2.1 m/s, remains constant in the runner. The overall efficiency is 0.85 and the speed is 500 rpm. The diameter at the discharge is half that at the inlet, and the discharge is radial. Take the reduction in flow area due to the vane thickness as 10%. Overall efficiency is 85%. Calculate the (a) power, (b) specific speed, (c) specific work, (d) diameter of runner at the inlet, (e) diameter of runner at the outlet, (f ) width of runner at the inlet, (g) width of runner at the outlet, (h) guide-vane angle, and (i) blade angle at the outlet in the rotor.

271

8. A run-of-the-river power house has a potential of a steady flow of 95 m3/s, and the head that can be arranged is 5.5 m of water. Design a Kaplan turbine that can be assumed to have an overall efficiency of 87%. The speed ratio and flow ratio may be taken as 2.1 and 0.75, respectively. The hub diameter is 0.35 times the tip diameter of the runner. The flow component of velocity remains constant. Calculate the (a) power of the turbine, (b) tip diameter, (c) hub diameter, (d) speed, (e) specific speed, (f ) specific work, (g) inlet blade angle at the tip, (h) inlet blade angle at the hub, (i) outlet blade angle at the tip, and (j) outlet blade angle at the hub. 9. A Kaplan turbine has a rated output of 2600 kW at 600 rpm, the head being 40 m of water. The speed ratio is 1.25. The overall and hydraulic efficiencies are 0.86 and 0.9, respectively. The hub diameter is 0.6 times the tip diameter. Calculate the (a) flow rate, (b) specific speed, (c) specific work, (d) tip diameter, (e) hub diameter, (f ) blade inlet and outlet angles at the tip, and (g) blade inlet and outlet angles at the hub. 10. The flow rate of water in a reaction turbine is 3.4 m3/s. The exit diameter of the turbine is 62 cm, which is also the inlet diameter of the draft tube connected to it. The diameter at the outlet of the draft tube is 92 cm. The length of the draft tube is 3 m. Calculate the head saved, due to the draft tube, if the efficiency of the draft tube is 90%. The end of the draft tube is 50 cm below the tailrace level. Take the atmospheric head as 10 m of water. If the saving in head is required to be 7.5 m, with the same efficiency and the same length, what should be the exit diameter of the draft tube? And in such a case, what is the half-cone-angle of the draft tube? If the atmospheric temperature is 33oC, is the cavitation likely to occur on the blade surfaces?

272

Turbomachines

S olutions A vailable

P roject- O riented Q uestions 1. There are 10 solved examples and 7 exercise problems (exclude problems 1, 2 and 10) in this chapter. Prepare a table to mention the values of all the parameters, such as head, flow rate, speed, diameter, specific speed, power, efficiency, etc. (whether these parameters are part of the data or part of the answers) so that one can have an overall view of the variety of the machines. Do they fit in to the values prescribed in Figs.6.2 and 6.3? A site has a head of

130 m of water and the flow rate of 3.5 m3/s. From the table which you have prepared, or otherwise, select and design a suitable turbine for the site. Assume suitable coefficients for the design. 2. At a project site, the head available is 160 m of water at a flow rate of 0.005 m3/s. Select and design a suitable turbine to generate power, assuming the required coefficients with justification and stating all the relevant parameters.

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5.

(b) (a) (c) (c) (c)

6. 7. 8. 9. 10.

(b) (c) (b) (a) (c)

Exercises 1. (a) 5.55 W, (b) 12.7 cm, (c) 7.9, (d) 24, (e) 11.5 mm, (f ) 14 mm 2. (a) 720 W, (b) 24.465 lps, (c) 15.86 cm, (d) 4.8 cm, (e) 1.54 cm, (f ) 14.38 cm, (g) 1.4cm, (h) 5, (i) 38.37o, (j) 73.6o, (k) 52o 3. (a) 28.2, (b) 1.2 m, (c) 3 jets, (d) 17.24 cm [Alternate solutions: (a) 24.43, (b) 1.2 m, (c) 4 jets, (d) 14.93 cm] 4. (a) 4886 kW, (b) 98%, (c) 92.3%, (d) 1.66 m3/s, (e) 28.73, (f) 16.2 cm 5. (a) 1464 kW, (b) 522.873 J/kg, (c) 0.73, (d) 92 cm, (e) 11.5 cm, (f ) 540 rpm, (g) 112, (h) 22.74o, (i) 95 cm, (j) 12 cm, (k) 0.88 6. (a) 634.9 kg/s, (b) 708.77 J/kg, (c) 49.22 cm, (d) 24.61 cm, (e) 4.1 cm, (f ) 80.2o, (g) 37.8o

7. (a) 36.48 kW, (b) 128.5, (c) 104.23 J/kg, (d) 39 cm, (e) 19.5 cm, (f ) 15.11 cm, (g) 30.22 cm, (h) 11.623o, (i) 22.36o 8. (a) 4460 kW, (b) 4.2 m, (c) 1.47 m, (d) 100 rpm, (e) 793, (f) 46.95 J/kg, (g) 21.61o, (h) 79.2o, (i) 19.65o, (j) 45.57o 9. (a) 7.7 m3/s, (b) 304; (c) 353.16 J/kg, (d) 1.115 m; (e) 0.669 m; (f ) 26.3o, 19.38o, (g) 71.25o, 30.4o 10. Head saved: 7.1 m; exit diameter in the second case: 1.02 m; half cone angle: 3.72o; at 33oC, the vapor pressure head is 0.5 m of water. At a saving of 7.5 m, the absolute pressure at exit of runner is 10 7.5  2.5 m, well above cavitation limits, hence safe.

7

Centrifugal Pumps

Learning Objectives After completing this chapter, you will be able to: v
Study the construction, working, and application of centrifugal pumps. v
Understand the various types of centrifugal pumps and the terminology used in their practice. v
Clearly understand the different heads and efficiencies that are generally employed to explain the use of centrifugal pumps. v
Correlate the efficiencies of centrifugal pump with those that were specified earlier with respect to any work-absorbing turbomachine.

v
Clearly understand the complete meaning of the specification of the centrifugal pump, such as its maximum suction lift and net positive suction head, in a given installation of the pump and the relevance of the terms. v
Know the need of priming of centrifugal pumps. v
Know under what circumstances the multiple units of the pumps are used, either in parallel or in series. v
Design a pump given the required specifications such as the head and the flow rate.

7.1 Introduction A centrifugal pump is one of the versatile examples of turbomachines. Perhaps the smallest centrifugal pump is the one that is employed in lifting water through a height of about 20 cm in a decorative water fall of a household drawing hall, circulating hardly a few milliliters of water per second. The input to such a pump could be in terms of a few milliwatts. On the higher side of the spectrum, there are centrifugal pumps for applications such as city water-supply, pulp handling in pulp-and-paper industries, crude oil handling in oil wells, etc. The power consumption of such pumps can be of the order of 1–1.5 MW per stage. The subject matter in this chapter is the study of centrifugal pump, its constructional features, different parts, its classification, and the various terms associated with its operation, such as minimum starting speed, net positive suction head (NPSH), cavitation, priming, and so on.

7.2 Centrifugal Pump A centrifugal pump is a turbomachine employed for lifting liquid from a lower level to a higher level. The input of the centrifugal pump is the mechanical energy in the form of torque on its rotating shaft. The output of the pump is the increased energy of the liquid in the form of the pressure or momentum (or both) of the liquid being lifted. Thus, the centrifugal pump converts mechanical energy into energy of the liquid. Centrifugal pumps handle a variety of liquid chemicals, different types of oils, slurries, etc. However, instead of the general term “liquid,” the term “water” is used in this chapter to explain the functions of the pump.

274

Turbomachines

7.2.1 Advantages of Centrifugal Pumps over Reciprocating Pumps While centrifugal pumps are the true examples of turbomachines, reciprocating pumps are the positivedisplacement machines referred to in Chapter 1. The advantages of centrifugal pumps over reciprocating pumps are recalled here to consolidate the ideas of comparison. Following are those advantages: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

The flow process is continuous and not periodic. The parts of centrifugal pumps are few in number. The parts of centrifugal pumps are simple, their manufacture requires less care, and hence they are less costly. The maintenance inventory of centrifugal pumps is very much limited; hence, the cost of inventory is also less. The hydraulic efficiency of centrifugal pumps is very high. The volumetric efficiency of centrifugal pumps is very high. The mechanical efficiency of centrifugal pumps is very high. The overall efficiency of centrifugal pumps is high. The operating cost of centrifugal pumps is low. Because of the purely rotary motion, there are no vibrations and higher speeds are possible. Large volumetric flow rates are possible in centrifugal pumps. Control of the flow of fluid is very easy in centrifugal pumps.

The only drawback is that in centrifugal pumps, very high pressure ratios are not possible.

7.3 Construction and Working of a Centrifugal Pump A schematic layout of an installation of a centrifugal pump is shown in Fig. 7.1. The main part of the centrifugal pump is its impeller (having the blades) enclosed in a casing and mounted on the shaft. The impeller and shaft are driven by the motor. The layout includes a suction pipe, with its lower end below the water level in the sump and upper end connected to the eye of the impeller. A foot-valve is connected at the lower end of the suction pipe. The foot-valve is basically a non-return valve, but also acts as a strainer to exclude any big solid particles from entering the suction pipe. The driving motor is the necessary prime mover, shown in the layout. When the motor drives the shaft and the impeller of the pump, the impeller exerts a centrifugal force on the water present between its blades. As a result, the water is forced to move radially outward. Thus, a vacuum, suction, or a “pull” is created in the center or eye of the impeller. Due to this, the water in the suction pipe is drawn into the impeller and the pumping continues. A priming valve is provided on the side of the delivery pipe or the casing to facilitate the priming. As the pumping continues, water moves into the spiral casing all around the periphery of the impeller. The spiral casing has its cross-sectional area continuously increasing. This increase of area may be just to accommodate the incremental quantity of water from the periphery of the impeller, or in addition, it may be to further diffuse the velocity of the water and increase its pressure (if the increasing of area is designed for that process and purpose). The different parts of a pump installation are clearly marked in Fig. 7.1. The parts are also listed in Table 7.1 along with their functions.

Centrifugal Pumps

275

Delivery valve Casing

Delivery

Delivery

Priming valve Coupling

Impeller

Shaft

Suction Vanes

Bearings

Motor

Base

Sump level

Foot-valve

Figure 7.1

Table 7.1

S. No.

Schematic view of a centrifugal pump installation.

Parts of a centrifugal pump installation and their functions

Part

Functions

1.

Motor

To provide power to the centrifugal pump

2.

Shaft

To transfer the power from the motor to the impeller

3.

Bearings

To support the shaft and to provide smooth running

4.

Impeller

To impart energy to the water

5.

Casing

To collect water from the periphery of the impeller and to carry it to the delivery flange (also to act as a diffuser to convert kinetic energy to pressure energy, if so designed)

6.

Delivery pipe

To deliver water from the pump to some tank

7.

Suction pipe

To draw water from the sump to the impeller

8.

Foot-valve

1. To act as a non-return valve 2. To prevent any solid particle of bigger size from entering the suction pipe and impeller

9.

Priming valve

To facilitate the process of priming (i.e., filling the impeller and casing with water)

(Continued)

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Table 7.1

Continued

S. No.

Part

Functions

10.

Delivery valve

1. To isolate the pump from the rest of the system for maintenance, repair, or even during shut-off 2. To control the flow

11.

Base

To provide support for the entire unit

7.4 Classification of Centrifugal Pumps Centrifugal pumps can be classified based on different criteria: 1.

Specific speed: One of the important basis for classification is the specific speed of the pump. The expression for the specific speed, Ns, is Ns 

N Q H 3/ 4

(7.1)

Depending on the required duty of the pump for any application (Q, H), its specific speed can be first determined. Based on this specific speed, the different types are as follows: (a) Low specific speed, (b) medium specific speed, and (c) high specific speed pumps. This classification results in different diameter ratios (D2/D1), as indicated in Table 7.2. Table 7.2

Classification based on the specific speed

D2

Ns

Flow Pattern

3.5 to2.0 2.0 to 1.5

º » Radial flow ¼

« 50 80 Medium ¬80 160 ­

1.5 to 1.3 1.3 to 1.1

º » Mixed flow ¼

High 110 – 500

1.0

Low

2.

D1 «10 30 ¬30 50 ­

Axial flow

Working head: The pumps are classified as low-head (up to 15 m), medium-head (15–40 m), and high-head (above 40 m) pumps. As the head increases, the casing of the pump also has to participate in the process of diffusion of kinetic energy into pressure energy. This means that diffuser blades are to be provided in the casing as shown in Fig. 7.2. The diffuser blades are the fixed blades in the casing, between the impeller and the volute casing, with their angles aligned with the absolute exit velocity of water from the impeller. The divergent passages between the diffuser blades convert kinetic energy into pressure energy of the water. The head generated is higher than that of the same impeller at the same speed without the diffuser blades.

Centrifugal Pumps

277

Delivery

Diffuser blades Impeller Spiral casing

Figure 7.2

3.

4.

Schematic diagram to show diffuser blades in the casing.

Type of casing: The casing can be of simple volute type or diffuser volute type. A third type of casing is of “vane-less” diffuser type. Even without the diffuser blades, the area of flow increases with increased radius, and the velocity decreases resulting in an increased pressure. Type of impeller: Depending on the number of entries, the impeller can be either single-entry type or double-entry type. The double-entry type is like two impellers kept back to back. Apart from this, there are other types of impellers such as closed-type, semi-open-type, or open-type impellers. The semi-open or open types are for applications such as sewage handling or handling of liquids which may have solid particles, or handling heavy and highly viscous slurries, etc. These impellers are shown in Fig. 7.3. Back shroud Vanes Front shroud

Hub

Eye

(a)

Figure 7.3

5.

6. 7.

(b)

(c)

Schematic diagrams to show the types of impellers: (a) Closed type, (b) semi-open type, and (c) open type (with stronger vanes).

Number of impellers: There can be two or more impellers on the same shaft. These are known as multi-stage pumps, each stage having one impeller. The casing is so designed that the outlet from one impeller is guided to the inlet of the next impeller. The direction of flow: The pumps are of the radial flow type, mixed flow type and axial flow type pumps, depending on the direction of flow of water in the impeller. Type of material of construction: This classification depends on the material of construction such as cast iron, alloy steel, or stainless steel, suitable for pumping water, corrosive chemicals, milk in the dairy industry, etc.

The above classifications are summarized in Table 7.3.

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Table 7.3

S. No.

Classification of centrifugal pumps

Basis of Classification

Different Types

1.

Specific speed

Low, medium, and high specific speed pumps

2.

Head

Low-, medium-, and high-head pumps

3.

Type of casing

Volute, diffusion volute, and vane-less diffusion pumps

4.

Type of impellers

1. Single-entry impeller, double-entry impeller 2. Closed-, semi-open-, and open-type impellers

5.

No. of impellers

Single-stage pumps, multi-stage pumps (2, 3, 4, … stages)

6.

Direction of flow

Radial flow type, mixed flow type, axial flow type pumps

7.

Type of material

Cast iron pumps, alloy steel pumps, stainless steel pumps

7.5 Different Heads of Centrifugal Pumps The study of a centrifugal pump is naturally with respect to the pumping of liquids. Therefore, the unit of energy, in general, is a column of liquid, that is, meters of liquid. Hence, the output of the centrifugal pump is in terms of meters of liquid or head units. With this background, various heads associated with an installation of a centrifugal pump are identified and these heads are as follows (shown in Fig. 7.4). All the heads are in meters of liquid or meters of water. 1. 2. 3. 4.

hs is suction head, that is the vertical distance from the sump level to the centerline of the pump. hd is delivery head, that is the vertical distance from the centerline of the pump to the exit point of the delivery pipe. hfs is frictional loss of head in the suction pipe, including the loss due to the Foot-valve, other pipe fittings such as bends, etc. hfd is frictional loss of head in the delivery pipe, including the loss due to the delivery valve, other pipe fittings, etc.

5.

Vd2 2 g is loss of head equivalent to the exit velocity.

6. 7.

h  hs  hd is the static head of a pump. hf  hfs  hfd is the total friction head.

Further, a head known as “manometric head, Hm” is defined as the head measured across the inlet and outlet of the pump. The manometric head represents the output of the pump. (The name might have been due to a manometer connected across the inlet and outlet of the pump, when the head generated was much less, sometime during the evolution of the pumps.) Now the pressure gauges are being mounted, one each at the inlet and outlet of the pump. The difference in their readings, suitably converted, is the manometric head.

Centrifugal Pumps

279

V 2d /2g hfd

Vd

hd

Pump hs

pa Sump

hfs

Foot-valve

Figure 7.4

Schematic layout of a centrifugal pump showing various heads.

Because the pump is required to work against the total of the static head, friction head, and delivery velocity head, one can write H m  h  hf  or

Vd2 2g

H m  hs  hd  hfs  hfd 

Vd2 2g

(7.2)

The head developed by the impeller is given by the Euler turbine equation, where the whirl component Vu1 is taken as zero. Hence HE 

WE g



U 2Vu 2 g

(7.3)

The difference between Hm and HE is the loss due to fluid friction, wall friction, eddies, etc. in the impeller. Referring back to the power flow diagram of Chapter 1 (Fig. 1.9), it may be noted that the manometric head Hm is the same as the output H, mentioned in the diagram. The net head Hn mentioned therein is the head developed by the impeller, equal to (U2Vu2/g).

7.6 Different Efficiencies of a Centrifugal Pump In Chapter 1, four types of efficiencies have been enumerated. These efficiencies are now reviewed for better clarity.

7.6.1 Mechanical Efficiency The mechanical efficiency (Gm) has been given as

Hm 

Rotor power Shaft power

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Turbomachines

 Hm 

Pr

(7.4)

Ps

Here, the shaft power is the input power supplied by the motor. Due to mechanical losses at the bearings, the impeller or rotor gets slightly less power. The mechanical efficiencies of the pumps are of the order of 98–99%. Higher capacity pumps have higher mechanical efficiencies.

7.6.2 Volumetric Efficiency The volumetric efficiency (Gv) has been given as

Hv 

m m  $m

(7.5)

where m is the mass flow rate, or equivalently, the volume flow rate, that is, the designed and intended flow rate to be available at the delivery end of the pump and $ m is the extra flow rate that goes through the impeller, absorbing extra power. The extra flow rate is the sum of the leakages at the glands and the return flow that occurs between the casing and the shrouds of the impeller, from its outlet end to the suction end. The values of volumetric efficiencies are of the order of 94–96%.

7.6.3 Hydraulic Efficiency or Manometric Efficiency The hydraulic efficiency (Gh) has been given as

Hh 

P wQH H   Pn wQH n H n

(7.6)

where H is the head that is finally overcome by the pump, namely, the manometric head Hm given by H  H m  h  hf 

Vd2 2g

The denominator Hn in Eq. (7.6) is the head developed by the impeller. This is equivalent to the specific work as per the Euler turbine equation: Hn  HE 

WE g



U 2Vu2 g

The difference between Hm and HE is due to (a) the fluid friction because of viscosity, (b) fluid-rotor friction, and (c) non-ideal flow of liquid in the impellers. Hence

Hh 

gH m U 2Vu2

The hydraulic and manometric efficiencies are the same

Hmano 

gH m U 2Vu2

(7.7)

Centrifugal Pumps

281

The manometric efficiency depends on the viscosity of liquids being pumped, the surface finish of the blades and shrouds, etc. The usual values of manometric efficiencies are of the order of 85–94%.

7.6.4 Overall Efficiency The overall efficiency has been given as

Ho  Hh s Hv s Hm In the light of the above discussion (Gh  Gmano), the overall efficiency can also be written as

Ho  Hmano s Hv s Hm

(7.8)

The overall efficiency is also known as the gross efficiency or actual efficiency.

7.7 Analysis of a Centrifugal Pump The analysis of a centrifugal pump is the same as that with the radial flow machines as discussed in Chapter 4. The pumps handle the liquids with constant densities (incompressible fluids) and the analysis continues on the same lines as earlier. The topics include the velocity triangles (Fig. 4.1), the specific work or head generated [Eqs. (4.1a)–(4.1c)], effects of the variations of the blade angles on the head and the reaction (Sections 4.3.1–4.3.3), the slip (Section 4.3.4), the losses (Section 4.3.5), and the characteristics (Section 4.3.6). The characteristics are the plots of the heads (H), losses, efficiency, power, and duty point on the variable base of the capacity flow rate. In referring to the topics mentioned in the above paragraph, the focus is on the “head generated,” as the equivalent quantity of the specific work. The head-capacity characteristic is actually the specific workcapacity characteristic with a conversion of only the units [W  J/kg  m2/s2, W/g  (m2/s2)/(m/s2)  m, head of fluid]. Sometimes, however, instead of the head generated, the pressure generated is taken as the ordinate while presenting the characteristic curve. Once again, this is also taken as the same characteristic with one more change of units converting the head into pressure difference. This can also be accepted as a true characteristic, but with a small difference: The pressure generated and the head generated can be equivalent only if the velocities at the inlet and the outlet are equal. If the suction and delivery pipes of a centrifugal pump are of different diameters, then the velocities in the pipes cannot be equal. With the parameters such as suction pipe diameter, impeller diameter, impeller blade geometry, spiral casing, and flow rate remaining the same, suppose that there are two alternate diameters possible for the delivery pipe, say pipe X with diameter DX and pipe Y with diameter DY and with DX  DY. Then the velocity in pipe X is more than that in pipe Y, and the pressure in pipe X is slightly less than that in pipe Y, when the two pipes are fitted (individually, one after the other) to the same pump. The specific work is the same in both cases, while the pressure difference is not. The pressure difference, hence, depends on the pipe diameter, with the same power input. This is to be very clearly understood, and the context is “same fluid, different delivery pipe diameters.” Further, there is an important difference in the terms “pressure created” and “head created.” This can be elaborated as follows: Consider an impeller of a specified tip diameter D2 and a specified blade angle A2 at the outlet running at a specified speed N rpm. (The entry of fluid is radial, @1  90o, Vu1  0.) The specific work is W  U2Vu2 J/kg ( m2/s2) and the head created is U2 Vu2/g (m2/s2)/(m/s2)  x meters of fluid. If

282

Turbomachines

water is the fluid, the head created is x meters of water; if gasoline is the fluid, the head created is x meters of gasoline; and if mercury is the fluid, the head created is x meters of mercury. In all the cases, the impeller is the same, the speed is the same, and therefore, the numerical value of x ( U2 Vu2/g) is the same. But x meters of water, x meters of gasoline, or x meters of mercury are not equal to one another. Hence, the pressure created is different for different fluids (depending on their densities) for the same impeller running at the same speed. As a result, the power inputs for different fluids are different. In contrast, if the output pressure generated is the same, with different fluids giving rise to different fluid columns, then the other parameters (D2, A2, N, W ) may be different. With the background of this discussion, the characteristic curves of the “head created” and the “pressure created” on the base of the flow rate can be different, but qualitatively the curves may still be of the same nature.

EXAMPLE 7.1 The blade angles at the inlet and outlet of the impeller of a centrifugal pump are 55o and 75o, and the corresponding diameters are 3 cm and 6 cm, respectively. The blade width at the outlet is 0.75 cm. The speed is 1500 rpm. The entry of water is radial without any whirl component. The flow component of fluid velocity remains constant in the impeller. Draw the velocity triangles and calculate (a) the specific work, (b) rate of flow, (c) power of the machine, and (d) the manometric head. The hydraulic efficiency may be taken as 0.85.

Solution: Given A1  55o, A2  75o, D1  0.03 m, D2  0.06 m, B2  0.0075 m, N  1500 rpm, @1  90o, Vu1  0, Vf1  Vf2, Gh  0.85. To find velocity triangles, W, Q, P, Hm, Gh  0.85. Blade velocity at inlet is P D1N P s 0.03 s 1500   2.356 m/s U1  60 60 Therefore, blade velocity at outlet is U 2  U1 s 2  4.712 m/s Now it is possible to draw the velocity triangles; these are shown in Fig. 7.5. Now Vf 1  U1 tan B1  2.356 s tan 55o  3.365  Vf 2  V1 Vr1

V2

Vr2

V1 =Vf1 =Vf2 90°

b1

55°

U1 = 2.356

Figure 7.5

Vu2 = 3.81 U2 = 4.712

Velocity triangles.

b2

75°

Centrifugal Pumps

283

Also we know that U 2 Vu2 

Vf2 tan B2



3.365  0.902 tan 75o

Therefore Vu2  U 2 0.902  4.712 0.902  3.81 (a) The specific work is W  U 2Vu2  4.712 s 3.81  17.953 m 2 /s 2 (or J/kg) (b) The flow rate is Q  P D2 B2Vf2  P s 0.06 s 0.0075 s 3.365  0.004757 m3 /s Therefore mass flow rate is m  4.757 kg/s (c) The power is P  Wm  17.953 s 4.757  85.4 W (d) The manometric head is ¥U V ´ H m  H E s Hh  ¦ 2 u 2 µ s Hh § g ¶ ¥ 17.953 ´ ¦ s 0.85  1.556 m § 9.81 µ¶ Comments: 1. This is a very small pump, handling only 4.575 kg/s of water, with a manometric head of only 1.556 m of water. However, the same pump, with all other data remaining the same, can be run at a speed of 3000 rpm (twice the 1500 rpm, given). The effect can be checked: U1  4.712 m/s, U2  9.424 m/s, Vf1  Vf2  V1  6.73 m/s, Vu2  7.621 m/s, W  71.82 J/kg, Hm  6.223 m of water, Q  9.5143 lps, P  683.2 W. These results can also be checked in the context of extrapolation of the performance or model studies. 2. The specific speed, Ns, of the pump is calculated as 65.75 (in both cases). 3. The same velocity triangles hold good with double the scale for 3000 rpm. 4. Suppose that this small pump is used to pump liquid mercury. The speed is 3000 rpm. What are the results? The velocity triangles, velocities, etc. do not change. The specific work remains as 71.82 J/kg, Hm  6.223 m of mercury, Q  9.5143 lps. Because the specific weight of mercury is 13.6 times that of water, the power is P  683.2 s 13.6  9.29 kW

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Turbomachines

EXAMPLE 7.2 The impeller of a centrifugal pump has the inlet and outlet diameters of 6 cm and 18 cm, respectively, with the blades bent backward at 70o to the wheel tangent at the outlet. The blade width at the outlet is 3 cm. The speed of the pump is 1000 rpm and the flow rate of water is 0.125 m3/s. The volute casing is designed to recover 25% of the outlet velocity head of the impeller. The losses in the impeller are equivalent to 20% of the velocity head at the outlet. The flow component is constant in the impeller and the entry of water is without any whirl component. Calculate (a) the pressure rise, (b) manometric efficiency of the pump, (c) power, and (d) specific speed.

Solution: Given D1  0.06 m, D2  0.18 m, A2  70o, B2  0.03 m, N  1000 rpm, Q  0.125 m3/s. Recovery  25% of Loss  20% of

V22 2g V22 2g

V1  Vf1  Vf2 , A1  90o, Vu1  0 To calculate $P, Gmano, P, Ns. The blade velocity at inlet is U1 

P D1N P s 0.06 s 1000   3.1416 m/s 60 60

The blade velocity at outlet is U 2  U1 s 3  9.4248 m /s The volume flow rate is Q  P D2 B2Vf2  P s 0.18 s 0.03 s Vf2  0.125 Therefore Vf2  7.368 m /s  Vf1  V1 The velocity triangle at outlet is drawn as in Fig. 7.6. V2 9.988

Vu2 = 6.7428

Vr2

b2

70°

U2 = 9.4248

Figure 7.6

Outlet velocity triangle.

