Solving System of two Differential Equations with one initial and one end condition

This is a code for the heat transfer in a heat exchanger. It solves the system of the two ODEs T_h_dx = ... and T_c_dx = ... for two initial conditions T_hot_in and T_cold_in.

How can this be solved, if T_cold_in should be the temperature at the end of the heat ecxhanger (x = length / end condition) and T_hot_in still the temperature at x = 0 (initial condition)?

Thank you!

% Functions for heat transfer to be solved by ode
    function T_calc = heat_transfer(~,T)
        T_h = T(1,1);
        T_c = T(2,1);

        T_h_dx = -1 / ((m_hot_plate  / 2) * c_water * R_width) * (T_h - T_c);         
        T_c_dx =  1 / ((m_cold_plate / 2) * c_water * R_width) * (T_h - T_c);      
        T_calc = [T_h_dx ; T_c_dx]; 
    end 
%__________________________________________________________________________                                                                                            
% solve ODE System
    [X, T_tot] = ode23(@(x,T) heat_transfer(x,T), [0 length],[T_hot_in T_cold_in]);

Answer 1

[Stephan]

Symbolic appears to work:

syms Th(x) Tc(x)
m_hp = 20;
m_cp = 5;
cw = 1;
Rw = 1;

eq(1) = diff(Th,x) == -1 / ((m_hp / 2) * cw * Rw) * (Th - Tc);
eq(2) = diff(Tc,x) ==  1 / ((m_cp / 2) * cw * Rw) * (Th - Tc);

conds = [Tc(10) == 20, Th(0) == 50];

sol = dsolve(eq,conds);

fplot(sol.Tc,[0 10])
hold on
fplot(sol.Th,[0 10])
hold off

[Daniel Schuhmacehr]

Thank you very much, that worked very well!

How could

How could be created two array out of sol.Th and sol.Tc with values between the limits 0 and 10?

[Stephan]

You could modify the conds and then solve in a loop and save the results in a cell array:

syms Th(x) Tc(x)
m_hp = 20;
m_cp = 5;
cw = 1;
Rw = 1;

solutions = cell(1,11);

eq(1) = diff(Th,x) == -1 / ((m_hp / 2) * cw * Rw) * (Th - Tc);
eq(2) = diff(Tc,x) ==  1 / ((m_cp / 2) * cw * Rw) * (Th - Tc);

for k=1:11
    conds = [Tc(10) == (k-1), Th(0) == 50]; % Tc(x=10) == 0...10 in 11 loop steps
    sol = dsolve(eq,conds);
    solutions{k} = sol;
end

to get the results from the cell arrray:

>> solutions{1}.Tc % for Tc(x=10) == 0
 
ans =
 
200/(exp(5) + 4) - (200*exp(-x/2)*exp(5))/(exp(5) + 4)
 
>> solutions{2}.Tc  % for Tc(x=10) == 1
 
ans =
 
(exp(5) + 200)/(exp(5) + 4) - (196*exp(-x/2)*exp(5))/(exp(5) + 4)

[Daniel Schuhmacehr]

Thanks again! And sorry, I made myself unclear.

Your first code gives a plot of Th and Tc over x. How can this be saved into an array instead. Doesn't matter which kind of array. I am not able to convert the syms into numbers. :/

[Stephan]

syms Th(x) Tc(x)
m_hp = 20;
m_cp = 5;
cw = 1;
Rw = 1;

eq(1) = diff(Th,x) == -1 / ((m_hp / 2) * cw * Rw) * (Th - Tc);
eq(2) = diff(Tc,x) ==  1 / ((m_cp / 2) * cw * Rw) * (Th - Tc);

conds = [Tc(10) == (20), Th(0) == 50];
sol = dsolve(eq,conds);

% You can make symbolic functions from the solutions
Th_sol(x) = sol.Th
Tc_sol(x) = sol.Tc

% and use them like you would use functions to write results into an array
x_vals = 0:0.1:10;
Th_vals = double(Th_sol(x_vals)); % double is not needed - but you want non symbolic...
Tc_vals = double(Tc_sol(x_vals));

% plot the array values
plot(x_vals,Tc_vals,x_vals,Th_vals)

[Daniel Schuhmacehr]

Thank you very much! Very helpful! The function double I was looking for!