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
Voltage sag: It is usually associated with system faults but can also be caused by energization of heavy loads or starting of large motors. Voltage sag is the short reduction in the RMS voltage between 0.1 to 0.9 pu for a duration of 0.5 cycle to 1 minute
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
Symmetrical breaking current: It is the rms value of a a.c component of the current in the pole at the instant of contact separation.
Asymmetrical breaking current: It is the rms value of the total current comprising the a.c and d.c components of the current in the pole at the instant of contact separation.
EXAMIANS