Question 7.58: A 50-kW, 250 V, 1200 rpm dc shunt motor when tested on no-lo...

A 50-kW, 250 V, 1200 rpm dc shunt motor when tested on no-load at 250 V draw an armature current of 13.2 A, while its speed is 1215 rpm. Upon conducting other tests it is found that R_{a} = 0.06 \Omega and R_{f} = 50 \Omega  while V_{b} (brush voltage drop) = 2 V.

Calculate the motor efficiency at a shaft load of 50 kW at rated voltage with a speed of 1195 rpm. Assume that the stray load loss is 1% of the output.

What would be the load for the motor to have maximum efficiency and what would be its value?

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No-load test

Armature input  =250 \times 13.2 =3300 W

P_{r o t}=P_{i 0}+P_{f w}=3300-0.06 \times(13.2)^{2}-2 \times 13.2

= 3263 W

As speed varies very little from no-load to full-load P_{r o t} almost remains constant.

On-load

P_{\text {out }}=50 kW (at shaft)

Let the armature current be I_{a}. We write the power balance equation.

250 I_{a}-0.06 I_{a}^{2}-2 I_{a}-3263-\underbrace{0.01 \times 50 \times 10^{3}}_{P_{s t}}=50 \times 10^{3}

Or                             0.06 I_{a}^{2}-248 I_{a}+53763=0

Solving we get

I_{a} = 229.6 A

P_{\text {in }}=250 \times 229.6+\frac{(250)^{2}}{50}=58650 W

\eta=\frac{50000}{58650} \times 100=85.25 \%

We shall assume that P_{st} remains mainly constant in the range of load, we are investigating. Then

\eta=\frac{250 I_{a}-0.06 I_{a}^{2}-2 I_{a}-3763\left(P_{\text {out }}+P_{s t}\right)}{250 I_{a}+250 \times 5}

For maximum efficiency                     \frac{d \eta}{d I_{a}}=0

Solving we get            I_{a}=284 A

Substituting this value of I_{a}   in expression, we get

\eta_{\max }=86.36 \%

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