single phase and three phase power

8/3/2019 Single Phase and Three Phase Power

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CHAPTER7Active, Reac tive,and Apparent Power

7.0 Introductionr;oncept of active, reactive, and apparentpower plays a major role in electric power tech-

nology.In effect. the transmission of electrical energyand the behavior o r ac machines are often easier tounderstand by working with POWCL rather than deal-ing with voltages and currents. The reader is thereforeencouraged to pay particular attention to this chapter.The terms active. reactive, and apparent power

apply to steady-state alternating current circuits inwhich the voltages and currents arc sinusoidal.They cannot he used to describe transient-state be-havior. nor can we apply them to dc circuits.

Our study begins with an analysis of the instan-taneous power in an uc circuit. \Ve then go on to de-fine the meaning of active and reactive power andhow to identify sources and loads. This is followedby a definition or apparent power, powerfactor. andthe power triangle. We then show how ac circuitscan he solved using these power concepts. In con-clusion. vector notation is used to determine theactive and reactive power in an ac circuit.

7.1 Instantaneous powerThe instantaneous power supplied to a device ISsimply the product of the instantaneous voltageacross its terminals times the instantaneous currentthat flows through it.

Instantaneous power is always expressed inwatts, irrespective of the type of circuit used. Theinstantaneous power may be positive or negative. Apositive value means that power flows into the de-vice. Conversely, a negative value indicates thatpower is flowing out of the device.

Example 7-1A sinusoidal voltage having a peak value of 162 Vand a frequency of60 Hz is applied to the terminalsof an ac motor. The resulting current has a peakvalue of 7.5 A and lags 50 behind the voltage.a. Express the voltage and current in terms of theelectrical angle < 1 > .

b. Calculate the value of the instantaneous voltageand current at an angle of 120.

13 4


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c. Calculate the value of the instantaneous powerat 120.

d. Plot the curve of the instantaneous power deliv-ered to the motor.

Solutiona. Let us assume that the voltage starts at zero andincreases positively with time. We can thereforewrite

e=Em sin < p =162 sin < pThe current lags behind the voltage by an anglet J = 50, consequently, we can writei=Jm sin ( d > - 8) =7.5 sin (e l) - 500)

h. A t < p = 1200 we havee = 162 sin 1200= 162 X 0.866=140.3 Vi=7.5 sin (1200 - 50) =7.5 sin 70= 7.5 X 0.94=7.05 A

C. The instantaneous power at 120C is

+1000 W

ACTIVE. REACTIVE. I1ND API~4RENT POWER 135

p = ei = 140.3 X 7.05 = + 989WBecause the power is positive, it tlows at thisinstant into the motor.

d. In order to plot the curve of instantaneouspower, we repeat procedures (b) and (c) abovefor angles ranging from q ) =0 to < p =360.Table 7 A lists part of the data used.

TABLE 7A VALUES OF e, i, AND P USED TO PLOTFIG.7.1

Angle Voltage Current Powert h 162 sin < I ) 7 .5 sin I ) - SOD) I'degrees volts amperes watts0 () -5.75 025 68.5 -3.17 -21850 124.1 0 075 156.5 3.17 497115 146.8 6.8 1000155 6R.5 7.25 497180 0 5.75 0205 -68.5 3.17 -218230 -124.1 () 0

1 - - - 1- -120 s- --I

115 V

F ig ure 7 .1Instantaneousvoltage, current and power in an ac circuit. (See Example 71.)


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1 : 1 6 r.LEeTRICAL MACHINES AN!) TRANSFORMERS

The vo ltage . curre n t. and in stan tan eous powe r areplo tte d in Fig. 7 .1. The powe r attain s a po s itiv e pe ako f + 1000 W and a n egativ e pe ak o f -218 W . Th e n eg-ativ e po we r m ean s th at po we r is actually flo win g fro mth e lo ad (rnotorj to th e source . Th is o ccurs during th ein te rv als ( ) - SO ". 18( )" - 230. and 3600 - 41(YA lthough a powe r flow from a dev ice con s ide re d to hea lo ad to a de v ice con s ide re d to he a source m ay seemto h e im po ss ible . it happe n s o fte n in ac circuits . There aso n is giv en in thc se ctio n s that fo llow .

W e also no te that th e po s itiv e pe aks o ccur at in -te rv als o f 1/120 s. Th is m ean s that th e fre que n cy o fth e powe r cycle is 120 Hz. wh ich is tw ice th e fre -que n cy o f th e v o ltage and curre n t th at produce th e

(a)

(b )

wattmeter(c)

1 : : - = > p load

Figure 7.2a. An ac voltage I,' produces an ac current I in this re-

sistive circuit.b. Phasors /:' and I are in phase.c. A wattmeter indicates Uwatts.d. The active power is composed of a series of posi-

tive power pulses,

power. A gain . th is phe nom enon is quite n o rm al: thfre que n cy o f ac powe r flow is always twice th e linfrequency,7.2 Active power*The s im ple ac circuit o f Fig. 7 ,2 a con sis ts o f a resis to r co nn e cte d to an ac ge n e rato r. The e ffe ctiv ev oltage an d curre nt are de sign ate d E an d I. respec-tiv e ly. and as we would expe ct in a re s istiv e circuit.phuso r s F an d 1 arc in phase (F ig. 7 .2 h ) , If we con -n ect a wattm ete r (Fig, 7 ,3) in to th e lin e . it w ill giv ea re ad in g P =EI w alls (Fig. 7 .2 c) .

To ge t a be tte r picture o f what go e s Oil in suchcircuit. we hav e drawn the s inuso idal curv e s o f Ean d 1 (F ig. 7 ,2 d) , The pe ak value s are re spe ctiv e ly~2E v olts an d ~21 am pe re s be cause . as s tate d pre v i-ously . E an d 1 ar e effective v alue s. B y m ultiplyin gth e in s tan tan e ous v alue s o f v o ltage and curren t awe did ill S e ctio n 7 .1. we obtain th e iIlS/UII/UIlCOIlSpow er in w atts .

Many person ....refer to active power as rco!/)(!\,.{'r or '1'11(,/W\\'C;: considering it (0 be more descr.ptivc. In this bookwe Lise the te rm .utiv 1}(}1I't>!: bccuu-c it co nfo rm -, to till'IE EE d es ign uu on .

Figure 7.3Example of a high-precision wattmeter rated 50 V, 100V, 200 V: 1 A, 5 A, The scale ranges from 0-50 W to O1000 W.(Courtesy of Weston Instruments)


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The power wave consists of a series of positivepulses that vary from zero to a maximum value of(02) X (-"j21) ~ 2E I =2P watts. The fact that (a)power is always positive reveals that it alwaysflows from the generator to the resistor. This is oneof the basic properties of what is called activepower: although it pulsates between zero and max-imum, it never changes direction. The direction ofpower flow is shown by an arrow P (Fig. 7.2c).The average power is clearly midway between

21' and zero, and so its value is P =2E!12 =E Iwalls. That is precisely the power indicated by thewattmeter.The two conductors leading to the resistor in Fig.

7.2a carry the acti vc power. However. unlike cur-rent flow. power does not tlow down one conductorand return by the other. Power flows over both con-ductors and. consequently. as far as power is con-cerned. we can replace the conductors by a singleline. as shown in Fig. 7.2c.In general. the line represents any transmission

line connecting two devices. irrespective of thenumber of conductors it may have.The generator is an active .I0UI'(C and the resistor

is an active load. The symbol for active power is Pand the unit is the watt (W). The kilowatt (kW) andmegawatt (MW) are frequently used multiples ofthe watt.

7.3 Reactive powerThe circuit of Fig. 7Aa is identical to the resistivecircuit (Fig. 7.2a) except that the resistor is now re-placed by a reactor XL' As a result. current I lags 90behind the voltage E (Fig. 7Ab).To see what really goes on in such a circuit, we

have drawn the waveforms for E and I and, by againmultiplying their instantaneous values. we obtainthe curve of instantaneous power (Fig. 7Ae). Thispower P consists of a series of identical positive andnegative pulses. The positive waves correspond toinstantaneous power delivered by the generator tothe reactor and the negative waves represent instan-timeous power del ivcrcd from the reactor to thegenerator. The duration of each wave correspondsto one quarter of a cycle of the line frequency. The

ACTIVE. RE;\CTlV/:', AN!) APPARENT POWfR 137

. . +F jX,t

(b),-------- .. I:L

1V2(e )

Figure 7.4a. An ac voltage E produces an ac current 1 in this in-ductive circuit.

b. Phasor 1 lags 90D behind I:.c. Reactive power consists of a series of positive and

negative power pulses.frequency of the power wave IS therefore againtwice the line frequency.Power that surges back and forth in this manner iscalled reactive power (symbol Q). to distinguish itfrom the unidirectional active power mentioned be-fore. The reactive power in Fig. 7 0 4 is also given bythe product Ef. However. to distinguish this powerfrom active power. another unit is used-the var. Itsmultiples are the kilovar (kvar) and megavar (Mvar).

Special instruments, called vunnctcrs. are avail-able to measure the reactive power in a circuit (Fig.7 .5 ) . . A . varrneter registers the product of thc effectiveline voltage E times the effective line current I timessin f) (where fl is the phase angle between I:and Z). . A .reading is only obtained when E and I arc out of phase:if they arc exactly in phase (or exactly I S O u out ofphase), the varmetcr reads zero.

Returning to l-ig. 7A. the dotted area under eachpulse is the energy. in joules. transported in oncdirection or the other. Clearly. the energy is deliv-ered in a continuous series of pulses of very shortduration. every positive pulse being followed by a


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133 EtECTRICAL MACHINES AND TNANSFORMERS

Figure 7.5Varmeter with a zero-center scale. It indicates positiveor negative reactive power flow up to 100 Mvars.

negative pulse. The energy flows back and forthbetween the generator and the inductor withoutever being used up.

What is the reason for these positive and nega-tive energy surges? The energy flows back and forthbecause magnetic energy is alternately being storedup and released by the reactor. Thus, when thepower is positive. the magnetic field is building upinside the coil. A moment later when the power isnegative. the energy in the magnetic field is de-creasing and f lowing back to the source.

We now have an explanation for the brief nega-tive power pulses in Fig. 7.1. In effect, they repre-sent magnetic energy, previously stored up in themotor windings. that is being returned to the source.

7.4 Definition of a reactive load andreactive sourceReactive power involves real power that oscillatesback and forth between two devices over a trans-mission line. Consequently. it is impossible to saywhether the power originates at one end or the lineor the other. Nevertheless. it is useful to assume thatsome devices generate reactive power while othersabsorb it. In other words. some devices behave likereactive sources and others like reactive loads.

