Induction Motor characteristics

Week 8 Challenge: Induction Motor Characteristics

Q1) How induction motor operation is similar to clutch?

Q2) Calculate the starting time of a drive with following parameters:

J=10 kg-m²,

Tm = 15 + 0.5wm

T_l=5+0.6ω_m

Q3) A drive has following equations for motor and load torques:

Tm=(1+2ω_m)

T_{l= 3} √ω_m

Obtain the equillibrium points and determine the stability.

ANS 1:

Working of Clutch:

In a manual transmission car, a clutch is a component that is needed to change the gear to transition from one speed range to another in a smooth manner. When the driver wants to change the gear in order to achieve a higher or lower speed, the driver has to first press the clutch pedal and make the gear transition while the clutch is pressed down. Once the gear change has taken place, the clutch can be released and the vehicle can continue to operate at the new speed range.

In manual transmission vehicles, to make a smooth gear change means to discontinue the power flow from the engine to the transmission. But it is not practical to switch off the engine for the purpose of a gear change while the vehicle is running. So, a clutch solves this issue as it can discontinue the engine power flow to the transmission without turning off the engine.

The main part of a clutch consists of a disc that is coated with high friction material on either side and it is connected to the engine flywheel. If an external force acts on the clutch disc, the flywheel will also turn the flywheel due to friction. So, when an external force is applied to to the clutch disc, it will get transmitted to the transmission system. This external force is provided by a pressure plate system, the cover of which is fixed firmly to the flywheel.

But for power disengagement, a diaphragm spring is used and it is housed between the pressure plate and its cover. The outer portion of the diaphragm spring is connected to the pressure disc. When the clutch is pressed, a hydraulic system converts the clutch motion to the centre of the diaphragm spring. If an external force acts on the centre of the diaphragm spring, the pressure disc will move away from the friction disc and discontinue power flow to the transmission. When the clutch pedal is released, the power flow continues flowing to the transmission.

Induction Motor & Clutch

So, the clutch is used to engage and disengage the power transmission from the driving shaft to the driven shaft. The driven shaft i.e. disc is located between the engine and the pressure plate. When the clutch is engaged, the driven disc is pressed against the flywheel transferring torque from the engine to the transmision. The driving disc i.e. clutch cover is rigidly bolted to the engine. When the clutch is engaged, it's inner part will press the pressure disc which in turn presses the driven disc to the flywheel. The 2 shafts are rotating at different speeds which causes a slip and clutch attempts to lock them together or release them. The slip that exists between the driving and driven shafts is given by:

`omega_2 = omega_1 (1-s)`

where,

driving shaft speed = `omega_1`

driven shaft speed = `omega_2`

slip = s

The operation of a clutch is similar to that of an induction motor in terms of the slip. In an induction motor, there is a slip that occurs between the speed of the rotating magnetic field i.e. the synchronous speed and the speed of the rotor. This slip is essential for the working of the induction motor because it is the rotating magnetic field exerts an electromagnetic force on the rotor causing it to rotate, producing a torque. This creates a small phase difference between the two. If they were rotating at the same speed, there would be no excitation of the rotor as there would be no interaction between the rotating magnetic field and the rotor. So, the speed of the rotating magnetic field i.e. synchronous speed is always greater than the speed of the rotor. Slip is defined mathematically with the equation below.

`Slip, s = (N_s - N_r) / N_s`

where,

Synchronous speed - `N_s`

Rotor speed - `N_r`

Slip - s

This can be rearranged to be written with the rotor speed as the subject:

`N_r = N_s (1-s)`

From this, we can see that the clutch and the induction motor have some similarities in their operation.

ANS 2:

To find the starting time of the induction motor drive, we need to first find the equilibrium point for the drive. This will be a common angular speed at which both the motor and the load will operate at. This common angular speed is also known as the steady state speed.

