Problem 2-23
The following figure shows a power system consisting of a three-phase 480-V, 60-Hz
generator supplying two loads through a transmission line with a pair of transformers at
either end.
a. Sketch the per-phase equivalent circuit of this power system.
b. With the switch open, find the real power P, reactive power Q, and apparent
power S supplied by the generator. What is the power factor of the generator?
c. With the switch closed, find the real power P reactive power Q, and apparent
power S supplied by the generator. What is the power factor of the generator?
d. What are the transmission losses (transformer plus transmission line losses) in
this system with the switch open? With the switch closed? What is the effect
of adding load 2 to the system?
Y
∆
∆
Y
G
480 V
T1
480/13,800 V
1000 kVA
R = 0.010 pu
X = 0.040 pu
T2
13,800/480 V
500 kVA
R = 0.020 pu
X = 0.085 pu
Load 1
ZL=0.5∠36.87oΩ
Y-connected
Load 2
ZL=-j0.833Ω
Y-connected
ZL = 1.5 + j10
Region 1
Sbase1=1000 kVA
VLbase1=480V
Region 2
Sbase2=1000 kVA
VL, base2= 13,800 V
Region 3
Sbase3=1000 kVA
VL, base3= 480 V
Solution: This problem can best be solved using the per-unit system of measurements.
The power system can be divided into three regions by the two transformers. If the perunit base quantities in Region 1 are chosen to be Sbase1 = 1000 kVA and VL,base1 = 480 V,
then the base quantities in Region 2 and 3 will be as shown above. The base impedance
of each region will be:
Z base1 =
3(Vφ1 )2
3(277 )2
=
= 0.238 Ω
Sbase1
1000 kVA
Z base 2 =
3(Vφ2 )2
3(7967 )2
=
= 190.4 Ω
Sbase 2
1000 kVA
Z base3 =
3(Vφ3 )2
3(277 )2
=
= 0.238 Ω
Sbase3
1000 kVA
a. To have the per-unit, per-phase equivalent circuit, we must convert each impedance
in the system to per unit on the base of the region in which it is located. The
impedance of transformer T1 is already in per unit to the proper base
R1, pu = 0.010
X1, pu = 0.040
The impedance of transformer T2 is already in per unit, but it is per-unit to the base of
transformer T2, so it must be converted to the base of the power system.
2
(R, X, Z)pu on base 2 = (R, X, Z)pu on base 1 (Vbase1 )2 (Sbase2 )
(Vbase2 ) (Sbase1 )
2
(7967 V ) (1000 kVA ) = 0.040
R 2, pu = 0.02
(7967 )2 (500 kVA )
(7967 V )2 (1000 kVA ) = 0.170
X 2, pu = 0.085
(7967 )2 (500 kVA )
The per unit impedance of the transmission line is
Zline, pu =
Zline
1.5 + j10
=
= 0.00788 + j 0.0525
Z base2
190.4
The per unit impedance of load 1 is
Zload1,pu =
Zload1 0.5∠36.87 o
=
= 1.681 + j1.261
Z base3
0.238
The per unit impedance of load 2 is
Zload2,pu =
- j 0.833
= − j 3.5
0.238
The per-unit, per-phase equivalent circuit is shown below
T1
Line
0.01+j0.04
0.0078+j0.052
T2
0.04+j17
L1
1.68+j
1.26
1∠0o
b. With the switch open, the equivalent impedance is
ZT = 0.010 + j0.040 + 0.00788 + j0.0525 + 0.040 + j0.170 + 1.681 + j1.261
= 2.312∠41.2o
The current is
I=
1∠0o
2.312∠41.2o
= 0.4325∠ − 41.2o A
The load voltage is
(
)
Vload, pu = I Zload = 0.4325∠ - 41.2o (1.681 + j1.261) = 0.909∠ − 4.3o
The actual load voltage is 0.909×480=436 V.
The power supplied to the load is
Pload, pu = I 2 R load = (0.4325)2 (1.681) = 0.314
Pload == Pload, pu Sbase = (0.314)(1000 kVA ) = 314 kW
The power supplied by the generator
PG, pu = VI cos θ = (1)(0.4325) cos 41.2o  = 0.325


Q G, pu = VIsinθ = (1)(0.4325) sin 41.2o  = 0.285


SG, pu = VI = (1)(0.4325) = 0.4325
Actual values are: PG = 325 kW; QG = 285 kVAR; SG = 432.5 kVA
L2
-j3.5
The power factor is: cos 41.2o = 0.752 lagging
c. With the switch closed, the equivalent circuit is
ZT = 0.010 + j0.040 + 0.00788 + j0.0525 + 0.040 + j0.170 +
(1.681 + j1.26)(− j3.5) = 2.698∠5.6o
1.681 + j1.261 − j 3.5
The current is
I=
1∠0o
2.698∠5.6o
= 0.371∠ − 5.6o
The load voltage is
Vload,pu = I Zload =  0.371∠ - 5.6o (2.627 − j 0.011) = 0.975∠ − 5.6o


Vload = (0.975)(480 ) = 468 V
The power supplied to the two loads is
Pload, pu = I 2 Rload = (0.371)2 (2.672) = 0.361
Pload = (0.361)(1000 kVA ) = 361 kW
The power supplied by the generator
PG, pu = VIcosθ = (1)(0.371) cos 5.6o = 0.369
QG,pu == VI sin θ = (1)(0.371) sin 5.6o = 0.0369
SG,pu = VI = 0.371
Multiply each value by 1000 kVA to get actual values: PG = 369 kW; QG = 36 kVAR; SG
= 371 kVA. The power factor of the generator is cos 5.6o = 0.995 lagging.
d. The transmission losses with the switch open are:
Pline,pu = (0.4325)2 (0.00788) = 0.00147
Pline = (0.00147 )(1000) = 1.47 kW
The transmission losses with the switch closed are
Pline,pu = (0.371)2 (0.00788) = 0.00108
Pline = (0.00108)(1000) = 1.08 kW
Load 2 improved the power factor of the system.