By definition*. a reactor is considered to be a reactive load that absorbs reactive power.Example 7-2A reactor having an inductive reactance of 4 0connected to the terminals of a 120 V ac generator(Fig. 7.6a).a. Calculate the value of the current in the reactorb. Calculate the power associated with the reactorc. Calculate the power associated with the ac

generatord. Draw the phasor diagram for the circuit

1 +120 V IL t jX [I 4Ut 30 A(a)

(b)

4'.I' "1 +

120 VQ

c - : : -3.6 kvar 30 A

F.120 V

(c) L A

Figure 7.6See Example 7.2.

Solutiona. Current in the circuit:

EJ =L X I.

120 V = 30A40b. Power associated with the reactor:

Q =E 1 = 120 X 30 = 3600 var = 3.6 kvarThis defiuinon is ill agreement with IEEE and lEe conventions


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This reactive power is absorbed by the reactor.C o Because the reactor absorbs 3.6 kvar of reactancepower, the ac generator must be supplying it.Consequently, the generator is a source of reactivepower: it delivers 3.6 kvar. The reactive power Qflows therefore in the direction shown (Fig. 7.6h).

d. The phasor diagram is shown in Fig. 7.6c.Current I, lags 90 behind voltage E .

This phasor diagram applies to the reactive load(the reactor) and the reactive source (the ac genera-tor) as well as the line connecting them.

7.5The capacitor and reactive powerSuppose now that we add a capacitor having a reac-tance of 4 nto the circuit of Fig. 7.6. This yields thecircuit of Fig. 7.7a. The current I e drawn by the ca-pacitor is I e = 120 V 1 4 n= 30 A and, as we wouldexpect, it leads the voltage by 90 (Fig. 7. 7b).The vector sum of I L and (. is zero and so the ac

generator is no longer supplying any power at all tothe circuit. However, the current in the reactor hasnot changed; consequently, it continues to absorb30 A X 120 V = 3.6 kvar of reactive power.Where is this reactive power coming from') It

can only come from the capacitor, which acts as a

(a )

+-4j

i120 V

(b )

Ie30 A

~----------------~ 120V

II.30 A


ACTIVE. RE4.CTlVE, AND APPANENT POWER 139

source of reactive power. The reactive power deliv-ered by the capacitor is equal to the current it car-ries times the voltage across its terminals, namelyQ ~ E(. = 120 V X 30 A = 3600 var = 3.6 kvarThe reactive power delivered by the capacitor is ex-pressed in vars or kilovars. Reactive power Q nowflows from the capacitor to the reactor.

We have arrived at a very important conclusion:a capacitor is a source o r reactive power: It acts asa reactive power source whenever it is part of asine-wave-based, steady-state circuit.

Let us take another step and remove the reactorfrom the circuit in Fig. 7.7a, yielding the circuit ofFig. 7.8a. The capacitor is now alone, connected tothe terminals of the ac generator. I t still carries a cur-rent of 30 A, leading the voltage E by 90 (Fig.7.8b). Consequently, the capacitor still acts as asource of reactive power, delivering 3.6 kvar. Wheredoes this power go? The answer is that the capacitordelivers reactive power to the very generator to

(a)

-4jQ


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140 IoL ECTRICAI, /vIA CHINE S ANIJ TRANSF ORM E RS

which it is connected! For most people, this takes alittle time to accept. How, we may ask, can a passivedevice like a capacitor possibly produce any power?The answer is that reactive power really representsenergy that, like a pendulum, swings back and forthwithout ever doing any useful work. The capacitoracts as a temporary energy-storing device repeatedlyaccepting energy for brief periods and releasing itagain. However, instead of storing magnetic energyas a reactor does, a capacitor stores electrostatic en-crgy (see Section 2.14).

If we connect a varrne ter into the circuit (Fig.7. '6c), it will give a negative reading of E 1 = - 3600var, showing that reactive power is indeed tlowingfrom the capacitor to the generator. The generator isnow bchaving like reactive load, but we sometimesprefer to call it a receiver of reactive power, which, ofcourse, amounts to the same thing. In summary, a ca-pacitive reactance a/wars generates reactive power.Example 7-3An ac generator G is connected to a group of R. L . Ccircuit elements (Fig. 7.9), The respective elementscarry the currents shown. Calculate the active andreacti ve power associated with the generator.

3.I

4Q


SolutionThe two resistors absorb active power given by

p =lR~ (142 X 4) + (16.122 X 2) =784 + 520 =1304 W

The 3 n reactor absorbs reactive power:Q = rx = 142 X 3 = 588 varL L

The 3.5 n capacitor generates reactive power:Qc = P X c = 202 X 3.5 = 1400 val'The R , L . C circuit generates a net reactive power

of 1400 - 588 = 812 val'This reactive power must be absorbed by the

generator; hence, as far as reactive power is con-cerned the generator acts as a load.

The active power absorbed by the resistors mustbe supplied by the generator; hence it is a source oactive power = 1304 W.

In conclusion, the ac generator is a source of ac-tive power (1304 W) and a receiver of reactivepower (812 var).

7.6 Distinction between activeand reactive powerThere is a basic difference between active and re-active power, and perhaps the most important thingto remember is that the one cannot be convertedinto the other. Active and reactive powers functionindependently of each other and, consequently.they can be treated as separate quantities in electriccircuits.

Both place a burden on the transmission line thatcarries them, but, whereas active power eventuallyproduces a tangible result (heat, mechanical power,light. etc.), reactive power only represents powerthat oscillates back and forth.

All ac inducti ve devices such as magnets, trans-formers, ballasts, and induction motors. absorb re-active power because one component of the currentthey draw lags 90 behind the voltage. The reactivepower plays a very important role because it pro-duces the ac magnetic field in these devices.

A building, shopping center, or city may he COIl-sidered to be a huge active/reactive load connectedto an electric utility system. Such load centers COI1-tain thousands of induction motors and other elec-tromagnetic devices that draw both reactive power(to sustain their magnetic fields) and active power(to do the useful work).

This leads us to the study of loads that absorbboth active and reactive power.


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7.7 Combined active and reactiveloads: apparent powerLoads that absorb hoth active power P and reactivepower Q may be considered to be made up of a resis-tance and an inductive reactance. Consider, for exam-pic. the circuit of Fig. 7.1 Oa in which a resistor and re-actor me connected to a source G. The resistor drawsa current I " , while the reactor draws a current 1 ' 1 'According to our definitions, the resistor is an

active load while the reactor is a reactive load.Consequently, I" is in phase with E while 1 '1 lags 90behind. The phasor diagram (Fig. 7.1 Ob) shows thatthe resultant line current I lags behind F by an an-gie e . Furthermore. the magnitude of I is given by

(a )

(b)

II'J\~.-I:/e I, IIIIII" J I(c )

[--,1 0 1I II IIl ~ jF ig u re 7 .1 0a. Circuit consisting of a source feeding an active and

reactive load.b. Phasordiagram of the voltage and currents.c. Activeand reactive power flow from source to load.

ACTlVt. REACTIVE. /\NDAPPAI?ENT POWER 14 1

The active and reactive power components Pand Q both flow in the same direction. as shown bythe arrows in Fig. 7.1 Oc. If we connect a wattmeterand a varmeter into the circuit, the readings willboth be positive, indicating P =Ell' watts and Q =Eft] vars, respectively.

Furthermore. if we connect an ammeter into theline. it will indicate a current of I amperes. As a re-sult. we are inclined to believe that the power sup-plied to the load is equal to E I watts. But this is ob-viously incorrect because the power is composed oran active component (watts) and a reactive compo-nent (vars), For this reason the product H I is calledapparent power. The symbol for apparent power is S .

Apparent power is expressed neither in watts norin vars, but in voltamperes. Multiples are the kilo-voltarnpere (kYA) and mcgavoltampere (MYA).7.8 Relationship between P , Q,

and 5Consider the single-phase circuit of Fig. 7.11 a com-posed of a source, a load, and appropriate meters.Let us assume that the voltmeter indicates E volts the ammeter indicates I amperes the wattmeter indicates +P watts the varmeter indicates +Q val'S

Knowing that P and Q are positive. we know thatthe load absorbs both active and reactive power.Consequently, the line current I lags behind Eelb byan angle 8.

Current I can be decomposed into two compo-nents II ' and 1 ' 1 ' respectively in phase. and in quad-rature. with phasor E (Fig. 7.llb). The numericalvalues of I and I can be found directly from the in-J ) qstrument readings

I" =PIEIq =Q IF

(7.1 )(7.2)

Furthermore. the apparent power S transmitted overthe line is given by S =F I, from which

1= SIE (7.3)


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142 FLECTRICAL /vIACH/NES AND TRANSFOR/v/ERS

(a)b

(b)

Figure 7.11a. Instruments used to measure E. I. P . and Q in a circuit.b. The phasor diagram can be deduced from the instrument readings.

Referri ng to the phasor diagram (Fig. 7 .11 b), it isobvious that

Consequently,

That is.(7.4 )

in whichS = apparent power lA)P ~ active power IWlQ = reacti vc power Ivar I

We can also calculate the value of the angle f I be-cause the tangent of f 1 is obviously equal to 1 / 1 1 "Thus, we have

e = arctan Ie/II' = arctan Q IP (7.5 )Example 7-4An alternating-current motor absorbs 40 kW of ac-tive power and 30 kvar of reactive power. Calculatethe apparent power supplied to the motor.Solution

(7.4)= \40" + 30"=50 kVA

Example 7-5A wattmeter and varmeter are connected into a 120 Vsingle-phase line that feeds an ac motor. They re-spectively indicate 1800 Wand 960 var.Calculatea. The in-phase and quadrature components II '

and 1 '1b. The line current Ic. The apparent power supplied by the sourced. The phase angle between the line voltage and

line currentSolutionReferring to Fig. 7.11, where the load is now a mo-tor, we havea. II ' =PIE = 1800/120 = 15 A

1 '1 = Q IE = 9601120 = SAb. From the phasor diagram we have

I = ViI" + I" = vi')" + 8"p q -= 17A

c. The apparent power isS = EI = 120 X 17 = 2040 VA

d. The phase angle U be twe e n E and I isf 1 = arctan Q I P = arctan 960/1 800=28.10

(7.1)(7.2)


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Example 7-6A voltmeter and ammeter connected into the induc-tive circuit of Fig. 7.4a give readings of 140 V and20 A, respectively.Calculatea. The apparent power of the loadh. The reactive power of the loadc. The a cti v c power of the loadSolilti Oila. The apparent power is

S = E I = 140 X 20=2ROOVA =2.8 kVA

b. The reactive power isQ =EI = 140 X 20= 2800 val' =2.8 kvar

If a vanneter were connected into the circuit, itwould give a reading of 2800 var.c. The active power is zero.If a wattmeter were connected into the circuit, it

would read zero.To recapitulate, the apparent power is 2800 VA,

but because the current is 90 out of phase with thevoltage, it is also equal to 2ROO var,

7.9 Power factorThe power factor of an alternating-current device orcircuit is the ratio of the active power P to the ap-parent power S. It is given by the equation

power factor = PIS (7.6)where

p ~ active power delivered or absorbed by thecircuit or device [W l

S = apparent power of the circuit or device [VAlPower factor is expressed as a simple number, or asa percentage.Because the acti ve power P can never exceed the

apparent powcr S, it follows that the power factorcan never be greater than unity (or 100 percent).

