MGVCL Exam Paper (30-07-2021 Shift 1)
Case 1: The PV connected to the DC bus was injecting 20 kW of power constituting of 1 kW of loss Case 2: The same PV is connected to the AC bus constituting to 1.5 kW loss. Find out the change in efficiency from Case 1 to Case 2?
In DC microgrid system, Efficiency α 1/Pin For 1 kW loss injecting power is 20 kW For 1.5 kW loss injecting power will be proportional to 30 kW So, Efficiency will be reduced 0.025
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
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
EXAMIANS