A 50 Hz, 3-phase induction motor has a full load speed of 1440 r.p.m. The number of poles of the motor are
(A) 4.
(B) 6.
(C) 12.
(D) 8.
Explanatory answer:
We know, N = (1-S)Ns
1440 = (1-S) Ns
Therefore, Ns = 1440/(1-S)
Ns = 120F/P = 120*50/P = 6000/P
always Ns will be closer to N. i.e. 1440
Let P=2, Ns=3000 rpm (not close to N)
Let P=4, Ns=1500 rpm (close to N)
Therefore, P=4 for N=1440
(A) 4.
(B) 6.
(C) 12.
(D) 8.
Explanatory answer:
We know, N = (1-S)Ns
1440 = (1-S) Ns
Therefore, Ns = 1440/(1-S)
Ns = 120F/P = 120*50/P = 6000/P
always Ns will be closer to N. i.e. 1440
Let P=2, Ns=3000 rpm (not close to N)
Let P=4, Ns=1500 rpm (close to N)
Therefore, P=4 for N=1440
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