Pragmatic Power

Pragmatic Power

Copyright © 2008 by Morgan & Claypool All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means—electronic, mechanical, photocopy, recording, or any other except for brief quotations in printed reviews, without the prior permission of the publisher. Pragmatic Power William J. Eccles www.morganclaypool.com ISBN: 9781598297980

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ISBN: 9781598297997

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DOI: 10.2200/S00145ED1V01Y200808DCS016 A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON DIGITAL CIRCUITS AND SYSTEMS # 16 Lecture #16 Series Editor: Mitchell Thornton, Southern Methodist University Series ISSN ISSN 1932-3166

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ISSN 1932-3174

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Pragmatic Power William J. Eccles Rose-Hulman Institute of Technology

SYNTHESIS LECTURES ON DIGITAL CIRCUITS AND SYSTEMS # 16

ABSTRACT Pragmatic Power is focused on just three aspects of the AC electrical power system that supplies and moves the vast majority of electrical energy nearly everywhere in the world: three-phase power systems, transformers, and induction motors. The reader needs to have had an introduction to electrical circuits and AC power, although the text begins with a review of the basics of AC power. Balanced three-phase systems are studied by developing their single-phase equivalents. The study includes a look at how the cost of “power” is affected by reactive power and power factor. Transformers are considered as a circuit element in a power system, one that can be reasonably modeled to simplify system analysis. Induction motors are presented as the most common way to change electrical energy into rotational energy. Examples include the correct selection of an induction motor for a particular rotating load. All of these topics include completely worked examples to aid the reader in understanding how to apply what has been learned. This short lecture book will be of use to students at any level of engineering, not just electrical, because it is intended for the practicing engineer or scientist looking for a practical, applied introduction to AC power systems. The author’s “pragmatic” and applied style gives a unique and helpful “nonidealistic, practical, and opinionated” introduction to the topic.

Keywords AC power, three-phase, transformer, induction motor

Contents 1.

Three-Phase Power: 3 > 3 ×1.................................................................................1 1.1 Review of AC Power............................................................................................ 1 1.1.1 AC Power Quantities............................................................................... 2 1.1.2 Example I................................................................................................. 4 1.1.3 Example II................................................................................................ 6 1.1.4 Example II Improved............................................................................... 8 1.2 Why Three?........................................................................................................ 11 1.3 Three-Phase Terminology.................................................................................. 15 1.4 Three-Phase Systems......................................................................................... 17 1.4.1 Dealing With Complexity...................................................................... 18 1.4.2 Single-Phase Equivalent......................................................................... 19 1.4.3 Example III............................................................................................ 20 1.4.4 Example IV............................................................................................ 21 1.4.5 Example V.............................................................................................. 22 1.5 Four Complete Examples................................................................................... 22 1.5.1 Example VI............................................................................................ 22 1.5.2 Example VII........................................................................................... 24 1.5.3 Example VIII.......................................................................................... 25 1.5.4 Example IX............................................................................................ 27 1.6 Summary............................................................................................................ 28

2.

Transformers: Edison Lost................................................................................. 31 2.1 Basic Transformer............................................................................................... 31 2.1.1 Example I: Ideal Transformer................................................................. 34 2.1.2 Example II: Ideal Again......................................................................... 36 2.2 Transformer Model............................................................................................ 38 2.2.1 Realistic Model....................................................................................... 38

vi  pragmatic power

2.2.2 Example III............................................................................................ 40 2.2.3 Example IV............................................................................................ 42 2.3 Transformer Testing........................................................................................... 45 2.3.1 Short-Circuit Test.................................................................................. 45 2.3.2 Open-Circuit Test.................................................................................. 47 2.4 Three-Phase Example........................................................................................ 49 2.4.1 Example V Using Y–Y........................................................................... 49 2.4.2 Example V Using ∆–∆............................................................................ 52 2.5 Summary............................................................................................................ 53 3.

Induction Motors: Just One Moving Part............................................................ 55 3.1 Rotating Fields................................................................................................... 55 3.1.1 Single-Phase Motor............................................................................... 55 3.1.2 Three-Phase Rotating Magnetic Field................................................... 57 3.1.3 Add a Rotor............................................................................................ 59 3.1.4 Slip, Speed, and Poles............................................................................. 61 3.2 Equivalent Circuit.............................................................................................. 62 3.2.1 Stator Equivalent Circuit........................................................................ 63 3.2.2 Rotor Equivalent Circuit........................................................................ 63 3.2.3 Complete Equivalent Circuit.................................................................. 65 3.3 Motor Nameplate............................................................................................... 66 3.4 Testing................................................................................................................ 69 3.4.1 DC Test.................................................................................................. 69 3.4.2 Blocked-Rotor Test................................................................................ 70 3.4.3 No-Load Test......................................................................................... 71 3.4.4 Complete Equivalent.............................................................................. 72 3.4.5 Comparison............................................................................................ 72 3.5 Energy Flow....................................................................................................... 72 3.5.1 Power, Torque, and Losses...................................................................... 73 3.5.2 Stored Energy......................................................................................... 74 3.6 Motor Curves..................................................................................................... 75 3.7 Overmotoring..................................................................................................... 77 3.8 Examples............................................................................................................ 80 3.8.1 Example I—Data Collection.................................................................. 80 3.8.2 Example II—Power Factor Adjustment................................................. 81

contents  vii

3.8.3 Example III—Using The Equivalent Circuit......................................... 82 3.8.4 Example IV—Motor Selection............................................................... 83 3.9 Single-Phase Motors.......................................................................................... 85 3.10 Other Motors..................................................................................................... 86 3.11 Summary............................................................................................................ 87



chapter 1

Three-Phase Power: 3 > 3 × 1 Three-phase power is more than just three single-phase power systems, so to study AC power, we need first to study three-phase systems, what they are, why we use them, and how they work. Threephase systems are by far the dominant method for delivering large blocks of energy from one place to another. To begin, let us review what it means to say “ac power” and refresh our memories about such terms as average power, reactive power, and power factor. You can also find material on this in Pragmatic Circuits: Frequency Domain (William J. Eccles, Morgan & Claypool, 2006, Chapter 8).

1.1

REVIEW OF AC POWER

We could treat AC power using functions of time such as sine and cosine. Figure 1.1 shows a load being driven by a voltage v(t) and a current i(t). The voltage and the current are sinusoids: � � v(t) = Vpeakcos 2p ft + qv � � i(t) = Ipeakcos 2p ft + qi . If we want to know the power, we calculate the product of these:

p(t) = v(t)i(t). If we calculate this product, we get what is known as instantaneous power, the power at any particular time. But this is not very useful, because we really need to know how much power is being delivered on the average, and more important, how much energy is being moved.

FIGURE 1.1: p(t) = v(t)i(t).

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From the calculation of p(t), we can derive this average value as well as define several other quantities that are very useful in dealing with AC power. All of these new quantities come from looking at p(t) for sinusoidal voltages and currents.

1.1.1 AC Power Quantities The average power comes from averaging the instantaneous power p(t) over one period of the AC waveform. For our sinusoidal voltage and current in Figure 1.1, p(t) is

� � � �� � �� p(t) = Vpeak cos 2p ft + qv Ipeakcos 2p ft + qi .

A few manipulations yield

p(t) =

VpeakIpeak � 2

� �� cos(qv − qi ) + cos 4p ft + (qv − qi ) .

If we now average this p(t) over one period, we get Pav:

Pav =

VpeakIpeak 2

cos (qv − qi ) .

Note that Vpeak is the peak value of the cosine wave, a value that makes mathematicians happy because that is the proper manner to describe the cosine. But it is not what we use in AC power. Instead, we use the root-mean-square (RMS) value, which tells exactly how to compute the value, namely, take the square root of the mean of the square of the function. For a sine wave, this comes out as Vpeak/√2 = 0.707Vpeak. Again, note that this is valid for sinusoidal waveforms; other waveforms have different RMS values. Using the RMS values and defining θv–i as the angle of the voltage θv minus the angle of the current θi, the average power P is

P = VRMSIRMScos (qv−i ) , where P is expressed in watts (W). P is sometimes referred to as average power or real power, but much more commonly, it is just plain power. If someone says simply, power, it means P. Now let us go back to the equation for p(t) and consider the second term, the one with 4π in it. This second term averages to zero because the average value of a sinusoid over any integral number of periods is zero. Yet this second term is still power, even though its average is zero. It is power that flows one way for a quarter of a cycle, then the other way for a quarter of a cycle, and so on. From this second term, we define reactive power Q:

Three-Phase Power: 3 > 3 × 1  

Q=

VpeakIpeak 2

sin (qv−i )

= VRMSIRMSsin (qv−i ) , where Q has the unit of volt-ampere-reactive (VAR). Note that Q is volts times amps, which is watts, but we use VAR to distinguish it from the average power P. Q delivers no energy on the average; it represents power that flows back and forth in the power system. Reactive power Q is arriving at a load for one-quarter of a cycle, leaving for a quarter, and so on. Hence, it represents no net flow of energy, but accounts for part of the current flowing in the system. Because of this, we must include it in power calculations. Q is positive for loads that are primarily inductive, that is, loads whose impedance has a positive phase angle. Q is negative for loads that are primarily capacitive. Reactive power Q is very important in energy transfer. For example, an induction motor is inductive (duh!) and requires current to supply the magnetic field that makes the motor work. Therefore, an induction motor requires reactive power Q to operate. To express this in another way, we must supply VARs as well as watts to make such a motor run and drive a rotating load. Because of these VARs, current supplied is larger than that needed for just the average power P. This increased current flows through the wires and transformers of the power system and increases the losses associated with running the motor. We will discuss more about this later. The power factor is defined as the cosine of the angle between the voltage and the current θv–i:

power factor = cos (qv−i ) . Power factor is unitless, but you must add a “unit” to it in the form of either “lagging” or “leading.” The cosine function is always going to produce a positive power factor, so this lagging–leading distinction is needed. How do you know which to use? A load that is inductive has a current that lags the voltage in time. Hence, the angle of the current θi is less than the angle of the voltage θv. We always say, “The current lags the voltage,” and never the other way around. Hence, the power factor is lagging. If the load is capacitive, the current leads the voltage and the power factor is leading. Another power quantity that is useful is the complex power S:

S = P + jQ. This quantity also needs a unit that is equivalent to watts, so we use the volt-ampere and the unit symbol VA. S is a complex number with a real part of P and an imaginary part of Q. The cosine of its angle is the power factor, lagging if Q is positive.

  Pragmatic power

FIGURE 1.2: Power triangle.

S is also the product of phasor voltage and phasor current:

S = VI ∗ , where the asterisk indicates the complex conjugate, replacing +j by −j in rectangular form or changing the sign of the angle in polar form. The magnitude of S is sometimes useful. It is called the apparent power and is just the product of the voltage and the current without regard to the phase angle between them. The power factor can be defined as the ratio of average power to apparent power:

power factor =

P . |S|

The power triangle is an easy way to visualize the relationships among P and Q and S and power factor. The triangle (Figure 1.2) recognizes that S is complex, with P on the real axis and Q on the imaginary axis. Moving Q to the right to the other end of P gives the triangle. The angle is the power-factor angle, the angle θv−i. Note that positive Q is upward and represents inductive reactive power. I will use the triangle to help out with my calculations in Example II (Section 1.1.4).

1.1.2 Example I The system in Figure 1.3 has two loads whose parameters are stated in terms of power. Find the total power P, the total reactive power Q, the total complex power S, the overall power factor, and the source current I. For the first load, P1 = 24 kW with a power factor of 0.86 lagging. If we look at the triangle and remember that the tangent of the angle is Q/P, then the reactive power is

Q1 = P1 tan cos−1 pf = 24 tan cos−1 0.86 = 14.24 kVAR

Three-Phase Power: 3 > 3 × 1  

FIGURE 1.3: Example I.

and the complex power is

S1 = (24 + j 14.24) kVA. I will do the same thing for the second load, which is given in terms of S2 = 20 kVA with a power factor of 0.72 lagging. The triangle tells me that Q and S are related by the sine of the power-factor angle:

P2 = |S2 | pf2 = 20 × 0.72 = 14.4 kW, Q2 = |S2 | sin cos−1 pf2 = 13.88 kVAR. Now, we use the fact that not only is P conserved, but so also are Q and S. This gives us the totals:

Ptotal = P1 + P2 = 24 + 14.4 = 38.4 kW, Qtotal = Q1 + Q2 = 14.24 + 13.88 = 28.12 kVAR, Stotal = Ptotal + jQtotal = 38.4 + j 28.12 = 47.6∠36.2◦ kVA, power factortotal =

Ptotal 38.4 = = 0.807 lagging. |Stotal | 47.6

The current is calculated from S = V × I  *:

Stotal 47.6 × 103 ∠36.2◦ = = 99.2∠36.2◦ A, V 480 I = 99.2∠ − 36.2◦ A.

I∗ =

I will carry the example one step further by finding the individual load currents. Their magnitudes will come from the equation P = |V||I|pf, and their angles will come from the power factor:

  Pragmatic power

|I1 | = |I2 | =

P1 24.0 × 103 = = 58.14 A, ∠I1 = cos−1 pf1 = cos−1 0.86 = 30.7◦ , V × pf1 480 × 0.86 P2 14.4 × 103 = = 41.67 A, ∠I2 = cos−1 pf2 = cos−1 0.72 = 43.9◦ , V × pf2 480 × 0.72

so I1 = 58.14∠ − 30.7◦ A, I2 = 41.67∠ − 43.9◦ A.

Note that the angles have to be chosen as negative because both loads have lagging power factors. As a check, let us see if the two currents add to the total current we have already found:

I = I1 + I2 = 99.2∠ − 36.2◦ A. Wow! It worked!

1.1.3 Example II Figure 1.4 shows a 2-hp motor that is 85% efficient, running on a 240-V supply with a power factor of 0.78 lagging. It is being fed by wires, each of which has an impedance of 0.5 + j0.8 Ω. We are to find the power delivered to the motor, the line current, the power lost in the line, the total P, Q, S, and power factor at the generator, the percent voltage regulation, and the percent efficiency. Notice that the generator voltage is not given. This is because the motor is designed to operate correctly on 240 V, so the generator voltage must be larger than that to compensate for the losses in the line. First, we need to put the motor’s specifications into AC power terms. One horsepower is 746 W, which is the shaft output of the motor. The motor is 85% efficient, so the power input is

Pload =

FIGURE 1.4: Example II.

2 × 746 = 1.755 kW 0.85

Three-Phase Power: 3 > 3 × 1  

and the reactive power input is

Qload = 1.755 tan cos−1 0.78 = 1.408 kVAR. From our knowledge of P comes the line current:

Pload = |V | |Iline | pfload , Iline =

1.755 × 103 ∠ − cos−1 0.78 = 9.375∠ − 38.7◦ A. 240 × 0.78

The loss in the line is due entirely to the resistive part of the line impedance. Resistors “don’t care” about phase angles, so I will use I-squared-R for the line loss, but with just the magnitude of the current:

Pline = 2Rline |Iline |2 = 87. 9 W. I need the generator voltage Vs for a couple of the steps that follow, so I will obtain it by using 240 plus the voltage drop in the line impedances:

Vs = 240 + Vline = 240 + (0.5 + j 0.8 + j 0.8 + 0.5)(9.375∠ − 38.7 ◦ ) = 256.7∠1.3◦ V.

