please use phyton program. Use numerical methods to find the temperature distribution inside

the concrete slab in the following simple $1$D heat transfer problem

and compare with the analytical solution. Analytical solution steps are below.

A large, thin concrete slab of thickness $L$ is "setting." Setting is an

exothermic process that releases heat volumetrically, $q^{}(\frac{W}{m^3}).$ The

outside surfaces are kept at the ambient temperature, so $T_w=T_{}.$Since we will solve for steady state case, $T$ is a function of $x$ only: $T=T(x)$

Step $2$ :

Write the heat conduction equation, and simplify it

$\frac{de{l}^{2}T}{del{x}^2}+$ubrace$(\frac{de{l}^{2}T}{del{y}^2}+\frac{de{l}^{2}T}{del{z}^2}$ubrace${)}_{\{\begin{array}{ccc}\text{:}[=0,& since& ]\\ T=T(y& or& z)\end{array}}+\frac{(q^{})}{k}=$ubrace$(\frac{1}{}\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}$ubrace${)}_{\{\begin{array}{ccc}\text{:}[=0,& since& ]\\ & \text{s}\text{t}\text{e}\text{a}\text{d}\text{y}& \end{array}}$

Therefore, since $T=T(x),$ the heat conduction equation is reduced to an o$.$d$.$e$.$:

$\frac{d^{2}T}{d{x}^2}=-\frac{(q^{})}{k}$

Step $3$ :

Obtain a general solution of the reduced heat conduction equation

Simply integrate the o$.$d$.$e$.$ in step $2$ twice and we get:

$T=-\frac{(q^{})}{2k}{x}^2+C_{1}x+C_2$

$T=-\frac{(q^{})}{2k}{x}^2+C_{1}x+C_2$

Step $4$ :

Identify and write the initial and boundary conditions

This is the trickiest part. We must have a good understanding of the heat conduction problem

Normally:

Boundary condition:

Temperatures at $2$ different locations $($for all imes$)$

Initial condition:

Temperature at one point in time $($for all locations$)$

In our examples, we have $2$ boundary conditions:

$T(x=0)=T_w,$ and $,T(x=L)=T_w$

Step $5$ :

Substitute the general solution in the boundary and initial conditions and solve for the

integration constants

This process gets very complicated in the transient and muti$-$dimensional cases

Numerical methods are often needed to solve the problem

The steady $1-$D problems are usually easy

In our example, we get:

$T_w=-0+0+C_{2}so{C}_2=T_w$

$T_w=-\frac{(q^{})L^2}{2k}+C_{1}L+$ubrace$(C_2$ubrace${)}_{?}=T_{w}so{C}_1=\frac{(q^{})L}{2k}$

Step $6$ :

Plug in the integration constants back to get the solution of the problem

In our example, we can obtain the following solution:

$T=-\frac{(q^{})}{2k}{x}^2+\frac{(q^{})L}{2k}x+T_w$

Step $7$ :

If heat flux at any point is needed, substitute $T$ into Fourier's law

In our example, we can obtain the following heat flux at the wall:

$q_{\text{w}\text{a}\text{l}\text{l}}=-k\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{x}}{|}_x=0=k[\frac{(q^{})}{k}x-\frac{(q^{})L}{2k}{]}_x=0=-\frac{(q^{})L}{2}$pppppp