GATE 2024 EE Q.55: solution and answer

GATE 2024 · EEQ.55 · 2 marks · NAT

A 5 kW, 220 V DC shunt motor has 0.5 Ω armature resistance including brushes. The motor draws a no-load current of 3 A. The field current is constant at 1 A. Assuming that the core and rotational losses are constant and independent of the load, the current (in amperes) drawn by the motor while delivering the rated load, for the best possible efficiency, is _______ (rounded off to 2 decimal places).

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Answer and solution

ANSWER

The motor draws about 27.29 A from the supply when it delivers its rated 5 kW output. This is the smaller root of the armature power-balance quadratic, and it is the one that gives the higher efficiency.

Official answer: 27.19 to 27.39Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

No-load armature current. The line current is $3 A$ and the shunt field takes $I_{f} = 1 A$ , so $I_{a 0} = 3 - 1 = 2 A$
Core and rotational losses. At no load the output is zero, so the power the armature takes in is used only by the core + rotational losses and the armature copper loss: $P_{r o t} = V I_{a 0} - I_{a 0}^{2} R_{a} = 220 (2) - (2)^{2} (0.5) = 440 - 2 = 438 W$ The question says this stays the same at every load.
Power balance at rated load. The rated output is $5 kW = 5000 W$ (shaft output). The power converted in the armature, $E_{b} I_{a} = V I_{a} - I_{a}^{2} R_{a}$ , has to supply this output plus $P_{r o t}$ : $220 I_{a} - 0.5 I_{a}^{2} = 5000 + 438 = 5438$
Solve the quadratic. $0.5 I_{a}^{2} - 220 I_{a} + 5438 = 0 \Rightarrow I_{a} = \frac{220 \pm 22 0 ^{2} - 4 ( 0.5 ) ( 5438 )}{2 ( 0.5 )} = 220 \pm 48400 - 10876 = 220 \pm 37524$ $37524 = 193.711 \Rightarrow I_{a} = 26.289 A or 413.711 A$
Choose the root with the best efficiency. Both roots give the same $5000 W$ output. The input $V I$ is smaller for the smaller current, so efficiency is higher there. The larger root would waste $0.5 (413.7)^{2} \approx 85.6 kW$ as armature copper loss, so it is not a practical operating point. Take $I_{a} = 26.289 A$ .
Current drawn from the supply. $I = I_{a} + I_{f} = 26.289 + 1 = 27.289 \approx 27.29 A$ Check: input $= 220 \times 27.289 = 6003.6 W$ and efficiency $= 5000/6003.6 \approx 83.3%$ .

KEY IDEA

Get the constant rotational loss from the no-load test: $P_{r o t} = V I_{a 0} - I_{a 0}^{2} R_{a}$ . At rated load, the power converted in the armature, $V I_{a} - I_{a}^{2} R_{a}$ , must equal the rated output plus $P_{r o t}$ . This gives a quadratic in $I_{a}$ with two roots. The smaller root is the one with the best efficiency. Add the field current to get the line current.

WATCH OUT FOR

Reading "best possible efficiency" as the maximum-efficiency condition (variable loss = constant loss). That calculation ignores the fixed 5 kW rated load and gives 30.60 A, or 37.28 A if the field copper loss is counted as a constant loss. Neither current delivers 5 kW. The phrase only tells you which root of the rated-load quadratic to pick. Other slips: forgetting to subtract the field current from the no-load line current, and giving the armature current (26.29 A) instead of the line current.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 2 Losses & efficiency questions, with notes