Answer to Question #223423 in Mechanical Engineering for Unknown346307

Question

A 10 pole, 50 Hz, Y connection 3-phase induction motor having a rating of 60 kW and 415V.

The slip of the motor is 5% at 0.6 power factor lagging. If the full load efficiency is 90%,

calculate:

(i) Input power

(ii) Line current and phase current

(iii) Speed of the rotor (rpm)

(iv) Frequency of the rotor

(v) Torque developed by the motor (if friction and windage losses is 0)

Answer 1

$1)\\ η=\frac{O/P \space power}{Input \space power}\\ \implies Input \space power=\frac{O/P \space power}{η}\\ Input \space power=\frac{60K}{0.9}=66.66kW\\ 2)\\ P= \sqrt{3}* V_2I_2 * cos \theta\\ \implies I_2= \frac{P}{\sqrt{3}*V_2 cos \theta}= \frac{66.66K}{\sqrt{3}*415 *0.6}=154.75 A\\ In \space star\\ I_2 = \sqrt{3} I_{ph}\\ \implies I_{ph}= \frac{I_2}{ \sqrt{3}}= 89.24 A \sqrt{3} \sqrt{3} \sqrt{3}\\ 3)\\ N= Ns(1-s)\\ Ns= \frac{120f}{P}= \frac{120*50}{10}=600 rpm\\ = 600(1-0.05)\\ =570 rpm\\ 4)\\ f_r=s*f=0.05*50=2.5 Hz\\ 5)\\ P=\tau \omega \\ \implies \tau = \frac{P}{\omega}= \frac{60K*60}{2\pi*570}= 1005.65 Nm$