ACTIVE, REACTIVE. AND APPARf,NT POWER 143

The power factor of a resistor is 100 percent be-cause the apparent power it draws is equal to the ae-tive power. On the other hand, the power factor ofan ideal coil having no resistance is zero, because itdoes not consume any active power.To sum up, the power factor of a circuit or device

is simply a way of stating what fraction of its ap-parent power is real, or active, power.

In a single-phase circuit the power factor is alsoa measure of the phase angle 0 between the voltageand current. Thus. referring to Fig. 7.11,

power factor = PIS= L1/t,'1= 1/1= cos fl

Consequently.power factor = cos fl = PIS (7.7)

wherepower factor = power factor of a single-phase

circuit or devicefl = phase angle between the

voltage and currentIf we know the power factor, we automaticallyknow the cosine of the angle between E and 1 and,hence, we can calculate the angle. The power factoris said to be lagging if the current lags behind thevoltage. Conversely, the power factor is said to beleading if the current leads the voltage.Example 7-7Calculate the power factor of the motor in Example7-5 and the phase angle between the line voltageand line current.Solution

power factor = PIS=IROO/2040=O.g~\2 or gR.2%(lagging)

cos fl = 0.RR2therefore. (J =2R. I 0


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144 l:LECTRICAL MACHINLSANJ) TRANSFORMERS

Example 7-8A single-phase motor draws a current of 5 A from a120 V, 60 Hz line. The power factor of the motor is65 percent.Calculatea. The active power absorbed by the motorb. The reactive power supplied by the lineSolutiona. The apparent power drawn by the motor is

Sill=L'l =120 X 5 =600 VAThe active power absorbed by the motor is

(7.7)= 600 > < (J.()5 = 390 W

b. The reactive power absorbed by the motor is(7.4)

=456 varNote that the motor draws even more reactivepower from the line than active power. This burdensthe line with a relatively large amount of nonpro-ductive power.

7.10 Power triangleThe=p2 + Q2 relationship expressed by Eq. 7.4,brings to mind a right-angle triangle. Thus, we canshow the relationship between S, P ; and Q graphi-cally by means of a power triangle. According toconvention, the following rules apply:I. Active power P absorbed by a circuit or deviceis considered to be positive and is drawn hori-zontally to the right

2. Active power P that is delivered by a circuit ordevice is considered to he negative and isdrawn horizontally to the left

3. Reactive power Q absorbed by a circuit or de-vice is considered to be positive and is drawnvertically upwards

Q(456 var)

P(390 W)

Figure 7.12Power triangle of a motor. See Example 7-8.

4. Reactive power Q that is delivered by a circuitor device is considered to be negative and isdrawn vertically downwardsThe power triangle for Example 7-8 is shown in

Fig. 7.12 in accordance with these rules. The powercomponents S, P . and Q look like phasors, but theyarc not. However. we can think of them as conve-nient vectors. The concept of the power triangle isuseful when solving ac circuits that comprise sev-eral active and reactive power components.7.11 Further aspects of sources

and loadsLet us consider Fig. 7.13a in which a resistor andcapacitor are connected to a source. The circuit issimilar to Fig. 7.10 except that the capacitor is a re-active source. As a result, reactive power f lowsfrom the capacitor to the source G while activepower flows from the source G to the resistor. Theactive and reactive power components thereforeflow in opposite directions over the transmissionline. A wattmeter connected into the circuit willgive a positive reading P = Ell' watts, but a varrne-ter will give a negative reading Q =El.; The sourceG delivers active power P but receives reactivepower Q . Thus, G is simultaneously an activesource and a rcacti vc load.


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(a ) G

(b )

(e)P = E J p Q = L /q

f : ' : ' : ' : : ' : ' : ' : ' : : ' { > PQ < 4 . . . : : : : : : . : . jFigure 7.13a. Source feeding an active and reactive (capacitive)

load.b. Phasor diagram of the circuit.c. The active and reactive powers flow in opposite di-

rections.

It m ay se em un usual to hav e two powe rs flowingin oppo s ite dire ctio n s o v e r th e sam e tran sm iss io nlin e . b ut we m ust again rem em be r that activ e powe rP is n o t th e sam e as a re activ e powe r Q and thate ach flow s in depe nde ntly o f' th e o th er.

S pe aking o f source s and lo ads . a de ceptiv e lys im ple e le ctrical o utle t. such as th e 120 Y re cep-tacle in a hom e , also de se rv e s our atte n tio n . A llsuch outle ts are ultim ate ly conn e cte d to th e hugealte rn ato rs that pow er th e e le ctrical tran sm iss io nand dis tributio n sys tem s . O dd as it m ay se em , ane le ctrical outle t can act n o t o n ly as an activ e o r re -activ e source (as we would expe ct) , but it m ayalso be hav e as an activ e o r re acti v e lo ad. W hatfacto rs de te rm in e whe th e r it w ill h e hav e in o n ew a y o r th e o th er'? It all depe nds upon th e type o f

ACTIVE. REACTIVE. AND APPARENT POWER 145

dev ice (o r de v ice s ) co nn e cte d to th e re ceptacle . Ifth e de v ice abso rbs activ e powe r. th e re ceptaclew ill pro v ide it; if th e de v ice de liv e rs activ e powe r.th e re ceptacle w ill re ce iv e it. In o th e r wo rds . as im ple re ceptacle outle t is at all tim es re ady to de -liv e r-o r accept-e ith e r activ e powe r P o r re ac-tiv e po we r Q in acco rdance with th e de v ice s con -n e cte d to it.

The sam e rem arks apply to any 3-phase 480 Yse rv ice e n tran ce to a facto ry o r th e te rm in als o f ah igh -powe r 345 kY tran sm iss io n lin e .Example 7-9A 50 j ,LF pape r capacito r is placed acro ss th e motorte rm in als in E xam ple 7~8.Calculatea. Th e re activ e powe r ge n e rate d by the capacito rb. The activ e powe r abso rbed by the m o to rc. The re activ e powe r abso rbed from th e lin ed. The n ew lin e curre n tSolutiona. The im pedance o f th e capacito r is

Xc = 1/(2 TI/C) (2 .11 )= 1/(2TI X ()O X .:'i0 X 10-(')= 53 n

The curre n t in th e capacito r isI =E/Xc = 120/53= 2 .261\

The re activ e powe r ge n e rate d by the capacito r isQc =EI'l -t- 120 X 2 .26

= 27 1 val 'b. The m oto r co n tin ue s to draw the sam e activ epowe r be cause it is s till fully lo aded .

Consequen t ly .

The m oto r also draws th e sam e re activ e powe ras be fo re . be cause n o th ing has take n place toch an ge its m agn etic fie ld. C on se que ntly,

Q m =456 var


13/49

146 EU:CTR1CAL MACH1Nl:S 11NDTRANSFORMERS

c. The motor draws 456 var from the line, but thecapacitor furnishes 271 var to the same line,The net reactive power drawn from the line is,therefore,

QL =Q m - Qc=456 - 271= 185 var

The active power drawn from the line isPI. =Pm=390W

d. The apparent power drawn from the line isSL =VP~ + Q 1

zr- V3902 + 1852= 432 V A

The new line current isII . =SIJE =4 3 2 1 1 2 0=3.6A

Thus, the line current drops from 5 A to 3.6 A byplacing the capacitor in parallel with the motor.This represents a big improvement because the linecurrent is smaller and the operation of the motor hasnot been changed in the least.The new power factor of the line is

cos < P I . =P I / S I . = 390/432=0.903 or 90.3%

C P I . = arcos 0.903 =25.5The power triangle is shown in Fig. 7.14. The re-

active power Qc generated by the capacitor isdrawn vertically downward. By comparing thispower triangle with that in Fig. 7.12, we can visu-ally observe the effect of the capacitor on the ap-parent power supplied by the line.

7.12 Systems comprisingseveral loads

The concept of active and reactive power enables usto simplify the solution of some rather complex cir-cuits. Consider, for example, a group of loads con-

.)",(600 VA)

Q,(271 var)

j Q "(456 var)r

Qr(185 var)I

P",(390 W)

Figure 7.14Power triangle of a motor and capacitor connected toan ac line. See Example 7-9.

nected in a very unusual way to a 380 V source (Fig.7 .15a). We wish to calculate the apparent power ab-sorbed by the system as well as the current suppliedby the source.

Using the power approach. we do not have toworry about the way the loads are interconnected.We simply draw a block diagram of the individualloads, indicating the direction (as far as the sourceis concerned) of active and reactive power tlow(Fig. 7.15b). Thus, because load A is inductive, itabsorbs reactive power; consequently, the 5 kvar ar-row tlows from the source to the load. On the otherhand, because load C represents a capacitor, itdelivers reactive power to the system. The 16 kvararrow is directed therefore toward the source.The distinct (and independent) nature of the

active and reactive powers enables us to add all theactive powers in a circuit to obtain the total activepower P . In the same way, we can add the reactivepowers to obtain the total reactive power Q. The re-sulting total apparent power S is then found by

(7.4)We recall that when adding reactive powers, we

assign a positive value to those that are absorbed bythe system and a negative value to those that aregenerated (such as by a capacitor). In the same way.


