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Unformatted text preview: CHAPTER 12  THREEPHASE CIRCUITS List of topics for this chapter :
Balanced Three—Phase Voltages
Balanced Wye—Wye Connection
Balanced Wye—Delta Connection
Balanced DeltaDelta Connection
Balanced Delta—Wye Connection
Power in a Balanced System
Unbalanced ThreePhase Systems
PSpice for ThreePhase Circuits
Applications Mm BALANCED THREE—PHASE VOLTAGES Problem 12.1 Given Van =1104d), Vab =
voltages, line voltages, and the phase sequence. Vab 40° , and (I) > 0° , determine all the phase The easiest way to completely solve for all of the elements of the balanced 3— phase voltages 1s to
do it graphically. All balanced 3 p—hase voltages generate the same shape. We start off by
sketching Vab, which has to be a horizontal line pointing to the right. Next we know that the value
of Van has its head connected to “a” either coming down (indicating its phase angle is negative)
or rising up from below (indicating that its phase angle is positive. . It is now easy to determine all of the values and the rotation. 203 Clearly the rotation is a_cb, or a negative phase sequence. Because this is a triangle with equal sides, the interior angles are 60° and the line—toneutral
voltages bisect the angle. Thus Van has to be at 30°. This then leads to the following; Van = 110430° V, Vbn = 1104150° V, Von = 1104—90°V and IVabl = 1106 = 190.52
therefore Vab = 190.5240° V, th = 190.52.4120° V, Vca = 190524—120o V
Problem 12.2 Is it possible to generate the effect of a balanced three—phase deltaconnected source with only two voltage sources? If so, how? Let us start with three sources arranged in a delta as shown below. 2004120° 20010° 2004120° Looking at this ﬁgure, we can see that KVL applies and going around the loop leads to a sum of
zero volts. Now, if we remove any single source, KVL still states that the voltage across the open
part of the circuit still is equal to the same voltage and phase of the source that was removed.
Thus, ou can enerate a erfect deltaconnected volta 9 source confi uration with 'ust two
sources. This will be expanded on later when we actually look at what happens to power
distribution when you have two sources. (Clearly the two remaining sources must make up for the
power the third source would have supplied.) Also, you will see later how this leads to being able
to measure power delivered in a 3phase system using only 2 wattmeters. Problem 12.3 Is it possible to generate the effect of a balanced three—phase wye—connected
source with only two voltage sources? Why or why not? Again, we can look at the sources arranged in a wye and see what would happen. If we were to
measure the lineto—line voltages, we would see a balanced 3—phase voltage system. Removing
any one of the sources actually takes away two linetoline voltages. Clearly, unlike the delta conﬁguration, you cannot deliver 3—phase voltage without 3 sources in a v_v1e conﬁguration. 204 2004120° 20040° 2004120°
+ BALANCED WYEWYE CONNECTION Problem 12.4 [12. 7] Obtain the line currents in the threephase circuit of Figure 12.1. . a la A 44040° 4401120° Figure 12.1 This is a balanced YY system. . 44040°VO IZY=6j8§2 205 Using the per—phase circuit shown above, 440.400
I = . ~ “44.453 13° A
a 6— J8 __ Ib =I 1120°: 44466.87°A
I ~14120° 444173.13°A WW BALANCED WYE—DELTA CONNECTION Problem 125 [12.15] In a wyedelta threephase circuit, the source is a balanced,
positive phase sequence with Van = 12010° V . It feeds a balanced load with Z A = 9 + j12 £2 per phase through a balanced line with ZL : 1+ 305 £2 per phase. Calculate the phase voltages
and currents in the load. Convert the Aconnected load to a Y—connected load and use perphase analysis. ZL Ia

.9 In
ZA .