Centrifugal Pumps

We have that U 2 Vu2   Vu 2

Vf2

tan 70o 7.368  U2  9.4248 2.682  6.7428 m/s tan 70o

The velocity at exit is V2  Vu2  Vf22  6.74282  7.3682  9.988 m/s Theoretical head developed in the impeller is U 2Vu2

9.4248 s 6.7428  6.478 m of water 9.81



g

If the recovery of the head in the volute casing and the loss in the impeller were absent, then Energy (head) at the inlet  Head developed  Energy (head) at the outlet  

p1 w



V12 2g

p2 p1 w

 

U 2Vu2 g

V12



V22 2g

p2



V22

w 2g UV  2 u2 g

9.9882  6.478 2 s 9.81  4.16 m of water 

7.6382

(a) The net pressure rise (head) is p2 p1 w

 0.25 s

9.9882 9.9882

0.20 s  4.16  0.25423 2 s 9.81 2 s 9.81  4.41423 m of water

This, in terms of pressure units is $p  4.41423(m) s 1000 (N /m3 )  4414.23 N /m 2  4.414 kPa (b) Manometric efficiency is

Hmano 

4.41423  68.18% 6.478

(c) The power of the pump is P  wQH  9810 s 0.125 s  7.94 kW

6.478 1000

285

286

Turbomachines

(d) Specific speed of the pump is Ns 

N Q 1000 0.125   87 H 3/ 4 6.4780.75

Comments: For the size of the impeller, the various heads are rather low. This is because the speed is low. If the speed were 1500 rpm, the calculations would be as follows: U1  4.712, U2  14.137, Vf2  7.368, Vu2  11.455, V2  13.62, (U2 Vu2/g)  16.5, ($p/w)  9.812, Net ($p/w)  10.285, Gmano  62.3%, P  20.23 kW, Ns  64.77.

EXAMPLE 7.3 A centrifugal pump is driven by an induction motor at 960 rpm. The flow rate of water is 50 lps against a head of 10 m. The flow velocity is constant at 6.5 m/s through the impeller. The blades are radial at the outlet and the losses are estimated as 15% of the output. Assume that water enters the rotor at @1  90o. Calculate (a) tip diameter of the impeller, (b) width of the blades at the outlet, (c) fluid delivery angle, (d) power, and (e) specific speed.

Solution: Given N  960 rpm, Q  50 lps, Hm  10 m, Vf1  Vf2  V1  6.5 m/s, A2  90o, losses

 15%, @1  90o. To calculate D2, B2, @2, P, Ns. The outlet velocity triangle can be drawn because A2 ( 90o) and Vf2 ( 6.5) are known as per data. It is as shown in Fig. 7.7. (At this stage, U2 is not known, and it will be known in the next two steps.) V2

Vr2 =Vf2 = 6.5

b2 = 90°

a2 U2 = 10.74

Figure 7.7

Outlet velocity triangle.

Because A2  90o, we have Vu2  U2. Therefore HE 

U 2Vu 2 g



U 22 g

Now, the hydraulic losses are 15%. Therefore, U 22 g

(1 0.15)  H m  10

Therefore outlet blade velocity is calculated as ¥ 10 s 9.81´ U2  ¦ § 0.85 µ¶

0.5

 10.74 m/s

Centrifugal Pumps

287

(a) To calculate tip diameter of impeller, we have U2 

P D2 N 60

Therefore tip diameter of impeller is D2 

U 2 s 60

PsN



10.74 s 60  0.2137 m  21.37 cm P s 960

(b) Now, volume flow rate is given. We have

P D2 B2Vf2  Q Therefore width of blade at the outlet is Q P D2Vf2 0.05  P s 0.2137 s 6.5  0.0115 m  11.5 mm

B2 

(c) Fluid delivery angle is ¥V ´ ¥ 6.5 ´ A 2  tan 1 ¦ f2 µ  tan 1 ¦  31.18o § 10.74 µ¶ §U2 ¶ (d) The power is P  wQH  9810 s 0.05 s

10  5.77 kW (1000 s 0.85)

(e) The specific speed is Ns 

N Q 960 s 0.05   38.17 H 3/ 4 100.75

EXAMPLE 7.4 The power input to a centrifugal pump is 50 kW at the shaft while running the pump at 1440 rpm. The impeller tip diameter is 30 cm and the blade width at the tip is 1.5 cm. The water flow rate is 110 lps. The vacuum gauge reading at the suction flange is –20 cm of mercury and at delivery flange the pressure gauge reading is 370 kPa. The blade outlet angle is 65o. A 2% slip (L  0.98) may be assumed. Calculate the (a) theoretical head, (b) ideal head, (c) hydraulic efficiency, (d) mechanical efficiency, (e) overall efficiency, and (f ) specific speed of the pump. Assume radial entry and constant flow velocity.

Solution: Given shaft power  50 kW, N  1440 rpm, D2  0.30 m, B2  0.015 m, Q  0.11 m3/s,

p1  –20 cm of mercury, p2  371 kPa, A2  65o, L  0.98, Gm  0.96. To calculate HE, Hi, Gh, Go, Ns. We know that Q  P D2 B2Vf2

288

Turbomachines

Therefore flow velocity is Q 0.11   7.78 m/s P D2 B2 (P s 0.3 s 0.015)

Vf2 

The peripheral blade velocity at outlet is U2 

P D2 N P s 0.3 s 1440   22.62 m/s 60 60

The velocity triangle at outlet is shown in Fig. 7.8. V2 Vf2 = 7.78

Vr1 65° b2

U2 = 22.62

Figure 7.8

Outlet velocity triangle.

We have Vu 2  U 2

Vf2 tan B2

 22.62

7.78  18.99 m/s tan 65o

(a) Theoretical head or Euler head is HE 

U 2Vu 2 g



22.62 s 18.99  43.787 m of water 9.81

(b) The ideal head is H i  H E s M  43.787 s 0.98  42.91 m of water (c) Now, to calculate the hydraulic efficiency, we have Manometric head (H m )  Exit head Inlet head  370 kPa ( 20 cm of mercury)  37 ( 0.2 s 13.6 )  39.72 m of water Hydraulic or manometric efficiency is

Hmano 

Hm Hi



39.72  92.57% 42.91

(d) To calculate the mechanical efficiency: Hi 1000 Mechanical efficiency (N m )   Shaft power Shaft power Rotor power

wQ s

Centrifugal Pumps

9810 s 0.11 s 

50

289

42.9 1000  46.29  92.58% 50

(e) The overall efficiency is

Ho  Hm s Hh s Hv (assume Hv  1.00 in the absence of data of leakage loss )  0.9258 s 0.9259  85.72% (f ) The specific speed is Ns 

N Q 1440 s 0.11 (Here H  H m )   30.18 3 / 4 H 39.720.75

EXAMPLE 7.5 The inlet and outlet diameters of the impeller of a centrifugal pump are 30 and 80 cm, respectively. The width of the impeller at the outlet is 12 cm. While running at 600 rpm, the pump delivers 3.89 m3/s at a head of 42 m of water. The water enters the impeller radially. The velocity of flow of water in the impeller is constant. The manometric efficiency of the pump is 85%. Calculate the blade angles at the inlet and outlet and the power of the driving motor if the mechanical efficiency is 97.5%.

Solution: Given D1  0.3 m, D2  0.8 m, B2  0.12 m, N  600 rpm, Q  3.89 m3/s, Hm  42 m of

water, @1  90o, Vu1  0, Vf1  Vf2, Gmano 0.85, Gm  0.975. We know that as it is required to calculate the blade angles, the velocity triangles are required. The blade velocity at the inlet is U1 

P D1N P s 0.3 s 600   9.425 m/s 60 60

The blade velocity at the outlet is U2 

P D2 N P s 0.8 s 600   25.133 m/s 60 60

Also, for volume flow rate, we have

P D2 B2Vf2  Q Therefore, the flow velocity is Vf2 

Q 3.89   12.898 m/s P D2 B2 P s 0.8 s 0.12

The velocity triangles are shown in Fig. 7.9.

290

Turbomachines

V1 = Vf1

V2

Vr1

Vr2

12.898 b2

b1 U2 = 25.133

U1 = 9.425

Figure 7.9

Velocity triangles.

We have

Hmano 

Hm HE



gH m U 2Vu 2

Therefore Vu 2 

gH m U 2 s Hmano



9.81 s 42  19.287 m/s 25.133 s 0.85

Blade angle at the inlet is, from the inlet triangle,

B1  tan 1 (12.898/9.425)  53.84o Blade angle at the outlet is, from outlet triangle,

B2  tan 1

12.898  65.62o (25.133 19.287 )

Theoretical head or Euler head is HE 

Hm

Hmano



42  49.41 m 0.85

Power developed in the rotor is P  wQH E  9810 s 3.89 s

49.41  1885.53 kW 1000

So Power of the driving motor 

1885.53 1885.53   1933.877 kW Hm 0.975

7.8 Minimum Starting Speed of a Centrifugal Pump The head developed by an impeller, while running at a certain speed, has three components, as given by the Euler turbine equation. Suppose that the delivery valve is closed when the pump is running. With the fluid velocity being zero, the head generated is only due to the centrifugal component (U 22 U12 )/ 2 g . A similar situation arises when the delivery valve is fully open, but the impeller runs very slowly. The head generated is so small that water rises in the delivery pipe to a height equal to the centrifugal head (U 22 U12 )/ 2 g . Increasing the speed of the impeller causes this height also to increase. When the centrifugal head equals the installed total head, the water just touches the upper end of the delivery pipe. At this speed of the impeller,

Centrifugal Pumps

291

the water just starts flowing. Further, since the water also starts having its velocities such as V1, V2, Vr1, and Vr2 in the impeller, the head developed in the impeller picks up and becomes equal to the manometric head. The minimum speed at which the pump just starts to deliver the water against the specified manometric head is known as the minimum starting speed of the pump. The water can be delivered only when the centrifugal head equals and exceeds the manometric head at the installation. Thus, at the minimum starting speed N, we have (U 22 U12 ) 2g

r Hm

2

¥ PN ´ 2 2 ¦§ µ ( D2 D1 ) 60 ¶  r Hm 2g 1/ 2

¨ 2 g s H s 3600 ·  N r © 2 2m ¸ 2 ª P ( D2 D1 ) ¹

(7.9)

EXAMPLE 7.6 A centrifugal pump has its impeller with inlet and outlet diameters of 35 cm and 70 cm, respectively. It is used in a setup with a manometric head of 28 m. Determine its minimum starting speed.

Solution: Given D1  35 cm, D2  70 cm, Hm  28 m. Now ®« 2 gH 3600 º® Minimum starting speed  ¬ 2 m 2 2 » ®­ P ( D2 D1 ) ®¼

0.5

« 2 s 9.81 s 28 s 3600 º ¬ 2 » ­ P (0.7 2 0.352 ) ¼  738.42 rpm

0.5

7.9 Maximum Suction Lift and Net Positive Suction Head The basic action of the centrifugal pump is due to the centrifugal force exerted on the water present in the passages between the blades of the impeller. Unless and until the water enters the impeller, the impeller cannot exert any force on it. The energy transfer from the impeller vanes to the water can occur only after the water enters the eye of the impeller. The movement of the water from the sump to the impeller is caused by the pressure acting on the water level in the sump. If this pressure is x meters of water, then the water can be pushed up the suction pipe by that pressure only up to a height of x meters. When the pressure acting on the sump level is due to the atmospheric head ha, the maximum suction head can only be ha. But this is only an ideal case. During pumping, because of the frictional losses, the suction head gets further reduced. In addition, during pumping, the dynamic head Vs2 / 2 g also has to be met.

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If the atmospheric head is ha and the frictional head in the suction pipe is hfs , then the maximum possible height up to which the water can be pushed up in the suction pipe is ha – hfs – Vs2 / 2 g . This is termed the maximum suction lift (MSL), given by MSL  ha hfs

V s2 2g

One more parameter affects this expression. At any point in the suction pipe, above the level of water in the sump, the pressure is lower than the atmospheric pressure. As the point under consideration moves up, the pressure at that point gets reduced. At some point, the pressure may reach the vapor pressure of the liquid at the prevailing temperature. Then the liquid may start to vaporize and the continuity of the body of the liquid may be disturbed due to the vapor bubbles, so formed. In order to avoid such occurrence, the vapor pressure head hv also has to be subtracted from the MSL and the expression becomes MSL  ha hfs

V s2 2g

hv

(7.10)

It may be observed that many factors affect the value of MSL: 1. 2.

3. 4.

Altitude of the location of the pump: The higher the altitude above the mean sea level, the lower the ha. Suction pipe parameters: The smoothness of surface, total length, pipe fittings, etc., affect hfs. The frictional loss in the pipe increases with the surface roughness, length, pipe fittings, and loss in the Foot-valve. Velocity of water in the suction pipe: The larger the diameter, the lower the velocity, and hence, the lower the loss. Ambient temperature at the location: This affects hv. As the local temperature increases, the vapor pressure head also increases.

A schematic diagram, shown in Fig. 7.10, indicates that a pump centerline can be at a height equal to the MSL above the sump water level. If the pump is installed at still higher levels, the vapor bubbles are sure to occur in the suction line. Continuous occurrence of such bubbles and their accumulation may cause the absence of liquid water in the impeller, and may eventually result in the complete stoppage of the pumping action. Hence, the upper limit of suction head hs is the MSL. Barometer hv hfs V 2s /2g

Upper limiting position of pump NPSH

ha MSL

Sump level

Figure 7.10

Any position of pump

hs

Maximum suction lift

pa

Schematic view of MSL and NPSH.

Centrifugal Pumps

293

The suction head, hs, can always be less than the MSL for the safe working of the pump. In such a case, the difference between the MSL and hs ensures that the pump is really working in safe limits. At this juncture, another parameter is identified as (MSL – hs ) or “Net Positive Suction Head” (NPSH) given by NPSH  MSL hs  NPSH  ha hfs

V s2 2g

hv hs

(7.11)

The value of NPSH is a sort of guarantee for the safe working of a pump. The factors that affect the value of the MSL, naturally, affect the value of the NPSH also. The lower values of the suction head, hs, indicate the higher values of the NPSH, which is better for the working of the pump.

EXAMPLE 7.7 A centrifugal pump is installed at a site where the atmospheric pressure and temperature are 100 kPa and 35oC, respectively. The pump has a maximum flow rate of 40 lps of water. The suction pipe of diameter 10 cm is 8 m long. The friction factor of the pipe is 0.005. The resistance of Foot-valve and other pipe fittings together is equivalent to 2 m length of the suction pipe. Determine the maximum height above the water level in the sump, where the pump can be installed. If the pump is actually installed 2.5 m above the sump water level, find the NPSH.

Solution: Atmospheric pressure head  100 kPa  10 m of water  ha Atmospheric temperature  35oC At 35oC, the vapor pressure of water (from steam tables)  0.05622 bar Therefore vapor pressure head  hv  0.5622 m of water Velocity of water in the suction pipe is Vs 

0.04 s 4 Q   5.1 m/s 2 ¥ P Ds ´ (P s 0.12 ) ¦§ 4 µ¶

Therefore the velocity head in the suction pipe is V s2 2g



5.12  1.3257 m of water 2 s 9.81

The friction loss of head is hfs 

4 flVs2 2 gDs



4 s 0.005 s 10 s 5.12  2.6514 m of water 2 s 9.81 s 0.1

Maximum possible suction lift is ha hfs hv

Vs2 2g

 10 2.6514 0.5622 1.3257  5.4607 m

The maximum height at which the pump can be installed  5.4607 m above the sump level

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Actual height of installation  2.5 m Therefore Net positive suction head  5.4607 – 2.5  2.9607 m of water

7.10 Cavitation Cavitation is a phenomenon that starts in the low-pressure regions in liquids, adjacent or close to the solid surfaces. If the low pressure is equal to or less than the vapor pressure of the liquid at the prevailing temperature, the liquid starts vaporizing. The vapor bubbles are formed on the solid surfaces. When, due to eddies and turbulence, the higher pressure liquid comes adjacent to such vapor bubbles, the pressure in the bubbles also increases and the vapor condenses. The bubbles collapse and the surrounding liquid rushes from all the directions to occupy the space vacated by the condensing vapors. This generates very high localized pressures on the surfaces, enough to even chip off the metal particles and cause pits and cavities. The overall process is called pitting or cavitation. Such cavitation occurs over a large number of points on a surface. It is accompanied by the vibrations due to the collapse of vapor bubbles and rush of water from all around occurring at a number of points. Cavitation, even from its starting, causes the loss of surface finish, and therefore increases losses. This reduces the efficiency of the machine. Continuous cavitation can cause even complete mechanical failure of the parts. Cavitation can occur in the low-pressure regions of the centrifugal pumps on the suction side. At the surface of the rotor blades, the water enters at low pressures. The energy from the blades is soon transferred to water. The process of formation of bubbles (due to the low pressures) and then their collapse (due to transfer of energy from impeller blades) is the onset of cavitation at the blade surfaces. In Fig. 7.11 the suction pipe of a centrifugal pump is shown. In this figure, the point of the inlet to the impeller from the suction pipe is at a height hs meters above the free surface level of water in the sump. If hs is equal to the MSL of Fig. 7.10, the cavitation is just about to start; if hs is more than the MSL, the cavitation is sure to occur; and if hs is less than the MSL, then there is a NPSH, guaranteeing that there is no cavitation. Cavitation is not a desirable phenomenon. Efforts to avoid cavitation can be made by preventing the formation of low-pressure regions by installing the pump at the level or below the level of sump, and providing smooth pipes for the suction, avoiding or reducing the pipe fittings such as bends. Cavitation damage can also be controlled by the use of high-strength materials (e.g., cast iron, cast steel, stainless steel, etc.). Possibility of cavitation hear Pump inlet

hs

Suction pipe pa Sump level

Figure 7.11

Cavitation.

Centrifugal Pumps

295

7.11 Priming Priming is the process of filling up the space in the suction pipe, impeller and casing by the same liquid as the one required to be pumped, before starting the pump. Unless the liquid exists in the body of the impeller, the impeller cannot create the head in terms of the column of that liquid even while running. Both the non-return valve (foot-valve) and the priming valve make it easier to prime the pump in the beginning.

7.12 Pumps in Series and in Parallel The procedure of designing a pump starts from the specification of the pump in terms of the required flow rate and the head to be developed. The two factors give rise to the specific speed of the pump. Hence, the specific speed becomes the important parameter in the design of a pump. However, there may be values of the head and flow rates that result in specific speeds which are outside the limits of those specified in Table 7.2. If attempts are made to design pumps with such values of specific speeds, the designs may not be alright. Instead of such designs, it is useful to manipulate the design by using multiple units of standard (or available) designs. Two such schemes are the pumps-in-series and pumps-in-parallel.

7.12.1 Pumps-in-Series This method is to meet the requirements of high heads and is shown in Fig. 7.12. The figure shows the schematic layout [Fig. 7.12(a)] and the characteristics of two identical pumps in series [Fig. 7.12(b)]. The resulting head, curve y, represents the values that are twice the head of a single pump, curve x. Pumps of different heads can also be operated in series, with a condition that the rated flow-capacities must be equal. In such a case, the heads are additive. In the case of pumps-in-series, individual pumps cannot be stopped or withdrawn from the total system. Multi-stage pumps with several impellers mounted on the same shaft and outlet from one impeller leading to the inlet of the next impeller are actually pumps-in-series. These are suitable for applications such as pumping of water to overhead tanks of high-rise buildings, boiler-feed pumps, deep-well pumps, pumps for oil rigs, etc.

H

y

x

(a)

Figure 7.12

(b)

Q

(a) Schematic view of two identical pumps in series;(b) Characteristic of combined operation.

7.12.2 Pumps-in-Parallel This method is to meet the requirements of large flow-rates and is shown in Fig. 7.13(a). The schematic layout shows two identical pumps operating in parallel. The characteristic of the combined operation is also shown in Fig. 7.13(b).

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Turbomachines

H

y

x

(a)

Figure 7.13

Q

(b)

(a) Schematic view of two identical pumps in parallel. (b) Characteristic of combined operation.

The resulting flow rate, curve y, represents the values that are twice the flow rate of a single pump, curve x. Pumps of different flow rates can also be operated in parallel, with a condition that the heads must be equal. The flow rates in such a case are additive. Individual pumps can be stopped or withdrawn from the system. The other pumps can continue the pumping.

EXAMPLE 7.8 A number of centrifugal pumps are available, with specifications of the one in Example 7.3, namely, Q  50 lps, Hm  10 m, and N  960 rpm. Calculate the number of pumps required to meet a demand of Hm  30 m of water and flow rate of 150 lps. Sketch the arrangement.

Solution: When three pumps, each of Hm  10 m, are connected in series, the total head becomes 30 m, but the flow rate is only 50 lps through the three pumps. In order to get 150 lps, therefore, three such assemblies are required in parallel. Hence, the total number of pumps required are 3 s 3  9. This assembly is shown in Fig. 7.14. 100 lps

50 lps

150 lps (30 m) Delivery

50 lps (20 m)

50 lps (20 m)

50 lps (20 m)

50 lps (10 m)

50 lps (10 m)

50 lps (10 m)

50 lps

Figure 7.14

100 lps

150 lps Suction

Assembly of pumps to meet requirements.

Centrifugal Pumps

297

Comments: This example is only to illustrate the principle of series and parallel operation of pumps. The answer, in principle, is quite alright. However, the installation (piping with various fittings) and operation (head loss, during running, in all the pipes) can make the layout quite complicated. Suppose a new single pump is to be designed for 30 m, 150 lps (and say 1440 rpm), the specific speed is 43.5, and a compact pump can be designed, with total power of, approximately, 50 kW.

EXAMPLE 7.9 The outlet diameter of the impeller of a centrifugal pump is restricted to 25 cm and width at the outlet to 1.5 cm. The blade angle at the outlet is 60o. The drive to the pump is from a motor at 1440 rpm. Find the number of identical stages required to pump water to a water tube boiler, operating at 4 MPa. The capacity of the boiler is 60 kg/s. The hydraulic efficiency of the pump is 80%. The entry of water to the impeller is radial, without any whirl component and the flow component remains constant in the impeller. If the overall efficiency of the setup is 75%, find the power of the motor to drive the pump.

Solution: Given D2  0.25 m, B2  0.015 m, A2  60o, N  1440 rpm, m  60 kg/s, Gh  0.8, G0  0.75. Now, blade velocity at outlet is U2 

P D2 N P s 0.25 s 1440   18.85 m/s 60 60

We have the mass flow rate as 60 kg/s, from which, the volume flow rate is Q  (60 kg /s )  0.06 m3 /s The flow velocity is given by Q 0.06 Vf2    5.1 m/s P D2 B2 P s 0.25 s 0.015 With U2  18.85, Vf2  5.1 and A2  60o, it is possible to draw the velocity triangle at outlet. The velocity triangle is shown in Fig. 7.15. V2 Vf2 = 5.1 Vu2 = 15.91

Vr2 60° b2

U2 = 18.85

Figure 7.15

Outlet velocity triangle.

Now Vu2  U 2

Vf2

tan B2 5.1  18.85  18.85 2.944 tan 60o  15.91 m/s

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Turbomachines

Theoretical head or Euler head is HE 

U 2Vu 2 g

We know that Gh  Hm/HE; therefore, ¥ 18.85 s 15.91´ H m  H E s Hh  ¦ µ¶ s 0.8  24.457 m of water § 9.81 Now, total head required is 4 MPa  400 m of water. Therefore, Total head 400 Number of stages    16.355 Head per stage 24.457 So adopt 17 stages. (17 impellers are mounted on the same shaft, driven by the motor, one impeller delivering to the inlet of the next impeller.) Now, with 17 impellers in series, Total head created  Head created by one impeller s Number of impellers  24.457 s17  415.769 m Equivalent power of this pumping process is Equivalent power 

wQH 9810 s 0.06 s 415.769   244.72 kW 1000 1000

Power of the motor, to drive this pump, with an overall efficiency of 0.75, is P  244.72 / 0.75  326.3 kW

EXAMPLE 7.10 It is proposed to design a pump to have a flow rate of 2 m3/s at a head of 30 m of water. The usual efficiencies that are assumed are as follows: Mechanical efficiency  0.98 and manometric efficiency  0.88. The usual speed (in rpm) to be adopted is 2880, 1440, 960, 735, or 590. (These are rpm of induction motors.) Assuming that the blade inlet angle is 45o and the outlet angle is 65o, find the possible dimensions of the impeller. The entry of water to the impeller is radial.

Solution: Given Q  2 m3/s, Hm  30 m, Gm  0.98, Gmano  0.88, A1  45o, A2  65o.

First, let us calculate specific speed. We have Ns 

N Q N 2  3/ 4  0.11 N H 3/ 4 30

With the different speeds given as part of data, we can calculate the specific speeds corresponding to each speed, as shown in Table 7.4. Table 7.4

N

2880

1440

960

735

590

Ns

316.8

158.4

105.6

80.85

64.9

Centrifugal Pumps

299

Generally, up to Ns b 105, the pumps are radial flow pumps; for 100  Ns  210, the pumps are mixed flow pumps; and for Ns  200, the pumps are axial flow pumps. Presently, therefore, the possible speeds are 960, 735, and 590. Also select D2/D1  2. As per data, A1  45o which implies U1  V1  Vf1  Vf2  U2/2 The general form of the exit velocity triangle is shown in Fig. 7.16. V2

Vr2

Vf2 = U2/2

b2 = 65° U2 = 20.884

Figure 7.16

Outlet velocity triangle.

Whirl component of velocity is given by Vu2  U 2  U2  U2

Vf2 tan B2 U2 2 tan 65o U2

4.289 1 ´ ¥  U 2 ¦1 µ § 4.289 ¶  0.7668 U 2 Manometric efficiency is given as 0.88. Therefore, we have gH m

 Hmano  0.88 U 2Vu 2  U 2 (0.7668 U 2 )0.88  9.81 s 30 Therefore 9.81 s 30 (0.88 s 0.7668)  U 2  20.887 m /s U 22 

Also U 2  P D2 N /60 Therefore

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Turbomachines

D2 N 

U 2 s 60

P

 398.855

Select N and workout further details. See Table 7.5. Table 7.5

N

960

735

590

D2

0.4155

0.5427

0.676

D1  D2 /2

D1

0.2078

0.2714

0.338

U2  20.884

U2

20.884

20.884

20.884

U1  U2 /2

U1

10.442

10.442

10.442

B2  Q/OD2 /Vf2  Q/OD2 U1

B2

0.147

0.1123

0.09

B 1  B2 s 2

B1

0.294

0.2246

0.18

D2  398.855/N

The alternate designs can be one of the above three speeds: 960 rpm, 735 rpm, or 590 rpm. The specific speed 316.8 (N  2880) belongs to the design of axial flow pump and that of 158.4 (N  1440) belongs to the design of mixed flow pump.

