Boundary value problem in heat conduction

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Boundary value problem in heat conduction

Boundary value problem in heat conduction

For steady state heat conduction the temperature distribution in one-dimension is governed by the Laplace equation:

$$ \nabla^2 T = 0$$

with boundary conditions that at $T(x=a) = T_A$ and $T(x=L) = T_B$. To solve this in Matlab, we need to convert the second order differential equation into a system of first order ODEs, and use the bvp5c command to get a numerical solution.

the analytical solution is not difficult here: $T =  T_A-\frac{T_A-T_B}{L}x$. but the bvp5c syntax takes a lot to get used to.

For this problem, lets consider a slab that is defined by x=0 to x=L, with T(x=0) = 100, and T(x=L) = 200. We want to find the function T(x) inside the slab.

To get the system of ODEs, let $T_1 = T$, and $T_2 = T_1'$. That leads to $T_2' = T_1''$ and the following set of equations:

$\frac{dT_1}{dx} = T_2$

$\frac{dT_2}{dx} = 0$

with boundary conditions $T_1(0) = 100$ and $T_1(L) = 200$. By inspection you can see that whatever value $T_2$ has, it will remain constant since the derivative of $T_2$ is equal to zero.

function main

we start by defining an initial grid to get the solution on

x = linspace(0,5,10);

Next, we use the bvpinit command to provide an initial guess using the guess function defined below

solinit = bvpinit(x,@guess);

we also need to define the ode function (see odefun below).

and a boundary condition function (see bcfun below).

now we can get the solution.

sol = bvp5c(@odefun,@bcfun,solinit)
sol = 

    solver: 'bvp5c'
         x: [0 0.5556 1.1111 1.6667 2.2222 2.7778 3.3333 3.8889 4.4444 5]
         y: [2x10 double]
     idata: [1x1 struct]
     stats: [1x1 struct]

sol is a Matlab struct with fields. the x field contains the x-values of the solution as a row vector, the y field contains the solutions to the odes. Recall that T1 is the only solution we care about, as it is the original T(x) we were interested in.

x = sol.x;
T1 = sol.y(1,:);
T2 = sol.y(2,:);
plot(x,T1); figure(gcf)

% for demonstration we add the analytical solution to the plot
hold on
Ta = 100; Tb = 200; L = 5;
plot(x,Ta - (Ta-Tb)/L*x,'r--')
legend 'Numerical bvp5c' 'analytical solution'

As we would expect, the solutions agree, and show that the temperature changes linearly from the temperature at x=0 to the temperature at x=L.

the stats field gives you some information about the solver, number of function evaluations and maximum errors

ans = 

    nmeshpoints: 10
         maxerr: 9.8021e-016
      nODEevals: 106
       nBCevals: 9

function Tinit = guess(x)
% this is an initial guess of the solution. We need to provide a guess of
% what the function values of T1(x) and T2(x) are for a value of x. In some
% cases, this initial guess may affect the ability of the solver to get a
% solution. In this example, we guess that T1 is a constant and average
% value between 100 and 200. In this example, it also does not matter what
% T2 is, because it never changes.
T1 = 150;
T2 = 1;
Tinit = [T1; T2]; % we return a column vector

function dTdx = odefun(x,T)
% here we define our differential equations
T1 = T(1);
T2 = T(2);

dT1dx = T2;
dT2dx = 0;

dTdx = [dT1dx; dT2dx];

function res = bcfun(T_0,T_L)
% this function defines the boundary conditions for each ode. In this
% example, T_O = [T1(0)  T2(0)], and T_L = [T1(L) T2(L)]. The function
% computes a residual, which is the difference of what the current solution
% at the boundary conditions are vs. what they are supposed to be.  We want
% T1(0) = 100; and T1(L)=200. There are no requirements on the boundary
% values for T2 in this example.
res = [100 - T_0(1);
       200 - T_L(1)];

% categories: ODEs
% tags: math, heat transfer
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