A 150 kW, 3000 V, 50 Hz, 6-pole star-connected induction motor has a starconnected slip-ring rotor with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.1 \Omega / \text { phase } and its per phase leakage inductance is 3.61 mH. The stator impedance may be neglected. Find (a) the starting current and torque on rated voltage with short-circuited slip-rings, and (b) the necessary external resistance to reduce the rated-voltage starting current to 30 A and the corresponding starting torque.
Question 9.11: A 150 kW, 3000 V, 50 Hz, 6-pole star-connected induction mot...
R_{2}=0.1 \Omega X_{2}=314 \times 3.61 \times 10^{-3}=1.13 \Omega
R_{2}^{\prime}=(3.6)^{2} \times 0.1=1.3 \Omega X_{2}^{\prime}=(3.6)^{2} \times 1.13=14.7 \Omega
(a) I_{s}=\frac{3000 / \sqrt{3}}{\sqrt{(1.3)^{2}+(14.7)^{2}}}=117.4 A
T_{s}=\frac{3}{\omega_{s}} \cdot \frac{V^{2} R_{2}^{\prime 2}}{R_{2}^{\prime 2}+X_{2}^{\prime 2}}
\omega_{s}=\frac{2 \pi \times 1000}{60}=104.7 rad / s
T_{s}=\frac{3}{104.7} \cdot \frac{(3000 / \sqrt{3})^{2} \times 1.3}{(1.3)^{2}+(14.7)^{2}}=513.1 Nm
(b) I_{s}=\frac{3000 \sqrt{3}}{\sqrt{\left(1.3+R_{e x t}^{\prime}\right)^{2}+(14.7)^{2}}}=30 A
R_{\text {ext }}^{\prime}=54.33
R_{e x t}=54.53 /(3.6)^{2}=4.21 \Omega
T_{s}=\frac{3}{104.7} \cdot \frac{(3000 \sqrt{3})^{2} \times(1.3+54.33)}{(1.3+54.33)^{2}+(14.7)^{2}}=1440 Nm
Remark By adding an external resistance in the rotor, starting current reduces by a factor of 117.4/30 = 3.91 while the starting torque increases by 1440/513.1 = 2.8.
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