Question

An 18.65 KW, 6pole, 50 Hz, 3-phase slip-ring induction motor runs at 960 rpm on full load with t rotor current per phase of 35 A. Allowing 1 KW for mechanical losses, find the resistance/phase; of 3-phase rotor winding.

Motor out put = 18.65 KW,

Mechanical losses = 1 KW.

Therefore

Mechanical power developed by motor Pm = 18.65 + 1 = 19.65KW.

Now

Ns = 120f / P = 120 x 50 / 60 – 1000 rpm.

Slip = 1000 – 960 / 1000 = 0.04

But

P2:Pm:RCL:1:1-S:S

Or P2 / Pm = 1/1-S, Pm /RCL=1-S/S.

But we know that

Rotor copper loss = S/1-S x Pm = 0.04 / (1- 0.04) x 19.65 = 0.819KW.= 819W.

Therefore