A 100 KW (output), 3300 V, 50 Hz, 3 phase star connected induction motor has a synchronous speed of 500 r.p.m. The full load slip is 1.8 % and F. L. power factor is 0.85. Stator copper loss=2440 W, Iron loss=3500 W, Rotational loss=1200 W. Calculate (i) the rotor copper loss (ii) the line current (iii) the full load efficiency.
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0
Answer:
The 21st term of an AP whose first two terms are – 3 and 4, is *
1 point
- (a) 17
(b) 137
(c) 143
(d)-143
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1
Answer:
P1 loss = 2440+3500 = 5940W
Pm=Pout+P3loss = 100+1200 = 101.200Kw
Pm=(1-s)Pg
101.200=(1-0.018)Pg
Pg=103.054Kw
Pin=Pg+P1
103.054+5940 = 108.994Kw
Pin =
i= 22.43 A
Efficiency= 100/108.99 *100 =91.7%
Explanation:
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