There is now enough information to obtain the total Q and S and overall power factor. Here is a neat place to run off the track, though! It is tempting to think of the power factor as the cosine of the angle of the current. But that is wrong—it is the cosine of the angle between the voltage and the current. Note that the angle of Vs is not zero! Here are the results, beginning with S = VI *:

Ss = (256.7∠1.3◦ )(9.375∠ + 38.7◦ ) = 2.408∠40.0◦ kVA, Ps = Re [Ss ] = 1.844 kW, Qs = Im [Ss ] = 1.548 kVAR, pfs = cos(40.0◦ ) = 0.766 lagging. That leaves percent voltage regulation and . . . oh, what is regulation? The easiest way to look at percent voltage regulation is to think of it as the percentage by which the voltage at the load will rise if the load is disconnected. Our generator voltage Vs is 256.8 V. If we disconnect the motor, its terminal voltage will rise from 240 to 256.8 V. Percent voltage regulation is that expressed as a percentage:

  Pragmatic power

% VR =

|Vs | − |Vload | 256.8 − 240 × 100 = × 100 = 7.0% . |Vload | 240

Percent efficiency is a little more obvious. It is the percentage of useful power divided by generated power, often stated as %η:

%h =

Pload 1.755 × 100 = × 100 = 95.2% . Ps 1.844

How about cost? Let us assume that this motor is part of a plant that operates 16 hours a day, 5 days a week, 50 weeks a year. Let us also assume that electricity costs 7¢ per kilowatt hour plus a charge of $10 per kVA/month for the peak kVA demand during the month. (The meter that measures kilowatt hours also “watches” for the peak kVA demand using a sliding average with a 15-min window.) Here are the numbers, which we will need for the next example:

kWh = (1.755)(16)(5)(50) = 7020 kWh/year, $kWh = (7020)(0.07) = $491.40 for 1 year; Sload = 1.755 + j 1.408 = 2.250∠38.7◦ kVA, $kVAmax = (2.250)(12)(10.00) = $270.00 for 1 year; Total annual cost = $491.40 + 270.00 = $761.40 per year.

1.1.4 Example II Improved You are a consulting engineer who has been asked to take a look at the power bill we just computed and see if you can reduce it. You look at the plant and decide that replacing the motor with a more efficient model would be very expensive and would not improve things very much. You cannot change P because that is what the motor requires to drive its load. You take another approach, which is to reduce the reactive power. You know that this will bring down the demand charge, but it should also reduce the current and hence the line losses. You can do this by adding negative reactive power, which is what capacitors absorb. But how much to add (subtract, really), to get the demand charge down? One obvious, but wrong, answer is to bring the power factor to unity by using capacitive reactance to exactly cancel the inductive reactance of the motor. This solution is wrong because of something that is beyond the scope of this text, namely, voltage regulation. If the power factor “gets past unity,” meaning that the load becomes capacitive, the power system has a hard time regulating the load voltage. In other words, if the load is capacitive, the 240 V at the load terminals wanders around too much for proper operation of the system.

Three-Phase Power: 3 > 3 × 1  

FIGURE 1.5: PF correction.

The usual solution is to improve the power factor, but not all the way to unity. You talk with a knowledgeable power system engineer, who recommends that you take the system to 0.9 lagging. Now let us get there. Look at the power triangle in Figure 1.5. It shows the original P and Q for this load ( just the load, not the lines, because the meter is at the load terminals). The new triangle has an angle whose cosine is 0.9. That triangle shows us the value of Qnew, the reactive power that will get the system to a power factor of 0.9:

Qnew = 1.755 tan cos−1 0.9 = 0.850 kVAR. The diagram also shows us that the capacitive load must “absorb” the difference:

Qc = Qnew − 1.408 = −0.558 kVAR. This says that you need a 240-V capacitor that will absorb 0.558 kVAR. Catalogs for such capacitors state specifications in kVARs and volts, so you pick the next larger capacitor that is rated at least 0.558 kVAR and will withstand at least 240 V. However, just to get a feeling for the size of this capacitor in terms of circuit elements, let us find C. Just as power delivered to a resistor is V 2 over R, the “power” delivered to a capacitor is V 2 over the capacitive reactance:

Qc =

|Vload |2 2402 so Xc = = −103.2 W. Xc −0.558 × 103

Because capacitive reactance is the impedance of a capacitor without the j, the value of capacitance needed at a frequency of 60 Hz is

Xc =

−1 −1 so C = = 25.7 m F. 2p fC (2p )(60)(−103.2)

10  Pragmatic power

I will repeat all the first set of calculations using the new reactive power and power factor:

Pload = 1.755 kW (unchanged), 1.755 × 103 = 8.125∠ − cos−1 0.9 = 8.125∠ − 25.8◦ A. 240 × 0.9 = 2Rline |Iline |2 = 66.0 W.

Iline = Pline

Vs = 240 + Vline = 240 + (0.5 + j 0.8 + j 0.8 + 0.5) (8.125∠ − 25.8◦ ) = 253.1∠1.8◦ V.

Ss = (253.1∠1.8◦ ) (8.125∠ + 25.8◦ ) = 2.056∠27.6◦ kVA, Ps = Re [Ss ] = 1.822 kW, Qs = Im [Ss ] = 0.953 kVAR, pfs = cos(27.6) = 0.886 lagging. 253.1 − 240 |Vs | − |Vload | 100 = 5.5%. 100 = |Vload | 240 Pload 1.755 %h = 100 = 96.3%. 100 = Ps 1.822

%VR =

Well, did things improve? The percent voltage regulation came down a little from 7.0% to 5.5%. Umm, that does not mean much because less than 10% is generally OK. The percent efficiency came up a little from 95.2% to 96.3%. Gee, that is not much of a change, either. Let us look at the power costs:

kWh = (1.755)(16)(5)(50) = 7020 kWh/year, $kWh = (7020)(0.07) = $491.40 for 1 year; Sload = 1.755 + j 0.850 = 1.950∠25.8◦ kVA, $kVAmax = (1.950)(12)(10.00) = $234.00 for 1 year; Total annual cost = $491.40 + 234.00 = $725.40 per year. Aha! We have saved about $36 per year! Is that good? Hard to say, because we have to consider the installed cost of the capacitors and how long they will last. That is an engineering economic analysis problem that is beyond what I want to do here. And yes, I will admit that this is a pretty trivial example. Increase the load to 200 hp, though, and the savings may be significant. (Another solution is “active” power factor correction, which is done using an electronic unit to control the power to the motor.)

Three-Phase Power: 3 > 3 × 1  11

1.2

WHY THREE?

Okay, you say, I have reviewed AC power now. I have even read through the example that has a 2hp motor in it. So what is all this stuff about three phases? What is wrong with one? And why did I read that three phases is better than three single phases? Well, let us see! Start by doubling the number of phases to two. Figure 1.6 shows a two-phase generator. It seems logical to space the two phases uniformly around a circle, so I have set the angle for the a phase at 0° and the angle for the b phase at −180°. I have gone around in the negative direction, because the cosine does not care. Here are the two voltages: √ va (t) = 120 2 cos(2p 60t) V, √ vb (t) = 120 2 cos(2p 60t − 180◦ ) V. These describe the usual 120-V household power where the frequency is 60 Hz (=2π 60 rad/s). But this setup is merely single-phase three-wire where the voltage from a to b is √ √ vab (t) = van + vnb = 120 2 (cos(2p 60t) + cos(2p 60t)) = 240 2 cos(2p 60t) V. (An arrangement sometimes used in control systems spaces two phases 90° apart, but we will not go into that here because it is not used for power systems.) Hmm, okay, two phases does not seem to work out, but let us not get discouraged. Let us try three phases. Figure 1.7 shows a three-phase generator. The voltages are placed uniformly around a circle, going around in the negative direction:

va (t) = Vm cos(2p 60t) V, vb (t) = Vm cos(2p 60t − 120◦ ) V, vc (t) = Vm cos(2p 60t − 240◦ ) V.

I have used Vm rather than numbers just to save some mess in the upcoming analysis. Put this generator into the simple three-phase circuit with a load of three equal resistors (Figure 1.8). Now, calculate the instantaneous power p(t) delivered to those resistors using V 2 over R. The analysis is messy trig; the outcome is surprising:

FIGURE 1.6: Two-phase generator.

12  Pragmatic power

FIGURE 1.7: Three-phase generator.

(Vm cos(2p 60t))2 W, R (Vm cos(2p 60t − 120◦ ))2 pB (t) = W, R (Vm cos(2p 60t − 240◦ ))2 pC (t) = W. R pA (t) =

Adding these three instantaneous powers involves the cosine squared:

p(t) = pA (t) + pB (t) + pC (t) � V2� = m cos2 (2p 60t) + cos2 (2p 60t − 120◦ ) + cos2 (2p 60t − 240◦ ) W. R After applying a few trig identities, the result is

FIGURE 1.8: p(t) calculation.

Three-Phase Power: 3 > 3 × 1  13

p(t) =

3Vm2 W. 2R

The surprise? That the instantaneous power is a constant! The instantaneous power delivered is not sinusoidal as it is in single-phase systems—it is steady. Be careful, though! This analysis required that the load be balanced, meaning all three legs of the load are the same. Now that we have one advantage of three-phase power, let us take a look at another. I have added current labels to the wires of this system (Figure 1.9). Let us apply Kirchoff ’s current law at the node N and determine iNn(t) in terms of the generator voltages.

iNn (t) = iaA (t) + ibB (t) + icC (t) =

Vm cos(2p 60t) Vm cos(2p 60t − 120◦ ) Vm cos(2p 60t − 240◦ ) + + A. R R R

Now I will apply some trig identities and get

iNn (t) = 0. Again, a surprise! In a three-phase balanced system the nN wire (the neutral ) carries no current! This says we can leave it out and build our balanced three-phase system with just three wires. How about one more surprise? It would seem that three wires would require more metal than two wires. Here are the ground rules for the calculations that follow: •

A single-phase system and a three-phase system are to supply the same load with the same total amount of power, so in Figures 1.10 and 1.11, the total power P is the same value. In the three-phase system, however, the load is split into three balanced phases.

FIGURE 1.9: Three-phase currents.

14  Pragmatic power

FIGURE 1.10: “Copper” calculation—1θ.

•

• •

The voltages between the conductors must be the same. In Figure 1.10, this voltage is Vs. Without proof at the present time, I will simply state that the generators in Figure 1.11 must supply Vs /√3 so that the voltage between the wires (excluding the neutral) is Vs. The cross-sectional area of a conductor must be proportional to the magnitude of the current so the wires all have the same current density (current per unit cross section). The load is the same distance from the generator in each system.

In the single-phase system, the power delivered to the load is

P1 = Vs I1 and the power delivered to the three-phase load is � � √ Vs P3 = 3 √ I3 = 3Vs I3 . 3 Equating the powers relates the wire currents:

P1 = Vs I 1 = P3 = √ I1 = 3I3 .

FIGURE 1.11: “Copper” calculation—3θ.

√

3Vs I3 ,

Three-Phase Power: 3 > 3 × 1  15

Calculating the ratio of each current to the cross-sectional area needed and applying the relationship between the currents gives

I1 I3 = 2, 2 p r1 pr √3 2 2 r1 = 3r3 . If the system is of length  , the metal volume for the wires in each system, with two wires for single phase and three for three phase, is � √ � � � Vol1 = 2 p r12 � = 2 p 3r32 �, � � Vol3 = 3 p r32 �. The surprise here is that the ratio of the total volume of metal for these systems is

Vol3 3 = √ = 86.6% . Vol1 2 3 A three-phase system with the same power delivery capability as a single-phase system requires 86.6% as much metal for the wires! How is that for reasons why three-phase? Nonpulsating power. Only three wires for threephase versus two for single phase. And 13.4% less metal for conductors. Not a bad deal! And that is why the vast majority of electrical power transmission is done via three-phase systems.

1.3

THREE-PHASE TERMINOLOGY

One of the most confusing things about three-phase systems is terminology. Certain terms have very specific meanings, which are somewhat different from what your intuition will tell you. There are also some rather strictly observed conventions that one follows when describing three-phase systems. Figure 1.12 shows a three-phase system with one three-phase generator and two loads. The generator is said to be “Y-connected.” Load 1 is “Y-connected;” load 2 is “∆-connected.” I have labeled a voltage and a current in each load as well as some quantities between the generator and the loads. Now, for some terms: •

The term “line” always refers to interconnecting wires between the generator and the load, labeled in the figure “Line.” This term never refers to the loads themselves.

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FIGURE 1.12: Three-phase terminology.

• • • •

The term “phase” always refers to legs of the load, labeled “1” and “2” in the figure. This term never refers to the interconnecting wires between the generator and the loads. In the “Line” portion of the figure, the voltages VAB, VBC, and VCA are line voltages and IaA is a line current. (The two other line currents are not labeled.) In the figure, VAN and VAB are phase voltages, whereas IAN and IAB are phase currents. At the generator in the figure, Van, Vbn, and Vcn are phase voltages.

There are two conventions: •

•

Three-phase system voltages are always stated as line voltages. These are VAB, VBC, and VCA in the “Line” portion of the figure. The phase voltages of the generator (Van, Vbn, and Vcn) are not used to describe the system. All voltages and currents are given as RMS values, never peak values.

The relationship between line and phase voltages can be obtained by applying Kirchoff ’s voltage law through Vbn, Van, and VAB. In a similar manner, Kirchoff ’s current law gives the relationship between phase currents in the ∆-connected load and the currents in the connecting lines.

Vbn − Van + VAB = 0

VAB = Van − Vbn , Iline = IAB − ICA.

Phasor diagrams (Figure 1.13) make finding the result easy.

Three-Phase Power: 3 > 3 × 1  17

FIGURE 1.13: Line versus phase values.

The results of this are •

For a Y-connected generator (the most common arrangement), the line and phase relationships are √ Vline Y = 3Vphase Y ∠ + 30◦

Iline Y = Iphase Y . •

This same rule applies to Y-connected loads. For ∆-connected loads, the line and phase relationships are

Vline D = Vphase D √ Iline D = 3Iphase D∠ − 30◦ . Finally (whew!), there is a matter of phase rotation. There are two ways to arrange the Y-connected generator, one with each phase 120° behind the previous one, and the other with each phase 120° ahead of the previous one. I have used positive rotation in all examples so far and will stick with that, which means Vbn lags Van by 120° and Vcn lags Vbn by 120°. Positive rotation is sometimes called a-b-c rotation. (Negative rotation is therefore a-c-b.) Is phase rotation important? Yes, any time your system involves rotating machines, because most three-phase motors run in opposite directions for a-b-c and a-c-b connections. It could be rather embarrassing to install a large three-phase fan and turn it on—running backward!

1.4

THREE-PHASE SYSTEMS

Let us go first for the big picture. The three-phase systems that provide electrical energy all over the world generally have seven major sections:

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• • • • • • •

Large generators to produce voltages in the 12- to 24-kV range, the larger the power, the higher the voltage. Transformers to step up the generation voltage to the transmission voltages for long-haul lines. Transmission lines, usually at voltages of at least 230 kV. Remember that we always state the line voltage, the voltage between the lines. Transformers to step down the transmission voltage to distribution voltages, generally at most a “few ten thousands.” Distribution lines to carry the energy through neighborhoods and sections of cities. Transformers to step down the distribution voltage to voltages that the customer can use, generally less than 600 V. Customer loads that “use”—and pay for— the electrical energy provided by the system.

Each of these sections has an inherent impedance. Generators have large coils of wire that present both resistance and inductance. Transformers are also coils of wire on iron cores that have both resistance and inductance. Wire lines have both resistance and inductance. Customers’ loads can generally be modeled by using resistance and inductance. In other words, the whole system looks kind of like a generator and a load with all sorts of Rs and Ls in between.

