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Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
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Mô tả chi tiết
(a) This generator is Y-connected, so L AI I= . At rated conditions, the line and phase current
in this generator is
I I 1000 kVA 251 A= = P = = at an angle of –36.87° A L
(3 3 2300 )V L
V
= = . The internal generated voltage of the machineThe phase voltage of this
machine is ⎞
T /V V 3 1328 V
is
= +E V +I I
AA ⎞ A S A
R jX
1328 0 (0.15 )( 251 36.87 A) (1.1 )( 251 36.87 A) EA = ° + & ° j+ & °
EA 1537 7.4 V= °
The input power to this generator is equal to the output power plus losses. The rated output power is
( )( ) OUT P 1000 kVA 0.8 800 kW= =
2 (
)= =
2
( )& =
CU 3 A A
P I 3R 251 A 0.15 28.4 kW
F&P = W 24 kW
coP re = 18 kW
stP ray = (assumed 0)
= + + + + = IN OUTP P CUPF&WPcoreP strayP 870.4 kW
η OUTP
100%
800 kW 100% 91.9%
= ⋅ = ⋅ =
INP 870.4 kW
(b) If the generator is loaded to rated kVA with lagging loads, the phase voltage is
φ
V
=1328 0 ° V and
the internal generated voltage is EA 1537 7.4 V= ° . Therefore, the phase voltage at noload would be
V⎞ 1537 0 V= ° . The voltage regulation would be:
1537 1328
RV 100% 15.7%
= ⋅ =
1328
(c) If the generator is loaded to rated kVA with leading loads, the phase voltage is
φ
V
the internal generated voltage is
=1328 0 ° V and
= +E V +I I
AA ⎞ A S A
R jX
1328 0 ( )0. (15 251 36.87 A) (1.1 )( 251 36.87 A)
EA = ° + & ° j+ & °
EA 1217 11.5=