Shunt motor losses and efficiency

At no load the motor input only supplies the iron, friction and windage losses (together called rotational loss) plus the small copper loss. This fixed (constant) loss is the same at 40 A load.

Equivalent circuit

 I_L --+--> Ia --[Ra=0.25]--(Eb)--+
       |                          |
       +--> Ish --[Rf=115]--------+--- V = 230 V

i) Iron and friction losses (from no-load test)

• Field current: I_sh = 230 / 115 = 2 A (constant)
• No-load armature current: I_a0 = 5 − 2 = 3 A
• No-load input = 230 × 5 = 1150 W
• Field copper loss = 230 × 2 = 460 W
• Armature copper loss (no load) = 3² × 0.25 = 2.25 W
• Iron + friction loss = 1150 − 460 − 2.25 = 687.75 W ≈ 688 W

ii) Efficiency at 40 A

• Input = 230 × 40 = 9200 W
• Armature current: I_a = 40 − 2 = 38 A
• Armature copper loss = 38² × 0.25 = 361 W
• Field copper loss = 460 W
• Constant (iron + friction) loss = 687.75 W
• Total losses = 361 + 460 + 687.75 = 1508.75 W
• Output = 9200 − 1508.75 = 7691.25 W (about 10.3 hp)

η = 7691.25 / 9200 = 0.836

Answer: (i) iron and friction losses = 687.75 W; (ii) efficiency = 83.6 %.