Question

Example 5.8. A 250 V shunt motor on no-load runs at 1000 r.p.m. and takes 5A. The total armature and shunt field resistances are 0.2 12 and 250 12 respectively. Calculate the speed when loaded and taking current of 50 A if armature reaction weakens the field by 3%.​

Answer 1

The field current is \frac{250}{250}=1

250

250

=1 A

Under no load condition the armature current is 5-1=45−1=4 A. Hence back emf is

250-4\times 0.2=249.2250−4×0.2=249.2 volts

Since the armature current is 4 A which is very small as compared to 50 A as far as armature reaction effect in concerned, hence we neglect the effect.

The back emf when 50 A is drawn by the motor

250-(50-1)\times 0.2=240.2250−(50−1)×0.2=240.2 volts

Now 249.2=K\Phi \times 1000249.2=KΦ×1000

240.2=0.97 KN

Hence \frac{0.97\, KN\Phi }{K\Phi \times 1000}=\frac{240.2}{249.2}

KΦ×1000

0.97KNΦ

=

249.2

240.2

or N=\frac{240.2}{249.2}\times \frac{1000}{0.97}\simeq 993.7N=

249.2

240.2

×

0.97

1000

≃993.7

\simeq 994≃994 rpm.

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