To find the steady state speed, we need to equate the load torque and motor torque equations:

Given Data:

J=10 kg-m²,

Tm = 15 + 0.5wm

T_l=5+0.6ω_m

`T_m = T_l`

`15 + (0.5*omega_m) = 5 + (0.6*omega_m)`

`0.1*omega_m = 100`

Therefore,

`omega_m = 100`

The steady state speed is calculated to be 100 radians per second and this has also been verified using a MATLAB program where the 2 torque-speed relations for the motor and the load were plotted and their intersection point was obtained. The script is shown below:

clear all
close all
clc

% Motor Torque - y1
% Motor Speed - x1
% Load Torque - y2

% Motor Speed Array

x = linspace(0, 200, 400);

y1 = 15 + 0.5*x;
y2 = 5 + 0.6*x;

% Plotting

plot(x, y1, 'linewidth', 2, 'color', 'b');
hold on
plot(x, y2, 'linewidth', 2, 'color', 'r');
title('Speed Vs. Torque')
xlabel('Motor Speed, rpm')
ylabel('Torque, Nm')
legend('Motor Torque', 'Load Torque')

This is the output plot of the program:

As we can see from the graph, the 2 curves intersect at a motor speed of 100 radians per seconds which verifies that this is the steady state speed from the first calculation.

Then, to find the starting time to reach the common angular speed, we need to solve a differential equation. To do this, we need to know the fundamental equation relating the motor torque and the load torque.

Newton's second law states that the resultant force is equal to the product of the mass in to the acceleration.

• `F = ma`

In the case of a motor, this force is the turning force called the torque, T and it is equal to the product of the moment of intertia, J in to the angular acceleration.

• `T = J * (domega_m)/dt`

But, the resultant torque is the difference between the motor torque and the load torque.

• `T_m - T_l = J * (domega_m)/dt`

where T is measured in Nm,

J is measured in `(kg)/m^2`

Angular speed is measured in radians per second

Using the third equation, we will get a differential equation that we need to solve to find the starting time:

Substituting the motor and torque load equations:

`(15 + 0.5*omega_m) - (5 + 0.6*omega_m) = 10 * (domega_m)/dt`

`10 - 0.1 = 10 * (domega_m)/dt`

`(domega_m)/dt = 1 - 0.01omega_m`

This is a first order differential equation and to solve it, I wrote a MATLAB program that uses the 'ode45' solver. This solver requires a function to be created and this was done in the second section of code.

The section of code below is the main program.

clear all
close all
clc

t_span = linspace(0, 600, 600);

w_0 = [0,0];

[t, w]  = ode45(@im2, t_span, w_0);

plot(t, w, 'linewidth', 2, 'color', 'red')
grid on
ylabel('Motor Speed, rad/s')
xlabel('Time, s')
title('Motor Speed Vs. Time')

The section of code below was written in a separate script as it is a function and is needed for the 'ode45' solver.

function [dw_dt] = im2(t, w)

dw_dt = 1 - 0.01*w;

end

The program will solve the differential equation and create an array of the different solutions for the time, t and the angular speed, w. Then, it will plot these results in a graph which is shown below:

We want to know the time taken from rest to attain the steady state speed which will mark the end of the transient period. In theory, the steady state speed is never achieved because it takes infinte time to do so. In practice, we want to know the time taken from rest to 95% of the steady state speed since there is negligible difference in the time duration after this point. As we can see from the graph above, to reach the angular speed of 95% of 100 radians per second i.e. 95 radians per second is approximately t = 299 or 300 seconds.

Using Integration

The time taken to attain steady speed can also be found mathematically using integration. This is shown below:

We know the fundamental equation:

`T_m - T_l = J * (dw_m)/dt`

Substituting the load and motor torque equations and collecting like terms, we get:

`(domega_m)/dt = 1 - 0.01omega_m`

Rearranging this equation and making 'dt' the subject, we will get:

`dt = 1/(1-0.01omega_m) * domega_m`

The integral limits that will be used will be the speed range for which we want to find the time taken to achieve. In this case integrating this differential from limits 0 to 95 radians per second, we can find the time taken to reach the steady state speed.