1.4.1 Dealing With Complexity How can we handle such complexity? We will not even try! But what we can do is simplify this by omitting the transformers (which we will study in Chapter 2) and then combine all of the impedances except the load into one impedance. Figure 1.14, which we will deal with in detail a little later, shows this. There are no transformers, and all the impedances from generator to (but not including) load are lumped into 0.2 + j0.5 Ω . That sounds okay so far, you think, but analysis of the system of Figure 1.14 looks like it would be messy, too. Sure, but if we remember that the neutral (nN ) carries no current if the system is balanced, we could draw in the neutral and then use the single node nN as our reference node for nodal analysis. There are only three “unknown” nodes: a, b, and c. We know the voltages at A, B, and C from the stated line voltage, and we now know how to convert line to phase voltage. Let us try what I have just said. A 480-V line voltage becomes a phase voltage of 480/√3 = 277 V. Let us make the A phase have a phase angle of 0° because no phase reference is otherwise specified. That gives us three node voltages at A, B, and C. At the generator end, the node voltages are the unknown generator voltages Van, Vbn, and Vcn. Now I can write three node equations . . . oh, wait! I can find Iline from my knowledge of the power absorbed by the load in the A phase and the A-phase voltage. Then I can find Van as VAN plus the voltage drop in the line . . . . But wait! The sys-

Three-Phase Power: 3 > 3 × 1  19

tem is balanced, and anything I calculate for one of the phases can be applied to the others by just shifting phase angles by 120° or 240°. And that is how I am going to proceed with the system analysis.

1.4.2 Single-Phase Equivalent I am going to reduce the system to just one phase, but with a few restrictions: • • •

Generators will be Y-connected. In fact, if the generator is ∆-connected and one of the gen­ erators is a bit different from the other two, there will be a wasteful circulating current. Loads can be either Y- or ∆-connected, but for my analysis they will always be balanced. Line impedances will be the same for all three phases. If the neutral is included in the system, its impedance will be taken as 0 because in a balanced system the neutral current is 0.

Here are the steps to transform a balanced three-phase system to its single-phase equivalent: 1. Draw a single-phase system consisting of a generator on the left, an empty box for a load on the right, and two connecting wires. Label a line current at the top. 2. Divide any voltages by √3 because voltages are always supposed to be given as line voltages, but we are drawing the single-phase equivalent. Label the voltages on your diagram. 3. In the top wire only, draw the line impedance, generally a resistance and a reactance. Label them with the values given. If a generator impedance is also given, draw that into the top line. Do not put any impedance in the bottom line because this is the neutral, which in a balanced three-phase system carries no current. 4. This step depends on how the load data are given: (a)  If the total load is stated in “power” terms such as “24 kW at 0.86 lagging,” divide the “power” number by 3 (but not the angle or the power factor) and write this load into the box you drew. It makes no difference whether the load is Y- or ∆-connected. (b)  If the load is given in “power” terms but “per phase,” write the load into the box. (c)  If the load is given as an “impedance per phase” and is Y-connected, write the stated impedance into the box. (d)  If the load is given as an “impedance per phase” and is ∆-connected, divide the impedance by three and write this into the box. 5. Solve the given problem by finding any requested quantities using your single-phase equivalent. 6. Transform the results from their single-phase values to the original three-phase circuit: (a)  The line currents stay the same. (b)  Multiply any voltage results by √3.

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FIGURE 1.14: Example III—3θ.

(c)  Multiply any “power” results (but not angles or power factor) by 3. (d)  Efficiency and voltage regulation are the same. 7. Answer any “three-phase” questions such as finding the phase current in the load using your knowledge of the relationship between line and phase quantities (Section 1.3). Now let us try all this!

1.4.3 Example III A three-phase, 480-V, Y-connected load absorbs 24 kW at 0.86 lagging through a system with line impedances of 0.2 + j0.5 Ω (See Figure 1.14). Draw the single-phase equivalent. Figure 1.15 shows the result: 1. I drew the generator on the left and a box on the right and then labeled the current Iline. 2. The load voltage becomes 480/√3 = 277 V. The generator voltage is simply Vs. 3. The top wire gets the line impedance as stated. 4. Because the load is stated in “power” terms, I divided by 3 and wrote the result into the box.

FIGURE 1.15: Example III—1θ equivalent.

Three-Phase Power: 3 > 3 × 1  21

5. Now I can solve whatever problem is given. 6. If my results yield a value for Vs, I multiply that by √3 to obtain the line voltage at the generator end of the three-phase system. If I am to find, say, Van, I use Vs as is. 7. A question such as “find VAN” is answered by dividing the line voltage by √3 and subtracting 30° from the angle.

1.4.4 Example IV A three-phase, 208-V, ∆-connected, 6-kW load is connected through lines with a line impedance of 0.3 + j0.8 Ω. The load power factor is 0.86 lagging. Draw the single-phase equivalent. Figure 1.16 shows the result. The load voltage is 208/√3 and the load is 6/3 = 2 kW. It makes no difference whether the load is Y- or ∆-connected.

FIGURE 1.16: Example IV—1θ equivalent.

1.4.5 Example V Two three-phase loads are driven by a 480-V system with line impedances of 0.3 + j1.0 W. One load is Y-connected with a per-phase impedance of 6 + j3 W. The other is ∆-connected with a per-phase impedance of 27 + j12 Ω (see Figure 1.17). Draw the single-phase equivalent.

FIGURE 1.17: Example V—3θ.

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FIGURE 1.18: Example V—1θ equivalent.

Figure 1.18 is the result. The load voltage is 480/√3, the Y-connected impedance and the line impedance are used as is. The ∆-connected impedance is divided by 3 before using.

1.5

FOUR COMPLETE EXAMPLES

That is enough talk about three-phase systems and analysis. Let us do some concrete examples using the process of transforming the system into its single-phase equivalent. Remember that this transformation works only if the system is balanced. If it is unbalanced, we cannot simplify it to the single-phase equivalent but have to write equations for the whole three-phase circuit. If you happen to think, “Gosh, writing three-phase system equations for large, complicated systems must be a huge chore,” you would be right. The folks who deal with large systems that are unbalanced but not badly so use a method called symmetrical components to reduce the work.

1.5.1 Example VI A balanced, 4800-V three-phase system feeds a ∆-connected load of 30 + j12 Ω/∅. The generator impedance is 0.1 + j0.6 Ω/∅ and the line impedance is 0.2 + j1.2 Ω/∅. Find the line current, total power P, reactive power Q, and complex power S delivered by the generator. Also, find the system’s voltage regulation and its efficiency. Figure 1.19 shows the single-phase equivalent. The load voltage is 4800/√3 = 2770 V. Both the generator and line impedances are shown. The ∆-connected load impedance has been divided by 3. First, I will find the line current and the power “lost” in the generator and line impedances:

Iline =

2770 = 238.8 − j95.5 = 257.2∠ − 21.8◦ A 10 + j4

Sline = (0.1 + j0.6 + 0.2 + j1.2) |Iline |2 = 19.8 + j119.1 kVA

Pline = Re[Sline ] = 19.8 kW, Qline = Im[Sline ] = 119.1 kVAR.

Three-Phase Power: 3 > 3 × 1  23

FIGURE 1.19: Example VI—1θ equivalent.

Using the line current, I will next find S for the load.

Sload = 2770I ∗line = 661.5 + j264.6 kVA. Adding these results answers the questions about total power:

Stotal = Sline + Sload = 681.3 + j383.7 kVA Ptotal = Re[Stotal ] = 681.3 kW, Qtotal = Im[Stotal ] = 383.7 kVAR. To find the voltage regulation, I need the generator voltage:

Vgen = 2770 + (0.1 + j0.6 + 0.2 + j1.2)Iline = 3014 + j401 = 3040∠7.6◦ V. Here is a good place to check your work. For a typical system, the phase angle of the generator voltage does not differ much from the phase angle of the load voltage. I have used 0° for the load voltage; the generator phase angle is just 7.6°, which looks OK. From the generator and load voltages I get the percent voltage regulation

% VR = 100

3040 − 2770 = 9.7% 2770

and from the power results I get the percent efficiency.

% h = 100 ×

Re[Sload ] 661.5 = 100 × = 97.1% Ptotal 681.3

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But I am not done! The results need to be transformed back to the three-phase system. The “power” numbers are multiplied by 3; the line current, the regulation, and the efficiency results are used as found:

Iline = 257.2∠ − 21.8◦ A

Ptotal = 3 × 681.3 kW = 2044 kW

Qtotal = 3 × 383.7 = 1151 kVAR

Stotal = Ptotal + jQtotal = 2346∠29.4◦ kVA

% VR = 9.7% % h = 97.1% .

1.5.2 Example VII A 208-V, Y-connected load absorbs 3.6 kW at 0.8 lagging. The feeders have an impedance of 0.3 + j0.6 Ω each. Find the line current, the generator voltage, the power factor at the generator, the voltage regulation, and the efficiency. The single-phase equivalent is Figure 1.20. Note the division of the voltage by √3 and the division of the load power by 3. I will find the line current first because that quantity is key to the rest of the problem:

Iline =

Pload ∠ − cos−1 pfload = 12.5∠ − 36.9◦ A. 120pfload

Next comes the generator voltage by adding the voltage drop in the line to the load voltage:

Vgen = 120 + (0.3 + j0.6)Iline = 127.5 + j3.75 V = 127.6∠1.7◦ V.

FIGURE 1.20: Example VII—1θ equivalent.

Three-Phase Power: 3 > 3 × 1  25

It is temping to use the generator’s phase angle of 1.7° for the power factor, but this would be wrong. The power-factor angle is the angle between the voltage and the current:

pfgen = cos (1.7◦ − (−36.9◦ )) = 0.782 lagging. Voltage regulation is

% VR = 100 ×

127.6 − 120 = 6.3% . 120

To find the efficiency, I need either the power delivered by the generator or the power dissipated in the line. I will use the line power by noting that power is I 2R, where the current is the magnitude of the line current. Only the resistance of the line can dissipate power, and resistances do not care about phase angles.

Pline = 0.3 |Iline |2 = 46.9 W, % h = 100

Pload 1200 = 96.2% . = 100 × Pload + Pline 1200 + 46.9

The only result that needs to be transformed to three-phase is the generator voltage, so I will multiply Vgen by √3 to give a generator voltage in the three-phase system of 221.0 V.

1.5.3 Example VIII A 500-hp, three-phase, ∆-connected, 2300-V motor is 84% efficient. Its power factor is 0.75 lagging. The lines feeding the motor have a line impedance of 0.2 + j0.5 Ω/∅. Find the generator voltage, the voltage regulation, and the efficiency. Also, assuming that the motor operates 100% of the time all year, find the annual cost of operation if energy costs 3.0¢ per kilowatt hour and the monthly demand charge is $10.00 per kVA. The capacity of the motor is the output power, because a designer wants to find the right motor for the load it is going to drive. The output horsepower is converted to input power by multiplying by 746 W/hp and dividing by the efficiency. Dividing this by 3 gives the per-phase power for the single-phase equivalent. Figure 1.21 is the equivalent.

(500)(0.746) = 444.0 kW 0.84 = 148.0 kW/phase.

Pload =

The line current and the generator voltage are

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FIGURE 1.21: Example VIII—1θ equivalent.

Iline =

Pload ∠ − cos−1 pfload = 148.6∠ − 41.4◦ A 1328pfload

Vgen = 1328 + (0.2 + j0.5)Iline = 1400∠1.5◦ V. The voltage regulation is

% VR = 100 ×

1400 − 1328 = 5.4% . 1328

The efficiency can be calculated using the power absorbed in the line and the load power:

Pline = 0.2 |Iline |2 = 4.42 kW %h = 100

Pload 148.0 = 97.1%. = 100 Pload + Pline 148.0 + 4.42

The cost of operation requires the “demand,” which is the magnitude of the complex power S for the motor. This is generally metered just as kilowatt hours are metered, and the maximum demand for the month is used. Usually the peak demand is calculated from a sliding average with a window of perhaps 15 min. For our motor, this demand is constant. The single-phase equivalent results need to be multiplied by 3:

$power = (148.0)(24)(365)(0.03)(3) = $116,683 per year

|Sload | =

Pload 148.0 = 197.3 kVA per phase = pfload 0.75

$demand = (197.3)(12)(10)(3) = $71,040 per year $total = 116,683 + 71,040 = $187,723 per year. (The demand-charge scheme in this example is just one way that utility companies assess VARs. Some companies charge for peak kVARs, some charge for low power factor, and some even ignore them in their tariffs.

Three-Phase Power: 3 > 3 × 1  27

1.5.4 Example IX Continue with example VIII by adding to the load sufficient capacitance to bring the load power factor up to 0.9 lagging. Then find the generator voltage, the voltage regulation, and the efficiency, as well as the new annual operating cost and how much is saved by adding the capacitors. From the total power for the motor that we already know, we can get the reactive power for the motor. Using this in the power triangle of Figure 1.22, I get the new reactive power for the combination of the motor and the added capacitor. Note that the real power P for the motor is not going to change because it still is a 500-hp motor that is 84% efficient.

Qload = Pload tan cos−1 pfload = 130.5 kVAR Qnew = Pload tan cos−1 pfnew = 148.0 tan cos−1 0.9 = 71.7 kVAR Qc = Qnew − Qload = 71.7 − 130.5 = −58.8 kVAR per phase The necessary capacitors would be purchased based on the required kVARs and the operating voltage. The new generator voltage, voltage regulation, and efficiency are

Sload new = Pload + jQnew = 148.0 + j 71.7 kVA Sload new ∗ I line = 111.5 + j54.0 = 123.9∠25.8◦ A new = 1328 Vgen new = 1328 + (0.2 + j0.5)Iline new = 1378∠1.9◦ V 1378 − 1328 = 3.7% 1328 = 0.2 |Iline new |2 3.1 kW

% VR = 100 × Pline new

% h = 100 ×

148 = 97.9% . 148 + 3.1

A major change—an improvement—is the current, which decreased from 148.6 to 123.9 A. This shows why utility companies have rate structures that discourage poor power factor in some way. In

FIGURE 1.22: Power triangle.

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this case, the current is over 16% less, which means line losses are about 30% less. Line losses are on the utility’s side of the meter. Figure 1.23 shows the completed system. The generator voltages are drawn as per-phase voltages so they have the same magnitudes as in the single-phase equivalent. Now for the cost. The energy cost does not change because the motor is still producing 500 hp. However, the demand is lower and the demand charge will be reduced.

$power = $116, 683 per year (unchanged) Pload 148.0 |Sload new | = = = 164.4 kVA per phase pfload new 0.9 $ demand = (164.4)(12)(10)(3) = $59,184 per year $ total = 116,683 + 59,184 = $175,867 per year. Gosh, that is an improvement of $11,856 per year, an improvement of 6.3%. Whether this saving justifies the cost of purchasing and installing capacitors is a problem for the accounting folks in the front office.

1.6

SUMMARY

We started this chapter by reviewing material for single-phase systems and then extended this to three phases. By restricting our systems to balanced ones, we can transform the three-phase system to its single-phase equivalent and then do all of the analysis on just that phase. Results can be transformed back to the original three-phase system.

FIGURE 1.23: Example IX—3θ with correction.