14/49

(a )

(b )

source

B 2kWD

16 kvar9 kvar

c

source

5 kvar< \ $ . . . . . . . . 19 kvar< ) )

8kW1 . . . . . . . . . . . . . - >

2kW< ' . 1 7 kvarL S t 916 kvar" " , " " ' > lf~! , . , . " . , < >8 kvar

F igu re 7 .1 5a. Example of active and reactive loads connected to

a 380V source.b. All loads are assumed to be directly connected to

the 380 V receptacle.

we assign a positive value to active powers that areabsorbed and a negative value to those that are gen-erated (such as by an alternator).Note that usually we cannot add the apparent

powers in various parts of a circuit to obtain thetotal apparent power S. We can only add them iftheir power factors are identical.Let us now solve the circuit of Fig. 7.15:I. Active powcr absorbed by the system:

P = (2 + R + 14) = -r-24 kW

!1CTIVE. REJ1CTIVE. !1ND 11PPJ1RENT POWER 14 7

2. Reactive power absorbed by the system:Q I = (5 + 7 + 8) = +20 kvar

3. Reactive power supplied by the capacitors:Q2 =(-9 - 16) ~ -25 kvar

4. Net reactive powcr Q absorbed by the system:Q =(+20 - 25) =-5 kvar

5. Apparent power of the system:S = V p2 + Q2 = V242 + ( - 5)2=24.5 kVA

6. Because the 380 Y source furnishes the appar-ent power. the line current is

I =SI E =24500/380 =64.5 A7. The power factor of the system is

cos W I . =P IS =24/24.5 =0.979 (leading)The 380 V source delivers 24 kW of active

power. but it receives 5 kvar of reactive power. Thisreactive power flows into the local distribution sys-tem of the electrical utility company. where it be-comes available to create magnetic fields. The mag-netic fields may be associated with distributiontransformers, transmission lines. or even the elec-tromagnetic relays of customers connected to thesame distribution system.

The powcr triangle for the system is shown in Fig.7.15c. It is the graphical solution to our problem.Thus. starting with the 5 kvar load. we progressivelymove from one device to the next around the system.While so doing. we draw the magnitude and direc-tion (up. down. left. right) of each power vector. tailto head. in accordance with the power of each devicewe meet. When the selection is complete. we candraw a power vector from the starring point to the endpoint. which yields the inclined vector having a valueof 24.5 kYA. The horizontal component of this vec-tor has a value of 24 kW and. because it is directed tothe right. we know that it represents power absorhedby the system. The vertical component of 5 kvar isdirected downward: consequently. it represents reac-tive power generated by the system.


15/49

148 EUXTRICAL MACHINES AND TRANSFORMERS

2kW

5 kvar

startingpoint

24kW- - - - - - - - - - - - - - - - - - - - - - - - - - ~ I5 kvar :

16 kvar

9 kvar

8kWFigure 7.1ScPower triangle of the system.

7.13 Reactive power withoutmagnetic fieldsWe sometimes encounter situations where loads

absorb reactive power without creating any mag-netic field at all. This can happen in electronicpower circuits when the current flow is delayedby means of a rapid switching device, such as athyristor.

Consider, for example, the circuit of Fig. 7.16 inwhich a 100 V . 60 Hz source is connected to a re-sistive load of Ion by means of a synchronous me-chanical switch. The switch opens and closes itscontacts so that current only flows during the latterpart of each half cycle. We can see, almost by intu-ition, that this forced delay causes the current to lagbehind the voltage. Indeed, if we connected awattmeter and varmeter between the source and theswitch, they would respectively read -I- 500 Wand-I- 31 8 var. This corresponds to a lagging power fac-tor (sometimes called displacement power factor)of 84.4 percent. The reactive power is associated

endpoint

8 kvar

14kW7 kvar

with the rapidly operating switch rather than withthe resistor itself. Nevertheless, reactive power isconsumed just as surely as if a reactor were presentin the circuit. This switching circuit will be dis-cussed in detail in Chapter 30.7.14 Solving AC circuits using

the power triangle methodWe have seen that active and reactive powers can beadded algebraically. This enahles LIS to solve somerather complex ac circuits without ever having todraw a phasor diagram or resorting to vector (j) no-tation. We calculate the active and reactive powersassociated with each circuit element and deduce thecorresponding voltages and currents. The followingexample demonstrates the usefulness of this powertriangle approach.Example 7-10 _In Fig. 7 .17a, the voltage between terminals I and 3is 60 Y .


16/49

(a )

~500W

(b )

Figure 7.16a. Active and reactive power flow in a switched resis-

tive load.b. The delayed current flow is the cause of the reac-

tive power absorbed by the system.

Calculatea. The current in each circuit elementh . The voltage between terminals I and 2c . The impedance between terminals I and 2SolutionWe know the impedances of the elements and that60 V exists between terminals 3 and I (Fig. 7.17b).We now proceed in logical steps, as follows:a. The current in the capacitor is

I e =60/5 =12Afrom which the reactive power generated is

Qc = 12 X 60 = -no varThe current in the resistor is

I R = 60/12 = 5 A

ACTIVE, REACIlVE. AND APPARENT POWER 14 9

60 V

0o-------------~----~(a)

S = 780 VAXL r-"--3 A

5Q

(b)

Figure 7.17a. Solving ac circuits by the power triangle method.b. Voltages and currents in the circuit. See Example7-10.

from which the active power absorbed isp =5 X 60 =300 W

The apparent power associated with temlinals 1-3:S =V p : > . + Q2 =V:lOO:>' + (-nO):> '=780 VA

The current I L must, therefore. beI L = SIE JI = 780/60 = 13 A

The voltage across the inductive reactance is23 = I X L = 13 X 8 = 104 V

The reactive power absorbed by the inductivereactance is

QL = 23 X II = 104 X 13=+ 1352 var


17/49

150 ELFCTRICAL MACHINES AND TRANSFORM FRS

The total reactive power absorbed by the circuit isQ=Q, + Qc = 1352 - 720= +632 var

The total active power absorbed by the circuit isP =30n W

The apparent power absorbed by the circuit is- -S =\p2 + Q" = \300" + 6322

= 700 VAb. The voltage of the line is therefore

E2l = . 111 , = 700113 = 53.9 Vc. The impedance between terminals 2-1 is

Example 7-11 _A single-phase 12.47 kV transmission line severalkilometers long feeds a load C from a substation(fig. 7.1 S). The line has a resistance of 2.4 nand areactance of ISH. Instruments at the substation in-dicate that the active and reactive power inputs tothe line are 3 MW and 2 Mvar, respectively.Co/ell/ore'a) the line current and its phase angle with respect

to the line voltage at the substationb) the active power absorbed by the loadc) the reactive power absorbed by the loadd) the line voltage at the loade) the phase angle between the voltage at the load

and that at the substationSolutiona) Apparent power delivered to the line:

S =V P" + Q2 =vi -t 2"= 3.60 MVALine current:

/= SE3600000 VA12470 V =289A

Power factor at the substationPFP = S

3MW3.6 MVA =0.833

1.25 Mvar 0.2MWuload Cubstation

15 n 2,4 Q12.47kV3MW2 Mvar

289 A 10.03 kV2.8MW0.75 Mvar

12.47kV10.03 kV

\ 33.6"\ ~ 289 A89 AFigure 7.18Voltages, currents and power. See Example 7-11.

Phase angle hetween the voltage and current at thesubstation:

8 =arccos 0.833 =33.6cb) Active power dissipated in theline:

PL = R P =2.4 X 2Sy2=0.2 X 106 =0.2 MW

Active power absorbed by the load:P, = P,uh - PL =3 MW - 0.2 MW

~ 2.8 MWc) Reactive power absorbed by the line:QL=Xr/2= 15x2892= 1.25X 10"= 1.25 MvarReactive power absorbed by the load:Q, = Q,uh - Q, =2 Mvar - 1.25 Mvar=0.75 Mvar

d) Apparent power at the load:Sc=VP l + Ql =\/Vl2 + 0.752=2.90 MVA

Voltage at the load end of the line:

E c ScI 28YA2.90MVA = 10.03 kV


18/49

Power factor at the load end of the line:2.XMW2.90 MY A =0.965 ou 96.5%

Phase angle bet ween the voltage and current at theload:O c = arccos 0.965 = 15.2

It follows that the phase angle between the voltageat the substation and that at the load is (33.6 -15.2") = 18.40.Fig. 7.18 summarizes the results of this analy-

sis. We could have found the same values usingvector algebra. However, on account of its sim-plicity. the power method of solving this problemis very appealing.

7.15 Power and vector notationIf vector notation is used to solve an ac circuit, wecan readily determine the active and reactive powerassociated with any component, including thesources. We simply multiply the phasor voltage Eacross the component by the conjugate (/*) of thecurrent that flows through iL" The vector productE l gives the apparent power Sin terms of P + . iQ .where P is the active power and Q the reactivepower absorbed (or delivered) by the component.A positive value for P or Q means that the com-

ponent absorbs active or reactive power, Negativevalues mean that the component delivers active orreactive power.When calculating the EF' vector product, it is

very important to follow a standard procedure in or-der to obtain the correct result, The procedure ap-plies to circuits that use the double subscript nota-tion or the sign notation (see Sections 2.4 and 2.5).Consider Figure 7.19a in which a circuit clement

Z is part of a larger "rest of circuit." We want to cal-culate the active and reactive power associated withelement Z . We note that current Illows from termi-nal (/ to terminal h. i.e. in the sequence aboConsequently, when calculating the product L'I"',the subscripts of voltage E must be written in thelfu current h as a v alue fLO. i ts c o n jug at e r=fL-O.

ACTIVE. REACTIVF.. AND APPARENT POWF.N lSI

f--- -o=J--"1 ~I 5 = Eabl*, ~ _ 1 _ = _ ~ _ = r ~ .': rest of circuit :L .: (a)

~~ 5=+11*~'- '1- ~",_ _ L _ _, ': rest of circuit :~ J (b )

f-rPI 5=-41*I ) ., .. ____ E 4- - - - - - - - ~ / )J - - - - J _ ., 'i rest of ci rcu it :L 1 (c)

Figure 7_19Method of writing power equations.same sequence ab (not ba), The apparent pown Sassociated with Z is therefore written

S =E"I/*I t would be incorrect to write S = EoJ.*

In Fig. 7.lllb. sign notation is used, and it is seenthat current I enters Z by the (+) terminal.Consequently, the apparent power is given by

S = -j F ; I / *The E1/* product is preceded by a (+) sign becausecurrent I is shown as entering by the ( 4 ) terminal ofelement 7 ..

I n the case of Fig. 7.19c. we write S = ..)./* be-cause the current enters Z by the (-) terminal.

