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Slip, rotor copper loss, shaft power and efficiency at 975 rpm

Short answer

Ns = 120×50/6 = 1000 rpm; s = (1000-975)/1000 = 0.025. Rotor input Pag = 40-1 = 39 kW; rotor copper loss = s×Pag = 0.025×39 = 0.975 kW. Shaft power = 39-0.975-2 = 36.025 kW; efficiency = 36.025/40 = 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 the friction and windage losses total 2 kW. Calculate: (i) The slip (ii) The rotor copper loss (iii) Shaft power and (iv) The efficiency.

Given: 500 V, 50 Hz, 6 poles, N=975N = 975 rpm, input Pin=40P_{in} = 40 kW, stator losses =1= 1 kW, friction and windage losses =2= 2 kW.

(i) Slip

Ns=120×506=1000 rpms=1000−9751000=0.025 (2.5%)\begin{aligned} N_s &= \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ s &= \frac{1000 - 975}{1000} = 0.025 \ (2.5\%) \end{aligned}

(ii) Rotor copper loss

Pag=Pin−stator losses=40−1=39 kWPcu2=sPag=0.025×39=0.975 kW=975 W\begin{aligned} P_{ag} &= P_{in} - \text{stator losses} = 40 - 1 = 39\ \text{kW} \\ P_{cu2} &= sP_{ag} = 0.025 \times 39 = 0.975\ \text{kW} = 975\ \text{W} \end{aligned}

(iii) Shaft power

Pm=Pag−Pcu2=39−0.975=38.025 kWPshaft=Pm−friction and windage=38.025−2=36.025 kW\begin{aligned} P_m &= P_{ag} - P_{cu2} = 39 - 0.975 = 38.025\ \text{kW} \\ P_{shaft} &= P_m - \text{friction and windage} = 38.025 - 2 = 36.025\ \text{kW} \end{aligned}

(iv) Efficiency

η=PshaftPin=36.02540=0.9006=90.06%\eta = \frac{P_{shaft}}{P_{in}} = \frac{36.025}{40} = 0.9006 = 90.06\% 
 40 kW in --> [-1 kW stator] --> 39 kW air gap
          --> [-0.975 kW rotor Cu] --> 38.025 kW
          --> [-2 kW F&W] --> 36.025 kW shaft

Answer: (i) s = 0.025; (ii) rotor Cu loss = 975 W; (iii) shaft power = 36.025 kW; (iv) η = 90.06%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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