Electrical Machines A fourpole 50Hz induction motor supplies

Electrical Machines

A four-pole. 50-Hz induction motor supplies 30kW to load at speed of 1470 rpm. 1) What is the motor\'s slip? 2) What is the induced torque in the motor in N.m under these conditions? 3) What will be the operating speed of the motor If its torque is halved? 4) How much power will be supplied by the motor when the torque is halved? 5) What is the relationship between induced torque and mechanical speed?

Solution

1)slip =(Ns-Nr)÷Ns

Ns = synchronous speed= (120*f)÷p= (120 *50)÷4= 1500rpm

, Nr - rotor speed=1470

S=0.02

2)induced torque = power induced in rotor ÷ angular speed

Power induced rotor= mech power ÷ (1-s)

= 30kw÷(1-0.02)=30.61kw

Angular speed=2Nr÷60=(2*1470)÷60=153.93nm

=

Torque induced= 30.61kw÷153.93nm=198.84

3)speed at torque half=use same formula for p= T*w by half the induced torque in previous condition..

4) here not mention copper loss so take rotor induced power as a input power. Power= torque induced* angular speed but Ws=angular speed =2Ns÷60; p= T/2*Ws

5) relation between torque and speed

Power= torque* angular speed