EENG 457 Power System Analysis I
HOMEWORK 2
1. A single-phase 50-kVA, 2400/240 V, 50 Hz transformer is used as a step-down transformer at the
load end of a line whose series impedance is Zline = 1.0 + j 2.0  . The equivalent series impedance of
the transformer is 1.2 + j 2.4  referred to the high-voltage (primary) side. The transformer is delivering
rated power at 0.8 power factor lagging and at rated secondary voltage. Determine
(a) the voltage at the primary side of the transformer.
(b) the voltage at the source end of the line.
(c) the real and reactive power delivered by the source.
2. Consider the single-phase transformer circuit shown below. Transformer ratings and reactances are:
Transformer A
Transformer B
: 30 kVA , 240 / 480 V, X = 0.10 p.u.
: 20 kVA , 460 / 115 V, X = 0.10 p.u.
The transmission line can be represented as a series impedance Zline =j 2 . The load impedance is
Zload = 0.9 + j 0.2 . Choose a base of 30 kVA, 240 V in the voltage source ( Vs ) circuit.
(a) Draw the impedance diagram for this circuit, with all impedances in per unit.
(b) The source voltage is given as VS =2200 V. Find the load current both in per-unit and
in amperes.
(c) Find the source current (Is) in per-unit and in amperes.
Is
+
Vs
A
transmission line
(Zline = j 2 Ω)
IL
B
Zload =0.9+j0.2 Ω
3-) A 500 MVA 200/20 kV three-phase Y-  transformer has an equivalent impedance of 1.2 + j 4  per
phase referred to the high-voltage side (Y). The transformer is supplying a three-phase load of 450
MVA, 0.8 power factor lagging at a terminal voltage of 20 kV on the low voltage side (  ). The primary is
supplied from a transmission line with an impedance of 0.8 + j 1.2  per phase. Determine
(a) the line-to-line voltage at the high-voltage terminals of the transformer.
(b) the total complex power delivered by the source at the sending end of the transmission line.
4-) A 50 MVA, 24-kV three-phase 50-Hz synchronous generator has a synchronous reactance of 8 
per phase and negligible armature resistance. The generator is delivering rated power at 0.8 power
factor lagging at rated terminal voltage to an infinite bus.
(a) Calculate the internal voltage Ef per phase and the power angle  .
(b) With the internal voltage held constant at the value in part (a), the shaft torque is reduced
until the generator delivers 28 MW. Find the armature current and the power factor at the
infinite bus.
5-) A three-phase synchronous generator, rated 18 kV and 150 MVA, has synchronous reactance of 1.5
per-unit. It is operated on an infinite bus of voltage 16 kV and delivers 120 MVA at 0.8 power-factor
lagging.
(a) Calculate the internal voltage Ef, the power angle  and the armature current.
(b) If the field current of the machine is reduced by 10%, while the prime mover power is
maintained constant, find the new value of  and the new power factor at the infinite bus.
SOLUTION
1. (a) Calculations in per-unit with base:
Sbase  50 kVA, Vbase  2400 V on high voltage side of transformer
(H )
 Z base

2
Vbase
24002

 115.2 
Sbase 50 103
 Line and transformer impedances in per-unit:
Z line  0.0087  j 0.0174 pu.,
Z tr  0.0104  j 0.0208 pu.
Current base on primary side: I base 
Sbase 50  103

 20.833 A
2400
Vbase
Zline
+
Ztr
I
V1
Vs
Vload
Load
(a)
Vload  1.000 pu., I  1.0  cos 1 (0.8)  1.0  36.870 pu.
 V1  Vload  Ztr I  1.0  (0.0104  j 0.0208) 1.0  36.870  1.0208  j 0.0104 pu.
 1.02090.580 pu.

V1  2400 1.0209  2450.2 V
(b)
Vs  V1  Zline I  1.0209  0.580  (0.0087  j0.0174) 1.0  36.87 0
 1.0382  j 0.0191  1.03841.0540 pu.

