A 200V, long-shunt compound-wound generator has a full-load output of 20kW. The various resistances are as follows; armature (including brush contact) 0.15 ohm, series field 0.025 ohm, interpole field 0.028 ohm, shunt field (including the field-regulator resistance) 115 ohm. The iron losses at full load are 780W, and the friction and windage losses 590W. Calculate the efficiency at full load.
Given
V = 200
KW = 20 = 20000 W
Ra = 0.15 ohm
Rsef = 0.025 ohm
Rif = 0.028 ohm
Rshf = 115 ohm
Li = 780 W
Lf = 590 W
To find
1. Efficiency
Solution
Shunt field current, Ishf = V / Rshf
= 200 / 115
= 1.74 A
We know that P = V x I
20000 = 200 x I
Load current = 100 A
Therefore Armature current I = 100 + 1.74 = 101.7 A
Cu loss in armature = I2R
= 101.72 x (0.15 + 0.025 + 0.028)
= 2099.6 W = 2.1 KW
Cu loss in shunt field = Ishf x V
= 1.74 x 200
= 348 W = 0.348 KW
Total Cu loss = 2.1 + 0.348 = 2.448 KW
Efficiency = output / input
= 20 / (20 + 2.448 + 0.59 + 0.78)
= 20 / 23.818
= 0.839
= 83.9 %
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