K eywords Impeller Casing Blade angles Suction pipe Delivery pipe Foot valve Delivery valve Specific speed Multi-stage pumps Static suction head Static delivery head Friction head Dynamic head Manometric head Euler head Ideal head Developed head

Mechanical efficiency Volumetric efficiency Hydraulic efficiency Manometric efficiency Overall efficiency Velocity triangles Radial entry Radial vanes at outlet Slip Losses Energy flow Maximum suction lift Net positive suction head Cavitation Priming Series operation Parallel operation

Centrifugal Pumps

301

S ummary 1. Different parts of pumps are (a) motor (power source); (b) shaft; (c) bearings (support); (d) impeller (energy transfer); (e) casing (spiral/ volute/diffusion); (f ) delivery pipe; (g) suction pipe; (h) Foot-valve (non-return); (i) priming valve; (j) delivery valve; (k) base. 2. Classification of centrifugal pumps is based on (a) specific speed; (b) head; (c) casing; (d) impeller; (e) direction of liquid flow; (f ) applications; (g) multi-staging. 3. A variety of heads in the practice of centrifugal pumps have been considered, followed by the various efficiencies (hs, hd, hfs, hfd, Hm, Gm, Gh  Gmano, Gv). 4. A number of performance parameters have also been studied, such as the minimum starting speed (MSS), maximum suction lift (MSL), net positive suction head (NPSH), cavitation, and priming.

5. Cavitation is due to low pressures, less than vapor pressure of water, 6. Priming is filling of pump with water (same liquid as being pumped). 7. The performances of pumps, when working in series or in parallel, have been considered taking into account their head-capacity characteristics as also their applications. 8. In-series: Heads are additive, Q is same. 9. In-parallel: Flow rates are additive, H is same; any pump can be isolated/disconnected. 10. It may be mentioned here that the effects of variation of the blade outlet angle on the reaction and the energy transfer have already been studied earlier in Chapter 4, including the overall performance of the pump.

I mportant E quations 1. Specific speed

7. Hmano 

N Q H 3/ 4

Ns 

2. H m  hs  hd  hfs  hfd  3. H E  4. Hm  5. Hv  6. Hh 

WE g



U 2Vu 2 g

Pr Ps m m  $m P wQH H   Pn wQH n H n

Vd2 2g

gH m U 2Vu 2

8. Ho  Hmano s Hv s Hm 9. Minimum starting speed 1/ 2

¨ 2 g s H s 3600 · N r © 2 2m ¸ 2 ª P ( D2 D1 ) ¹ 10. MSL  ha hfs

V s2 2g

11. NPSH  ha hfs

hv

V s2 2g

hv hs

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Turbomachines

M ultiple- C hoice Q uestions 1. The flow of energy in a centrifugal pump is from the motor to (a) shaft to bearings to impeller to liquid (b) shaft to rotor to spiral casing to liquid (c) shaft to rotor to blades to liquid (d) shaft to glands to blades to liquid 2. The purpose of the bearings in a centrifugal pump installation is (a) to support the rotor (b) to support the shaft (c) to reduce the friction (d) to reduce the leakage 3. The purpose of the Foot-valve is to (a) support the suction pipe (b) prevent the liquid from reverse flow (c) fill the liquid during priming (d) all the above 4. The manometric head is (a) the vertical distance between water levels in the sump and delivery tank (b) the head measured by a manometer in the suction pipe (c) the head measured by a manometer in the delivery pipe (d) the head measured between suction and delivery flanges 5. Difference between the manometric head and the Euler head is (a) the loss in the suction side (b) the loss in the delivery side (c) the loss in the suction  delivery side (d) the loss in the impeller 6. For a centrifugal pump, the occurrence of slip is (a) good (b) bad (c) harmless (d) a loss 7. In this question, pick the wrong statement. (a) NPSH depends on the atmospheric pressure. (b) NPSH depends on the atmospheric temperature. (c) NPSH depends on the delivery pipe diameter. (d) NPSH depends on the suction pipe diameter.

8. Diffusion blades are required for the purpose of (a) increasing the power output (b) increasing the output pressure (c) increasing the output velocity (d) decreasing the power input 9. In a centrifugal pump, when A2  90o, the output head is (a) proportional to the speed (b) proportional to the square of the speed (c) proportional to the cube of the speed (d) independent of the speed 10. In a centrifugal pump, the flow rate is (a) proportional to the speed (b) proportional to the square of the speed (c) proportional to the cube of the speed (d) independent of the speed 11. In a centrifugal pump, the losses are (a) proportional to the speed (b) proportional to the square of the speed (c) proportional to the cube of the speed (d) independent of the speed 12. For a good design, at the design point, the slope of the head-flow rate curve must be (a) positive (b) negative (c) zero (d) it depends on various factors 13. As per design norms, the ratio outlet diameter/ inlet diameter (a) must increase as the specific speed increases (b) must decrease as the specific speed increases (c) does not depend on the specific speed (d) none of these 14. Taking that all the centrifugal pumps are well designed, their overall efficiency (a) increases as the capacity flow rate of the pumps increases (b) decreases as the capacity flow rate of the pumps increases (c) does not depend on the capacity flow rate (d) none of these

Centrifugal Pumps

15. Taking that all the centrifugal pumps are well designed, their overall efficiency (a) increases as the specific speed increases (b) decreases as the specific speed increases (c) does not depend on the specific speed (d) none of these 16. For a constant manometric head, the minimum starting speed of a pump (a) increases as the outlet diameter of the impeller increases (b) decreases as the outlet diameter of the impeller increases (c) does not depend on the outlet diameter of the impeller (d) none of these 17. When the centrifugal pumps work in series, (a) their heads are additive (b) their flow rates are additive (c) both (a) and (b) (d) none of these

303

18. When the centrifugal pumps work parallel, (a) their heads are additive (b) their flow rates are additive (c) both (a) and (b) (d) none of these 19. When centrifugal pumps have to work series, (a) their heads have to be equal (b) their flow rates have to be equal (c) both their heads and flow rates have to equal (d) none of these 20. When pumps have to work in parallel, (a) their heads have to be equal (b) their flow rates have to be equal (c) both their heads and flow rates have to equal (d) none of these

in

in

be

be

R eview Q uestions 1. With a neat diagram, explain the construction and working of a centrifugal pump. (Refer Section 7.3) 2. List the parts of a centrifugal pump, mentioning their functions in each case. (Table 7.1) 3. How are centrifugal pumps classified? What are the basic criteria for the classification? (Refer Section 7.4, with Table 7.3) 4. With a neat diagram, show the different heads that are used in the study of the centrifugal pump. (Refer Section 7.5) 5. Define the different efficiencies as the performance criteria of centrifugal pumps. List the factors that influence the values of such efficiencies. (Refer Section 7.6) 6. Define the minimum starting speed of a centrifugal pump. Derive an expression to calculate the minimum starting speed. (Refer Section 7.8)

7. Define the maximum suction lift. State the expression to calculate it. What factors affect its values? (Refer Section 7.9) 8. Define the NPSH, as referred to the installation of a centrifugal pump. State an expression to find its value. What factors affect its values? Which one is desirable, a high NPSH or a low NPSH? (Refer Section 7.9) 9. What is cavitation? Explain the phenomenon of cavitation. State some methods of eliminating or reducing cavitation. (Refer Section 7.10) 10. Explain the process of priming of pumps. Why is it required? (Refer Section 7.11) 11. Why multiple pumps are required? Under what circumstances, do we use multiple pumps? Write a note on operation of multiple pumps. (Refer Section 7.12)

304

Turbomachines

E xercises 1. The diameters of the impeller of a centrifugal pump are 20 cm and 40 cm at the inlet and outlet, respectively. The blades of the impeller are bent backward at an angle of 40o at the outlet. The radial velocity of flow is constant at 6.8 m/s. The water enters the impeller in the radial direction. The total head to overcome is 61.5 m of water. The speed is 1440 rpm. Calculate the following: (a) Blade angle at the inlet, (b) angle and magnitude of outlet velocity, (c) specific work, (d) manometric efficiency, and (e) power required to drive the pump if the width of the impeller passage at the outlet is 2 cm and the overall efficiency is 88%. 2. The blades of the impeller of a centrifugal pump are bent backward at 35o to the tangent of the wheel rim. The outer diameter of the impeller is 30 cm and the speed is 980 rpm. The measured head across the inlet and outlet of the pump is 15 m. Find (a) loss of head in the impeller (b) hydraulic efficiency. (c) Also calculate the power required to drive the pump if the flow rate is 15 kg/s. The entry of water is radial and the flow velocity remains constant at 3.5 m/s. 3. In a performance test of a centrifugal pump, it was found that the maximum efficiency occurred when the flow rate was 45 lps, at a head of 30 m, while running at 1440 rpm. The impeller diameter of the pump was 28 cm. Calculate the number of similar stages and the diameter of the impeller of a multi-stage unit to deliver 85 lps against a head of 130 m at a speed of 960 rpm. 4. The vertical height of the outlet of the delivery pipe of a centrifugal pump is 22 m, above the centerline of the pump. The length of the delivery pipe is 32 m, including the effective lengths of all the pipe fittings, and its diameter is 7.5 cm. The suction side head and the losses are negligible. A flow rate of 1500 lps

is required. The overall efficiency is estimated as 80%. Calculate the power output of the motor driving the pump, if f  0.009, and the specific speed of the pump if the speed is 1440 rpm. 5. The diameters of the impeller of a centrifugal pump are 0.4 and 1.0 m at the inlet and outlet, respectively. The speed is 300 rpm. The outlet angle of the blades is 30o. The radial component of velocity is constant and is equal to 6.5 m/s. The head across the impeller is 6 m, and the flow rate of water is 1 m3/s. Calculate (a) hydraulic efficiency, (b) power required to drive the pump. (c) Also determine the minimum starting speed. 6. The head measured at the outlet flange of a centrifugal pump is 20 m of water above that at the inlet. The flow rate is 0.09 m3/s. The impeller diameters are 20 and 40 cm. The blades are bent backward with an angle of 40o at the outlet. The width of the impeller at the outlet is 20 mm. The losses in the impeller are 1.5 m, and the gain in the diffuser blades is 3.5 m. The entry of water is radial. The flow velocity remains constant in the impeller. Calculate the following: (a) Head at the outlet of the impeller, (b) head developed by the impeller, (c) hydraulic or manometric efficiency, (d) speed, and (e) angle of the diffuser blades. 7. A pump is required to be designed for a flow of 0.1 m3/s at a manometric head of 60 m. A manometric efficiency of 85% and a mechanical efficiency of 95% can be assumed. The effect of the vane thickness is to reduce the flow area by 10%. The proposed speed is 1500 rpm. The velocity of flow remains constant at 2.5 m/s and the blade angle at the outlet is 55o. If a three-stage unit is preferred, calculate the following: (a) Impeller diameter, (b) width of impeller at the outlet, and (c) overall powe required to drive the pump.

Centrifugal Pumps

8. In a centrifugal pump, the width of the impeller passage is 2.5 cm at the outer diameter of 45 cm. The thickness of blades in the impeller accounts for 10% reduction in the flow area. The manometric efficiency is 82%, and the overall efficiency is 78%. The whirl velocity at the exit is 24 m/s. The speed is 1440 rpm. Calculate the following: (a) Head generated, (b) flow rate, (c) exit vane angle, and (d) power required to drive the pump. 9. The following observations were recorded during a trial in a centrifugal pump installation. Static suction head  5.3 m, static delivery head  26 m, frictional loss of head in suction pipe  1 m, frictional loss of head in delivery pipe  6.3 m, diameter of suction pipe  10 cm, diameter of delivery pipe  7.5 cm, flow rate  0.045 m3/s, manometric efficiency  82%, overall efficiency  76%, room temperature

 29oC, and barometer reading  10.01 m of water. Calculate the following: (a) Manometric head, (b) head developed in the impeller, (c) power drawn by the pump, and (d) NPSH. 10. A centrifugal pump delivers 0.04 m3 of water per second through a vertical height of 35 m. The suction and delivery pipes are of equal diameter of 10 cm, and their total length inclusive of effects of pipe fittings is 55 m. The pipe friction coefficient f is 0.006. In the impeller, the blade width at the outlet is 1/10th of the diameter, and the thickness of blades accounts for 8% of loss of flow area. The blades are bent backward, making an angle of 65o with the wheel tangent. The hydraulic efficiency is 80% and the specific speed is 30. Calculate the following: (a) Manometric head, (b) speed, and (c) impeller diameter.

S olutions A vailable

P roject- O riented Q uestions 1. This is a small project having three parts: Part (a), Part (b) and Part (c). Part (a): Enlist the design parameters and performance parameters of a centrifugal pump. Explain, to the extent possible, the effects of variation of such parameters (e.g., in the impeller, the width of the blade, B2, affects the flow rate; higher the width, higher is the flow rate; and hence higher power; also, it does not affect the head created, … .). Part (b): Prepare a table which shows the values of parameters [as in Part (a) above] in the solved examples and the exercise problems of this chapter. If some values are not available in the problems (either as part of data or part of answers), calculate them wherever possible.

Part (c): Make an attempt to reinforce your explanations of Part (a), citing the table of Part (b), wherever possible. 2. Two pumps are required to be designed: one for a fire hose and another for pumping water to a town water supply scheme in a remote place where electrical power supply is snapped. Both require power supply from mobile units and therefore the power of the prime mover is limited to an upper limit of 60 kW. Discuss the parameters which distinguish the design of the two pumps. Substantiate your discussion with some numerical values that you might use in the design.

A nswers Multiple-Choice Questions 1. (c) 2. (b)

305

3. (b) 4. (d)

306

5. 6. 7. 8. 9. 10. 11. 12.

Turbomachines

(d) (c) (c) (b) (b) (b) (b) (b)

13. 14. 15. 16. 17. 18. 19. 20.

(b) (d) (a) (b) (a) (b) (b) (a)

Exercises 1. (a) 24.27o, (b) 17.13o and 23.08 m/s, (c) 665.35 J/kg, (d) 90.68%, (e) 129.2 kW 2. (a) 1.31 m, (b) 92%, (c) 2.4 kW 3. 5, 39.62 cm 4. 14.92 kW, 12.35 5. (a) 84.16%, (b) 80.23 kW, (c)226 rpm 6. (a) 16.5 m, (b) 18 m, (c) 91.67%, (d) 744.5 rpm, (e) 17.55o

7. (a) 20.5 cm, (b) 6.2 cm, (c) 72.89 kW 8. (a) 83 m, (b) 0.25447 m3/s, (c) 38.86o, (d) 265.6 kW 9. (a) 43.888 m, (b) 53.522 m, (c) 25.493 kW, (d) 1.63656 m 10. (a) 53.772 m, (b) 2978.6 rpm, (c) 17.2 cm

8

Fans, Blowers, and Compressors

Learning Objectives After completing this chapter, you will be able to: v
Understand the constructional and working features of radial and axial flow compressors. v
Analyze the working of radial and axial flow compressors. v
Identify the factors influencing the working of radial and axial flow compressors. v
Find the influences of various parameters on the performance of radial and axial flow compressors.

8.1

v
Understand the mechanism of diffusion of velocity to create pressure and hence find the diffusion efficiency in radial and axial flow compressors. v
Understand the phenomena of surging, choking, and stalling as they occur in the process of compression in radial and axial flow compressors.

Introduction

Power-absorbing turbomachines that handle gaseous or compressible fluids, such as air, ammonia, freons, etc., are classified as fans, blowers, and compressors on the basis of the pressure ratios they create. 1. The pressure ratios of fans hardly exceed 1.07 (i.e., a few centimeters of water). Fans are used to supply air for circulation or ventilation. 2. Blowers have their pressure ratios of the order of 1.5–2. Blowers are used to provide air for ventilation of large areas, to supply combustion air to the furnaces, to fluidize combustion beds, and so on. 3. Compressors create the overall pressure ratios of the order of about 2.5–10. Compressors are employed to circulate refrigerants in refrigeration plants of large sizes and to supply compressed air to the combustion chambers of gas turbine power plants or aircraft engines. Blowers and compressors are of two types: (a) Radial flow type and (b) axial flow type. Adoption of multiple stages in radial flow type machines involves some constructional complexities. Axial flow type machines can be arranged in multiple stages very easily, with less cost. Their working is with low losses and high efficiencies. Multi-stage axial flow compressors (AFCs) are very compact and are best suited for turboengines of aircrafts. Their pressure ratios are of the order of 6–8, sometimes even up to 10. Unlike centrifugal pumps, blowers and compressors are run at high speeds to increase the mass flow rates. Especially, the compressors employed in gas turbine plants are directly coupled to turbines and therefore the speeds can reach as high as 30000 rpm or more, thus increasing their air-handling capacities.

8.2 Radial Flow Compressors or Centrifugal Compressors A radial flow compressor (RFC) is shown in Fig. 8.1. The blades of the impeller are radial, because the loading on the blades is comparatively less, and the fabrication of the impeller can be at low costs. An inducer

308

Turbomachines

section is provided in the radial blades, as shown, to avoid entry shocks. At the outlet of the impeller, there is a diffuser ring with diffuser blades. When the fluid flows through the passages between these diffuser blades, its velocity is converted into its pressure. Even without the blades, due to the increased diameter and increased area of flow, the passage acts as a diffuser. Such a passage is known as a vaneless diffuser. The fluid further flows into a volute casing to finally enter the delivery pipe. The volute casing has continuously increasing area of flow. The fluid enters the volute casing from the entire periphery of the impeller. Hence, the increase of area of flow in the volute casing is to accommodate the incremental flow of fluid from around the impeller, so that the velocity of fluid does not vary in the volute casing. Alternately, if the rate of increase of area is more than the rate required to maintain a constant velocity, then, the casing may further contribute to the diffusion of velocity into the increase of pressure. Delivery

Inducer section

Shaft

Impeller vanes

Diffuser vanes Volute casing

Figure 8.1

Schematic view of a radial flow compressor (RFC).

8.2.1 Stage Velocity Triangles, Inlet Flow Conditions, and Pre-Whirl For the purpose of analysis, it is assumed that the suction side has a suction pipe, long enough to have a uniform velocity in its cross-section. A blade is separately shown in Fig. 8.2. In the inlet section, at different radii (that is, radius at a, b, c and d) the blade velocities (U1a ,U1b , ) are different and the fluid inlet velocity (V1) is the same. The corresponding velocity triangles are also shown in Fig. 8.2. It may be observed that the inlet blade angles [(A1)r, r  a, b, …] vary over the radius. V1 Vr1 a b c d

Figure 8.2

V1

Vr1 V 1 (b1)b

(b1)a U1a

U1b

Vr1 V 1 (b1)c U1c

Vr1 (b1)d U1d

Inlet velocity triangles at different radii.

Now, the impeller shown in Fig. 8.1 or Fig. 8.2, has the inlet section or “eye” with a diameter corresponding to the point “a”. The effect of variation of this diameter is to be considered in some details.

Fans, Blowers, and Compressors

309

In any velocity triangle of Fig. 8.2, at any radius such as at a, b, c or d, the following equation holds: Vr12  U 12  V 12 In this equation, U 12

¥ PD N ´ ¦ 1 µ § 60 ¶

2

¥ ´ 4Q V 12  ¦ µ 2 2 § P ( D1 Dh )¶

and

2

In these equations for U12 and V12, D1 is the impeller eye diameter corresponding to point “a” and Dh is the hub diameter. It is seen that, when D1 is increased, the quantity corresponding to U 12 increases and the quantity corresponding to V 12 decreases in the expression for Vr12 For the specified or constant values of the flow rate Q, hub diameter Dh and speed N, if the differential of Vr1 with respect to D1 (i.e., d Vr1 /dD1) is equated to zero, a value of D1 is obtained for which Vr1 is a minimum. This means that at any value of D1, other than this value, Vr1 is more than the minimum possible value. It is desirable to adopt this value of D1 for the diameter of the eye as its optimum diameter, during the design process to get a minimum Vr1, for the following reason: Compressors generally have very high speeds. Hence, along with blade velocities, the fluid and relative velocities also have very high values. In the process, if the values of relative velocity Vr1 approach the sonic velocity (or exceed it), then there is the possibility (or certainty) of occurrence of shocks or shock waves at the inlet of the impeller. This is not desirable because such shock waves give rise to discontinuities in the flow field of the fluid, further resulting in losses. It is desirable, therefore, to have as low a value of Vr1 as possible for the specified conditions of the flow rate, speed, temperature, etc. Apart from deciding upon the optimum value of D1 to get minimum Vr1, a second method to reduce Vr1 is to use inlet guide vanes (IGV) so as to shift the fluid flow at an angle and introduce a pre-whirl in the direction of the blade velocity. In Fig. 8.3(a), the inlet velocity V1 is axial, without the presence of IGV. In Fig. 8.3(b), the inlet velocity triangle is with the presence of inlet guide vanes (IGV) to direct the flow so that the inlet velocity V1 is at the designed angle. The result is that Vr1 gets reduced. In Fig. 8.3(b), it looks as if the base of the velocity triangle, U1, is shifted backward, keeping the vertex at the same point. The pre-whirl, so introduced, reduces the specific work, by a quantity which is equal to U1Vu1 (whereas, without the inlet guide vanes, this U1Vu1 would be zero). But the undesirable effects of shocks are avoided. It may also be mentioned here that in multi-stage units, the first stage is more like later stages because of the pre-whirl. In later stages the pre-whirl is likely to exist (or, designed so) due to the preceding stages. V1

V1

Vr1

Vr2 b2 = 90°

a1 = 90° (a)

V2

Vr1

U1

U1 U1

(c)

U2

(b) With IGV

Figure 8.3

Without IGV

Velocity triangles for an RFC: (a) Inlet tip, (b) inlet tip with and without the presence of inlet guide vanes (IGV), (c) outlet: A2  90o.

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Turbomachines

Shown in Fig. 8.3(c) is the outlet velocity triangle. At the outlet, because the blades are radial, A2  90o, the whirl component of the fluid velocity becomes equal to the blade velocity, that is, Vu2  U2. Hence, the work done on the fluid can be stated as follows, from Euler’s equation (changing the negative sign) WE  WE 

and

U 2Vu2



gc U 22 gc

U 22 gc

U1Vu1 gc

(without IGV )

(8.1a)

(with IGV )

(8.1b)

In Chapter 4, topics on general velocity triangles, effects of the blade outlet angle A2 on the specific work, degree of reaction, the slip, the losses, the head capacity relationship, and the performance characteristics were studied. All these topics may be recalled for this chapter also. Although the general analysis includes the variation of outlet blade angle A2, it may be mentioned here that in RFCs, the exit blade angle is almost always 90o, that is, the blades are radial at the outlet. In fact, the entire blade is radial, with an inducer section at the inlet to accommodate the axial flow of fluid and to avoid the entry shocks.

8.3

Compression Process

The compression process is shown in its different forms on an enthalpy–entropy diagram in Fig. 8.4. h E 02


p02

02 p01

01

Figure 8.4

s

Compression processes.

Referring to Fig. 8.4: 1. 2. 3.

Process 01–E is the Euler process, WE  U22/gc. Process 01–02 is the adiabatic process, Wi  h02 – h01  cp(T02 – T01). Process 01–02a is the isentropic process, Wis  h02a – h01  cp(T02a – T01).

Now, it can be realized that the work required for the isentropic process is the least; the work for the adiabatic process is more than the isentropic work; and the theoretical or Euler work is still more. However, the actual work required, W, taking into account the frictional losses, eddies, turbulence (and such other dissipative processes), is even more than the Euler work.

8.3.1 Slip in the Compression Process Slip was explained in Chapter 4 with the details of causes and effects. The coefficient of slip (L) has values of the order of 0.9, 0.95, etc., for the compressors. For the process shown in Fig. 8.4, we have

Fans, Blowers, and Compressors

Wi

M

WE



h02 h01 U 22 / g c



c p (T02 T01 ) U 22 / g c

311

(8.2)

Also Wi  MWE

8.3.2 Isentropic Compression Efficiency The isentropic compression efficiency (Gc), was elaborated in Chapter 4. In the present context, it is defined as the ratio of the isentropic work to the actual work, also equal to Gt–t Wis

Hc 

W



h02a h01 W



c p (T02a T01 )

(8.3)

W

8.3.3 Pressure Coefficient or Loading Coefficient The pressure coefficient or loading coefficient (Xp) is defined as the ratio of the isentropic work for a given pressure ratio, at a blade-tip speed U2, to the work for maximum possible or Euler pressure ratio at the same tip speed U2. Thus

Yp  Yp 

Isentropic work at the tip speed U 2 Euler work at the tip speed U 2 Wis

(8.4)

WE

The denominator WE  U22/gc is the Euler work input with the assumption that the blades are radial (A2  90o, Vu2  U2). Even if the blades are not radial, this quantity is taken as a standard of reference. Also

Yp 

Wis WE



h02a h01 U 22 / g c



c p (T02a T01 ) U 22 / g c

8.3.4 Power Input Factor or Work-Done Factor A work-done factor (7) is defined as the ratio of the actual work supplied to the theoretical work supplied. Thus 7 or

7

Actual work input Theoretical work input W Wi

(8.5)

The various work interactions Wis, Wi, WE, and W are shown in Fig. 8.5 just to indicate their relative magnitudes and relationships with each other, but no scale is implied in the figure.

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Turbomachines

WE

Wi

W Wis

Figure 8.5

Comparative compressor works.

In Fig. 8.5, Wis is the isentropic work, 01–02a, Wi is the work from 01–02, WE is the work from 01–E, W is the actual input work, Gc  Wis/W, L  Wi/WE, Xp Wis/WE, 7  W/Wi. It can be seen that

Yp 

Wis WE



Wis W

s

W Wi s Wi WE

 Y p  Hc 7M

(8.6)

Only 7 has a value more than 1, such as 1.035, 1.04, etc. All other coefficients are less than 1.

8.3.5 Analysis of Diffuser Flow: Diffuser Efficiency At the exit of the rotor, the fluid has some kinetic energy depending on the degree of reaction of the rotor. In the diffuser ring (as shown in Fig. 8.1), only the transformation of energy takes place in the fluid from its kinetic form to its pressure form. There is no transfer of energy from or to the fluid in the diffuser ring. The process that takes place in the diffuser (or stator) is shown in Fig. 8.6. T or h

p02 p03

02

2

03 p3

3 3

2

V2 2gc

V3 2gc

3
p2 2 S

Figure 8.6

Diffusion process (p03 < p02 due to losses in the diffuser).

