A 4-pole induction motor with inertia of 0.1 kgm² 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)

The correct answer is 1733.

Based on the provided graph, the speed of the 4-pole induction motor increases linearly from 1000 RPM at $t=4\text{s}$ to 1500 RPM at $t=8\text{s}$. This linear change occurs over a duration of 4 seconds ($\Delta t=4\text{s}$).

1. Equations of Motion and Power
The dynamic equation for the motor-load system is given by:
$J\frac{d\omega }{dt}=T_e-T_L$
where:
• $J=0.1{\text{kg·m}}^2$ is the rotor inertia,
• $T_L=2\text{N·m}$ is the constant load torque,
• $T_e$ is the electromagnetic torque developed by the motor,
• $\omega$ is the angular speed of the motor in rad/s.

Rearranging for electromagnetic torque:
$T_e=J\frac{d\omega }{dt}+T_L$

The mechanical power output (neglecting motor losses, which represents the input electrical power) is:
$P_e=T_e\omega =J\omega \frac{d\omega }{dt}+T_L\omega$

2. Finding Total Energy Consumed
The energy consumed during the speed change is the time integral of the power from $t=4\text{s}$ to $t=8\text{s}$:
$E={\int }_4^8{P}_{e}dt={\int }_4^8(J\omega \frac{d\omega }{dt}+T_L\omega )dt$
Splitting this into kinetic energy change ($E_k$) and energy consumed by the load ($E_L$):
$E=J{\int }_{{\omega }_1}^{{\omega }_2}\omega d\omega +T_L{\int }_4^8\omega dt$

Step 2a: Calculating Kinetic Energy Change ($E_k$)
Convert the initial and final speeds from RPM to rad/s:
${\omega }_1=\frac{2\pi \times 1000}{60}\approx 104.72\text{rad/s}$
${\omega }_2=\frac{2\pi \times 1500}{60}\approx 157.08\text{rad/s}$
Now evaluate the kinetic energy term:
$E_k=\frac{1}{2}J({\omega }_2^2-{\omega }_1^2)$
$E_k=\frac{1}{2}\times 0.1\times ({157.08}^2-{104.72}^2)\approx 685.39\text{J}$

Step 2b: Calculating Energy Dissipated by Load Torque ($E_L$)
Since the speed increases linearly with time, we can determine the average angular speed:
${\omega }_{\text{avg}}=\frac{{\omega }_1+{\omega }_2}{2}=\frac{104.72+157.08=261.80}{2}=130.90\text{rad/s}$
The energy transferred to the load over the 4-second interval is:
$E_L=T_L\times {\omega }_{\text{avg}}\times \Delta t$
$E_L=2\times 130.90\times 4=1047.20\text{J}$

Step 2c: Total Energy ($E$)
Adding both energy components:
$E=E_k+E_L=685.39+1047.20=1732.59\text{J}$
Rounding to the nearest integer gives 1733 Joules.