A 250 V, 25 kW shunt motor has maximum efficiency of 89% at shaft load of 20 kW and speed of 850 rpm. The field resistance is 125 Ω. Calculate the rotational loss and armature resistance. What will be the efficiency, line current and speed at an armature current of 100 A?

Question

A 250 V, 25 kW shunt motor has maximum efficiency of 89% at shaft load of 20 kW and speed of 850 rpm. The field resistance is 125 [latex] \Omega [/latex]. Calculate the rotational loss and armature resistance. What will be the efficiency, line current and speed at an armature current of 100 A?

Accepted Answer

Total loss, [latex] P_{L}=\frac{P( ext { shaft })}{\eta}-P( ext { shaft })=\left(\frac{1}{\eta}-1 ight) P( ext { shaft }) [/latex]

[latex] P_{L}=\left(\frac{1}{0.89}-1 ight) imes 20=2.472 kW ext { or } 2472 W[/latex]

Power input, [latex] P_{ ext {in }}=20+2.472=22.472 kW[/latex]

Line current, [latex] I_{L}=\frac{22.472}{250} imes 10^{3}=[/latex] 89.89 A

[latex] I_{f}=\frac{250}{125}=[/latex] 2 A

[latex] I_{a}=89.89-2=[/latex] 87.89 A

At maximum efficiency

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

[latex] R_{a}=\frac{1236}{(89.89)^{2}}=0.153 \Omega [/latex]

[latex] P_{s h}=250 imes 2=[/latex] 500 W

[latex] herefore [/latex] [latex] P_{r o t}=1236-500=[/latex] 736 W

[latex] E_{a}=250-89.89 imes 0.153=236.25 V [/latex]

At speed, n = 850 rpm

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

Armature copper loss, [latex] P_{c}=(100)^{2} imes 0.153=[/latex] 1530 W

[latex] P_{L}=P_{c}+P_{r o t}+P_{s h} [/latex]

= 1530 + 1236 = 2766 W

[latex] P_{ ext {in }}=250 imes I_{L}=250 imes(100+2)=[/latex] 25.5 kW

[latex] \eta=\left(1-\frac{2.766}{25.5} ight) imes 100=[/latex] 89.15

[latex] E_{a}=250-100 imes 0.153=234.7 V [/latex]

[latex] n \propto E_{a}[/latex], constant [latex] I_{f} [/latex]

[latex] herefore [/latex] [latex] n=850 imes \frac{234.7}{236.25}=844.4 rpm [/latex]

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

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