`t = int_0^95 1/(1-0.01omega_m)*domega_m`

`t = -100ln|omega_m - 100| + C`

Under initial conditions (t = 0 and speed = 0), the constant is 0. We have to substitute the limits in to the equation and subtracting the lower limit term from the upper limit term. Since, we cannot find the natural logarithm of negative values, we need to take the modulus of those values.

`t = [-100ln(5)] - [-100ln(100)]`

`t = [-160.9] - [-460.5]`

`t = 299.62 s`

So, we can see that we obtain the same value for the time taken to achieve the steady state speed from rest as we did for the graphical method.

ANS 3:

To find the equilibrium points and the stability of the drive, we need to first equate the equations of the motor and load torques.

The 2 equations are:

Motor Torque : `T_m = 1 + (2*omega_m)`

Load Torque : `T_l = 3*sqrt(omega_m`

By equating the two:

`1 + (2*omega_m) = 3*sqrt(omega_m)`

To get rid of the square root, both sides are squared. Then, the brackets are expanded and the like terms are collected and a quadratic equations is obtained:

`4*(omega_m)^2 - (5*omega_m) + 1 = 0`

Solving this quadratic equation, we will obtain 2 solutions:

`omega_m = 0.25`

`omega_m = 1`

This is verified by a MATLAB program I created which plots the curves of both motor and load's torque-speed characteristics:

clear all
close all
clc


% Motor Speed Array

x = linspace(0, 5, 20);


% Motor Torque

y1 = 1 + 2*x;


% Load Torque

y2 = 3*(x.^0.5);


% Plotting

plot(x, y1, 'linewidth', 2)
hold on
plot(x, y2, 'linewidth', 2)
title('Speed Vs. Torque characteristics')
xlabel('Speed, rpm')
ylabel('Torque, Nm')
legend('Motor Torque', 'Load Torque')

The output graph of this script is shown below:

As we can see from the graph, the 2 curves intersect at 2 points which means there are 2 solutions. The 2 curves intersect exactly at x = 0.25 rad/s and x =1 rad/s.

Now that we have the 2 equilibrium points, we need to determine at what speed the drive is stable.

The stability condition is:

`(dT_l)/(domega_m) - (dT_m)/(domega_m) > 0`

In other words, if the difference between the differential equations of the load and torque is greater than 0, the condition is satisfied.

If this condition is satisfied for the given equilibrium points, then it means that the induction motor is stable to operate at those speeds.

To use this condition, we need to find the differentiate the load torque and motor torque equations with respect to the angular speed.

First, we will differentiate the load torque equation:

`(dT_l)/(domega_m) = 3*0.5 * omega_m^(0.5-1)`

`(dT_l)/(domega_m) = 3/2 * omega_m^-0.5`

Then, we will differentiate the motor torque equation:

`(dT_m)/(domega_m) = 2`

The differential of the motor torque equation is just a constant, 2.

So, the stability equation for this drive is:

`(3/2*(omega_m)^-0.5) - 2 > 0`

For this equation, we will substitute the 2 equilibrium points and check if the condition is satisfied.

• For `omega_m = 0.25`

`= 3/2 * (0.25)^-0.5 - 2`

`= 3`

So, when the angular speed is 0.25 rad/s, the stability condition is satisfied, because the difference in the differential between the motor and the load torque is greater than 0. Now we can check when the angular speed is 1.

• For `omega_m = 1`

`= 3/2 * (1)^-0.5 - 2`

`= 1.5 - 2`

`= -0.5`

So when the angular speed is 1 rad/s, the stability condition is not satisfied, because the difference in the differential between the motor and the load torque is lesser than 0.

So, the induction motor drive is only stable at 0.25 rad/s.