Pragmatic Power 1598297988, 9781598297980

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Pragmatic Power

Copyrght © 2008 by Morgan & Claypool All rghts reserved. No part of ths publcaton may be reproduced, stored n a retreval system, or transmtted n any form or by any means—electronc, mechancal, photocopy, recordng, or any other except for bref quotatons n prnted revews, wthout the pror permsson of the publsher. Pragmatc Power Wllam J. Eccles www.morganclaypool.com ISBN: 9781598297980

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ISBN: 9781598297997

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DOI: 10.2200/S00145ED1V01Y200808DCS016 A Publcaton n the Morgan & Claypool Publshers seres SYNTHESIS LECTURES ON DIGITAL CIRCUITS AND SYSTEMS # 16 Lecture #16 Seres Edtor: Mtchell Thornton, Southern Methodst Unversty Series ISSN ISSN 1932-3166

prnt

ISSN 1932-3174

electronc

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

SYNTHESIS LECTURES ON DIGITAL CIRCUITS AND SYSTEMS # 16

ABSTRACT Pragmatic Power s focused on just three aspects of the AC electrcal power system that supples and moves the vast majorty of electrcal energy nearly everywhere n the world: three-phase power systems, transformers, and nducton motors. The reader needs to have had an ntroducton to electrcal crcuts and AC power, although the text begns wth a revew of the bascs of AC power. Balanced three-phase systems are studed by developng ther sngle-phase equvalents. The study ncludes a look at how the cost of “power” s affected by reactve power and power factor. Transformers are consdered as a crcut element n a power system, one that can be reasonably modeled to smplfy system analyss. Inducton motors are presented as the most common way to change electrcal energy nto rotatonal energy. Examples nclude the correct selecton of an nducton motor for a partcular rotatng load. All of these topcs nclude completely worked examples to ad the reader n understandng how to apply what has been learned. Ths short lecture book wll be of use to students at any level of engneerng, not just electrcal, because t s ntended for the practcng engneer or scentst lookng for a practcal, appled ntroducton to AC power systems. The author’s “pragmatc” and appled style gves a unque and helpful “nondealstc, practcal, and opnonated” ntroducton to the topc.

KEYWORDS AC power, three-phase, transformer, nducton motor

Contents 1.

Three-Phase Power: 3 > 3 ×1 ................................................................................1 1.1 Revew of AC Power ........................................................................................... 1 1.1.1 AC Power Quanttes .............................................................................. 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 Termnology ................................................................................. 15 1.4 Three-Phase Systems ........................................................................................ 17 1.4.1 Dealng Wth Complexty ..................................................................... 18 1.4.2 Sngle-Phase Equvalent ........................................................................ 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 Basc Transformer .............................................................................................. 31 2.1.1 Example I: Ideal Transformer ................................................................ 34 2.1.2 Example II: Ideal Agan ........................................................................ 36 2.2 Transformer Model ........................................................................................... 38 2.2.1 Realstc Model ...................................................................................... 38

vi

PRAGMATIC POWER

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

Induction Motors: Just One Moving Part ........................................................... 55 3.1 Rotatng Felds .................................................................................................. 55 3.1.1 Sngle-Phase Motor .............................................................................. 55 3.1.2 Three-Phase Rotatng Magnetc Feld .................................................. 57 3.1.3 Add a Rotor ........................................................................................... 59 3.1.4 Slp, Speed, and Poles ............................................................................ 61 3.2 Equvalent Crcut ............................................................................................. 62 3.2.1 Stator Equvalent Crcut ....................................................................... 63 3.2.2 Rotor Equvalent Crcut ....................................................................... 63 3.2.3 Complete Equvalent Crcut ................................................................. 65 3.3 Motor Nameplate .............................................................................................. 66 3.4 Testng ............................................................................................................... 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 Equvalent ............................................................................. 72 3.4.5 Comparson ........................................................................................... 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 Overmotorng .................................................................................................... 77 3.8 Examples ........................................................................................................... 80 3.8.1 Example I—Data Collecton ................................................................. 80 3.8.2 Example II—Power Factor Adjustment ................................................ 81

CONTENTS

3.8.3 Example III—Usng The Equvalent Crcut ........................................ 82 3.8.4 Example IV—Motor Selecton .............................................................. 83 3.9 Sngle-Phase Motors ......................................................................................... 85 3.10 Other Motors .................................................................................................... 86 3.11 Summary ........................................................................................................... 87

vii

1

Three-Phase Power: 3 > 3 × 1 CHAPTER 1

Three-phase power s more than just three sngle-phase power systems, so to study AC power, we need irst to study three-phase systems, what they are, why we use them, and how they work. Threephase systems are by far the domnant method for delverng large blocks of energy from one place to another. To begn, let us revew what t means to say “ac power” and refresh our memores about such terms as average power, reactve power, and power factor. You can also ind materal on ths n Pragmatic Circuits: Frequency Domain (Wllam J. Eccles, Morgan & Claypool, 2006, Chapter 8).

1.1

REVIEW OF AC POWER

We could treat AC power usng functons of tme such as sne and cosne. Fgure 1.1 shows a load beng drven by a voltage v(t) and a current i(t). The voltage and the current are snusods: � � 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 ths product, we get what s known as instantaneous power, the power at any partcular tme. But ths s not very useful, because we really need to know how much power s beng delvered on the average, and more mportant, how much energy s beng moved.

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

2

PRAGMATIC POWER

From the calculaton of p(t), we can derve ths average value as well as deine several other quanttes that are very useful n dealng wth AC power. All of these new quanttes come from lookng at p(t) for snusodal voltages and currents.

1.1.1 AC Power Quantities The average power comes from averagng the nstantaneous power p(t) over one perod of the AC waveform. For our snusodal voltage and current n Fgure 1.1, p(t) s

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

A few manpulatons yeld

p(t) =

VpeakIpeak � 2

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

If we now average ths p(t) over one perod, we get Pav:

Pav =

VpeakIpeak 2

cos (qv − qi ) .

Note that Vpeak s the peak value of the cosne wave, a value that makes mathematcans happy because that s the proper manner to descrbe the cosne. But t s not what we use n AC power. Instead, we use the root-mean-square (RMS) value, whch tells exactly how to compute the value, namely, take the square root of the mean of the square of the functon. For a sne wave, ths comes out as Vpeak/√2 = 0.707Vpeak. Agan, note that ths s vald for snusodal waveforms; other waveforms have dfferent RMS values. Usng the RMS values and deinng θv–i as the angle of the voltage θv mnus the angle of the current θi, the average power P s

P = VRMSIRMScos (qv−i ) , where P s expressed n watts (W). P s sometmes referred to as average power or real power, but much more commonly, t s just plan power. If someone says smply, power, t means P. Now let us go back to the equaton for p(t) and consder the second term, the one wth 4π n t. Ths second term averages to zero because the average value of a snusod over any ntegral number of perods s zero. Yet ths second term s stll power, even though ts average s zero. It s power that lows one way for a quarter of a cycle, then the other way for a quarter of a cycle, and so on. From ths second term, we deine reactve power Q:

THREE-PHASE POWER: 3 > 3 × 1 3

Q=

VpeakIpeak 2

sin (qv−i )

= VRMSIRMSsin (qv−i ) , where Q has the unt of volt-ampere-reactve (VAR). Note that Q s volts tmes amps, whch s watts, but we use VAR to dstngush t from the average power P. Q delvers no energy on the average; t represents power that lows back and forth n the power system. Reactve power Q s arrvng at a load for one-quarter of a cycle, leavng for a quarter, and so on. Hence, t represents no net low of energy, but accounts for part of the current lowng n the system. Because of ths, we must nclude t n power calculatons. Q s postve for loads that are prmarly nductve, that s, loads whose mpedance has a postve phase angle. Q s negatve for loads that are prmarly capactve. Reactve power Q s very mportant n energy transfer. For example, an nducton motor s nductve (duh!) and requres current to supply the magnetc ield that makes the motor work. Therefore, an nducton motor requres reactve power Q to operate. To express ths n another way, we must supply VARs as well as watts to make such a motor run and drve a rotatng load. Because of these VARs, current suppled s larger than that needed for just the average power P. Ths ncreased current lows through the wres and transformers of the power system and ncreases the losses assocated wth runnng the motor. We wll dscuss more about ths later. The power factor s deined as the cosne of the angle between the voltage and the current θv–i:

power factor = cos (qv−i ) . Power factor s untless, but you must add a “unt” to t n the form of ether “laggng” or “leadng.” The cosne functon s always gong to produce a postve power factor, so ths laggng–leadng dstncton s needed. How do you know whch to use? A load that s nductve has a current that lags the voltage n tme. Hence, the angle of the current θi s 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 s laggng. If the load s capactve, the current leads the voltage and the power factor s leadng. Another power quantty that s useful s the complex power S:

S = P + jQ. Ths quantty also needs a unt that s equvalent to watts, so we use the volt-ampere and the unt symbol VA. S s a complex number wth a real part of P and an magnary part of Q. The cosne of ts angle s the power factor, laggng f Q s postve.

4

PRAGMATIC POWER

FIGURE 1.2: Power trangle.

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

S = VI ∗ , where the astersk ndcates the complex conjugate, replacng +j by −j n rectangular form or changng the sgn of the angle n polar form. The magntude of S s sometmes useful. It s called the apparent power and s just the product of the voltage and the current wthout regard to the phase angle between them. The power factor can be deined as the rato of average power to apparent power:

power factor =

P . |S|

The power trangle s an easy way to vsualze the relatonshps among P and Q and S and power factor. The trangle (Fgure 1.2) recognzes that S s complex, wth P on the real axs and Q on the magnary axs. Movng Q to the rght to the other end of P gves the trangle. The angle s the power-factor angle, the angle θv−i. Note that postve Q s upward and represents nductve reactve power. I wll use the trangle to help out wth my calculatons n Example II (Secton 1.1.4).

1.1.2 Example I The system n Fgure 1.3 has two loads whose parameters are stated n terms of power. Fnd the total power P, the total reactve power Q, the total complex power S, the overall power factor, and the source current I. For the irst load, P1 = 24 kW wth a power factor of 0.86 laggng. If we look at the trangle and remember that the tangent of the angle s Q/P, then the reactve power s

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

THREE-PHASE POWER: 3 > 3 × 1 5

FIGURE 1.3: Example I.

and the complex power s

S1 = (24 + j 14.24) kVA. I wll do the same thng for the second load, whch s gven n terms of S2 = 20 kVA wth a power factor of 0.72 laggng. The trangle tells me that Q and S are related by the sne 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 s P conserved, but so also are Q and S. Ths gves 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 =

38.4 Ptotal = = 0.807 lagging. |Stotal | 47.6

The current s 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 wll carry the example one step further by indng the ndvdual load currents. Ther magntudes wll come from the equaton P = |V||I|pf, and ther angles wll come from the power factor:

6

PRAGMATIC POWER

|I1 | =

P1 24.0 × 103 = = 58.14 A, ∠I1 = cos−1 pf1 = cos−1 0.86 = 30.7◦ , V × pf1 480 × 0.86

|I2 | =

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 negatve because both loads have laggng power factors. As a check, let us see f 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 Fgure 1.4 shows a 2-hp motor that s 85% eficent, runnng on a 240-V supply wth a power factor of 0.78 laggng. It s beng fed by wres, each of whch has an mpedance of 0.5 + j0.8 Ω. We are to ind the power delvered to the motor, the lne current, the power lost n the lne, the total P, Q, S, and power factor at the generator, the percent voltage regulaton, and the percent eficency. Notce that the generator voltage s not gven. Ths s because the motor s desgned to operate correctly on 240 V, so the generator voltage must be larger than that to compensate for the losses n the lne. Frst, we need to put the motor’s specicatons nto AC power terms. One horsepower s 746 W, whch s the shaft output of the motor. The motor s 85% eficent, so the power nput s

Pload =

FIGURE 1.4: Example II.

2 × 746 = 1.755 kW 0.85

THREE-PHASE POWER: 3 > 3 × 1 7

and the reactve power nput s

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

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

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

The loss n the lne s due entrely to the resstve part of the lne mpedance. Resstors “don’t care” about phase angles, so I wll use I-squared-R for the lne loss, but wth just the magntude 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 wll obtan t by usng 240 plus the voltage drop n the lne mpedances:

Vs = 240 + Vline = 240 + (0.5 + j 0.8 + j 0.8 + 0.5)(9.375∠ − 38.7 ◦ ) = 256.7∠1.3◦ V. There s now enough nformaton to obtan the total Q and S and overall power factor. Here s a neat place to run off the track, though! It s temptng to thnk of the power factor as the cosne of the angle of the current. But that s wrong—t s the cosne of the angle between the voltage and the current. Note that the angle of Vs s not zero! Here are the results, begnnng wth 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 regulaton and . . . oh, what s regulaton? The easest way to look at percent voltage regulaton s to thnk of t as the percentage by whch the voltage at the load wll rse f the load s dsconnected. Our generator voltage Vs s 256.8 V. If we dsconnect the motor, ts termnal voltage wll rse from 240 to 256.8 V. Percent voltage regulaton s that expressed as a percentage:

8

PRAGMATIC POWER

% VR =

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

Percent eficency s a lttle more obvous. It s the percentage of useful power dvded by generated power, often stated as %η:

%h =

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

How about cost? Let us assume that ths motor s part of a plant that operates 16 hours a day, 5 days a week, 50 weeks a year. Let us also assume that electrcty costs 7¢ per klowatt hour plus a charge of $10 per kVA/month for the peak kVA demand durng the month. (The meter that measures klowatt hours also “watches” for the peak kVA demand usng a sldng average wth a 15-mn wndow.) Here are the numbers, whch we wll 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 consultng engneer who has been asked to take a look at the power bll we just computed and see f you can reduce t. You look at the plant and decde that replacng the motor wth a more eficent model would be very expensve and would not mprove thngs very much. You cannot change P because that s what the motor requres to drve ts load. You take another approach, whch s to reduce the reactve power. You know that ths wll brng down the demand charge, but t should also reduce the current and hence the lne losses. You can do ths by addng negatve reactve power, whch s what capactors absorb. But how much to add (subtract, really), to get the demand charge down? One obvous, but wrong, answer s to brng the power factor to unty by usng capactve reactance to exactly cancel the nductve reactance of the motor. Ths soluton s wrong because of somethng that s beyond the scope of ths text, namely, voltage regulaton. If the power factor “gets past unty,” meanng that the load becomes capactve, the power system has a hard tme regulatng the load voltage. In other words, f the load s capactve, the 240 V at the load termnals wanders around too much for proper operaton of the system.