ZY =“"‘=3+_]4
3
V 12040°
I =—“‘——=——.———,=19.9314— 48.37° a 2 +2 (3+j4)+(1+j0.5) But Ia =IAJBJ3‘43oo
I _19.9314—.48.37°
AB ” Jig—30° 13c = 11.514138.4°1A
ICA = 115141015? A = 11.511  18.37° A VAB = IAB ZA = (11.514—18.37°)(15453.l3°) 206 VAB = 172.6234.76° V
. VBC = 172.6485.24° V
VCA = 172.64154.8°V W BALANCED DELTA—DELTA CONNECTION Problem 12.6 [12.1 7] For the AA circuit of Figure 12.2, calculate the phase and line
currents. a A 17340° V
30 Q . 1734120° V j10 g2 1734—120° V Figure 12.2 ZA 2 30+j10 = 31.62418.43° The phase currents are I —y—*‘£———1£4)°——547z 1843°A
AB ‘ zA " 316248.430 ‘_'__'_'__ IBC = 1A3; 120° = 5.474 138.43° A
ICA = IAlezo° = 5.474101.57° A The line currents are . 1 =IAB—ICA=IABJ§z30° a 207 Ia = 5.47854  48.43° = 9.4741  48.43° A Ib = 1,4 — 120° = 9.4744  l68.43° A
Ic : 134120° = 9.474471.57° A , __ BALANCED DELTAWYE CONNECTION Problem 12.7 [12.23] In a balanced threephase AY circuit, the source is connected in
the positive phase sequence, with Vab : 220420° V and ZY = 20 + j15 Q. Find the line currents. Vab430° 2204—10° 1“ = JEZY = J3(20+j15) 121 : 508144687o A 1b = 1,4 420°: 5.0814 166.87° A
10 = I,4120° = 5.081473.13° A POWER IN A BALANCED SYSTEM Problem 12,8 Given a deltaconnected source with linetoline voltages of 100 volts—rms
and a deltaconnected load of 10ohm resistors, calculate the power absorbed by each resistor and the power delivered by each source. 10040° 10 Q
1004120° 1001120° As we can see from the circuit, there is 100 volts across each 100ohm resistor. 208 . P4, = (100)2/10 = 1000 watts (per phase) So a total of 3000 watts is delivered to the load and supplied by the source, with each individual
voltage source supplying 1000 watts. Problem 12.9 Given V4, = 100 volts for the load in Figure 12.3, determine R1 and R2 so that the sources see a balanced load. Find I1 and I 2. Figure 12.3 . We can combine the two circuits as follows since each corresponding resistor is in parallel with
each other. 2020 10R1
10+R 30R,
30+R, Obviously, the 20120 is equal to 10 ohms. This means that the others need to combine into 10
ohms if possible. 10R1
10+R1 is= 10 ifand only ifR1 = 99 30R
—2=10 ———> 30R2=300+10R2
30+R2 or R2 = 300/(30—10) = 159 . Clearly, since the load appears balanced to the sources, 11 = 0. 209 12 can be found by summing all of the currents ﬂowing into it from the second circuit (see Figure .
12.3). It is easier if we assume that V20 = VAN = 10040°, V10 = VBN = 1004120", and V30 = VCN = 1004120". 12 =£11+ VBN + vCN 210040 +1004120 +1004—120 20 10 i 30 20 10 3O = 5 + 101120° +3.333z—1200 = 5 + (—5 + J8.66) +(~1.667 — J2.887) = —l.667 + J5.773 = 6.0091106.1° A W. UNBALANCED THREEPHASE SYSTEMS Problem 12.10 Given a balanced wyeconnected source of 200 volts each, calculate the
power delivered to each of the resistors in the unbalanced wye—connected load in Figure 12.4. 109 209 SQ Figure 12.4 We start by labeling the circuit and arbitrarily placing the sources. c
C
a
20 Q
n — Source
neutral
B Let the voltage from a to the source neutral, n, be Van = 20040°
Likewise, VbI1 = 2001120" and Vcn = 2001—120° . 210 . This can easily be solved if we look at it as a simple circuits problem with only one unknown
node voltage, VN (measured with respect to n). We now can write a nodal equation at node N. VN—‘Van+VN_Vbn+VN—Vcn =0
10 5 20 Solving for VN we get, VN = 599+E4120°+§9b120° 7 7 7
= 57.14 + (—14.285 ~j24.74) + (—57.14 +j98.98)
= —14.28 +j74.24 = 75.641009o Now that we have VN, we can either ﬁnd the currents through the resistors or voltages across
them and then calculate the power. Solving for the voltageswe get, vAN = Vaﬁ—VN = 200—(—14.28+j74.24) = 214.28 —j74.24 = 226.84—19.11° [I VBN = Vbn — VN —100 +j173.21 — (—14.28 +j74.24) = ~85.72 +j98.97 . = l30.93£l30.9° VCN = Vcn—VN = —100—j173.21—(—14.28 +j74.24) = ~85.72—j247.45
= 261.884—109.11° Now to calculate the power. IVANI2 (226.8)2
P10 = ————— : ~——— 2 5.144 kwatts 10 10 : IVBNIZ : (130.93)2 P5 = 3.429 kwatts
5 5
2
V 2
P20 =l C”! = —(261‘88) = 3.429 kwatts 20 20 211 Problem 12.11 Given a balanced wyeconnected source of 200 volts each, calculate the
power delivered to each of the resistors in the unbalanced wyeconnected load in Figure 12.5.
