obtaining p-v curves using Matlab

Question

0 个投票

hi, everyone, I am trying to Obtain P-V curves using Matlab can anyone help me through it, please

4 个评论
显示 2更早的评论隐藏 2更早的评论

Shahabullah Amin 2018-4-28

it's plotting the result in the not solving area which I have shown in the picture with the circle

Shahabullah Amin 2018-4-28

请先登录，再进行评论。

请先登录，再回答此问题。

Follow Question

Answer 1

Walter Roberson 2018-4-28

在 MATLAB Online 中打开

1 个投票

clc;
clear all
syms X
z=0.1+0.5*1j;
Vs=1;
A=1;
a1=real(A); a2=imag(A);
A=a1+a2*1j;
B=z;
b1=real(B); b2=imag(B);
C=0;
D=A;
fi=acos(1);
K1=a1*(b2-b1*tan(fi))+a2*(b1+b2*tan(fi));
K2=a1*(b1+b2*tan(fi))+a2*(b1-b2*tan(fi));
deltarcrit=(pi/4)+0.5*atan(K2/-K1);
Vrcrit=Vs/(2*(a1*cos(deltarcrit)+a2*sin(deltarcrit)));
K3=b1*cos(deltarcrit)+b2*sin(deltarcrit);
K4=a1*cos(deltarcrit)+a2*sin(deltarcrit);
Prcrit=((Vs^2)*(2*K3*K4-(a1*b1+a2*b2)))/((b1^2+b2^2)*4*K4);
Vr=[];
for P=0.1:0.01:1
        Qr=P*tan(fi);
        P1=a1^2+a2^2;
        P2=2*P*(a1*b1+a2*b2)+2*Qr*(a1*b2+a2*b1)-Vs^2;
       P3=((b1+b2).^2)*(P^2+Qr^2);
       equation=P1*(X^2)+P2*X+P3;
        these_roots = roots([P1 P2 P3]);
        mask = any(imag(these_roots) ~= 0,2);
        these_roots(mask,:) = nan;
        Vr=[Vr these_roots];
  end
        Pr=(0.1:0.01:1);
plot(Pr,Vr.')
display(Prcrit)

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Walter Roberson 2018-4-28

在 MATLAB Online 中打开

The area that it does not draw is the area where the roots go complex.

If you change the P loop to

        syms P
        Qr=P*tan(fi);
        P1=a1^2+a2^2;
        P2=2*P*(a1*b1+a2*b2)+2*Qr*(a1*b2+a2*b1)-Vs^2;
       P3=((b1+b2).^2)*(P^2+Qr^2);
       equation=P1*(X^2)+P2*X+P3;
       Vr = solve(equation, X);

then because you do not change anything other than P in the loop, you can get the general form, which is

equation = (9*P^2)/25 + X^2 + X*(P/5 - 1)

and then Vr is

   1/2 - (5^(1/2)*(-(7*P - 5)*(P + 1))^(1/2))/10 - P/10
 (5^(1/2)*(-(7*P - 5)*(P + 1))^(1/2))/10 - P/10 + 1/2

That has a term

(-(7*P - 5)*(P + 1))^(1/2)

so the equation is real-valued if -(7*P - 5)*(P + 1) is positive. When P is positive (as is the case in your for loop), P+1 is always positive. So real or imaginary is going for the root is going to have a boundary when 7*P - 5 becomes 0, which is P = 5/7 which is about 0.714285714285 . Below that you have real roots; above that you have only imaginary roots.

请先登录，再进行评论。