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Max-torque speed and efficiency at 1350 rpm

Short answer

Effective ratio K = 180/380, referred R2'=0.2674 Ω, X'=1.337 Ω, total X=3.837 Ω. Speed at max torque: sm = R2'/√(R1²+X²)=0.0691, Nm = 1396 rpm. At 1350 rpm, s=0.1: input 18.48 kW, output 13.51 kW, efficiency 73.1%.

A 380 V, 4-pole, 50 Hz, 3 phase, slip ring induction motor has a star connected stator winding and a star connected rotor winding. At standstill the voltage between the two slip rings is 180 V. The stator impedance is 0.5+j2.5 ohm. The rotor resistance and reactance at standstill are 0.06 ohm and 0.3 ohm respectively. The stator to rotor turn ratio is 1:1. The motor consumes 500 watts at no-load. Calculate: i) Speed at which the motor develops the maximum torque ii) Efficiency of the motor when it is running at 1350 rpm

Given: VL=380V_L = 380 V (star), 4 poles, 50 Hz, slip-ring voltage at standstill =180= 180 V (star), Z1=0.5+j2.5 ΩZ_1 = 0.5 + j2.5\ \Omega, R2=0.06 ΩR_2 = 0.06\ \Omega, X2=0.3 ΩX_2 = 0.3\ \Omega, no-load input =500= 500 W, N=1350N = 1350 rpm.

Assumption: the stated 1:1 turns ratio conflicts with the measured 380 V / 180 V; the measured voltages are used to find the effective ratio K=180/380K = 180/380, and the rotor values are referred to the stator. The approximate equivalent circuit is used (no-load current neglected in the series branch), and the 500 W no-load loss is taken as the constant (core + mechanical) loss.

Rotor values referred to stator

K=180380=0.47368,K2=0.22438R2′=R2K2=0.060.22438=0.2674 ΩX2′=X2K2=0.30.22438=1.3370 ΩX=X1+X2′=2.5+1.337=3.837 Ω\begin{aligned} K &= \frac{180}{380} = 0.47368, \qquad K^2 = 0.22438 \\ R_2' &= \frac{R_2}{K^2} = \frac{0.06}{0.22438} = 0.2674\ \Omega \\ X_2' &= \frac{X_2}{K^2} = \frac{0.3}{0.22438} = 1.3370\ \Omega \\ X &= X_1 + X_2' = 2.5 + 1.337 = 3.837\ \Omega \end{aligned}

(i) Speed at maximum torque

sm=R2′R12+X2=0.26740.25+14.723=0.26743.8695=0.0691Nm=(1−0.0691)×1500=1396.3 rpm\begin{aligned} s_m &= \frac{R_2'}{\sqrt{R_1^2 + X^2}} = \frac{0.2674}{\sqrt{0.25 + 14.723}} = \frac{0.2674}{3.8695} = 0.0691 \\ N_m &= (1 - 0.0691) \times 1500 = 1396.3\ \text{rpm} \end{aligned}

(ii) Efficiency at 1350 rpm

s=1500−13501500=0.1,R2′s=2.674 Ωs = \frac{1500 - 1350}{1500} = 0.1, \qquad \frac{R_2'}{s} = 2.674\ \Omega Vph=3803=219.39 VZ=(0.5+2.674)+j3.837=3.174+j3.837=4.980 ΩI1=219.394.980=44.06 A\begin{aligned} V_{ph} &= \frac{380}{\sqrt{3}} = 219.39\ \text{V} \\ Z &= (0.5 + 2.674) + j3.837 = 3.174 + j3.837 = 4.980\ \Omega \\ I_1 &= \frac{219.39}{4.980} = 44.06\ \text{A} \end{aligned} Pin=3I12(R1+R2′/s)=3×44.062×3.174=18 483 WPcu1=3×44.062×0.5=2912 WPag=3×44.062×2.674=15 572 WPcu2=sPag=1557 WPm=Pag−Pcu2=14 014 WPout=Pm−500=13 514 Wη=13 51418 483=0.7312=73.1%\begin{aligned} P_{in} &= 3I_1^2(R_1 + R_2'/s) = 3 \times 44.06^2 \times 3.174 = 18\,483\ \text{W} \\ P_{cu1} &= 3 \times 44.06^2 \times 0.5 = 2912\ \text{W} \\ P_{ag} &= 3 \times 44.06^2 \times 2.674 = 15\,572\ \text{W} \\ P_{cu2} &= sP_{ag} = 1557\ \text{W} \\ P_m &= P_{ag} - P_{cu2} = 14\,014\ \text{W} \\ P_{out} &= P_m - 500 = 13\,514\ \text{W} \\ \eta &= \frac{13\,514}{18\,483} = 0.7312 = 73.1\% \end{aligned}

(Note: 1350 rpm lies beyond the maximum-torque speed, in the high-slip region, which is why the efficiency is low.)

Answer: (i) maximum torque at about 1396 rpm (sm=0.0691s_m = 0.0691); (ii) efficiency at 1350 rpm ≈73.1%\approx 73.1\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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