Ssyn99.162kVA=SsynPinFpsyn=Fpsyn0.8=Since the new motors operate at 0.8 power factor, the apparent power will bePin79.33kW=PinPoutη=Pout100 hp⋅=For the new motors, the power out is 100 hp or 74.6 kW (2 times 50 hp), so the power in is the power out divided by efficiencySold166.5151.337i+kVA=SoldSloadejθold⋅⋅=θoldacos Fpold(29=b. We can find the total load with the new motors by adding the two loads together as complex numbers IL270.633A=ILSload3VL⋅=a. The line current can be found from the three-phase power formulaSolutionη94%=VL480 V⋅=Fpold0.74=Sload225 kVA⋅=Given:kVARkW≡kVAkW≡Define: P2-25. A machine shop is fed by a 300 KVA, 480 Volt, three-phase transformer bank that has a load consisting of many small motors. The total shop load is 225 KVA and the power factor is 0.74 lagging. (a) What is the line current in the feeder from the transformer to the facility?
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