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Maximum torque and torque at 1430 rpm of slip ring motor

Short answer

Refer stator impedance to rotor side (K=0.5 gives R₁'=0.2 Ω, X=1 Ω). Using given starting torque as reference, Tmax/Tst=5.525, so Tmax≈55.25 Nm (slip 0.0784). At 1430 rpm (slip 0.04667), torque ratio=4.954, so torque≈49.5 Nm.

A 400 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 200 V. The stator impedance is 0.8+j3 ohm. The rotor resistance and standstill reactance are 0.08 ohm and 0.25 ohm respectively. The motor produces a torque of 10 Nm at starting. Calculate: (i) Maximum torque that can be developed by the motor. (ii) Calculate the torque developed by the rotor at 1430 rpm.

Given: VL=400V_L = 400 V (star), 4 poles, 50 Hz, slip-ring voltage at standstill =200= 200 V (star), Z1=0.8+j3 ΩZ_1 = 0.8 + j3\ \Omega, R2=0.08 ΩR_2 = 0.08\ \Omega, X2=0.25 ΩX_2 = 0.25\ \Omega, Tst=10T_{st} = 10 N·m.

Method: Refer the stator impedance to the rotor side and use torque ratios (the supply voltage cancels). The 10 N·m starting torque is used as the reference value.

Step 1: Transformation ratio and referred impedance

K=E2E1=200400=0.5R1′=K2R1=0.25×0.8=0.2 ΩX1′=K2X1=0.25×3=0.75 ΩX=X1′+X2=0.75+0.25=1.0 Ω\begin{aligned} K &= \frac{E_2}{E_1} = \frac{200}{400} = 0.5 \\ R_1' &= K^2 R_1 = 0.25 \times 0.8 = 0.2\ \Omega \\ X_1' &= K^2 X_1 = 0.25 \times 3 = 0.75\ \Omega \\ X &= X_1' + X_2 = 0.75 + 0.25 = 1.0\ \Omega \end{aligned}

Torque (with V′V' = stator voltage referred to rotor):

T=3ωs⋅V′2(R2/s)(R1′+R2/s)2+X2T = \frac{3}{\omega_s}\cdot\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}

Ns=120×50/4=1500N_s = 120 \times 50/4 = 1500 rpm.

(i) Maximum torque

Maximum torque occurs when R2/sm=R1′2+X2R_2/s_m = \sqrt{R_1'^2 + X^2}:

sm=R2R1′2+X2=0.080.04+1=0.081.0198=0.0784\begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.08}{\sqrt{0.04 + 1}} = \frac{0.08}{1.0198} = 0.0784 \end{aligned} Tmax=3ωs⋅V′22(R1′+R1′2+X2)T_{max} = \frac{3}{\omega_s}\cdot\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}

Ratio with starting torque (s=1s = 1):

TmaxTst=(R1′+R2)2+X22R2(R1′+R1′2+X2)=(0.28)2+122×0.08×(0.2+1.0198)=1.07840.19517=5.525\begin{aligned} \frac{T_{max}}{T_{st}} &= \frac{(R_1' + R_2)^2 + X^2}{2R_2\left(R_1' + \sqrt{R_1'^2+X^2}\right)} \\ &= \frac{(0.28)^2 + 1^2}{2 \times 0.08 \times (0.2 + 1.0198)} \\ &= \frac{1.0784}{0.19517} = 5.525 \end{aligned} Tmax=5.525×10=55.25 N⋅mT_{max} = 5.525 \times 10 = 55.25\ \text{N·m}

(It occurs at N=1500(1−0.0784)≈1382N = 1500(1 - 0.0784) \approx 1382 rpm.)

(ii) Torque at 1430 rpm

s=1500−14301500=0.04667,R2s=0.080.04667=1.714 Ωs = \frac{1500 - 1430}{1500} = 0.04667, \qquad \frac{R_2}{s} = \frac{0.08}{0.04667} = 1.714\ \Omega TTst=R2/sR2⋅(R1′+R2)2+X2(R1′+R2/s)2+X2=10.04667×1.0784(0.2+1.714)2+1=21.43×1.07844.6645=4.954\begin{aligned} \frac{T}{T_{st}} &= \frac{R_2/s}{R_2}\cdot\frac{(R_1'+R_2)^2 + X^2}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1}{0.04667}\times\frac{1.0784}{(0.2 + 1.714)^2 + 1} \\ &= 21.43 \times \frac{1.0784}{4.6645} = 4.954 \end{aligned} T=4.954×10=49.54 N⋅mT = 4.954 \times 10 = 49.54\ \text{N·m}

Note: if the stator impedance were neglected, Tmax/Tst=(R22+X22)/(2R2X2)=1.72T_{max}/T_{st} = (R_2^2+X_2^2)/(2R_2X_2) = 1.72 only; including Z1Z_1 (as given) changes the result a lot, so it has been included.

Answer: (i) Tmax≈55.25T_{max} \approx 55.25 N·m (at slip 0.0784); (ii) torque at 1430 rpm ≈49.5\approx 49.5 N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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