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.
For the cut-in speed is the point at which the wind turbine is able to generate power. Between the cut-in speed and the rated speed, where the maximum output is reached, the power output will increase cubically with wind speed. Pmax directly propotional to cubical of wind speed. So, by double the speed of wind power will be increased by 8 times.
1.6 inches = 1.6 inches x 1,000 mils per inch = 1,600 mils 0.25 inch = 0.25 inch x 1,000 mils per inch = 250 mils Area = 1,600 x 250 = 400,000 square mils
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