Question

A 100 kW, 460 V shunt generator was run as a motor on no load at its rated voltage and speed. The total current taken was 9.8 A, including a shunt current of 2.7 A. The resistance of the armature circuit at normal working temperature was 0.11 Ω. Calculate the efficiencies at: (16)

(i) Full load

(ii) Half load.

Answer 1

A 100 kW, 460 V shunt generator was run as a motor on no load at rated voltage and speed. Total current 9.8 A, shunt current 2.7 A, armature resistance 0.11 ohm.

Field current If = 2.7 A. Field copper loss = V x If = 460 x 2.7 = 1242 W.
No-load armature current = 9.8 - 2.7 = 7.1 A.
No-load armature copper loss = 7.1^2 x 0.11 = 50.4 x 0.11 = 5.5 W.
No-load input = 460 x 9.8 = 4508 W.
Constant losses (iron + friction + windage + field copper) = no-load input - no-load armature copper loss = 4508 - 5.5 = 4502.5 W.
(This includes the field copper loss of 1242 W and the rotational losses of 3260 W.)

(i) Full load:

Output = 100 kW. Load current = 100000/460 = 217.4 A.
Armature current Ia = load current + field current = 217.4 + 2.7 = 220.1 A.
Armature copper loss = 220.1^2 x 0.11 = 48,444 x 0.11 = 5329 W.
Total losses = 4502.5 + 5329 = 9831.5 W.
Input = 100000 + 9831.5 = 109,831.5 W.
Efficiency = 100000/109831.5 = 0.9105 = 91.05%.

(ii) Half load:

Output = 50 kW. Load current = 50000/460 = 108.7 A.
Armature current Ia = 108.7 + 2.7 = 111.4 A.
Armature copper loss = 111.4^2 x 0.11 = 12,410 x 0.11 = 1365 W.
Total losses = 4502.5 + 1365 = 5867.5 W.
Input = 50000 + 5867.5 = 55,867.5 W.
Efficiency = 50000/55867.5 = 0.895 = 89.5%.

So the efficiency is 91.05% at full load and 89.5% at half load.

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