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Steady-state analysis of a slab

Problem description: Consider the slab shown for a heat transfer analysis. The boundary conditions are characterized by the ambient temperatures T₁, T₂ and the surface coefficients $\alpha_1$, $\alpha_2$. Establish the steady-state solution for the surface coefficients being functions of the wall temperatures T_w1, T_w2.

Mesh: Use the following element types:
ITE=6--see appendix A.1, four quadrilaterals
ITE=4--see appendix A.2, eight triangles
ITE=56--see appendix B.1, four hexahedra.
ITE=61--see appendix B.2, four semi-loofs.
ITE=71--see appendix B.4, two hexahedra, two semi-loofs.

Material properties: $\lambda=20$ W/mK.

Boundary conditions: $T_1=-20^\circ$C, $T_2=25^\circ$C.

$\alpha_1=\alpha(T_{w1})$
$\alpha_2=\alpha(T_{w2})$

$$\alpha(T)= \left\{\begin{array}{ll} 30 \mbox{W/m$^2$ K} & \... .../m$^2$ K} & \mbox{for}~T > 0^\circ\mbox{C} \end{array}\right.$$

Solution: Since the program does not support discontinuous distribution of the surface coefficients we must use some close but continuous approximation to the $\alpha$-function, for instance

$$\alpha(T)= \left\{\begin{array}{ll} 30 & \mbox{for}~T \le 0... ...T\in (0,0.1)\\ 10 & \mbox{for}~T \ge 0.1 \end{array}\right.$$

In the PMD tabular format we simply enter

T:	0	0.1
$\alpha$:	30	10

Furthemore, it is necessary for the nonlinear solver to start with an initial estimate of the temperature field, e.g. $T^{(0)}(x)\equiv 25=\mbox{\it constant\/}$.

It should be noted that the wall temperatures T_w1<0, T_w2>0, thus $\alpha_1=30$, $\alpha_2=10$. Once the surface coefficients become known we can easily compute the heat resistance

$$\frac{1}{k}=\frac{1}{\alpha_1}+\frac{\mbox{{\it thickness}}}{\lambda} +\frac{1}{\alpha_2}= 0.1833\,\mbox{[m$^2$ K/W]}$$

and the heat flux

$$q=k(T_1-T_2)=-245.45\,\mbox{[W/m$^2$ ]}$$

Hence we obtain the exact solution

$$T_{w1}=-11.81^\circ\mbox{C}~,~~~T_{w2}=0.45^\circ\mbox{C}$$

verifying the FEM results.