MGVCL Exam Paper (30-07-2021 Shift 3)
A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed.
The ratio of full load current to short circuit current = 1/6 Xsc = j/(1/6) External reactance required = j*((1/6) - 0.06)) = j*0.106 pu Full load current = (40*1000)/(√3*15) = 1539.6 A Per unit reactance = j*0.106 = (I*Xr)/V j*0.15 = (1539.6*Xb)/((15/√3)*1000)) = 0.60 ohm
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (1.1/0.6)*cosδ = (1.1/0.6)*0.5 = 0.91 M = (H*s)/(πf) = 4/(50π)
EXAMIANS