If we want to, we can determine the apparentpower associated with the "rest o r circuit" (roc).Thus, in Fig. 7.19a. because the current circulatesfrom b to a in the roc. we would write:


19/49

152 ELFCTRICAL MACHINES AND TRANSFORMERS

Similarly, in Fig. 7.19


20/49

a 45 U c-~I

28Q

bFigure 7.21Solvingan ac circuit using vector notation.

b. The magnitude and phase of the voltage acrossthe resistor and across the reactancec. The active and reactive power associated withthe resistor. the reactance. and the source

SolutionC \. Applying Kirchhoff's voltage law (see Section2.32). we obtain

-Lan + 45 1 + j281 =0-159L65 + 1(45 + j28) = 0

15 9 L 65 1 ~ 45 + j28Transforming the denominator into polar coor-dinates. we obtain

amplitude = \452 + 282 = 53phase angle ~ arctan 28/45 c--; .11.89"

hence 45 + j28 = 53L31 .89'159 L 65and so 1 =- --- - = 3 L (650 - 31.89)5 .1 1 31 .89

=3L3.1.110b . Voltage across the resistor is

E"c =45 1=45 X 3L.13.llc= 135L33.llo

ACTIVE. REACTIVE. AN}) APPARtNT POWER 153

Voltage across the reactance isch=j281

= j28 X 3L33.11 0= 84L(33.11 0 + 9()O)=84LI2.l.llo

c. The conjugate 1* of the current 1 is1*=3L-33 .110

The apparent power associated with the resistor is

= (135L33.11 0) (3/- -33.1 n=405LO=405 (cos 00 + j sin 0)= 405 (I + jO )= 405

Thus. the resistor absorbs only real power (405 W)because there is no j component in Sr.The apparent power associated with the reac-

ranee is

= (84LI23.1I) (3L-33.1 l")= 52L90~ 252 (cos 90 + j sin 9()U)~ 252 (0 + j I)= j252

Thus. the reactance absorbs only reactive power(252 var).

The apparent power associated with the source is

= -(159!65)(31_-33.ln=-477L(65 - .13.11)= -477 L31.89~ -477 (cos 31.89 + j sin 31.89)=-477 (0.849 +.i 0.528)= -405 - j 252


21/49

1 5 4 ELECTR1CAL MACHINES AND 7RANSFORMERS

The active and reactive powers are both negative,which proves that the source delivers an activepower of 405 Wand a reactive power of 252 var.7.16 Rules on sources and loads

(sign notation)We are often interested in determining whether a de-vice is an active/reactive source or an active/reactiveload without making a complete mathematical analy-sis, such as we performed in Section 7.15. To enableus to positively identify the nature of the source orload. consider Fig. 7.22 in which a device A carries aline current I. The device is part of a circuit.The voltage between the terminals is E . and one

of the terminals bears a (+) sign. The phase anglebetween E and 1 can have any value. As a result, Ican be decomposed into two components. Ip and 1'1that are respectively parallel to, and at right anglesto E . Let II ' be the component of I that is parallel toE . It will therefore be either in phase with, or 1800out of phase with E . Similarly, 1'1can he either 90behind or 900 ahead of E .

The circuit diagram, together with the phasor re-lationships between E and I. enable us to statewhether a device is an active load or an activesource. The following rule applies":I. A device is an active load whena. voltage E and component Ip are in phase andb. the line current I is shown as entering the(+) terminal.

Otherwise. the device is an active source.The following rule also applies:2. A device is a reactive load when

a. component Ie ) lags 900 behind voltage E andb. line current I is shown as entering the ( ~)terminal.

Otherwise. the device is a reactive source.Based on these rules, and observing the phasor rela-tionships in Fig. 7.22, we deduce that device A is anactive load because I" is in phase with E . Also. deviceA is a reactive source because 1'1is 900 ahead of E .These rulc-, arc in agreement with IEEE and lEe conventions.

Figure 7.22Device A may be an active/reactive source oractive/reactive load depending upon the phasor rela-tionship between E and I.7.17 Rules on sources and loads

(double subscript notation)We can also tell whether a device is an active sourceor active load when double subscript notation isused. Consider Figure 7.23 in which a device A car-ries a current I flowing in the direction shown. Thevoltage between terminals a and b is E"I1' The fol-lowing rule applies:3. A device is an active load when:

a. voltage Eah and component I" are in phase ow lb. line CUITent I is shown as entering terminal a.

Otherwise. the device is an active source.The following rule also applies:4. A device is a reactive load when

a. current 1 4 lags 900 behind voltage Eah and

electricaldevice A

Figure 7.23Same circuit as in Fig. 7.22 except that double-subscript notation is used.


22/49

b. line current 1 is shown as entering by termi-nal a.

Otherwise. the device is a reactive source.Based on these rules, and observing the phasor rela-tionships in Fig. 7.23, we deduce that device A is anactive source because 1pis 1800 out of phase with Eao .Also. device A is a reactive load because 1 4 lags

90" behind e.;Questions and ProblemsPractical level7-1 What is the unit of active power? reactive

power? apparent power'?7-2 A capacitor of 500 kvar is placed in parallelwith an inductor of 400 kvar. Calculate the

apparent power of the group.7-3 Name a static device that can generate reac-

tive power.7-4 Name a static device that absorbs reactive

power.7-5 What is the approximate power factor, in

percent. of a capacitor? of a coil? of an in-candescent lamp?

7-6 The current in a single-phase motor lags50 behind the voltage. What is the powerfactor of the motor')

intermediate level7-7 A large motor absorbs 600 kW at a power

factor of 90 percent. Calculate the apparentpower and reactive power absorbed by themachine.

7-8 A 200 f.LFcapacitor is connected to a 240 V,flOHz source. Calculate the reactive powerit generates.

7-9 A IOU resistor is connected across a 120 V,flO Hz source. Calculate:a. The active power absorbed by the resistorb. The apparent power absorbed by the resistorc. The peak power absorbed by the resistord. The duration of each positive power pulse

7-10 A Ion reactance is connected to a 120 V,e o Hz line. Calculate:

ACTIVIo', REACTIVE, AND APPARENT POWER 155

a. The reactive power absorbed by the reactorb. The apparent power absorbed by the reactorc. The peak power input to the reactord. The peak power output of the reactore. The duration of each positive power pulse

7-II Using the rules gi ven in Sections 7.16 and7,17, determine which of the devices in Fig.7.24a through 7,24f acts as an active (or re-active) power source.

7-12 A single-phase motor draws a current of12 A at a power factor of 60 percent.Calculate the in-phase and quadrature com-ponents of current 1 1 ' and I q with respect tothe line voltage.

7 -13 A single-phase motor draws a current or I flA from a 240 V, 60 Hz line, A wattmeterconnected into the line gives a reading of2765 W. Calculate the power factor of themotor and the reactive power it absorbs.

7-14 If a capacitor having a reactance of 30 n isconnected in parallel with the motor ofProblem 7-13, calculate:a. The active power reading: of the wattmeterb. The total reactive power absorbed by the ca-pacitor and motor

c. The apparent power of the ae lined. The line currente. The power factor of the motor/capacitorcombination

7-15 Using only power triangle concepts(Section 7.14) and without drawing anyphasor diagrams, find the impedance of thecircuits in Fig. 7.25.

7-16 An induction motor absorbs an apparentpower of 400 kVA at a power factor of 80percent. Calculate:a. The active power absorbed by the motorb. The reactive power absorbed by the motorc. What purpose does the reactive power serve

7 -17 A circuit composed of a 12 D resistor inseries with an inductive reactance of 5 ncarries an ac current of lOA. Calculate:a. The active power absorbed by the resistorb. The reactive power nbsorbed by the inductorc. The apparent power of the circuitd. The power factor of the circuit


23/49

1- ~I

----~E E1-(b)

i~ f + 0lIIEIe)

(a l

Id)

Figure 7.24See Problem 7-11.

10 S1

(a) (b)


7- I 8 A coil having a resistance of 5nand an in-ductance of 2 H carries a direct current of20 A. Calculate:a. The active power absorbedh. The reactive power absorbed

Advanced level7-19 A motor having a power factor of 0.8 ab-

sorbs an active power of 1200 W. Calculatethe reactive power drawn from the line.

7-20 In Problem 7-13, if we place a capacitor of500 var in parallel with the motor, calculate:a. The total active power absorbed by the systemh. The apparent power of the systemc. The power factor of the system

156

1-

E(e)

1-Q 1 + C )~E I

(f)

6S1

(e)

7-21 A coi I havi ng a reactance of 10!land a re-sistance of 2 n is connected in parallel witha capacitive reactance of Ion. If the sup-ply voltage is 200 V, calculate:a. The reactive power absorbed by the coilh. The reactive power generated hythe capacitor

c. The active power dissipated by the coild. The apparent power of the circuitThe power factor at the terminals of a120 Y source is O .h lagging (Fig. 7.26).Without using phasor diagrams,calculate:a. The value of I:h. The impedance or the load Z

7-22


24/49

H H


7-23 In Figs. 7.27a and 7.27b, indicate the mag-nitude and direction of the active and reac-tive power flow. (Hint: Decompose I into Ipand ([' and treat them independently.)

Industria { appli cation7-24 !\ single-phase capacitor has a rating of

30 kvar, 4RO Y, 60 Hz. Calculate its capaci-tance in microfarads.

7-25 In Problem 7-24 calculate:a. The peak voltage across the capacitor whenit is connected to a 460 V source

b. The resulting energy stored in the capacitorat that instant. in joules

7-26 Safety rules state that one minute after a ca-pacitor is disconnected from an ac line, thevoltage across it must be 50 Y or less, Thedischarge is done by means of a resistorthat is permanently connected across thecapacitor terminals. Based on the dischargecurve of a capacitor, calculate the dischargeresistance required, in ohms, for the capaci-tor in Problem 7-24. Knowing the resis-tance is subjected to the service voltagewhen the capacitor is in operation. calculateits wattage rating.

7-27 A 13.2 kY. 60 Hz single-phase line con-nects a substation to an industrial load. Theline has a resistance of 2.4 nand a reac-tance of 12 n. The metering equipment atthe substation indicates that the line voltageis 12.5 kY and that the line is drawing 3 MWof active power and 2 Mvar of reactivepower. Calculate:a. The current flowing in the line

ACTIVE. REACTIVE, AND APPARENT POWER 157

1- 1-

E> 120 V

(a) (bl


b. The active and reactive power consumed bythe line

c. The active, reactive and apparent power ab-sorbed by the load

d. The voltage across the load7-2R A 2 hp. 230 V. 1725 r/min 60 Hz single-

phase washdown duty motor, manufacturedby Baldor Electric Company. has the fol-lowing characteristics:Fullload current: 11.6 Aefficiency: 75.5%power factor: 74%weight: 80 Iba. Calculate the active and reactive power ab-sorbed by this machine when it operates atlull load.

b. If a 40 microfarad capacitor is connectedacross the motor terminals. calculate the linecurrent feeding the motor.

c. Will the presence of the capacitor affect thetemperature of the motor'?

7-29 A single-phase heater absorbs 4 kW on a 240Y line, A capacitor connected in parallel withthe resistor deli vel'S 3 kvar to the line.a. Calculate the value of the linc current.b. If the capacitor is removed, calculate thenew line current.


