A dc shunt motor rated 10 kW connected to 250 V supply is loaded to draws 35 A armature current running at a speed of 1250 rpm. Given $R_{a} = 0.5 \Omega$

(a) Determine the load torque if the rotational loss is $500 \Omega$ .

(b) Determine the motor efficiency if the shunt field resistance is $250 \Omega$ .

(c) Determine the armature current for the motor efficiency to be maximum and its value. What is the corresponding load torque and speed?

Question

A dc shunt motor rated 10 kW connected to 250 V supply is loaded to draws 35 A armature current running at a speed of 1250 rpm. Given [latex] R_{a} = 0.5 \Omega [/latex]

(a) Determine the load torque if the rotational loss is [latex]500 \Omega [/latex].

(b) Determine the motor efficiency if the shunt field resistance is [latex] 250 \Omega [/latex].

(c) Determine the armature current for the motor efficiency to be maximum and its value. What is the corresponding load torque and speed?

Accepted Answer

[latex] E_{a}=250-0.5 imes 35=232.5 V[/latex]

Electromagnetic power, [latex] P_{e}=E_{a} I_{a}=P_{out}[/latex] (gross)

[latex] P_{ ext {out }}( ext { gross })=232.5 imes 35=[/latex] 8137.5 W

[latex] P_{ ext {rot }}=[/latex]500 W

[latex] P_{ ext {out }} ext { (net) }=8137.5-500=[/latex] 7637.5 W

Speed [latex] \omega=\frac{2 \pi}{60} imes 1250=130.9 rad / s [/latex]

Load torque, [latex] T_{L}=\frac{7637.5}{130.9}=58.35 Nm [/latex]

(b) [latex] I_{f}=\frac{250}{250}=[/latex] A

[latex] I_{L}=35+1=36[/latex] A

[latex] P_{ ext {in }}=250 imes 36=[/latex] 9000 W

Efficiency, [latex] \eta=\frac{P_{ ext {out }}( ext { net })}{P_{ ext {in }}}=\frac{7637.5}{9000}=8486 \% [/latex]

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

Or [latex] P_{k}=500+250 imes 1=750 W [/latex]

For maximum efficiency [latex] I_{a}^{2} R_{a}=P_{k} [/latex]

Or [latex] I_{a}=\sqrt{\frac{750}{0.5}}=38.73 A [/latex]

Total loss [latex] =2 imes 750=[/latex] 1500 W

[latex] I_{L}=38.73+1=[/latex] 39.73 A

[latex] P_{i n}=250 imes 39.73 A =[/latex] 9932.5 W

[latex] \eta_{\max }=1-\frac{1500}{9932.5}=84.9 \% [/latex]

Speed [latex] E_{a}=250-0.5 imes 38.73=[/latex] 230.64V

[latex] n=1250 imes \frac{230.64}{232.5}=1240 [/latex] rpm (marginally different)

[latex] \omega=\frac{2 \pi}{60} imes 1240=[/latex]129.85 rad / s

[latex] P_{ ext {out }}( ext { gross })=230.64 imes 38.73= [/latex] 8932.7 W

[latex] P_{ ext {out }}( ext { net })=8932.7-500=[/latex] 8432.7 W

[latex] T_{L}=\frac{8432.7}{129.85}=84.94 Nm [/latex]

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

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