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
The ratio of full load current to short circuit current = 1/5 Xsc = j/(1/5) External reactance required = j*((1/5) - 0.05)) = j*0.15 pu Full load current = (30*1000)/(√3*11) = 1574.6 A Per unit reactance = j*0.15 = (I*Xr)/V j*0.15 = (1574.6*Xb)/((11/√3)*1000)) = 0.60 ohm
Tank - Name of power station - capacity 1 - Vindhyachal Thermal Power Station - 4,760MW. 2 - Mundra Thermal Power Station - 4620 MW 3 - The Sasan Ultra Mega power plant - 3960 MW 4 - The Tiroda thermal power plant - 3300 MW
EXAMIANS