Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz lin

Question

Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging. The second machine uses 27 kVA at 0.93 power factor leading. The lighting load consumes 13 kW. What is the current in the line feeding the three loads? What kVAR value of capacitance connected across the line is necessary to obtain a power factor of 0.98 lagging? With this capacitor installed, what will be the current flowing in the line feeding the four loads?

Explanation / Answer

voltage for all three instruments remain the same=220 V. for the first machine, current=(power/(voltage*power factor))=224.466 A with a power factor of 0.81 lagging=224.466*(0.81 - j sin(arccos(0.81)) =181.82 - j131.63 for the second machine, current=kva rating/voltage=122.727 A with power factor of 0.93 leading.=122.727*(0.93 + j sin(arccos(0.93)) =114.14 + j45.11 for lighting load,the power factor is 1 i.e.it is resistive. hence current=59.1 A total current=taking the vector sum of all three currents=355.06- j 86.52 hence supply current is 365.45 A with a power factor of 0.9715 lagging. (b.)to bring power factor to 0.98 ,let the kvar rating be x. without capacitor, the total power=220*(355.06- j 86.52)=78.113*10^3 - j 19034.4 if we add x kVar more, total kVar= x+19.034 total kW=78.113 power factor is given by arctan(kVar/kW) hence tan(arccos(0.98))=(x+19.034)/78.113 solving we get, x=3.172

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