Vs  2400 1.0384  2492.2 V
(c) The complex power supplied by the source:
S  Vs I *  1.03841.0540  1.036.870  1.038437.9240 pu.  0.8191  j 0.6382 pu.
 Ps  50  0.8191  40.956 kW
2-)
Qs  50  0.6382  31.91 kVar
(a)
Base voltages,
Vb, B  480 V
Base impedances,
Zb, B 
, Vb,C  120 V
4802
1202

7.68

,
Z

 0.48 
b
,
C
30 103
30 103
Impedances in per-unit
0.9  j 0.2
Z load 
 1.875  j 0.4167
0.48
p.u.,
ZTL 
2
XT ,A
 30   240 
 0.1    
  0.1
 30   240 
j2
 j 0.2604
7.68
p.u
2
p.u,
X T ,B
 30   460 
 0.1    
  0.1378
 20   480 
Impedance diagram,
j0.1
j0.2604
j0.1378
+
Vs
1.875+j0.4167
p.u
(b)
Vs  (220 / 240)0 p.u.  0.91670 p.u.
IL 
0.9167
Z total
Z total  1.875  j 0.9149
I L  0.4394 p.u.,
I base (C ) 
p.u. 
I L  0.3949  j 0.1927 p.u.
Sb
30  103

 250 A  I L  0.4394  250  109.85 A
Vb (C )
120
(c)
Ibase ( A) 
Sb
30 103

 62.5 A  I L  0.4394  62.5  27.46 A
Vb ( A)
480
3-) (a) Calculations in per-unit with base:
Sbase  500 MVA, Vb( H )  200 kV on high-voltage side of transformer,
Vb( L )  20 kV on low-voltage side of transformer
(H )
 Z base

(Vb( H ) ) 2 2002

 80 ,
Sbase
500
(L)
Z base

(Vb( L ) ) 2 202

 0.80 
Sbase
500
 Line and transformer impedances in per-unit:
Z line  0.01  j 0.015 pu., Z tr  0.015  j 0.05 pu.
Load impedance:
Z load 
2
Vload
202

 0.888 
Sload
450

Z load  0.888  36.87   0.711  j 0.533 
 Z load  1.111  36.87 pu.
Zline
Impedance diagram:
+
I
VH
Vs
Let Vload  1.00
pu. 
Ztr
I
Vload
Vload
 0.9  36.87
Zload
 VH  Vload  Ztr I  1.0378  j 0.0279 pu.

Load
pu.
VH  207.64 kV line-to-line.
(b)
Vs  VH  Z line I  1.0531  j 0.0333 pu.
 S source  Vs I *  0.7402  j 0.5927
pu.
S source  370.12  j 296.33 MVA
4-)
Let Vt 
24
3
0 kV  13.860 kV line-to-neutral
Ia 
50
 1.203 kA  I a  1.203  36.87 kA
3  24
 E f  Vt  jX s I a  13.86  j8  1.203  36.87 kV  19.634  j 7.698 kV line-to-neutral
Generator current,
E f  36.53 kV line-to-line,
  tan 1 (7.698 /19.634)  21.4
(b)
P
E f . Vt
Xs
sin( ) 
28
MW,
3
E f  E f   Vt  jI a X s
, Let I a  I a 
E f cos( )  13.86  8 I a cos( )
E f sin( )  8 I a sin( )
   39.522

28  8


E f  21.09 kV from part (a)    sin 1 
  14.8
 3  21.09  13.86 
Ia 
 tan( ) 
21.09sin( )
 0.825
21.09 cos( )  13.86
21.09sin( )
 1.0582 kA
8sin( )
5-)
Let Sbase  150 MVA,
Vbase  18 kV,
120
 0.8 pu.  Vt I a 
150
Real power: P  0.64 pu.
SG 
Ia 

Vt 
16
0 pu.  0.88880 pu.
18
0.8
 0.9 pu., I a  0.9  36.87 pu
0.8888
(a)
E f  Vt  jX s I a  0.8888  j1.5  0.9  36.87  1.699  j1.08 pu.  2.01332.45 pu.
   32.45 .
(b)
E f  0.9  2.013  1.8117 pu.
From part(a), prime mover power  P  0.64 pu. 
Vt E f
Xs
sin( )
0.64  1.5
 0.5963    36.6  E f  1.811736.6 pu.  1.454  j1.08
0.8888  1.8117
E f  Vt 0.454  j1.08
 Ia 

 0.72  j 0.3027 pu.  cos( )  0.922
jX s
j1.5
 sin( ) 