Page 115 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 115
SOLUTION If the load is suddenly removed, V = E A . The internal generated voltage E A at full load was
2814 V, so V = 2814 V after the load is removed. Since the generator is Y-connected,
V 3V 4874 V when the load is removed.
T
4-20. What are the electrical losses in this generator at rated conditions?
SOLUTION The current flow at rated conditions is I 90.2 A , so the electrical losses are
A
P 3I 2 R 3 90.2 A 2 0.7 17.1 kW
CU A A
4-21. If this machine is operating at rated conditions, what input torque must be applied to the shaft of this
generator? Express your answer both in newton-meters and in pound-feet.
SOLUTION To get the applied torque, we must know the input power. The input power to this generator is
equal to the output power plus losses. The rated output power and the losses are
P 1 MVA 0.9 900 kW
OUT
P 3I 2 R 3 90.2 A 2 0.7 17.1 kW
CU A A
P 10 kW
F&W
P core 8 kW
P (assumed 0)
stray
P P P P P P 935.1 kW
IN OUT CU F&W core stray
Therefore, the applied torque is
P 935.1 kW
IN 7441 N m
APP
m 1200 r/min 2 rad 1 min
1 r 60 s
7.04 P 7.04 935.1 kW
or APP 5486 lb ft
n m 1200 r/min
4-22. What is the torque angle of this generator at rated conditions?
SOLUTION From the calculations in Problem 4-17, 26.3 .
4-23. Assume that the generator field current is adjusted to supply 3200 V under rated conditions. What is the
static stability limit of this generator? (Note: You may ignore R A to make this calculation easier.) How
close is the full-load condition of this generator to the static stability limit?
SOLUTION From Problem 4-17, the phase voltage of this generator at rated conditions is V 1850 0
and the internal generated voltage at rated conditions is E 2814 26.3 V . Therefore, the static
A
stability limit is
3 VE 3 1850 V 2814 V
P A 995 kW
MAX
X S 15.7
The maximum possible power of this generator is about twice the full-load rated power.
109