Three-Phase Power: 3 > 3 × 1  29

In Section 1.4, I described a typical power system that included transformers to raise and lower voltages for transmission, distribution, and customer. Throughout the three-phase analysis in this chapter, though, I have omitted the transformer. That comes in Chapter 2, where we will see that the transformer can be modeled in terms of basic circuit elements and included in the analysis of more complete three-phase systems. • • • •

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chapter 2

Transformers: Edison Lost Fortunately! Nikola Tesla (1858–1943) came to the United States in 1884 and began working as an assistant to Thomas Edison (1847–1931). Edison had already invented the light bulb and had started building a large direct-current (DC) system in New York. Edison asked Tesla to debug the system and promised to pay him well for his work. Tesla did improve the system, saving Edison considerable money, but Edison refused to pay Tesla. Edison was wedded to the idea of using DC as the power source to light everyone’s homes and businesses, arguing that alternating current (AC) was dangerous. He staged elaborate electrocutions of animals using AC and noted that AC was used for executions at Sing Sing, the state prison. Tesla left Edison and went on to devise the basics of AC power systems, demonstrating that large blocks of power could be moved by raising the voltage very high, thereby reducing the required current so smaller wires could be used. Edison’s DC system, on the other hand, would require very large cables and, one writer has said, a generating station for each square mile of city. The last of Edison’s DC systems did not quit until November 14, 2007, though, when the Pearl Street Station that served old buildings on New York’s Upper East and West Sides was shut down after 125 years of operation. One of the keys to Telsa’s AC system is the transformer, which makes it possible to generate at a modest voltage, step the voltage up (and the current down, therefore) to a much higher voltage for transmission, and then step the voltage back down (and the current back up) for distribution to consumers. (Tesla also invented the induction motor to use this available AC power, but that is another chapter.) George Westinghouse (1846–1914) pushed Tesla’s ideas, purchased European-made transformers, and built a hydro-powered AC system in western Massachusetts. From there on, Westinghouse and Edison engaged in a bitter war, but Westinghouse ultimately won. For this, he received the Edison Medal (irony?) from the American Institute of Electrical Engineers, a forerunner of the Institute of Electrical and Electronics Engineers.

2.1

BASIC TRANSFORMER

A transformer looks so simple that it is perhaps hard to believe that it was a major key in the development of the AC power system. It is nothing more than a couple of coils of wire wound around a core of iron. Figure 2.1 shows the basic structure.

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Figure 2.1: Transformer.

It is such a simple-looking device but it played a big part in Tesla’s developments. How does it work? Without going very deeply into magnetic field theory. . . . •   The left coil has an AC current flowing through it. Ampere’s circuital law states that this current creates an alternating magnetic flux, shown as Φ in Figure 2.1. •   This flux “links” the right coil, passing through its windings. •   Faraday’s law says that a changing flux creates a voltage across the windings of the coil that it links. •   Voila! Applying a voltage V1 on the left drives current I1 through the left coil, creating flux Φ in the iron core, linking the right coil, and developing a voltage V2 that is ready to do work when connected to something that draws current I2. Figure 2.1 represents an ideal transformer, which means that it is lossless. What does this mean? •   Wires have zero resistance so that there is no power dissipated in the wires. •   The flux is contained entirely in the iron core; none escapes. All the energy in this magnetic field “makes it” from one coil to the other. •   The iron presents no “resistance” to having its magnetic domains flipped back and forth in step with the AC current that is developing the flux. How about real transformers? Modern transformers almost meet these conditions. Wire losses and core losses are small enough that most transformers dissipate only a few percent of the energy that passes through them. Alright, enough of the physics! How does this contraption work? Electrically, that is. First, let us use the more schematic symbol for the transformer. Figure 2.2 shows the usual way of drawing

transformers: edison lost  33

Figure 2.2: Transformer symbol.

a transformer. The core is reduced to the two vertical lines, which means that this is an iron-core transformer (as opposed to air core). The voltages and currents are the same as in the first drawing. The turns ratio is stated, which is often given as “a:1” or “1:a” instead of specifying actual turns. A transformer with a turns ratio of 1000:250 is more likely to be labeled “4:1.” The relationships between V1 and V2 and between I1 and I2 follow from Ampere’s and Faraday’s laws (which I will not do here):

N2 V2 = , V1 N1

I2 N1 = . I1 N2

These state simply that the voltage transfers as the turns ratio, the current as the inverse of the turns ratio. Fine, but the goal of using a transformer in a power system is to move power through the system. So what happens in the ideal transformer when we consider power? Let us look at power as simply the product of voltage and current as if phase angles were all zero. The power delivered to the left side of the transformer is

P1 = V1 I 1 and the power leaving the right side is

P2 = V2 I2 . If I use the ratios for voltage and current, I get � �� � N2 N1 P2 = V1 I1 = V1 I 1 . N1 N2 What goes in comes out! Remember, though, that this is an ideal transformer so it is lossless.

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Before going further, I need to introduce two terms: primary and secondary. These are terms for convenience to help us determine which way energy is moving. We generally mean that the primary is the side to which power is delivered and the secondary is where power “comes out.” Note that the transformer knows no such restriction and can support power flow in either direction.

2.1.1 Example I: Ideal Transformer Before tackling more realistic transformers, let us look at a circuit involving the ideal transformer. The circuit of Figure 2.3 includes an ideal 5:1 transformer. The two R-L impedances represent connecting line (wire) resistance and inductance. The 5-W resistor represents the load to be served. The generator voltage VS = 120 ∠ 0°V. The goal is to find the load voltage VL. I could solve this problem by writing equations on the primary side that include the transformer voltage and current (V1 and I1 of Figure 2.2), then writing equations on the secondary side that include the transformer voltage and current there (V2 and I2), and finally using the transformer ratios we have just seen. That process will yield several equations. But I am lazy and I would be happier with a simpler method. The simpler method is to do what is called “reflecting,” meaning that we can “pass through” the impedances and sources from one side of the transformer to the other. So, before I proceed with the example, I need to show what “reflect” means: •   Looking back at the equations of the previous section, I see that the voltage on, say, the “2” side reflects to the left as

V1 = V2

N1 . N2

•   Similarly, the current on the “2” side reflects as

I1 = I2

Figure 2.3: Example I.

N2 . N1

transformers: edison lost  35

Figure 2.4: Example I: reflected to primary.

•   Remembering that impedance is voltage divided by current, I will see what happens to the impedance as it reflects from the “2” to the “1” side:

Z2 =

V2 , I2

�

� N1 � �2 V2 N1 V1 N2 � � Z1 = = . = Z2 N I1 N2 2 I2 N1

I n words, this shows that an impedance on one side (here, the “2” side) reflects to the other side (here, the “1” side) as the turns ratio squared. Take a look at Figure 2.4 where I have reflected the three impedances and VL from the secondary (right) side through the transformer to the primary (left) side. The two resistances and the reactance have been multiplied by 52 and the voltage has been multiplied by 5. Now I can write a simple Ohm’s law equation to solve for I and from that obtain 5VL:

I=

120∠0◦ = 0.889∠ − 5.8◦ A, 3 + j5 + 6.25 + j8.75 + 125

5VL = (125)(0.889∠ − 5.8◦ ) = 111.2∠ − 5.8◦ V. I get the final result by reflecting the load voltage VL back to the secondary, which means dividing by the turns ratio:

VL = 5VL /5 = 22.2∠ − 5.8◦ V. To illustrate that this reflection process can be used in either direction, I am going to take the same example but reflect the primary side of the circuit through the transformer to the secondary side. I will divide the source voltage by the turns ratio, multiply the current by the turns ratio, and divide the impedances by the turns ratio squared. Figure 2.5 shows the result of this.

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Figure 2.5: Example I: reflected to secondary.

Now, let us find I and VL again:

5I =

120∠0◦ /5 = 4.446∠ − 5.8◦ A, 0.12 + j 0.2 + 0.25 + j 0.35 + 5

I = 0.889∠ − 5.8◦ A,

VL = (5)(4.446∠ − 5.8◦ ) = 22.2∠ − 5.8◦ V. Note that the current through the 5-W load is 5I, not I in Figure 2.5.

2.1.2 Example II: Ideal Again Here is one more example with the ideal transformer and lossy wires, this time with the load stated in power terms rather than as an impedance. The process will be the same, reflecting the secondary to the primary and then finding what is required. The difference, though, is in how the load is reflected. A load is a load! So the 50-kW load remains 50 kW no matter which side of the transformer it is on. Remember that power in equals power out for an ideal transformer. The load’s voltage and current both change when we reflect the load, but their product does not. Figure 2.6 is the circuit. We are to find Iline, Vgen, percent efficiency, and percent voltage regulation.

Figure 2.6: Example II.

transformers: edison lost  37

Figure 2.7: Example II: reflected to primary.

I reflect the secondary into the primary, the resistance and the reactance are multiplied by 27.52, the load voltage is multiplied by 27.5, and the load remains unchanged. The outcome is shown in Figure 2.7. I am going to use the load power to find the magnitude of the current Iline and the power factor to find its phase angle:

P load = |Vload | |I line | pf, 50 × 103 = 4.564 A, |I line | = (13.2 × 103 ) (0.83) ∠I line = −cos−1 pf = −33.9◦, I line = 4.564∠ − 33.9◦ A. Note the choice of the minus sign on the angle because the power factor is lagging so the current must lag the voltage. The generator voltage is the load voltage (taken with a phase angle of 0) plus the voltage drop in the impedances of the line:

Vgen = 13.2 × 103 + (30 + j50 + 75.6 + j151.3) Iline = 14.12∠2.0◦ kV. The losses in the line are due only to the resistances and the magnitude of the current, so the percent efficiency is

% h = 100 ×

50 × 103

(30 + 75.6) (4.564)2 +50 × 103

= 95.8% .

The percent voltage regulation is

% VR = 100 ×

14.12 − 13.2 = 7.0%. 13.2

38  Pragmatic power

2.2

TRANSFORMER MODEL

So far, the transformer has been portrayed as perfect. All it does is sit in the system happily reducing the voltage and increasing the current—or the other way around. It is happy because it is very “green” conscious, wasting no energy itself. Unrealistic, right? Yup! The transformer is no more perfect than anything else we use or do. The good news, though, is that the transformer is a very efficient device, dissipating only about 5% of the energy passing through. We can make a rather decent model of the transformer that includes the effects of its losses and take into account other “real world” characteristics at the same time.

2.2.1 Realistic Model Figure 2.8 is a good model of a real transformer, yet fairly simple. In the previous section, I listed the attributes of the ideal transformer. The circuit elements in the model in Figure 2.8 represent the nonidealness of the transformer. Let us look at each of the added elements to see how they reasonably account for what is happening in a real transformer. Let us start on the left and work across: •   R1 represents the losses in the wire of the primary. •   The reactance X1 is an inductance that accounts for the fact that the magnetic flux created by the primary coil does not quite stay entirely in the iron core. Some of the flux leaks out, and some of the flux does not even link all the turns of the winding. Hence, the primary winding looks to some extent like an inductor. •   Just as the winding introduces losses, so does the core, represented by the parallel resistance Rc. Iron is a great magnetic material with magnetic domains that flip from one direction to another quite easily. But not without some resistance. It takes some energy to flip the domains back and forth in step with the AC applied to the transformer. Rc accounts for this loss.

Figure 2.8: Transformer model.

transformers: edison lost  39

Figure 2.9: Transformer model reflected.

•   As domains are flipped one way or another, energy is being stored in the core in the same way that energy is stored in the magnetic field of an inductor. Hence, the core presents some inductance, called the magnetizing reactance, represented by the parallel reactance Xm. •   The transformer symbol in the model represents an ideal transformer. All of the nonideal characteristics are represented by the other six circuit elements. •   The secondary coil has characteristics similar to those of the primary, namely, resistive losses represented by R2 and leakage represented by X2. I suppose it is time to get worried. What started off as a simple, ideal device is now complicated by the addition of six circuit elements. It is beginning to look like the analysis of a circuit containing a real transformer is going to be hard work. Do not worry—things are not that bad because we can make some simplifications that reduce the number of elements in the model and also simplify the analysis. Remember that we learned how to reflect an impedance from one side of a transformer to the other? Let us do that here. Let us take the two elements on the secondary side and reflect them to the primary. Figure 2.9 shows this, with a prime designating the values of these elements after reflecting them through the transformer. Okay, but if you buy this as a simplification, I have some snake oil for sale, too! This does not simplify anything, really, until we learn more about the numerical values of these elements. I am

Figure 2.10: Transformer model reflected and simplified.

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Figure 2.11: Transformer model with numbers.

going to just state how these values relate and ask you to take it on faith until we look at numbers for an actual transformer, please. Here goes: If RC >> R1 or R2 and Xm >> X1 or X2, then I can move the parallel elements to the left. When I do this, R1, X1, R2′, and X2′ are all in series and can be combined into just two elements. Figure 2.10 shows the result of doing all this. This turns out to be a neat simplification that can be carried further when we actually embed a transformer in a circuit with other circuit elements. But how do I justify it? You saw my “much greater than” conditions, but how can we just move the parallel elements? The justification comes by looking at the currents. The current through the series R + jX is large because it is the current that is delivering energy to the right. The current through the parallel elements is much, much smaller, perhaps only about 1% of the current through R and jX. Therefore, moving the parallel elements does not really change the total current and hence does not change very much the effects those elements have on the accuracy of the model. Figure 2.11 shows our “adjusted” model with actual numbers. These numbers are typical for a 10-kVA transformer designed to transform 13.2 kV to 480 V. Notice how the numbers do meet my conditions for moving the parallel elements—they are large, given in kW, not W, which means that they are a thousand times larger than the series elements. How do we get these numbers? There are some simple measurements that we can make on a transformer that yield good results. We will look at transformer testing in the next section. Before that, though, let us use this model in two examples.

2.2.2 Example III A 480-V 10-kW load that operates with unity power factor is to be fed from a 13.2-kV supply using the 10-kVA transformer just modeled. The system is shown in Figure 2.12. Our job is to find the voltage needed at the generator, Vgen, and the overall efficiency of the system. Start by reflecting the load through the transformer to the primary. The only number that changes is the voltage. A 10-kW load is the same on either side of the transformer, so only the voltage of 480 V changes to the primary voltage of 13.2 kV. I will be lazy and not even draw the rearranged circuit.

transformers: edison lost  41

Figure 2.12: Example III.

The current through the reflected load is a good starting place, because from that I can get the voltage drop across the two series elements. That will let me get to the generator voltage. Recalling that P = |V||I|power factor,

I load =

P load 10 × 103 = = 0.7576∠0◦ A, 3 |Vload | pf 13.2 × 10 × 1

Vgen = 13.2 × 103 + (0.7576∠0◦ )(518 + j 1221) = 13.62∠3.9◦ kV.

To find efficiency, I need to know either the power delivered by the source or the losses in the circuit. If I want to do this using the source power, I will need the line current, which is the load current plus the current through the parallel elements. While doing this, I will get to see how much of a factor this parallel branch current is relative to the load current and get a feeling for whether moving the parallel branch to the left was a good approximation.

Iparallel =

Vgen 3

+

Vgen

581 × 10 j694 × 103 = 30.58∠ − 36.0◦ mA.

Let us note here that this current, flowing through the parallel R-L combination, is only about 4% of the load current, so moving it from one side of the series R-L to the other (Figure 2.10) is not a significant change. Now that I have the current through the parallel branch, I can get the line current and from that, the power supplied by the generator:

I line = I load + I parallel = 0.7576∠0◦ + 30.58 × 10−3 ∠ − 36.0◦ Pgen

= 0.7825∠ − 1.3◦ A, � � = �Vgen � |I line | cos (qv − qi )

= (13.62)(0.7825)cos(3.9◦ − (−1.3◦ )) = 10.61 kW.