Explain the difference between this problem and Problem 12.10. 109 209 SQ Figure 12.5 First of all, this problem is a lot easier to solve since the neutral connection means that each phase
of the wyeconnected source is across only its corresponding resistor, unlike the previous
problem. In fact, there are no unknown node voltages in this circuit. Stated mathematically, VAN = Van, VBN = Vbn, and VCN = Vcn
Thus, P10 = (200)2/10 =2 kwatts
P5 = (200)2/5 = 4 kwatts P20 = (200)2/20 = 1 kwatt om PSPICE FOR THREEPHASE CIRCUITS Problem 12.12 Solve Problem 12.8 using PSpice. The circuit was set up as the following
schematic, calculating the current through each resistor and the voltage across each resistor.
Since this is a resistance circuit, there is no phase angle between the voltage and the current. So
all we need to do to calculate power is to multiply the magnitudes of the currents and voltages.
Since it is a balanced circuit, each source supplies the same power equal to the power being
absorbed by each resistor. After storing the circuit, we run Simulate and then examined the output ﬁle. The results yield the following: P10 = 100x10 = 1 kwatt, for a total of 3 kwatts (both delivered and absorbed) It is interesting to note that a small resistor needed to be placed in the voltage loop to allow
PSpice to run properly. 212 PHASE=yesu IPRINT
ACMAG=1 00 _ _
p _
AcpHAse=o _ y = Sﬁéiyéis AC=yes PHASE=yes = MAG=yes PHASE=yes AC=VES
Ac=yes
PRWT Problem 12.13 Solve Problem 12 10 using PSpice. We develop the following schematic in
PSpice. Note, we are making more measurements than necessary to solve this problem. However these extra measurements will allow us to check the results of the voltage calculations
in 12.10. IPRINT MAG=V99 Ac=yes
g PHASEzyes  MAG=yes
’—"' PHASE=yes MAGzyes
‘PR'NT PHASE=yes R1 After saving the schematic, we run Simulate and open the output ﬁle. The results allow us to determine the voltages and currents in the circuit. We then can calculate the power delivered to
each resistor. V10 = 22681—1911" and I10 = 22684—1911o Therefore P10 = (226.8)(22.68) = 5.144 kwatts (please note that cose
. = 1 since these are resistors) 213 V5 = 130.91130.9° and 15 = 26.194130.9°
Therefore P20 = (130.9)(26.l9) = 3.428 kwatts
V20 = 261.94—109.1° and 120 = 13.094—109.1° Therefore P20 = (261 .9)(13.09) = 3.428 kwatts
Going back to the solution obtained by hand, we see that we have virtually the same values of branch voltages. PSpice also gave us the value of VNn = 79.594100.9°, which agrees. This
shows the value of using PSpice to check results obtained by hand. Problem 12.14 [12.49] Given the circuit in Figure 12.6, use PSpice to determine currents I21A and voltage VBN. 24040‘) V 4 o j3 Q A 10 Q j15 Q 109 leQ The schematic is shown below. ACMAG=24UV
ACPHASE=U R1 3H
ACMAG=24UV
ACPHASE=~1 20
MM
4
3H
ACMAG=24UV
CPHASE=120
m
4
V3 3H 214 Pseudocomponents IPRINT and PRINT are inserted to measure 13A and VBN. In the AC Sweep
box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is
. simulated, we get an output ﬁle which includes FREQ VM(2) VP(2)
1.592 E~01 2.308 E+02 —1.334 E+02 FREQ IM(V__ PRINTZ) IP(V_PRINT 2)
1.592 E~01 1.115 E+01 3.699 E+01
from’which V
I21A = 11.15437° A , VBN = 230.81  133.4° V
APPLICATIONS Problem 12.15 [12.59] For the circuit displayed in Figure 12.7, ﬁnd the wattmeter . readings. z=10+j30§2 Figure 12.7 215 Consider the circuit as shown below. 2401—60° V 2404—120° V Z =10+j30 = 31.62171.57° 240460° Ia =m=7.59Z131.57 ‘
