Question 7.61: A 250 V, 25 kW shunt motor has maximum efficiency of 89% at ...

Total loss,              P_{L}=\frac{P(\text { shaft })}{\eta}-P(\text { shaft })=\left(\frac{1}{\eta}-1\right) P(\text { shaft })

Power input,             P_{\text {in }}=20+2.472=22.472 kW

Line current,                  I_{L}=\frac{22.472}{250} \times 10^{3}= 89.89 A

I_{f}=\frac{250}{125}= 2 A

I_{a}=89.89-2= 87.89 A

At maximum efficiency

I_{a}^{2} R_{a}=P_{r o t}+P_{s h}=2472 / 2= 1236 W

P_{s h}=250 \times 2= 500 W

\therefore                                                 P_{r o t}=1236-500= 736 W

At speed,                                         n = 850 rpm

Now shaft load is raised till armature current becomes I_{a} = 100 A

Armature copper loss,               P_{c}=(100)^{2} \times 0.153= 1530 W

= 1530 + 1236 = 2766 W

P_{\text {in }}=250 \times I_{L}=250 \times(100+2)= 25.5 kW

\eta=\left(1-\frac{2.766}{25.5}\right) \times 100= 89.15

n \propto E_{a}, constant   I_{f}

\therefore                                                            n=850 \times \frac{234.7}{236.25}=844.4 rpm