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The percent efficiency is

%h = 100 ×

10 = 94.2%. 10.61

One way to check this is to figure out the losses instead of the power supplied by the generator. Sometimes this is easier to do, too. The losses are represented by just the two resistance elements. I know the current through the series resistance and I know the voltage across the parallel resistance, so I will use I-squared-R and V-squared-over-R to get the power lost. Note that phase angles do not enter into this calculation because a resistor’s voltage and current are in phase.

P loss = (518) (|I load |)2 + = 617 W. %h = 100 ×

(|Vgen )2

581 × 103

10 = 94.2%. 10 + 0.617

Close enough!

2.2.3 Example IV One more example will illustrate transformer power calculations using the move-the-parallel-branch approximation. The circuit of Figure 2.13 includes between the terminals a 50-kVA, 13.2-kV/480-V transformer. Measurements and the approximation of Figure 2.10 have yielded series values of 90 and j220 W and parallel values of 120 and j200 kW. The load here is a 480-V, 50-kW load with a power factor of 0.83 lagging. This load is fed from the secondary by a line with an impedance of 0.1 + j0.2 W. The transformer’s primary is fed by a line with an impedance of 30 + j50 W. Our job is the usual: find the necessary generator voltage, the overall efficiency of the circuit, and the voltage regulation.

Figure 2.13: Example IV.

transformers: edison lost  43

Figure 2.14: Example IV simplified.

Simplifying the circuit seems like a good idea, so I will first reflect the load and its feeder to the primary. After reflection, the load remains 50 kW at 0.83 lagging, but its voltage becomes 13.2 kV. The line impedance is multiplied by the ratio squared, which yields 75.6 + j151.3 W. I could add that result to 90 + j220 W, but I am going to take the approximation one step further. I will move the parallel elements all the way to the left, past 30 + j50 W. After all, if some moving is good, more should be better! When I make this move, then all three R-L series combinations simply add. The result is shown in Figure 2.14. Now the problem is not much different from the previous example. First, the load current:

|I load | =

50 × 103

(13.2 × 103 )(0.83)

= 4.564 A,

∠I load = −cos−1 0.83 = −33.9◦, I load = 4.564∠ − 33.9◦ A.

Now the generator voltage:

Vgen = 13.2 × 103 + (4.564∠ − 33.9◦ )(195.6 + j421.3) = 15.05∠4.2◦ kV.

Next, power loss: 2

P loss = (195.6)(4.564) + = 5.96 kW.

� �2 15.05 × 103 120 × 103

Finally, percent efficiency and percent voltage regulation:

50 = 89.4%. 50 + 5.96 15.05 − 13.2 = 14.0%. % VR = 100 × 13.2 % h = 100 ×

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Figure 2.15: Example IV not quite so simplified.

This is a good place to see how much the move-the-parallel-branch approximation changes the final results. I am going to rework the example with the secondary load and feeder impedance reflected to the primary but without moving the parallel elements further to the left. Figure 2.15 shows this not-so-simplified circuit. The load current is the same as before, namely, 4.564 ∠ - 33.9°A . From this, I find the voltage across the parallel branches:

Vparallel = 13.2 × 103 + (4.564∠ − 33.9◦ )(165.6 + j 371.3) = 14.81∠3.8◦ kV.

The line current is the sum of the load current and the current through the parallel branches:

I line = 4.564∠ − 33.9◦ +

14.81∠3.8◦ × 103

= 4.703∠ − 33.6◦ A.

120 × 10

3

+

14.81∠3.8◦ × 103 j 220 × 103

Now get the generator voltage by adding the voltage drop across the series R-L elements and the voltage across the parallel branches:

Vgen = 14.81∠3.8◦ × 103 + (4.703∠ − 33.6◦ )(30 + j 50) = 15.07∠4.2◦ kV.

(The other approach yielded 15.05 kV.) The power losses are

P loss = (165.6) (4.564)2 + = 5.94 kW,

(14.81 × 103 )2 120 × 103

+ (4.703)2 (30)

which is slightly smaller than the 5.96-kW losses computed after moving the parallel branches. The percent efficiency is the same:

transformers: edison lost  45

% h = 100 ×

50 = 89.4% . 50 + 5.92

In other words, moving the parallel branches around makes very little difference in the results but simplifies the computations enough to make the move worthwhile. Two notes about the “real world:” (1) The numbers on my sample transformers are for rather lossy transformers; real transformers generally are better, meaning losses are smaller. (2) Practicing engineers often completely omit the parallel R-L branch from model calculations except for very large transformers. Now that we have used models for a couple of transformers, it is time to see how we obtain data for those models. The next section will show how.

2.3

TRANSFORMER TESTING

Our transformer model is rather simple, series resistance and inductance representing the actual coil, parallel resistance representing core loss, and parallel inductance representing magnetization effects. But how can we get numerical values for these? The coils and the core are all there is, and we need to somehow measure these parameters from the outside. The approach to getting these numbers is in a way like a physician’s determining what is going on in your heart by observing voltages from outside the body. We simply have to find combinations of external voltage and current measurements that give us a handle on the elements themselves. The method relies on the fact that the numerical values of the two parallel elements in the model are many times the numerical values of the series elements. This “many times” is several hundred. Hence, we can devise tests that “see” rather clearly the series elements without influence of the parallel ones and vice versa. There are two parts to a transformer test: the short-circuit test and the open-circuit test. These tests are done with knowledge of the rated characteristics of the transformer, namely, the rated current and the rated voltage. Both of these can be obtained from the nameplate data. A transformer has a stated kVA rating as well as a stated voltage ratio. For example, one of the transformers in the previous section has been a 10-kVA, 13.2-kV/480-V transformer. From these data, we can obtain the rated current for both sides of the transformer.

2.3.1 Short-Circuit Test I will start with the short-circuit test. This is normally done on the high-voltage side of the transformer because the rated current on this side is smaller than on the low-voltage side. The shortcircuit test requires three meters: an ammeter, a wattmeter, and a voltmeter. Figure 2.16 shows how the meters are arranged. Note that the low-voltage side of the transformer is shorted out, which is where this test gets its name. Let us see what data we collect on our 10-kVA, 13.2-kV/480-V transformer.

46  Pragmatic power

Figure 2.16: Short-circuit transformer test.

First we need to determine rated current on the high-voltage side:

|Srated | on the high-voltage side Vrated 10 × 103 = = 0.758 A. 13.2 × 103

Irated =

Now we adjust the supply voltage Vs until the ammeter reads 0.758 A. Note that rated current is now flowing in the coil of the high-voltage side, which means that rated current is also flowing in the short circuit on the low-voltage side. That short-circuit current is

Ishort-circuit =

13.2 × 103 0.758 = 20.85 A, 480

although we do not need to know that for this test. Once we have the current set to the rated value, we read the meters. Suppose the outcomes are

Isc = 0.758 A, Psc = 298 W, Vsc = 1005 V. From these three pieces of data, we can calculate the values of the series elements R and X. But, you might ask, how can you ignore the two parallel elements? First, they are going to be several hundred times the sizes of R and X. Second, the applied voltage (1005 V) is much smaller than the rated voltage of 13.2 kV, so the currents through the parallel elements and the power dissipated in them will be much smaller than they would be at rated voltage. Here are the three steps in the calculations for R and X based on the data from the shortcircuit test:

transformers: edison lost  47

1.   Calculate the power factor:

pfsc =

Psc 298 = = 0.391, Vsc I sc (1005)(0.758)

∠I sc = − cos−1 0.391 = −67.0 ◦. 2.  Determine the phasor current:

|Isc | = 0.758A, ∠Isc = −cos−1 0.391 = −67.0◦ , Isc = 0.758∠ − 67.0◦ A.  ote that the angle must be chosen to be negative because the current must lag the voltage N in an inductive circuit. 3.  Compute the series impedance in rectangular form, completely ignoring the parallel elements in the model: Vsc 1005 Zseries = = = 1326∠67.0◦ W Isc 0.758∠ − 67.0◦

= 518 + j1221 W, R = 518 W, X = 1221 W.

There! That takes care of the series elements, so we are half done with the tests.

2.3.2 Open-Circuit Test The open-circuit test uses the same meter arrangement, but it is usually done on the low-voltage side of the transformer as shown in Figure 2.17. Notice that the terminals on the opposite side are not connected to anything. To begin, we set the source voltage to the rated voltage on the low-voltage side, so we set Vs = 480 V. If we want to check the transformer’s turns ratio, we would measure the voltage Vo. Now we read the three meters. Suppose the outcomes are

Figure 2.17: Open-circuit transformer test.

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Voc = 480 V, Poc = 397 mW, Ioc = 1.078 mA. (The voltage across the primary terminals, Vo, is now 13.2 kV, so safety is a major factor in these tests.) The calculations are just about the same as before: 1.    Calculate the power factor:

pfoc =

Poc 397 × 10−3 = = 0.767. Voc Ioc (480)(1.078 × 10−3 )

2.  Determine the phasor current in rectangular form:

|Ioc | = 1.078 mA, ∠Isc = −cos−1 0.767 = −39.9◦ , Ioc = 1.078∠ − 39.9◦ = 0.826 − j0.692 mA. 3.  From the real and imaginary parts of the open-circuit current, calculate Rc and Xm:

Voc 480 = 581 kW, = Re[Ioc ] 0.826 × 10−3 Voc 480 Xm = = 694 kW. = −Im[Ioc ] 0.692 × 10−3 Rc =

Now we have the values for the four elements of our transformer’s model, Figure 2.18. Why are these tests made from opposite sides of the transformer? Would not it be easier to do all this with everything set up on just one side? Sure. But there is a reason for the two choices. The short-circuit test is made from the high-voltage side because this requires a smaller current. The

Figure 2.18: Results of transformer tests.

transformers: edison lost  49

open-circuit test is made from the low-voltage side because this requires a smaller voltage, which is safer. The high-voltage terminals can be completely guarded against accidental contact.

2.4

THREE-PHASE EXAMPLE

Example V is a three-phase system that is to supply a three-phase, ∆-connected load from a 13.2-kV feeder. The load is rated at 480 V, 25 kW at a power factor of 0.9 lagging. The lines to the load have an impedance of 0.2 + j0.5 W/∅. We are to provide the transformer and then find the required primary voltage, the overall efficiency, and the voltage regulation.

2.4.1 Example V Using Y–Y Figure 2.19 shows the complete three-phase system. However, once I get some parameters established, I will reduce this to its single-phase equivalent for my calculations. The total load is 25 kW at 0.9 lagging. This means the total power is

Stotal =

Ptotal 25 = = 27.78 kVA. pf 0.9

This is a balanced three-phase system, so I can use three 10-kVA, 13.2-kV/480-V single-phase transformers that I chose to arrange Y–Y. (Remember that the stated voltage for three-phase systems is always the line voltage, the voltage between the lines, so the coils are seeing less than rated voltage.) Each load in the circuit is 25/3 kW at 0.9 lagging.

Figure 2.19: Example V.

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Figure 2.20: Example V: single-phase equivalent.

I will start by converting this circuit to its single-phase equivalent. The load is unchanged, but its voltage per phase must be divided by √3. The line impedance is the same for one line. The single-phase equivalent is Figure 2.20. The next step is to simplify this by reflecting the load and its line impedance through the transformer as shown in Figure 2.21. The load current is P load 8.33 × 103 |I load | = = = 1.215 A, |Vload | pf (7620)(0.9)

∠I load = −cos−1 0.9 = −25.8◦ , I load = 1.215∠ − 25.8◦ A.

The supply voltage Vs does not depend on the parallel elements:

Vs = 7620 + (1.215∠ − 25.8◦ )(518 + j1221 + 151 + j 378) = 9303∠ 8.6◦ V.

The losses are

Ploss = 1.2152 (518 + 151) +

Figure 2.21: Example V: move to primary.

93032 581 × 103

= 1.137 kW,

transformers: edison lost  51

so the efficiency is

%h = 100 ×

8.33 = 88.0%. 8.33 + 1.137

Finally, the voltage regulation is

% VR = 100 ×

9303 − 7620 = 22.1%. 7620

The job is not finished, though, because there are two things to do. First, I need to convert the supply voltage (the voltage on the primary size of the Y–Y transformers) back to the three-phase system:

√ Vs = 9303 3 = 16.11 kV. Second, there is a question of whether this load is exceeding the rated capacity of the transformers. Huh? How can that be, because each phase of the load is only 8.33 kW and the transformers are rated at 10 kVA? Yes, but we are operating them at below their rated voltage, so the current could be above the rated value. The load current on the primary side is

|Iload | = 1.215 A, so the total power on the primary side of one of these transformers is

|Stransformer | = 7620 × 1.215 = 9.26 kVA. Whew! That is less than 10 kVA per phase, so everything looks okay. But things are not all that good. Look back at the efficiency (88.0%) and the percent voltage regulation (22.1%). Those are rather poor numbers. Now, consider the primary current rating:

|Irated | =

|Srated | 10 × 103 = = 0.758 A |Vrated | 13.2 × 103

Our primary current of 1.215 A exceeds this by 60%!

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What is the problem? Wrong transformer! The transformer is underused. It is rated at 13.2 kV/480 V and we are using it at 7620/277. Same ratio, but the current is higher, whereas the voltage is lower. Higher current means higher losses in the series elements. Now what?

2.4.2 Example V Using D–D If I am stuck with these 10-kVA transformers, then I should use them at somewhere near their rated voltage and current. Because the load requires a line voltage of 480 V, I can connect the transformers D–D so their secondary voltage is 480 V. Figure 2.22 is the same load with its feeders, but now supplied by three 10-kVA transformers arranged as D–D. As before, I will convert this to its single-phase equivalent. The only difference from what we saw before is that the delta-connected source also needs to be transformed. This is done by dividing the voltage by √3 as usual and dividing the impedances of the transformer model by 3. Figure 2.23 is the result. The calculations proceed as before with the following results:

I load = 1.215∠ − 25.8◦ A,

Vs = 7620 + (1.215∠ − 25.8◦ )(173 + j 407 + 151 + j378) = 8418∠4.7◦ V,

Ploss = 1.2152 (173 + 151) +

84182 = 0.844 kW, 194 × 103

8.33 = 90.8%, 8.33 + 0.844 8418 − 7620 % VR = 100 × = 10.5%. 7620 % h = 100 ×

Figure 2.22: Example V as ∆–∆.

transformers: edison lost  53

Figure 2.23: Example V: move to primary again.

Not bad! Efficiency is up a little and voltage regulation is greatly improved. All that remains is to convert the supply voltage back to the three-phase circuit:

√ Vs = 8418 3 = 14.58 kV. But have I exceeded any ratings? I must first transform the circuit back to the three-phase delta: √ � � �Iprimary � = 1.215/ 3 = 0.701 A

We have already found the rated current is 0.758 A, so the current is okay. How about the primary kVA for one transformer?

|Stransformer | = (14.58 × 103 )(0.701) = 10.22 kVA That is about 2% larger than the 10-kVA rating but it is acceptable as long as the transformer operates with adequate cooling so that it stays within temperature specifications. Can we do even better? Not much without finding a more efficient transformer.

2.5

SUMMARY

If Nikola Tesla and George Westinghouse had not strongly disagreed with Thomas Edison, we might not have needed this chapter. The transformer is the key to the power system we have today. The ideal transformer, a lossless device, is not real, but the modern transformer is pretty close to it, dissipating only about 5% of the power passing through. An actual transformer can be modeled very well with only four elements: series resistance to account for the wire losses, series inductance to account for leakage flux, parallel resistance to represent losses in the iron core, and parallel inductance to represent the inductive effect of magnetizing iron. This simple model lends itself very well to a straightforward testing scheme that yields values of the four elements without a great deal of work. The short-circuit and open-circuit tests require only standard instruments and a facility that can provide the voltages and currents (and safety!)

