A composite wall consist of three materials, as shown in fig. Convection heat transfer takes place on inner surface of the wall with $T_\infty=800^0c$ and $h = 30W/{m^2}^0c$. The outer surface temperature $T_0= 20^0c$. Determine temperature distribution in the wall.

(i) Number of nodes and elements

(ii) Element matrix equation

$$ \begin{Bmatrix} \begin{bmatrix} h_cA & 0 \\ 0 & 0 \end{bmatrix} + \frac{kA}{h_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \end{Bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

For element 1

h$_c$ = 30 W/m$^2$; A = 1m$^2$; K = 25; h$_e$ = 0.3

$ \therefore \frac{KA}{h_e} = \frac{25 \times 1}{0.3} = 83.33 $

$$ \begin{Bmatrix} \begin{bmatrix} 30 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 83.33 & -83.33 \\ -83.33 & 83.33 \end{bmatrix} \end{Bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

$$ \begin{bmatrix} 113.33 & -83.33 \\ -83.33 & 83.33 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

For element 2

h$_c$ = 0; A = 1m$^2$; K = 30; h$_e$ = 0.2

$ \therefore \frac{KA}{h_e} = \frac{30 \times 1}{0.2} = 150 $

$$ \begin{bmatrix} 150 & -150 \\ -150 & 150 \end{bmatrix} \begin{Bmatrix} \theta_2 \\ \theta_3 \end{Bmatrix} = \begin{Bmatrix} Q_2 \\ Q_3 \end{Bmatrix} $$

For element 3

h$_c$ = 30 W/m$^2$; A = 1m$^2$; K = 70; h$_e$ = 0.15

$ \therefore \frac{KA}{h_e} = \frac{70 \times 1}{0.15} = 466.67 $

$$ \begin{Bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 30 \end{bmatrix} + \begin{bmatrix} 466.67 & -466.67 \\ -466.67 & 466.67 \end{bmatrix} \end{Bmatrix} \begin{Bmatrix} \theta_3 \\ \theta_4 \end{Bmatrix} = \begin{Bmatrix} Q_3 \\ Q_4 \end{Bmatrix} $$

$$ \begin{bmatrix} 466.67 & -466.67 \\ -466.67 & 466.67 \end{bmatrix} \begin{Bmatrix} \theta_3 \\ \theta_4 \end{Bmatrix} = \begin{Bmatrix} Q_3 \\ Q_4 \end{Bmatrix} $$

(iii) Global matrix equation

$$ \begin{bmatrix} 113.33 & -83.33 & 0 & 0 \\ -83.33 & 233.33 & -150 & 0 \\ 0 & -150 & 616.67 & -466.67 \\ 0 & 0 & -466.67 & 496.67 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \\ \theta_4 \\ \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

Imposing boundary condition,

$ \theta_1 = 800^\circ C \hspace{0.5cm} \theta_4 = 20^\circ C \\ \theta_2 = \theta_3 = 0 \,\, (For \,\, balancing)$

$$ \begin{bmatrix} 113.33 & -83.33 & 0 & 0 \\ -83.33 & 233.33 & -150 & 0 \\ 0 & -150 & 616.67 & -466.67 \\ 0 & 0 & -466.67 & 496.67 \end{bmatrix} \begin{Bmatrix} 800 \\ \theta_2 \\ \theta_3 \\ 20 \\ \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ 0 \\ 0 \\ Q_4 \end{Bmatrix} $$

Now,

(113.33 x 800) - 83.33 $\theta_2$ = $\theta_1$ ...(1)

(-83.33 x 800) + 233.33 $\theta_2$ - 150 $\theta_3$ = 0 ...(2)

(-150 $\theta_2$ + 616.67 $\theta_3$ - 466.67 x 20) = 0 ...(3)

(-466.67 $\theta_3$ + 496.67 x 20) = $\theta_4$ ...(4)

By solving equation (2) and (3), we get