2) A 1,000 kVA, 3,200V, single phase 60Hz, 24 pole alternator/generator has an s

Question

2) A 1,000 kVA, 3,200V, single phase 60Hz, 24 pole alternator/generator has an stator winding re-sistance of 0.1 ohm from line-to-neutral. The rotor is not a permanent magnet but a separately pow-ered electromagnet (field windings) that takes 100 A at 120 V. The power to run the filed is consid-ered a loss. The friction losses, including windage, are 15 kW. The core losses are 20 kW. a) Determine the output (line ) current based on kVA and ouput voltage. b) Calculate the total I2R resistance loss. c) Calculate the efficiency. (93.3 % answer) (Generator input = output + losses

Explanation / Answer

Given data :

1000 KVA, 3200 volts, 1 ph, 60 Hz, 24 pole, stator winding resistance = 0.1 ohm,

Field voltage = 120 V, field current =100A, Friction & windage loss=15 kw, core losses = 20 kw

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a) Line current = I L = P/(3 x VL ) = 1000 KVA/(3 x 3200) = 180.42 A

The phase voltage of this machine is 3200/3 = 1847.52

The internal generated voltage of the machine is Ea = V ph + I L * R = 1847.52 +(180.42)(0.1) =1865.65 V = 3231.34 (Line voltage)

b) The input power required to the machine = output power + losses

output power = (1000 KVA) (0.8 PF) =800 KW

copper losses = 3 IL2 R = 3 x 180.422 x0.1 = 9.766 KW

P( F&W)Friction & windage loss=15 kw

core losses = 20 kw

P stray = (assumed 0)

P filed losses = 120 x100 = 12 kw

P input = P output + P (fw) +P (core) + P( copper) + P( stray)+ P (field) = 800+15+20+9.766+ 12 = 856.766

efficiency = output/input = 800/856.766 = 93.37 %

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