54  Pragmatic power

needed. Once the model parameters have been determined, the model can be used to represent the transformer and permit circuit analysis using our usual techniques. Yet even though Edison lost, a strange thing has been happening more and more in recent years. We are running more equipment on DC! How so? Well, your computer and your digital TV and probably your stereo and certainly your DVD player are all devices that internally use DC power. Even stranger is that we run them from a DC supply—almost. We buy an uninterruptible power supply (UPS) so that power glitches do not kill our computers. But what is a UPS but a DC supply that takes in AC, charges batteries, produces DC, and then converts the DC back to AC. Where next? Tesla rides again! He invented the induction motor, which in its three-phase form is a spectacularly simple machine—one moving part. Just about every significant load that requires a rotating shaft is driven by a three-phase induction motor. Hence, in a study of AC power, we will take a good look at this induction motor in the next chapter. • • • •

55

chapter 3

Induction Motors: Just One Moving Part Nikola Tesla’s alternating current (AC) system beat Edison’s direct current (DC) system not only because of the transformer but also because of the induction motor, which Tesla invented. Tesla’s patent (No. 416,195, December 3, 1889) has a drawing of a three-phase motor, but his claim is for “two or more energizing-circuits through which alternating currents differing in phase are caused to pass.” The beauty of Tesla’s invention does not lie in the various arrangements of the parts of the motor so much as in the fact that the three-phase induction motor has just one moving part. You will work a long time to develop a rotating prime mover with fewer! This chapter is going to stick with the three-phase induction motor. This motor is, by far, the most common method of transforming electrical energy into rotational energy. The motor is simple, rugged, and surprisingly efficient. We will start by looking at how the motor works and develop an equivalent circuit for it. Then we will look at details such as nameplate data, testing, and energy flow. We will end with some example calculations and a discussion of why a designer should not “over-motor.”

3.1

ROTATING FIELDS

The three-phase induction motor creates a rotating magnetic field that drags an armature around with it. To explain this, though, I am going to start with a simple one-phase structure and see how that could work.

3.1.1 Single-Phase Motor The two poles of the motor in Figure 3.1 are energized by a current flowing such that the top pole is magnetic north and the bottom pole is south. These two poles form the stator, which is the nonmoving part of the motor. For the moment, assume that this magnetization is being done by a steady current. The moving part in Figure 3.1 is the rotor, which in this illustration is a permanent magnet. If the rotor is started in the position shown in the drawing, it will experience a clockwise torque

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FIGURE 3.1: Clockwise torque.

because, as we learned in high school physics, “like poles repel, unlike poles attract.” This rotor, however, is not going to get very far as shown in Figure 3.2. The “unlike poles” will get lined up and the torque will cease. But suppose we reverse the magnetic field of the poles as the rotor is arriving. Figure 3.3 shows this and assumes that the rotor has some inertia that carries the magnet past the lower pole so that the new pole arrangement can continue to provide torque.

FIGURE 3.2: Torque continues.

introduction motors: just one moving part  57

FIGURE 3.3: Field reversal.

Now let us provide the magnetic field of the stator using AC. Then the poles switch back and forth between N–S and S–N. It sounds like the rotor ought to follow that reversing field. Actually, it will—but it will not start by itself, which is not good.

3.1.2 Three-Phase Rotating Magnetic Field Tesla’s invention was for a “multiphase motor.” The most common arrangement is three phases. Figure 3.4 shows the motor with two pairs of poles added. The three pairs are spaced uniformly around the circumference of the stator. Each pair of poles is energized by one phase of a three-phase system. In the arrangement shown in the drawing, not only are the poles spaced 120° apart around the circumference, but they are energized by successive phases of the power system that are also spaced 120° apart. (That is an observer standing in there to help us later.) All three pairs of poles are the same, so the magnetomotive forces (mmf ) developed by the pairs of poles all have the same magnitude. The mmfs for the three pairs differ because they are excited by different phases of the power line and because they are in different positions around the circumference. The equations below all have the same mmf magnitude (F ). The angle θ is measured clockwise starting at the north pole of the a phase. The observer is stationed at an angle θ relative to the north pole for the a phase. Hence, relative to the b phase’s north pole, the observer is “behind” by 120°. For the c phase, the observer is 240° “behind.” The magnetic flux Φ generated by each of the pole pairs is

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FIGURE 3.4: Three-phase field.

Fa = F cos(w t)cosq , Fb = F cos(w t − 120◦ )cos(q − 120◦ ), Fc = F cos(w t − 240◦ )cos(q − 240◦ ).

I will skip all the trig relationships to get to the final result:

Ftotal =

3 Fcos(w t − q ). 2

This says two things. At any instant, the amplitude of the flux is a constant and the flux is distributed around the stator as cos(−θ) is spread around the circumference. If the observer stands still as shown in Figure 3.4, the observer sees a magnetic flux that seems to vary in amplitude as cos(ωt) varies. But more important, this field is rotating uniformly around the stator. No matter where the observer is standing, the flux appears to vary in amplitude as cos(ωt) varies. An observer standing at θ = 0° will see a flux varying as cos(ωt). An observer standing at θ = 10° will see a flux varying as cos(ωt – 10°). In other words, this second observer will see the same flux as the first observer, but 10° later. Hence, the flux is rotating clockwise around the stator in step with the frequency of the power line. At 60 Hz, the flux rotates clockwise around the stator at 60 rev/s, which is 3600 rpm.

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FIGURE 3.5: Add a rotor.

The direction of rotation is established by the angles for the three phases. Here, they are 0°, –120°, and –240°. This order is called abc rotation. If it were the other direction, it would be called acb rotation.

3.1.3 Add a Rotor Now that we have magnetic flux that is rotating around the stator, let us add a rotor and see what happens to it. Figure 3.5 shows the added rotor. There is just one catch. If the rotor is standing still, it sees the flux as simply oscillating at the power-line frequency, just as the observer did. Therefore, because the rotor experiences no net torque, so it just sits there doing nothing useful. That is no better than our one-pole-pair motor in Figure 3.3. One way we can help the rotor to develop torque is to get it spinning at the same speed that the flux is rotating around the stator. At 60 Hz, we must set it to spinning at 3600 rpm, a condition that develops torque by following the peak of the flux around the stator. As we require more torque from this motion (which is presumably driving some type of load through a shaft), the rotor will lag behind the peak by a larger and larger angle. If we require too much torque, the rotor will “break loose” from the torque peak and will not produce any more net torque. A motor fabricated this way is called a synchronous motor. This type of motor is not very common, but is sometimes used where a load requires torque at exactly the synchronous speed. If we add

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FIGURE 3.6: Add a working rotor.

three more pairs of poles to the stator of Figure 3.4, thereby spacing the phases 60° apart instead of 120°, we find that a 60-Hz magnetic field rotates around the stator at a synchronous speed of 1800 rpm. If we add three more pairs, the synchronous speed becomes 1200 rpm, and so on. Let us add a rotor that works. Instead of just a magnet, this rotor is going to have coils of wire in its circumference. But these coils do not go “around the rotor.” Instead, they go lengthwise. In Figure 3.6, the coil marked X on the left side of the rotor goes around the end of the rotor and comes back as X′ on the right side. The coil's ends are all tied together to form numerous shortcircuited loops. Using this new rotor arrangement, we have a situation where the three-phase supply creates a rotating field around the stator. This changing field induces a current in the rotor windings. This rotor current acts with the stator field to develop a circumferential torque. This torque tries to make the rotor “catch up” with the rotating stator field. But the rotor cannot catch up with the stator field! If there is no relative rotation, there is no induced current. If there is no induced current, there is no interaction between the rotor and the stator. If there is no interaction, there is no torque. To work, the rotor must slip relative to the rotation of the stator flux. Figure 3.7 shows a three-phase induction motor with part cutaway so some of the stator winding and some of the rotor with its winding are visible. The winding is very simple because it is just cast aluminum bars in slots in the rotor. The bars are shorted together at each end. Figure 3.8 shows a portion of these bars without the iron of the rotor. The picture shows the short circuit at one end of the bars (The short at the other end has been cut off.). Notice the protrusions sticking out of the end that stir the air and provide some cooling inside the motor.

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FIGURE 3.7: Rotor cutaway view.

3.1.4 Slip, Speed, and Poles The rotor in an induction motor cannot travel as fast as the rotating field or there would be no torque. The rotor is said to slip relative to the rotation of the field. To define this slip, however, we need first to look at the rotational speed of the field. A motor is described by stating the number of poles per phase. The three-phase motor in Figure 3.6 has six distinct poles, but there are just two per phase. Hence, this motor is said to be a “two-pole motor.” Poles always appear in pairs.

FIGURE 3.8: Aluminum rotor bars without iron.

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Synchronous speed is the speed at which the magnetic field of the stator rotates around the stator. If it were not for slip, this is the speed at which the rotor would revolve. The synchronous speed for a three-phase motor is

ns =

60f , p/2

where f is the frequency of the power system and p is the number of poles per phase. In North America, the frequency is 60 Hz, so the synchronous speed is

ns =

3600 . p/2

Slip s is the fraction by which the actual rotor speed nm differs from the synchronous speed ns:

s=

ns − nm . ns

For example, a two-pole motor whose rotor is spinning at 3420 rpm has a slip of

s=

3600 − 3420 = 0.05, 3600

which is normally expressed as a percentage: “5% slip.” Suppose we have a six-pole motor operating with slip of 4%. The motor speed nm is

ns =

3600 = 1200, 6/2

nm = 1200 (1 − 0.04) = 1152 rpm. Three-phase induction motors commonly operate with 2–5% slip.

3.2

EQUIVALENT CIRCUIT

An equivalent circuit is helpful in understanding how energy flows through the induction motor. The equivalent circuit is going to be a per-phase equivalent just as we did with balanced three-phase systems in Chapter 1. The equivalent is on the Y basis, which means that voltages are phase voltages. It is easy to forget that phase voltages are line voltages divided by √3. When we developed an equivalent circuit for the transformer, we took into account two types of losses: copper and core. We also included leakage and magnetization reactances. We need all four of these in the induction-motor equivalent as well. The magnetic part of the induction motor is quite different from that of the transformer, though. The induction motor has an airgap that increases by a large amount the magnetic reluctance of the magnetic part of the system. In addition,

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the frequency of the induced current in the rotor is much less than the line frequency because of slip. I am going to develop the model in two steps, considering the stator first and then the rotor.

3.2.1 Stator Equivalent Circuit The per-phase equivalent circuit for the stator of the induction motor needs to include copper losses, leakage reactance, core (iron) losses, and magnetization reactance. Recall from the transformer model that leakage reactance represents magnetic flux that does not stay in the iron and that magnetization reactance represents energy stored in the magnetic domains of the iron. (See Section 2.2.1.) The circuit shown in Figure 3.9 is the single-phase equivalent of the stator of the induction motor. Note that the components are similar to those in the transformer model. Voltage E2 shown on the right is the voltage that is induced into the rotor by the stator.

3.2.2 Rotor Equivalent Circuit The rotor is more complicated because slip is involved. Slip determines the frequency of the current in the rotor coil. When the rotor is completely stopped, the slip is 1 and the rotor “sees” 60 Hz from the stator. When the rotor is going at synchronous speed, the slip is 0 and the rotor “sees” 0 Hz from the stator. The motor produces no torque when the slip is 0. Figure 3.10 shows one form of the rotor’s equivalent circuit. It includes both the “copper” loss of the rotor winding (which is commonly aluminum) and the reactance of the rotor coil. Two things stand out about this model. First, reactance X2 is the 60-Hz reactance so that we have everything referenced to a common frequency. The frequency in the rotor winding, however, is

FIGURE 3.9: Single-phase stator equivalent.

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FIGURE 3.10: Rotor equivalent.

much lower. If the slip is 0.03, the rotor frequency is 0.03 × 60 = 1.8 Hz. This is why the reactance is shown as jsX2. Second, the voltage induced in the rotor winding depends on the slip, too. Hence the rotor voltage is shown as sE2. This model is being viewed from the standpoint of the rotor itself. We need the model as viewed from the stator circuit. This is similar to what we did with the transformer model, namely, reflecting the secondary model into the primary. To make the transition to a model as seen from the stator, I first need the rotor current:

I2 =

sE2 E2 = . R R2 + jsX2 2 + jX2 s

Now I can model the rotor slightly differently: However, there is something missing in this model. How does power that the motor produces get through from the stator to the rotor? The model shown in Figure 3.11 represents the rotor as

FIGURE 3.11: Rotor equivalent adjusted.

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FIGURE 3.12: Rotor with rotation power.

seen from the stator. In this model, the only element that could absorb power is R2/s. Power delivered to this resistor must represent both the output power of the motor and the copper losses of the rotor. This power is the power delivered across the airgap by the stator:

Pairgap = |I2 |2

R2 . s

However, R2 itself, not R2/s, represents the copper losses:

PCu = |I2 |2 R2 . The motor’s output power must therefore be the power delivered across the airgap less the power lost in the winding: � � 1−s Pmotor = Pairgap − PCu = |I2 |2 R2 . s Now I can finish the rotor model by leaving R2 as the copper loss and adding another resistor to represent the power delivered to the shaft. Figure 3.12 shows the complete rotor model. The resistor that represents shaft power has an arrow through it to indicate that it is a variable that is controlled by slip s.

3.2.3 Complete Equivalent Circuit Now I can combine the stator and the rotor by “reflecting” the rotor impedance across the airgap to the stator. The induced voltage E2 remains the same because I divided it by s earlier. The rest of the elements in the rotor model acquire a prime to indicate they are the reflected values. Figure 3.13 shows the complete single-phase equivalent of the induction motor. When I finished the model of the transformer, I said that I could generally move the parallel branch further to the left because the values of Rc and Xm were generally much, much larger

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FIGURE 3.13: Complete single-phase equivalent with rotor reflected.

than the series resistance and reactance. This assumption is not a good one for induction motors because Xm is not “much, much larger.” It includes a large airgap that was not present in the transformer. We will make use of this model a little later, but before doing that, I am going to take a look at one real motor and its nameplate. I am also going to show how we can test a motor to find out what numbers should go into the equivalent circuit.

3.3 MOTOR NAMEPLATE The motor in Figure 3.14 is hidden in an equipment room not far from my office. It has been driving an HVAC fan for about 20 years. The only service that it is likely to have required is replacement of the belts from time to time.

FIGURE 3.14: The motor.

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FIGURE 3.15: Motor’s nameplate redrawn.

With the help of a flashlight and a mirror, I was able to read all the information on the nameplate, which I have reproduced in Figure 3.15. Motor nameplates bear standard information about the motor. What is written there is specified by the National Electrical Manufacturers Association (NEMA) standard for motor nameplates, the “NEMA Standards Publication Motors and Generators,” MG1-1998. NEMA was formed by industry authorities in 1926 to provide “a forum for the standardization of electrical equipment, enabling consumers to select from a range of safe, effective, and compatible electrical products.” Let us go through this nameplate line by line: •

•

• • •

Model is the manufacturer’s assigned model number. A motor with a model number similar to this one appears in Marathon Electric’s catalog today, although the design has been improved to increase efficiency. Frame is a code for the critical physical dimensions of the motor. The simplest way to describe frame is to say that any motor with this number will fit as a replacement. This applies to mounting holes, shaft position and diameter, keyways, and so on. If the motor in the picture needs to be replaced, any motor with a 213T frame should fit. Type is the manufacturer’s designation. Des stands for Design. There are four standard NEMA designs, which have four different torque-versus-speed characteristics. We will look at these later. Ph is Phase, and this is a three-phase motor.

