The document describes developing a program to plot the reactive power of an induction generator at different speeds and for three injected voltages: 20V lagging 100 degrees, 0V, and 20V leading 80 degrees. The program calculates the reactive power Qt at various rotor speeds from 960rpm to 1560rpm. The results show that the induction generator can either supply or consume reactive power depending on the injected voltage magnitude and phase. This reactive power control improves grid flexibility compared to other wind turbine types.

1. %%%Spencer Minder
%%%0420537
%%%EE 551
%HW 6
% Develop a program to plot the reactive power at the terminals of the
induction
% generator versus speed. Test your program using the following data:
%
% ns=1200 rpm
% r2'=0.01 ohm
% r1=0.01 ohm
% x1=0.1 ohm
% x2'=0.1 ohm
% xm=5 ohm
%
% Va1=690 V (line-to-line). Use Va1 as a reference vector
%
% Plot the reactive power for the three injected voltages:
%
% va2'= 20V at 100 degree lagging
% va2'=0 V
% va2'=20 at 80 degree leading
%
% the speed range should be from 0.8 ns to 1.3 ns
%Slip will vary depending on the speed
%Speed of the rotor will vary from 960rpm to 1560rpm
ns=1200;
nr=960:1:1560;
s=(ns-nr)/ns;
%Finding the Thevenin circuit values
r2prime=0.01;
r1=0.01;
x1=0.1;
x2prime=0.1;
xm=5;
Vll=690;
Va1_LN=690/sqrt(3);
%Finding thevenin impedance Zth
zth=((xm*1j)*(r1+x1*1j))/(r1+(x1+xm)*1j);
%Finding thevenin reactance xth
xth=imag(zth);
%Finding thevenin resistance Rth
rth=real(zth);
%Finding equivalent reactance Xeq
xequiv=xth+x2prime;
%Finding thevenin voltage Vth
vth=((xm*1j)*(Va1_LN))/(r1+(x1+xm)*1j);
%Solving for the Three different cases of Va2'
%First injected voltage 20V <100degrees lagging (converting to phasor form)
%Note: cosd and sind allows the use of degrees instead of radians
va2prime1=20*(cosd(-100)+sind(-100)*1j);

2. %second injected voltage 0V (converting to phasor form)
va2prime2=0;
%Third injected voltage 20V <80degrees leading (converting to phasor form)
va2prime3=20*(cosd(80)+sind(80)*1j)
%Finding the currents and Voltages needed to solve for reactive power
%Note have to find these values for each different case
%%%%Case 1
%Found on page 25 of chapter 8 in the book Ia2' is as follows
Ia2prime1=((va2prime1./s)-vth)./((rth+(r2prime./s))+xequiv*1j);
%Magnetizing voltage was found on page 49 of chapter 8
Vm1=(va2prime1./s)-(Ia2prime1.*((r2prime./s)+x2prime*1j));
Im1=Vm1/(xm*1j);
Ia1=Ia2prime1-Im1;
%Solving for reactive power Qt
%complex conjugate of Ia1
Ia1conj=conj(Ia1);
%Solving for terminal voltage of Va1
%Using the equation on page 27 of chapter 8
%Ia1=Ia2'+(Va1/jxm)
%Finding the reactive power
Qt1=3*imag(Va1_LN.*Ia1conj);
%%%%Case 2
Ia2prime2=((va2prime2./s)-vth)./((rth+(r2prime./s))+xequiv*1j);
%Magnetizing voltage was found on page 49 of chapter 8
Vm2=(va2prime2./s)-(Ia2prime2.*((r2prime./s)+x2prime*1j));
Im2=Vm2/(xm*1j);
Ia1_2=Ia2prime2-Im2;
%Solving for reactive power Qt
%complex conjugate of Ia1
Ia1_2conj=conj(Ia1_2);
%Solving for terminal voltage of Va1
%Using the equation on page 27 of chapter 8
%Ia1=Ia2'+(Va1/jxm)
%Finding the reactive power
Qt2=3*imag(Va1_LN.*Ia1_2conj);
%%%%Case 3
Ia2prime3=((va2prime3./s)-vth)./((rth+(r2prime./s))+xequiv*1j);
%Magnetizing voltage was found on page 49 of chapter 8
Vm3=(va2prime3./s)-(Ia2prime3.*((r2prime./s)+x2prime*1j));
Im3=Vm3/(xm*1j);
Ia1_3=Ia2prime3-Im3;
%Solving for reactive power Qt

3. %complex conjugate of Ia1
Ia1_3conj=conj(Ia1_3);
%Solving for terminal voltage of Va1
%Using the equation on page 27 of chapter 8
%Ia1=Ia2'+(Va1/jxm)
%Finding the reactive power
Qt3=3*imag(Va1_LN.*Ia1_3conj);
%Scaling the reactive power and putting it in terms of MVar we divide by
%10^6
plot(Qt1/10^6,nr,Qt2/10^6,nr,Qt3/10^6,nr);
grid on
title('Reactive power vs Rotor speed for injected Voltages');
xlabel('Reactive Power (MVAr)');
ylabel('Speed(rpm)');
(20 <‐100⁰) V
(20 <80⁰) V
(0<0⁰) V

4. Results:
Reactive power control from type 3 systems adds a huge amount of flexibility
to power grid. First off there is no need for Var compensation like in other
types of wind turbines(type I & II). Also this allows the wind farm to
either supply reactive power or consume unwanted reactive power to the grid.
Looking at the graphical results of the 3 different test cases we can get an
idea of what happens to the reactive power as different voltages are injected
into the system and at what speeds. For 20V <80deg(leading) the system is
consuming reactive power below synchronous speed(1200rpm) and supplying
reactive power above this speed. For the cases 0V<0deg the system consumes
reactive power with the max at synchronous speed(1200rpm).
In the case of 20V <-100deg(lagging), the Machine is consuming reactive power
and more so in the when the speed of the machine is above 1200 rpm.
By changing the value of the injected voltage into the type 3 system, this
changes the reactive power generated by the induction machine. This is done
by either changing the magnitude of the injected voltage (ie 20V or 0V) or
changing the angle between the Voltage and the current from leading to
lagging(Va1 and Ia1*). This allows more ranges of operation for the machine
whether you want to generate more reactive power or to consume reactive power
(ie where reactive power is positive). This has distinct advantages over
type I and type II wind turbines.