25/49

CHAPTER8Three-Phase Circuits

8.0 IntroductionE lectric power is generated, transmitted. and dis-tributed in the form of 3-phase power. Homesand small establishments are wired for single-phasepower. but this merely represents a tap-off from thebasic 3-phasc system. Three-phase power is pre-ferred over single-phase power for several impor-tant reasons:a. Three-phase motors, generators. and transform-ers are simpler. cheaper, and more efficient

b. Three-phase transmission lines can delivermore power for a given weight and cost

c. The voltage regulation of 3-phase transmissionlines is inherently betterA knowledge of 3-phase power and 3-phase cir-

cuits is. therefore. essential to an understanding ofpower technology. Fortunately, the basic circuittechniques used to solve single-phase circuits can bedirectly applied to 3-phase circuits. Furthermore. wewill see that most 3-phase circuits can be reduced toelementary single-phase diagrams. In this regard,we assume the reader is familiar with the previouschapters dealing with ac circuits and power.

8.1 Polyphase systemsWe can gain an immediate preliminary understand-ing of polyphase systems by referring to the commongasoline engine.A single-cylinder engine having onepiston is comparable to a single-phase machine. Onthe other hand. a 2-cylinder engine is comparable toa 2-phase machine. The more common 6-cylinderengine could be called a 6-phase machine. In6-cylinder engine identical pistons move up anddown inside identical cylinders, but they do nomove in unison. They are staggered in such a way ato deliver power to the shaft in successive pulsesrather than at the same time. As the reader may knowfrom personal experience. this produces a smootherrunning engine and a much smoother output torque.Similarly, in a 3-phase electrical system. the

three phases are identical, but they deliver power adifferent times. As a result. the total power flow ivery smooth. Furthermore. because the phases arcidentical, one phase may be used to represent thebehavior of all three.Although we must beware of carrying analogies

too far, the above description reveals that a 3-phasesystem is basically composed of three single-phase

158


26/49

systems that operate in sequence. Once this basicfact is realized, much of the mystery surrounding3-phase systems disappears.

8.2 Single-phase generatorConsider a permanent magnet NS revolving at con-stant speed inside a stationary iron ring (Fig. 8.1).The magnet is driven by an external mechanicalsource. such as a turbine. The ring (or stator) re-duces the reluctance of the magnetic circuit: conse-quently. the flux density in the air gap is greater thanif the ring were absent. A multiturn rectangular coilhaving terminals a, 1 is mounted inside the ring butinsulated from it. Each turn corresponds to two con-ductors. one in each slot.

stator winding A

Figure 8.1A single-phase generator with a multiturn coil embed-dedin two slots. At this instant Ea1 is maximum (+).

As the magnet turns, it sweeps across the con-ductors. inducing a voltage in them according to theequation:

E =BII'.01 (2.25)whereinE " I ~ instantaneous voltage induced in the coil IV I

H = instantaneous flux density cutting across theconductors in the slots IT]

I = length of conductors lying in the magneticfield IInI

r ~ peripheral speed of the revolving poles [m/s]The SUIll of the voltages induced in all the con-

ductors appears across the terminals. The terminal

THREE-PHASE CIRCUITS 159

stator winding A

Figure 8.2At this instant Ea1 ~ 0 because the flux does not cutthe conductors of winding A .

voltage f.'al is maximum when the poles are in theposition or Fig. 8.1 because the flux density isgreatest at the center of the pole. On the other hand.the voltage is zero when the poles are in the positionof Fig. 8.2 because nux does not cut the conductorsat this moment.

I f we plot Eal as a function of the angle of rota-tion. and provided the N. S poles are properlyshaped, we obtain the sinusoidal voltage shown illFig. 8.3." Suppose the alternating voltage has a peakvalue of 20 V . Machines that produce such voltagesare called alternating-current generators or S\'I1-chronous generators. The particular machine shownin Fig. 8.1 is called a single-phosc generator:

v+ 20

10t.,I 0 0-10

- 20


27/49

160 E UXTRICA/_ /V IACIIINE S AND TRANSF OI?M E RS

8.3 Power output of a single-phasegenerator

If a re s is to r is co nn e cte d acro ss te rm in als a, Ia cur-ren t w ill n ow and the re s is to r will h eat up (Fig. 8A).T he cu rre nt fa is in phase with th e v o ltage and. co n -se que n tly. th e in s tan tan e ous powe r is com po sed o f aseries o f po s itiv e pulse s . as shown in Fig. 8.S. Theav e rage powe r is o n e -half th e pe ak powe r. Th is e le c-trical power is de riv e d from the m echan ical powe rpro v ided by the turbin e driv ing th e ge n e rato r. A s are sult. th e turbin e must de liv e r its m echan ical e n -e rgy in pulse s . to m atch th e pulse d e le ctrical output.Th is se ts up mechanical v ibratio ns w ho se frequencyis tw ice th e e le ctrical fre que n cy. Con seque n tly. th ege n e rato r will v ibrate and le nd to be no isy.8.4 Two-phase generatorUsing th e sam e s ingle -phase ge n e rato r. le t us mounta se cond winding (B) on th e s tato r. iden tical to

R load on phase A

Figure 8.4Single-phase generator delivering power to a resistor.winding A . but displaced from it by a mechan ica langle o r ( JO ( Fig . 8 .6 a) .

A s th e m agn e t rotates, s in uso idal v o ltage s are in -duced in each w inding. They obv ious ly hav e th esam e m agn itude and frequen cy but do no t re achth e ir m axim um value at th e sam e tim e. In e ffe ct. atthe moment when th e m agn e t o ccupie s th e po sitio nshown in Fig. 8.6 a. v o ltage E, 1 passe s th rough itsm axim um po sitiv e v alue . whe re as v o ltage Eh2 isze ro . Th is is be cause th e flux on ly cuts across th econducto rs in slo ts 1 and a at th is in s tan t. Howev e r.afte r th e ro to r has m ade on e quarte r- turn (o r Y O O ) .

v

- 10

- 20

peak power c:o L 'a fa -= Pm

instantaneous power of phase A

Figure 8.5Graph of the voltage, current, and power when thegenerator is under load.

voltage E "I be com es ze ro and vo ltage Eb2 attain s itsm axim um po sitiv e v alue . The two vo ltage s areth e re fo re out o f phase by Y O . T he y are re pre se nte dby curv e s in Fig. 8.6b and by phasors in F ig. 8.6 c.No te that E,I I leads Eb2 be cause it re ache s its pe akpo sitiv e v alue be fo re Lb2 docs .

Th is m ach in e is calle d a two-phase generatorand th e s tato r w indings are re spe ctiv e ly calle dphose A an d phose B.Example 8-/The ge n e rato r shown in Fig. 8.6 a ro tate s at 6000r/rn in and ge n e rate s an e ffe ctiv e s inuso idal v o ltageo f 17 0 V pe r w inding.Co/cu/olea. The pe ak vo ltage acro ss e ach phaseb. The output fre que n cyc. The tim e in te rv al co rre sponding to a phase an -

gie o f 90Solutiona. The pe ak vo ltage pe r phase is

EIll = . . J 2 L = IAI4 X 170~ 240 V

(2 .6 )


28/49

b. One cycle is completed every time the magnetmakes one turn. The period of one cycle is

T ~ 1 /6000 min= 60/6000 s = 0.0 I s=10 ms

The frequency isf= IIT= 1/0.01 =100Hz

c. A phase angle of 90 corresponds to a time in-terval of one quarter-revolution, or 10 rns/4 =2.5 ms. Consequently, phasor Eh2 lags 2.5 msbehind phasor E " I'

stator winding A

(a )

(b )

t;;Ea' I IEb2~~~ rIX " II" ,

f-9 1 \ _ 18~ ~70,-;--~60,- 450-:angle of rotat ion e" \, ,

THREE-PH/1SE CIRCUnS 1 6 1

8.5 Power output of a 2-phasegenerator

Let us now connect two identical resistive loadsacross phases A and B (Fig. X.7a). Currents i, and t;will tlow in each resistor. They are respectively inphase with E "I and Eh2. The currents are, therefore,90 out of phase with each other (rig. X.7b). Thismeans that I " reaches its maximum value one quarter-period before II>docs. Furthermore, the generatornow produces a 2-phase power output.The instantaneous power supplied to each resis-

tor is equal to the instantaneous voltage times theinstantaneous current. This yields the two powerwaves shown in Fig. XX Note that when the powerof phase A is maximum. that of phase B is zero, andvice versa. If we add the instantaneous powers ofboth phases, we discover thai the resultant p()wer isconstant and equal to the peak power Pill of one

(a)

load on phase Br:nEa,( C ) Lb (b)Eb2

Figure8.6a. Schematic diagram of a 2-phase generator.b. Voltages induced in a 2phase generator.c. Phasor diagram of the induced Voltages.

R load onphase A

Figure 8.7a. Two-phase generator under load.b. Phasor diagram of the voltages and currents.


29/49

162 ELlTTRICAt MACfllNES AN/) TRANSFORMERS

phase.* In other words. the total power output of the2-phase generator is the same at every instant. As aresult. the mechanical power needed to drive thegenerator is also constant. A 2-phase generator docsnot vibrate and so it is less noisy. As an importantadded benefit. it produces twice the power outputwithout any increase in size. except for the additionof an extra winding.

8.6 Three-phase generatorA l-phase generator is similar to a 2-phase generator.except that the stator has three identical windings in-stead of two. The three windings a-I, b-Z, and c.-3 areplaced at 120G to each other. as shown in Fig. ~.9a.

When the magnet is rotated at constant speed.the voltages induced in the three windings have thesame effective values. but the peaks occur at differ-ent times. At the moment when the magnet is in theposition shown in Fig. ~.9a. only voltage Eal is at itsmaximum positive value.

Voltage Ebc. will reach its positive peak afterthe rotor has turned through an angle of 120 (or

The rcrm I'/w.le is used to designate different things.Convequently, it h;IS to be read in context to be understood.The following examples show some of the ways in whichthe word phas is used.