Process 2–3 is the actual diffusion from the static pressure p2 to the static pressure p3, with some increase of entropy. For any actual process, there is an ideal or isentropic process, and it is presently 2–3q for the diffusion of the same quantity of kinetic energy. A diffuser efficiency is defined as the ratio of actual rise of static pressure to the isentropic rise of static pressure, for the diffusion of a specified quantity of kinetic energy. Thus diffuser efficiency is

Fans, Blowers, and Compressors

Hd  

313

Actual rise of static pressure Isentropic rise of static pressure p3 p2 p3aa p2

For the ideal or isentropic process, the total pressure remains constant. Thus p2 

RV 32 RV 22  p3aa  2 gc 2 gc

Therefore p3aa p2 

(V 22 V 32 )R 2 gc

For a limited change in the density Q

R A2V2  R A3V3 V3

or

V2



A2 A3

Therefore V22 p3aa p2 

¨ ¥ A ´2· ©1 ¦ 2 µ ¸ R © § A3 ¶ ¸ ª ¹ 2 gc

Substituting this expression, for (p3q – p2) in the denominator of the defining equation for diffuser efficiency, we get:

Hd 

( p3 p2 )2 g c ¨ ¥ A ´2· ©1 ¦ 2 µ ¸ RV 22 © § A3 ¶ ¸ ª ¹

(8.7)

For a compressible flow, where the density is variable, the diffuser efficiency is stated in terms of enthalpy changes (that can be converted to pressure changes using the isentropic relationships). The diffuser efficiency is defined as the ratio of the isentropic enthalpy rise corresponding to the actual rise of static pressure, to the enthalpy rise corresponding to the isentropic rise of static pressure for the diffusion of a specified quantity of the kinetic energy: ¨¥ T ´ · c pT2 ©¦ 3a µ 1¸ c p (T3a T2 ) h h ©ª§ T2 ¶ ¸¹ Hd  3a 2  2  2 2 (V 2 V 32 )2 g c h3aa h2 (V 2 V 3 )2 g c

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Turbomachines

· ¨¥ p ´ (G 1)/G c pT2 ©¦ 3 µ

1¸ ¸ ©§ p2 ¶ ¹ ª  Hd  2 ¨ ¥A ´ ·V2 ©1 ¦ 2 µ ¸ 2 © § A3 ¶ ¸ 2 g c ª ¹

(8.8)

The diffuser efficiencies stated in Eqs. (8.7) and (8.8) have to be used with some discretion. The diffusion of kinetic energy depends also on the rate of change of areas (from A2 to A3), that is, the angle of expansion of areas and the shape of the areas. The general values of the diffuser efficiencies are of the order of 0.7–0.9 when the cone angles of expansion of areas are less than 20o.

8.3.6 Stage Work and Stage Efficiency It may be noticed at this juncture that in Fig. 8.4, the process that can be subjected to the analysis or calculation is the isentropic process 01–02a. The other processes have to be correlated with the isentropic process only by getting the values of factors such as slip, efficiency, and work-done factors from experimental work. The end states of an isentropic process can be connected by the following isentropic relation: T02a T01

¥p ´  ¦ 02 µ § p01 ¶

(G 1)/G

As a result, the actual stage work required for a compression process from pressure p01 to pressure p02 can be calculated by knowing the values of the compression efficiency Gc. The usual values of this efficiency are of the order of 0.7–0.8. The stage work is given by W 

Wis

Hc



c p (T02a T01 )

Hc

c pT01 ¨¥ p02 ´ © W  Hc ©¦§ p01 µ¶ ª

(G 1)/G

·

1¸ ¸ ¹

(8.9)

·

1¸ ¸ ¹

(8.10)

Alternately, if W is known, the stage efficiency is given by c pT01 ¨¥ p02 ´ © Hc  W ©¦§ p01 µ¶ ª

(G 1)/G

8.3.7 Pressure Developed in the Rotor and Compressor From Eq. (8.10), the pressure developed for a prescribed input work can be written as p02 p01

¥W s H ´ c ¦  1µ § c pT01 ¶

G /(G 1)

(8.11)

Fans, Blowers, and Compressors

315

where W can be taken as equal to 7LU22/gc. For a known initial pressure (p1 or p01) and a prescribed actual input work W, the total pressure p02 at the outlet of the rotor can be determined using Eq. (8.11). Then, the static pressure p2 at the rotor outlet can be calculated when the exit fluid velocity V2 is known. After this, the diffuser efficiency Gd can be used to find the static pressure at the exit of the diffuser. For example, from Eq. (8.7) ¨ ¥ A ´2· RV 2 p3 p2  ©1 ¦ 2 µ ¸ Hd s 2 2 gc © § A3 ¶ ¸ ¹ ª

(8.12)

and from Eq. (8.11) p02

¥WH ´ c  p01 ¦  1µ § c pT01 ¶

G /(G 1)

(8.13)

Finally p2  p02

RV22 2 gc

(8.14)

Now, using Eq. (8.13) in Eq. (8.14), that is, substituting for p02 from Eq. (8.13), in Eq. (8.14), and then using the value of p2, so obtained, in Eq. (8.12), we get the expression for the exit pressure as ¥WH ´ c p3  p01 ¦  1µ § c pT01 ¶

G /(G 1)

« ¥ A ´ 2 º· RV 22 ¨© ® ® 1 Hd ¬1 ¦ 2 µ » ¸ 2 gc © A § 3 ¶ ®¸ ¼¹ ­® ª

(8.15)

Equation (8.15) states the value of the exit pressure of the compressor in terms of the initial pressure, rotor exit velocity, and values of Gc, Gd, cp, A2/A3, etc. However, instead of the combined expression [Eq. (8.15)], at a stretch, it is suggested that Eqs. (8.7) [or (8.8)], (8.10), and (8.11) may be used in steps to get the outlet pressure.

8.3.8 Performance of a Radial Flow Compressor The performance curves of RFCs are the plots of the parameters such as specific work, power, efficiency, etc. on a base of flow rate. These curves have been discussed in Chapter 4, Section 4.3, with Figs. 4.11–4.14. The same may be referred to once again with energy transfer suitably converted to the pressure rise parameter, wherever required.

EXAMPLE 8.1 The diameter ratio of the impeller of a centrifugal compressor is 2 and the pressure ratio is 4. At a speed of 12000 rpm, the flow rate is 10 m3/s of free air. The isentropic efficiency of the compressor is 84%. The blades are radial at the outlet and the entry is radial at the inlet. The velocity of flow remains constant at 60 m/s through the impeller. Calculate (a) power input to the machine, (b) the impeller diameters at the inlet and outlet, and (c) the blade angle at the inlet. The suction is from the atmosphere at 100 kPa and 300 K.

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Turbomachines

Solution: We have D2/D1  2, p02 /p01  4, N  12000 rpm, Q  10 m3/s, Gis  0.84, A2  90o, @1  90o, Vf1  Vf2  60 m/s, p01  100 kPa, T01  300 K.

(a) To calculate the power input or specific work, first, we have to calculate the actual temperature at the end of compression. We know that the temperature at the end of isentropic compression process can be calculated by: T02a T01

¥p ´  ¦ 02 µ § p01 ¶

(G 1)/G

 40.4/1.4  1.486

Therefore, T02a  T01 s 1.486 That is, temperature at the end of isentropic process is T02a  300 s 1.486  445.8 K Now we have ( T02a – T01 ) / ( T02 – T01)  Gis  0.84 as per data. Therefore, T02 T01 

( 445.8 300)  173.56 0.84

Therefore actual outlet temperature is T02  T01  (T02 T01 ), that is, T02  300  173.56  473.56 K The specific work is now given by W  U 2 s Vu2 = U 22  c p ( $T )  1004 s (473.56 300)  174254 J/kg or m 2 /s2 Therefore U 2  (174254)0.5  417.44 m/s Power input to the machine per unit mass flow rate is P  W  U 22  174.254 kW per kg /s flow (b) Impeller outlet diameter is D2 

U 2 s 60

PN



417.44 s 60  0.6644 m P s 12000

Also, Inlet diameter  D1  D2/2  0.6644/2  0.3322 m (c) To calculate the blade inlet angle, we have the blade velocity at inlet (because D2/D1 2), U1 

U2  208.72 m /s 2

Fans, Blowers, and Compressors

317

Inlet blade angle is ¥V ´ ¥ 60 ´ B1  tan 1 ¦ f 1 µ  tan 1 ¦  16.04o § 208.72 µ¶ § U1 ¶

EXAMPLE 8.2 The impeller of a centrifugal compressor has the inlet and outlet diameters of 0.3 and 0.6 m, respectively. The intake is from the atmosphere at 100 kPa and 300 K, without any whirl component. The outlet blade angle is 75o. The speed is 10000 rpm and the velocity of flow is constant at 120 m/s. If the blade width at the intake is 6 cm, calculate (a) specific work, (b) exit pressure, (c) mass flow rate; and (d) power required to drive the compressor if the overall efficiency can be assumed as 0.7.

Solution: We have D1  0.3 m, D2  0.6 m, p01  100 kPa, T01  300 K, Vu1  0, A2  75o, N  10000 rpm, Vf1  Vf2  120 m/s, B1  0.06 m.

(a) First, to calculate the specific work, blade velocities are found as follows: We know that the blade velocity at inlet, U1, is given by U1 

P D1N P s 0.3 s 10000   157.08 m /s 60 60

The blade velocity at the outlet, U2, is (because D2  D1 s 2) U 2  U1 s 2  157.08 s 2  314.16 m /s Now, the velocity triangle at exit is drawn as shown in Fig. 8.7. V2 Vr2

Vu2 = 282.01

b2 = 75°

U2 = 314.16

Figure 8.7

Outlet velocity triangle for Example 8.2.

Also, from the triangle, (Fig. 8.7), we have the whirl component Vu2: Vu2  U 2

Vf2

tan B2 120  314.16  282.01 m /s tan75o

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Turbomachines

Specific work is given by W  U 2Vu 2  314.16 s 282.01  88596.3 J/kg or m 2 /s2 (b) To calculate the exit pressure, we have: W  c p ($T )  c p (T02 T01 ) ¥T ´  c pT01 ¦ 02 1µ § T01 ¶ ¨¥ p ´  1004 s 300 ©¦ 02 µ ©§ p01 ¶ ª

So

0.4 /1.4

¨¥ p ´ 88596.3  1004 s 30 s ©¦ 02 µ ©§ p01 ¶ ª

·

1¸ ¸ ¹

0.4 /1.4

·

1¸ ¸ ¹

Therefore p02  2.466 p01 Hence the exit pressure is p02  2.466 s p01  (2.466 s 100) kPa  246.6 kPa (c) Also, to calculate the mass flow rate, first, we have the volume flow rate given by: Volume flow rate  P D1B1Vf 1  P s 0.3 s 0.06 s 120  6.786 m3 /s We know that pV  mRT  Therefore, the mass flow rate is pV RT 100 s 1000 s 6.786  287 s 300  7.88 kg /s

m 

(d) We have Theoretical power input  Wm  88596.3 s 7.88  698.138 kW With an overall efficiency of 0.7, the actual power required is P  698.138 / 0.7  997.34 kW

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319

EXAMPLE 8.3 A centrifugal compressor runs at 15000 rpm and produces a stagnation pressure ratio of 4 between the impeller inlet and outlet. Stagnation conditions of air at the intake are 100 kPa and 300 K. The absolute velocity at the compressor intake is without any whirl component. At the exit of the impeller, the flow component of the velocity is 135 m/s and the blades are radial. The total-to-total efficiency of the compressor is 0.78. Draw the velocity triangles and find the blade angle at the inlet. Also compute the slip and slip coefficient. The rotor diameter at the exit is 0.58 m and that at the inlet is 0.25 m.

Solution: We have N  15000 rpm, p02/p01  4, p01  100 kPa, T01  300 K, Vu1  0, Vf2  135 m/s, Gt-t  0.78, D2  0.58 m, D1  0.25 m. We know that the blade velocity at inlet, U1, is:

P D1N P s 0.25 s 15000   196.35 m /s 60 60

U1  The blade velocity at outlet, U2, is U2 

P D2 N P s 0.58 s 15000   455.53 m /s 60 60

Also, by isentropic relation for the ideal process, we have: T02a T01

¥p ´  ¦ 02 µ § p01 ¶

(G 1)/G

 4 0.4/1.4  1.486

Therefore T02a  T01 s 1.486  300 s 1.486  445.8 K The actual stagnation temperature rise is given by

 T02

(T02a T01 )

( 445.8 300)  186.923 Ht-t 0.78  T01  186.923  300  186.923  486.923 K

T02 T01 



Now specific work is W  c p (T02 T01 )  1004 s 186.923 Also W  U 2Vu2 Therefore, the actual whirl component at outlet, Vu2, is Vu2 

1004 s 186.923 455.53

 411.983 m /s

Because blades are radial at exit, we have the theoretical whirl component, Vu2’ as:

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Vu2a  U 2  455.53 The velocity triangles at inlet and outlet are shown in Fig. 8.8. From the velocity triangle, we have Slip  Vu2a Vu 2  455.53 411.983  43.547 m /s

V1

C1

Vr1

b1

34.51°

U1 = 196.35

Figure 8.8

Slip V2

C Vr2

Vu2 = 411.983 A

B U2 = 455.53 ABC: Exit velocity triangle without slip ABC1: Exit velocity triangle with slip

Velocity triangles for Example 8.3.

Now Slip coefficient 

Vu 2 411.983   0.9044 Vu2a 455.53

Blade angle at inlet is ¥V ´ ¥ 135 ´ B1  tan 1 ¦ f 1 µ  tan 1 ¦  34.51o § 196.35 µ¶ § U1 ¶ Comments: 1. 2.

The slip calculation may be particularly noted. The example could be extended to find power if either the width of the blades were given (at the inlet or outlet) or a mass flow rate were given. Suppose that the width at the inlet is 5 cm. Then Volume flow rate  P D1B1Vf 1  P s 0.25 s 0.05 s 135  5.3 m3 /s

Also, pV RT 100 s 103 s 5.3  287 s 300  6.1556 kg /s

Mass flow rate  m 

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So power is P  Wm  1004 s 186.923 s 6.1556  1155.232 kW

8.4

Matching of Compressor and System Characteristics: Surging in Compressors

Centrifugal compressors are employed to supply air to combustion zones in furnaces, fluidized beds, positive ventilation systems, etc. In some designs of air-conditioning systems, the compressors are also employed for circulation of refrigerants, where large volume flow-rates at comparatively lower pressure ratios are required. All these types of utilities are known as the “load” on the compressor system. Any variation of the load can be a load line, which is a plot of the load head versus load flow rate. It is to be expected that a load line may have a positive slope, with higher flow rates demanding higher heads. It is very much desirable that a compressor be matched with its utility or load so that the operation is smooth. Any disturbance of the load (temporary change in either the head or flow rate) should not hamper the working of the system. The system when disturbed must be able to return to the stable duty point or design point. The operation of a centrifugal compressor is shown, along with a load line, in two separate cases: (a) With a negative slope of the H–Q characteristic (Figs. 8.9) and (b) With a positive slope of the characteristic (Fig. 8.10). Case (a): Negative slope of the characteristic (Fig. 8.9): Suppose that the disturbance is in the form of reduced flow rate, being less than the flow rate at the duty point. It can be seen that there is an excess of head available, such as Hm – Hn. This excess head causes the flow rate to increase back to the design value. Alternately, if the disturbance is the increase of flow rate, then there is a deficit of available head, such as Hj – Hk. This deficit head causes the flow rate to decrease back to the design value. In either way, the compressor continues to work at the stable duty point. H or (Δ P) m n

Load line j k Characteristic

Q Duty point

Figure 8.9

Stable working of a compressor.

Case (b): Positive slope of the characteristic (Fig. 8.10): Suppose that the disturbance is the reduction of flow rate. It can be seen that the available head Hn is less than the required head Hm, and this deficit head causes a further reduction of the flow rate. This continues and the flow

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rate reaches practically a zero value. At this point, the load side gets drained and suddenly the flow rate shoots up. Even when the initial disturbance is an increased flow rate, due to excess head, the flow rate shoots up. At that time, the compressor reaches its upper limit, gets choked, and the flow rate suddenly returns to its initial point and the cycle repeats. A cycle may take approximately 1 or 2 s depending on the overall capacity of the machines. H or (Δ P) j k

m

Characteristic Load line

n

Q Duty point

Figure 8.10

Working of a compressor with likely occurrence of surge.

The periodic stoppage and revival of fluid flow is known as surging. During surging, the fluid flow fluctuates between zero and peak flows. Surging is harmful for the machine, with consequent vibration, and is not desirable. Surging during the working is avoided by maintaining the working range in the characteristic with negative slope. In a centrifugal compressor or an RFC, the forward bent blades are likely to give rise to H–Q characteristics with a large portion of positive slopes. Therefore, forward bent blades are avoided in the design. While the preferred shapes of vanes are backward bent ones, radial vanes may be accepted as design alternates because of the ease of fabrication and lower costs. In both backward bent vanes and radial vanes, the useful portions of their characteristics, for normal operations, are only those with negative slopes. A typical set of characteristics of a compressor, with speed as a parameter, is shown in Fig. 8.11. The points of peak efficiencies of all the lines are joined by a broken line x. To the left of line x, another broken line y is drawn. This line separates the characteristics into portions with positive and negative slopes. As such, line y sets the limits of operation of the compressor on the lower side to avoid surging. H or (Δ P) y

x z

Q

Figure 8.11

Limits of surging and choking on compressors.

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323

To the right of line x, there is another line z beyond which the characteristics droop sharply. This drooping nature is due to the choking of the flow passages (increased flow rates have higher losses, and hence the sharp fall in pressure or head). Once again, line z sets the limits of operation on the higher side, due to the highly unstable operations. The operation of the compressor, therefore, has to be only in the regions between lines y and z.

8.5 Axial Flow Compressors AFCs have large frontal areas coupled with high speed of the rotor. The large frontal area facilitates high volume flow rates or mass flow rates of the fluid. Multiple rotors can be mounted on a common shaft, and stator blade rings are mounted on the stator, so as to be between the rotor blade rings. This results in a very compact multi-stage compressor with high pressure ratios between the inlet and outlet. Casing Inlet Outlet

Inlet velocity Inlet pressure

Figure 8.12

Stator

Rotor

Stator

Rotor

Stator

Rotor

Shaft

Outlet velocity Outlet pressure

Schematic diagram of a three-stage axial flow compressor (AFC).

A three-stage AFC is shown in Fig. 8.12 along with the development view of the rotor and stator blades and the corresponding velocity and pressure diagrams. Being a compressor, the blade geometries are essentially reaction-oriented. Hence, the rotor blades while imparting energy to the fluid also transform some kinetic energy to pressure energy. After each ring of rotor blades, there is a ring of stator blades acting as a diffuser ring, where some more kinetic energy of the fluid is converted into pressure energy. One ring of rotor blades followed by one ring of stator blades constitutes one stage. At the outlet of a stage, the velocity is low and the pressure is high. The fluid gets ready to enter the next stage rotor, and the processes keep repeating. As the pressure of the fluid increases, its specific volume decreases. Hence, for the same order-of-magnitude flow velocities, the area of flow has to be decreased for a given mass flow rate.

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8.5.1 Stage Velocity Triangles The velocity triangles of an axial flow machine were discussed in Chapter 4 and the details may be recalled for further discussion. Topics such as slip, degree of reaction, effect of A2 etc., are of significance in the design of blades. C V1

Vr1

D2 D1 M1 M2 V2 V2 Vr2

Vr2 a1 = 90° b2

A

N1 N2

Set of blades, D1

Figure 8.13

U

B b2

Set of blades, D2

Velocity triangles and blade profiles.

The velocity triangles of the rotor blades are shown in Fig. 8.13. The triangles have a common base. The inlet velocity triangle is ABC. There are two outlet velocity triangles ABD1 and ABD2, of which only one has to be chosen at a time. The purpose of showing the two outlet triangles is to compare the profiles of the blades that are also shown in Fig. 8.13 with one set of blades each for one outlet velocity triangle. Thus, there are the two sets, D1 and D2, of the blades. The blades of set D2 have a larger outlet blade angle (A2) than that of set D1. The following points may be noted: 1. 2.

The specific work of the blade set D1 is less than that of the blade set D2 (because the whirl component Vu2 of set D1 is less than that of set D2). The degree of reaction of blade set D1 is more than that of blade set D2 (because Vru mean of set D1, BN1, is more than that of set D2, i.e., BN2).

The above two comparisons of the specific work and the degree of reaction suggest that as the blade outlet angle A2 increases, the specific work increases but the degree of reaction decreases. For a compressor, these two tendencies are in opposite directions: The increase of specific work is desirable but the decrease of the degree of reaction is not desirable. Now, in order that the work per kilogram of fluid may be more and the machine may be more compact, if the outlet blade angle is increased, then the degree of reaction is reduced with a lower outlet pressure. But more important is yet another effect: The flow separation may take place. Blade set D2 of Fig. 8.13 is once again shown in Fig. 8.14 to indicate the onset of separation. Once separation takes place, the entire working of the blades gets disrupted and the blades stall. This is not a desirable phenomenon. To avoid the stall, it is a usual practice to maintain a minimum degree of reaction of 0.5. Then the remaining 50% of the specific work is in the kinetic form and this can be converted into the increase of pressure in stator blades. The net result is that the fluid velocity at the end of the stage is almost the same as that at the entry to the stage, and the increase of pressure is equivalent to the specific work. These processes continue stage after stage.

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Blades

Points of separation Flow

Figure 8.14

Separation in set of blades D2 of Fig. 8.13: Blades of higher specific works.

8.5.2 Stage Work, Stage Efficiency, and Performance The concepts of Euler work, slip, pressure coefficient, work-done factor, and actual work input are almost the same as discussed with respect to the RFC above. Some differences are explained as follows: 1.

2.

3.

The design for high specific work blades may give rise to separation and stall. Therefore, one should not aim at a high specific work, high efficiency, and compact compressor. Instead, the reaction must be limited to 50%. The value of work-done factor, 7, is less than 1, about 0.7–0.8, in sharp contrast to 1.03–1.04 of RFC. This is because the Euler work of AFC is calculated at the tip of the blades (WE  U 22 /2gc). However, the absorption of the energy by fluid takes place more at mid-height stretches of the blade than at ends due to the non-uniform distribution of the axial velocity of the fluid over the height of the blades. As a result, the actual work done on the fluid is less than the theoretical work done as determined at the tip speed. Referring back to Fig. 8.5, it is interesting to see W less than Wi with respect to AFC. The efficiency and pressure rise of AFC have a fall on either side of the peak value at the design point, in contrast to that of the RFC. This is shown in Fig. 8.15. h, Δ p

Useful stretch

Radial flow Useful stretch

Axial flow Q

Figure 8.15

Useful portions of performance.

The useful stretch of the curve of axial flow machine can be seen as a narrow strip, compared to somewhat wider stretch of the curve of radial flow machine. In fact, the limits due to surge and choking in an AFC make up a narrow strip shown in Fig. 8.16, compared to those of RFC, as shown earlier in Fig. 8.9. An AFC is very “sensitive” for the load variation. In spite of this, it is preferred to an RFC because of its compact construction, and in particular, its application to aircraft engines where the load is generally uniform.

326

Turbomachines Δp N4 N3 N2 Surging limit

N1 Choking limit Q

Figure 8.16

Limits of surging and choking on an AFC.

EXAMPLE 8.4 In a multi-stage axial flow air compressor, the initial state of air is at 1 bar, 20°C and the final state is 5 bar, 270°C. Calculate the overall isentropic and polytropic efficiencies of the compressor. If the actual temperature rise per stage is limited to 16°C, calculate the number of stages required, assuming that the polytropic efficiency is the stage efficiency.

Solution: We have p01  1 bar, T01  20  273  293, p0 n 1  5 bar, T0 n 1  270  273  543. A temperature-entropy diagram for the compression processes (isentropic and actual) is shown in Fig. 8.17. T p0n + 1 = 5

543 464

p01 = 1

293

s

Figure 8.17

Temperature–entropy diagram for Example 8.4.

Let there be n stages (n to be found). From the isentropic relaton (ideal compression process), we have: T0n 1 T01

¥p ´  ¦ 0n 1 µ § p01 ¶

(G 1)/G

 (5)0.4/1.4

Fans, Blowers, and Compressors

Therefore, T0n  1  293 s (5)0.4/1.4  464 K . This is shown in Fig. 8.17. Isentropic overall efficiency is

Ho 

464 293  76.4% 543 293

For the polytropic process we have T0n 1 T01

¥p ´  ¦ 0n 1 µ § p01 ¶

( n 1)/n

Taking logarithms and rearranging we get

n 1  n

¥T ´ ln ¦ 0n 1 µ § T01 ¶ ¥ ln ¦ §

¥ 543 ´ ln ¦ § 293 µ¶  0.3833  ¥ 5´ p0n 1 ´ ln ¦ µ § 1¶ p01 µ¶

Polytropic efficiency is (G 1) 0.4 G Hp   1.4  74.53% (n 1) 0.3853 n Now, as per data, the actual rise of temperature, per stage, is given as: $Tact  16 Therefore ( $T )is  $Tact s 0.7453  16 s 0.7453  11.9248 Considering the first stage we have, T02  T01  11.9248, that is, T02  293  11.9248  304.9248. Therefore, for the first stage, the pressure ratio is p02 p01

¥T ´  ¦ 02 µ § T01 ¶

G /(G 1)

¥ 304 . 9248 ´ ¦ § 293 µ¶

3.5

 1.14984

327

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If there are n stages then n

¥ p02 ´ p0n 1 ¦¦ µµ  p01 § p01 ¶ Therefore, ¥p ´ ¥ 5´ ln ¦ 0n 1 µ ln ¦ µ § p01 ¶ § 1¶  11.52 n  ¥ p02 ´ ¥ 1.14984 ´ ln µ ¦§ ln ¦ µ 1 ¶ § p01 ¶ Therefore, the number of stages, n  12.

EXAMPLE 8.5 A 16-stage AFC is to have an overall pressure ratio of 6.3. Tests have shown that a stage efficiency of 89.5% can be obtained. The intake conditions are 1 bar and 288 K. Find the overall efficiency, polytropic efficiency, and pre-heat factor.

Solution: We have Pr  6.3, n  16, Gst  0.895, T01  288, p01  1 bar. We have the stage pressure ratio, pr, given by pr  (Pr )1/n  (6.3)1/16  1.1219 Now

Hst 

( pr )(G 1)/G 1 [(G 1)/G ]–(1/Hp )

( pr )

 0.895 

1

(1.1219)0.2857

1

02857 /Hp

1 ¥ 1.12190.2857 1´ 0.2857 /Hp  (1.1219) ¦ µ¶  1  1.0373276 § 0.895 0.2857 ln(1.0373276 )    0.31861 Hp ln(1.1219) (1.1219)

So the polytropic efficiency is

Hp  89.67% The overall efficiency is

Hc 

(6.3)0.2857 1  0.8675  86.75% (6.3)0.2857 /0.8967 1

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The pre-heat factor is PF 

Hc 86.75   0.9693 Hst 89.5

EXAMPLE 8.6 A compressor draws air from atmosphere at 1 bar and 290 K at a velocity of 70 m/s. The isentropic efficiency of the compressor is 72%. The stagnation pressure ratio is 3. Find the stagnation pressure at the exit and the power of the driving motor if the mechanical efficiency is 95% for a flow rate of 30 kg/min.

