Example: Three Phase TrnasfomersWe have a small power system as shown in the figure:
generator, two transformers, distribution line, and load. Ratings on the equipment are shown. The
generator voltage is 13.8kV.
a. Find the load voltage without the transformers.
b. Find the load voltage WITH the tranformers.
c. Find the generator load real and reactive power.
6
MVA  10  V A
j  1
Vg  13.8 kV
VT1L  13.8 kV
SgR  50 MVA
VT1H  VT1L  10 3  239.023 kV
VT2H  VT1H  239.023 kV
Xg  2.7 Ω
VgLN 
Zload  ( 4  j  1 )  Ω
Vg
3
VT2L  VT2H
VgLN  7.967 kV
1
Xline  100  Ω
XT1  0.07 Ω
 13.8 kV
10 3
XT2  0.07 Ω
Without the transformers, the line and load and generator are in series. Using voltage division on a
line-to-neutral per phase circuit.
Vload  VgLN
Z
Zload
Vload  ( 91.239  311.929i) V
load  j  Xline

Vload  324.998 V
Voltage Regulation
Vout_noload  VgLN
Vregulation 
Vout_noload

arg Vload  73.696 deg
 Vload
Vload
3
Vregulation  2.352  10 %
This is not good. Let's include the transformers. We must make a reflection of the line impedance.
Let's reflect everything to the generator side. With identical transformers, the load impedance
reflects across both transformers, cancelling out the effect.
VT1H
NT1 
VT1L
NT1  17.321
Xline
Xline_eq 
NT1
NT2  NT1
Again, using voltage division,



 load  j  XT2  j  Xline_eq  j  XT1 
Zload
Vload  VgLN 
Z
Vload  ( 7.662  0.83i) kV

Vload  7.706kV
Vregulation 

arg Vload  6.184deg
VgLN  Vload
Vload
Vregulation  3.386%
2
Xline_eq  0.333Ω
Find the load power.
Sload
 Vload 
 3 
2
Sload  ( 41.922  10.481i ) MVA

Zload
Find the generator power.
Vload
Iload 
Zload


3
Iload  1.754  10  646.022i A
NT1
Igen  Iload
NT2

3

Igen  1.754  10  646.022i A
Calculate the combined real and reactive power.

Sgen  3  VgLN Igen
Sgen  ( 41.922  15.441i ) MVA