Solución - prof.usb.ve.

Solución Tarea No. 3 CT-4311 (10/07/09)
Programa principal
maq.m
%principal
global a Ld Lq Ldf Ldad Lqaq L LW R invL
Re=0.01;
Rf=0.01;
Rad=0.02;
Raq=0.02;
R=diag([Re,Re,Rad,Raq,Rf],0);
Ld=1;Lq=0.6;Lf=1.05;Ldf=0.9;
Lad=1;Laq=0.6;
Ldad=0.93;Lqaq=0.53;Lfad=0.85;Lqad=0.80;Ldaq=0.40;
L=[Ld,0,Ldad,0,Ldf;0,Lq,0,Lqaq,0;Ldad,0,Lad,0,Lfad;0,Lqaq,0,Laq,0;Ldf,0,Lfad,0,Lf];
invL=inv(L);
LW=[0,­Lq,0,­Ldaq,0;Ld,0,Lqad,0,Ldf;0,0,0,0,0;0,0,0,0,0;0,0,0,0,0];
a=exp(j*2*pi/3);
tspan=[0,1000];
i0=[0,0,0,0,0,0,0];
[T,Y]=ode45(@modmaq,tspan,i0);
figure(1)
plot(T/377,Y(:,1))
xlabel('t [s]')
ylabel('Corriente i\_d [pu]')
grid
print ­dpng figura1
figure(2)
plot(T/377,Y(:,2))
xlabel('t [s]')
ylabel('corriente i\_q [pu]')
grid
print ­dpng figura2
figure(3)
plot(T/377,Y(:,3))
xlabel('t [s]')
ylabel('corriente i\_ad [pu]')
grid
print ­dpng figura3
figure(4)
plot(T/377,Y(:,4))
xlabel('t [s]')
ylabel('corriente i\aq [pu]')
grid
print ­dpng figura4
figure(5)
plot(T/377,Y(:,5))
xlabel('t [s]')
ylabel('corriente de campo i\_f [pu]')
grid
print ­dpng figura5
1/4 (13/07/09)
figure(6)
plot(T/377,Y(:,6))
xlabel('t [s]')
ylabel('velocidad angular \omega_m [pu]')
grid
print ­dpng figura6
Te=zeros(length(T));
ia=zeros(length(T));
for k=1:length(T)
Te(k)=((Ld*Y(k,1)+Ldf*Y(k,5)+Ldad*Y(k,3))*Y(k,2))­((Lq*Y(k,2)+Lqaq*Y(k,4))*Y(k,1));
ia(k)=sqrt(2/3)*real((Y(k,1)+j*Y(k,2))*exp(j*Y(k,7)));
end
figure(7)
plot(T/377,Te)
xlabel('t [s]')
ylabel('Par eléctrico T\_e [pu]')
grid
print ­dpng figura7
figure(8)
plot(T/377,ia)
xlabel('t [s]')
ylabel('Corriente de la fase a [pu]')
grid
print ­dpng figura8
Modelo de la máquina:
modmaq.m
function ve=modmaq(t,y)
global a Ld Lq Ldf Ldad Lqaq L LW R invL
Te=((Ld*y(1)+Ldf*y(5)+Ldad*y(3))*y(2))­((Lq*y(2)+Lqaq*y(4))*y(1));
Tm=(0.3+(0.7*((y(6))^2)));
if (t>400) % Sincronización con corriente de campo
Vf=1/100;
else
Vf=0;
end
if (t>600) % Aplicación del cortocircuito en bornes
VE=0;
else
VE=1.154700538379252*[1,a,a^2]*[sin(t);sin((t)­(2*pi/3));sin((t)+(2*pi/3))];
end
VEr=VE*exp(­j*y(7));
Vd=real(VEr);Vq=imag(VEr);
RWL=R+LW*y(6);
I=invL*([Vd;Vq;0;0;Vf]­(RWL*[y(1);y(2);y(3);y(4);y(5)]));
ve=zeros(7,1);
ve(1)=I(1);
ve(2)=I(2);
ve(3)=I(3);
ve(4)=I(4);
ve(5)=I(5);
ve(6)=(Te­Tm)/500;
ve(7)=y(6);
2/4 (13/07/09)
Resultados:
3/4 (13/07/09)
4/4 (13/07/09)