Studying Induction Motor characteristics using MATLAB

1.Discuss about the equivalent circuit network of induction motor MATLAB model.

One way to analyse and understand the operation of an induction motor is by the use of an equivalent circuit.

But first we to need understand about few parameters

Slip

Induction motors are asynchronous. The stator magnetic field rotates at the motors synchronous speed (n_s). The rotor can never rotate at synchronous speed, otherwise there would be no induced current.

The difference between the motors synchronous speed(`n_s`) and actual asynchronous speed(`n_r`) is known as the slip.

therefore, Slip `S=(n_s-n_r)/n_s`

Frequency

From magnetic winding theory, the relationship between frequency in the stator (f), number of pole pairs (p) and the synchronous speed is given by

`f=n_sp`

Euqivalent Induction Motor Circuit

In the equivalent circuit R₁ represents, the resistance of the stator winding and X₁ the stator leakage reactance (flux that does not link with the air gap and rotor). Magnetising reactance required to cross the air gap is represented by X_m and core losses (hysteresis and eddy current) by R_c.

An ideal transformer of N1 and N2 turns respectively represents the air gap. For the rotor side, the induced emf is affected by the slip (as the rotor gains speed, slip reduces and less emf is induced). The rotor resistance and reactance are represented by R₂ and X₂; with X₂ being dependant on the frequency of the inductor rotor emfs.

The rotor circuit, the current I₂ is given by:

`I_2=(sE_2)/sqrt((R_2)^2+(sX_2)^2`

`I_2=(E_2)/sqrt((R_2/s)^2+(X_2)^2`

this circuit can be further simplified and represented as

The reffered values are calculated by multiplying them by `k^2`, where k is the effective stator/rotor turns ratio.

Torque equation

`T=3/(omega_(ms))[(V^2((R_2)/S))/((R_1+R_2/S)^2+(X_1+X_2)^2]]`

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2. 3 phase, 50 Hz induction motor, represented by equivalent circuit constants X1 = X2 = 0.1 ohm and R1 = R2 = 0.2 ohm is operated at half of rated voltage and frequency. Calculate the ratio of starting torque at half voltage & frequency to rated values.

Given
Frequency - 50 Hz

Stator inductance X1 = 0.1

Rotor inductance X2 = 0.1

Stator resistance R1 = 0.2 ohm

Rotor resistance R2 = 0.2 ohm

No of phases - 3

Torque `T=3/(omega_(ms))[(V^2((R_2)/S))/((R_1+R_2/S)^2+(X_1+X_2)^2]]`

now,

Rated frequency `f` = 50Hz

Rated synchronous speed

`omega_(m)^' = (120*f)/p`

`omega_(m)^' = (120*50)/p`

`omega_(m)^' = (6000)/p`

Half rated frequency `f^'` = 25Hz

Half rated synchronous speed

`omega_(ms)^' = (120*f^')/p`

`omega_(ms)^' = (120*25)/p`

`omega_(ms)^' = (3000)/p`

Rated torque

`T=3/(omega_(ms))[(V^2(R_2))/((R_1+R_2)^2+(X_1+X_2)^2]]`

`T=3/(omega_(ms))[(V^2(0.2))/((0.2+0.2)^2+(0.1+0.1)^2]]`

`T=(3*V^2)/(omega_(ms))`

Half rated torque

`T^'=1/2*3/(omega_(ms))[(V^2(R_2))/((R_1+R_2)^2+(X_1+X_2)^2]]`

`T^'=1/2*3/(omega_(ms))[(V^2*0.2)/((0.2+0.2)^2+(0.1+0.1)^2]]`

`T^'=(1.5*V^2)/(omega_(ms))`

Now, to find ratio

`T^'/T=((1.5*V^2)/(omega_(ms)))/((3*V^2)/(omega_(ms)))`

`T^'/T=1/2=0.5`

ratio of starting torque at half voltage & frequency to rated values is 0.5

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3. Make a MATLAB script file which will plot speed torque characteristics for frequency control method.

clear all
close all
clc

% Data inputs
p = 4; % No of poles 
s=linspace(0,1,100); % slip
R1 = 0.2; % stator resistance
R2 = 0.2; % rotor resistance
X1 = 0.1; % stator inductance
X2 = 0.1; % rotor inductance
V = 400/sqrt(3); % voltage per phase
f = linspace(0.5,2,5); % frequency

for i=1:5
    % synchronous speed rad/sec
    wms = (120*f(i))/p;
    for j=1:100
        % torque N-m
        wm(j)=(1-s(j)).*(wms);
        d=((R1+((R2)./(s(j)))).^2)+((X1+X2)^2);
        T(j)=((3/wms)*((V^2)*(R2)./s(j)./(d)));
    end
    
    %plotting
    plot(wm, T,'linewidth',2)
    hold on
    title('Motor speed vs Torque')
    xlabel('Speed in rad/sec')
    ylabel('Torque in N-m')
    grid on
end

Output

By seeing the graph we can observe that with increasing frequency the speed of the motor also increases.

But torque reduces with increasing speed.

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4. Explain in brief about operation of BLDC motor.

Brushless DC motors

A brushless DC motor consists of a rotor in the form of a permanent magnet and stator in the form of polyphase armature windings. It differs from the conventional dc motor in such that it doesn’t contain brushes and the commutation is done using electrically, using an electronic drive to feed the stator windings.

Basically a BLDC motor can be constructed in two ways- by placing the rotor outside the core and the windings in the core and another by placing the windings outside the core. In the former arrangement, the rotor magnets act as an insulator and reduce the rate of heat dissipation from the motor and operate at low current. It is typically used in fans. In the latter arrangement, the motor dissipates more heat, thus causing an increase in its torque. It is used in hard disk drives.

4 Pole 2 Phase motoroperation

Working principle

The magnetic field generated by the permanent magnets interacts with the field induced by the current in the stator windings, creating a mechanical torque. The electronic switching circuit or the drive switches the supply current to the stator so as to maintain a constant angle of 0 to 90 degrees between the interacting fields. Hall Sensors are mostly mounted on the stator or on the rotor. When the rotor passes through the hall sensor, based on the North or South Pole, it generates a high or low signal. Based on the combination of these signals, the winding to be energized is defined. In order to keep the motor running, the magnetic field produced by the windings should shift position, as the rotor moves to catch up with the stator field.