# Mathcad - 07-13 - S in 3 V L ⋅ I L ⋅ = note that...

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S in 73.3 kVA = F p P in S in = F p 0.815 = b. At half load the output is 30 hp P out2 30 hp = P out2 22.371kW = P in2 P out2 η = P in2 23.901kW = 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 42.458kVAR = 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 48.723 kVA = and F p2 P in2 S in2 = F p2 0.491 = rpm 1 min kVA kW Problem 7-13: Solution kVAR kW HP 75 hp = V L 460 V = I L 92 A = η 93.6 % = RPM 1160 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 55.927kW = P in P out η = P in 59.752kW = the apparent power can be found from the input voltage and current. Since it is a three-phase motor:

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Unformatted text preview: S in 3 V L ⋅ I L ⋅ = note that although the efficiency of the motor was assumed to remain high at 93% when the motor is half loaded, the power factor becomes very poor at half load. c. Since the motor's frequency is 60 hz, and its stated speed is 1180 rpm, we can assume that the synchronous speed is 1200 rpm, which would mean a 6 pole motor. Alternatively, we can calculate the number of poles at rated speed RPS RPM 60 sec min ⋅ = RPS 19.333 1 sec = revolutions per second RPS 2 f ⋅ P = so P 2 f ⋅ RPS = P 6.207 = poles Obviously, the number of poles must be an even integer, so this is a six pole motor...
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Mathcad - 07-13 - S in 3 V L ⋅ I L ⋅ = note that...

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