GATE 2022 EE Q.63: solution and answer

GATE 2022 · EEQ.63 · 2 marks · NAT

A 4-pole induction motor with inertia of $0.1 kg-m^{2}$ drives a constant load torque of $2 Nm$ . The speed of the motor is increased linearly from $1000 rpm$ to $1500 rpm$ in $4$ seconds as shown in the figure below. Neglect losses in the motor. The energy, in joules, consumed by the motor during the speed change is ______. (round off to nearest integer)

GATE 2022 EE question 63: A speed-vs-time graph. — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

ANSWER

The energy consumed by the motor during the speed change is 1733 J.

Official answer: 675 to 700, or 1725 to 1740Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Convert the given speeds to angular velocity: $ω_{1} = \frac{2 π \times 1000}{60} = 104.720 rad/s$ and $ω_{2} = \frac{2 π \times 1500}{60} = 157.080 rad/s$ .
From the graph, the speed rises linearly only over the interval $t = 4 s$ to $t = 8 s$ (a duration of $4 s$ ), so the average angular speed during this ramp is $ω_{a v g} = \frac{ω _{1} + ω _{2}}{2} = 130.900 rad/s$ .
The angle turned during the ramp is $θ = ω_{a v g} \times t = 130.900 \times 4 = 523.599 rad$ .
Energy delivered to overcome the constant load torque over this angle: $E_{l o a d} = T_{L} θ = 2 \times 523.599 = 1047.198 J$ .
Change in the rotor's stored kinetic energy as the speed rises: $Δ K E = \frac{1}{2} J (ω_{2}^{2} - ω_{1}^{2}) = \frac{1}{2} (0.1) (157.08 0^{2} - 104.72 0^{2}) = 685.389 J$
Since losses in the motor are neglected, the total energy the motor consumes during the speed change equals the sum of the load work and the kinetic-energy change: $E = E_{l o a d} + Δ K E = 1047.198 + 685.389 = 1732.587 J$ .
Rounding to the nearest integer: $E \approx 1733 J$ .

KEY IDEA

During the linear speed ramp, the motor's electrical output energy (losses neglected) must equal the sum of (1) the mechanical work done against the constant load torque over the angle turned, and (2) the increase in the rotor's stored kinetic energy: $E = T_{L} θ + \frac{1}{2} J (ω_{2}^{2} - ω_{1}^{2})$ , where $θ$ is the angle turned only during the 4 s ramp.

WATCH OUT FOR

Forgetting that the motor must supply BOTH the energy absorbed by the constant load torque (work done against the load, $T_{L} θ$ ) AND the change in the rotor's kinetic energy ( $Δ K E$ ) during the acceleration — computing only one of the two terms is the usual error. A second common slip is using the wrong duration (4 s, the ramp portion only, not 8 s, the full graph) for $θ$ .

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 17 Induction machines questions, with notes

GATE 2022 EE Q.63: solution and answer

Answer and solution

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