Solution: We have p1 1 bar, T1 290 K, V1  70 m/s, Gis  72%, p02 / p01  3, m  30 kg/min 

0.5 kg/s, Gmech  95%. Stagnation temperature at the inlet is

T01  T1 

V12 702  290   292.44 K 2 g ccp 2 s 1004

Now, applying the isentropic relation between the static and stagnation states, (knowing temperatures at both states) we can write: ¥T ´  ¦ 01 µ p1 § T1 ¶ p  01  1.03 p1 p01

G /(G 1)

¥ 292.44 ´ ¦ § 290 µ¶

1.4 /0.4

 1.02976

p01  1.03 bar (because, p1 is 1 bar) Stagnation pressure ratio is 3, as per data. Therefore, stagnation pressure at outlet, p02, is p02  p01 s 3  3.09 bar From isentropic relation, we can now get ideal stagnation temperature at outlet T02a: T02a T01

¥p ´  ¦ 02 µ § p01 ¶

(G 1)/G

 30.4/1.4  1.3687

Therefore the isentropic exit temperature is T02a  290 s 1.3687  400.27 K Now (Hc )t-t 

T02a T01

T02 t 01 ( 400.27 290)  0.72  (T02 290)

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Therefore, the actual exit temperature T02 is: T02  290 

( 400.27 290)  443.15 K 0.72

Specific work is given by W  c p (T02 T01 )  1.004 (443.15 290)  153.7626 kJ/kg At 95% mechanical efficiency and for a mass flow rate of 0.5 kg/s, the power of motor is P

Wm 153.7626 s 0.5   80.93 kW 0.95 0.95

EXAMPLE 8.7 An overhead fan with a vertical shaft is used to draw air through a small cooling tower from its top. The rotor blades have the tip and hub diameters of 2.0 and 0.5 m, respectively. The speed is 500 rpm. The blades are so designed that the blade loading is uniform through its length (i.e., the specific work is the same at any section of the blade). At the tip, the blade outlet angle is 12o. The flow velocity is constant through the rotor, at 10.5 m/s. Calculate (a) the blade inlet angle at the tip, (b) blade inlet and outlet angles at the hub, (c) the rise in total pressure, (d) mass flow rate, and (e) power required to run the rotor. The total-to-total efficiency may be taken as 0.85 and mechanical efficiency as 0.9. The inlet to rotor is at 288 K and 1 bar.

Solution: This is an axial flow type induced draft fan, at the top of the cooling tower. We have Dtip 

2.0 m, Dhub  0.5 m, N  500 rpm, (A2)tip 12o, Vf1  Vf2  10.5 m/s, (Hc )t-t  0.85, Gm  0.9, T01  288, p01  1 bar. Now Blade velocity at tip P Dt N P s 2 s 500 (U1 )tip  (U 2 )tip    52.36 m /s 60 60 Blade velocity at hub (U1 )hub  (U 2 )hub 

P Dh N P s 0.5 s 500   13.09 m /s 60 60

(a) Inlet blade angle at the tip is (B1 )t  tan 1

10.5  11.34o 52.36

(B1 )h  tan 1

10.5  38.73o 13.09

(b) Inlet blade angle at the hub is

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There are totally four velocity triangles, at inlet and outlet at tip; and at inlet and outlet at hub. These are shown in Fig. 8.18. Also, the blade profiles are shown at both tip and hub. Blade hub end

83.2°

Rotation 38.73°

Flow Blade tipend

Vr2

Vr1 V1 =Vf1 V2 =Vf2

b1 = 11.34° b2 = 12° Ut = 52.36 Velocity triangles at the tip V1 =Vf1 =Vf2

V2 b1

Vr2 b = 38.73° 1 b 2 = 83.2°

Uh = 13.09 Velocity triangles at the hub

Figure 8.18

Velocity triangles and blade profile for Example 8.7.

At the tip, we have the whirl components Vu1  0; and Vu2 given by Vu2  U 2

Vf 2

tan B2 10.5  52.36 tan 12o  2.96 m /s

Therefore the specific work is W  U 2Vu2  52.36 s 2.96  155 J/kg The specific work remains the same through the length of the blade. Therefore, at the hub (U 2Vu2 )hub  155 J/ kg Therefore, at the hub, the whirl component is Vu2 

155  11.84 m /s 13.09

Outlet blade angle at the hub is 10.5 ¨ · B2  tan 1 © ¸  83.2o ª (13.09 11.84 ) ¹

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Turbomachines

(c) Now, to calculate the rise in total pressure, we have: Specific work  Change in total enthalpy ($h)o  155 J/kg Therefore the change in total temperature, including the effect of the given efficiency, is ( $h ) ( $T )o  c p s (Hc )t-t 

155  0.182 1004 s 0.85

Hence, T02  T01  ($T)o  288  0.182  288.182 K. Now, knowing T01, T02 and p01, it is possible to use isentropic relation to get outlet stagnation pressure, p02: p02 p01

¥T ´  ¦ 02 µ § T01 ¶

G /(G 1)

¥ 288  0.182 ´ ¦ µ¶ § 288

3.5

 1.0022

So p02  1.0022 bar. Rise in total pressure  0.0022 bar  2.2 cm of water. (d) Now, to calculate the mass flow rate, first we have volume flow rate:

P ( Dt2 Dh2 ) 4 P (22 0.52 )   2.945 m 2 4

Area of flow 

We have Volume rate of flow  Area s Velocity  2.945 s 10.5  30.923 m3 /s Therefore, the mass flow rate is calculated as: pV 30.923 s 100 s 103 Mass flow rate    37.41 kg /s RT 287 s 288 (e) The power required is P

Wm 155 s 37.41   6.443 kW Hm 0.9

EXAMPLE 8.8 An AFC draws air from atmosphere at 1 bar and 15oC at the rate of 20 kg/s. The tip diameter of the rotor is 60 cm and the speed is 12000 rpm. The blade angles at the inlet and outlet are 40o and 70o, respectively. The air enters the rotor axially without any whirl component. The axial flow component of the velocity of air remains constant through the rotor and stator blades. The diffuser blades, following the

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333

rotor, make the flow again axial for the next stage. The total-to-total efficiency for the rotor is 85% and the mechanical efficiency of the drive is 98%. The work-done factor is 0.88 and the diffusion efficiency is 80%. Calculate the following: (a) The static pressure ratio across the rotor. (b) The static pressure ratio across the stage. (c) The degree of reaction. (d) The power input.

Solution: We have p01  1 bar, T01  15oC  288 K, m  20 kg/s, D  0.6 m, N  12000 rpm, A1 

40o, A2  70o, V1  Vf1  Vf2  V3, (Hc )t-t  0.85, Gm  0.98, 7  0.88, Gd  0.8. Calculate (a) p2/p1, (b) p3/p1, (c) R, and (d) P. The problem is worked out, with reference to the tip of the rotor. We know, the blade velocity is, U1  U 2  U 

P DN P s 0.6 s 12000   377 m /s 60 60

Now, because Vu1  0, we have V1 U

 tan B1

Therefore inlet velocity is V1  U tan B1  377 tan 40o  316 . 34 m /s  Vf1  Vf2  V3 The velocity triangles can now be drawn as shown in Fig. 8.19. V1 316.34

Vr1

V2

90°

Vr2 70°

Stator a2 = 50.38° U

a 2 Vu2 = 261.86 a 2 = 50.38°

40°

70° = b 2

40°

U = 377 Velocity triangles for rotor

Figure 8.19

Velocity triangles for the rotor. (There is no velocity triangle for the stator.)

From the outlet velocity triangle, we have, for the whirl component at outlet, Vu2: Vu2  U 2

Vf 2

tan B2 316.34  377 tan 70o  261.86 m /s

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Turbomachines

So, specific work is W  U 2Vu2  377 s 261.86  98721.22 J/kg Change in total enthalpy is ( $h )o  W s 7  98721.22 s 0.88  86874.674 J/kg Therefore the change in total temperature is ( $T )o  

( $h )o c p s (Hc )t-t

[inclusive of the effect of (Hc )t -t ]

86874. 674 (1004 s 0.85)

 101.8o

The stagnation temperature at the outlet is T02  T01  101.8  389.8 K  Total temperature at diffuser exit T03 (because there is no energy transfer in the diffuser) Now, from isentropic relation, we have p02 p01

¥T ´  ¦ 02 µ § T01 ¶

G /(G 1)

¥ 389.8 ´ ¦ § 288 µ¶

3.5

 2.88

Therefore p02  p01 s 2.88  2.88 bar  p03 From outlet velocity triangle, we have Exit velocity at the outlet of rotor is V2  Vu22  Vf22  261.862  316.342  410.66 m /s Static temperature at rotor exit is T2  T02

V 22 2c p

410.662 2 s 1004  305.82 K  389.8

Static temperature at the stage exit is T3  T03

V

2 3

2c p

316.342 2 s 1004  339.96 K  389.8

Diffuser efficiency is Gd  0.86  (T3a – T2) / (T3 – T2), where state 2 is rotor exit; state 3 is actual stage exit and state 3a is isentropic stage exit, corresponding to actual stage exit. (Fig. 8.6 may be referred). Now

Hd 

(T3a 305.82) (339.96 305.82 )

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335

Therefore T3a  335.18 K Now, the temperatures T2, T3, T3¾,T01, T03 are all determined; the isentropic relations can be used to find the corresponding pressures. Thus, p3 p03

¥T ´ ¦ 3 µ § T03 ¶

G /(G 1)

¥ 339.96 ´ ¦ § 389.8 µ¶

3.5

 0.62

Therefore p3  p03 s 0.62  2.88 s 0.62  1.7856 bar Again, p2 p3aa

¥T ´ ¦ 2µ § T3a ¶

G /(G 1)

¥ 305.82 ´ ¦ § 335.18 µ¶

3.5

 0.7255

Hence p2  p3aa s 0.7255  1.295 bar In a similar way as above, at the inlet to the rotor also, T1 and p1 are determined as follows: T1  T01 p1 p01

V 12 2c p

¥T ´ ¦ 1 µ § T01 ¶

 288

G /(G 1)

316.342  238.16 K 2 s 1004

¥ 238.16 ´ ¦ § 288 µ¶

3.5

 0.51427

Therefore p1  p01 s 0.51427  0.51427 bar (a) The static pressure ratio across the rotor is p2 1.295   2.518 p1 0.51427 (b) Static pressure ratio across the stage is p3 p1



1.7856  3.4721 0.51427

(c) Degree of reaction in the velocity triangles is, R

Vru mean U



377  115.14  0.6527 2 s 377

(d) Power to drive the compressor is Wm 98721.22 s 20 P   2015 kW Hm 0.98 s 1000

336

Turbomachines

Comments: 1.

2. 3.

4.

8.6

A step-by-step procedure is adopted, instead of using combined (and complex) equations, to find the pressures and temperatures. The process equations are only the isentropic relations and definitions of efficiencies or work-done factors. The “next stage” is very much the same as the present stage because the exit velocity from the stage is equal in magnitude and direction (V3 y V1) to the inlet velocity to the next stage. As the pressure is more at the exit of the present stage (i.e., entry to the next stage), the specific volume of air is reduced, the volume flow rate is less and therefore the area of flow has to be reduced. The area is reduced by decreasing the height of the blades. The degree of reaction is 0.6527. This means 65% pressure generation is in the rotor. In the stator, therefore, the conversion is of kinetic component only. The design can be accepted as such. However, the design can be modified to have 50% reaction also.

Stalling of the Blades

The word stall is used here to mean the loss of effective operation. The blades of the compressor are said to stall when their operation becomes suddenly ineffective. Stalling can be explained with respect to the functioning of a single blade or airfoil in a stream of fluid, but the same holds good with respect to the series of blades mounted on a rotor. Experimental results on cascades of blades with different blade profiles (refer Box) show the common nature of variation of the coefficients of lift and drag. A schematic representation of such a variation is shown in Fig. 8.20, on the base of angle of incidence i. It may be noticed that as the incidence keeps increasing from negative to positive values, the coefficient of lift increases slowly, but before it stabilizes at its peak value, the coefficient of drag (until then at low and stable values) suddenly starts increasing.

CL

CD

CL

CD

Stall −i

O +i Incidence i

Figure 8.20

Results of tests on blades.

This is obvious because as incidence becomes positive, the tendency for separation on the backside of the blade (convex side) increases. The drag increases and the blade stalls. This situation where the smooth function of blade ceases and the flow disturbance occurs is the stall situation. When this happens to the blades on the periphery of the rotor, the blades of the compressor are said to have stalled. Even when the rotor rotates, the fluid flow or the compression of the fluid or pressure rise does not occur.

Fans, Blowers, and Compressors

BOX 8.1    

337

Some Preliminary Terms With Respect to a Blade in a Fluid Stream

/LIWIRUFH)RUFHRQDEODGHLQDGLUHFWLRQSHUSHQGLFXODUWRWKHÁXLGÁRZGLUHFWLRQ 'UDJIRUFH)RUFHRQDEODGHLQDGLUHFWLRQSDUDOOHOWRWKHÁXLGÁRZGLUHFWLRQ P9/JF 5HIHUHQFHSUHVVXUH $5HIHUHQFHDUHD Lift force

Drag force V : Velocity r : Density

Figure 8.21

An airfoil in a uniform flow field, with lift and drag forces.

 &RHIÀFLHQWRIOLIW&//LIWIRUFH$Q9JF  &RHIÀFLHQWRIGUDJ&''UDJIRUFH$Q9JF 1RZFRQVLGHUDQDVVHPEO\RIEODGHVRUD´FDVFDGHµRIEODGHVDVVKRZQLQ)LJ$OVRFRQVLGHURQH EODGHRXWRIWKHVHULHVRIEODGHV UHSUHVHQWHGE\LWVPHDQOLQHRU´FDPEHUOLQHµ/LQH$$LVWDQJHQWWRWKH FDPEHUOLQHDWWKHOHDGLQJHGJHRUQRVHRIWKHDLUIRLO1RZFRQVLGHUWKHVWUHDPOLQHVRIWKHPRYLQJÁXLG 7KHPRYHPHQWLVUHODWLYHWKHÁXLGPD\PRYHEODGHPD\EHVWDWLRQDU\RUWKHEODGHPD\PRYHLQWKH VWDJQDQWÁXLG ,IWKHVWUHDPOLQHVKLWWKHFRQFDYHVLGHRIWKHEODGHDVVKRZQLQWKHÀJXUHWKHDQJOHRI LQFLGHQFHLLVWDNHQDVSRVLWLYH,IWKHVWUHDPOLQHVKLWWKHFRQYH[VLGHWKHDQJOHRILQFLGHQFHLVQHJDWLYH 7KHDQJOHRILQFLGHQFHDQGLWVPHDVXUHPHQWDUHVKRZQLQWKH)LJ

A

Camber line of blade (mean line)

−i A +i

Figure 8.22

Stream lines of fluid with a cascade of blades.

338

Turbomachines

K eywords Axial flow compressor Comparison of radial and axial flow compressors Diffuser blades Diffuser efficiency Effect of A2 on R Effect of A2 on W Fans, blowers, compressors H–Q relationship Inducer section Isentropic compression Lift and drag Losses Performance

Power input factor Pressure coefficient Pressure ratios Radial flow compressors Slip Stage efficiency Stage performance Stage velocity triangles Stage work, stage efficiency Stalling of blades Surging and choking Vaneless diffuser WE, Wi, Wis, W

S ummary 1. A brief distinction is made between the fans, blowers, and compressors, along with their applications. 2. Pressure ratio is of the order of 1.07 for fans, 1.5 for blowers and 2 – 2.5 for compressors. 3. Fans are generally single stage, axial flow machines, without the stator, for ventilation only. 4. Blowers and compressors are: (a) Radial flow compressors (RFC), for ventilation and circulation. (b) Axial flow compressors (AFC), for combustion chambers, fluidization, etc. (multi-stages) 5. A schematic layout of an RFC is studied and the representative velocity triangles are shown. 6. RFCs have radial blades (with inducer sections) for ease of fabrication and lower costs. 7. AFCs are high-end machines, for gas turbine power plants and air-craft engines. 8. The compression process, as it occurs in the RFCs, is studied in detail, including the slip, the isentropic efficiency, the pressure coefficient, and the work-done factor. The various works are identified and quantified and are related to each other.

9. Wis  Wi  WE  W. 10. Compression efficiency, Gc  Wis/W; Gc has values of the order of 0.8. 11. Coefficient of slip, L  Wi /WE ; L has values of the order of 0.9. 12. Work done factor, 7  W/Wi; 7 has values of the order of 1.04 for RFC and 0.8 for AFC. 13. Coefficient of pressure, Xp  Wis/WE; Xp  Gc7L; Xp has values of the order of 0.75 for RFC and 0.6 for AFC. 14. The process of diffusion of velocity into pressure, in the diffuser, and the diffuser efficiency are studied in detail. Because of diffusion, p03  p02. The stage work, stage efficiency, and pressure developed are also detailed. 15. The performance of the radial flow compressor is mapped. 16. The phenomena of surging and choking are explained with their causes, effects, and the limitations they impose on the working of the radial flow compressor. 17. Surging is a phenomenon in which the flow fluctuates between zero and full capacity in an RFC; it is likely to occur due to positive slope of the characteristic. Choking is a phenomenon in which the flow is too much that

Fans, Blowers, and Compressors

18.

19. 20.

21.

cannot be handled by the machine; and hence the pressure ratio falls steeply. Surging imposes the limitation on the lower flow-limit of characteristic; choking imposes the upper flow-limit of characteristic. In AFCs, this stretch is very narrow; in RFCs, the tolerable stretch is comparatively wider. With a schematic layout of an AFC,, its construction and working are studied. The stage work, stage efficiency, and performance of AFC are discussed in parallel with those of radial flow compressor. A distinction is drawn between the radial and

axial flow compressors, especially with respect to their working and useful stretch of performance. 22. The mechanism of stalling of the blades is explained with the causes and effects. 23. Stalling is the loss of performance of blades; it occurs due to flow separation taking place at higher angles of incidence. Stall depends on the blade geometry, operating speed, fluid velocity, mass flow rate and so on.

I mportant E quations 1. WE 

U 2Vu2 gc



U 22 gc

2. Coefficient of slip

M

Wi

WE 3. Isentropic compression efficiency W Hc  is W 4. Pressure coefficient/loading coefficient W Y p  is WE

· ¨¥ p ´ (G 1)/G c pT2 ©¦ 3 µ

1¸ ¸ ©§ p2 ¶ ¹ ª 8. Hd  ¨ ¥ A ´2· ©1 ¦ 2 µ ¸V 22 © § A3 ¶ ¸ ª ¹ 2 gc c pT01 ¨¥ p02 ´ © Hc ©¦§ p01 µ¶ ª

(G 1)/G

9. W 

·

1¸ ¸ ¹

c pT01 ¨¥ p02 ´ © W ©¦§ p01 µ¶ ª

(G 1)/G

10. Hc 

·

1¸ ¸ ¹

5. Work-done factor 7

W Wi

6. Xp  Gc7L 7. Diffuser efficiency

Hd 

( p3 p2 )2 g c ¨ ¥ A ´2· ©1 ¦ 2 µ ¸ RV 22 © § A3 ¶ ¸ ª ¹

339

11.

p02 p01

¥W s H ´ c ¦  1µ § c pT01 ¶

G /(G 1)

340

Turbomachines

M ultiple- C hoice Q uestions 1. In an RFC, the radial blades are employed (a) to avoid surging (b) to avoid stalling (c) to reduce cost (d) to increase efficiency 2. The purpose of inducer section in RFCs is (a) to have a high specific work (b) to increase the flow rate of fluid (c) to avoid surging (d) to have a smooth intake of flow 3. The slip in the RFCs is the reduction of (a) specific work (b) reaction (c) speed (d) utilization factor 4. The power input factor or work-done factor is (a) ideal work/actual work (b) actual work/theoretical work (c) isentropic work/actual work (d) ideal work/Euler work 5. For the diffusion of a specified quantity of kinetic energy, the diffusion efficiency is (a) isentropic rise in static pressure/actual rise in static pressure (b) isentropic rise in dynamic pressure/actual rise in dynamic pressure (c) actual rise in static pressure/isentropic rise in static pressure (d) actual rise in dynamic pressure/isentropic rise in dynamic pressure

6. Diffusion efficiency depends on (a) the rate of change of pressure (b) the rate of change of velocities (c) the rate of change of temperature (d) the rate of change of areas 7. Surging in an RFC is due to (a) the positive slope of the load line (b) the negative slope of the load line (c) the positive slope of the characteristic (d) the negative slope of the characteristic 8. AFCs have (a) purely impulse stages (b) purely reaction stages (c) impulse and reaction stages (d) alternate stages of impulse and reaction 9. Stalling of the blades means (a) loss of drag (b) loss of lift (c) separation of flow (d) loss of performance 10. Pick the wrong statement. (a) An AFC has larger frontal areas than an RFC. (b) An AFC has higher pressure ratios than an RFC. (c) An AFC has higher a weight-to-power ratio than an RFC. (d) In an AFC, it is easier to have multi-staging than in an RFC.

R eview Q uestions 1. Distinguish between fans, blowers, and compressors. (Refer Section 8.1) 2. Explain the working of an RFC. What is the purpose of the inducer section in RFCs? (Refer Section 8.2) 3. Draw a typical set of velocity triangles of an RFC and state the expression for the specific work. (Refer Section 8.2.1)

4. Describe the various losses in the impellers of RFCs and AFCs. (Refer Section 4.3.5) 5. With respect to a compression process, define the following: Slip, isentropic compression efficiency, pressure coefficient or loading coefficient, and power input factor. Establish a relation between them. (Refer Section 8.3)

Fans, Blowers, and Compressors

6. With an enthalpy–entropy diagram, describe the diffusion process. Derive an expression for the diffuser efficiency when the change in density is limited. (Refer Section 8.3.5) 7. Derive an expression for the stage efficiency in terms of initial temperature and pressure ratio. (Refer Section 8.3.6) 8. What are the performance curves of RFCs? Illustrate and explain. (Refer Section 8.3.8) 9. What is surging? When is it likely to occur during the operation of RFCs? (Refer Section 8.3)

341

10. Discuss surging and choking in an RFC. How do these phenomena affect the working of an RFC? (Discuss with respect to Fig. 8.11) 11. With a neat sketch, explain the working of an AFC. (Refer Section 8.5) 12. Draw a typical set of velocity triangles and corresponding blade shapes for an AFC. (Refer Section 8.5.1) 13. Explain the features of reaction in an AFC. Why the degree of reaction is limited to 50%? (Refer Section 8.5.1) 14. Explain the phenomenon of stalling of the blades. (Refer Section 8.6)

E xercises 1. A centrifugal blower has an impeller of outer diameter of 60 cm. The speed is 6000 rpm. The air at 100 kPa and 27°C enters the impeller without any whirl component. The radial component of velocity remains constant at 95 m/s. If the degree of reaction is required to be 0.6, calculate the outlet blade angle, the exit pressure, and the power input. Take the totalto-total efficiency as 70%. 2. The stagnation pressure ratio of a centrifugal compressor is 2.5 between the inlet and outlet of its impeller when the speed is 8000 rpm. The intake may be taken as still air from atmosphere at 100 kPa and 27oC. The entry of air is radial. At the exit of the impeller, the blades are radial. The total-to-total efficiency is 80%. If the impeller diameter at the outlet is 85 cm and if the flow component of velocity is constant at 160 m/s, calculate the specific work, power, and the slip factor. 3. A centrifugal fan draws air from still atmosphere at 100 kPa, 20°C while the speed is 580 rpm. The rise in the static pressure is 2 kPa and the power consumed is 44 kW for a flow rate of 3 kg/s. Suppose that the suction side conditions are changed to 98 kPa and 200oC and the

speed is changed to 490 rpm. Calculate the exit static pressure, the flow rate, and the power. 4. In a centrifugal compressor running at 7200 rpm, the impeller tip diameter is 90 cm and the mass flow rate is 40 kg/s. The suction is from still atmosphere, at 100 kPa, 300 K. A slip factor of 92% is to be accounted. For a static pressure ratio of 2.4, calculate the exit air temperature, the total-to-total efficiency, the pressure coefficient, and the power. 5. A compressor draws air from still atmosphere at 100 kPa, 310 K, at a velocity of 120 m/s. The isentropic efficiency of compression is 70%. The stagnation pressure ratio is 2.5. Find the stagnation pressure at the exit and the power of the driving motor. Assume a mechanical efficiency of 92% and a flow rate of 1 kg/s. 6. The overall pressure ratio of a six-stage AFC is 6. Its isentropic efficiency is 0.92. The mass flow rate is 4.5 kg/s and the suction is from free air at 100 kPa and 302 K. All the stages are similar. The degree of reaction is 0.5 in all the stages. The axial flow velocity remains constant at 120 m/s and the mean blade speed is 275 m/s. Calculate the polytropic efficiency, power, and the blade angles at the inlet and outlet.

342

Turbomachines

7. The overall stagnation pressure ratio in a multistage AFC is 6. The inlet is from atmosphere at 100 kPa and 290 K. The isentropic efficiency is 84%. The blade angles are 40o and 80o at the inlet and outlet, respectively. The degree of reaction is 0.5. The mean blade speed is 180 m/s and the velocity of flow remains constant. Calculate the polytropic efficiency and the number of stages.

8. Air at 100 kPa and 310 K enters a 50% reaction AFC that has a flow coefficient of 0.6. The total-to-total efficiency is 82%. The totalto-total pressure ratio is 1.7. If the coefficient of pressure is 0.6 and the work-done factor is 0.85, calculate the mean blade speed, and rotor blade angles at the inlet and outlet. Also find the power input for a flow rate of 10 kg/s.

S olutions A vailable

P roject- O riented Q uestions 1. Boilers in a thermal power plant frequently work with forced draft/induced draft systems. A forced draft blower supplies air to the combustion zone, maintaining a higher pressure zone below the fire grates. An induced draft fan creates lower pressure zones, somewhere in the flue-gas path, near the chimney (after the combustion gases transfer the heat to boiler, then to the economizer, etc.). 2. Conduct a small survey to find the range of higher or lower pressures mentioned above.

Also make an estimate of air requirement for the combustion, on the basis of unit output (say, per MWh) or unit input (per tonne of coal burnt) so that the flow rate of air (expected duty of the fans/blowers) can be readily reckoned. Check the parameters of any one fan/blower of the solved/unsolved problems of this chapter to suit with a power plant of a specified capacity.

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5.

(c) (d) (a) (b) (c)

6. 7. 8. 9. 10.

(d) (c) (c) (d) (c)

Exercises 1. 2. 3. 4.

68.35o, 125.1 kPa, 28.426 kW/(kg/s) 112.67 kJ/kg, 112.67 kW/(kg/s), 0.889 98.8668 kPa, 1.539 kg/s, 16.11 kW 405.5 K, 80.82%, 0.743, 4236 kW

5. 6. 7. 8.

270.834 kPa, 136.8 kW 93.73%, 980.4 kW, 30.54o, 59.25o 87.44%, 10 439.3 m/s, 41.57o, 61.65o, 579 kW

9

Power-Transmitting Turbomachines

Learning Objectives After completing this chapter, you will be able to: v
Study the constructional features of a powertransmitting device and be familiar with the terminology associated with it. v
Analyze the working of a power-transmitting device. v
Study the characteristics of a power-transmitting device.

9.1

v
Understand the advantages of a power-transmitting device over other conventional mechanical devices. v
Study the suitability of a power-transmitting device and its applications.