THREE-PHASE POWER: 3 > 3 × 1 9

FIGURE 1.5: PF correcton.

The usual soluton s to mprove the power factor, but not all the way to unty. You talk wth a knowledgeable power system engneer, who recommends that you take the system to 0.9 laggng. Now let us get there. Look at the power trangle n Fgure 1.5. It shows the orgnal P and Q for ths load ( just the load, not the lnes, because the meter s at the load termnals). The new trangle has an angle whose cosne s 0.9. That trangle shows us the value of Qnew, the reactve power that wll get the system to a power factor of 0.9:

Qnew = 1.755 tan cos−1 0.9 = 0.850 kVAR. The dagram also shows us that the capactve load must “absorb” the dfference:

Qc = Qnew − 1.408 = −0.558 kVAR. Ths says that you need a 240-V capactor that wll absorb 0.558 kVAR. Catalogs for such capactors state specicatons n kVARs and volts, so you pck the next larger capactor that s rated at least 0.558 kVAR and wll wthstand at least 240 V. However, just to get a feelng for the sze of ths capactor n terms of crcut elements, let us ind C. Just as power delvered to a resstor s V 2 over R, the “power” delvered to a capactor s V 2 over the capactve reactance:

Qc =

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

Because capactve reactance s the mpedance of a capactor wthout the j, the value of capactance needed at a frequency of 60 Hz s

Xc =

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

10

PRAGMATIC POWER

I wll repeat all the irst set of calculatons usng the new reactve 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 = 100 = 5.5%. |Vload | 240 1.755 Pload %h = 100 = 100 = 96.3%. Ps 1.822

%VR =

Well, dd thngs mprove? The percent voltage regulaton came down a lttle from 7.0% to 5.5%. Umm, that does not mean much because less than 10% s generally OK. The percent eficency came up a lttle from 95.2% to 96.3%. Gee, that s not much of a change, ether. 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 consder the nstalled cost of the capactors and how long they wll last. That s an engneerng economc analyss problem that s beyond what I want to do here. And yes, I wll admt that ths s a pretty trval example. Increase the load to 200 hp, though, and the savngs may be sgnicant. (Another soluton s “actve” power factor correcton, whch s done usng an electronc unt to control the power to the motor.)

THREE-PHASE POWER: 3 > 3 × 1 11

1.2

WHY THREE?

Okay, you say, I have revewed AC power now. I have even read through the example that has a 2hp motor n t. So what s all ths stuff about three phases? What s wrong wth one? And why dd I read that three phases s better than three sngle phases? Well, let us see! Start by doublng the number of phases to two. Fgure 1.6 shows a two-phase generator. It seems logcal to space the two phases unformly around a crcle, 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 n the negatve drecton, because the cosne 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 descrbe the usual 120-V household power where the frequency s 60 Hz (=2π 60 rad/s). But ths setup s merely sngle-phase three-wre where the voltage from a to b s √ √ vab (t) = van + vnb = 120 2 (cos(2p 60t) + cos(2p 60t)) = 240 2 cos(2p 60t) V.

(An arrangement sometmes used n control systems spaces two phases 90° apart, but we wll not go nto that here because t s not used for power systems.) Hmm, okay, two phases does not seem to work out, but let us not get dscouraged. Let us try three phases. Fgure 1.7 shows a three-phase generator. The voltages are placed unformly around a crcle, gong around n the negatve drecton:

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 n the upcomng analyss. Put ths generator nto the smple three-phase crcut wth a load of three equal resstors (Fgure 1.8). Now, calculate the nstantaneous power p(t) delvered to those resstors usng V 2 over R. The analyss s messy trg; the outcome s surprsng:

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 W. pC (t) = R pA (t) =

Addng these three nstantaneous powers nvolves the cosne 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 applyng a few trg denttes, the result s

FIGURE 1.8: p(t) calculaton.

THREE-PHASE POWER: 3 > 3 × 1 13

p(t) =

3Vm2 W. 2R

The surprse? That the instantaneous power s a constant! The nstantaneous power delvered s not snusodal as t s n sngle-phase systems—t s steady. Be careful, though! Ths analyss requred that the load be balanced, meanng 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 wres of ths system (Fgure 1.9). Let us apply Krchoff ’s current law at the node N and determne iNn(t) n 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 wll apply some trg denttes and get

iNn (t) = 0. Agan, a surprse! In a three-phase balanced system the nN wre (the neutral ) carres no current! Ths says we can leave t out and buld our balanced three-phase system wth just three wres. How about one more surprse? It would seem that three wres would requre more metal than two wres. Here are the ground rules for the calculatons that follow: •

A sngle-phase system and a three-phase system are to supply the same load wth the same total amount of power, so n Fgures 1.10 and 1.11, the total power P s the same value. In the three-phase system, however, the load s splt nto three balanced phases.

FIGURE 1.9: Three-phase currents.

14

PRAGMATIC POWER

FIGURE 1.10: “Copper” calculaton—1θ.

•

• •

The voltages between the conductors must be the same. In Fgure 1.10, ths voltage s Vs. Wthout proof at the present tme, I wll smply state that the generators n Fgure 1.11 must supply Vs /√3 so that the voltage between the wres (excludng the neutral) s Vs. The cross-sectonal area of a conductor must be proportonal to the magntude of the current so the wres all have the same current densty (current per unt cross secton). The load s the same dstance from the generator n each system.

In the sngle-phase system, the power delvered to the load s

P1 = Vs I1 and the power delvered to the three-phase load s � � √ Vs P3 = 3 √ I3 = 3Vs I3 . 3 Equatng the powers relates the wre currents:

P1 = Vs I 1 = P3 = √ I1 = 3I3 .

FIGURE 1.11: “Copper” calculaton—3θ.

√

3Vs I3 ,

THREE-PHASE POWER: 3 > 3 × 1 15

Calculatng the rato of each current to the cross-sectonal area needed and applyng the relatonshp between the currents gves

I3 I1 = 2, 2 p r1 pr √3 2 2 r1 = 3r3 . If the system s of length  , the metal volume for the wres n each system, wth two wres for sngle phase and three for three phase, s � √ � � � Vol1 = 2 p r12 � = 2 p 3r32 �, � � Vol3 = 3 p r32 �. The surprse here s that the rato of the total volume of metal for these systems s

Vol3 3 = √ = 86.6% . Vol1 2 3 A three-phase system wth the same power delvery capablty as a sngle-phase system requres 86.6% as much metal for the wres! How s that for reasons why three-phase? Nonpulsatng power. Only three wres for threephase versus two for sngle phase. And 13.4% less metal for conductors. Not a bad deal! And that s why the vast majorty of electrcal power transmsson s done va three-phase systems.

1.3

THREE-PHASE TERMINOLOGY

One of the most confusng thngs about three-phase systems s termnology. Certan terms have very specic meanngs, whch are somewhat dfferent from what your ntuton wll tell you. There are also some rather strctly observed conventons that one follows when descrbng three-phase systems. Fgure 1.12 shows a three-phase system wth one three-phase generator and two loads. The generator s sad to be “Y-connected.” Load 1 s “Y-connected;” load 2 s “∆-connected.” I have labeled a voltage and a current n each load as well as some quanttes between the generator and the loads. Now, for some terms: •

The term “lne” always refers to nterconnectng wres between the generator and the load, labeled n the igure “Lne.” Ths term never refers to the loads themselves.

16

PRAGMATIC POWER

FIGURE 1.12: Three-phase termnology.

• • • •

The term “phase” always refers to legs of the load, labeled “1” and “2” n the igure. Ths term never refers to the nterconnectng wres between the generator and the loads. In the “Lne” porton of the igure, the voltages VAB, VBC, and VCA are line voltages and IaA s a line current. (The two other lne currents are not labeled.) In the igure, VAN and VAB are phase voltages, whereas IAN and IAB are phase currents. At the generator n the igure, Van, Vbn, and Vcn are phase voltages.

There are two conventons: •

•

Three-phase system voltages are always stated as line voltages. These are VAB, VBC, and VCA n the “Lne” porton of the igure. The phase voltages of the generator (Van, Vbn, and Vcn) are not used to descrbe the system. All voltages and currents are gven as RMS values, never peak values.

The relatonshp between lne and phase voltages can be obtaned by applyng Krchoff ’s voltage law through Vbn, Van, and VAB. In a smlar manner, Krchoff ’s current law gves the relatonshp between phase currents n the ∆-connected load and the currents n the connectng lnes.

Vbn − Van + VAB = 0 VAB = Van − Vbn , Iline = IAB − ICA. Phasor dagrams (Fgure 1.13) make indng the result easy.

THREE-PHASE POWER: 3 > 3 × 1

17

FIGURE 1.13: Lne versus phase values.

The results of ths are •

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

Iline Y = Iphase Y . •

Ths same rule apples to Y-connected loads. For ∆-connected loads, the lne and phase relatonshps are

Vline D = Vphase D √ Iline D = 3Iphase D∠ − 30◦ . Fnally (whew!), there s a matter of phase rotaton. There are two ways to arrange the Y-connected generator, one wth each phase 120° behnd the prevous one, and the other wth each phase 120° ahead of the prevous one. I have used positive rotaton n all examples so far and wll stck wth that, whch means Vbn lags Van by 120° and Vcn lags Vbn by 120°. Postve rotaton s sometmes called a-b-c rotaton. (Negatve rotaton s therefore a-c-b.) Is phase rotaton mportant? Yes, any tme your system nvolves rotatng machnes, because most three-phase motors run n opposte drectons for a-b-c and a-c-b connectons. It could be rather embarrassng to nstall a large three-phase fan and turn t on—runnng backward!

1.4

THREE-PHASE SYSTEMS

Let us go irst for the bg pcture. The three-phase systems that provde electrcal energy all over the world generally have seven major sectons:

18

PRAGMATIC POWER

• • • • • • •

Large generators to produce voltages n the 12- to 24-kV range, the larger the power, the hgher the voltage. Transformers to step up the generaton voltage to the transmsson voltages for long-haul lnes. Transmsson lnes, usually at voltages of at least 230 kV. Remember that we always state the lne voltage, the voltage between the lnes. Transformers to step down the transmsson voltage to dstrbuton voltages, generally at most a “few ten thousands.” Dstrbuton lnes to carry the energy through neghborhoods and sectons of ctes. Transformers to step down the dstrbuton voltage to voltages that the customer can use, generally less than 600 V. Customer loads that “use”—and pay for— the electrcal energy provded by the system.

Each of these sectons has an nherent mpedance. Generators have large cols of wre that present both resstance and nductance. Transformers are also cols of wre on ron cores that have both resstance and nductance. Wre lnes have both resstance and nductance. Customers’ loads can generally be modeled by usng resstance and nductance. In other words, the whole system looks knd of lke a generator and a load wth all sorts of Rs and Ls n between.

1.4.1 Dealing With Complexity How can we handle such complexty? We wll not even try! But what we can do s smplfy ths by omttng the transformers (whch we wll study n Chapter 2) and then combne all of the mpedances except the load nto one mpedance. Fgure 1.14, whch we wll deal wth n detal a lttle later, shows ths. There are no transformers, and all the mpedances from generator to (but not ncludng) load are lumped nto 0.2 + j0.5 Ω . That sounds okay so far, you thnk, but analyss of the system of Fgure 1.14 looks lke t would be messy, too. Sure, but f we remember that the neutral (nN ) carres no current f the system s balanced, we could draw n the neutral and then use the sngle node nN as our reference node for nodal analyss. There are only three “unknown” nodes: a, b, and c. We know the voltages at A, B, and C from the stated lne voltage, and we now know how to convert lne to phase voltage. Let us try what I have just sad. A 480-V lne 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 s otherwse specied. That gves 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 wrte three node equatons . . . oh, wat! I can ind Ilne from my knowledge of the power absorbed by the load n the A phase and the A-phase voltage. Then I can ind Van as VAN plus the voltage drop n the lne . . . . But wat! The sys-

THREE-PHASE POWER: 3 > 3 × 1

19

tem s balanced, and anythng I calculate for one of the phases can be appled to the others by just shftng phase angles by 120° or 240°. And that s how I am gong to proceed wth the system analyss.

1.4.2 Single-Phase Equivalent I am gong to reduce the system to just one phase, but wth a few restrctons: • • •

Generators wll be Y-connected. In fact, f the generator s ∆-connected and one of the generators s a bt dfferent from the other two, there wll be a wasteful crculatng current. Loads can be ether Y- or ∆-connected, but for my analyss they wll always be balanced. Lne mpedances wll be the same for all three phases. If the neutral s ncluded n the system, ts mpedance wll be taken as 0 because n a balanced system the neutral current s 0.

