Question

A 750 kw, 11 kv, 3-phase, star connected synchronous motor has a synchronous reactance of 35 /phase and negligible resistance. _________ kv is the excitation emf per phase when the motor is operating on full-load at 0.8 pf leading. Its efficiency under this condition is 93%

Answer 1

Explanation:

o/p=750 kw.

n=93%

I/p=750/0.93=

p=√VL IL cosfi

Answer 2

Answer:

|E| = 7.981 KV

Explanation:

Efficiency=0.93

Efficiency=0.93Input power=750/0.93=806.45 KW

I = 806.45/√3*11*0.75

Implies, I=56.47 A at 0.75 p. f. leading,

V= 11000/√3 = 6351 V

Therefore,

→ → →

E= V - jI*X

Implies,

6351 - j35*56.47∠cos^-1(0.75)

Therefore,

|E| = 7.981 KV

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