Question 7.59: A 600 V dc shunt motor drives a 60 kW load at 900 rpm. The s...

P_{\text {out }}  =60 kW

=10.59 W

I_{L} =\frac{70.59 \times 10^{3}}{600}= 117.65 A

I_{f} =\frac{600}{100}= 6 A

I_{a} =117.65-6= 111.65 A

E_{a} =600-111.65 \times 0.16= 582.14 V

n =900 rpm (given)

(a)                             P_{L} =I_{a}^{2} R_{a}+P_{\text {rot }}+P_{\text {sh }}

10.59 \times 10^{3} =(111.65)^{2} \times 0.16+P_{\text {rot }}+600 \times 6

Or                           P_{\text {rot }} =4995 W \text { or } 4.995 kW

(b) No load

Armature resistance loss can be ignored

Rotational loss = P_{rot} = 4995 W (loss in nearly independent of speed)

Input power, P_{0} \approx P_{ rot }= 4995 W

At no load I_{ao} R_{a}  drop can be neglected. Therefore

E_{a o} \approx 600 V

Speed regulation =\frac{927.6-900}{900} \times 100=3.07 \%

(c)                                       E_{a}=K_{a} \Phi \omega

Or                             K_{a} \Phi=6.177 (constant as \Phi is constant)

Electromagnetic torque                  T=K_{a} \Phi I_{a}

600 =6.177 I_{a}

Or                          I_{a}=97.13 A