Answer to Question #230235 in Electrical Engineering for KHANYA

Question

A 150 KW ,3000 v, 50 Hz, 6-pole-star-connected induction motor has a start connected slip-ring with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.2 Ω/phase and its per phase/leakage reactance is 14.7 Ω/phase. The stator impedance may be neglected. Find the starting current and starting torque at rated voltage with short circuited slip–ring. 

Answer 1

$X_2= 2 \pi *50*5.61*10^{-3}=1.13 \Omega\\ K= 1/3.6, R_2'= R_2/K^2 = 3.6^2*0.1 = 1.3 \Omega\\ X_2= 2 \pi *50*3.61*10^{-3}=1.13 \Omega ; X_2' =3.6^2*1.13 = 14.7 \Omega\\ So, I_{st}= \frac{V}{(R_2')^2+(X_2')^2}\\ I_{st}= \frac{3000/\sqrt{3}}{\sqrt{(1.3)^2+(14.7)^2}}\\ I_{st}=117.4A\\ N_s= 120*50/6=1000 rpm=50/3 rps\\ T= \frac{3}{2 \pi N_s}* \frac{V^2R_2'}{(R_2')^2+(X_2')^2}\\ T=\frac{3}{2 \pi* 50/3}* \frac{(3000/\sqrt3)^2*1.3}{(1.3)^2+(14.7)^2}\\ T=513 Nm$