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I am solving 20 nonlinear equation using fsolve to find 20 variables' value. But i previously know one of variable's value. That variable is x(17) in eqn F(17). x(17)=1200. Based on this fixed variable i want to find remaining variables' value.

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chirag  rohit
chirag rohit on 1 Jan 2019
Answered: Alex Sha on 28 Feb 2020
function F = root16d(x)
Qs_ref=0;
Pw_ref=0.322;
F(1) = -0.5*((x(1)*x(2))-(x(3)*x(4)))-Qs_ref; %Qr=0.5*((x(5)*x(8))-(x(7)*x(6))
F(2) = 0.5*((x(1)*x(4))+(x(3)*x(2)))+0.5*((x(5)*x(6))+(x(7)*x(8)))+Pw_ref;
%% Algebric Equation
F(3) = 0.5*((x(1)*x(9))+(x(3)*x(10)))+0.5*((x(1)*x(4))+(x(3)*x(2)))-0.5*((x(11)*x(15))+(x(12)*x(16)));
F(4) = 0.5*((x(1)*x(2))-(x(3)*x(4)))+0.5*((x(1)*x(10))-(x(3)*x(9)))+0.5*((x(11)*x(16))-(x(12)*x(15)));
%% SS of Tr line
delta=0.1049;
Rl=0.02; we=2*pi*60;
Xl=0.5; wb=2*pi*60;
Xc=0.75*0.5;
EB=1;
Ebq=EB*cos(delta);
Ebd=EB*sin(delta);
%
F(5) = -(Rl*wb/Xl)*x(9)-we*x(10)-(wb/Xl)*x(13)+(wb/Xl)*(x(1)-Ebq);
F(6) = we*x(9)-(Rl*wb/Xl)*(x(10))-(wb/Xl)*x(14)+(wb/Xl)*(x(3)-Ebd);
F(7) = (wb*x(9)*Xc)-(we*x(13));
F(8) = (wb*x(10)*Xc)+(we*x(13));
%% two aglebric equation
Xtg=0.3;
F(13)= x(1)-x(11)+(Xtg*x(16));
F(14)= x(3)-x(12)-(Xtg*x(15));
F(15)= x(15)-x(4)-x(9);
F(16)= x(16)-x(2)-x(10);
%% Power flow equation for DC capacitor placed between GSC and RSC
C=50*14000e-6;%C=1;
F(17)= (100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*1200));%100e6**x(17)
%% State Space model of Generator Turbine shaft model
Dt=0; Ht=4.29;
Dtg=1.5; Hg=0.9; Ktg=0.15;
Xm=3.95279; Tw=0.322/(0.75);
Wbase=2*pi*60; Te=0.5*Xm*(((x(4)+x(6))*x(8))-((x(2)+x(8))*x(6)));
F(18)= ((-Dt-Dtg)/(2*Ht))*x(18)+((Dtg/(2*Ht))*x(19))-(x(20)/(2*Ht))+(Tw/(2*Ht));
F(19)= ((Dtg/(2*Hg))*x(18))+((-Dt-Dtg)/(2*Hg))*x(19)+(x(20)/(2*Hg))-(Te/(2*Hg));
F(20)=(Ktg*Wbase)*x(18)-(Ktg*Wbase)*x(19);
%% State Space model of DFIG
Rs = 0.00488; Wb= 2*pi*60%2*pi*60;
Rr = 0.00549; We= 2*pi*60;%2*pi*60;
Xls = 0.09231; Wr= 2*pi*60*x(19);%2*pi*60*x(19)
Xlr = 0.09955;
Xm = 3.95279;
Xss = Xls+Xm; Xrr = Xlr+Xm;
F(9)=x(2)*((Xm^2*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (We*(Xlr + Xm)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(8)*((We*Xm*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xlr + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(1)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(5))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rs*Wb*x(4)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rr*Wb*Xm*x(6))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(10)= x(6)*((We*Xm*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xlr + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(4)*((Xm^2*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (We*(Xlr + Xm)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(3)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(7))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rs*Wb*x(2)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rr*Wb*Xm*x(8))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(11)=x(2)*((We*Xm*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + x(8)*((We*Xm^2)/(Xlr*Xls + Xlr*Xm + Xls*Xm) - ((Xlr + Xm)*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(5)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(1))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rr*Wb*x(6)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rs*Wb*Xm*x(4))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(12)=(Wb*x(7)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - x(6)*((We*Xm^2)/(Xlr*Xls + Xlr*Xm + Xls*Xm) - ((Xlr + Xm)*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(4)*((We*Xm*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - (Wb*Xm*x(3))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rr*Wb*x(8)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rs*Wb*Xm*x(2))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
  1 Comment
Walter Roberson
Walter Roberson on 2 Jan 2019
Please do not close questions that have an Answer.
I spent a number of hours investigating this for you, and it is frustrating to have my work vanish.

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Accepted Answer

Walter Roberson
Walter Roberson on 1 Jan 2019
x(17) does not appear in your function, so setting a particular value for it does not make any difference.
What does make a difference is that you have 20 equations in 19 variables (since x(17) is unused). That is either going to have no solutions or an infinite number of solutions. My tests suggest that this particular set of equations has no consistent solutions.
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More Answers (1)

Alex Sha
Alex Sha on 28 Feb 2020
The equation F(17) seems to be meanless:
F(17)= (100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*x(17)))
equal to:
(100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*x(17)))=0
equal to:
(100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))=0
equal to:
(0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16)))=0
equal to:
x(5)*x(6)+x(7)*x(8)+x(11)*x(15)+x(12)*x(16)=0
so x(17) actually do nothing, it leads to 19 variables but 20 equations, and there is no idea solution.

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