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•

• • • • • • • • • • •

• •

• •

•

Code specifies the locked-rotor kVA of the motor. “Locked-rotor” means that the shaft is totally restrained so it cannot turn. Code H means this motor has a locked-rotor kVA in the range of 7.1–8.0 kVA per rated horsepower. Because this is a 7 ½-hp motor, it could absorb as much as 60 kVA when full voltage is applied with the rotor completely locked. Ins Cl is Insulation Class and specifies the maximum temperature that the winding can properly withstand. Class H insulation is limited to 180°C. Encl is the description of the Enclosure of the motor. This motor is drip-proof, meaning that water dripping down on it will not get inside. It is not waterproof, though. Duty is CONTinous, but the motor must be operated within specifications, especially load and temperature. Max Amb is the maximum ambient temperature, which refers to the air temperature around the motor. Shaft End Brg is the code for the ball bearing at the shaft end. A 307 bearing has a 35-mm bore, an 80-mm outside diameter, and a thickness of 21 mm. Opp End Brg is the bearing at the other end, 30 × 62 × 16 mm. Volt specifies the operating voltage, here dual voltages selected by how the motor’s multiple windings are connected to the power source. Amp specifies the full-load current corresponding to the chosen operating voltage. RPM is the shaft speed at full rated load. For this motor, rated slip is 2.78%. Hz is 60. NEMA Nom Eff is the NEMA rated efficiency. All motor losses must be included. The percent stated must be no larger than “the average efficiency of a large population of motors of the same design.” A modern energy-efficient motor of this size should be at least 88.5% efficient by NEMA standards. Nom pf is the nominal power factor at full rated load. Max Cap kVAR is the largest capacitor bank, measured in kilovars, that can be used to correct the power factor of this motor. This capacitor bank will correct the power factor to unity. HP is the full-load horsepower. Standard medium-sized motors come with ratings of 1, 1.5, 2, 3, 5, 7.5, 10, etc. In other words, you cannot order a 6-hp motor. See Section 3.8.4. SF is the service factor, which specifies by how much the motor can be overloaded and still operate continuously. Medium-sized motors usually have a service factor of 1.15, which means they can operate at 15% above rated full load provided other nameplate conditions are met (especially ambient temperature). Corr Amp gives the full-load current values when the power factor is corrected to unity.

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3.4

TESTING

Transformer tests in the previous chapter measured voltage, current, and power under two conditions, short-circuit and open-circuit. These tests yielded enough information to calculate the values of the transformer model’s circuit elements. Motor testing is only slightly different and is the equivalent of the short-circuit and open-circuit transformer tests. The output of a motor is through its shaft. The two basic tests are done with the shaft completely blocked so it cannot rotate and with the shaft completely unloaded. We measure three-phase voltage, current, and power under each of these conditions. However, we need one additional test, DC resistance, so we can obtain all the circuit parameters. In the following description of these tests, I am using data from a 460-V, three-phase, 60-Hz, 50-hp, 1770-rpm induction motor. This particular motor is a modern, energy-efficient motor. All data have been taken using three-phase metering, measuring line voltage, line current, power for all three phases, and power factor. Because the motor’s equivalent circuit is the Y equivalent, we will have to adjust the collected data to the single-phase Y model.

3.4.1 DC Test A DC resistance test provides an estimate of the copper-loss resistance in the stator part of the model. At DC, the only resistance seen from the motor terminals is R1 because both X1 and Xm look like short circuits at DC and because there is no induction across the airgap. The DC measurements are made between the line terminals of the motor. This means that the measurement sees two legs of the winding in series. Hence, the resistance measured will be twice the value of R1 in the model. The DC data are

Vdc = 5.0 V, Idc = 28.4 A. The steps to obtain R1 are: 1. Compute the total DC resistance between two phase terminals:

Rdc =

Vdc 5.0 = = 0.1761 W. Idc 28.4

2.  Divide this resistance by 2 to obtain R1:

R1 = Rdc /2 = 0.1761/2 = 0.088 W.

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3.4.2 Blocked-Rotor Test During the blocked-rotor test, the shaft is firmly blocked so that it cannot rotate. Then the motor input current is increased to its rated value. Because the input voltage will still be quite low, well below rated value, model elements Rc and Xm will have minimal effect. As the slip is now 100%, the value of the output resistor R2′(1 − s)/s = 0, a short circuit. Figure 3.16 shows the equivalent circuit under these test conditions with Rc, Xm, and the output resistor omitted. The data collected for the motor under test are

V = 114.8 V, I = 58.0 A (rated current), P = 1589 W, pf = 13.78% lagging. Blocked-rotor calculations proceed as follows: 1. Convert data to single-phase Y values: √ Vph = 114.8/ 3 = 66.28 V,

Iph = 58.0∠ − cos−1 pf = 58.0∠ − 82.08◦ A,

Pph = 1589/3 = 529.7 W.

2.  Calculate the impedance, which is R1 + jX1 + R2′ + jX2′: Vph 66.28 Zb = = = 1.143∠82.08◦ = 0.1575 + j 1.132 W, Iph 58.0∠ − 82.08◦

R1 + R�2 = Re[Zb ] = 0.1575 W,

X1 + X �2 = Im[Zb ] = 1.132 W.

3.  Find the individual values by using the DC value of R1 and assuming X1 and X2′ are equal:

R�2 = 0.1575 − R1 = 0.1575 − 0.088 = 0.0695 W, X1 = X�2 = 1.132/2 = 0.566 W.

FIGURE 3.16: Equivalent during blocked-rotor test.

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3.4.3 No-Load Test During the no-load test, the motor shaft is free, so the motor turns at almost the synchronous speed. The slip is close to 0, so the load “resistor” is essentially an open circuit and the rotor current is very small. This makes the parallel Rc–Xm combination the dominant impedance. The no-load test is done by applying rated voltage to the motor and measuring current and power. Figure 3.17 shows the equivalent circuit. The no-load data collected for this motor are

V = 460.0 V (rated voltage), I = 21.5 A, P = 464.6 W, pf = 2.71% lagging. The calculations for the no-load test are very similar to the locked-rotor calculations: 1. Convert data to single-phase Y values:

√ Vph = 460.0/ 3 = 265.6 V, Iph = 21.5∠ − cos−1 pf = 21.5∠ − 88.45◦ A,

Pph = 464.6/3 = 154.9 W.

2.  Find the values of Rc and Xm:

Iph = 21.5∠ − 88.45◦ = 0.5816 − j 21.49 A, Vph 265.6 � �= Rc = = 456.7 W, 0.5816 Re Iph Vph 265.6 � � = = 12.36 W. Xm = 21.49 −Im Iph

FIGURE 3.17: Equivalent during no-load test.

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FIGURE 3.18: “Textbook” tests compared with manufacturer’s test.

3.4.4 Complete Equivalent Figure 3.18 shows the Y-equivalent circuit derived from the DC, locked-rotor, and no-load tests. The circuit shows two sets of numbers. The upper number on each element is the result of the tests just performed; the lower number is the value the manufacturer obtained from more extensive tests.

3.4.5 Comparison How well does this model the motor? One way of checking is to see what values the model yields for the input current, input power, and output power under full-load conditions. Table 3.1 compares the results taken from three sources: the manufacturer’s model of the motor, the text book model in Figure 3.18, and the data obtained by actually testing the motor at rated speed (1770 rpm). The results of all this are within about 10% of the actual operating values. Part of the discrepancy in the text model is a result of assuming that X1 and X2′ are equal. Additional motor tests indicate that this split should be about 40:60. Another discrepancy is introduced by the assumption that, during the no-load test, R1 and X1 can be ignored.

3.5

ENERGY FLOW

We have been developing a single-phase Y-equivalent circuit to model the behavior of the threephase induction motor. So far, all we have are the numbers that go into the equivalent and a knowledge of what the circuit parameters represent in the motor itself. This model is useful for looking at the flow of energy through the motor. We know that a portion of what goes into the machine gets lost inside and does not appear at the output. Motor efficiencies are not bad, ranging from about 85% to 95%, but they are not 100%. If you could invent one at 100%, you would be famous!

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table 3.1:  Performance comparison of models a

Parameter

Manufacturer’s a model

Text model

Actual test

Line voltage

460 V

460 V

460 V

Line current

59.94∠–33.3° A

63.99∠–32.0° A

60.02∠–33.3° A

Power factor

83.6% lagging

84.8% lagging

83.6% lagging

Power input

39.93 kW

43.23 kW

39.98 kW

Power output

37.99 kW 50.92 hp

41.09 kW 55.1 hp

37.30 kW 50.0 hp

Efficiency

95.10%

95.10%

93.30%

b

a

Model power-output data ignore mechanical and windage losses.

b

Actual test data include mechanical and windage losses.

3.5.1 Power, Torque, and Losses Let us look at what happens to energy by adding a descriptive graphic that shows the primary energy flow. Figure 3.19 is the equivalent circuit with the power flow added. The drawing shows the motor divided into its physical parts: the incoming power line, the stator winding, the airgap, the rotor, and the shaft. Power through the motor is depicted as a line of diminishing thickness. Let us look at the power along the way:

FIGURE 3.19: Single-phase equivalent with three-phase power flow.

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• • • •

•

•

Pin is the input power given in three-phase terms (all power in this drawing is three-phase). V1 is the line voltage and I1 is the line current. Losses in the stator are copper loss, represented by R1, and core loss, represented by Rc. What is not lost in the stator is passed along to the airgap. The torque developed in the airgap is related to the airgap power by the synchronous speed ωs (in radians per second). In the rotor, the developed torque is now operating at the rotor speed ωm , which is slightly slower than the synchronous speed. Hence, the power developed for the rotor is diminished. This loss is represented by rotor losses, primarily via R2′. The magnitude of this loss is the slip times the airgap power. The power actually delivered to the shaft is the developed power. This power, however, does not entirely make it out of the motor. A portion is lost in the mechanical parts of the motor, mainly windage and bearings. Most induction motor rotors include fins of some kind to stir the air inside the motor to aid in cooling the windings and it takes some power to move this air. The fins are visible in the photographs (Figures 3.7 and 3.8). What comes out as shaft power and shaft torque is what is left after subtracting the various losses.

A motor is rated by its output horsepower, not its input power. Therefore, a 50-hp motor that is 90% efficient will have an input power of 55.55 hp. To look at this another way, a rating of 50 hp means the output power is 37.3 kW (746 W/hp). If the motor is 90% efficient, the power input from the power line must be 41.4 kW, leaving 4.1 kW to be dissipated by the motor itself. If the motor temperature is not to rise more than allowed, the motor must be in an environment where it can dissipate that heat. The manufacturer of the motor I have been using as an example in Section 3.4 has furnished some temperature data. After running the motor for more than five hours at full load in an ambient air temperature of 29°C, the air leaving the motor was at 46°C. Both bearings were at about 40°C. The frame of the motor reached 55°C. The hottest spot in the windings was about 66°C, which is well below the maximum level for its insulation class (180°C).

3.5.2 Stored Energy The induction motor is a large iron-core inductor. As such, it stores magnetic energy in the same way that any other inductor does. The stored energy for an inductor is 1/2Li(t)2 at any instant. When we shut off an operating motor, this stored energy must go somewhere. The running motor also has stored energy in the form of the inertial energy of the rotation system.

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Although going into detail about this stored energy is beyond where I want to go, several points are useful to know: •

•

•

The stored energy can be used at shutdown to help stop the motor. As the power line is disconnected, short the coils. The stored energy is dissipated in the shorted resistance of the motor, bringing the motor to a rapid halt. Electric lawnmowers use this to stop the blade quickly. If the power supply is accidentally short-circuited near the motor, the short must dissipate both the electrical energy associated with the short circuit and also the stored energy in the nearby motor. Moreover, protective devices such as circuit breakers must be able to clear such a fault. Using a DC ohmmeter to test the windings leaves some 1/2Li2 stored energy behind when the meter is disconnected. This energy can show up as a high—and perhaps dangerous— voltage across the terminals or an arc as the meter leads are taken away.

3.6 MOTOR CURVES If you want to make the shaft of some piece of machinery go around, you need torque. Torque is what an induction motor produces. But not just any torque. There is a relationship between the power output of the motor and the torque, of course, and between the torque the motor produces and the speed of the shaft. More especially, there is a relationship between the output torque and the slip. The curves of torque versus slip tell a great deal about an induction motor. Although there are numerous possible shapes for these curves, there are not very many common ones. NEMA classifies induction motors into number “design” categories. The NEMA design code is specified on the motor’s nameplate. Recall that the nameplate in Section 3.3 showed “Des B” to designate that this motor has a torque–slip curve that follows the NEMA curve B. Figure 3.20 shows four of the NEMA torque–slip curves. They all give torque as a percentage of full-load torque, and speed as slip percentage. Before looking at these curves in detail, let us look at just the Design B curve in Figure 3.21. Several things that we can learn will apply to the other curves as well. There are four important points along the Design B torque–speed curve of Figure 3.21. I am going to work from right to left in the direction of increasing slip. •

Normal full-load torque (100%) occurs at a fairly small slip, usually no more than 5%, depending on the particular design. Note how straight the curve is in this region. It is close

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FIGURE 3.20: NEMA induction motor torque curves.

FIGURE 3.21: NEMA Design B torque curve.

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•

•

•

enough to linear that we are going to assume it is linear between synchronous speed (s = 0) and full-load speed. Breakdown torque is the largest torque the motor can sustain without “slipping over the top of the hill.” If the breakdown torque is exceeded, the motor’s speed drops quickly. Reaching this torque requires a considerable overload. The pull-up torque is the lowest spot on the left portion of the torque–speed curve. It is the minimum value of torque that the motor can produce as its speed is run up from starting to operating. The starting torque is the torque that gets a shaft turning. It has to overcome not only inertia but some “stickiness” that some loads present when they are standing still.

Now let us look at the general characteristics of these designs. Note that the general operating range of these motors is at the far right of the graph, generally with a slip of no more than 5%. •

•

• •

Design A has the highest breakdown torque, and the relation between torque and slip in the normal operating range is very steep. This means that the shaft speed does not vary a great deal with torque, even at more than 100% of full load. It is a good general-purpose motor but generally has a higher starting current than other common designs. (Starting current cannot be determined from these curves but rather comes from other information.) Design B is the most common motor sold. Its curve has a fairly steep slope for the normal range of slip, which generally is not more than 5%. This design’s breakdown torque is among the lowest, so it does not take too kindly to large overloads. Starting current is modest, however, and the motor can often be line-started, meaning that it can be started by switching on full line voltage without anything special to limit the current. Design C has a high starting torque and a modest starting current. It is not a good choice for loads that may at times exceed full load and it is not as efficient as A and B. Design D has a very different curve compared to the others. It produces a very high starting torque without a high starting current. The normal range of operating slip is very broad.

3.7 OVERMOTORING This section has a simple message: Do not overmotor! In other words, if a particular job requires a 10-hp motor, do not “throw in a safety factor” and choose a 20-hp motor. Match the motor to the load, specifying a motor that is just large enough to provide the torque required. Why? I will show several motor curves that will help you understand this statement. Figure 3.22 shows the torque–slip curve for a particular motor. It is most similar to a Design A motor but has higher starting and breakdown torques.

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FIGURE 3.22: Percent rated torque.

FIGURE 3.23: Power factor.

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FIGURE 3.24: Percent efficiency.

