Solving non linear equations
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Ghanim
el 2 de Feb. de 2024
Comentada: Ghanim
el 6 de Feb. de 2024
Hi all,
The folloiwng code solves non linear equations for T1, T2, T3 and T4 as well as for J1, J2 and J3. I am only interested on the tempreture:
It returns an array solution that includes several answers for each T. How Can I obtain the exact solution (one single soution) for each T?
syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
Je = 330.136;
A2 = 449200;
A4 = 519000;
Fms = 0.305;
Fm1 = 0.45;
Fme = 0.245;
F1s = 0.610;
F1m = 0.389;
eps = 0.85;
K2 = 15;
L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;
eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];
sol = solve(eqs, vars);
T1_val = real(double(sol.T1))
T2_val = real(double(sol.T2))
T3_val = real(double(sol.T3))
T4_val = real(double(sol.T4))
0 comentarios
Respuesta aceptada
Torsten
el 3 de Feb. de 2024
Editada: Torsten
el 3 de Feb. de 2024
Each of the 16 quadruples (T1(i),T2(i),T3(i),T4(i)) (i = 1,...,16) constitutes a solution for your system of equations.
You must check which of the 16 quadruples are physical. E.g. the first quadruple (with the corresponding values for J1, J2 and J3 printed) would be
syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
Je = 330.136;
A2 = 449200;
A4 = 519000;
Fms = 0.305;
Fm1 = 0.45;
Fme = 0.245;
F1s = 0.610;
F1m = 0.389;
eps = 0.85;
K2 = 15;
L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;
eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];
sol = solve(eqs, vars);
J1_val = real(double(sol.J1(1)))
J2_val = real(double(sol.J2(1)))
J3_val = real(double(sol.J3(1)))
T1_val = real(double(sol.T1(1)))
T2_val = real(double(sol.T2(1)))
T3_val = real(double(sol.T3(1)))
T4_val = real(double(sol.T4(1)))
Más respuestas (1)
Walter Roberson
el 3 de Feb. de 2024
syms J1 J2 J3 T1 T2 T3 T4
Jm = 5077.12;
Js = 301.32;
Je = 330.136;
A2 = 449200;
A4 = 519000;
Fms = 0.305;
Fm1 = 0.45;
Fme = 0.245;
F1s = 0.610;
F1m = 0.389;
eps = 0.85;
K2 = 15;
L2 = 0.03;
eq1 = (Jm - Js)*(A2*Fms) + (Jm - J1)*(A2*Fm1) + (Jm - Je)*(A2*Fme) == 0;
eq2 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (J1 - Js)*(A4*F1s) + (J1 - Jm)*(A4*F1m) == 0;
eq3 = -(5.67e-8*T1^4 - J1)*(A4*eps)/(1-eps) + (T1-T2)*K2*A4/L2 == 0;
eq4 = -(T1 - T2)*K2/L2 + (5.67e-8*T2^4 - J2)*eps/(1-eps) == 0;
eq5 = -(5.67e-8*T2^4 - J2)*eps/(1-eps) + (J2 - J3) == 0;
eq6 = -(J2-J3) + (5.67e-8*T3^4 - J3)*eps/(1-eps) + 185.95 == 0;
eq7 = -(5.67e-8*T3^4 - J3)*eps/(1-eps) + (T3 - T4)*K2/L2 == 0;
eqs = [eq1, eq2, eq3, eq4, eq5, eq6, eq7];
vars = [J1, J2, J3, T1, T2, T3, T4];
sol = solve(eqs, vars);
vals = double(subs([T1, T2, T3, T4], sol));
valid_vals = vals(all(vals > 0, 2),:)
Ver también
Categorías
Más información sobre Partial Differential Equation Toolbox en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!