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Mathcad - 07-10 - S in 3 V L ⋅ I L ⋅ = note that...

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S in 56.569kVA = F p P in S in = F p 0.85 = b. At half load the output is 30 hp P out2 30 hp = P out2 22.371kW = P in2 P out2 η = P in2 24.055kW = We can find the reactive power in part a since we know S and P Q mot S in 2 P in 2 - = Q mot 29.757kVAR = with this reactive power and the real power in at half-load, we can calculate the apparent power: S in2 P in2 2 Q mot 2 + = S in2 38.264kVA = and F p2 P in2 S in2 = F p2 0.629 = rpm 1 min kVA kW Problem 7-10: Solution kVAR kW HP 60 hp = V L 460 V = I L 71 A = η 93 % = RPM 1180 rpm = f 60 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 efficiency P out HP = P out 44.742kW = P in P out η = P in 48.11kW = the apparent power can be found from the input voltage and current. Since it is a three-phase motor:
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