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6-pole induction motor: slip, losses, efficiency

Short answer

Ns=120×50/6=1000 rpm; s=(1000-975)/1000=0.025 (2.5%). Air-gap power Pg=40-1=39 kW; rotor copper loss sPg=0.975 kW. Net mechanical power=39-0.975=38.025 kW; output=38.025-2=36.025 kW=48.29 HP. Efficiency=36.025/40×100=90.06%.

The power input to a 500 V, 50 Hz, 6-pole, 3-phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction loss is 2 kW. Calculate: (a) slip (b) rotor copper loss (c) output HP (d) efficiency.

Given: 500 V, 50 Hz, 6-pole, Nr=975N_r = 975 rpm, input Pin=40P_{in} = 40 kW, stator loss =1= 1 kW, friction loss =2= 2 kW.

(a) Slip

Ns=120×506=1000 rpm,s=1000−9751000=0.025=2.5%N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-975}{1000} = 0.025 = 2.5\%

(b) Rotor copper loss

Pg=40−1=39 kWP_g = 40 - 1 = 39\ \text{kW} Pcu2=sPg=0.025×39=0.975 kWP_{cu2} = sP_g = 0.025\times 39 = 0.975\ \text{kW}

(c) Output HP

Pm=Pg−Pcu2=39−0.975=38.025 kWP_{m} = P_g - P_{cu2} = 39 - 0.975 = 38.025\ \text{kW} Pout=38.025−2=36.025 kW=36025746=48.29 HPP_{out} = 38.025 - 2 = 36.025\ \text{kW} = \frac{36025}{746} = 48.29\ \text{HP}

(d) Efficiency

η=36.02540×100=90.06%\eta = \frac{36.025}{40}\times 100 = 90.06\%

Answer: (a) s=2.5%s = 2.5\%; (b) rotor copper loss =0.975= 0.975 kW; (c) output =36.03= 36.03 kW ≈48.29\approx 48.29 HP; (d) η=90.06%\eta = 90.06\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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