MGVCL Exam Paper (30-07-2021 Shift 1)
A balanced delta connected load of 8+j6 Ω per phase is connected to a 415 V, 50 Hz, 3-phase supply lines. If the input power factor is improved to 0.9 by connecting a star connected capacitor bank, then the required kVAR of the bank is
Current flow through the circuit, I = V/Z = 400/10 = 41.5 A Power factor = 8/10 = 0.8 After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged. I1*cosφ1 = I2*cosφ2 41.5*0.8 = I2*0.9 I2 = 36.88 A Q1 (before connecting capacitor) = √3*VL*IL*sinφ1 = √3*415*41.5*sin(36.86) = 26.847 kVAR Q2 (after connecting capacitor) = √3*VL*IL*sinφ2 = √3*415*36.88*sin(25.84) = 15.867 kVAR kVAR supplied by capacitor bank = Q1 - Q2 = 26.847 - 15.867 = 10.98 kVAR
For small turbine starts generating power 12.6 kph (3.5 m/s) is the typical cut-in speed. At 36–54 kph (10–15 m/s) produces maximum generation power. At 90 kph (25 m/s) maximum, the turbine is stopped or braked.
Two-part tariff: In this type of tariff, the total bill is divided into two parts. The first one is the fixed charge and the second is the running charge. The fixed charge is because of the maximum demand and the second charge depends on the energy consumption by the load.
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