I. The current is 0111of"l'ir ll.l(, with the voltage (refers to pha-xor diagram)

2 The three 1 ' / 1 1 1 . 1 ( ' . 1 ' of a transmission line (the three couduc-tors 01 the line)

3. The /'/W.I('-/O-I'/IIIS(' voltug (thc line voltage)4. The 1 ' / 1 1 1 . 1 ' ( ' .I(,{/II('I/C(' (the order in which the phasors follow

each other)'i. The lnrrncd-otn plru, (the burned-out winding of a -'-phase

machine)Ii. The 3 -1 ' / 1 1 1 1 ' ( ' m/lage (the line voltage of a -'-phase system)7. The 3-/,/w.\{' current: arc unbalanced (the currents in a -'-phase

line or machine are unequal and not displaced at 12()")X. The l'/w.I('-.lili!II/"{/IISj(i/"li/{'/" (a device that can change the

phase lUlgk or the output voltage with respect 10 the inputvoltage)

Y . The l'/wse-IO-I'/II/Si'.!(III!t (a short-circuit between two lineconductors i

1 0. P/W\{'-lo-gmlllldj(III/1 (a short-circuit between a line orw in din g an d gnHlllli)

II. The l'/ws{'.1 arc /lId",/a//{{'t! (the line voltages. Of the linecurrents. are unequal or not displaced at 12()" 10 eachother)

o 180 2700~ angle of rotation

Total instantaneous power output

Figure B .BPower produced by a 2-phase generator.

one-third of a turn). Similarly, voltage E"o, will at-tain its positive peak after the rotor has turnedthrough 2400 (or two-thirds of a turn) from its ini-tial position.

Consequently. the three stator voltages-E"I'Eho. and Ed-arc respectively out of phase by120. They arc shown as sine waves in Fig. 8.9b.and as phasors in Fig. 8.9c.

8.7 Power output of a 3-phasegenerator

Let us connect the three windings of the generator tothree identical resistors. This arrangement requiressix wires to deliver power to the individual single-phase loads (Fig. 8.1 Oa). The resulting currents 1 . , . l h 'and I" are respectively in phase with voltages E"I 'E b2 and Ed' Because the resistors are identical. thecurrents have the same effective values. but they aremutually out of phase by 1200 (Fig. 8.1 Ob), The factthat they arc out of phase simply means that theyreach their positive peaks at different times.


30/49

(a )

+Er

(b) 0

Ea! Eb2 H - Ec3 t-+-, ~-..,1\

I

1\ !,0 .~20 r-~1? f-36? 1~80 1-

,,.-

1 1

(c )

F ig ure 8 .9a. Three-phase generator.b. Voltages induced in a 3-phase generator.c. Phasor diagram of the induced voltages.

THREE-PlIASE CIRCUITS 163

The instantaneous power supplied to each resis-tor is again composed of a power wave that surgesbetween zero and a maximum value Pill' However.the power peaks in the three-resistors do not occurat the same time, due to the phase angle between thevoltages. If we add the instantaneous powers of allthree resistors, we discover that the resulting poweris constant, as in the case of a 2-phase generator.However. the total output of a 3-phas~ generator hasa magnitude of 1.5 P'll' Because the electrical out-put is constant. the mechanical power required todrive the rotor is also constant. and so a 3-phasegenerator does not vibrate, Furthermore. the powerflow over the transmission line, connecting the gen-erator to the load, is constant.Example 8-2 _The 3-phase generator shown in Fig. ~. lOa is con-nected to three 20 n load resistors. If the effectivevoltage induced in each phase is 120 V. calculatethe following:a. The power dissipated in each resistorh. The power dissipated in the 3-phase loadc. The peak power Pill dissipated in each resistord. Thc total 3-phase power compared to P'llSolutiona. Each resistor behaves as a single-phase load

connected to an effective voltage of 120 V . Thepower dissipated in each resistor is, therefore,

(a)

F igu re 8 .1 0a. Three-phase, 6-wire system.b. Corresponding phasor diagram.

(b)


31/49

164 ELECTRICAL MACHINES AND TRANSFORMERS

(a)

Figure 8.11a. Three-phase, 4-wire system.b. Line currents in a 3-phase, 4-wire system.

P = E"IR = 1202120=now

b. The total power dissipated in the 3-phase load(all three resistors) is

PT =3 P = 3 X no=2160W

This power is absolutely constant from instantto instant.

c. The peak voltage across one resistor isE ", = -\2 = 'i2 X 120=169.7 V

The peak current in each resistor is1m = En,lR = 169.7/20= 8.485 A

Figure 8.12Three-phase, 3-wire system showing source and load.

+ lmax IT' ' 7 b 1 - ! = r 8 - ~T1f i+P+-R11~ 1- '~ v' -, ' - r = t = lt-- ~ ; 8 - 1 =,\tl "\ Ij f~;" -I, ~ " r- j? + r~20t-- 2 , ~ J r-3~?....J_ 1~80t r-;\ f-H + '+1\ i/" " /f H - , ~ B l f J. : . H - T... ,_ . .l..... L.

o

- I m ax

(bl

The peak power in each resistor isr; = E" , /1 l l = 169.7 X 8.485=1440W

d. The ratio of r, to r . ; isPTI r; = 216011440

= 1.5Thus. whereas the power in each resistor pul-

sates between 0 and a maximum of 1440 W, the to-tal power for all three resistors is unvarying andequal to 2160 W.8.8 Wye connectionThe three single-phase circuits of Fig. 8.10 areelectrically independent. Consequently, we canconnect the three return conductors together toform a single return conductor (Fig. 8.lla), Thisreduces the number of transmission line conduc-tors from 6 to 4. The return conductor, called neu-tral conductor (or simply neutral), carries the sumof the three currents (I" + Ih + (.).At first it seemsthat the cross section of this conductor should bethree times that of lines a, b. and c. However, thediagram of Fig. 8.11 b clearly shows that the slimof/he three return currents is ;:em at cl'ery in-stant, For example, at the instant corresponding to240, t; = Illlax and Ib = i,= -0.5 In",, ' makingI " + Ib + I e =0, We arrive at the same result (and


32/49

much more simply) by taking the sum of the pha-sors (/" + Ih + /J in rig. 8. lOb. The sum is clearlyzero.We can, therefore, remove the neutral wire al-

together without in any way affecting the voltagesor currents in the circuit (Fig. 8.12).ln one strokewe accomplish a great saving because the numberof line conductors drops from six to three!However, the loads in Fig. 8.11 a must be identicalin order to remove the neutral wire, I f the loads arenot identical, the absence of the neutral conductorproduces unequal voltages across the three loads.The circuit of Fig. S.12-composed of the

generator, transmission line, and load-is calleda 3-p/wse. 3-wire system. The generator. as wellas the load. are said to be connected in wye. be-cause the three branches resemble the letter Y.For equally obvious reasons. some people preferto use the term connected in star.The circuit of Fig. S.lla is called a I-phase.

-i-wire system. The neutral conductor in such asystem is usually the same size or slightly smallerthan the line conductors. Three-phase, 4-wire sys-tems are widely used to supply electric power tocommercial and industrial users. The line conduc-tors are often called phases, which is the sameterm applied to the generator windings.B.9 Voltage relationshipsConsider the wye-connected armature windings of a3-phase generator (Fig. 8.13a). The induced voltagein each winding has an effective value L'LN repre-sented by the length of each phasor in the diagram inFig. R.13b. Knowing that the line-to-neutral voltagesare represented by phasors L 'a ,, ' Eh". and Een the ques-tion is. what are the line-to-line voltages L'a!>' Ehe, andE c a ' ? Referring to Fig. 8.13a, we can write the fol-lowing equations, based on Kirchhoff's voltage law:

Eab = Ean + Enh O ~ . I )~ Ea " ~ E lm (8.1 )

Ehe = Lhl1 + e; (S.2)= e.; ~ Eell (8.2)

Eea = s. ; + En" (S.3)~ E e l l ~ E"" (S.3)

THREE-PHASE CIRCUITS 165

a

c b

(a) (b)

(e )

(d)

Figure 8,13a. Wye-eonnected stator windings of a 3-phase gen-

erator.b. Line-to-neutral voltages of the generator.c. Method to determine line voltage Fabd. Line voltages Eab, Ebc, and Eca are equal and dis-placed at 1200


33/49

!66 ELfTTRICAL MACIilNES AND TRANSFORMERS

Referring first to Eq. g.!. we draw phasor E "h ex-actly as the equation indicates:

The resulting phasor diagram shows that linevoltage E"h leads I:'"" by 30 (Fig. 8.l3c). Usingsimple trigonometry. and based upon the fact thatthe length of the line-to-neutral phasors is LLN' wehave the following:

length E ,. of phasor E"b =2 X Eu" cos 30EI. = 2 X EI.N ' < 1 3 / 2=< 1 3 E u .. ;

The line-to-line voltage (called line voltage) istherefore '>/3 times the line-to-neutral voltage:

(g.4)where

F . [ ~ effccti ve value of the line voltage [V]ELN = effective value of the line-to-neutral

voltage [VI.,,;3 = a constant [approximate value = 1.731

Due to the symmetry of a 3-phase system. we con-clude that the line voltage across any two generatorterminals is equal to '< 1 3 E LN. The truth of this canbe seen by referring to Fig. 8.13d. which shows allthree phasors: E"". ElK' and Lea' The phasors aredrawn according to Eqs. 8.1. 8.2. and 8.3. respec-tively. The line voltages are equal in magnitude andmutually displaced by 1200To further clarify these results. Fig. 8.14 shows

the voltages between the terminals of a 3-phasegenerator whose line-to-neutral voltage is 100 V .l'he line voltages are all equal to 100 '1 / 3. or173 Y . The voltages between lines a. b. c. consti-tute a :I-phase system. but the voltage betweenany two lines (a and b. band c. band n. etc.) isnevertheless an ordinary single-phase voltage.

Example 8-3A 3-phase 60 Hz generator. connected in wye, gen-erates a line (line-to-line) voltage of 23 900 Y .

Figure 8.14Voltages induced in a wye-connected generator.Calculatea. The line-to-neutral voltageb. The voltage induced in the individual windingsc. The time interval between the positive peak

voltage of phase A and the positive peak ofphase B

d. The peak value of the line voltageSolutiona. The line-to-neutral voltage is

ELN = E,/ '


34/49

and 8.12. In other words, the line voltage is ~ 3times the line-to-neutral voltage.Example 8-4 _The generator in Fig. 8.12 generates a line voltageof 865 y, and each load resistor has an impedanceof 50 n.Calculatea. The voltage across each resistorb. The current in each resistorc. The total power output of the generatorSolutiona. The voltage across each resistor is

t'LN =E, /,j 3 =865/,) 3=500Y

(8.4)

b . The current in each resistor is!= ELN /R = 500/50= lO A

All the line currents arc. therefore, equal to 10 A.c. Power absorbed by each resistor is

P = E IN ! = 500 X 10= 5000W

The power delivered by the generator to allthree resistors is

P = 3 X 5000 = 15 kW

B.10Delta connectionA3-phase load is said to he balanced when the linevoltages are equal and the line currents are equal.This corresponds to three identical impedancesconnected across the 3-phase line. a condition thatis usually encountered in 3-phase circuits.The three impedances may he connected in wye

(as we already have seen) or in delta (Fig. 8.15a).The line voltages arc produced by an external gen-erator (not shown).Let us determine the voltage and current rela-

tionships in such a delta connection,';' assuming aresistive load. The resistors are connected across

THREE-FHASf:- CIRCUITS 167

I.a -.

b Ib

I,Ie

(a)

(b)

--------....;

Figure 8.15a. Impedances connected in delta.b. Phasor relationships with a resistive load.the line; consequently, resistor currents 1[ , 1 2 ' and 1\are in phase with the respective line voltages E"I)'E bc and E e,," Furthermore, according to Kirchhoff'slaw. the line currents are given by

I " =[ - 1 1I" = 2 - I [I e = 1 - 1 2

,', The connection is", named hecuuxe it resembles the Greekletter tl.

