Sin56.569kVA=FpPinSin=Fp0.85=b. At half load the output is 30 hpPout230 hp⋅=Pout222.371kW=Pin2Pout2η=Pin224.055kW=We can find the reactive power in part a since we know S and PQmotSin2Pin2-=Qmot29.757kVAR=with this reactive power and the real power in at half-load, we can calculate the apparent power:Sin2Pin22Qmot2+=Sin238.264kVA=and Fp2Pin2Sin2=Fp20.629=rpm1min≡kVAkW≡Problem 7-10: SolutionkVARkW≡HP60 hp⋅=VL460 V⋅=IL71 A⋅=η93 %⋅=RPM1180 rpm⋅=f60 Hz⋅=a. To find the power factor we need the input power and apparent power. The input power can be found from the output power and the efficiencyPoutHP=Pout44.742kW=PinPoutη=Pin48.11kW=the apparent power can be found from the input voltage and current. Since it is a three-phase motor:
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