Introduction

It may be recalled from the basic classification of fluid machines that under the category of turbomachines, there are the three types, namely, power-producing, power-absorbing, and powertransmitting machines The power-transmitting types of turbomachines are those in which there are two separate shafts, one at the input end and the other at the output end. It is like a coupling, a clutch, or a gear box, but without the positive drive between the input and output shafts. The drive is through a fluid, some type of oil, which circulates continuously between the suitably shaped rotors of the inlet and outlet ends of the shaft. The oil receives the energy from the rotor of the input shaft, then moves over to the output end, and imparts the energy to the rotor of the output shaft. This chapter is an attempt to study such a device.

9.2

Working Principle of a Power-Transmitting Turbomachine

Let us perform a simple experiment in which two table fans are kept facing each other. Let one fan be connected to the power mains and the second fan be totally disconnected from the power mains. Now let the first fan be switched ON, as shown in Fig. 9.1. As the first fan keeps gaining speed, it creates a flow of air. The second fan, facing this air flow, also starts to pick up speed. Thus, it is seen that the shaft of the second fan gets its speed because its blades get energy from the flowing air. The first fan acts like a blower and the second fan acts like a turbine. The shafts of the two fans are not connected by any mechanical coupling, but they are now connected by the fluid (air). The power is transmitted from the first shaft to the second by means of three elements in the system, the blower, the turbine and the medium (fluid) flowing between them.

344

Turbomachines

Blower

Figure 9.1

Turbine

Two fans coupled by air flow.

The whole assembly of Fig. 9.1 is a “coupling.” The capacity of this coupling is very small and the efficiency of the transmission is perhaps too small due to the dispersal of air in all directions and losses. One simple conduit, a cylindrical enclosure between the two rotors as shown in Fig. 9.2, can increase the efficiency of the transmission very much, but still the capacity of the system remains very low. If instead of air the fluid is some liquid or oil (the blower is now a pump), the capacity can also be increased. Naturally, the oil has to be recirculated.

Cylindrical conduit (no dispersal of air)

Figure 9.2

Two fans of Fig. 9.1 now enclosed in a conduit.

A power-transmitting turbomachine is a compact unit, having an input shaft working on an impeller of a pump and the pumped oil driving the runner of a turbine, with its shaft acting as the output shaft. Thus, the working principle of a power-transmitting turbomachine is that the input shaft imparts energy to the liquid in a pumping unit and the liquid imparts that energy to the rotor in a turbine unit and returns to the pumping unit. A fixed member between the pumping and turbine units may (or may not) be present depending on the various purposes of the device. The oil, employed as the medium, must essentially be non-corrosive, non-foaming, non-toxic, and chemically stable with lubricating properties and preferably with high density and low viscosity in order that the capacity rating of the device could be high with compact size and low running losses.

9.3 Construction and Working of a Power-Transmitting Turbomachine The schematic diagram of a power-transmitting turbomachine is shown in Fig. 9.3(a). The power input is at shaft S1 that has an impeller P keyed onto it. The oil enters the impeller at P1 and leaves at P2. The oil then enters the blades of casing C. These stationary blades guide the oil (at the designed angle) to the inlet of the turbine blades, T, at T1. The energy of the oil is absorbed by the blades of the turbine, and then the oil leaves the runner at T2. The oil is now ready to enter the inlet of the pump, and the processes repeat continuously.

Power-Transmitting Turbomachines

345

C

P P2

T

T1 T

P P1 T2

S2

S1

S2

S1

(a)

Figure 9.3

(b)

Schematic diagram of a power-transmitting turbomachine (a) with stator blades and (b) without stator blades.

In order to get a clear and better idea of the device, impeller P is shaped like the impeller of a centrifugal pump and turbine unit T is shown like a Francis runner in Fig. 9.3(a). The purpose of the blades in the casing may be stated as directing the oil from the outlet of the impeller of the pump to the turbine runner. Different orientations of these casing blades (stator blades) can vary the speed of the driven shaft, resulting in different values of the output torques. Alternately, for equal torques, the stator blades can be totally absent, and the blades of the impeller and runner can be so shaped as to make the flow of the fluid from the impeller directly to the inlet of the turbine blades. This is shown in Fig. 9.3(b).

9.4 Analysis of Power-Transmitting Turbomachines Suppose that the speeds of the input and output shafts are Ni and No, respectively. The speed ratio Nr is stated as Nr 

No Ni

(9.1)

Similarly, let the torques be Si and So at the input and output shafts, respectively. The torque ratio Sr is stated as

Tr 

T o

Ti

(9.2)

346

Turbomachines

The input torque is negative and therefore the value of the torque ratio becomes positive. The following cases arise: 1. 2. 3. 4.

Sr  , that is, So  |Si|: This is the case of a simple fluid coupling, as shown in Fig. 9.3(b). Sr  1, that is, So  |Si|: This is the case of a torque multiplier. Sr  1, that is, So  |Si|: This is the case of a torque divider. S  0, that is, So  0: The output torque is zero and this is the case of a stalled output shaft.

The efficiency of a power-transmitting turbomachine is the ratio of the output power to the input power, as stated by

H

2P N oT o 2P N iT i

 Nr s Tr

(9.3)

The efficiency of any device is always less than 1. Hence, one can conclude the following: 1. 2.

If Sr is equal to 1, then Nr has to be less than 1. If Nr is equal to 1, then Sr has to be less than 1.

The above conclusions are the results of the assumption that there are the losses, and therefore, the efficiency is less than 1, with a consequent reduction of either the speed or the torque in the output shaft. The losses can be visualized as the power losses (or the energy losses). A simple energy balance can be stated as follows Energy input Energy loss in dissipation  Energy output or in the rate form as Power input Power losses  Power output That is

2ONiSi Power losses  2ONoSo The “power losses” in the above equation can also be written in the same form as other terms, so that one can write

2P N iT i 2P N T  2P N oT o or

N i T i N T  N oT o

(9.4)

where S is the loss of torque. This torque S can be associated with a speed that may be either the input speed Ni or the output speed No. Suppose that N in Eq. (9.4) is Ni that is, input speed. Then one can write

N i T i N i T  N oT o Hence substituting for NoSo in Eq. (9.3), that is,

H 

N oT o

N iT i

we get, efficiency as

H

N iT i N iT

N iT i



Ti  T Ti

(9.5)

Power-Transmitting Turbomachines

347

Suppose that N in Eq. (9.4) is No that is, output speed. Then one can write

N i T i N oT  N oT o

N iT i  N o (T  T o )

or Hence substituting for NiSi in Eq. (9.3), that is,

H 

N oT o

N iT i

we get efficiency,

H 

N oT o N o (T  T o )

H

To T  To

(9.6)

In Eqs. (9.4)–(9.6), the loss of torque S is due to the dissipation of energy in the fluid. The variation (increase or decrease) of torque can also be designed to get a torque multiplier or torque divider as required. This is done by introducing the stationary blades with different outlet angles. Hence, Eqs. (9.4)–(9.6) require some modifications. In such a case, the casing or stator also participates in the process of conversion of torque. The equipment is the one shown in Fig. 9.3(a). Therefore, in addition to the two torques Si and So, now there is another torque, Sf , of the fixed blades. An internal balance of the torques gives rise to the equation

Ti  To  Tf  0

or

To T 1 f Ti Ti Tr  1 

or

Tf Ti

(9.7)

Again, the following cases arise: 1. 2. 3.

Rr  1: This implies that Sf  0; the fixed blades do not participate in the process. This is a simple coupling. Rr  1: This implies that (Sf /Si)  0; because Si is negative, Sf also must be negative. The fixed blades exert a torque in the same direction as that of the input torque (the impeller torque). Rr  1: This implies that (Sf /Si)  0; Si is already negative. Hence, Sf is required to be positive. The fixed blades exert a torque in a direction opposite to the direction of the impeller torque (the same direction as that of the turbine).

With the above background, the study of the fluid coupling and the torque converter can be taken up now.

9.4.1

Fluid Coupling

A fluid coupling is shown in Fig. 9.3(b), where there are only two elements: (a) The pumping element and (b) the turbine element. The interacting faces of the elements guide the oil flow in the axial direction. The flow from the pump to the turbine is at a larger radius. The return flow from the turbine to the pump

348

Turbomachines

is at a smaller radius. The velocity triangles can be drawn corresponding to the two elements, in order to determine the interaction of the torques/energy/power. As the flow is parallel to the axis, the velocity triangles are on a plane that is parallel to the axis. For the case of the fluid coupling, Sr is equal to 1 and Nr is less than 1. The blade velocities of the turbine blades are, therefore, less than those of the pump blades. It has to be remembered now that the analysis is with respect to the flow of fluid in the two (or three) elements of the whole device, that is, the pumping unit (the stator unit) and the turbine unit. Naturally, the velocity triangles become the reference for analysis. Also, the velocity triangles are as practiced with reference to the respective machines. It is quite possible that the outlet of the pump impeller and the inlet of the turbine runner may be in the radial direction [as in Fig. 9.3(a)], or in the axial direction [as in Fig. 9.3(b)]. But these possibilities do not alter the analysis because the torque interactions depend on the tangential components of the fluid velocities. The notations U, V, and Vr continue with their meanings, that is, the blade velocity, absolute velocity of fluid, and the velocity of fluid relative to the blade, respectively. Subscripts 1 and 2 continue to mean the inlet and outlet conditions, respectively. The additional subscripts p and t signify the pump and the turbine, respectively. Just as there are different designs of the pumps or the turbines with varying parameters (D1, D2, A1, A2, @1, @2, etc.), there are various designs of fluid coupling and torque converters. Apart from that, the power input at the pumping unit must essentially be higher than the power output at the turbine unit. Vf1t = Vf2p

Vf2p = Vf1p C

b2p Vr2p

F b1t

B

Vr1t

E V2p

U2p

V1t = V2p

U1t U2p > U1t D

A Pump outlet

Turbine inlet Vf2t = Vf1t

Vf1p = Vf2t Vr1p c

b b1p U1p

V1p = V2t Pump inlet

Figure 9.4

a

Vr2t f U1p > U2t

e b2t U2t

V2t

d

Turbine outlet

General velocity triangles of fluid coupling.

The general velocity triangles for a fluid coupling are shown in Fig. 9.4. It is to be noted that: 1.

The speed of the turbine shaft is less than the speed of the pumping shaft. Therefore, U1t is less than U2p at the same radius, that is, the mean radius at the outlet of the pump or the inlet of the turbine (r2 p  r1t ).

2.

Both U1t and U2p are at a larger radius compared to the velocities at the other ends, that is, at a smaller radius (r1p  r2 t ).

Power-Transmitting Turbomachines

3. 4. 5. 6.

349

The flow components are the same, that is, Vf 1t  Vf 2 p . In a similar way, U2t is less than U1p at the same radius. Both U2t and U1p are at a smaller radius. The flow components are the same, that is, Vf 1p  Vf 2 t .

Suppose that at the exit of the pump, the blades are radial, that is, A2p  90o. Then, Vu 2 p  U 2 p . From the velocity triangles, the pump torque is given by

T p  m (r1pVu1p r2 pVu2p )  m (r1pU 2t r2 pU 2p ) and turbine torque is

T t  m (r1tU 2p r2 tU 2t )  m (r2 pU 2p r1pU 2t ) The magnitudes of the two torques are the same. St is the negative of Sp. Now U 2t U 2p



2P r2 t N t 2P r2 p N p



r1p N o r2 p N i



r1p r2 p

s Nr

Hence ¨

T p  T t  m

¥

´

r1p r 22p(2P N p ) ©1 ¦ µ © § r2 p ¶ ª

2

· Nr ¸ ¸ ¹

(9.8)

It can be concluded from Eq. (9.8) that 1. 2.

The torque is a function of the ratio of radii (r1p /r2 p ). The torque is proportional to r 22p.

3.

If the speed ratio Nr is increased, the torque is decreased. This means that the torque impressed on the output shaft is maximum when the turbine is stationary. Usually, the driven members in a machine require the highest torque when they are about to start (i.e., from the stand-still condition), to overcome the inertia. The above features of the fluid system, therefore, are an ideal solution to the transmission problems in an automobile. Further, when the speed ratio Nr reaches 1, it may be noticed that the torque reduces to the minimum. This is because the centrifugal head created by the pump is nearly balanced by the head created by the turbine. The torque is also proportional to the mass flow rate of the fluid. This is how, when the fluid is air, the capacity of transmission is very low. A high-density fluid can transmit higher torques with a specified volume flow rate.

4.

5.

The operating characteristics can now be explained with reference to Fig. 9.5. The characteristics are drawn on a base of the pump speed (Np or Ni). Initially, both the pump and turbine are stationary. As the pump starts and picks up the speed, the turbine continues to remain stationary until a critical pump speed Nc. At this speed of the pump, the torque is just sufficient to overcome the friction on the turbine side. Once this speed is exceeded by the pump, the turbine also picks up the speed slowly and the output torque keeps increasing. The speed ratio is zero until the pump speed reaches the critical speed.

350

Turbomachines

A “slip” is defined as (Np Nt)/Np. Until the pump speed is Nc, the slip is 1, and as the turbine starts, the slip decreases. When the slip is down to about 5%, the torque ratio becomes unity. Output torque

Pump torque, Tp

Efficiency

100% 95%

tr = 1.0

Slip = 1.0

1.0 Efficiency Slip 5% Slip 0

Nc 0

0 Np = Ni

0.95 0

Figure 9.5

Nr

1.0

Operating point

Characteristics of fluid coupling.

A feature to be noted is that the whole stretch of the x-axis of Fig. 9.5 represents the acceleration process of the turbine or the output shaft, and therefore, it represents the unsteady-state condition. A steady-state condition is reached only when the slip stabilizes at 4%–5%. A fluid coupling is designed with all its parameters ( B1p , B2p , B1t , B2t , N, D1, D2, etc.) for a particular capacity of the coupled power. This means that at other operating conditions, the efficiency of the coupling is very low. An efficiency line is also drawn in Fig. 9.5. The efficiency starts at a value of zero at the critical speed Nc of the impeller and rises to a peak value at the rated conditions (the value of peak efficiency depends on various parameters: Viscosity, density, and flow rate of the fluid, speed, radius, etc.). But after reaching the maximum value, the efficiency falls sharply if the speed ratio were to further increase and reach unity.

EXAMPLE 9.1 The output of a car engine is 80 kW at 6000 rpm. This engine drives the primary side (input shaft) of a fluid coupling. The secondary side (output shaft) of the coupling runs at 5700 rpm, at steady-state cruising. Calculate the output power and efficiency of the coupling.

Solution: In a fluid coupling, the torques at the input and output sides are equal. Torque at the input shaft is Ti 

Power ¥ 2P N ´ ¦§ µ 60 ¶

80000 ¥ 2P s 6000 ´ ¦§ µ¶ 60  127.324 N m 

Torque at the output shaft  So  127.324 N m.

Power-Transmitting Turbomachines

Power at the output shaft is 2P N oT o 60 The efficiency of the coupling 



351

2P s 5700 s 127.324  76 kW (60 s 1000)

76  95%. 80

9.4.2 Torque Converter While there is no need of stationary blades in a fluid coupling, an assembly of fixed blades is required in a torque converter between the pump element and the turbine element. This assembly, in the form of a ring, can be at the exit of the impeller and before the entry to the turbine element in the fluid flow path, as shown in Fig. 9.3(a). It can also be at the turbine exit before the fluid enters the pump inlet. This type of the torque converter is shown in Fig. 9.6. Theoretically, the fixed-blade assemblies can be at both the places.

T 1 P2

T T2

P P1

C Output end S2

Input end S1

Fixed stator

Figure 9.6

Schematic diagram of a torque converter.

Consider the device shown in Fig. 9.3(a). The velocity triangles corresponding to this torque converter are shown in Fig. 9.7. Two separate situations are illustrated: (a) One with fixed blades giving rise to torque multiplication and (b) the other with fixed blades giving rise to torque reduction. The orientation of the fixed blades is different for the two cases. Only for the sake of comparison and to show the difference, the same figure (Fig. 9.7) is used.

352

Turbomachines

Suppose that the load on the turbine is more. The turbine is required to overcome more torque. When the torque is required to be increased, the speed gets reduced. The fixed blades are so oriented that the velocity of fluid is now DF1, and the blade velocity is DE1. The area of flow is now more, and mass flow rates are also more. The fixed blades have an effect of adding the reaction torque, and the net torque on the turbine is also increased. Vf1t = Vf2p

Vf2p = Vf1p

Vr2p

C

F

B

E

U2p

E1 U1t

A

F1

E2

V2p Pump outlet

D

F2

Turbine inlet e2

Vr1p C

e1 U1p

V1p = V2t Vf1p = Vf2t Pump inlet

Figure 9.7

e

b

a

U2t

f V2t Vf2t

d

Turbine outlet

Typical velocity triangles of torque converter.

If the torque on the turbine is reduced (that is, when the device is torque divider), the process goes on with the blade speed increased to DE2, the fluid velocity being DF2, and the stator exerting torque in the opposite direction to that of the pump impeller. Instead of a single unit, it is possible to have multiple units of the impellers, turbines, and stator rings, so that the multiplication of torque can be even to five times. In a single unit, the multiplication may not be more than about 1.5–1.8. There is always a transition zone with unsteady-state operation of the device. But once this is stabilized (as in automobiles), the input–output elements can be mechanically locked, so that the entire device can be bypassed to avoid the losses in the device. The characteristics of a torque converter are shown in Fig. 9.8. This torque converter is rated for the multiplication of the input torque three times. The nature of the characteristics remains the same, even when the multiplication factor is different, such as 2 or 2.5; the scale of Sr on the right of the figure may get altered. The fluid coupling has its efficiency increasing, starting from zero at a critical speed of the impeller. But torque converters have better efficiencies in the lower speed regions. Hence, it is a practice to retain the device as a torque converter in the initial working conditions (higher efficiency, higher torque) and then shift the conditions of working as a fluid coupling to take the advantage of higher efficiencies of the coupling at higher loads. Such combined characteristics are shown in Fig. 9.9.

Power-Transmitting Turbomachines 1.0

353

3.0

0.8 h h

tr

tr

0 0.65 Nr

0

Figure 9.8

0.95

Characteristics of torque converter. 3.0 TC, tr FC, h 2.0 tr

h

FC, tr

1.0

TC, h TC 0

FC 0.95

Nr

Figure 9.9

Operation of fluid system first as TC and then as FC.

EXAMPLE 9.2 A torque converter has an input torque of magnitude 200 Nm. The stator exerts a torque equal to 80 Nm in the same direction as the input side. The speeds of the input and output shafts are 3000 rpm and 1800 rpm, respectively. Calculate the output torque and the ratio of power output to power input.

Solution: We have Torque output  Torque input  Stator torque  200  80  280 N m Ratio of output power to input power is 2P N oT o 2P N iT i



(1800 s 280)  84% (3000 s 200)

354

Turbomachines

EXAMPLE 9.3 The pumping element of a fluid coupling has a mean outlet diameter of 25 cm and a mean inlet diameter of 18 cm. The blade width is 1.5 cm at the outlet. The blades are at 90o to the blade velocity at the outlet. The inlet blade angle is 45o. The speeds of the input and output shafts are 1440 rpm and 1350 rpm, respectively. The oil in the system has a specific gravity of 0.8. The oil enters the impeller without any whirl component. The areas of flow remain constant throughout. Calculate (a) the angles of the turbine blades at the inlet and outlet, (b) mass flow rate of the oil, (c) the torque transmitted, and (d) the efficiency of the coupling. Also find (e) the absolute velocity of fluid from the pump to the turbine and its angle.

Solution: The stated data are shown in Fig. 9.10. B2p = 1.5 cm

A2p Outlet area, pump

P T

B1p = 2.08 cm 25 cm dia

A1p Inlet area, pump

18 cm dia

Figure 9.10

Fluid coupling.

Area of flow at the pump outlet is A2 p  P D2p B2p  P s 0.25 s 0.015  0.01178 m 2 Area of flow at the pump inlet is A1p  A2 p  0.01178 m 2 . We have that 0.01178  P s 0.18 s B1p So the blade width at the pump inlet is B1p 

0.01178  0.0208 m  2.08 cm P s 0.18

Blade velocity at the pump inlet is U1p 

P s D1p s N p

Blade velocity at the pump outlet is

60



P s 0.18 s 1440  13.57 m /s 60

Power-Transmitting Turbomachines

U 2p 

P s 0.25 s 1440  18.85 m /s 60

The velocity of fluid at the pump inlet is V1p  Vf1p  U1p  13.57 m /s (because B1p  45o) Mass flow rate is m  R A1pVf1p  (0.8 s 1000) s 0.01178 s 13.57  127.88 kg /s The velocity triangles of pump element and turbine element are shown in Fig. 9.11. Vf1t = 13.57 = Vf2p

13.57 Vr2p = Vf2p

Vr1t

U2p = 18.85

U1t = 17.67

V2p

V1t = V2p

Pump outlet

Turbine inlet

b1p = 45°

b2t

U1p = 13.57

U2t = 12.7

V1p = Vf1p = Vf2t = 13.57 Pump inlet

Figure 9.11

Vf2t = 13.57 Turbine outlet

Velocity triangles for Example 9.3.

Tangential force by the impeller is m (Vu1p Vu2p ) ¥¦ kg s m  N´µ  127.88(0 18.85) § s

s

¶

 2410.6 N Torque input to the pump is

(&Vu1p  U 2p )

355

356

Turbomachines

¥ 0.25 ´ T p  2410.6 s ¦  301.326 N m § 2 µ¶ Power input is 2P N pT p

Pp 

60 2P s 1440 s 301.326  Nm /s 60  45439 W

Torques are equal. So

T t  T p  301.326 Nm Power output is Pt 

2P N tT t

60 2P s 1350 s 301.326  60  42600 W

Efficiency of the coupling is Pt Pp



42600  93.75% 45439

The loss in the power (45439 42600) W, that is, 2839 W, is lost to the oil in the system, which in turn gets heated up. This heat is required to be dissipated by some cooling arrangement. Blade velocity at the turbine inlet is U1t 

P s 0.25 s 1350  17.67 m /s 60

Blade velocity at the turbine outlet is U 2t 

P s 0.18 s 1350  12.7 m /s 60

Turbine torque (output) is

T t  T p  301.326 N m  m (r1tVu1t r2 tVu2t ) Therefore

Power-Transmitting Turbomachines

357

301.326 (&Vu 2 t  0) m s r1t 301.326  127 .88 s 0.25 ´ ¥ ¦§ µ¶ 2  18.85 m /s

Vu1t 

(a) From the velocity triangles, inlet blade angle of turbine blades is 13.57 ¨ · B1t  tan 1 © ¸  85o ª (18.85 17.67 ) ¹ and, outlet blade angle of turbine blades is ¥ 13.57 ´ B2t  tan 1 ¦  46.9o § 12.7 µ¶ (b) (c) (d) (e)

The mass flow rate of fluid is 127.88 kg/s. The torque transmitted is 301.326 N-m. The efficiency of transmission is 93.75%. Absolute velocity of oil at the outlet of pump, inlet of turbine is V2p  V1t  (18.852  13.57 2 )1/2  23.23 m /s

Fluid angle at the outlet of pump, inlet of turbine is ¥ 13.57 ´ A 2p  A1t  tan 1 ¦  35.75o § 18.85 µ¶ (The detailed calculations are required for the complete design. But if only efficiency is required, then it can be obtained as Nt/Np, that is, 1350/1440, equal to 93.75%.)

EXAMPLE 9.4 The fluid coupling of Example 9.3 is required to be converted into a torque converter so that the torque is increased to 1.25 times the input torque by introducing fixed blades (stator blades) between the outlet of the pump and the inlet of the turbine. The diameters at the pump outlet and turbine inlet continue to be 25 cm as earlier. The other diameter at the inlet to pump or outlet from turbine also remains the same at 18 cm. The entry and exit of fixed blades are radial. Assume that the efficiency is now 90%. Calculate (a) the fixed-blade angles at the inlet and outlet, (b) the turbine blade angles at the inlet and outlet, and (c) the speed of the output shaft. Also determine (d) the ratio of outlet to inlet velocity of oil in the fixed blades.

Solution: The data of the pump element of Example 9.3 are not altered, except that the outlet is now in the radially outward direction instead of axial direction. Hence, the following results of Example 9.3 hold good. Torque input  Sp  301.326 N m

358

Turbomachines

Mass flow rate of oil  m  127.88 kg/s Power input  Pp  45.439 kW V 2p  23.23 m/s, A 2p  35.75o Now, the power output at the turbine shaft is Pt  Pp s Efficiency  45.439 s 0.9  40.9 kW Also Fixed-blade inlet angle @1p  Fluid angle at the pump outlet @2p Therefore

A1f  35.75o Torque at the output shaft is

T t  T p s 1.25  301. 326 s 1.25  376.66 N m At the turbine Pt 

2P N tT t

60 2P s N t s 376.66  40900  60 Therefore Nt 

40900 s 60  1036.9 rpm (2P s 376.66 )

Blade speed at the inlet to turbine is U1t 

P s 0.25 s 1036.9  13.57 m /s 60

Power at turbine is Pt  m(U1tVu1t U 2 tVu 2 t )  40900  127.88 (13.57 s Vu1t 0)

(because Vu2t  0)

Therefore Vu1t 

40900  23.56 m /s (127.88 s 13.57 )

With this, it is possible to draw the velocity triangles of the turbine (inlet and outlet), as shown in Fig. 9.12.

Power-Transmitting Turbomachines Vf1t = 13.57

53.64° b1t Vu1t = 23.56

V1t = V2f = 27.19 b2t = 54.25°

U1t 13.57

U2t =9.77 Vr2t

a1t = 29.94°

V2t = Vf2t = 13.57

Figure 9.12

Velocity triangles of turbine for Example 9.4.

Velocity of oil at the inlet to turbine is V1t  (23.562  13.57 2 )  27.19 m /s Fluid angle at the inlet is ¥ 13.57 ´ A1t  tan 1 ¦  29.94o § 23.56 µ¶ Therefore, the fixed-blade angle at the outlet is

A 2 f  A1t  29.94o Blade angle at the turbine inlet is 13.57 ¨ · B1t  tan 1 © ¸  53.64o

( 23 . 56 13 . 57 ) ª ¹ Blade velocity at the turbine outlet is U 2t 

P s 0.18 s 1036.9  9.77 m /s 60

Now Vf2t  Vf1p  13.57 Turbine blade angle at the outlet is (refer Fig. 9.12, outlet velocity triangle) ¥ 13.57 ´ B2 t  tan 1 ¦  54.25o § 9.77 µ¶ Ratio of outlet velocity to inlet velocity in the fixed blades is V2f V1f



V1t V2p



27.19  1.17 23.23

359

360

Turbomachines

Thus, we have: (a) Fixed blade angles: inlet, 35.75o; outlet, 29.94o; (b) Turbine blade angles: inlet, 53.64o; outlet, 54.25o; (c) Speed of turbine shaft: 1036.9 rpm; and (d) Ratio of outlet velocity to inlet velocity in the fixed blades: 1.17. (Note that the velocity of oil in the fixed blades is increased by 17% by area adjustments, and the inlet and outlet angles of fixed blades are 35.75o and 29.94o, respectively. The fixed blades can exert reaction torques only when the angles and velocities are different at the inlet and outlet.)

