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
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
In bias differential protection for transformer, There are two basic settings required to setup the percentage biased differential protection. 1. Minimum pickup 2. The slope.
The minimum pick-up covers the transformer magnetizing current typically 1-4% of transformer rating. A pick up of 0.2 to 0.3 times the TAP is recommended for general cases.
The slope is defined by determining the potential sources of differential current.
EXAMIANS