Suppose I have a load that requires just 200 N-m for normal operation. That is about half the rated torque, so the slip is cut in two to about 1.5%. Hmm, that does sound okay. After all, the motor will not be working as hard and it should last longer. But look at the power factor curve in Figure 3.23. The full-load power factor is 88% lagging. If we “run light” with a full-load slip of just 1.5%, the power factor gets considerably worse. It looks to me like it falls to around 70% lagging. This will increase your load’s reactive power and you will most likely pay the power company for low power factor. The efficiency also decreases. The motor was designed to operate near the peak of its efficiency curve. Note in Figure 3.24 that this is about 91%. As the slip gets smaller, the efficiency decreases. You will require more power and you will pay the power company for this, too. Figure 3.25 shows about the only thing that is good about overmotoring—the current is reduced. But you have already seen that the decrease in current does not come with a lower power bill due to poorer power factor and lower efficiency. So do not overmotor! Remember that induction motors specify a service factor, often 1.15. This means that the motor can be run continuously at a load larger than full load. This rate indicates that it can sustain a 15% overload continuously, provided you give the motor proper ambient conditions. If, for example, the maximum allowed ambient temperature is 40°C, you should make sure that there is good air flow around the motor and that air stays below 40°C.

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FIGURE 3.25: Line current.

There are not a lot of common motor sizes. In the “medium horsepower” range, which comprises motors between 1 and 500 hp, common ratings are 1, 1.5, 2, 3, 5, 7.5, 10, 15, 20, 25, and so on. For example, if you have a 6-hp load, a 5-hp motor will not do, even at 15% over, so a 7.5-hp motor is required. Likewise, an 8-hp load can be handled properly by a 7.5-hp motor.

3.8

EXAMPLES

Here are several examples to bring all of this together.

3.8.1 Example I—Data Collection A 50-hp, three-phase, 60-Hz, 460-V induction motor has a full-load operating speed of 1770 rpm. The full-load current is 59.5 A at 85% power factor. Rotational losses have been measured at 1200 W. 1. Find the number of poles. The synchronous speed must be a submultiple of 3600 for a 60-Hz motor and 1800 is the first submultiple larger than 1770.

ns = 1800 = poles = 4. 2.  Find the slip at rated load.

3600 poles/2

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s=

1800 − 1770 × 100 = 1.667% . 1800

3.  Find the output torque. The number of watts per horsepower is 746.

Pout = 746 × 50 = 37.3 kW,

rpm 1770 2p = 2p = 185.4 rad/s, 60 60 Pout 37.3 × 103 = = = 201.2 N-m. wm 185.4

wm = Tout

4.  Find the input power.

Pin =

√

3(460)(59.5)(0.85) = 40.3 kW

5.  Find the efficiency.

% efficiency =

Pout 37.3 × 100 = × 100 = 92.6% Pin 40.3

6. Find the electrical losses in the rotor. I will follow the energy flow in Figure 3.19 from right to left, deriving the developed power from the output plus the rotational losses. From that, I will get the airgap power and then the rotor losses.

Pdevel = Pout + Protation = 37.3 + 1.2 = 38.5 kW = (1 − s)Pairgap

38.5 = 39.2 kW 1 − 0.01667 = sPairgap = (0.01667)(39,200) = 653 W

Pairgap = Protor loss

7.  Find the stator losses. Pstator = Pin − Pairgap = 40.3 − 39.2 = 1100 W

3.8.2 Example II—Power Factor Adjustment Specify the capacitance necessary to improve to 92% the power factor of the motor of Example I when operating at full load.

Pin = 40.3 at 0.85 lagging Qin = Pin tan cos−1 (pf ) = 40.3 tan cos−1 0.85 = 24.98 kVAR Qnew = 40.3 tan cos−1 0.92 = 17.17 kVAR Qcap = Qnew − Qin = 17.17 − 24.98 = −7.81 kVAR

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This capacitance is the total required, which must be split among the three phases.

Qc = −7.81/3 = −2.60 kVAR/phase at 460 V.

3.8.3 Example III—Using the Equivalent Circuit A 10-hp, 60-Hz, 220-V, four-pole, three-phase induction motor drives a particular load with 5% slip. Rotational losses are 180 W. This motor’s single-phase-equivalent parameters are shown below (Figure 3.26). Use this model in the calculations that follow. R1 = 0.25 Ω X1 = 1.2 Ω

R2′ = 0.20 Ω X2′ = 1.2 Ω

Rc = 200 Ω Xm = 20 Ω

1. One node equation for E2 and two auxiliary equations will yield the necessary circuit data. (V1 = 220/√3 V.)

0.20 (1 − 0.05) W = 3.8 W 0.05 =0

Rload =

E2 − V1 E2 E2 E2 + + + 0.25 + j 1.2 200 j 20 0.20 + j 1.2 + 3.8 Solving, E2 = 103.5∠ − 11.8◦ V V1 − E2 I1 = = 27.15∠ − 38.7◦ A 0.25 + j 1.2 E2 I2 = = 24.78∠ − 28.5◦ A 0.20 + j 1.2 + 3.8

FIGURE 3.26: Single-phase equivalent for example III.

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2. Find the power input, the power output, and the horsepower output. Because the model is just one phase, values of power derived from the model must be tripled for three-phase results.

√ 3 |Vline | |I1 | pf = 3(220)(27.15)cos(−38.7◦ ) = 8.074 kW, � � � � 2 103.52 2 |E2 | 2 = 3 R1 |I1 | + = 3 0.25(27.15) + = 714 W, Rc 200 � � � � = 3 R2 |I2 |2 = 3 0.2(24.78)2 = 368 W,

Pin = Pstator Protor loss

√

Pdevel = Pin − Pstator − Protor loss = 8.074 − 0.714 − 0.368 = 6.99 kW, Pout = Pdevel − Protation = 6.99 − 0.180 = 6.81 kW,

hpout = Pout /746 = 9.13 hp.

3.  Find the percent of full load, the overall efficiency and the input power factor.

hpout × 100 = 91.3% , 10 Pout × 100 = 84.3% , % efficiency = Pin % full load =

pfin = cos (−38.7) = 78.0% lagging.

4. Find the starting current. This is done by noting that the resistor representing output power to the shaft becomes 0 because the slip is 1. So in the original node equation, replace 3.8 by 0 in the last fraction. Then solve for E2 and find I1. The result is

E2 = 61.4∠1.3◦ V Istart = I1 |Rload =0 = 53.5∠ − 79.4◦ A.

3.8.4 Example IV—Motor Selection A certain machine in a plant requires a continuous torque of 80 N-m at 1200 rpm. The peak starting torque is twice the running torque. This machine is to be driven by a 440-V, three-phase, 60-Hz induction motor with low slip, because the load is to be kept as near to 1200 rpm as possible. Specify the appropriate induction motor. The general choices are 5, 7.5, 10, 15, 20, and 25 hp, with Design A, B, C, or D characteristics.

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1. Do the easy one first, number of poles.

ns = 1200 =

3600 , so p = 6 poles. p/2

2. Because slip will be small, using the synchronous speed instead of the slightly lower motor speed should not make much difference.

1200 2p = 125.7 rad/s, 60 ≈ ws Tout = 125.7 × 80 = 10.05 kW, Pout = 13.5 hp. = 746

ws = Pout hpout

 his output is too large for a 10-hp motor with a service factor of 1.15. A 15-hp motor is T required. 3. Use the design curves of Figure 3.20 to determine which design to specify. Doing this requires that we know the “100% torque” output of the motor, which is the torque for a rated load of 15 hp. Again, assume that the slip is small.

T100% =



(746) (15) = 89.0 N-m. 125.7

4. The specified load requires a starting torque of 160 N-m (2 × 80). The 100% full-load torque for the 15-hp motor is 89.0 N-m, so 160 N-m is 180% of the motor's full-load torque. Although Design B is the most commonly available motor, its starting torque is only about 160% of the full-load torque. The next choice is Design C, which is similar to Design B but has a higher starting torque, about 210%. Hence, the best choice appears to be Design C. 5. The chosen motor is running at (80/89)100 = 90% of full rated load. The full-load slip for a Design C motor on the graph of Figure 3.20 appears to be about 3%. We can assume that the rightmost portion of the curve is linear, so the motor will be driving the load at about 0.90 × 3 = 2.7%, which is 1168 rpm.

The result of all this is that you specify a 440-V, three-phase, 60-Hz, six-pole, Design C induction motor. Once you have this information, you can see what various manufacturers offer and check the design using the manufacturer’s torque–speed curve for the motor chosen. This design ignores an important question, though. What is the torque-versus-speed characteristic of the load over the full range of speeds from starting to full load? This is beyond the scope of this text, but it can be answered by plotting the torque–speed curve of the load on the same

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torque–speed graph as the motor. If the load’s curve is below the motor’s curve except where they cross at the operating values, the motor specified is probably the right choice.

3.9

SINGLE-PHASE MOTORS

This chapter has been devoted to three-phase induction motors because they provide by far the bulk of rotating-energy delivery. The are several good reasons for this: smooth rotating power, a single moving part, very simple structure. However, many applications that need rotating energy are in places that do not have three-phase power. Homes are a significant example. Not that three-phase power is not available for homes—it is expensive and it is not really needed. Single-phase induction motors can do the smaller jobs very well. So what is the difference between three-phase and single-phase induction motors? The biggest difference is that the field the stator creates is not rotating smoothly around the stator. Instead, it just changes direction, albeit smoothly, from “up” to “down” and back again. A rotor will not follow this field from a standstill—the torque just pulsates with no net rotation. But if a rotor can be optimized to go at the right speed, the motor delivers net torque, pulsating, to be sure, but net. The basic structure of the single-phase induction motor is very similar to that of the threephase motor. One pair of poles provides a synchronous speed of 3600 rpm (at 60 Hz) just as with three-phase motors. Two pairs provide 1800 rpm, and so on. The rotor is similar to that of its three-phase big brother—shorted coils. The single-phase motor exhibits slip in the same way. The problem is to get things started. There are four different ways of getting the necessary net torque to get the rotor going. All four involve the same fundamental idea: provide a second set of poles whose current is out of phase with the first set. If this second set of poles is placed with its axis electrically offset by the same angle as its current is offset, a rotating field is created that provides torque to move the rotor. The four ways of doing this are shaded pole, split phase, capacitor start, and capacitor start and run, often called split capacitor. Let us discuss briefly these four options, remembering that in each case the goal is a rotating stator field that will give the rotor net torque. •

•

The shaded-pole motor provides the “off-center” poles by wrapping a shorted coil of wire around just a portion of the main poles. This shorted coil provides additional inductance for that portion of the pole, thereby adding an additional phase angle to the angle of the magnetic field created by the main poles. The generated torque is not large, but it is enough to begin rotation. The result is a motor with low starting torque and usually not much power, either. Slip is high (10% or more) and efficiency is very low. The split-phase motor uses two windings: one for running and one for starting. These windings are on two separate sets of poles, generally spaced 90 electrical degrees apart. The

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•

•

wire size and the number of turns on these poles are set up so that the running poles have high reactance and low resistance, whereas the starting poles exhibit the opposite condition. The motor is started with all poles connected. The large difference in reactance and resistance provides a large phase shift between the sets of poles and hence a net starting torque. As the motor begins to come up to speed, a centrifugal switch cuts off the starting winding, leaving the running winding to drive the rotor. These motors can be reversed by reversing the current in one of the two sets of poles. They have moderate starting torque and are fairly efficient (more than 50%). Ceiling fans are an example of this type of motor. The Casablanca XLP motor is an 18-pole split-phase motor, so its synchronous speed is 3600/(18/2) = 400 rpm, but it runs more slowly than that. Capacitor-start motors have roughly the same structure as the split-phase motor, except that a capacitor in series with the starting winding provides the necessary phase shift. The winding must be switched off as the motor speed increases. Characteristics are similar to the split-phase motor, but the starting torque is higher. After they have started, they behave much as the split-phase motor does. Capacitor motors are often easy to recognize—the capacitor is often in a separate cylindrical housing on the outside of the motor. Capacitor-start capacitor-run motors leave the capacitor in the circuit after starting. Sometimes they use two capacitors, one just for starting and one for both starting and running. Although these are a little more complicated than the capacitor-start motors, they have a better power factor.

All but the shaded-pole motor have a noticeable disadvantage over the three-phase induction motor: additional parts. The most common point of failure of the capacitor motors is the capacitor. If a capacitor motor will not start but still draws current, the odds-on bet is capacitor failure. Motor capacitors are electrolytic capacitors, which we generally think of as being polarized. Motor capacitors are made to be nonpolarized. The second common point of failure is the centrifugal switch, which is designed to snap open when the rotor reaches a set speed, generally somewhere about 25–30% of normal running speed. There is one other common small single-phase motor, the “universal” motor. It is really a DC motor with brushes and a commutator. It is called universal because it will run on both AC and DC, although its common use is on AC in such machines as vacuum cleaners and power drills.

3.10 OTHER MOTORS We have not looked at DC motors because I have been focusing on larger machines driven by the AC power system. There are millions of DC motors in use that we are ignoring. Most of these are in low-to-medium power applications, such as battery-operated toys and portable tools and automo-

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bile starters. Large DC motors power so-called diesel locomotives, which are really diesel-electric. The diesel prime mover drives a DC generator and this powers DC motors on the locomotive’s trucks. We have learned that you cannot really control the speed of an induction motor. All you can do is select from what is available based on the number of pole pairs. Speed control can be achieved, however, by using variable-speed drives. For these, electronics provides a frequency different from that of the power line. So a six-pole motor could have a synchronous speed of, say, 1100 rpm rather than 1200 by reducing the supplied frequency to (1100/1200)60 = 55 Hz. The same electronics can also provide additional control such as starting.

3.11

SUMMARY

Tesla’s invention is pure simplicity. Its smoothly rotating magnetic field drags the rotor around to provide a smooth torque at the shaft. The torque–speed curves of standard motors are just about linear in their usual operating range. The single-phase equivalent circuit is reasonably simple. In short, the induction motor is a remarkable electrical device, one of the oldest still in common use. Yet most of us do not appreciate it. One important lesson in this chapter is to avoid overmotoring. Choosing a motor larger than what is required leads to wasteful power use and lowered efficiency. No “safety margin” is needed in most applications. The right-sized motor in the right environment will do its job efficiently and is likely to perform continuously for many years. In this chapter, we have gone through the way in which the three-phase induction motor works, developed an equivalent circuit for it, and laid out energy flow through the machine. We have also learned about the standard nameplate data and how a motor is tested to obtain the parameters for the equivalent circuit. Finally, we have presented some examples of motor applications and of choosing a motor. • • • •

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Author Biography William J. Eccles has been a professor of electrical and computer engineering at Rose-Hulman Institute of Technology since 1990 (except for one year at Oklahoma State). He retired in 1990 as Distinguished Professor Emeritus after 25 years at the University of South Carolina. He founded the Department of Computer Science at that university and served at one time or another as head of four different departments, Computer Science, Mathematics and Computer Science, and Electrical and Computer Engineering, all at South Carolina, and Electrical and Computer Engineering at Rose-Hulman. Most of his teaching has been in circuits and in microprocessor systems. He has published Microprocessor Systems: A 16-Bit Approach (Addison-Wesley, 1985) and numerous monographs on circuits, systems, microprocessor programming, and digital logic design. Bill has also published Pragmatic Circuits: DC and Time Domain, Pragmatic Circuits: Frequency Domain, Pragmatic Circuits: Signals and Filters, and Pragmatic Logic, four texts in this Synthesis Lectures in Digital Circuits and Systems series. Bill and his wife Trish have two children and three grandchildren. Bill is also a conductor (appropriate for an electrical engineer) on the Whitewater Valley Railroad, a tourist line in Connersville, IN. He is a registered professional engineer and an amateur radio operator.