A 10 kW, 250 V, dc shunt motor with an armature resistance of 0.8 Ω and a field resistance of 275 Ω takes 3.91 A, when running light at rated voltage and rated speed. (a) What conclusions can you draw from the above data regarding machine losses? (b) Calculate the machine efficiency as a

Question

A 10 kW, 250 V, dc shunt motor with an armature resistance of [latex] 0.8 \Omega [/latex] and a field resistance of [latex] 275 \Omega [/latex] takes 3.91 A, when running light at rated voltage and rated speed.

(a) What conclusions can you draw from the above data regarding machine losses?

(b) Calculate the machine efficiency as a generator when delivering an output of 10 kW at rated voltage and speed and as a motor drawing an input of 10 kW. What assumption if any do you have to make in this computation?

(c) Determine the maximum efficiencies of the machine when generating and when motoring.

Accepted Answer

(a) Shunt field loss, [latex] P_{s h}=\frac{(250)^{2}}{275}=227.3 W [/latex]

Rotational loss [latex] =P_{r o t}=250 imes 3.91-(3.91)^{2} imes 0.8=965 W [/latex]

(b) Generator

[latex]I_{L} =\frac{10 imes 10^{3}}{250}=40 A ; I_{f}=\frac{250}{275}=0.91 A [/latex]

[latex] I_{a} =40+0.91=40.91 A [/latex]

[latex] P_{L} =965+227.3+(40.91)^{2} imes 0.8=2.53 kW [/latex]

[latex] \eta_{G} =1-\frac{2.53}{10+2.53}=79.8 \% [/latex]

Motor

[latex] I_{a} =I_{L}-I_{f}=40-0.91=39.1 A [/latex]

[latex] P_{L} =965+227.3+(39)^{2} imes 0.8=415 kW [/latex]

[latex] \eta_{M} =1-\frac{2.415}{10}=75.85 \% [/latex]

Assumption: Stray-load loss has been neglected.

(c) The condition for maximum efficiency is

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

[latex] 0.8 I_{a} =965+227.3=1192.3 W [/latex]

or [latex] I_{a} =38.6 A [/latex]

Total loss [latex] P_{L} =2 imes 1192.3=2384.6 W [/latex]

Generator

[latex] I_{L} =I_{a}-I_{f}=38.6-0.91=37.69 A [/latex]

[latex] P_{ ext {out }} =250 imes 37.69=9422.5 W [/latex]

[latex] \eta_{G}(\max ) =I-\frac{2384.6}{9422.5+2384.6}=79.8 \% [/latex]

Motor

[latex] I_{L} =I_{a}+I_{f}=38.6+0.91=39.51 A[/latex]

[latex] P_{ ext {in }} =250 imes 39.51=9877.5 W [/latex]

[latex] \eta_{M}(\max ) =1-\frac{2384.6}{9877.5}=75.85 \% [/latex]

The MATLAB program for the Example 7.57 is shown below.

Question 7.57: A 10 kW, 250 V, dc shunt motor with an armature resistance o...

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