English, asked by tarunpuppala9440, 6 months ago

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.

Answers

Answered by ankitverma999999999
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

Answered by Elsharkawy811
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 = \sqrt{3} VlIlCosQ

i= 22.43 A

Efficiency= 100/108.99 *100 =91.7%

Explanation:

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