The circular rod depicted in Figure has an outside diameter of 60 mm, length 1m, and is perfectly insulted on its circumference. The left half of the cylinder is aluminium, for which $k_x =200 W/m. ^0C$ and the right half is copper having $k_x=389 W/m.^0C$ The extreme right end of the cylinder is maintained at a temperature of 80 $^0C$, while the left end is subjected to a heat input rate 4000 $W/m^2$. Using four equal length elements, Determine the steady state temperature distribution in the cylinder.

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

Marks: 10M

Year: Dec 2016

$\hspace{2cm}A=\frac{\pi}{4}60_2$

$\hspace{2.2cm}=2826mw^2$

$\hspace{2.2cm}=0.002826w^2$

For element 1 & 2 :

$k_1=k_2=\frac{KA}{L}$

$\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $=\frac{260\times0.002826}{0.25}$

$\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$=2.26

$\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$

$hspace{2cm}\therefore k_1=$ $\begin{bmatrix} \ 2.26 & -2.26 \\ \ -2.26 & 2.26 \\ \end{bmatrix}_2^1$ \&$k_2=$ $\begin{bmatrix} \ 2.26 & -2.26 \\ \ -2.26 & 2.26 \\ \end{bmatrix}_1^2$

For element 3 & 4 :

$k_3=k_4 :\frac{KA}{L}$

$\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$

$=\frac{389\times0.002826}{0.25}$

$\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$=4.397 $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$

$\hspace{0.3cm}\therefore k_3=$

$\begin{bmatrix} \ 4.397 & -4.397 \\ \ -4.397 & 4.397 \\ \end{bmatrix}_4^3$ & $k_4=$

$\begin{bmatrix} \ 4.397 & -4.397 \\ \ -4.397 & 4.397 \\ \end{bmatrix}$

Global stiffnes matrix,

[k]= $\begin{bmatrix} \ 2.26 & -2.26 & 0 & 0 & 0 \\ \ -2.26 & 4.52 & -2.26 & 0 & 0 \\ \ 0 & -2.26 & 6.657 & -4.397 & 0 \\ \ 0 & 0 & -4.397 & 8.794 & -4.397 \\ \ 0 & 0 & 0 & -4.397 & 4.397 \\ \end{bmatrix}$

Golbal matrix equation,

$\hspace{1.6cm}$[K]{T}={Q}

$\begin{bmatrix} \ 2.26 & -2.26 & 0 & 0 & 0 \\ \ -2.26 & 4.52 & -2.26 & 0 & 0 \\ \ 0 & -2.26 & 6.657 & -4.397 & 0 \\ \ 0 & 0 & -4.397 & 8.794 & -4.397 \\ \ 0 & 0 & 0 & -4.397 & 4.397 \\ \end{bmatrix}$

$\begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \ T_4 \\ \ T_5 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \ Q_3 \\ \ Q_4 \\ \ Q_5 \\ \end{Bmatrix}$

Boundary conditions :

$\hspace{1.6cm}Q_1=4000 w/w_2$

$\hspace{5cm} \& T_5 = 80^0 C$

$\hspace{1.6cm}Q=4000\times0.002826$

$\hspace{2cm}=11.304w$

$\therefore$

$\begin{bmatrix} \ 2.26 & -2.26 & 0 & 0 & 0 \\ \ -2.26 & 4.52 & -2.26 & 0 & 0 \\ \ 0 & -2.26 & 6.657 & -4.397 & 0 \\ \ 0 & 0 & -4.397 & 8.794 & -4.397 \\ \ 0 & 0 & 0 & -4.397 & 4.397 \\ \end{bmatrix}$

$\begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \ T_4 \\ \ 80 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ 11.304 \\ \ 0 \\ \ 0 \\ \ 0 \\ \ Q_5 \\ \end{Bmatrix}$

Accounting for the knowntemp at node 5 the first four equation can be written as,

$\begin{bmatrix} \ 2.26 & -2.26 & 0 & 0 \\ \ -2.26 & 4.52 & -2.26 & 0 \\ \ 0 & -2.26 & 6.66 & -4.40 \\ \ 0 & 0 & -4.40 & 8.80 & \\ \end{bmatrix}$ $\begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \ T_4 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ 11.304 \\ \ 0 \\ \ 0 \\ \ 352.0 \\ \end{Bmatrix}$

$4 \times4$ matrix is not solve using calculator

$\therefore$ convert it into $3\times3$

$therefore R_2 \rightarrow R_1+R_2$

$\begin{bmatrix} \ 2.26 & -2.26 & 0 & 0 \\ \ -2.26 & 4.52 & -2.26 & 0 \\ \ 0 & -2.26 & 6.66 & -4.40 \\ \ 0 & 0 & -4.40 & 8.80 & \\ \end{bmatrix}$

$\begin{Bmatrix} \ T_1 \\ \ T_2 \\ \ T_3 \\ \ T_4 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ 11.304 \\ \ 11.304 \\ \ 0 \\ \ 352.0 \\ \end{Bmatrix}$

Solve for unknown $T_2 T_3 \& T_4$

$\hspace{1cm} \therefore T_2 = 90.14^0C$

$\hspace{1.2cm}T_3=85.15^0C$

$\hspace{1.2cm}T_4=85.57^0C$

$\hspace{.9cm}substitule T_2,T_3, \&T_4,$

$\hspace{1.2cm}\therefore T_1=95.15^0C$

$Q_5=-11.38 w$

$\hspace{0.3cm}\frac{-11.30}{A}=\frac{-11.38}{0.002826}=-3998.7 w/m^2$