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
As per the key statistics data of IEA With electricity consumption reached to 1,164 TWh in the country till November 2019, India ranked third among the top ten electricity consuming countries across the globe.
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
There is one IEC guideline, number is IEC 599, Statement of guideline is that "If the ratio of carbon dioxide and Carbon monoxide in Dissolve gas analysis test result is more than 11, it is predict that the condition of paper insulation inside the transformer is poor." For healthy condition this ratio will be between 4 to 11.
EXAMIANS