(8.5)(8.6)(8.7)


35/49

168 r;UCTRIC;\L MACHINES AND TRANSFORMERS

Let the current in each branch of the delta-connectedload have an effective value I;. which corresponds tothe length of phasors I,. 12 , 11 , Furthermore. let theline currents have an effective value 1 1 . . which cor-responds to the length of phasors I" . lb' l.: Referringfirst to Eq. l{'5. we draw phasor I ., exactly as theequation indicates. The resulting phasor diagramshows that I" leads I, by 30 (Fig. 8.15b). Using sim-ple trigonometry. we can now write

Ir. ~ 2 X I; cos 30= :2 X I; -\ 3!2=j 3 I ;

The line current is therefore'; 3 times greater thanthe current in each branch of a delta-connected load:

(8.8)where

Ir . = effective value of the line current lA]I ; = effective value of the current in one

branch of a della-connected load IA J'- i 3 = a constant Iapproximate value = 1.73]

The reader can readily determine the magnitude andposition of phasors I" and I" and thereby observe thatthe three line currents arc equal and displaced by 120.

Table 8A summarizes the basic relationships be-tween the voltages and currents in wye-connectedand delta-connected loads. The relationships arevalid for any type of circuit element (resistor. ca-pacitor. inductor. motor winding. generator wind-ing. ctc.) as long as the elements in the three phasesare identical. In other words, the relationships inTable 8A apply to any balanced 3-phase load.

Example 8-5Three identical impedances are connected in deltaacross a 3-phase 550 V line (Fig. 8. 15c). If the linecurrent is lOA. calculate the following:a. The current ill each impedanceb. The value of each impedance InSolutiona. The current in each impedance is

a 10 A--550 V

b 10 A--I,

lO Ac

Figure 8.15cSee Example 8.5.

I; = IO! -V3 = 5 .7 7 Ab. The voltage across each impedance is 550 V.

Consequently.z =til;~50!5.77= 95 n

8.11 Power transmittedby a 3-phase line

The apparent power supplied by a single-phase linis equal to the product of the line voltage E times thline current I. The question now arises: What is thapparent power supplied by a 3-phase line havingline voltage E and a line current I:

If we refer to the wye-connected load of Fig8.16a, the apparent power suppl ied to each branch

E5 =-=XII. \3

The apparent power supplied to all three branchesis obviously three times as great." Consequently, thtotal apparent power is

ES = X I X 3 =\ 3 E I\3In .I-phase balanced circuits. we can add the apparent powers of the three phases because they have identical powerfactors. If the power factors are not identical. the apparentpowers cannot be added.


36/49

THREE-PHASr: CiRCUlI'S 16')

Wye connectionTABLE 8A VOLTAGE AND CURRENT RELATIONSHIPS IN 3-PHASE CIRCUITS

Delta connectionA

c [-Figure 8.16aImpedances connected in wye.o The current in each element is equal to the linecurrent 1 .

o The voltage across each element is equal to theline voltage E divided by > 1 3 .

o The voltages across the elements are 120" out ofphase.

o The currents in the elements are 120D out ofphase.

Ec [-

Figure 8.16bImpedances connected in deltao The current in each element is equal to the linecurrent I divided by -03.

o The voltage across each element is equal to theline voltage F :.

o The voltages across the elements arc 1200 out ofphase.

o The currents in the elements are 1200 out ofphase.

In the case of a delta-connected load (Fig. 8.16b j.the apparent power supplied to each branch is

IS =EX--/ V3which is the same as for a wye-connected load.Consequently. the total apparent power is also thesame.We therefore have

S =~3 E 1 (8.9)where

S = total apparent power deli vcred by a3-phase line [VAl

F = effective line voltage [VI1= effective line current IA I

. J 3 = a constant [approximate value = 1.731

8.12 Active, reactive, and apparentpower in 3-phase circuits

The relationship between acti ve power P . reactivepower Q . and apparent power S is the same for bal-anced 3-phase circuits as for single-phase circuits.We therefore have

(8. I ( ))and

cos e =P IS (8.11 )where

S =total 3-phasc apparent power IVA IP = total 3-phase active power IW IQ =total 3-phase reactive power Ivar I

cos fl =power factor of the 3-phase loadfl =phase angle between the line current

and the line-to-neutral voltage n


37/49

170 I:"L E CTRICAL M ACH INE S AND 7RANSF ORM E Rc'\'

Example 8-6 _A 3-phase motor, connected to a 440V line, drawsa line current of 5 A. If the power factor of themotor is 80 percent, calculate the following:a. The total apparent powerb. The total active powerc. The total reactive power absorbed by the machineSolutiona, The total apparent power is

S=-J 3 E I =-J 3 X 440 X 5=3811VA=:UI kVA

b. The total active power isp = S cos e = 3.81 X O.SO=3.05 kW

c. The total reactive power is

=2.2S kvar8.13 Solving 3-phase circuitsA /7([/({/1('('d 3-phasc load may be considered to becomposed of three identical single-phase loads.Consequently. the easiest way to solve such a cir-cuit is to consider only one phase. The followingexamples illustrate the method to be employed.Example 8-7 _Three identical resistors dissipating a total power of3000 Ware connected in wye across a 3-phase550 V line (Fig. 8.17).

Calculatea. The current in each lineb. The value of each resistorSolutiona. The power dissipated by each resistor is

p = 3000 W13 =1000 WThe voltage across the terminals of each resistor is

E = 550 V N 3 ~ 318 V

E

550 V3-phase

60 Hz line

Figure 8.17See Example 8-7.

The current in each resistor isI = P I E = IOOOW/318V=3.15A

The current in each line is also 3.15 A.b. The resistance of each element is

R = Ell = 31813. 15 = IOJ nExampk8-8 __In the circuit or Fig. 8. IS. calculate the following:a. The current in each lineb. The voltage across the inductor terminalsSolutiona. Each branch is composed of an inductive reactance

X I. =4 nin series with a resistance R =3 fl.Consequently, the impedance of each branch is

Z = V42 + 32 zr: 5 n (2.12The voltage across each branch is

ELN = EL / - - J 3 ir-: 440 V / - - J 3 = 254 VThe current in each circuit element is

1 =F.LN/Z = 254/5 = 50.S A(50.8 A is also the line current.)

440 V3phase line

Figure 8.18See Example 8"8.


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b. The voltage across each inductor ist:=XL =50.8 X 4=203.2 V

Example 8-9 _A 3-phase 550 V, 60 Hz line is connected to threeidentical capacitors connected in delta (Fig. 8.19).If the line current is 22 A, calculate the capacitanceo f each capacitor.SolutionThe current in each capacitor is

1-IL /-!3 - 22A/> /3 ~ 12.7 AVoltage across each capacitor =550 VCapacitive reactance Xc of each capacitor is

Xc = E L I I = 550112.7 = 43.3 nThe capacitance of each capacitor is

c=1/2TI{Xc= 1/(2TI X 60 X 43.3)=61 .3 f1 .F

(2.11 )

550 V3 phase60 Hz ~

~f---Figure 8.19See Example 8-9.

8.14 Industrial loadsIn most cases, we do not know whether a particular3-phase load is connected in delta or in wye. For ex-ample, 3-phase motors, generators, transformers,capacitors, and so on, often have only three externalterminals, and there is no way to tell how the inter-nal connections are made. Under these circum-stances, we simply assume that the connection is inwye, (A wye connection is slightly easier to handlethan a delta connection.)

THRE E -P H ASE CIRCUITS 171

In a wye connection the impedance per phase isunderstood to bc the line-to-neutral impedance. Thevoltage per phase is simply the line voltage dividedby - ! 3. Finally, the current per phase is equal to theline current.

The assumption of a wye connection can bemade not only for individual loads, but for entireload centers such as a factory containing motors,lamps, heaters, furnaces. and so forth. We simplyassume that the load center is connected in wye andproceed with the usual calculations.Example 8-10A manufacturing plant draws a total of 415 k VAfrom a 2400 V (line-to-line) 3-phase line (Fig.8.20a). If the plant power factor is 87.5 percent lag-ging, calculate the following:a. The impedance of the plant. per phaseb. The phase angle between the line-to-neutral

voltage and the line currentc. The complete phasor diagram for the plantSolutiona. We assume a wye connection composed of

three identical impedances 7. (Fig. 8.20b).The voltage per branch is

[; = 2400/>/ 3=1386 V

The current per branch isI=SI([;..J3)= 415 000/(2400 ~ 3)=IOOA

(8.9)

The impedance per branch isZ = Ell = 13861100=13.9 n

b. The phase angle f) between the line-to-neutralvoltage ( 1386 V) and the corresponding linecurrent (100 A) is given by

cos e =power factor =0.875e =29

(8.11 )


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172 EIECTRICM" MACHINES AND TRANSFORMERS

2400V I : 415kva3-phase line c

C>-------{f p z87.5%(a)Earl

b~---L~Z~_t-~ncO----l Z

(b)

(c)Figure 8.20a. Power input to a factory. See Example 8-10.b. Equivalent wye connection of the factory load.c. Phasor diagram of the voltages and currents.

The current in each phase lags 29 behind theline-to-neutral voltage.

c. The complete phasor diagram is shown in Fig.8.20c. In practice, we would show only onephase; for example. Eoll' I" , and the phase anglebetween them.

Example 8-11A 5000 hp wye-connected motor is connected to a4000 V (line-to-line), 3-phase, 60 Hz line (Fig. S.21).

1J ~ 93%Fp = 90%

4000 V3phase c=) P I

3594 HP

-H-1800 kvar

Figure 8.21Industrial motor and capacitor. See Example 8-11 .A delta-connected capacitor bank rated at ISOO kvaris also connected to the line. If the motor produces anoutput of 3594 hp at an efficiency of 93 percent anda power factor of 90 percent

single phase and three phase power

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