Here are the steps to transform a balanced three-phase system to ts sngle-phase equvalent: 1. Draw a sngle-phase system consstng of a generator on the left, an empty box for a load on the rght, and two connectng wres. Label a lne current at the top. 2. Dvde any voltages by √3 because voltages are always supposed to be gven as lne voltages, but we are drawng the sngle-phase equvalent. Label the voltages on your dagram. 3. In the top wre only, draw the lne mpedance, generally a resstance and a reactance. Label them wth the values gven. If a generator mpedance s also gven, draw that nto the top lne. Do not put any mpedance n the bottom lne because ths s the neutral, whch n a balanced three-phase system carres no current. 4. Ths step depends on how the load data are gven: (a) If the total load s stated n “power” terms such as “24 kW at 0.86 laggng,” dvde the “power” number by 3 (but not the angle or the power factor) and wrte ths load nto the box you drew. It makes no dfference whether the load s Y- or ∆-connected. (b) If the load s gven n “power” terms but “per phase,” wrte the load nto the box. (c) If the load s gven as an “mpedance per phase” and s Y-connected, wrte the stated mpedance nto the box. (d) If the load s gven as an “mpedance per phase” and s ∆-connected, dvde the mpedance by three and wrte ths nto the box. 5. Solve the gven problem by indng any requested quanttes usng your sngle-phase equvalent. 6. Transform the results from ther sngle-phase values to the orgnal three-phase crcut: (a) The lne currents stay the same. (b) Multply any voltage results by √3.

20

PRAGMATIC POWER

FIGURE 1.14: Example III—3θ.

(c) Multply any “power” results (but not angles or power factor) by 3. (d) Eficency and voltage regulaton are the same. 7. Answer any “three-phase” questons such as indng the phase current n the load usng your knowledge of the relatonshp between lne and phase quanttes (Secton 1.3). Now let us try all ths!

1.4.3 Example III A three-phase, 480-V, Y-connected load absorbs 24 kW at 0.86 laggng through a system wth lne mpedances of 0.2 + j0.5 Ω (See Fgure 1.14). Draw the sngle-phase equvalent. Fgure 1.15 shows the result: 1. I drew the generator on the left and a box on the rght and then labeled the current Ilne. 2. The load voltage becomes 480/√3 = 277 V. The generator voltage s smply Vs. 3. The top wre gets the lne mpedance as stated. 4. Because the load s stated n “power” terms, I dvded by 3 and wrote the result nto the box.

FIGURE 1.15: Example III—1θ equvalent.

THREE-PHASE POWER: 3 > 3 × 1 21

5. Now I can solve whatever problem s gven. 6. If my results yeld a value for Vs, I multply that by √3 to obtan the line voltage at the generator end of the three-phase system. If I am to ind, say, Van, I use Vs as s. 7. A queston such as “ind VAN” s answered by dvdng the lne voltage by √3 and subtractng 30° from the angle.

1.4.4 Example IV A three-phase, 208-V, ∆-connected, 6-kW load s connected through lnes wth a lne mpedance of 0.3 + j0.8 Ω. The load power factor s 0.86 laggng. Draw the sngle-phase equvalent. Fgure 1.16 shows the result. The load voltage s 208/√3 and the load s 6/3 = 2 kW. It makes no dfference whether the load s Y- or ∆-connected.

FIGURE 1.16: Example IV—1θ equvalent.

1.4.5 Example V Two three-phase loads are drven by a 480-V system wth lne mpedances of 0.3 + j1.0 Ω. One load s Y-connected wth a per-phase mpedance of 6 + j3 Ω. The other s ∆-connected wth a per-phase mpedance of 27 + j12 Ω (see Fgure 1.17). Draw the sngle-phase equvalent.

FIGURE 1.17: Example V—3θ.

22

PRAGMATIC POWER

FIGURE 1.18: Example V—1θ equvalent.

Fgure 1.18 s the result. The load voltage s 480/√3, the Y-connected mpedance and the lne mpedance are used as s. The ∆-connected mpedance s dvded by 3 before usng.

1.5

FOUR COMPLETE EXAMPLES

That s enough talk about three-phase systems and analyss. Let us do some concrete examples usng the process of transformng the system nto ts sngle-phase equvalent. Remember that ths transformaton works only f the system s balanced. If t s unbalanced, we cannot smplfy t to the sngle-phase equvalent but have to wrte equatons for the whole three-phase crcut. If you happen to thnk, “Gosh, wrtng three-phase system equatons for large, complcated systems must be a huge chore,” you would be rght. The folks who deal wth large systems that are unbalanced but not badly so use a method called symmetrcal 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 mpedance s 0.1 + j0.6 Ω/∅ and the lne mpedance s 0.2 + j1.2 Ω/∅. Fnd the lne current, total power P, reactve power Q, and complex power S delvered by the generator. Also, ind the system’s voltage regulaton and ts eficency. Fgure 1.19 shows the sngle-phase equvalent. The load voltage s 4800/√3 = 2770 V. Both the generator and lne mpedances are shown. The ∆-connected load mpedance has been dvded by 3. Frst, I wll ind the lne current and the power “lost” n the generator and lne mpedances:

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 equvalent.

Usng the lne current, I wll next ind S for the load.

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

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

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

% VR = 100

3040 − 2770 = 9.7% 2770

and from the power results I get the percent eficency.

% h = 100 ×

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

24

PRAGMATIC POWER

But I am not done! The results need to be transformed back to the three-phase system. The “power” numbers are multpled by 3; the lne current, the regulaton, and the eficency 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 laggng. The feeders have an mpedance of 0.3 + j0.6 Ω each. Fnd the lne current, the generator voltage, the power factor at the generator, the voltage regulaton, and the eficency. The sngle-phase equvalent s Fgure 1.20. Note the dvson of the voltage by √3 and the dvson of the load power by 3. I wll ind the lne current irst because that quantty s key to the rest of the problem:

Iline =

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

Next comes the generator voltage by addng the voltage drop n the lne 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 equvalent.

THREE-PHASE POWER: 3 > 3 × 1 25

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

pfgen = cos (1.7◦ − (−36.9◦ )) = 0.782 lagging. Voltage regulaton s

% VR = 100 ×

127.6 − 120 = 6.3% . 120

To ind the eficency, I need ether the power delvered by the generator or the power dsspated n the lne. I wll use the lne power by notng that power s I 2R, where the current s the magntude of the lne current. Only the resstance of the lne can dsspate power, and resstances do not care about phase angles.

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

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

The only result that needs to be transformed to three-phase s the generator voltage, so I wll multply Vgen by √3 to gve a generator voltage n the three-phase system of 221.0 V.

1.5.3 Example VIII A 500-hp, three-phase, ∆-connected, 2300-V motor s 84% eficent. Its power factor s 0.75 laggng. The lnes feedng the motor have a lne mpedance of 0.2 + j0.5 Ω/∅. Fnd the generator voltage, the voltage regulaton, and the eficency. Also, assumng that the motor operates 100% of the tme all year, ind the annual cost of operaton f energy costs 3.0¢ per klowatt hour and the monthly demand charge s $10.00 per kVA. The capacty of the motor s the output power, because a desgner wants to ind the rght motor for the load t s gong to drve. The output horsepower s converted to nput power by multplyng by 746 W/hp and dvdng by the eficency. Dvdng ths by 3 gves the per-phase power for the sngle-phase equvalent. Fgure 1.21 s the equvalent.

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

Pload =

The lne current and the generator voltage are

26

PRAGMATIC POWER

FIGURE 1.21: Example VIII—1θ equvalent.

Iline =

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

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

% VR = 100 ×

1400 − 1328 = 5.4% . 1328

The eficency can be calculated usng the power absorbed n the lne and the load power:

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

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

The cost of operaton requres the “demand,” whch s the magntude of the complex power S for the motor. Ths s generally metered just as klowatt hours are metered, and the maxmum demand for the month s used. Usually the peak demand s calculated from a sldng average wth a wndow of perhaps 15 mn. For our motor, ths demand s constant. The sngle-phase equvalent results need to be multpled 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 n ths example s just one way that utlty companes assess VARs. Some companes charge for peak kVARs, some charge for low power factor, and some even gnore them n ther tarffs.

THREE-PHASE POWER: 3 > 3 × 1

27

1.5.4 Example IX Contnue wth example VIII by addng to the load suficent capactance to brng the load power factor up to 0.9 laggng. Then ind the generator voltage, the voltage regulaton, and the eficency, as well as the new annual operatng cost and how much s saved by addng the capactors. From the total power for the motor that we already know, we can get the reactve power for the motor. Usng ths n the power trangle of Fgure 1.22, I get the new reactve power for the combnaton of the motor and the added capactor. Note that the real power P for the motor s not gong to change because t stll s a 500-hp motor that s 84% eficent.

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 capactors would be purchased based on the requred kVARs and the operatng voltage. The new generator voltage, voltage regulaton, and eficency 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 mprovement—s the current, whch decreased from 148.6 to 123.9 A. Ths shows why utlty companes have rate structures that dscourage poor power factor n some way. In

FIGURE 1.22: Power trangle.

28

PRAGMATIC POWER

ths case, the current s over 16% less, whch means lne losses are about 30% less. Lne losses are on the utlty’s sde of the meter. Fgure 1.23 shows the completed system. The generator voltages are drawn as per-phase voltages so they have the same magntudes as n the sngle-phase equvalent. Now for the cost. The energy cost does not change because the motor s stll producng 500 hp. However, the demand s lower and the demand charge wll be reduced.

$power = $116, 683 per year (unchanged) 148.0 Pload = 164.4 kVA per phase = |Sload new | = 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 s an mprovement of $11,856 per year, an mprovement of 6.3%. Whether ths savng justies the cost of purchasng and nstallng capactors s a problem for the accountng folks n the front ofice.

1.6

SUMMARY

We started ths chapter by revewng materal for sngle-phase systems and then extended ths to three phases. By restrctng our systems to balanced ones, we can transform the three-phase system to ts sngle-phase equvalent and then do all of the analyss on just that phase. Results can be transformed back to the orgnal three-phase system.

FIGURE 1.23: Example IX—3θ wth correcton.

THREE-PHASE POWER: 3 > 3 × 1

29

In Secton 1.4, I descrbed a typcal power system that ncluded transformers to rase and lower voltages for transmsson, dstrbuton, and customer. Throughout the three-phase analyss n ths chapter, though, I have omtted the transformer. That comes n Chapter 2, where we wll see that the transformer can be modeled n terms of basc crcut elements and ncluded n the analyss of more complete three-phase systems. • • • •

31

CHAPTER 2

Transformers: Edison Lost Fortunately! Nkola Tesla (1858–1943) came to the Unted States n 1884 and began workng as an assstant to Thomas Edson (1847–1931). Edson had already nvented the lght bulb and had started buldng a large drect-current (DC) system n New York. Edson asked Tesla to debug the system and promsed to pay hm well for hs work. Tesla dd mprove the system, savng Edson consderable money, but Edson refused to pay Tesla. Edson was wedded to the dea of usng DC as the power source to lght everyone’s homes and busnesses, argung that alternatng current (AC) was dangerous. He staged elaborate electrocutons of anmals usng AC and noted that AC was used for executons at Sng Sng, the state prson. Tesla left Edson and went on to devse the bascs of AC power systems, demonstratng that large blocks of power could be moved by rasng the voltage very hgh, thereby reducng the requred current so smaller wres could be used. Edson’s DC system, on the other hand, would requre very large cables and, one wrter has sad, a generatng staton for each square mle of cty. The last of Edson’s DC systems dd not qut untl November 14, 2007, though, when the Pearl Street Staton that served old buldngs on New York’s Upper East and West Sdes was shut down after 125 years of operaton. One of the keys to Telsa’s AC system s the transformer, whch makes t possble to generate at a modest voltage, step the voltage up (and the current down, therefore) to a much hgher voltage for transmsson, and then step the voltage back down (and the current back up) for dstrbuton to consumers. (Tesla also nvented the nducton motor to use ths avalable AC power, but that s another chapter.) George Westnghouse (1846–1914) pushed Tesla’s deas, purchased European-made transformers, and bult a hydro-powered AC system n western Massachusetts. From there on, Westnghouse and Edson engaged n a btter war, but Westnghouse ultmately won. For ths, he receved the Edson Medal (rony?) from the Amercan Insttute of Electrcal Engneers, a forerunner of the Insttute of Electrcal and Electroncs Engneers.

2.1

BASIC TRANSFORMER

A transformer looks so smple that t s perhaps hard to beleve that t was a major key n the development of the AC power system. It s nothng more than a couple of cols of wre wound around a core of ron. Fgure 2.1 shows the basc structure.

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PRAGMATIC POWER

FIGURE 2.1: Transformer.

It s such a smple-lookng devce but t played a bg part n Tesla’s developments. How does t work? Wthout gong very deeply nto magnetc ield theory. . . . • The left col has an AC current lowng through t. Ampere’s crcutal law states that ths current creates an alternatng magnetc lux, shown as Φ n Fgure 2.1. • Ths lux “lnks” the rght col, passng through ts wndngs. • Faraday’s law says that a changng lux creates a voltage across the wndngs of the col that t lnks. • Vola! Applyng a voltage V1 on the left drves current I1 through the left col, creatng lux Φ n the ron core, lnkng the rght col, and developng a voltage V2 that s ready to do work when connected to somethng that draws current I2. Fgure 2.1 represents an deal transformer, whch means that t s lossless. What does ths mean? • Wres have zero resstance so that there s no power dsspated n the wres. • The lux s contaned entrely n the ron core; none escapes. All the energy n ths magnetc ield “makes t” from one col to the other. • The ron presents no “resstance” to havng ts magnetc domans lpped back and forth n step wth the AC current that s developng the lux. How about real transformers? Modern transformers almost meet these condtons. Wre losses and core losses are small enough that most transformers dsspate only a few percent of the energy that passes through them. Alrght, enough of the physcs! How does ths contrapton work? Electrcally, that s. Frst, let us use the more schematc symbol for the transformer. Fgure 2.2 shows the usual way of drawng

TRANSFORMERS: EDISON LOST

33

FIGURE 2.2: Transformer symbol.

a transformer. The core s reduced to the two vertcal lnes, whch means that ths s an ron-core transformer (as opposed to ar core). The voltages and currents are the same as n the irst drawng. The turns rato s stated, whch s often gven as “a:1” or “1:a” nstead of specfyng actual turns. A transformer wth a turns rato of 1000:250 s more lkely to be labeled “4:1.” The relatonshps between V1 and V2 and between I1 and I2 follow from Ampere’s and Faraday’s laws (whch I wll not do here):

N2 V2 = , V1 N1

I2 N1 = . I1 N2

These state smply that the voltage transfers as the turns rato, the current as the nverse of the turns rato. Fne, but the goal of usng a transformer n a power system s to move power through the system. So what happens n the deal transformer when we consder power? Let us look at power as smply the product of voltage and current as f phase angles were all zero. The power delvered to the left sde of the transformer s

P1 = V1 I 1 and the power leavng the rght sde s

P2 = V2 I2 . If I use the ratos for voltage and current, I get � �� � N1 N2 P2 = V1 I1 = V1 I 1 . N1 N2 What goes n comes out! Remember, though, that ths s an deal transformer so t s lossless.

