Examples:
Example
1
Prepare a per-phase schematic of the system shown below in Figure B4.7 and show all
impedances in per-unit on a 100 MVA, 154 kV base in the transmission line circuit.
Necessary data for this problem are as follows
G1:
50 MVA, 13.8 kV, X = 0.15 per-unit
G2:
20 MVA, 14.4 kV, X = 0.15 per-unit
T1:
60 MVA, 13.2/161 kV, X = 0.10 per-unit
T2:
25 MVA, 13.2/161 kV, X = 0.10 per-unit
Load: 25 MVA, 0.80 pf lag
T1
T2
20  j80
G1
G2
10  j 40
10  j 40
Load
Solution
Base kV in the Transmission Line = 154 kV
Base kV in G1 and G2 = 154  13 .2  12 .63 kV
161
Note: Once the Base kV is specified in the transmission line circuit, the Base kV in all other
circuits is determined by the transformation ratio of the appropriate transformers. In this
example T1 and T2 have the same transformation ratio. Hence the Base kV in G1 and G2 are
equal. If the transformation ratios were not the same then the appropriate transformation
ratios should be used to determine the base voltage.
2
G1 : X  0.15 
100  13.8 

  0.3583 per  unit
50  12.63 
G2 : X  0.15 
100  14.4 

  0.9755 per  unit
20  12.63 
2
2
2
2
2
100  161 
100  13.2 
T1 : X = 0.10 


  0.10 
  0.18216 per  unit
60  154 
60  12.63 
100  161 
100  13.2 
T2 : X = 0.10 


  0.10 
  0.4372 per  unit
25  154 
25  12.63 
Base Impedance in Transmissi on Line Circuit =
Z T.Line 
(154x10 3 ) 2
 237.16
100x10 6
20  j80
 0.084  j 0.3373 per  unit
237.16
Base Impedance in Load Circuit =
(12.63x10 3 ) 2
 1.595
100x10 6
10  j 40
 6.269  j 25.075 per  unit
1.595
Load = 25(0.8 + j0.6) = 20 + j15 MVA
Z D. Line 
2
Ru 
Vload (100 x10 6 )
20 x10 6  (12.63 x10 3 ) 2
2
,
Xu 
Vload (100 x10 6 )
15 x10 6  (12.63 x10 3 ) 2
j 0 . 3583
G1
j 0 .18216
6.269
0 . 08
j 0 . 3373
j 25.075
j 25.075
j 0 .4372
6.269
Load
Ru
Xu
j 0 .9755
G2