Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

(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 no￾load 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=