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PRAGMATIC POWER

Before gong further, I need to ntroduce two terms: primary and secondary. These are terms for convenence to help us determne whch way energy s movng. We generally mean that the prmary s the sde to whch power s delvered and the secondary s where power “comes out.” Note that the transformer knows no such restrcton and can support power low n ether drecton.

2.1.1 Example I: Ideal Transformer Before tacklng more realstc transformers, let us look at a crcut nvolvng the deal transformer. The crcut of Fgure 2.3 ncludes an deal 5:1 transformer. The two R-L mpedances represent connectng lne (wre) resstance and nductance. The 5-Ω resstor represents the load to be served. The generator voltage VS = 120 ∠ 0°V. The goal s to ind the load voltage VL. I could solve ths problem by wrtng equatons on the prmary sde that nclude the transformer voltage and current (V1 and I1 of Fgure 2.2), then wrtng equatons on the secondary sde that nclude the transformer voltage and current there (V2 and I2), and inally usng the transformer ratos we have just seen. That process wll yeld several equatons. But I am lazy and I would be happer wth a smpler method. The smpler method s to do what s called “relectng,” meanng that we can “pass through” the mpedances and sources from one sde of the transformer to the other. So, before I proceed wth the example, I need to show what “relect” means: • Lookng back at the equatons of the prevous secton, I see that the voltage on, say, the “2” sde relects to the left as

V1 = V2

N1 . N2

• Smlarly, the current on the “2” sde relects as

I1 = I2

FIGURE 2.3: Example I.

N2 . N1

TRANSFORMERS: EDISON LOST

35

FIGURE 2.4: Example I: relected to prmary.

• Rememberng that mpedance s voltage dvded by current, I wll see what happens to the mpedance as t relects from the “2” to the “1” sde:

Z2 =

V2 , I2

�

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

I n words, ths shows that an mpedance on one sde (here, the “2” sde) relects to the other sde (here, the “1” sde) as the turns rato squared. Take a look at Fgure 2.4 where I have relected the three mpedances and VL from the secondary (rght) sde through the transformer to the prmary (left) sde. The two resstances and the reactance have been multpled by 52 and the voltage has been multpled by 5. Now I can wrte a smple Ohm’s law equaton to solve for I and from that obtan 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 inal result by relectng the load voltage VL back to the secondary, whch means dvdng by the turns rato:

VL = 5VL /5 = 22.2∠ − 5.8◦ V. To llustrate that ths relecton process can be used n ether drecton, I am gong to take the same example but relect the prmary sde of the crcut through the transformer to the secondary sde. I wll dvde the source voltage by the turns rato, multply the current by the turns rato, and dvde the mpedances by the turns rato squared. Fgure 2.5 shows the result of ths.

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PRAGMATIC POWER

FIGURE 2.5: Example I: relected to secondary.

Now, let us ind I and VL agan:

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-Ω load s 5I, not I n Fgure 2.5.

2.1.2 Example II: Ideal Again Here s one more example wth the deal transformer and lossy wres, ths tme wth the load stated n power terms rather than as an mpedance. The process wll be the same, relectng the secondary to the prmary and then indng what s requred. The dfference, though, s n how the load s relected. A load s a load! So the 50-kW load remans 50 kW no matter whch sde of the transformer t s on. Remember that power n equals power out for an deal transformer. The load’s voltage and current both change when we relect the load, but ther product does not. Fgure 2.6 s the crcut. We are to ind Ilne, Vgen, percent eficency, and percent voltage regulaton.

FIGURE 2.6: Example II.

TRANSFORMERS: EDISON LOST

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FIGURE 2.7: Example II: relected to prmary.

I relect the secondary nto the prmary, the resstance and the reactance are multpled by 27.52, the load voltage s multpled by 27.5, and the load remans unchanged. The outcome s shown n Fgure 2.7. I am gong to use the load power to ind the magntude of the current Ilne and the power factor to ind ts 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 choce of the mnus sgn on the angle because the power factor s laggng so the current must lag the voltage. The generator voltage s the load voltage (taken wth a phase angle of 0) plus the voltage drop n the mpedances of the lne:

Vgen = 13.2 × 103 + (30 + j50 + 75.6 + j151.3) Iline = 14.12∠2.0◦ kV. The losses n the lne are due only to the resstances and the magntude of the current, so the percent eficency s

% h = 100 ×

50 × 103 (30 + 75.6) (4.564)2 +50 × 103

= 95.8% .

The percent voltage regulaton s

% VR = 100 ×

14.12 − 13.2 = 7.0%. 13.2

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PRAGMATIC POWER

2.2

TRANSFORMER MODEL

So far, the transformer has been portrayed as perfect. All t does s st n the system happly reducng the voltage and ncreasng the current—or the other way around. It s happy because t s very “green” conscous, wastng no energy tself. Unrealstc, rght? Yup! The transformer s no more perfect than anythng else we use or do. The good news, though, s that the transformer s a very eficent devce, dsspatng only about 5% of the energy passng through. We can make a rather decent model of the transformer that ncludes the effects of ts losses and take nto account other “real world” characterstcs at the same tme.

2.2.1 Realistic Model Fgure 2.8 s a good model of a real transformer, yet farly smple. In the prevous secton, I lsted the attrbutes of the deal transformer. The crcut elements n the model n Fgure 2.8 represent the nondealness of the transformer. Let us look at each of the added elements to see how they reasonably account for what s happenng n a real transformer. Let us start on the left and work across: • R1 represents the losses n the wre of the prmary. • The reactance X1 s an nductance that accounts for the fact that the magnetc lux created by the prmary col does not qute stay entrely n the ron core. Some of the lux leaks out, and some of the lux does not even lnk all the turns of the wndng. Hence, the prmary wndng looks to some extent lke an nductor. • Just as the wndng ntroduces losses, so does the core, represented by the parallel resstance Rc. Iron s a great magnetc materal wth magnetc domans that lp from one drecton to another qute easly. But not wthout some resstance. It takes some energy to lp the domans back and forth n step wth the AC appled to the transformer. Rc accounts for ths loss.

FIGURE 2.8: Transformer model.

TRANSFORMERS: EDISON LOST

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FIGURE 2.9: Transformer model relected.

• As domans are lpped one way or another, energy s beng stored n the core n the same way that energy s stored n the magnetc ield of an nductor. Hence, the core presents some nductance, called the magnetzng reactance, represented by the parallel reactance Xm. • The transformer symbol n the model represents an deal transformer. All of the nondeal characterstcs are represented by the other sx crcut elements. • The secondary col has characterstcs smlar to those of the prmary, namely, resstve losses represented by R2 and leakage represented by X2. I suppose t s tme to get worred. What started off as a smple, deal devce s now complcated by the addton of sx crcut elements. It s begnnng to look lke the analyss of a crcut contanng a real transformer s gong to be hard work. Do not worry—thngs are not that bad because we can make some smplicatons that reduce the number of elements n the model and also smplfy the analyss. Remember that we learned how to relect an mpedance from one sde of a transformer to the other? Let us do that here. Let us take the two elements on the secondary sde and relect them to the prmary. Fgure 2.9 shows ths, wth a prme desgnatng the values of these elements after relectng them through the transformer. Okay, but f you buy ths as a smplicaton, I have some snake ol for sale, too! Ths does not smplfy anythng, really, untl we learn more about the numercal values of these elements. I am

FIGURE 2.10: Transformer model relected and smplied.

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PRAGMATIC POWER

FIGURE 2.11: Transformer model wth numbers.

gong to just state how these values relate and ask you to take t on fath untl 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 ths, R1, X1, R2′, and X2′ are all n seres and can be combned nto just two elements. Fgure 2.10 shows the result of dong all ths. Ths turns out to be a neat smplicaton that can be carred further when we actually embed a transformer n a crcut wth other crcut elements. But how do I justfy t? You saw my “much greater than” condtons, but how can we just move the parallel elements? The justicaton comes by lookng at the currents. The current through the seres R + jX s large because t s the current that s delverng energy to the rght. The current through the parallel elements s much, much smaller, perhaps only about 1% of the current through R and jX. Therefore, movng 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. Fgure 2.11 shows our “adjusted” model wth actual numbers. These numbers are typcal for a 10-kVA transformer desgned to transform 13.2 kV to 480 V. Notce how the numbers do meet my condtons for movng the parallel elements—they are large, gven n kΩ, not Ω, whch means that they are a thousand tmes larger than the seres elements. How do we get these numbers? There are some smple measurements that we can make on a transformer that yeld good results. We wll look at transformer testng n the next secton. Before that, though, let us use ths model n two examples.

2.2.2 Example III A 480-V 10-kW load that operates wth unty power factor s to be fed from a 13.2-kV supply usng the 10-kVA transformer just modeled. The system s shown n Fgure 2.12. Our job s to ind the voltage needed at the generator, Vgen, and the overall eficency of the system. Start by relectng the load through the transformer to the prmary. The only number that changes s the voltage. A 10-kW load s the same on ether sde of the transformer, so only the voltage of 480 V changes to the prmary voltage of 13.2 kV. I wll be lazy and not even draw the rearranged crcut.

TRANSFORMERS: EDISON LOST

41

FIGURE 2.12: Example III.

The current through the relected load s a good startng place, because from that I can get the voltage drop across the two seres elements. That wll let me get to the generator voltage. Recallng 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 ind eficency, I need to know ether the power delvered by the source or the losses n the crcut. If I want to do ths usng the source power, I wll need the lne current, whch s the load current plus the current through the parallel elements. Whle dong ths, I wll get to see how much of a factor ths parallel branch current s relatve to the load current and get a feelng for whether movng the parallel branch to the left was a good approxmaton.

Iparallel =

Vgen 3

+

Vgen

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

Let us note here that ths current, lowng through the parallel R-L combnaton, s only about 4% of the load current, so movng t from one sde of the seres R-L to the other (Fgure 2.10) s not a sgnicant change. Now that I have the current through the parallel branch, I can get the lne current and from that, the power suppled 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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PRAGMATIC POWER

The percent eficency s

%h = 100 ×

10 = 94.2%. 10.61

One way to check ths s to igure out the losses nstead of the power suppled by the generator. Sometmes ths s easer to do, too. The losses are represented by just the two resstance elements. I know the current through the seres resstance and I know the voltage across the parallel resstance, so I wll use I-squared-R and V-squared-over-R to get the power lost. Note that phase angles do not enter nto ths calculaton because a resstor’s voltage and current are n phase.

P loss = (518) (|I load |)2 +

(|Vgen )2 581 × 103

= 617 W. %h = 100 ×

10 = 94.2%. 10 + 0.617

Close enough!

2.2.3 Example IV One more example wll llustrate transformer power calculatons usng the move-the-parallel-branch approxmaton. The crcut of Fgure 2.13 ncludes between the termnals a 50-kVA, 13.2-kV/480-V transformer. Measurements and the approxmaton of Fgure 2.10 have yelded seres values of 90 and j220 Ω and parallel values of 120 and j200 kΩ. The load here s a 480-V, 50-kW load wth a power factor of 0.83 laggng. Ths load s fed from the secondary by a lne wth an mpedance of 0.1 + j0.2 Ω. The transformer’s prmary s fed by a lne wth an mpedance of 30 + j50 Ω. Our job s the usual: ind the necessary generator voltage, the overall eficency of the crcut, and the voltage regulaton.

FIGURE 2.13: Example IV.

TRANSFORMERS: EDISON LOST

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FIGURE 2.14: Example IV smplied.

Smplfyng the crcut seems lke a good dea, so I wll irst relect the load and ts feeder to the prmary. After relecton, the load remans 50 kW at 0.83 laggng, but ts voltage becomes 13.2 kV. The lne mpedance s multpled by the rato squared, whch yelds 75.6 + j151.3 Ω. I could add that result to 90 + j220 Ω, but I am gong to take the approxmaton one step further. I wll move the parallel elements all the way to the left, past 30 + j50 Ω. After all, f some movng s good, more should be better! When I make ths move, then all three R-L seres combnatons smply add. The result s shown n Fgure 2.14. Now the problem s not much dfferent from the prevous example. Frst, 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) +

� �2 15.05 × 103 120 × 103

= 5.96 kW. Fnally, percent eficency and percent voltage regulaton:

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

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PRAGMATIC POWER

FIGURE 2.15: Example IV not qute so smplied.

