Question 7.60: A dc shunt motor rated 10 kW connected to 250 V supply is lo...

Electromagnetic power,             P_{e}=E_{a} I_{a}=P_{out}  (gross)

P_{\text {out }}(\text { gross })=232.5 \times 35= 8137.5 W

P_{\text {rot }}=500 W

P_{\text {out }} \text { (net) }=8137.5-500= 7637.5 W

Speed              \omega=\frac{2 \pi}{60} \times 1250=130.9 rad / s

Load torque,                     T_{L}=\frac{7637.5}{130.9}=58.35 Nm

(b)                           I_{f}=\frac{250}{250}=  A

I_{L}=35+1=36 A

P_{\text {in }}=250 \times 36= 9000 W

Efficiency,           \eta=\frac{P_{\text {out }}(\text { net })}{P_{\text {in }}}=\frac{7637.5}{9000}=8486 \%

(c) Constant loss,             P_{c}=P_{r o t}+P_{s h} ; P_{s h} is neglected

Or                           P_{k}=500+250 \times 1=750 W

For maximum efficiency                             I_{a}^{2} R_{a}=P_{k}

Or                                      I_{a}=\sqrt{\frac{750}{0.5}}=38.73 A

Total loss =2 \times 750= 1500 W

I_{L}=38.73+1= 39.73 A

P_{i n}=250 \times 39.73 A = 9932.5 W

Speed                               E_{a}=250-0.5 \times 38.73= 230.64V

n=1250 \times \frac{230.64}{232.5}=1240 rpm (marginally different)

\omega=\frac{2 \pi}{60} \times 1240=129.85 rad / s

P_{\text {out }}(\text { gross })=230.64 \times 38.73= 8932.7 W

P_{\text {out }}(\text { net })=8932.7-500= 8432.7 W