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Solving non linear equations

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Ghanim
Ghanim 2024-2-2
评论： Ghanim 2024-2-6
Ran in:
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))
T1_val = 16×1
684.3838 684.3838 684.3838 684.3838 0 0 0 0 0 0
T2_val = real(double(sol.T2))
T2_val = 16×1
666.9262 666.9262 666.9262 666.9262 -17.4576 -17.4576 -17.4576 -17.4576 -17.4576 -17.4576
T3_val = real(double(sol.T3))
T3_val = 16×1
456.1979 -0.0000 -0.0000 -456.1979 508.4679 508.4679 42.8456 42.8456 -42.8456 -42.8456
T4_val = real(double(sol.T4))
T4_val = 16×1
439.1121 -17.0857 -17.0857 -473.2836 491.3822 491.3822 25.7598 25.7598 -59.9313 -59.9313

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采纳的回答

Torsten
Torsten 2024-2-3
编辑：Torsten 2024-2-3
Ran in:
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)))
J1_val = 1.0899e+04
J2_val = real(double(sol.J2(1)))
J2_val = 9.6771e+03
J3_val = real(double(sol.J3(1)))
J3_val = 948.2529
T1_val = real(double(sol.T1(1)))
T1_val = 684.3838
T2_val = real(double(sol.T2(1)))
T2_val = 666.9262
T3_val = real(double(sol.T3(1)))
T3_val = 456.1979
T4_val = real(double(sol.T4(1)))
T4_val = 439.1121

更多回答（1 个）

Walter Roberson
Walter Roberson 2024-2-3
Ran in:
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),:)
valid_vals = 1×4
684.3838 666.9262 456.1979 439.1121

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