Ths s a good place to see how much the move-the-parallel-branch approxmaton changes the inal results. I am gong to rework the example wth the secondary load and feeder mpedance relected to the prmary but wthout movng the parallel elements further to the left. Fgure 2.15 shows ths not-so-smplied crcut. The load current s the same as before, namely, 4.564 ∠ − 33.9°A . From ths, I ind 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 lne current s the sum of the load current and the current through the parallel branches:

I line = 4.564∠ − 33.9◦ +

14.81∠3.8◦ × 103 120 × 10

3

+

14.81∠3.8◦ × 103 j 220 × 103

= 4.703∠ − 33.6◦ A. Now get the generator voltage by addng the voltage drop across the seres 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 yelded 15.05 kV.) The power losses are

P loss = (165.6) (4.564)2 +

(14.81 × 103 )2 120 × 103

+ (4.703)2 (30)

= 5.94 kW, whch s slghtly smaller than the 5.96-kW losses computed after movng the parallel branches. The percent eficency s the same:

TRANSFORMERS: EDISON LOST

% h = 100 ×

45

50 = 89.4% . 50 + 5.92

In other words, movng the parallel branches around makes very lttle dfference n the results but smplies the computatons enough to make the move worthwhle. Two notes about the “real world:” (1) The numbers on my sample transformers are for rather lossy transformers; real transformers generally are better, meanng losses are smaller. (2) Practcng engneers often completely omt the parallel R-L branch from model calculatons except for very large transformers. Now that we have used models for a couple of transformers, t s tme to see how we obtan data for those models. The next secton wll show how.

2.3

TRANSFORMER TESTING

Our transformer model s rather smple, seres resstance and nductance representng the actual col, parallel resstance representng core loss, and parallel nductance representng magnetzaton effects. But how can we get numercal values for these? The cols and the core are all there s, and we need to somehow measure these parameters from the outsde. The approach to gettng these numbers s n a way lke a physcan’s determnng what s gong on n your heart by observng voltages from outsde the body. We smply have to ind combnatons of external voltage and current measurements that gve us a handle on the elements themselves. The method reles on the fact that the numercal values of the two parallel elements n the model are many tmes the numercal values of the seres elements. Ths “many tmes” s several hundred. Hence, we can devse tests that “see” rather clearly the seres elements wthout nluence of the parallel ones and vce versa. There are two parts to a transformer test: the short-crcut test and the open-crcut test. These tests are done wth knowledge of the rated characterstcs of the transformer, namely, the rated current and the rated voltage. Both of these can be obtaned from the nameplate data. A transformer has a stated kVA ratng as well as a stated voltage rato. For example, one of the transformers n the prevous secton has been a 10-kVA, 13.2-kV/480-V transformer. From these data, we can obtan the rated current for both sdes of the transformer.

2.3.1 Short-Circuit Test I wll start wth the short-crcut test. Ths s normally done on the hgh-voltage sde of the transformer because the rated current on ths sde s smaller than on the low-voltage sde. The shortcrcut test requres three meters: an ammeter, a wattmeter, and a voltmeter. Fgure 2.16 shows how the meters are arranged. Note that the low-voltage sde of the transformer s shorted out, whch s where ths test gets ts name. Let us see what data we collect on our 10-kVA, 13.2-kV/480-V transformer.

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PRAGMATIC POWER

FIGURE 2.16: Short-crcut transformer test.

Frst we need to determne rated current on the hgh-voltage sde:

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

Irated =

Now we adjust the supply voltage Vs untl the ammeter reads 0.758 A. Note that rated current s now lowng n the col of the hgh-voltage sde, whch means that rated current s also lowng n the short crcut on the low-voltage sde. That short-crcut current s

Ishort-circuit =

13.2 × 103 0.758 = 20.85 A, 480

although we do not need to know that for ths 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 peces of data, we can calculate the values of the seres elements R and X. But, you mght ask, how can you gnore the two parallel elements? Frst, they are gong to be several hundred tmes the szes of R and X. Second, the appled voltage (1005 V) s much smaller than the rated voltage of 13.2 kV, so the currents through the parallel elements and the power dsspated n them wll be much smaller than they would be at rated voltage. Here are the three steps n the calculatons for R and X based on the data from the shortcrcut 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. Determne 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 negatve because the current must lag the voltage N n an nductve crcut. 3. C ompute the seres mpedance n rectangular form, completely gnorng the parallel elements n the model: 1005 Vsc = 1326∠67.0◦ W = Zseries = Isc 0.758∠ − 67.0◦

= 518 + j1221 W, R = 518 W, X = 1221 W. There! That takes care of the seres elements, so we are half done wth the tests.

2.3.2 Open-Circuit Test The open-crcut test uses the same meter arrangement, but t s usually done on the low-voltage sde of the transformer as shown n Fgure 2.17. Notce that the termnals on the opposte sde are not connected to anythng. To begn, we set the source voltage to the rated voltage on the low-voltage sde, so we set Vs = 480 V. If we want to check the transformer’s turns rato, we would measure the voltage Vo. Now we read the three meters. Suppose the outcomes are

FIGURE 2.17: Open-crcut transformer test.

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PRAGMATIC POWER

Voc = 480 V, Poc = 397 mW, Ioc = 1.078 mA. (The voltage across the prmary termnals, Vo, s now 13.2 kV, so safety s a major factor n these tests.) The calculatons are just about the same as before: 1. Calculate the power factor:

pfoc =

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

2. Determne the phasor current n 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 magnary parts of the open-crcut 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, Fgure 2.18. Why are these tests made from opposte sdes of the transformer? Would not t be easer to do all ths wth everythng set up on just one sde? Sure. But there s a reason for the two choces. The short-crcut test s made from the hgh-voltage sde because ths requres a smaller current. The

FIGURE 2.18: Results of transformer tests.

TRANSFORMERS: EDISON LOST

49

open-crcut test s made from the low-voltage sde because ths requres a smaller voltage, whch s safer. The hgh-voltage termnals can be completely guarded aganst accdental contact.

2.4

THREE-PHASE EXAMPLE

Example V s a three-phase system that s to supply a three-phase, ∆-connected load from a 13.2-kV feeder. The load s rated at 480 V, 25 kW at a power factor of 0.9 laggng. The lnes to the load have an mpedance of 0.2 + j0.5 Ω/∅. We are to provde the transformer and then ind the requred prmary voltage, the overall eficency, and the voltage regulaton.

2.4.1 Example V Using Y–Y Fgure 2.19 shows the complete three-phase system. However, once I get some parameters establshed, I wll reduce ths to ts sngle-phase equvalent for my calculatons. The total load s 25 kW at 0.9 laggng. Ths means the total power s

Stotal =

Ptotal 25 = = 27.78 kVA. pf 0.9

Ths s a balanced three-phase system, so I can use three 10-kVA, 13.2-kV/480-V sngle-phase transformers that I chose to arrange Y–Y. (Remember that the stated voltage for three-phase systems s always the lne voltage, the voltage between the lnes, so the cols are seeng less than rated voltage.) Each load n the crcut s 25/3 kW at 0.9 laggng.

FIGURE 2.19: Example V.

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PRAGMATIC POWER

FIGURE 2.20: Example V: sngle-phase equvalent.

I wll start by convertng ths crcut to ts sngle-phase equvalent. The load s unchanged, but ts voltage per phase must be dvded by √3. The lne mpedance s the same for one lne. The sngle-phase equvalent s Fgure 2.20. The next step s to smplfy ths by relectng the load and ts lne mpedance through the transformer as shown n Fgure 2.21. The load current s 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 prmary.

93032 581 × 103

= 1.137 kW,

TRANSFORMERS: EDISON LOST

51

so the eficency s

%h = 100 ×

8.33 = 88.0%. 8.33 + 1.137

Fnally, the voltage regulaton s

% VR = 100 ×

9303 − 7620 = 22.1%. 7620

The job s not inshed, though, because there are two thngs to do. Frst, I need to convert the supply voltage (the voltage on the prmary sze of the Y–Y transformers) back to the three-phase system:

√ Vs = 9303 3 = 16.11 kV. Second, there s a queston of whether ths load s exceedng the rated capacty of the transformers. Huh? How can that be, because each phase of the load s only 8.33 kW and the transformers are rated at 10 kVA? Yes, but we are operatng them at below ther rated voltage, so the current could be above the rated value. The load current on the prmary sde s

|Iload | = 1.215 A, so the total power on the prmary sde of one of these transformers s

|Stransformer | = 7620 × 1.215 = 9.26 kVA. Whew! That s less than 10 kVA per phase, so everythng looks okay. But thngs are not all that good. Look back at the eficency (88.0%) and the percent voltage regulaton (22.1%). Those are rather poor numbers. Now, consder the prmary current ratng:

|Irated | =

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

Our prmary current of 1.215 A exceeds ths by 60%!

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PRAGMATIC POWER

What s the problem? Wrong transformer! The transformer s underused. It s rated at 13.2 kV/480 V and we are usng t at 7620/277. Same rato, but the current s hgher, whereas the voltage s lower. Hgher current means hgher losses n the seres elements. Now what?

2.4.2 Example V Using ∆–∆ If I am stuck wth these 10-kVA transformers, then I should use them at somewhere near ther rated voltage and current. Because the load requres a lne voltage of 480 V, I can connect the transformers ∆–∆ so ther secondary voltage s 480 V. Fgure 2.22 s the same load wth ts feeders, but now suppled by three 10-kVA transformers arranged as ∆–∆. As before, I wll convert ths to ts sngle-phase equvalent. The only dfference from what we saw before s that the delta-connected source also needs to be transformed. Ths s done by dvdng the voltage by √3 as usual and dvdng the mpedances of the transformer model by 3. Fgure 2.23 s the result. The calculatons proceed as before wth the followng 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 ∆–∆.

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53

FIGURE 2.23: Example V: move to prmary agan.

Not bad! Eficency s up a lttle and voltage regulaton s greatly mproved. All that remans s to convert the supply voltage back to the three-phase crcut:

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

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

|Stransformer | = (14.58 × 103 )(0.701) = 10.22 kVA That s about 2% larger than the 10-kVA ratng but t s acceptable as long as the transformer operates wth adequate coolng so that t stays wthn temperature specicatons. Can we do even better? Not much wthout indng a more eficent transformer.

2.5

SUMMARY

If Nkola Tesla and George Westnghouse had not strongly dsagreed wth Thomas Edson, we mght not have needed ths chapter. The transformer s the key to the power system we have today. The deal transformer, a lossless devce, s not real, but the modern transformer s pretty close to t, dsspatng only about 5% of the power passng through. An actual transformer can be modeled very well wth only four elements: seres resstance to account for the wre losses, seres nductance to account for leakage lux, parallel resstance to represent losses n the ron core, and parallel nductance to represent the nductve effect of magnetzng ron. Ths smple model lends tself very well to a straghtforward testng scheme that yelds values of the four elements wthout a great deal of work. The short-crcut and open-crcut tests requre only standard nstruments and a faclty that can provde the voltages and currents (and safety!)

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PRAGMATIC POWER

needed. Once the model parameters have been determned, the model can be used to represent the transformer and permt crcut analyss usng our usual technques. Yet even though Edson lost, a strange thng has been happenng more and more n recent years. We are runnng more equpment on DC! How so? Well, your computer and your dgtal TV and probably your stereo and certanly your DVD player are all devces that nternally use DC power. Even stranger s that we run them from a DC supply—almost. We buy an unnterruptble power supply (UPS) so that power gltches do not kll our computers. But what s a UPS but a DC supply that takes n AC, charges batteres, produces DC, and then converts the DC back to AC. Where next? Tesla rdes agan! He nvented the nducton motor, whch n ts three-phase form s a spectacularly smple machne—one movng part. Just about every sgnicant load that requres a rotatng shaft s drven by a three-phase nducton motor. Hence, n a study of AC power, we wll take a good look at ths nducton motor n the next chapter. • • • •

55

CHAPTER 3

Induction Motors: Just One Moving Part Nkola Tesla’s alternatng current (AC) system beat Edson’s drect current (DC) system not only because of the transformer but also because of the nducton motor, whch Tesla nvented. Tesla’s patent (No. 416,195, December 3, 1889) has a drawng of a three-phase motor, but hs clam s for “two or more energzng-crcuts through whch alternatng currents dfferng n phase are caused to pass.” The beauty of Tesla’s nventon does not le n the varous arrangements of the parts of the motor so much as n the fact that the three-phase nducton motor has just one movng part. You wll work a long tme to develop a rotatng prme mover wth fewer! Ths chapter s gong to stck wth the three-phase nducton motor. Ths motor s, by far, the most common method of transformng electrcal energy nto rotatonal energy. The motor s smple, rugged, and surprsngly eficent. We wll start by lookng at how the motor works and develop an equvalent crcut for t. Then we wll look at detals such as nameplate data, testng, and energy low. We wll end wth some example calculatons and a dscusson of why a desgner should not “over-motor.”

3.1

ROTATING FIELDS

The three-phase nducton motor creates a rotatng magnetc ield that drags an armature around wth t. To explan ths, though, I am gong to start wth a smple one-phase structure and see how that could work.

3.1.1 Single-Phase Motor The two poles of the motor n Fgure 3.1 are energzed by a current lowng such that the top pole s magnetc north and the bottom pole s south. These two poles form the stator, whch s the nonmovng part of the motor. For the moment, assume that ths magnetzaton s beng done by a steady current. The movng part n Fgure 3.1 s the rotor, whch n ths llustraton s a permanent magnet. If the rotor s started n the poston shown n the drawng, t wll experence a clockwse torque

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PRAGMATIC POWER

FIGURE 3.1: Clockwse torque.

because, as we learned n hgh school physcs, “lke poles repel, unlke poles attract.” Ths rotor, however, s not gong to get very far as shown n Fgure 3.2. The “unlke poles” wll get lned up and the torque wll cease. But suppose we reverse the magnetc ield of the poles as the rotor s arrvng. Fgure 3.3 shows ths and assumes that the rotor has some nerta that carres the magnet past the lower pole so that the new pole arrangement can contnue to provde torque.

FIGURE 3.2: Torque contnues.

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57

FIGURE 3.3: Feld reversal.

Now let us provde the magnetc ield of the stator usng AC. Then the poles swtch back and forth between N–S and S–N. It sounds lke the rotor ought to follow that reversng ield. Actually, t wll—but t wll not start by tself, whch s not good.

