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
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
The magnitude of earth fault current for a given fault position within a winding demands upon the winding connections and method of neutral grounding. Earth fault protection for an electric motor is provide by instantaneous relay having a setting of approximately 30% of motor rated current in the residual circuits of two CTs and with ground wire.
EXAMIANS