Week-9 Challenge: Induction Motor-2

1. Given below is the equivalent circuit of an induction motor with stator and rotor circuits separated. The frequency of current in both circuits are different. Suppose the frequency of the rotor circuit is f2 and frequency of stator circuit f1 and let the slip value be s.

f2=s*f1

This means the frequency of the rotor circuit will be varying by a factor of s with respect to f1.

In the following figure, V1 is the applied voltage to the stator and E1 is the induced emf. R1 and X1 are stator resistance and reactance respectively. I1 is the current flowing through the stator circuit. Io is the no load current which splits into Im which is the magnetizing component and Iw which is the core loss current. Xo is pure inductive reactor and Ro a non-inductive resistor.

E2 is the induced emf in the rotor circuit, R2 and X2 are the rotor resistance and reactance respectively.

Given below is the equivalent circuit with reference to the stator. This is made possible by making the frequencies of stator and rotor circuits equal. Here, the resistance is split into two parts: R'2 and R'2(1-s/s). R'2(1-s/s) is the electrical equivalent of the mechanical power used up by the rotor and R'2 indicates the pure copper loss of rotor.

2. Given X1=X2= 0.1 ohm and R1=R2= 0.2 ohm.

The equation for torque is:

T=(3/Wms)[((V^2(R2/s))/((R1+R2/s)^2+(X1+X2)^2)]

where T=torque, WmsWms= synchronous speed, V=voltage per phase, s=slip, R1=stator resistance,

R2=rotor resistance, X1=stator reactance and X2= rotor reactance.

Since its the ratio of starting torques we need to find out we can say that slip, s=1. At half of rated voltage and frequency, let voltage be V and synchronous speed be Wms. At rated voltage and frequency, let voltage be 2V and frequency be 2Wms. Substituting the values of X1, X2, R1 and R2 the equations become:

Thus we can say that the ratio of starting torques is 0.5

3. Given below is the MATLAB code for plotting speed-torque characteristics for frequency control method.

clear all
close all 
clc
p=4; % number of poles are set at 4
s=linspace(0,1,100); % slip value
X1=1.106; % stator reactance
X2=0.464; % rotor reactance
R1=0.641; % stator resistance
R2=0.332; % rotor resistance 
k=linspace(0.7,2,5); % k is for controlling frequency
V=400/sqrt(3); % voltage per phase
for j=1:5
f=50.*k(j);
Wms=(120*f)/p; % synchronous speed calculation
for i=1:100  
Wm(i)=(1-s(i))*(Wms); % rotor speed calculation
denom=((R1+((R2)./(s(i)))).^2)+((X1+X2)^2); % denominator for torque calculation
T(i)=((3/Wms)*((V^2)*(R2)./s(i))./(denom)); % torque calculation
T(1)=0;
end
plot(Wm,T) % plotting rotor speed and torque 
hold on
xlabel('Speed(rpm)')
ylabel('Torque(N-m)')
end

CODE EXPLANATION

The speed-torque characteristics is plotted based on the following equation:

T=(3/Wms)[((V^2(R2/s))/((R1+R2/s)^2+(X1+X2)^2)]

where T=torque, Wms= synchronous speed, V=voltage per phase, s=slip, R1=stator resistance,

R2=rotor resistance, X1=stator reactance and X2= rotor reactance.

The characteristics are plotted by varying the slip of the rotor using a variable 's'. The speed-torque characteristics for different frequencies are plotted by multiplying a variable 'k' with 50 Hz frequency.

The speed-torque characteristics plot is shown below. As we can see, the speed increases with increasing frequency but the value of torque comes down.

4.Given below is the schematic diagram of a BLDC motor

In a BLDC motor the rotor is a permanent magnet and the stator has a coil arrangement similar to the image above. The coils are energized by applying a DC current and they are magnetized. The coils will be magnetized with an opposite polarity to the permanent magnet of the rotor. For example, A1 will be a south pole and A2 will be north pole when the position of the rotor is similar to the above image. Then C2 will become a south pole and C1 a north pole which makes the rotor move to the left. Thus the rotor keeps on rotating as long as the coils are magnetized.

If a coil has a particular polarity at an instant, it will have a reversed polarity when it is energized again. For example, A1 will become a north pole to attract the south pole of the permanent magnet rotor and similary A2 will become a south pole.