9.5

Comparison of Fluid Systems with Mechanical Systems

Fluid coupling and torque converters have come as improvements over their earlier counterparts of mechanical systems of clutch and gear boxes. The prime movers, especially the internal combustion engines, are generally the low speed–low torque and high speed–high torque devices, whereas the loads require high torques at lower speeds. The development of fluid coupling and torque converters has been with reference to automobiles. When an automobile starts from the parked condition, the inertia of the whole vehicle requires very high torques to start moving, and as the speed is picked up, lower torques are sufficient. The nature of the engine torque is just the opposite: Low torque at lower speed, high torque at higher speed. The clutch and the gear box are meant to convert the torque–speed characteristics of the internal combustion engines to suit the vehicle demand. Fluid coupling and torque converters do this job in a very smooth way without any shock loads. In the operation of mechanical system, the operator (driver of the vehicle) has to be knowledgeable and skilled to change the gears at the right speeds, or else, the efficiencies of both engine and system are likely to suffer. The operation of a fluid system is almost automatic, without the requirement of human intervention. The step-wise change of gears produces jerks to the passengers, but in fluid systems, the operation is very smooth. Even when the brakes are required to be applied (either to stop the vehicle or decrease the speed), the process is simple with fluid systems, because there is no positive drive from the engine. However, a sudden pickup of speed, as required during starting or overtaking other vehicles, may not be good in fluid systems, as in mechanical systems. All these features are summarized in Table 9.1. Table 9.1

Comparison of fluid systems with mechanical systems

Mechanical Systems 1. 2. 3. 4. 5. 6. 7.

Consist of mechanical clutches and gear boxes. More weight, more space. Speed ratios are in steps. Operation is sometimes with jerks. Skill of the operator is very important. Operator fatigue is more. Sometimes, the braking effort has to overcome engine torque also. 8. Higher reliability for sudden acceleration, during overtaking, etc. 9. Overall efficiencies are high at all speeds.

Fluid Systems 1. 2. 3. 4. 5. 6. 7.

Consist of fluid couplings and torque converters. Less weight, less space. Speed ratios are continuous. Operation is very smooth. Skill is of secondary importance. Operator fatigue is much less. Braking is very smooth, without any positive drive from the engine. 8. Sudden acceleration (during overtaking) may not be reliable. 9. The highest possible efficiency is lower than mechanical systems.

Power-Transmitting Turbomachines

361

The number of components, their manufacturing accuracies, their costs, etc. are more or less comparable for both mechanical and fluid systems. There is not much difference between the two systems in this regard.

9.5.1

Applications

Automotive transmission systems make up the largest field of applications of fluid coupling and torque converters. It may even be said that the evolution of power-transmitting turbomachines is an outcome of the research in the automotive field. In addition, the applications include the sectors of large power requirements, such as locomotives, marine, earth-moving equipment, belt conveyors, and so on. Torque multiplication is almost an integral part of the design of such equipment.

EXAMPLE 9.5 A torque converter has the following details: Pump: Speed is 1500 rpm. Inlet is axial. Pitch-line diameter or mean diameter is 30 cm. Blade width at the inlet is 2 cm. Blade inlet angle A1p is 50o. Outlet is radial. Outlet diameter is 40 cm. Blade outlet angle A2p is 80o (backward bent). Oil enters the blades without any whirl component. Turbine: Inlet is radially inward; inlet diameter is 40 cm. Inlet blade angle A1t is 75o. Outlet is axial. Pitch-line diameter or mean diameter is 30 cm. Fixed blades guide the oil from the pump outlet to the turbine inlet. The velocity of oil in the fixed blades is increased by 10% by designing the areas. The specific gravity of oil is 0.82. The flow components of the velocity of oil remain the same throughout. Calculate: (a) (b) (c) (d) (e) (f ) (g) (h) (i) ( j)

The mass flow rate of oil. Fixed-blade angle at the inlet. Fixed-blade angle at the outlet. Turbine blade outlet angle A2t. Input torque Sp. Input power Pp. Output torque St. Output power Pt. Efficiency. Torque ratio Sr.

362

Turbomachines

Solution: Pump inlet: Blade velocity is U1p 

P s 0.3 s 1500  23.56 m /s 60

Flow component is Vf1p  U1p tan B1p  23.56 s tan 50o  28.1 m /s Flow area is A1p  P s 0.3 s 0.02  0.01885 m 2 (a) We now have: mass flow rate as m  R A1pVf1p  (0.82 s 1000) s 0.01885 s 28.1  434.3417 kg /s The velocity triangle at pump inlet is shown in Fig. 9.13. b1p = 50° U1p 23.56 V1p = Vf1p = 28.1

Figure 9.13

Velocity triangle at pump inlet for Example 9.5.

Pump outlet: Flow velocity is Vf2p  Vf1p  28.1 m /s Blade velocity is U 2p 

P s 0.4 s 1500  31.416 m /s 60

Whirl component of velocity is Vu2p  U 2p

Vf2p tan B2p

28.1 tan 80o  26.46 m /s

 31.416

Power-Transmitting Turbomachines

The velocity triangle at pump outlet is shown in Fig. 9.14. Fluid angle is ¥ 28.1 ´ A 2p  tan 1 ¦  46.72o § 26.46 µ¶ Fluid velocity is V2p  (26.462  28.12 )1/2  38.6 m /s Vf2p = 28.1 Vr2p 80°

U2p Vu2p

Figure 9.14

V2p

Velocity triangle at pump outlet for Example 9.5.

Fixed blades (stator blades): Inlet fluid velocity is V1f  V2p  38.6 m /s Outlet fluid velocity is V2f  V2p s 1.1  38.6 s 1.1  42.46 m /s (b) Fixed blade angle at inlet is

B1f  A 2p  46.72o The velocity triangle at inlet to turbine is shown in Fig. 9.15. Vf1t = 28.1 75° Vu1t = 31.83

V1t U1t 24.3

Figure 9.15

a1t

Velocity triangle at turbine inlet for Example 9.5.

Turbine inlet: Fluid inlet velocity is V1t  V2f  42.46 m /s

363

364

Turbomachines

Flow velocity is Vf1t  Vf2p  28.1 m /s Fluid angle is ¥V ´ ¥ 28.1 ´ A1t  sin 1 ¦ f1t µ  sin 1 ¦  41.44o § 42.46 µ¶ § V1t ¶ (c) Therefore, fixed blade angle at outlet  @1t  41.44o. Whirl component is Vu1t  V1t cos 41.44o  31.83 m /s Blade velocity is U1t  Vu1t

Vf1t tan B1t

 31.83

28.1  24.3 m /s tan 75o

Speed of the turbine is Nt 

U1t s 60

P s D1t



24.3 s 60  1160 rpm P s 0.4 b2t = 57° U2t = 18.22

V2t = 28.1

Figure 9.16

Velocity triangle at turbine outlet for Example 9.5.

Turbine outlet: Blade velocity is U 2t 

P s 0.3 s 1160  18.22 m /s 60

Flow velocity is Vf2t  Vf1t  28.1 m /s It is possible to draw velocity triangle at turbine outlet, as shown in Fig. 9.16. (d) Turbine outlet blade angle is ¥ 28.1 ´ B2t  tan 1 ¦  57o § 18.22 µ¶ (e) Input torque is  2pVu2p T p  mr  434.3417 s 0.2 s 26.46  2298.5 N m

Power-Transmitting Turbomachines

365

(f ) Input power is Pp  2P N pT p  2 s P s 1500 s 2298.5  361.05 kW (g) Output torque is  1tVu1t T t  mr  434.3417 s 0.2 s 31.83  2765 N m (h) Output power is Pt  2P N tT t  2 s P s 1160 s 2765  335.88 kW (i) Efficiency is

H

Pt Pp



335.88  93% 361.05

(j) Torque ratio is

Tt 2765   1.2 T p 2298.5

S ummary 1. The basic features of power-transmitting turbomachines are studied in detail. 2. The components are: input shaft, impellerblades, turbine blades and output shaft in a fluid coupling. In a torque converter, the stator blades are designed to increase or decrease the torque. 3. The fluid is non-corrosive, non-foaming, nontoxic, chemically stable oil, with high density, low viscosity and lubricating properties. 4. The working principles and the constructional details are studied, including some quantitative details.

5. Energy-flow (or power-flow) is from input shaft to impeller, to fluid, to turbine blades, to output shaft. 6. Fluid coupling: torque remains same, speed is reduced. 7. Torque converter: torque can be increased or decreased (respectively with decrease or increase of speed). 8. The characteristics of the machines are studied, so as to know how the component performances are matched. 9. There is a critical speed of impeller, at which the turbine just starts running.

366

Turbomachines

10. At lower speeds, the device is designed to run as a torque converter; then near design speeds, to run as a coupling, in a combined mode of operation. 11. The design of the components (blade angles, radii etc) is for full load conditions; at other speeds, the performance is transient or unsteady in nature.

12. A brief comparison of power-transmitting devices with the mechanical systems is made. 13. Except in the reliability of sudden acceleration, the fluid systems score over the mechanical systems with respect to all other factors.

M ultiple- C hoice Q uestions 1. The energy flow in a torque converter is (a) input shaft–fluid–impeller–turbine–output shaft (b) input shaft–impeller–turbine–output shaft (c) input shaft–turbine–impeller–output shaft (d) input shaft–impeller–fluid–turbine–output shaft 2. The most suitable condition for fluid coupling is (a) Nr  1 (b) Sr  1 (c) Sr  1 (d) Nr  1 3. In a power-transmitting turbomachine, the entry and exit of fluid in the pump element are at (a) equal radii (b) smaller radius to larger radius (c) larger radius to smaller radius (d) none of these 4. In a power-transmitting turbomachine, the entry and exit of fluid in the turbine element are at (a) equal radii (b) smaller radius to larger radius (c) larger radius to smaller radius (d) none of these 5. The stalling of a torque converter is due to (a) the input torque less than the fixed-blade torque (b) the negative input torque (c) the input torque more than the fixed-blade torque

6.

7.

8.

9.

10.

(d) the input torque less than the friction torque In a torque multiplier, (a) the fixed-blade torque is zero (b) the fixed-blade torque is in the same direction as the input torque (c) the fixed-blade torque is in the opposite direction of the input torque (d) the fixed-blade torque is more than the input torque Choose the wrong statement: A fluid must be (a) chemically stable (b) of high density (c) of high viscosity (d) non-toxic Choose the wrong statement: The transmitted torque is (a) proportional to the radius of impeller at the outlet (b) proportional to the square of radius of impeller at the outlet (c) proportional to the speed of the impeller. (d) proportional to the mass rate of flow of fluid In a torque converter, as the input speed varies from zero to rated value, (a) the slip starts from zero and reaches 95% (b) the slip starts from 1.0 and reaches 0 (c) the slip starts from 1.0 and reaches 5% (d) the slip starts from 1.0 and remains constant In a fluid coupling, as Nr varies from zero to 0.95,

Power-Transmitting Turbomachines

(a) the efficiency starts from zero and increases continuously (b) the efficiency starts from zero, increases, and then decreases

367

(c) the efficiency starts from 1 and then decreases continuously (d) none of these

R eview Q uestions 1. Define a power-transmitting turbomachine. (Refer Section 9.2) 2. Explain the working principle of a powertransmitting turbomachine. (Refer Section 9.2) 3. Enlist the desirable properties of a fluid medium in power-transmitting turbomachines. (Refer Section 9.2) 4. Explain the working of a power-transmitting turbomachine. (Refer Section 9.2; use any one figure out of Figs. 9.3(a), 9.3(b), or 9.6, use corresponding explanation) 5. Draw the characteristics of a fluid coupling and explain. (Refer Fig. 9.5 with related explanation)

6. Explain how the fixed blades give rise to increase (or decrease) the torque at the output shaft. (Refer Section 9.4; Start from the internal balance of torques) 7. Discuss the characteristics of a torque converter. (Refer Fig. 9.8, with related explanation) 8. Explain how the characteristics of a fluid system (fluid coupling and torque converter) can be integrated for higher efficiencies. (Refer Fig. 9.9 with related explanations) 9. Compare fluid transmission systems with mechanical transmission systems. (Refer Section 9.5) 10. State the applications of torque converters. (Refer Section 9.6)

E xercises 1. A fluid coupling is required to be designed to drive a conveyor belt. The input shaft is driven at a speed of 600 rpm by a motor. Speed of the output shaft is required to be 510 rpm. The power output of the motor is 48 kW. For the coupling, calculate (a) input torque, (b) output power, and (c) efficiency. 2. It is proposed to fabricate a small fluid coupling (which is like the one shown in Fig. 9.10). The pumping element has a mean inlet diameter of 7.5 cm, a mean outlet diameter of 11 cm, and the blades have inlet and outlet angles of 60o and 75o (backward bent), respectively. The speeds of the input and output shafts are to be 1000 and 900 rpm, respectively. The blade width at the inlet is 1.0 cm. The velocity of flow of fluid remains constant at inlet and outlet. The

oil in the coupling has a specific gravity of 0.85. The fluid velocity at the inlet to the pump is at 75oto the plane of rotation of impeller. Draw the velocity triangles at the inlet and outlet of the pump impeller and at the inlet and outlet of the turbine rotor. Calculate (a) Turbine blade angles at inlet and outlet, (b) mass flow rate, (c) torque transmitted, (d) power at inlet, (e) power at outlet, and (f ) efficiency. 3. The inlet blade angle of the pump element of a torque converter is 40o, the outlet blade angle is 50o, and the blade is bent backward. The mean radius at inlet is 12 cm and that at outlet is 16 cm. The blade width at inlet is 2 cm. The pump speed is 3000 rpm. The flow component of fluid velocity at inlet is 25 m/s that remains constant throughout the system. The stator

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blades alter the flow path of fluid by an angle of 15o so as to accelerate the fluid. The fluid is oil with a specific gravity of 0.85. The efficiency of the device is 85%. Draw the velocity triangles. Determine (a) mass flow rate of fluid, (b) torque at input shaft, (c) input power, (d) output power, (e) speed of the output shaft, (f ) output torque, and (g) ratio of output torque to input torque. 4. The pump of a torque converter has its inlet blade angle of 40o, outlet blade angle of 65o bent backward. The mean radius at inlet is 10 cm, that at outlet is 15 cm, and the blade width at inlet is 2 cm. The pump speed is 3000 rpm and the flow component of fluid velocity is 20 m/s. The efficiency of the device is 85%. Fluid is oil with a specific gravity of 0.85. The stator blades alter the flow path of fluid by an

angle of 10o so as to decelerate the flow. Draw the velocity triangles. Determine: (a) mass flow rate, (b) input torque, (c) input power, (d) output power, (e) output speed, (f ) output torque, and (g) torque ratio. 5. The input power to a torque converter is 30 kW, at a speed of 4500 rpm. Let the flow of fluid from the turbine outlet to the pump inlet be “axial, without whirl component”. Design the torque converter so that the output torque is twice the input torque. (Hint: The data and solutions of the examples/exercise problems can be tabulated for an overall assessment/evaluation, so that the inlet and outlet diameters of the pumping unit can be assumed as a starting point for this problem. All other dimensions and angles can then be calculated.)

S olutions A vailable

P roject- O riented Q uestions 1. The input power to a torque converter is 30 kW, at a speed of 4500 rpm. Let the flow of fluid from the turbine outlet to the pump inlet be “axial, without whirl component.” Design the torque converter so that the output torque is twice the input torque (Hint: The data and

solutions of the examples/exercise problems can be tabulated for an overall assessment/ evaluation, so that the inlet and outlet diameters of the pumping unit can be assumed as a starting point for this problem. All other dimensions and angles can then be calculated).

A nswers Multiple-Choice Questions 1. 2. 3. 4. 5.

(d) (c) (b) (c) (d)

6. 7. 8. 9. 10.

(b) (c) (a) (c) (a)

Exercises 1. 764 N-m; 40.8 kW; 85% 2. 81.85o,63.81o; 9.3 kg/s; 0.9378 N-m; 98.2 W; 88.38 W; 90% 3. 320.45 kg/s; 1198 N-m; 376.36 kW; 319.9 kW; 1703 rpm; 1793.8 N-m; 1.497

4. 213.63 kg/s; 1049.32 N-m; 329.654 kW; 280.2 kW; 4044 rpm; 661.6 N-m; 0.63

Glossary [The Glossary terms listed below are intended to provide a second revision (after Summary in each chapter) for the students. For their university examinations, students should go through the exhaustive list.] Absolute velocity: It is the velocity of the fluid and is considered independent of blade movement. The momentums of fluid and the components of fluid velocity are all calculated with respect to the absolute velocity of the fluid. It may be considered that the inlet absolute fluid velocity is just before the actual entry of fluid into the rotor and the absolute outlet fluid velocity is just after the fluid exits the rotor. Blade angles: These are the angles between the blade velocity vector and the tangent to the mean line of the blade. Blade angles are identified at inlet (A1) and outlet (A2). Blade velocity: It is tangential to the circular path of the blade, as the blade rotates about the axis of the shaft. Any point on the circular path can be chosen for the tangent, but a convenient point is that which has a horizontal tangent. For the purpose of analysis, blade velocities are required at the inlet to the blade (entry of fluid) and outlet from the blade (exit of fluid). In an axial flow machine, at both inlet and outlet, the blade velocities are equal. Blades or vanes: These are the parts that interact with the fluid. Blades are generally of airfoil shapes. Their design is guided by the requirement of area changes of passages between them. In impulse machines, the flow area is required to be constant and therefore blade shapes are not like airfoils. Blowers: These are the machines that create a pressure ratio of the order of 1.5–2.0. Volume flow rates of blowers are higher than fans. Blowers also supply air for ventilation of large areas, air for combustion chambers, for fluidizing reaction beds, and so on. Camber: It is the mean line of the blade. Cavitation: In brief, it is the formation of cavities on a surface due to a sequence of events. These events are as follows: 1. Reduction of pressure in a body of water, adjacent to a surface, at the discharge end of a reaction turbine, or at the suction end of a centrifugal pump. Both occurrences are on the surface of rotor blades because of the installation of the machine above tailrace (in case of turbine) or sump (in case of pump). 2. The reduction of pressure is to the level of “below the vapor pressure” of the water. 3. The water vaporizes and forms bubbles of vapor on the surface of rotor blades. 4. Due to eddies, water at slightly higher pressure, from other regions, comes in contact with the vapor bubbles. 5. The bubbles collapse due to pressure impressed and condensation of vapor. 6. Water from all around rushes into the space of vapor. 7. The force is so much that it can chip off the metal particles from the solid surface, causing cavities.

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Cavitation can be prevented by not allowing the pressure to go below the level of vapor pressure of the water. This is by installing the machines as near to the tailrace or sump as possible. High-strength materials can also be used to resist the cavitation damage. Choking: It is a sharp fall in the pressure created in a compressor due to increased flow rate and consequent increased losses. The flow passages get practically choked and the flow rate almost drops to zero in such a case. Compounding of steam turbines: It is the splitting of the process of transfer of energy, from steam to the rotor, between multiple rows of rotor blades. The purpose is to avoid the sources of inefficiency of single expansion, which are as follows: 1. 2. 3. 4.

Convergent–divergent nozzles, employed for single expansion are less efficient. Process of energy transfer is less efficient due to exit losses. Frictional loss in single expansion is more. Very high speed of rotor, associated with single expansion, causes more losses and is less efficient; materials of blades also have to be of superior strength. 5. High speeds of single expansion require speed reduction (with losses) even at utility end. Pressure compounding and velocity compounding are two separate methods of compounding. Compressors: These are the machines that generate pressure ratio of the order of 2.5 and more. Multistage compressors can have a pressure ratio as high as 10. Axial flow compressors are multistage units with higher efficiencies. Degree of reaction: It is the ratio of energy transferred due to the change of pressure in the rotor to the total energy transferred in the rotor. The total energy transferred has two parts: one part is due to the change in pressure, which is the reaction component, and the other part is due to the change in velocity, which is the kinetic component. DeLaval turbine: It is an impulse-type steam turbine where the expansion of steam takes place in a single row of nozzles and the resulting kinetic energy of steam is absorbed in a single row of rotor blades. Since it is not possible for a single row of rotor blades to absorb the kinetic energy to a reasonable extent, the utilization factor of DeLaval turbine is low, compared to other types of turbines. Delivery head of a centrifugal pump: It is the vertical distance between the centerline of the pump and the end of the delivery pipe, plus the frictional head in the delivery pipe and losses in the delivery valve. Delivery valve in a centrifugal pump setup: It is a control valve to regulate the flow. It is also used to isolate the pump, that is, to disconnect the pump for maintenance or repair. Diffuser blades: These are fixed to the stator in a compressor and convert the kinetic energy of fluid into pressure energy. The flow area between the blades increases, thereby reducing the velocity of air (see also diffuser). Diffuser: It is a flow passage in which the fluid decelerates. Gain in pressure is due to the reduction in kinetic energy. Flow passage between blades can act like diffusers, as in the compressor. Draft tube: It is a flow passage employed for the purpose of recovering some head, which otherwise would be lost due to installation of a turbine above the tailrace. Draft tube is fitted at the discharge end of the turbine and the other end is invariably submerged under the tailrace level. Draft tube is a divergent tube with half-cone angle of about 4o–5o, which decreases the exit velocity and also saves some exit loss. Different types of draft tubes are Moody’s, Elbow type, Split exit type, etc.

Glossary

371

Drag: It is the force experienced by airfoil in the direction of the fluid flow. Dynamic action: It is the interaction between the fluid and the blades of rotor, when the fluid is in motion and is not confined in a space. The fluid gives the energy (or gains the energy) as it moves. Dynamic head: It is the head that is equivalent to the velocity of the flow. Energy flow diagram: It is a diagrammatic representation of the flow of energy in a turbomachine, including the components of energy and losses. The rate process of energy is the power flow; hence, energy flow diagram can also be considered as power flow diagram. Euler head: It is the theoretical head, as calculated by the Euler equation, depending on the geometrical and operating features of the machine or blades. Euler head is calculated with one main assumption that fluid flow in the impeller is “Vane congruent.” Fans: These are the axial flow machines to impart only velocity (kinetic energy) to the air. The pressure created is only a few centimeters of water, that is, the pressure ratio is hardly 1.05–1.07, and so on. The purpose of fans is only circulation of air or exhaust of used air. Usually, fans do not have a casing at all. Flow coefficient: It is proportionality constant and is used to correlate the flow of fluid to the operating and geometrical parameters of a machine. (O1  Q/ND3.) Fluid coupling: It is one type of power-transmitting turbomachine in which the input and output torques are equal, with output speed less than the input speed. This is a substitute for mechanical clutch. Hence, fluid coupling can be used to couple or decouple the shafts, just like the clutch. However, the operation of fluid coupling is much smoother than the clutch. Fluid parameters: These are pressure, head, enthalpy, fluid angle, etc. Foot valve: It is a non-return valve in a centrifugal pump installation, fixed at the inlet end of the suction pipe. Its function is to hold the water in the body of the pump, including the suction pipe, during the process of priming and after the pump is shut down. A second function of foot valve is to prevent the entry of debris or solids into the suction pipe. Francis turbine: It is a reaction-type hydraulic turbine suitable for medium heads (50 250 m) and medium specific speeds (60–300). The main components of Francis turbine are the volute casing, guide vanes, rotor, and draft tube. Speed control is by regulating the flow, which is done by guide-vane control. The draft tube creates a sub-atmospheric pressure at the discharge side of the rotor, thereby increasing the effective head acting on the turbine. Friction head: It is the loss of head due to friction when the fluid flows in a pipe. In a turbine installation, friction head of penstock pipes is deducted from the available gross head. In a pump installation, friction head is added to the geometrical head (to get the total head, which is to be overcome by the pump). Governing of turbines: It is a process by which the speed of the turbine is maintained constant for the purpose of generating electrical power (in the coupled generator) at constant frequency. Governing is done by adjusting the input to a turbine (flow rate of water) to suit the output. In Pelton turbine, this is done by working on the spear in the nozzle, and in Francis or Kaplan turbines, this is done by adjusting the guide vanes. In steam turbines, the governing process is either by nozzle control or by throttle valve control. Guide vanes: These are the blades that direct the flow of fluid in a designed direction. Guide vanes are fixed to the stator. Guide vanes can also be anchored to the stator so that their orientation can be altered as required, as in Francis or Kaplan turbines, where generating velocity is also a part of their function. In compressors, the inlet guide vanes impart a prewhirl to the inlet fluid stream.

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Head coefficient: This is the proportionality constant and is used to correlate the head of the fluid to the operating and geometrical parameters of a machine (O2  H/N 2D2). Head of water: It is the height of the water available for a turbine or height to which water is required to be pumped by a pump. Impeller: It is the rotor of a work-absorbing turbomachine, such as pump or compressor. Impeller has blades on it and it also transfers energy from shaft to the fluid. Impulse: It is action solely due to the velocity. In impulse turbine, the rotor gets energy from fluid due to the change in velocity only. There is no change in the pressure of the fluid, as it flows through the blade passages of the rotor. DeLaval steam turbine and Pelton turbine are examples of impulse turbines. Inducer section: It is that portion of the blade of a centrifugal blower which is slightly curved so as to prevent entry shock to air at the inlet. While the entire blade is radial, a small portion is given a curvature so that the air enters the impeller smoothly. Infinitesimal stage or small-stage efficiency or polytropic efficiency: This efficiency is found out by assuming that there is a stage in which the increase (or decrease) of pressure is very small ( dp). This hypothetical stage has an expression for efficiency, which is independent of the pressure ratio, initial/final temperature, and such other parameters; the efficiency is totally in terms of index of compression (expansion) and ratio of specific heats. Hence, this efficiency characterizes the process of compression (or expansion). Kaplan turbine: It is a reaction-type hydraulic turbine, suitable for low heads (5–75 m) or high specific speeds (300 and above). The main components of Kaplan turbine are the volute casing, guide vanes, rotor, and draft tube. One main feature of Kaplan turbine is that its rotor blades can swing, about their own axes (duly acted upon by governing mechanism) even when the turbine is running. The turbine, with blades that are integral with hub, is known as propeller turbine. Speed control of Kaplan turbine is by regulating the flow, by guide vane control along with rotor blade control. Lift: It is the force experienced by airfoil in a direction perpendicular to the direction of fluid flow. Loading coefficient or pressure coefficient: It is the ratio of isentropic work at a given tip speed to Euler work at the same tip speed in a compressor. Machine parameters: They are the physical dimensions and ratios between such dimensions, angles of blades, etc. (e.g., D1, D2, A1, A2, d, B1, B2, {). Manometric head of a centrifugal pump: It is the head measured across the inlet and outlet flanges of the pump. Manometric head represents the output head of the pump. Maximum suction lift (MSL): It is the maximum possible height of installation of a centrifugal pump, above the sump level, so that the cavitation is just prevented, or the pressure at the suction end of impeller is just the vapor pressure of water. This depends on the local ambient pressure, temperature, suction pipe losses, and velocity in the suction pipe. Model studies: This is a method of evaluating the performance of a designed prototype machine, by first having a scaled-down model, then conducting tests on the model, and finally extrapolating the performance of the model to that of prototype machine. Multijet Pelton turbine: When the flow rate available is more, giving rise to specific speeds in the range of 30–70, multiple jets on the same Pelton wheel can be arranged, so as to share the flow between them. Thus, the cost of multiple units of Pelton wheels (and generators) can be saved by employing multiple jets, on the same wheel.

