GATE 2023 EE Q.48: solution and answer

GATE 2023 · EEQ.48 · 2 marks · NAT

A three phase 415 V, 50 Hz, 6-pole, 960 RPM, 4 HP squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by 20%, the resultant speed of the motor in RPM (neglecting the stator leakage impedance and rotational losses) is __________ (Round off to the nearest integer).

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Answer and solution

ANSWER

The resultant speed is 760 RPM.

Official answer: 760Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Synchronous speed at rated 50 Hz, 6-pole: $n_{s 1} = \frac{120 \times 50}{6} = 1000 RPM$ .
Rated slip: $s_{1} = \frac{n _{s 1} - n _{1}}{n _{s 1}} = \frac{1000 - 960}{1000} = 0.04$ .
Neglecting the stator leakage impedance, the per-phase equivalent circuit reduces to the rotor branch $R_{2} / s$ in series with $X_{2}$ across the supply voltage $V$ : $T = \frac{3}{ω _{s}} \cdot \frac{V ^{2} ( R _{2} / s )}{( R _{2} / s ) ^{2} + X _{2}^{2}}$ Since the rated slip (0.04) is small, $(R_{2} / s)^{2} ≫ X_{2}^{2}$ , giving the low-slip approximation $T \approx \frac{3 V ^{2} s}{ω _{s} R _{2}}$ , i.e. $T \propto \frac{V ^{2} s}{f}$ (using $ω_{s} \propto f$ ).
The problem specifies that supply voltage and frequency are BOTH reduced by 20%, so $V \propto f$ is given directly by the data (not an added assumption). Hence $V^{2} \propto f^{2}$ , and $T \propto \frac{f ^{2} s}{f} = f s$ — torque is proportional to the slip frequency.
A constant torque load therefore requires the slip frequency to stay the same: $s_{1} f_{1} = s_{2} f_{2}$ .
New frequency: $f_{2} = 0.8 \times 50 = 40 Hz$ (voltage is also $0.8$ of rated, consistent with the given data).
Solve for the new slip: $s_{2} = \frac{s _{1} f _{1}}{f _{2}} = \frac{0.04 \times 50}{40} = 0.05$ .
New synchronous speed: $n_{s 2} = \frac{120 \times 40}{6} = 800 RPM$ .
Resultant speed: $n_{2} = n_{s 2} (1 - s_{2}) = 800 \times (1 - 0.05) = 760 RPM$ .

KEY IDEA

Neglecting stator impedance, the low-slip torque approximation gives $T \propto \frac{V ^{2} s}{f}$ ; since $V$ and $f$ are both scaled down by the same 20% (V/f held constant), torque becomes proportional to the slip frequency $s f$ , so a constant-torque load keeps $s f$ unchanged across the frequency change.

WATCH OUT FOR

Assuming the slip stays at its rated value (0.04) and only recomputing the new synchronous speed with the reduced frequency, forgetting that a constant-torque load forces the slip to change (via constant slip-frequency $s f$ ) when both V and f drop together.

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GATE 2023 EE Q.48: solution and answer

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