Page 61 - Solutions Manual to accompany Electric Machinery Fundamentals
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I I 27.57 A 2.757 A
line load
10 10
and the losses in the transmission line are
P loss line 2 R line I 2.757 A 2 30 228 W
(d) The efficiency of this power system is
P P 190 kW
out 100% out 100% 100% 99.9%
P in P out P loss 190 kW 0.228 kW
(e) Transmission losses have decreased by a factor of more than 80 when the transformers were added
to the system.
2-15. An autotransformer is used to connect a 12.6-kV distribution line to a 13.8-kV distribution line. It must
be capable of handling 2000 kVA. There are three phases, connected Y-Y with their neutrals solidly
grounded.
(a) What must the N / N SE turns ratio be to accomplish this connection?
C
(b) How much apparent power must the windings of each autotransformer handle?
(c) What is the power advantage of this autotransformer system?
(d) If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings
be?
SOLUTION
(a) The transformer is connected Y-Y, so the primary and secondary phase voltages are the line
voltages divided by 3 . The turns ratio of each autotransformer is given by
V H N N SE 13.8 kV/ 3
C
V L N C 12.6 kV/ 3
12.6 N C 12.6 N SE 13.8 N
C
12.6 N SE 1.2 N
C
Therefore, N C / N SE 12.6 /1.2 10.5 .
(b) The power advantage of this autotransformer is
S IO N C N SE N C 10.5N C 11.5
S W N C N C
so 1/11.5 of the power in each transformer goes through the windings. Since 1/3 of the total power is
associated with each phase, the windings in each autotransformer must handle
2000 kVA
S 63.5 kVA
W
10.5
3
(c) As determined in (b), the power advantage of this autotransformer system is 11.5.
(d) The voltages across each phase of the autotransformer are 13.8/ 3 = 7967 V and 12.6 / 3 = 7275
V. The voltage across the common winding ( N C ) is 7275 kV, and the voltage across the series winding
( N SE ) is 7967 kV – 7275 kV = 692 V. Therefore, a single phase of the autotransformer connected as an
ordinary transformer would be rated at 7275/692 V and 63.5 kVA.
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