Glossary

373

Multistage pumps: They are pumps-in-series or impellers-in-series. The same flow rate passes through all the impellers. All the impellers are keyed onto same shaft. The heads created in each stage add up to the total head of the multistage pump. These are sometimes known as the “deep-well pumps.” Net positive suction head (NPSH): It is the difference between the maximum possible suction height (MSL) and the actual height of installation of the pump above the sump level (hs). (Thus, NPSH  MSL hs). Higher the value of NPSH, lower is the height of installation of pump and safer is the pump from cavitation troubles. Nozzle: It is a flow passage in which the fluid accelerates. The gain in the kinetic energy of fluid is because of conversion of pressure energy (of fluids) or enthalpy (of gases or steam). Convergent nozzles generate subsonic velocities. Convergent–divergent nozzles generate supersonic velocities. Nozzles are a part of stator in the impulse machines. Flow passages between blades also act as nozzles. Operating parameters: Operating parameters of a turbomachine are the rotor speed, velocities of blades, changes in pressures, head generated, flow rates, losses, efficiencies, and so on. Parallel operation of pumps: Parallel operation of pumps is done to meet the requirement of higher flow rates. The delivery pressures or heads must be same for all the pumps; and the flow rates are additive. Any time, for any maintenance work, a pump can be isolated and removed with the flow rate of that pump reduced in the total flow; other pumps can continue to work. Parsons turbine: It is a reaction-type steam turbine, where the degree of reaction is 0.5. Steam first expands in stator blade and, then with the resultant velocity, enters the rotor blades. In the rotor blades, the steam again expands. But the velocity so generated is absorbed by the rotor blades, along with the velocity at the inlet. Therefore, the velocity of steam at the exit of rotor blades is equal to the velocity at the entry to the stator blades. The two rows (stator  rotor) together form a stage. Depending on the total pressure drop available between supply pressure and exit pressure (boiler to condenser), there can be many stages of Parsons turbine in succession. Pelton turbine: It is an impulse-type hydraulic turbine suitable for high heads (150 m and above) and low specific speeds (5–35). The main components are the nozzle (to generate a high speed jet) and a rotor with Pelton double cups. A spear in the nozzle is used to regulate the flow. A deflector helps in diverting the jet in opposite direction, when the turbine is required to be stopped. Positive displacement machines (PDMs): They are those in which the fluid is confined in spaces such as cylinder–piston assemblies. The process of compression or expansion is due to the movement of boundary (i.e., piston). Power coefficient: It is the proportionality constant used to correlate the power of the fluid to the operating and geometrical parameters of a machine. (O3  P/N 3D5). Power flow diagram: See energy flow diagram. Power input factor or workdone factor: It is the ratio of actual work supplied to the theoretical work supplied in a compressor. Power-transmitting turbomachine: It is a device, which transfers power from an input shaft to an output shaft through a fluid medium, without any mechanical coupling. It can be a fluid coupling with output torque equal to input torque, but at slightly reduced speed. It can also be a torque converter, either increasing the torque (at reduced speed) or decreasing the torque (at increased speed). However, in all cases, there are losses involved; and the output power is always less than the input power.

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Preheat in multistage compressors: When the fluid is compressed in multiple stages, the entry of fluid to a subsequent stage from the previous stage is at a slightly higher temperature, due to the actual compression process. The isentropic compression in the subsequent process requires more work input, due to the divergent constant pressure lines. Hence, when there are two or more stages, the sum of isentropic works (for each stage) is more than the isentropic work of compression of single stage (through the same initial and final pressure). This increased input is said to be due to preheat effect. Pressure coefficient or loading coefficient: It is the ratio of isentropic work at a given tip speed to Euler work at the same tip speed in a compressor. Pre-whirl: It is a designed whirl component imparted to the fluid at the inlet to a compressor, in order to restrict the relative velocity to subsonic values, to avoid occurrence of shocks. Priming of pumps: It is the process of filling up of impeller casing and suction pipe of a pump setup, by the liquid to be pumped, before starting the pump. If the pump is not primed, the impeller is not able to create the suction; and the pump runs dry. When the foot valve is leakproof, it is sufficient to prime the pump once, after the installation; and it is not necessary to prime it again and again. If the pump is installed at a level lower than the sump level, the priming is automatic; and there is no need of any special steps to prime the pump. Rateau turbine: It is an impulse-type steam turbine. Expansion of steam is only in stator blades, with the passage between the stator blades acting as nozzles. There is no expansion of steam in the rotor blades. The difference between DeLaval and Rateau turbines is that the pressure ratio (of expansion) in the Rateau turbine is far less than in the DeLaval turbine; as a result, the velocity generated is less, losses are also less. The total expansion of steam (boiler pressure to condenser pressure) is shared in multiple stages of Rateau turbines. Reaction: It is an action, solely due to the expansion of fluid in rotor passages. The velocity is generated in the blade passages of rotor and is simultaneously absorbed by the rotor. Pressure of fluid drops as the fluid flows through the reaction blades. The reaction blades are so shaped that the passages between them act like nozzles. Parsons turbine is an example of reaction turbine. Reheat and reheat factor (RF) in multistage turbines: When the fluid is expanded in multiple stages, the entry of fluid to a subsequent stage from the previous stage is at a slightly higher temperature due to actual expansion process, that is, due to irreversibilities, eddies, turbulence, and such other dissipative processes. However, due to the higher temperature at the entry (to the subsequent stage), the fluid is able to give out more expansion work. When there are two or more stages of expansion, the sum of isentropic works (from each stage) is more than the isentropic work of a single-stage expansion (through the same initial and final pressure limits). This increased output is said to be due to reheat effect. The reheat effect is quantified by a “reheat factor”, defined as the ratio of sum of the isentropic works to the single-stage work. RF is about 1.04, 1.05. Factors affecting RF are no. of stages, initial state, and stage efficiency. Relative velocity: Relative velocity of a fluid with respect to the blade is tangential to the blade profile. For the purpose of analysis, the relative velocities are required at the inlet and outlet of the rotor. Hence, the relative velocities at the inlet and outlet are tangential to the blade profile at the inlet and outlet, respectively. In ideal case, that is, vane congruent flow, the relative velocity, is tangential to the stream lines of fluid in the rotor. Runner: It is the rotor of a work-producing turbomachine such as turbine. Runner has blades on it. Runner transfers energy from fluid to shaft.

Glossary

375

Similarity: This is like proportionality: 1. Geometric similarity between two machines is that the dimensions of the two machines are proportional. 2. Kinematic similarity between two machines is that the velocities in the machines are proportional. 3. Dynamic similarity between two machines is that the forces in the machines are proportional. Similarities are the prerequisites for model testing. “Similitude” comprises all the three similarities. Similitude: See similarity. Slip: It is the deviation of actual flow from the theoretical flow, that is, vane congruent flow. This deviation is accounted for by a slip factor or the coefficient of slip. Slip is due to factors such as friction, turbulence, inertia, finite thickness of blades, and so on. Specific work: It is the work interaction between the fluid and the rotor blades, calculated for 1 kg of fluid, as it flows from the inlet of the rotor to the outlet. (Units are J/kg, m-N/kg, m2/s2 – all having the same magnitude.) Specific work is positive, if the energy flow is from the fluid to the blades. Speed ratio: It is the ratio of the blade velocity at the inlet to the fluid velocity at the inlet. In impulse machines, speed ratio is of the order of 0.45–0.49. In reaction machines, speed ratios may be from 0.8 to 0.98. Speed ratio is some sort of synthesis of fluid parameters (head, velocity), machine parameters (diameter of rotor), and operating parameter (speed). Hence it represents a design factor. Stagnation state: Stagnation state of a moving fluid is the state reached by the fluid, when it is brought to rest abruptly, adiabatically, without any energy transfer to or from the fluid. The initial kinetic energy of the fluid gets converted into pressure or temperature (and hence enthalpy) of fluid. Stalling of the blades: This means the loss of performance of the blades due to increased drag and sudden fall of lift. Even when the compressor runs, during stall, no pressure builds up. Static state: It is the state of fluid characterized by static properties, that is, properties measured by any static probes. Suction head of a centrifugal pump: It is the vertical distance between the sump level and centerline of the pump, plus the frictional head in the suction pipe and losses in the foot valve. Surging: It is a phenomenon, which occurs in compressors and in which the fluid flow fluctuates repeatedly, between zero and peak values in quick succession, due to the positive slope of the pressure flow rate characteristic of the compressor. When the load line of the system and the characteristic of the compressor cross at design point, if a small disturbance has a tendency to grow, then the surging takes place. If the small disturbance at the design point gets balanced, then the working becomes stable and smooth, without any surging. Compressors are designed such that their pressure–capacity–flow rate characteristics always have a negative slope over a stretch around the design point. Torque converter: It is one type of power-transmitting turbomachine, in which the output torque is either more or less than the input torque; the two types are known as torque multiplier or torque divider, respectively. The torque multiplier is like a gear box of an automobile. The function of the gear box is to increase the torque in steps; but in torque multiplier, the increase is continuous, in a smooth jerk-less operation. Turning losses: These are the losses which occur in the turbomachines when the fluid does not smoothly glide over the blade surface during the transit of fluid through the rotor. At the design point of performance, the fluid has a smooth transit, as characterized by the velocity triangle. At other points, the fluid either

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“impinges” on the blades or “separates” from the blades, because the three velocities (of the triangle) do not form the triangle. The more the point of operation shifts from the design point, more is this form of loss. Utilization factor: Utilization factor of a turbine is the ratio of actual specific work to the specific work inclusive of exit losses:   W/(W  V22 /2). The exit velocity and exit losses are a necessary part of the working of a turbine, which cannot be avoided. Well-designed turbines can have utilization factors as high as 0.85. Vane congruent flow: It is a theoretical fluid flow, in which the stream lines are identical with the shape of the blades (without the effects of eddies, turbulence, friction, boundary layer, etc.). Velocity triangle: This represents a perfect balance between the blade velocity, absolute fluid velocity, and the relative fluid velocity (with respect to blade). During such a balance, the fluid smoothly glides over the surface of the blade without any shock. The analysis is with respect to such a perfect balance and therefore with respect to the velocity triangles. When the balance is disturbed, the losses take place, as in the case of off-design performance. Work done factor or power input factor: It is the ratio of actual work supplied to the theoretical work supplied in a compressor. Zero-work impeller: It is a particular combination of diameter ratio of rotor, angles of blades, and the operating speed, when there is no energy transfer between the fluid and the blades, even when the rotor runs. The same impeller, when run at other speeds, can have the energy transfer (either positive or negative).

Bibliography 1. D. G. Shepherd (1956). Principles of Turbomachinery. New York: Macmillan. 2. V. Kadambi and M. Prasad (1977). An introduction to energy conversion. In Turbomachinery, Vol. III. New York: Wiley Eastern Ltd. 3. S. L. Dixon and C. A. Hall (2010). Fluid Mechanics and Thermodynamics of Turbomachinery, 6th Edn. New Delhi: Elsevier. 4. A. J. Stepanoff (1957). Centrifugal and Axial Flow Pumps. New York: Wiley. 5. J. Lal (1992). Hydraulic Machines. New Delhi: Metropolitan Book Co. Pvt. Ltd. 6. G. Gopalakrishnan and D. Prithviraj (2008). A Treatise on Turbomachines. Chennai, India: SciTek Publications (India) Pvt. Ltd. 7. D. B. Spalding and E. H. Cole (1958). Engineering Thermodynamics. London: ELBS Edition. 8. M. Pauken (2011). Thermodynamics for Dummies. Hoboken, NJ: Wiley Publishing, Inc.

Index A absolute velocity, 85, 87 actual compression process, 50–51 actual shaft work, 58 actual work of compression, 53 adiabatic efficiency, 50 alternate form of Euler turbine equation, 99–102 angular momentum equation, 98 angular velocity, 98 axial components of fluid velocities, 99 axial flow, 5, 35–36, 106 axial flow compressors (AFC), 151 definition of, 323 stage work, stage efficiency, and performance, 325–326 velocity triangles, 324–325 axial flow machine, 88, 106 absolute velocity of fluid, 145 blade angles, effect of, 150–151 blade shapes, effect of, 142 fan laws, 155–156 losses in blade passages, 155 reaction in, 143–147 runner blades, flow analysis, 154–155 velocity triangles, 141–142, 145 axial flow type rotor, 38, 87

B backward bent vanes, 139 Bernoulli’s equation, 101 blade angles, 88 blade inlet angle, 127 blade outlet angle, effect of energy transfer, 125–128 performance, 135 reaction, 129–130 blade passage efficiency, 197 blade velocity, 86

C carryover efficiency, 179, 196 cavitation, 251–252, 294 centrifugal compressors, 307–310 centrifugal energy component, 101

centrifugal pumps, 5–6 advantages of, 274 analysis of, 281–282 classification of, 276–278 construction of, 274–276 definition of, 273 different heads of, 278–279 efficiencies of, 279–281 hydraulic and manometric efficiencies, 280 maximum suction lift and net positive suction head, 291–293 mechanical efficiency, 279–280 minimum starting speed of, 290–291 overall efficiency, 281 parts of, 275–276 volumetric efficiency, 280 working of, 274–276 choking on compressors, 322–323, 325–326 coefficient of slip, 136–137, 154–155 coefficients of lift and drag, 336–337 compounding of steam turbine, 169–172 advantages of, 169 kinetic energy utilization, 169–170 pressure-compounded reaction turbine, 170–171 pressure-velocity compounding, 172 velocity compounding, 171 compressible fluids, 47 in multi-stage compressor, 61 compression process, 73–75 actual, 50–51 actual temperature rise, 73–74 actual work of compression, 53 ideal, 50 isentropic, 50–53 isentropic compression efficiency, 75 isentropic stage efficiency, 73 matching of compressor and system characteristics, 321–323 of perfect gas, 61 confined spaces, 3 constant pressure, 60 O -constants, 32–33 Curtis, Charles, 171

Curtis stage, 184–190 absolute velocity of steam, 185 axial components of absolute steam velocities, 185 losses in, 188–190 rotor blade angles, 185 specific work of rotor blade ring, 188 velocity triangles, 184, 186–187 Curtis turbine, 171

D degree of reaction, 103, 129–130 de Laval, Gustafe, 164 De Laval turbine, 164–167 blade efficiency, 165–167 Curtis stage, 184–190 expression for specific work, 166 kinetic energy, 164 losses due to fluid friction and turbulence in, 167 maximum rotor efficiency, 165 nozzle efficiency, 167 Parsons stages, 178–181 Rateau stages, 172–175 rotor blade angles, 165 rotor efficiency, 166 stage efficiency, 167–168 vane efficiency, 165 velocity triangles, 167 design point, 140 diffuser efficiency, 312–314 diffusers, 5 dimensional analysis of turbomachine, 24–26 efficiencies of the model and prototype machine, 26 flow rate, 26 head coefficient, 24–25 power coefficient, 25 scale factor, 26 volumetric flow rate of a fluid, 24 draft tube, 247–249 analysis of, 249–250 drag force, 337 duty point, 140 dynamic action, 3, 102

380

Index

E efficiencies, 4, 15 of centrifugal pumps, 279–281 enthalpies in, 52, 57 hydraulic efficiency, 226 hydraulic or adiabatic efficiency, 16, 18 infinitesimal-stage efficiency, 66–70 isentropic efficiency, 50, 56–57 isothermal efficiency, 50 mechanical efficiency, 4, 15–16, 18 nozzle efficiency of steam turbine, 196 overall efficiency, 4, 16, 18, 34 point of maximum efficiency, 34 polytropic efficiency, 66–70 of power-transmitting turbomachine, 346 small-stage efficiency, 66–70 static-to-static efficiency, 52 static-to-total efficiency, 52 total-to-static efficiency, 52 total-to-total efficiency, 52 volumetric efficiency, 4, 16, 18 efficiency-flow rate curve, 141 elevation component, 101 energy consumption of mankind, 1 energy flow, process of, 3 energy of fluids, 8–10 energy transfer in a turbomachine, 102–105 by compounding, 170 head-capacity relationship, 139–141 possible designs, 103 pre-whirl or pre-rotation, effect of, 141 ratio of the components of energy transferred, 103 reaction components, 103 reaction in rotation, 105 energy-transfer process, 3 enthalpy, 9–10 enthalpy-drop component, 101 enthalpy-entropy diagram, 56, 60, 62 with an m-stage compression, 75 Euler turbine equation, 97, 99, 124, 136, 166 alternate form of, 99–102 components of energy transfer, 101 energy equation of relative velocities, 101–102 in pressure-velocity-compounded turbine, 172 exit blade angle, 127 exit losses, 52 exit velocity of fluid, 106, 136

expansion process, 75–78 actual drop in temperature, 75–77 overall isentropic expansion efficiency, 77 overall pressure ratio, 77 stage-wise isentropic heat, 77

F fan laws, 155–156 finite-stage efficiency, 73–78 first fan law, 156 first law of thermodynamics, 12–14, 100 fluid angles, 88 fluid mechanics, classification of, 1–2 fluid systems vs mechanical systems, 360–361 Francis turbine, 6 analysis of, 236–237 casing in, 234–235, 238 construction of, 234–236 design parameters, 238–240 efficiencies of, 237 guide vanes, 234–235, 238–239 horizontal shaft installations, 235 rotor, 235, 239–240 velocity triangles, 236 vertical shaft installations, 235

G guide vanes, 5–6, 38, 238–239, 309

H head–capacity relationship, 139–141 head coefficient, 24–25 head energy, 11 hydraulic efficiency, 226, 255 hydraulic or adiabatic efficiency, 16 hydraulic turbines. see also Francis turbine; Kaplan turbine; Pelton turbine characteristics of, 266–267 classification of, 220–222 governing of, 263–266 project site, 219–220 selection of, 220–222 hydro turbomachines, 5

I ideal compression process, 50 ideal expansion process, 50, 57 ideal isentropic process, 56–57 ideal isentropic work, 50

ideal isothermal work, 50 impeller, 5, 35 flow analysis of, 135–137 losses in blade passages, 137–138 impulse machines, 5 pressure of fluid between inlet and outlet of rotor, 144 impulse momentum equation, 97 inertia effects on fluid, 136 infinitesimal actual process, 66 infinitesimal-stage compression process, 67 infinitesimal-stage efficiency, 66–70 inlet guide vanes (IGV), 309 inlet velocity, 106 inlet velocity triangle, 100 instantaneous relative velocity of the fluid, 87 internal combustion engines, 18 isentropic compression process, 50–53 isentropic efficiency, 50, 56–57 isentropic expansion, 195 isentropic expansion efficiency, 64 isentropic expansion process, 56–58 isentropic process, 50 isothermal efficiency, 50

K Kaplan turbine analysis of, 255 construction of, 254 design parameters, 256–258 efficiencies of, 255 kinetic component of the energy transfer, 101 kinetic energy de Laval turbine, 164 of a fluid, 9 Parsons turbine, 168 Pelton turbine, 223 kinetic energy utilization, 169–170

L laws of perfect gases, 47 lift force, 337 linear momentum equation, 97 losses in turbomachine, 137–138 in blade passages, 137–138, 155 Curtis stage, 188–190 Parsons stages, 179 in power-transmitting turbomachine, 346 Rateau stages, 174–175

Index

M magnitude of energy transfer, 136 mass flow rate, 61, 195 mechanical efficiency, 4, 15, 18 meters of liquid, 11, 14 mixed flow machines, 5 multi-jet Pelton turbines, 223 multi-stage air compressor, 60 actual total work, 62 compressible fluids, 61 isentropic compression efficiency, 63 pre-heat effect in, 62–63 re-heat effect in, 63–64 multi-stage expansion process, 64 multi-stage machine, 60

N Newton’s second law, 8, 97 non-dimensional specific speed, 33 nozzle efficiency of steam turbine, 196

O orientation of the imaginary plane, 87 outlet velocity triangle, 100 overall efficiency, 4, 16, 18, 34 overall isentropic efficiency, 60–64, 73–78

P Parson, Charles, 168 Parsons stage, 169 Parsons stages, 178–181 areas of flow, 178 carryover efficiency, 179 degree of reaction, 178 design of blades, 178 enthalpy drop in rotor blade passages, 179 losses in, 179 nozzle efficiency, 179 stage efficiency, 180 steady flow energy equation, 179 Parsons turbine, 168–169 Curtis stage, 184–190 kinetic energy, 168 Parsons stages, 178–181 Rateau stages, 172–175 Pelton turbine, 7 analysis of, 224–226 construction of, 222–224 design parameters, 227–228 efficiencies of, 226–227

kinetic energy of water jet, 223 overall efficiency, 227 rotor, 222 shut-off valve, 223 volume flow rate, 226 performance of a machine, 135 peripheral velocity, 85 point of maximum efficiency, 34 polytropic efficiency, 66–70 for expansion process or turbine process, 68 polytropic process, 50 positive displacement machines, 3–4 potential energy of a fluid, 9 power-absorbing turbomachines, 17–18 power coefficient, 25 power-consuming machine, 14 power-flow diagrams, 18–19 power-flow rate curve, 141 power-generating turbomachine power flow in, 15 power-transmitting turbomachine analysis of, 345–350 construction and working of, 344–345 efficiency of, 346 fluid coupling, 348–350 power losses, 346 torque converter, 351–353 working principle of, 343–344 pre-heat effect in multi-stage compressor, 62–63 pressure-compounded impulse turbine, 170 pressure-compounded reaction turbine, 170–171 pressure-drop component, 101 pressure energy of a fluid, 9 pressure–velocity compounding, 172 pre-whirl or pre-rotation, effect of, 141, 308–310 priming, 295 propeller turbine. see Kaplan turbine prototype machines, 23–24 pumps-in-parallel, 295–296 pumps-in-series, 295 purely axial flow machine, radial components in, 98

R radial component of absolute velocity of fluid, 98 radial flow, 106 radial flow compressor (RFC), 307–310

381

diffuser efficiency, 312–314 isentropic compression efficiency, 311 performance of, 315 power input factor or work-done factor, 311–312 pressure coefficient or loading coefficient, 311 pressure developed in rotor and compressor, 314–315 slip in, 310–311 stage work and stage efficiency, 314 radial flow machine energy transfer, effect of blade outlet angle on, 125–128 general analysis of, 123–124 head–capacity relationship, 139–141 impeller, flow analysis of, 135–137 losses in blade passages of impeller, 137–138 performance, effect of blade outlet angle on, 135 pre-whirl or pre-rotation, effect of, 141 reaction, effect of blade outlet angle on, 129–130 velocity triangles for, 124–125 radial flow machines, 5 radial flow rotor, 87 Rateau, Auguste, 170 Rateau stages, 172–175 losses in nozzles and blade passages, 174–175 of a pressure-compounded steam turbine, 173 total output in, 174 velocity triangles for, 173, 175 volume flow rate, 174 Rateau turbine, 170 reaction in axial flow machine, 143–147 defined, 129 reaction machines, 5, 145 reference velocity, 238 re-heat effect in multi-stage compressor, 63–64 reheat factor for expansion process, 70 steam turbine, 203–204 relative velocity, 85, 101 reversible isothermal process, 50 Reynolds number, 33 rotor, 5 of Francis turbine, 235, 239–240

382

Index

of Pelton turbine, 222 rotor drum, 87–88 rotor efficiency of steam turbine, 197 rotor vanes, 14 runner, 5, 35

S second fan law, 156 second law of thermodynamics, 14–19 single-stage impulse turbine. see De Laval turbine single-stage isentropic work, 70 single-stage process, 50 single-stage reaction turbine. see Parsons turbine skin friction effects, 14 skin friction losses, 137 slip factor, 136–137, 350 small isentropic process, 67 small-stage efficiency, 66–70 for expansion process or turbine process, 68 in terms of polytropic efficiencies, 68–70 specific speed of the pump, 33–38 in different systems of units, 37 values, 36 specific work of a turbomachine, 13, 99, 136 blade angles, effect of, 150–151 degree of reactions, effect of, 150–151 ideal, 136 speed ratio, 107–109 stage efficiency, 60–64, 167 De Laval turbine, 167–168 steam turbine, 197 stagnation state of a moving fluid, 48 properties of fluid, 48 symbols of, 48 stalling of blades, 336 static state of a fluid, 47 static-to-static efficiency, 52 static-to-total efficiency, 52 stator, 5 stator efficiency, 197 steady flow energy equation (SFEE), 12 steady-flow energy equation (SFEE), 85, 97 steam turbine, 7–8 blade efficiency, 168–169 blade heights and volume flow rate, 195

carryover efficiency of, 196 characteristic of efficiency versus load for a steam turbine, 210 classification of, 164–169 compounding of, 169–172 De Laval turbine, 164–167 enthalpy drop of steam in nozzles, 167 enthalpy drop of the steam in, 163 enthalpy drop per kg of steam, 211 enthalpy–entropy chart, 210 governing of, 208–209 mass flow rate, 195 nozzle control in, 208 nozzle efficiency of, 196 performance characteristics, 209–211 reheat factor, 203–204 rotor efficiency, 197 single-stage reaction turbine, 168–169 specific steam consumption (SSC ), 211 stage efficiency, 167–168, 197 stator efficiency or blade passage efficiency, 197 throttle control in, 208–209 vane efficiency, 197 volume flow rate of steam, 195 steam turbines, 57 surging in compressors, 321–323

T tangential components of fluid velocities, 99 tangential flow machines, 5 thermal energy, 9–10 thermal turbomachines, 5 thermodynamics of turbomachine, 50 third fan law, 156 throttling process, 208–209 torque converter, 351–353 fluid systems vs, 360–361 torque of the machine, 97 total-to-static efficiency, 52, 57 total-to-total efficiency, 52, 57 transfer of energy, 3 turbine process, 75–78 turbomachine classification of, 4–8 defined, 2 dimensional analysis, 24–26 fluid flow path, 5 generalities, 3 non-dimensional parameters, 32–33 observations, 3 positive displacement machines, 3–4

power-generating, 15 prototype machines, 23–24 Reynolds number and, 33 specific speed of the pump, 33–38 thermodynamics of, 50 unit and specific quantities, 29–31 work output of, 13 turning losses, 137

U unit quantities of turbomachine, 29–31 specific flow, 30 specific power, 30–31 unit flow, 29–30 unit power, 30 unit speed, 29 utilization factor, 16 utilization factor of a turbine, 105–107

V vane-congruent flow, 86–87, 100, 135 losses due to, 137 vaned rotor, 5 vane efficiency, 165, 197 vaneless diffuser, 308 vanes, 5 velocities in a generalized rotor, 98 velocity compounding, 171 velocity distribution of fluid, 135 velocity triangles, 85–92, 308–310 axial flow machine, 141–142, 145 for axial flow machines, 91–92 Curtis stage, 184, 186–187 De Laval turbine, 167 Francis turbine, 236 general, 100 guidelines in understanding, 86 impulse machine, 150 for radial flow machines, 90–91 Rateau stages, 173, 175 rotor of Parsons turbine, 168 velocities in, 85 volume flow rate, 61, 174, 195, 226 volumetric efficiency, 4, 16, 18

W work-absorbing type of turbomachines, 106 work output of turbomachine, 13

Z zero-work impeller, 126–127