Page 49 - Solutions Manual to accompany Electric Machinery Fundamentals




V 13,800 0 V       36.87  362.3  j 1.9   A 0.38   14,330 1.9 V
P
There is a voltage drop of 14 V under these load conditions. Therefore the voltage regulation of the
transformer is

14,330 13,800
VR   100%  3.84%
13,800
The transformer copper losses and core losses are

P CU  I S 2 EQ,S   R 362.3 A 2   0.38   49.9 kW
  2  2

V
P core  P  14,330 V  91.5 kW
R C 2244 

Therefore the efficiency of this transformer at these conditions is

P 4000 kW
  OUT  100%   96.6%


P OUT  P  CU P core 4000 kW 49.9 kW 91.5 kW
2-10. A three-phase transformer bank is to handle 500 kVA and have a 34.5/11-kV voltage ratio. Find the
rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent
power) if the transformer bank is connected to (a) Y-Y, (b) Y-, (c) -Y, (d) -, (e) open-, (f) open-
Y—open-.

SOLUTION For the first four connections, the apparent power rating of each transformer is 1/3 of the total
apparent power rating of the three-phase transformer. For the open- and open-Y—open- connections,
the apparent power rating is a bit more complicated. The 500 kVA must be 86.6% of the total apparent
power rating of the two transformers, so 250 kVA must be 86.6% of the apparent power rating of a single
transformer. Therefore, the apparent power rating of each transformer must be 288 kVA.
The ratings for each transformer in the bank for each connection are given below:
Connection Primary Voltage Secondary Voltage Apparent Power Turns Ratio
Y-Y 19.9 kV 6.35 kV 167 kVA 2.50:1
Y- 19.9 kV 11.0 kV 167 kVA 1.44:1
-Y 34.5 kV 6.35 kV 167 kVA 4.33:1
- 34.5 kV 11.0 kV 167 kVA 2.50:1
open- 34.5 kV 11.0 kV 288 kVA 2.50:1
open-Y—open- 19.9 kV 11.0 kV 288 kVA 1.44:1

Note: The open-Y—open- answer assumes that the Y is on the high-voltage side; if the Y is on the low-
voltage side, the turns ratio would be 4.33:1, and the apparent power rating would be unchanged.

2-11. A 100-MVA 230/115-kV -Y three-phase power transformer has a per-unit resistance of 0.015 pu and a
per-unit reactance of 0.06 pu. The excitation branch elements are R  100 pu and X M  20 pu .
C
(a) If this transformer supplies a load of 80 MVA at 0.8 PF lagging, draw the phasor diagram of one
phase of the transformer.
(b) What is the voltage regulation of the transformer bank under these conditions?

(c) Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer.
Calculate all the transformer impedances referred to the low-voltage side.
(d) Determine the losses in the transformer and the efficiency of the transformer under the conditions of
part (b).

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