MGVCL Exam Paper (30-07-2021 Shift 1)
A single phase motor connected to 400 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging.
For single phase induction motor, P = VIcosφ P = 400*30*0.7 P = 8400 W Q1 = P*tan(φ) = 8400*1.020 Q = 8568.82 VAR Now power factor change to 0.9 So, φ' = 25.84 degree Q = 8400*tan(25.84) Q = 4067.95 Qnet = Q - Q' Qnet = 8568.82 - 4067.95 Qnet = 4500.8 C = V²/(2ᴨf) C = (400*400)/(2ᴨ*50) C = 8.95*10⁻⁵ C = 89.5 μF
The magnitude of earth fault current for a given fault position within a winding demands upon the winding connections and method of neutral grounding. Earth fault protection for an electric motor is provide by instantaneous relay having a setting of approximately 30% of motor rated current in the residual circuits of two CTs and with ground wire.
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
EXAMIANS