MGVCL Exam Paper (30-07-2021 Shift 3)
A single phase motor connected to 400 V, 50 Hz supply takes 25 A at a power factor of 0.75 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.95 lagging.
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
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
The capacity factor is the ratio of an actual electrical energy output over a given period of time to the maximum possible electrical energy output over that period. Nuclear power plants are at the high end of the range of capacity factors
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