3.1.2 Three-Phase Rotating Magnetic Field Tesla’s nventon was for a “multphase motor.” The most common arrangement s three phases. Fgure 3.4 shows the motor wth two pars of poles added. The three pars are spaced unformly around the crcumference of the stator. Each par of poles s energzed by one phase of a three-phase system. In the arrangement shown n the drawng, not only are the poles spaced 120° apart around the crcumference, but they are energzed by successve phases of the power system that are also spaced 120° apart. (That s an observer standng n there to help us later.) All three pars of poles are the same, so the magnetomotve forces (mmf ) developed by the pars of poles all have the same magntude. The mmfs for the three pars dffer because they are excted by dfferent phases of the power lne and because they are n dfferent postons around the crcumference. The equatons below all have the same mmf magntude (F ). The angle θ s measured clockwse startng at the north pole of the a phase. The observer s statoned at an angle θ relatve to the north pole for the a phase. Hence, relatve to the b phase’s north pole, the observer s “behnd” by 120°. For the c phase, the observer s 240° “behnd.” The magnetc lux Φ generated by each of the pole pars s

58

PRAGMATIC POWER

FIGURE 3.4: Three-phase ield.

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 wll skp all the trg relatonshps to get to the inal result:

Ftotal =

3 Fcos(w t − q ). 2

Ths says two thngs. At any nstant, the ampltude of the lux s a constant and the lux s dstrbuted around the stator as cos(−θ) s spread around the crcumference. If the observer stands stll as shown n Fgure 3.4, the observer sees a magnetc lux that seems to vary n ampltude as cos(ωt) vares. But more mportant, ths ield s rotatng unformly around the stator. No matter where the observer s standng, the lux appears to vary n ampltude as cos(ωt) vares. An observer standng at θ = 0° wll see a lux varyng as cos(ωt). An observer standng at θ = 10° wll see a lux varyng as cos(ωt – 10°). In other words, ths second observer wll see the same lux as the irst observer, but 10° later. Hence, the lux s rotatng clockwse around the stator n step wth the frequency of the power lne. At 60 Hz, the lux rotates clockwse around the stator at 60 rev/s, whch s 3600 rpm.

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59

FIGURE 3.5: Add a rotor.

The drecton of rotaton s establshed by the angles for the three phases. Here, they are 0°, –120°, and –240°. Ths order s called abc rotaton. If t were the other drecton, t would be called acb rotaton.

3.1.3 Add a Rotor Now that we have magnetc lux that s rotatng around the stator, let us add a rotor and see what happens to t. Fgure 3.5 shows the added rotor. There s just one catch. If the rotor s standng stll, t sees the lux as smply oscllatng at the power-lne frequency, just as the observer dd. Therefore, because the rotor experences no net torque, so t just sts there dong nothng useful. That s no better than our one-pole-par motor n Fgure 3.3. One way we can help the rotor to develop torque s to get t spnnng at the same speed that the lux s rotatng around the stator. At 60 Hz, we must set t to spnnng at 3600 rpm, a condton that develops torque by followng the peak of the lux around the stator. As we requre more torque from ths moton (whch s presumably drvng some type of load through a shaft), the rotor wll lag behnd the peak by a larger and larger angle. If we requre too much torque, the rotor wll “break loose” from the torque peak and wll not produce any more net torque. A motor fabrcated ths way s called a synchronous motor. Ths type of motor s not very common, but s sometmes used where a load requres torque at exactly the synchronous speed. If we add

60

PRAGMATIC POWER

FIGURE 3.6: Add a workng rotor.

three more pars of poles to the stator of Fgure 3.4, thereby spacng the phases 60° apart nstead of 120°, we ind that a 60-Hz magnetc ield rotates around the stator at a synchronous speed of 1800 rpm. If we add three more pars, the synchronous speed becomes 1200 rpm, and so on. Let us add a rotor that works. Instead of just a magnet, ths rotor s gong to have cols of wre n ts crcumference. But these cols do not go “around the rotor.” Instead, they go lengthwse. In Fgure 3.6, the col marked X on the left sde of the rotor goes around the end of the rotor and comes back as X′ on the rght sde. The col's ends are all ted together to form numerous shortcrcuted loops. Usng ths new rotor arrangement, we have a stuaton where the three-phase supply creates a rotatng ield around the stator. Ths changng ield nduces a current n the rotor wndngs. Ths rotor current acts wth the stator ield to develop a crcumferental torque. Ths torque tres to make the rotor “catch up” wth the rotatng stator ield. But the rotor cannot catch up wth the stator ield! If there s no relatve rotaton, there s no nduced current. If there s no nduced current, there s no nteracton between the rotor and the stator. If there s no nteracton, there s no torque. To work, the rotor must slip relatve to the rotaton of the stator lux. Fgure 3.7 shows a three-phase nducton motor wth part cutaway so some of the stator wndng and some of the rotor wth ts wndng are vsble. The wndng s very smple because t s just cast alumnum bars n slots n the rotor. The bars are shorted together at each end. Fgure 3.8 shows a porton of these bars wthout the ron of the rotor. The pcture shows the short crcut at one end of the bars (The short at the other end has been cut off.). Notce the protrusons stckng out of the end that str the ar and provde some coolng nsde the motor.

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61

FIGURE 3.7: Rotor cutaway vew.

3.1.4 Slip, Speed, and Poles The rotor n an nducton motor cannot travel as fast as the rotatng ield or there would be no torque. The rotor s sad to slip relatve to the rotaton of the ield. To deine ths slp, however, we need irst to look at the rotatonal speed of the ield. A motor s descrbed by statng the number of poles per phase. The three-phase motor n Fgure 3.6 has sx dstnct poles, but there are just two per phase. Hence, ths motor s sad to be a “two-pole motor.” Poles always appear n pars.

FIGURE 3.8: Alumnum rotor bars wthout ron.

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PRAGMATIC POWER

Synchronous speed s the speed at whch the magnetc ield of the stator rotates around the stator. If t were not for slp, ths s the speed at whch the rotor would revolve. The synchronous speed for a three-phase motor s

ns =

60f , p/2

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

ns =

3600 . p/2

Slp s s the fracton by whch the actual rotor speed nm dffers from the synchronous speed ns:

s=

ns − nm . ns

For example, a two-pole motor whose rotor s spnnng at 3420 rpm has a slp of

s=

3600 − 3420 = 0.05, 3600

whch s normally expressed as a percentage: “5% slp.” Suppose we have a sx-pole motor operatng wth slp of 4%. The motor speed nm s

ns =

3600 = 1200, 6/2

nm = 1200 (1 − 0.04) = 1152 rpm. Three-phase nducton motors commonly operate wth 2–5% slp.

3.2

EQUIVALENT CIRCUIT

An equvalent crcut s helpful n understandng how energy lows through the nducton motor. The equvalent crcut s gong to be a per-phase equvalent just as we dd wth balanced three-phase systems n Chapter 1. The equvalent s on the Y bass, whch means that voltages are phase voltages. It s easy to forget that phase voltages are lne voltages dvded by √3. When we developed an equvalent crcut for the transformer, we took nto account two types of losses: copper and core. We also ncluded leakage and magnetzaton reactances. We need all four of these n the nducton-motor equvalent as well. The magnetc part of the nducton motor s qute dfferent from that of the transformer, though. The nducton motor has an argap that ncreases by a large amount the magnetc reluctance of the magnetc part of the system. In addton,

INTRODUCTION MOTORS: JUST ONE MOVING PART

63

the frequency of the nduced current n the rotor s much less than the lne frequency because of slp. I am gong to develop the model n two steps, consderng the stator irst and then the rotor.

3.2.1 Stator Equivalent Circuit The per-phase equvalent crcut for the stator of the nducton motor needs to nclude copper losses, leakage reactance, core (ron) losses, and magnetzaton reactance. Recall from the transformer model that leakage reactance represents magnetc lux that does not stay n the ron and that magnetzaton reactance represents energy stored n the magnetc domans of the ron. (See Secton 2.2.1.) The crcut shown n Fgure 3.9 s the sngle-phase equvalent of the stator of the nducton motor. Note that the components are smlar to those n the transformer model. Voltage E2 shown on the rght s the voltage that s nduced nto the rotor by the stator.

3.2.2 Rotor Equivalent Circuit The rotor s more complcated because slp s nvolved. Slp determnes the frequency of the current n the rotor col. When the rotor s completely stopped, the slp s 1 and the rotor “sees” 60 Hz from the stator. When the rotor s gong at synchronous speed, the slp s 0 and the rotor “sees” 0 Hz from the stator. The motor produces no torque when the slp s 0. Fgure 3.10 shows one form of the rotor’s equvalent crcut. It ncludes both the “copper” loss of the rotor wndng (whch s commonly alumnum) and the reactance of the rotor col. Two thngs stand out about ths model. Frst, reactance X2 s the 60-Hz reactance so that we have everythng referenced to a common frequency. The frequency n the rotor wndng, however, s

FIGURE 3.9: Sngle-phase stator equvalent.

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PRAGMATIC POWER

FIGURE 3.10: Rotor equvalent.

much lower. If the slp s 0.03, the rotor frequency s 0.03 × 60 = 1.8 Hz. Ths s why the reactance s shown as jsX2. Second, the voltage nduced n the rotor wndng depends on the slp, too. Hence the rotor voltage s shown as sE2. Ths model s beng vewed from the standpont of the rotor tself. We need the model as vewed from the stator crcut. Ths s smlar to what we dd wth the transformer model, namely, relectng the secondary model nto the prmary. To make the transton to a model as seen from the stator, I irst need the rotor current:

I2 =

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

Now I can model the rotor slghtly dfferently: However, there s somethng mssng n ths model. How does power that the motor produces get through from the stator to the rotor? The model shown n Fgure 3.11 represents the rotor as

FIGURE 3.11: Rotor equvalent adjusted.

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65

FIGURE 3.12: Rotor wth rotaton power.

seen from the stator. In ths model, the only element that could absorb power s R2/s. Power delvered to ths resstor must represent both the output power of the motor and the copper losses of the rotor. Ths power s the power delvered across the argap by the stator:

Pairgap = |I2 |2

R2 . s

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

PCu = |I2 |2 R2 . The motor’s output power must therefore be the power delvered across the argap less the power lost n the wndng: � � 1−s Pmotor = Pairgap − PCu = |I2 |2 R2 . s Now I can insh the rotor model by leavng R2 as the copper loss and addng another resstor to represent the power delvered to the shaft. Fgure 3.12 shows the complete rotor model. The resstor that represents shaft power has an arrow through t to ndcate that t s a varable that s controlled by slp s.

3.2.3 Complete Equivalent Circuit Now I can combne the stator and the rotor by “relectng” the rotor mpedance across the argap to the stator. The nduced voltage E2 remans the same because I dvded t by s earler. The rest of the elements n the rotor model acqure a prme to ndcate they are the relected values. Fgure 3.13 shows the complete sngle-phase equvalent of the nducton motor. When I inshed the model of the transformer, I sad 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 sngle-phase equvalent wth rotor relected.

than the seres resstance and reactance. Ths assumpton s not a good one for nducton motors because Xm s not “much, much larger.” It ncludes a large argap that was not present n the transformer. We wll make use of ths model a lttle later, but before dong that, I am gong to take a look at one real motor and ts nameplate. I am also gong to show how we can test a motor to ind out what numbers should go nto the equvalent crcut.

3.3

MOTOR NAMEPLATE

The motor n Fgure 3.14 s hdden n an equpment room not far from my ofice. It has been drvng an HVAC fan for about 20 years. The only servce that t s lkely to have requred s replacement of the belts from tme to tme.

FIGURE 3.14: The motor.

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FIGURE 3.15: Motor’s nameplate redrawn.

Wth the help of a lashlght and a mrror, I was able to read all the nformaton on the nameplate, whch I have reproduced n Fgure 3.15. Motor nameplates bear standard nformaton about the motor. What s wrtten there s specied by the Natonal Electrcal Manufacturers Assocaton (NEMA) standard for motor nameplates, the “NEMA Standards Publcaton Motors and Generators,” MG1-1998. NEMA was formed by ndustry authortes n 1926 to provde “a forum for the standardzaton of electrcal equpment, enablng consumers to select from a range of safe, effectve, and compatble electrcal products.” Let us go through ths nameplate lne by lne: •

•

• • •

Model s the manufacturer’s assgned model number. A motor wth a model number smlar to ths one appears n Marathon Electrc’s catalog today, although the desgn has been mproved to ncrease eficency. Frame s a code for the crtcal physcal dmensons of the motor. The smplest way to descrbe frame s to say that any motor wth ths number wll it as a replacement. Ths apples to mountng holes, shaft poston and dameter, keyways, and so on. If the motor n the pcture needs to be replaced, any motor wth a 213T frame should it. Type s the manufacturer’s desgnaton. Des stands for Desgn. There are four standard NEMA desgns, whch have four dfferent torque-versus-speed characterstcs. We wll look at these later. Ph s Phase, and ths s a three-phase motor.