2404120° Ib = 7.594 191.57° = 3152471570 Ic Z+ 2404 — 60——2404 —120° = 0
240 c = m = 7H59410843 I I1 = Ia —Ic =13.1464101.57°
I2 = II) +Ic = 13.1461138.43° P1 = Re[V11; ]= Re[ (2404  60°)(13.1464101.57°) ] = 2360 w P2 = Re[v2 1;]: Re[(240z120°)(13.1464 138.43°) ] = 632.8 w P1 = 2360 w  P2 =  632.8 w Problem\12.16 Given two threephase sources with linetoline voltages of 110 volts nns, is
it possible to use standard 100 watt light bulb(s) to see if the sources are in phase with each other?
Is it possible to see if they have the same rotation? More than one light bulb can be used. Actually, only one light bulb is necessary. We need to make two assumptions before we start. The ﬁrst is that both sources have exactly the same frequency. This is possible if both sources are
driven by the same mechanical system. The second is that they share the same ground. 216 Consider the following representation of the two sources. Now we connect the light to terminal a. Then we connect it to A, B, and C of the other source. If
the sources are out of phase with each other, the light will light at least dimly when connected to
each of the terminals. If the sources are reasonably close to being in phase with each other, there
will be one terminal where the light will not light. Let us assume that it is the connection between a and A. If we now connect the light to b and B and the light does not light, then the two sources have the
same rotation. It the light lights, then they have opposite rotations. Once the rotations have been
determined, the two sources can be connected in parallel. Problem 12.17 Given an unbalanceddelta connected load, show how you can use two
wattrneters to determine the power delivered to the load. Calculate the voltages and currents each meter sees and determine the power delivered to the load. The circuit is shown below. 10040° 20 Q
1004120° 1001120° 217 This can be a difficult problem to solve unless we remember what we found as an answer to
Problem 12.2. It is possible to represent a balanced delta source with only two sources. That tells us how to hook up the meters. We can solve this USing either hand calculations or PSpice. Let us use PSpice since we will need
to make calculations twice in order to check our results, although the hand calculations are not that difﬁcult. First, we design the schematic and save it. We then simulate and get the results shown here.
P10 : lOOXlO : lkwatt 100x20 = 2 kwatts [I P5 100x5 = 500 watts [I P20 PTotal = 1000 + 2000 + 500 = 3.5 kwatts AC: 88
MAG=xVes @—
PHASE=yes
_ IPRINT IPRINT
ﬁgyﬁAGSELUUD E MAGWES PHASE=yes Ac: es
MAG=¥es ACPHASE=—1QIV2 PHASE=yE§SU
CMAG=1DU
ACPHASE=120 PHASE=yes MAG=yes PHASE=yes AC=YES
AC=yes
IPRINT Now look at the following circuit. If we assume that V3 is not in the circuit for the moment, we
see that V1 and V2 would supply all the power. So we place an IPRINT in series with V1 and an
IPRINT in series with V2. Then all we need to do is measure the Voltages V1 and V2. (Actually
in practice there are line losses and these voltages would be measured at the load.) 218 IPRINT ea PHASE=yes AC=yes
MAG=yes
ACMAG=1DU PHASE=yes
ACPHASE=U PHASE=yes AC=yes
IPRINT Ac=yes
PHASE=yes
MAG=yes After we get the results from PSpice, we can make the following calculations. . P1 = 100x13.23cos(~120°+160.9°) = lkwatt 100x26.46cos(120° — 139.1°) = 2.5 kwatts ll P2 PTotal = P1 + P2 = 3.5 kwatts (the answer checks”) 219 ...
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