Solve taking too long, how to optimize code?

Basically, what the title asks. Code is here:

k = 180;
length = .05;
b = .01;
T0 = 180;
Tinf = 25;
h = 25;
Tsurr = 290;
epsilon = 0.9;
theta = atan((b/2)/length);
stef = 5.67*10^(-8);
xchange = b;

syms T1 T2 T3 T4 T5

eqn1 = (1-0.5*(xchange/length))*(T0 - T1) + (1-1.5*(xchange/length))*(T2-T1) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T1) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T1+273)^4) == 0;
eqn2 = (1-1.5*(xchange/length))*(T1 - T2) + (1-2.5*(xchange/length))*(T3-T2) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T2) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T2+273)^4) == 0;
eqn3 = (1-2.5*(xchange/length))*(T2 - T3) + (1-3.5*(xchange/length))*(T4-T3) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T3) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T3+273)^4) == 0;
eqn4 = (1-3.5*(xchange/length))*(T3 - T4) + (1-4.5*(xchange/length))*(T5-T4) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T4) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T4+273)^4) == 0;
eqn5 = 2*k*(xchange/2)*tan(theta)*((T4-T5)/xchange) + 2*h*((xchange/2)/cos(theta))*(Tinf - T5) + 2*epsilon*stef*((xchange/2)/cos(theta))*(Tsurr^4 - (T5 + 273)^4 ) == 0;

sol = solve([eqn1, eqn2, eqn3, eqn4, eqn5], [T1, T2, T3, T4, T5]);

T1sol = double(sol.T1)
T2sol = double(sol.T2)
T3sol = double(sol.T3)
T4sol = double(sol.T4)
T5sol = double(sol.T5)

Also, how long would this theoretically take to solve?

Thanks.

Answer 1

[Matt J]

Well, one solution is clearly,

T1=T2=T3=T4=T5=(Tsurr-273)

as you can verify by direct subsitution.

[Walter Roberson]

This is not one of the 1024 roots that my code comes up with.

Answer 2

[Alan Weiss]

Another point of view in case Matt's solution is not satisfactory: work completely in floating point computation rather than symbolic math computation. If you use the Problem-Based Optimization Workflow, you will basically have the same equations as before, simply using optimization variables rather than symbolic variables.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

Answer 3

[Walter Roberson]

k = 180;
length = .05;
b = .01;
T0 = 180;
Tinf = 25;
h = 25;
Tsurr = 290;
epsilon = 0.9;
theta = atan((b/2)/length);
stef = 5.67*10^(-8);
xchange = b;
syms T1 T2 T3 T4 T5
eqn1 = (1-0.5*(xchange/length))*(T0 - T1) + (1-1.5*(xchange/length))*(T2-T1) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T1) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T1+273)^4) == 0;
eqn2 = (1-1.5*(xchange/length))*(T1 - T2) + (1-2.5*(xchange/length))*(T3-T2) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T2) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T2+273)^4) == 0;
eqn3 = (1-2.5*(xchange/length))*(T2 - T3) + (1-3.5*(xchange/length))*(T4-T3) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T3) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T3+273)^4) == 0;
eqn4 = (1-3.5*(xchange/length))*(T3 - T4) + (1-4.5*(xchange/length))*(T5-T4) + ((h*xchange^2)/(k*length*sin(theta)))*(Tinf - T4) + ((epsilon*stef*xchange^2)/(k*length*sin(theta)))*(Tsurr^4 - (T4+273)^4) == 0;
eqn5 = 2*k*(xchange/2)*tan(theta)*((T4-T5)/xchange) + 2*h*((xchange/2)/cos(theta))*(Tinf - T5) + 2*epsilon*stef*((xchange/2)/cos(theta))*(Tsurr^4 - (T5 + 273)^4 ) == 0;

E1 = [eqn1, eqn2, eqn3, eqn4, eqn5];
%{
partial2 = solve(E1(1), T2)
E2 = subs(E1(2:end), T2, partial2)
partial3 = solve(E2(1), T3)
E3 = subs(E2(2:end), T3, partial3)
partial4 = solve(E3(1), T4)
E4 = subs(E3(2:end), T4, partial4)
partial5 = solve(E4(1), T5)
E5 = subs(E4(2:end), T5, partial5)
%}

partials = solve(E1(1:4), [T2, T3, T4, T5]);
E5 = subs(lhs(E1(end))-rhs(E1(end)), partials);

At this point you can expand(E5) to get a plain polynomial in T1.

According to Maple, that plain polynomial is in degree 1024.

You should be able to take the following steps, but they take too much time for me to be able to run them and show the output here

tic
P = sym2poly(E5);
R = roots(P);
mask = imag(R) == 0;

T1_sol = R(mask);
T2_sol = double(subs(partials.T2, T1, T1_sol));
T3_sol = double(subs(partials.T3, T1, T1_sol));
T4_sol = double(subs(partials.T4, T1, T1_sol));
T5_sol = double(subs(partials.T5, T1, T1_sol));
toc

T1_sol
T2_sol
T3_sol
T4_sol
T5_sol

[Walter Roberson]

T1_sol =

-9069.32334661296
-919.882898288149
177.024444908286
-24.4166503435627


T2_sol =

27738.4267958556
-2336.41842690512
174.081315968202
-287.461625512644


T3_sol =

7095829.16820471
-4126.23669567827
171.16801338191
-657.54975176523


T4_sol =

4.81617135399466e+16
-2960.81576251875
168.277608495836
-1280.43369676003


T5_sol =

3.06585398161644e+56
3425.69627315686
165.36383335676
-3127.23531356248