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•

• • • • • • • • • • •

• •

• •

•

Code species the locked-rotor kVA of the motor. “Locked-rotor” means that the shaft s totally restraned so t cannot turn. Code H means ths motor has a locked-rotor kVA n the range of 7.1–8.0 kVA per rated horsepower. Because ths s a 7 ½-hp motor, t could absorb as much as 60 kVA when full voltage s appled wth the rotor completely locked. Ins Cl s Insulaton Class and species the maxmum temperature that the wndng can properly wthstand. Class H nsulaton s lmted to 180°C. Encl s the descrpton of the Enclosure of the motor. Ths motor s drp-proof, meanng that water drppng down on t wll not get nsde. It s not waterproof, though. Duty s CONTnous, but the motor must be operated wthn specicatons, especally load and temperature. Max Amb s the maxmum ambent temperature, whch refers to the ar temperature around the motor. Shaft End Brg s the code for the ball bearng at the shaft end. A 307 bearng has a 35-mm bore, an 80-mm outsde dameter, and a thckness of 21 mm. Opp End Brg is the bearing at the other end, 30 × 62 × 16 mm. Volt species the operatng voltage, here dual voltages selected by how the motor’s multple wndngs are connected to the power source. Amp species the full-load current correspondng to the chosen operatng voltage. RPM s the shaft speed at full rated load. For ths motor, rated slp s 2.78%. Hz s 60. NEMA Nom Eff s the NEMA rated eficency. All motor losses must be ncluded. The percent stated must be no larger than “the average eficency of a large populaton of motors of the same desgn.” A modern energy-eficent motor of ths sze should be at least 88.5% eficent by NEMA standards. Nom pf s the nomnal power factor at full rated load. Max Cap kVAR s the largest capactor bank, measured n klovars, that can be used to correct the power factor of ths motor. Ths capactor bank wll correct the power factor to unty. HP s the full-load horsepower. Standard medum-szed motors come wth ratngs of 1, 1.5, 2, 3, 5, 7.5, 10, etc. In other words, you cannot order a 6-hp motor. See Secton 3.8.4. SF s the servce factor, whch species by how much the motor can be overloaded and stll operate contnuously. Medum-szed motors usually have a servce factor of 1.15, whch means they can operate at 15% above rated full load provded other nameplate condtons are met (especally ambent temperature). Corr Amp gves the full-load current values when the power factor s corrected to unty.

INTRODUCTION MOTORS: JUST ONE MOVING PART

3.4

69

TESTING

Transformer tests n the prevous chapter measured voltage, current, and power under two condtons, short-crcut and open-crcut. These tests yelded enough nformaton to calculate the values of the transformer model’s crcut elements. Motor testng s only slghtly dfferent and s the equvalent of the short-crcut and open-crcut transformer tests. The output of a motor s through ts shaft. The two basc tests are done wth the shaft completely blocked so t cannot rotate and wth the shaft completely unloaded. We measure three-phase voltage, current, and power under each of these condtons. However, we need one addtonal test, DC resstance, so we can obtan all the crcut parameters. In the followng descrpton of these tests, I am usng data from a 460-V, three-phase, 60-Hz, 50-hp, 1770-rpm nducton motor. Ths partcular motor s a modern, energy-eficent motor. All data have been taken usng three-phase meterng, measurng lne voltage, lne current, power for all three phases, and power factor. Because the motor’s equvalent crcut s the Y equvalent, we wll have to adjust the collected data to the sngle-phase Y model.

3.4.1 DC Test A DC resstance test provdes an estmate of the copper-loss resstance n the stator part of the model. At DC, the only resstance seen from the motor termnals s R1 because both X1 and Xm look lke short crcuts at DC and because there s no nducton across the argap. The DC measurements are made between the lne termnals of the motor. Ths means that the measurement sees two legs of the wndng n seres. Hence, the resstance measured wll be twce the value of R1 n the model. The DC data are

Vdc = 5.0 V, Idc = 28.4 A. The steps to obtan R1 are: 1. Compute the total DC resstance between two phase termnals:

Rdc =

Vdc 5.0 = = 0.1761 W. Idc 28.4

2. Dvde ths resstance by 2 to obtan R1:

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

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3.4.2 Blocked-Rotor Test Durng the blocked-rotor test, the shaft s irmly blocked so that t cannot rotate. Then the motor nput current s ncreased to ts rated value. Because the nput voltage wll stll be qute low, well below rated value, model elements Rc and Xm wll have mnmal effect. As the slp s now 100%, the value of the output resstor R2′(1 − s)/s = 0, a short crcut. Fgure 3.16 shows the equvalent crcut under these test condtons wth Rc, Xm, and the output resstor omtted. 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 calculatons proceed as follows: 1. Convert data to sngle-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 mpedance, whch s R1 + jX1 + R2′ + jX2′: Vph 66.28 = = 1.143∠82.08◦ = 0.1575 + j 1.132 W, Zb = Iph 58.0∠ − 82.08◦

R1 + R�2 = Re[Zb ] = 0.1575 W, X1 + X �2 = Im[Zb ] = 1.132 W.



3. Fnd the ndvdual values by usng the DC value of R1 and assumng 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: Equvalent durng blocked-rotor test.

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71

3.4.3 No-Load Test Durng the no-load test, the motor shaft s free, so the motor turns at almost the synchronous speed. The slp s close to 0, so the load “resstor” s essentally an open crcut and the rotor current s very small. Ths makes the parallel Rc–Xm combnaton the domnant mpedance. The no-load test s done by applyng rated voltage to the motor and measurng current and power. Fgure 3.17 shows the equvalent crcut. The no-load data collected for ths motor are

V = 460.0 V (rated voltage), I = 21.5 A, P = 464.6 W, pf = 2.71% lagging. The calculatons for the no-load test are very smlar to the locked-rotor calculatons: 1. Convert data to sngle-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. Fnd 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: Equvalent durng no-load test.

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

3.4.4 Complete Equivalent Fgure 3.18 shows the Y-equvalent crcut derved from the DC, locked-rotor, and no-load tests. The crcut shows two sets of numbers. The upper number on each element s the result of the tests just performed; the lower number s the value the manufacturer obtaned from more extensve tests.

3.4.5 Comparison How well does ths model the motor? One way of checkng s to see what values the model yelds for the nput current, nput power, and output power under full-load condtons. Table 3.1 compares the results taken from three sources: the manufacturer’s model of the motor, the text book model n Fgure 3.18, and the data obtaned by actually testng the motor at rated speed (1770 rpm). The results of all ths are wthn about 10% of the actual operatng values. Part of the dscrepancy n the text model s a result of assumng that X1 and X2′ are equal. Addtonal motor tests ndcate that ths splt should be about 40:60. Another dscrepancy s ntroduced by the assumpton that, durng the no-load test, R1 and X1 can be gnored.

3.5

ENERGY FLOW

We have been developng a sngle-phase Y-equvalent crcut to model the behavor of the threephase nducton motor. So far, all we have are the numbers that go nto the equvalent and a knowledge of what the crcut parameters represent n the motor tself. Ths model s useful for lookng at the low of energy through the motor. We know that a porton of what goes nto the machne gets lost nsde and does not appear at the output. Motor eficences are not bad, rangng from about 85% to 95%, but they are not 100%. If you could nvent one at 100%, you would be famous!

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TABLE 3.1: Performance comparson of models a

PARAMETER

MANUFACTURER’S a MODEL

TEXT MODEL

ACTUAL TEST

Lne voltage

460 V

460 V

460 V

Lne current

59.94∠–33.3° A

63.99∠–32.0° A

60.02∠–33.3° A

Power factor

83.6% laggng

84.8% laggng

83.6% laggng

Power nput

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

Eficency

95.10%

95.10%

93.30%

b

a

Model power-output data gnore mechancal and wndage losses.

b

Actual test data nclude mechancal and wndage losses.

3.5.1 Power, Torque, and Losses Let us look at what happens to energy by addng a descrptve graphc that shows the prmary energy low. Fgure 3.19 s the equvalent crcut wth the power low added. The drawng shows the motor dvded nto ts physcal parts: the ncomng power lne, the stator wndng, the argap, the rotor, and the shaft. Power through the motor s depcted as a lne of dmnshng thckness. Let us look at the power along the way:

FIGURE 3.19: Sngle-phase equvalent wth three-phase power low.

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PRAGMATIC POWER

• • • •

•

•

Pn s the nput power gven n three-phase terms (all power n ths drawng s three-phase). V1 s the lne voltage and I1 s the lne current. Losses n the stator are copper loss, represented by R1, and core loss, represented by Rc. What s not lost n the stator s passed along to the argap. The torque developed n the argap s related to the argap power by the synchronous speed ωs (n radans per second). In the rotor, the developed torque s now operatng at the rotor speed ωm , whch s slghtly slower than the synchronous speed. Hence, the power developed for the rotor s dmnshed. Ths loss s represented by rotor losses, prmarly va R2′. The magntude of ths loss s the slp tmes the argap power. The power actually delvered to the shaft s the developed power. Ths power, however, does not entrely make t out of the motor. A porton s lost n the mechancal parts of the motor, manly wndage and bearngs. Most nducton motor rotors nclude ins of some knd to str the ar nsde the motor to ad n coolng the wndngs and t takes some power to move ths ar. The ins are vsble n the photographs (Fgures 3.7 and 3.8). What comes out as shaft power and shaft torque s what s left after subtractng the varous losses.

A motor s rated by ts output horsepower, not ts nput power. Therefore, a 50-hp motor that s 90% eficent wll have an nput power of 55.55 hp. To look at ths another way, a ratng of 50 hp means the output power s 37.3 kW (746 W/hp). If the motor s 90% eficent, the power nput from the power lne must be 41.4 kW, leavng 4.1 kW to be dsspated by the motor tself. If the motor temperature s not to rse more than allowed, the motor must be n an envronment where t can dsspate that heat. The manufacturer of the motor I have been usng as an example n Secton 3.4 has furnshed some temperature data. After runnng the motor for more than ive hours at full load n an ambent ar temperature of 29°C, the ar leavng the motor was at 46°C. Both bearngs were at about 40°C. The frame of the motor reached 55°C. The hottest spot n the wndngs was about 66°C, whch s well below the maxmum level for ts nsulaton class (180°C).

3.5.2 Stored Energy The nducton motor s a large ron-core nductor. As such, t stores magnetc energy n the same way that any other nductor does. The stored energy for an nductor s 1/2Li(t)2 at any nstant. When we shut off an operatng motor, ths stored energy must go somewhere. The runnng motor also has stored energy n the form of the nertal energy of the rotaton system.

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Although gong nto detal about ths stored energy s beyond where I want to go, several ponts are useful to know: •

•

•

3.6

The stored energy can be used at shutdown to help stop the motor. As the power lne s dsconnected, short the cols. The stored energy s dsspated n the shorted resstance of the motor, brngng the motor to a rapd halt. Electrc lawnmowers use ths to stop the blade quckly. If the power supply s accdentally short-crcuted near the motor, the short must dsspate both the electrcal energy assocated wth the short crcut and also the stored energy n the nearby motor. Moreover, protectve devces such as crcut breakers must be able to clear such a fault. Usng a DC ohmmeter to test the wndngs leaves some 1/2Li2 stored energy behnd when the meter s dsconnected. Ths energy can show up as a hgh—and perhaps dangerous— voltage across the termnals or an arc as the meter leads are taken away.

MOTOR CURVES

If you want to make the shaft of some pece of machnery go around, you need torque. Torque s what an nducton motor produces. But not just any torque. There s a relatonshp 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 especally, there s a relatonshp between the output torque and the slp. The curves of torque versus slp tell a great deal about an nducton motor. Although there are numerous possble shapes for these curves, there are not very many common ones. NEMA classies nducton motors nto number “desgn” categores. The NEMA desgn code s specied on the motor’s nameplate. Recall that the nameplate n Secton 3.3 showed “Des B” to desgnate that ths motor has a torque–slp curve that follows the NEMA curve B. Fgure 3.20 shows four of the NEMA torque–slp curves. They all gve torque as a percentage of full-load torque, and speed as slp percentage. Before lookng at these curves n detal, let us look at just the Desgn B curve n Fgure 3.21. Several thngs that we can learn wll apply to the other curves as well. There are four mportant ponts along the Desgn B torque–speed curve of Fgure 3.21. I am gong to work from rght to left n the drecton of ncreasng slp. •

Normal full-load torque (100%) occurs at a farly small slp, usually no more than 5%, dependng on the partcular desgn. Note how straght the curve s n ths regon. It s close

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FIGURE 3.20: NEMA nducton motor torque curves.

FIGURE 3.21: NEMA Desgn B torque curve.

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•

•

•

77

enough to lnear that we are gong to assume t s lnear between synchronous speed (s = 0) and full-load speed. Breakdown torque s the largest torque the motor can sustan wthout “slppng over the top of the hll.” If the breakdown torque s exceeded, the motor’s speed drops quckly. Reachng ths torque requres a consderable overload. The pull-up torque s the lowest spot on the left porton of the torque–speed curve. It s the mnmum value of torque that the motor can produce as ts speed s run up from startng to operatng. The startng torque s the torque that gets a shaft turnng. It has to overcome not only nerta but some “stckness” that some loads present when they are standng stll.

Now let us look at the general characterstcs of these desgns. Note that the general operatng range of these motors s at the far rght of the graph, generally wth a slp of no more than 5%. •

•

• •

3.7

Desgn A has the hghest breakdown torque, and the relaton between torque and slp n the normal operatng range s very steep. Ths means that the shaft speed does not vary a great deal wth torque, even at more than 100% of full load. It s a good general-purpose motor but generally has a hgher startng current than other common desgns. (Startng current cannot be determned from these curves but rather comes from other nformaton.) Desgn B s the most common motor sold. Its curve has a farly steep slope for the normal range of slp, whch generally s not more than 5%. Ths desgn’s breakdown torque s among the lowest, so t does not take too kndly to large overloads. Startng current s modest, however, and the motor can often be lne-started, meanng that t can be started by swtchng on full lne voltage wthout anythng specal to lmt the current. Desgn C has a hgh startng torque and a modest startng current. It s not a good choce for loads that may at tmes exceed full load and t s not as eficent as A and B. Desgn D has a very dfferent curve compared to the others. It produces a very hgh startng torque wthout a hgh startng current. The normal range of operatng slp s very broad.

OVERMOTORING

Ths secton has a smple message: Do not overmotor! In other words, f a partcular job requres a 10-hp motor, do not “throw n a safety factor” and choose a 20-hp motor. Match the motor to the load, specfyng a motor that s just large enough to provde the torque requred. Why? I wll show several motor curves that wll help you understand ths statement. Fgure 3.22 shows the torque–slp curve for a partcular motor. It s most smlar to a Desgn A motor but has hgher startng 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 eficency.

