A composite wall consists of three materials, as shown in figure,

The outer temperature $T_0 = 20$ $^0C$ Convection hear transfer takes place on the inner surface of the wall with $T_\infty = 800$ $^0C$ and h = 30 W / $m^2 $ $^0C$.Determine temperature distribution in the wall.

$$k_1 = 25 W/ m.^0C$$

$$k_2 = 30 W/ m.^0C$$

$$k_3 = 70 W/ m.^0C$$

Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 10M

Year: May 2015

Elemental Stiffness Matrix,

$K_1 = \frac{K_1A}{L_1}\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} + \begin{bmatrix} \ h.A & 0 \\ \ 0 & 0 \\ \end{bmatrix} \\ K_1 = \frac{25 \times 1}{0.3} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} + \begin{bmatrix} \ 30 x 1 & 0 \\ \ 0 & 0 \\ \end{bmatrix} \\ K_1 = \begin{bmatrix} \ 113.33 & -83.33 \\ \ -83.33 & 83.33 \\ \end{bmatrix}$

$K_2 = \frac{K_2A}{L_2} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{30 x 1}{0.2}\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \begin{bmatrix} \ 150 & -150 \\ \ -150 & 150 \\ \end{bmatrix}$

$K_3 = \frac{K_3A}{L_3} \begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \frac{70 x 1}{0.15}\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix} = \begin{bmatrix} \ 466.67 & -466.67 \\ \ -466.67 & 466.67 \\ \end{bmatrix}$

$K = \begin{bmatrix} \ 113.33 & -83.33 & 0 & 0 \\ \ -83.33 & 233.33 & -150 & 0 \\ \ 0 & -150 & 616.67 & -466.67 \\ \ 0 & 0 & -466.67 & 466.67 \\ \end{bmatrix}$

$Q = \begin{Bmatrix} \ h.A.T_\infty \\ \ 0 \\ \ 0 \\ \ 0 \\ \end{Bmatrix} = \begin{Bmatrix} \ 30 \times 800 \\ \ 0 \\ \ 0 \\ \ 0 \\ \end{Bmatrix} = \begin{Bmatrix} \ 24000 \\ \ 0 \\ \ 0 \\ \ 0 \\ \end{Bmatrix}$

[K] {T} = {Q}

$\begin{bmatrix} \ 113.33 & -83.33 & 0 & 0 \\ \ -83.33 & 233.33 & -150 & 0 \\ \ 0 & -150 & 616.67 & -466.67 \\ \ 0 & 0 & -466.67 & 466.67 \\ \end{bmatrix} \begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \ 20 \\ \end{Bmatrix} = \begin{Bmatrix} \ 24000 \\ \ 0 \\ \ 0 \\ \ 0 \\ \end{Bmatrix}$

$\begin{bmatrix} \ 113.33 & -83.33 & 0 \\ \ -83.33 & 233.33 & -150 \\ \ 0 & -150 & 616.67 \\ \ 0 & 0 & -466.67 \\ \end{bmatrix} \begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \end{Bmatrix} = \begin{Bmatrix} \ 24000 \\ \ 0 \\ \ 9333.4 \\ \end{Bmatrix}$

$T_1 = 319.8^0 C\\ T_2 = 146.91^0 C\\ T_3 = 50.87^0 C$