GATE 2018 EE Q.54: solution and answer

GATE 2018 · EEQ.54 · 2 marks · NAT

The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are $R_{1} = R_{2}^{'} = X_{l 1} = X_{l 2}^{'} = 12 Ω$ , $X_{M} = 240 Ω$ and $s$ is the slip. At no-load, the motor speed can be approximated to be the synchronous speed. The no-load lagging power factor of the motor is____________ (up to 3 decimal places).

GATE 2018 EE question 54: Single-phase induction motor per-phase equivalent circuit (double revolving field model). — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

The no-load lagging power factor is 0.106.

Official answer: 0.104 to 0.112Source: Official answer key, GATE 2018 EE (IIT Guwahati)

At no-load, the motor runs essentially at synchronous speed, so the slip s → 0.
Forward-field branch: as s → 0, the rotor term $\frac{R _{2}^{'}}{2 s} \to \infty$ (open circuit), so this branch reduces to just the magnetizing reactance $j \frac{X _{M}}{2} = j \frac{240}{2} = j 120 Ω$ .
Backward-field branch: as s → 0, $2 - s \to 2$ , so the rotor resistance term becomes $\frac{R _{2}^{'}}{2 ( 2 - s )} = \frac{12}{4} = 3 Ω$ , in series with $j \frac{X _{l 2}^{'}}{2} = j 6 Ω$ , giving $3 + j 6 Ω$ ; this is in parallel with $j \frac{X _{M}}{2} = j 120 Ω$ .
Compute the backward-branch impedance: $Z_{b} = \frac{( j 120 ) ( 3 + j 6 )}{3 + j 6 + j 120} = \frac{- 720 + j 360}{3 + j 126} = 2.720 + j 5.779 Ω$
Total input impedance seen at no-load: $Z = R_{1} + j X_{l 1} + j \frac{X _{M}}{2} + Z_{b} = 12 + j 12 + j 120 + (2.720 + j 5.779) = 14.720 + j 137.779 Ω$
Magnitude: $∣ Z ∣ = 14.72 0^{2} + 137.77 9^{2} = 138.564 Ω$
No-load power factor (lagging, since the net reactance is inductive): $cos ϕ = \frac{R}{∣ Z ∣} = \frac{14.720}{138.564} = 0.106$

KEY IDEA

At no-load the slip s → 0, so in the forward-field branch of the double-revolving-field equivalent circuit the rotor term $R_{2}^{'} / (2 s)$ tends to infinity (open circuit), leaving only the magnetizing reactance $j X_{M} /2$ in the forward branch; in the backward-field branch, $2 - s \to 2$ , so it stays a finite parallel combination that must be computed and added to the series stator and forward-branch impedances to get the total no-load impedance and hence the power factor.

WATCH OUT FOR

Forgetting that the backward-field branch does NOT vanish or simplify away at no load (only the forward branch's rotor resistance term goes to infinity because of s → 0 in the denominator; the backward branch's denominator is $2 - s \to 2$ , a finite number), so students either drop the backward branch entirely (giving a much higher, wrong power factor) or incorrectly send $2 - s \to 0$ as well.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 17 Induction machines questions, with notes

GATE 2018 EE Q.54: solution and answer

Answer and solution

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