Suppose I have a load that requres just 200 N-m for normal operaton. That s about half the rated torque, so the slp s cut n two to about 1.5%. Hmm, that does sound okay. After all, the motor wll not be workng as hard and t should last longer. But look at the power factor curve n Fgure 3.23. The full-load power factor s 88% laggng. If we “run lght” wth a full-load slp of just 1.5%, the power factor gets consderably worse. It looks to me lke t falls to around 70% laggng. Ths wll ncrease your load’s reactve power and you wll most lkely pay the power company for low power factor. The eficency also decreases. The motor was desgned to operate near the peak of ts eficency curve. Note n Fgure 3.24 that ths s about 91%. As the slp gets smaller, the eficency decreases. You wll requre more power and you wll pay the power company for ths, too. Fgure 3.25 shows about the only thng that s good about overmotorng—the current s reduced. But you have already seen that the decrease n current does not come wth a lower power bll due to poorer power factor and lower eficency. So do not overmotor! Remember that nducton motors specfy a servce factor, often 1.15. Ths means that the motor can be run contnuously at a load larger than full load. Ths rate ndcates that t can sustan a 15% overload contnuously, provded you gve the motor proper ambent condtons. If, for example, the maxmum allowed ambent temperature s 40°C, you should make sure that there s good ar low around the motor and that ar stays below 40°C.

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FIGURE 3.25: Lne current.

There are not a lot of common motor szes. In the “medum horsepower” range, whch comprses motors between 1 and 500 hp, common ratngs are 1, 1.5, 2, 3, 5, 7.5, 10, 15, 20, 25, and so on. For example, f you have a 6-hp load, a 5-hp motor wll not do, even at 15% over, so a 7.5-hp motor s requred. Lkewse, an 8-hp load can be handled properly by a 7.5-hp motor.

3.8

EXAMPLES

Here are several examples to brng all of ths together.

3.8.1 Example I—Data Collection A 50-hp, three-phase, 60-Hz, 460-V nducton motor has a full-load operatng speed of 1770 rpm. The full-load current s 59.5 A at 85% power factor. Rotatonal losses have been measured at 1200 W. 1. Fnd the number of poles. The synchronous speed must be a submultple of 3600 for a 60-Hz motor and 1800 s the irst submultple larger than 1770.

ns = 1800 = poles = 4. 2. Fnd the slp at rated load.

3600 poles/2

INTRODUCTION MOTORS: JUST ONE MOVING PART

s=

81

1800 − 1770 × 100 = 1.667% . 1800

3. Fnd the output torque. The number of watts per horsepower s 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. Fnd the nput power.

Pin =

√

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

5. Fnd the eficency.

% efficiency =

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

6. F nd the electrcal losses n the rotor. I wll follow the energy low n Fgure 3.19 from rght to left, dervng the developed power from the output plus the rotatonal losses. From that, I wll get the argap 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. Fnd the stator losses. Pstator = Pin − Pairgap = 40.3 − 39.2 = 1100 W

3.8.2 Example II—Power Factor Adjustment Specfy the capactance necessary to mprove to 92% the power factor of the motor of Example I when operatng 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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PRAGMATIC POWER

Ths capactance s the total requred, whch must be splt 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 nducton motor drves a partcular load wth 5% slp. Rotatonal losses are 180 W. Ths motor’s sngle-phase-equvalent parameters are shown below (Fgure 3.26). Use ths model n the calculatons 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 equaton for E2 and two auxlary equatons wll yeld the necessary crcut 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: Sngle-phase equvalent for example III.

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83

2. F nd the power nput, the power output, and the horsepower output. Because the model s just one phase, values of power derved from the model must be trpled 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 0.25(27.15) + = 3 R1 |I1 | + = 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. Fnd the percent of full load, the overall eficency and the nput power factor.

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

pfin = cos (−38.7) = 78.0% lagging.

4. F nd the startng current. Ths s done by notng that the resstor representng output power to the shaft becomes 0 because the slp s 1. So n the orgnal node equaton, replace 3.8 by 0 n the last fracton. Then solve for E2 and ind I1. The result s

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

3.8.4 Example IV—Motor Selection A certan machne n a plant requres a contnuous torque of 80 N-m at 1200 rpm. The peak startng torque s twce the runnng torque. Ths machne s to be drven by a 440-V, three-phase, 60-Hz nducton motor wth low slp, because the load s to be kept as near to 1200 rpm as possble. Specfy the approprate nducton motor. The general choces are 5, 7.5, 10, 15, 20, and 25 hp, wth Desgn A, B, C, or D characterstcs.

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

ns = 1200 =

3600 , so p = 6 poles. p/2

2. B ecause slp wll be small, usng the synchronous speed nstead of the slghtly lower motor speed should not make much dfference.

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

ws = Pout hpout

hs output s too large for a 10-hp motor wth a servce factor of 1.15. A 15-hp motor s T requred. 3. Use the desgn curves of Fgure 3.20 to determne whch desgn to specfy. Dong ths requres that we know the “100% torque” output of the motor, whch s the torque for a rated load of 15 hp. Agan, assume that the slp s small.

T100% =



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

4. T he specied load requres a startng torque of 160 N-m (2 × 80). The 100% full-load torque for the 15-hp motor s 89.0 N-m, so 160 N-m s 180% of the motor's full-load torque. Although Desgn B s the most commonly avalable motor, ts startng torque s only about 160% of the full-load torque. The next choce s Desgn C, whch s smlar to Desgn B but has a hgher startng torque, about 210%. Hence, the best choce appears to be Desgn C. 5. The chosen motor s runnng at (80/89)100 = 90% of full rated load. The full-load slp for a Desgn C motor on the graph of Fgure 3.20 appears to be about 3%. We can assume that the rghtmost porton of the curve s lnear, so the motor wll be drvng the load at about 0.90 × 3 = 2.7%, whch s 1168 rpm.

The result of all ths s that you specfy a 440-V, three-phase, 60-Hz, sx-pole, Desgn C nducton motor. Once you have ths nformaton, you can see what varous manufacturers offer and check the desgn usng the manufacturer’s torque–speed curve for the motor chosen. Ths desgn gnores an mportant queston, though. What s the torque-versus-speed characterstc of the load over the full range of speeds from startng to full load? Ths s beyond the scope of ths text, but t can be answered by plottng the torque–speed curve of the load on the same

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85

torque–speed graph as the motor. If the load’s curve s below the motor’s curve except where they cross at the operatng values, the motor specied s probably the rght choce.

3.9

SINGLE-PHASE MOTORS

Ths chapter has been devoted to three-phase nducton motors because they provde by far the bulk of rotatng-energy delvery. The are several good reasons for ths: smooth rotatng power, a sngle movng part, very smple structure. However, many applcatons that need rotatng energy are n places that do not have three-phase power. Homes are a sgnicant example. Not that three-phase power s not avalable for homes—t s expensve and t s not really needed. Sngle-phase nducton motors can do the smaller jobs very well. So what s the dfference between three-phase and sngle-phase nducton motors? The bggest dfference s that the ield the stator creates s not rotatng smoothly around the stator. Instead, t just changes drecton, albet smoothly, from “up” to “down” and back agan. A rotor wll not follow ths ield from a standstll—the torque just pulsates wth no net rotaton. But f a rotor can be optmzed to go at the rght speed, the motor delvers net torque, pulsatng, to be sure, but net. The basc structure of the sngle-phase nducton motor s very smlar to that of the threephase motor. One par of poles provdes a synchronous speed of 3600 rpm (at 60 Hz) just as wth three-phase motors. Two pars provde 1800 rpm, and so on. The rotor s smlar to that of ts three-phase bg brother—shorted cols. The sngle-phase motor exhbts slp n the same way. The problem s to get thngs started. There are four dfferent ways of gettng the necessary net torque to get the rotor gong. All four nvolve the same fundamental dea: provde a second set of poles whose current s out of phase wth the irst set. If ths second set of poles s placed wth ts axs electrcally offset by the same angle as ts current s offset, a rotatng ield s created that provdes torque to move the rotor. The four ways of dong ths are shaded pole, splt phase, capactor start, and capactor start and run, often called splt capactor. Let us dscuss brely these four optons, rememberng that n each case the goal s a rotatng stator ield that wll gve the rotor net torque. •

•

The shaded-pole motor provdes the “off-center” poles by wrappng a shorted col of wre around just a porton of the man poles. Ths shorted col provdes addtonal nductance for that porton of the pole, thereby addng an addtonal phase angle to the angle of the magnetc ield created by the man poles. The generated torque s not large, but t s enough to begn rotaton. The result s a motor wth low startng torque and usually not much power, ether. Slp s hgh (10% or more) and eficency s very low. The splt-phase motor uses two wndngs: one for runnng and one for startng. These wndngs are on two separate sets of poles, generally spaced 90 electrcal degrees apart. The

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PRAGMATIC POWER

•

•

wre sze and the number of turns on these poles are set up so that the runnng poles have hgh reactance and low resstance, whereas the startng poles exhbt the opposte condton. The motor s started wth all poles connected. The large dfference n reactance and resstance provdes a large phase shft between the sets of poles and hence a net startng torque. As the motor begns to come up to speed, a centrfugal swtch cuts off the startng wndng, leavng the runnng wndng to drve the rotor. These motors can be reversed by reversng the current n one of the two sets of poles. They have moderate startng torque and are farly eficent (more than 50%). Celng fans are an example of ths type of motor. The Casablanca XLP motor s an 18-pole splt-phase motor, so ts synchronous speed s 3600/(18/2) = 400 rpm, but t runs more slowly than that. Capactor-start motors have roughly the same structure as the splt-phase motor, except that a capactor n seres wth the startng wndng provdes the necessary phase shft. The wndng must be swtched off as the motor speed ncreases. Characterstcs are smlar to the splt-phase motor, but the startng torque s hgher. After they have started, they behave much as the splt-phase motor does. Capactor motors are often easy to recognze—the capactor s often n a separate cylndrcal housng on the outsde of the motor. Capactor-start capactor-run motors leave the capactor n the crcut after startng. Sometmes they use two capactors, one just for startng and one for both startng and runnng. Although these are a lttle more complcated than the capactor-start motors, they have a better power factor.

All but the shaded-pole motor have a notceable dsadvantage over the three-phase nducton motor: addtonal parts. The most common pont of falure of the capactor motors s the capactor. If a capactor motor wll not start but stll draws current, the odds-on bet s capactor falure. Motor capactors are electrolytc capactors, whch we generally thnk of as beng polarzed. Motor capactors are made to be nonpolarzed. The second common pont of falure s the centrfugal swtch, whch s desgned to snap open when the rotor reaches a set speed, generally somewhere about 25–30% of normal runnng speed. There s one other common small sngle-phase motor, the “unversal” motor. It s really a DC motor wth brushes and a commutator. It s called unversal because t wll run on both AC and DC, although ts common use s on AC n such machnes as vacuum cleaners and power drlls.

3.10

OTHER MOTORS

We have not looked at DC motors because I have been focusng on larger machnes drven by the AC power system. There are mllons of DC motors n use that we are gnorng. Most of these are n low-to-medum power applcatons, such as battery-operated toys and portable tools and automo-

INTRODUCTION MOTORS: JUST ONE MOVING PART

87

ble starters. Large DC motors power so-called desel locomotves, whch are really desel-electrc. The desel prme mover drves a DC generator and ths powers DC motors on the locomotve’s trucks. We have learned that you cannot really control the speed of an nducton motor. All you can do s select from what s avalable based on the number of pole pars. Speed control can be acheved, however, by usng varable-speed drves. For these, electroncs provdes a frequency dfferent from that of the power lne. So a sx-pole motor could have a synchronous speed of, say, 1100 rpm rather than 1200 by reducng the suppled frequency to (1100/1200)60 = 55 Hz. The same electroncs can also provde addtonal control such as startng.

3.11

SUMMARY

Tesla’s nventon s pure smplcty. Its smoothly rotatng magnetc ield drags the rotor around to provde a smooth torque at the shaft. The torque–speed curves of standard motors are just about lnear n ther usual operatng range. The sngle-phase equvalent crcut s reasonably smple. In short, the nducton motor s a remarkable electrcal devce, one of the oldest stll n common use. Yet most of us do not apprecate t. One mportant lesson n ths chapter s to avod overmotorng. Choosng a motor larger than what s requred leads to wasteful power use and lowered eficency. No “safety margn” s needed n most applcatons. The rght-szed motor n the rght envronment wll do ts job eficently and s lkely to perform contnuously for many years. In ths chapter, we have gone through the way n whch the three-phase nducton motor works, developed an equvalent crcut for t, and lad out energy low through the machne. We have also learned about the standard nameplate data and how a motor s tested to obtan the parameters for the equvalent crcut. Fnally, we have presented some examples of motor applcatons and of choosng a motor. • • • •

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Author Biography William J. Eccles has been a professor of electrcal and computer engneerng at Rose-Hulman Insttute of Technology snce 1990 (except for one year at Oklahoma State). He retred n 1990 as Dstngushed Professor Emertus after 25 years at the Unversty of South Carolna. He founded the Department of Computer Scence at that unversty and served at one tme or another as head of four dfferent departments, Computer Scence, Mathematcs and Computer Scence, and Electrcal and Computer Engneerng, all at South Carolna, and Electrcal and Computer Engneerng at Rose-Hulman. Most of hs teachng has been n crcuts and n mcroprocessor systems. He has publshed Microprocessor Systems: A 16-Bit Approach (Addson-Wesley, 1985) and numerous monographs on crcuts, systems, mcroprocessor programmng, and dgtal logc desgn. Bll has also publshed Pragmatic Circuits: DC and Time Domain, Pragmatic Circuits: Frequency Domain, Pragmatic Circuits: Signals and Filters, and Pragmatic Logic, four texts n ths Synthess Lectures n Dgtal Crcuts and Systems seres. Bll and hs wfe Trsh have two chldren and three grandchldren. Bll s also a conductor (approprate for an electrcal engneer) on the Whtewater Valley Ralroad, a tourst lne n Connersvlle, IN. He s a regstered professonal engneer and an amateur rado operator.