# Mathcad - 02-22 - For the first motor: power out power...

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this is the magnitude only of the complex power S in1 101.318 KVA = Q in1 S in1 sin θ ( 29 = Q in1 60.791KVAR = For second motor: P out2 50 hp = F p2 0.9 = η 2 0.9 = θ 2 acos F p2 ( 29 = P in2 P out2 η 2 = θ 2 25.842deg = P in2 41.428kW = sin θ 2 - ( 29 0.436 - = S in2 P in2 F p2 = S in2 46.031 KVA = Q in2 S in2 sin θ 2 - ( 29 = Q in2 20.064 - KVA = Problem 2-22: Solution Two motors connected to a 480 volt, three-phase supply. The first motor is a 100 hp induction motor with efficiency of 92%, and power factor of 0.8 lagging. The second motor is a 50 hp synchronous motor operating at an efficiency of 90% and power factor of 0.9 leading a. For each motor, calculate the three-phase real power input to the machine (in KW) and the three-phase reactive power input to the machine (in KVAR). KVA kW KVAR kW

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Unformatted text preview: For the first motor: power out power factor efficiency P out1 100 hp = F p1 0.8 = 1 0.92 = P in1 P out1 1 = acos F p1 ( 29 = 36.87deg = P in1 81.054kW = sin ( 29 0.6 = S in1 P in1 F p1 = b. Find the power factor of the combination of the two machines. The total real power is the sum of the two machines' real powers and the total reactive power is the sum of the two machines' reactive powers P tot P in1 P in2 + = Q tot Q in1 Q in2 + = P tot 122.482kW = Q tot 40.726KVA = reactive F ptot cos atan Q tot P tot = from the power triangle, tan ( 29 Q P = F ptot 0.949 =...
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## Mathcad - 02-22 - For the first motor: power out power...

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