Power Systems The sequence components of the fault current are as follows : Ipositive = j1.5 pu, Inegative = − j0.5 pu, Izero = − j pu Which fault is this? LL. LLG. LLLG. LG. LL. LLG. LLLG. LG. ANSWER DOWNLOAD EXAMIANS APP
Power Systems Which of the following power stations is mainly used to cover peak load on the system? pumped storage hydro power plant nuclear plant coal based thermal power plant gas based thermal power plant pumped storage hydro power plant nuclear plant coal based thermal power plant gas based thermal power plant ANSWER DOWNLOAD EXAMIANS APP
Power Systems A steam power generation has an overall efficiency of 20 %. 0.6 Kg of coal is burnt per kWh of electrical energy generated. Calculated the calorific value of fuel 7000 KCal / Kg. 7562 KCal / Kg. 7166.67 KCal / Kg. 7268.45 KCal / Kg. 7000 KCal / Kg. 7562 KCal / Kg. 7166.67 KCal / Kg. 7268.45 KCal / Kg. ANSWER DOWNLOAD EXAMIANS APP
Power Systems Maximum power transfer in a transmission line can be obtained by either 1 or 2 increasing voltage level reducing reactance None of these either 1 or 2 increasing voltage level reducing reactance None of these ANSWER DOWNLOAD EXAMIANS APP
Power Systems A lossless power system has to serve a load of 250 MW. There are two generators(G1 and G2) in the system with cost curves C1 and C2 respectively defined as follows: C1 = PG1+0.055P²G1 C2 = 3 PG2 + 0.03 P²G2 Where PG1 and PG2 are the MW injections from generators G1 and G2 respectively. Thus, the minimum cost dispatch will be PG1 = 250 MW and PG2 = 0 MW PG1 = 150 MW and PG2 = 100 MW PG1 = 0 MW and PG2 = 250 MW PG1 = 100 MW and PG2 = 150 MW PG1 = 250 MW and PG2 = 0 MW PG1 = 150 MW and PG2 = 100 MW PG1 = 0 MW and PG2 = 250 MW PG1 = 100 MW and PG2 = 150 MW ANSWER DOWNLOAD EXAMIANS APP
Power Systems The zero sequence current of a generator for line to ground fault is j3 pu. Then the current through the neutral during the fault is j1 pu j6 pu j9 pu j3 pu j1 pu j6 pu j9 pu j3 pu ANSWER DOWNLOAD EXAMIANS APP
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