Consider the slab shown for a heat transfer analysis. The boundary conditions are characterized by the ambient temperatures $T_1,$ $T_2$ and the surface coefficients $\alpha_1,$ $\alpha_2.$ Establish the steady-state solution for the surface coefficients being functions of the wall temperatures $T_{w1},$ $T_{w2}.$

$\lambda=20\text{ W/mK}.$

$T_1=-20\text{ $^\circ$C},$ $T_2=25\text{ $^\circ$C}.$ $\alpha_1=\alpha(T_{w1}),$ $\alpha_2=\alpha(T_{w2}),$ $\alpha(T)=\left\{\begin{align} 30\text{ W/m}^2\text{K}\quad\text{for}\quad T\le0\ ^\circ\text{C} \\ 10\text{ W/m}^2\text{K}\quad\text{for}\quad T>0\ ^\circ\text{C} \end{align}\right..$

Since PMD does not support discontinuous distribution of the surface coefficients we must use some close but continuous approximation to the $\alpha$-function, for instance $$\alpha(T)=\left\{\begin{array}{lll} 30 & \text{for} & T\le0 \\ 30-200T & \text{for} & T\in(0,0.1) \\ 10 & \text{for} & T\ge0.1 \end{array}\right.$$ In the PMD tabular format we simply enter $$\begin{array}{ccc} T: & 0 & 0.1 \\ \alpha: & 30 & 10 \end{array}$$ Furthemore, it is necessary for the nonlinear solver to start with an initial estimate of the temperature field, e.g., $$T^{(0)}(x)\equiv25=\text{const.}$$

• Quadrilaterals, BEAM6T1
• Triangles, BEAM4T1
• Hexahedra, BEAM56T1
• Semi-loofs, BEAM61T1
• Transition elements, BEAM71T1

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 $$\frac{1}{k}=\frac{1}{\alpha_1}+\frac{\text{thickness}}{\lambda}+\frac{1}{\alpha_2}=0.1833\text{ m$^2$K/W}$$ and the heat flux $$q=k(T_1-T_2)=-245.45\text{ W/m$^2$}.$$ Hence we obtain the exact solution $$T_{w1}=-11.81\text{ $^\circ$C},\quad T_{w2}=0.45\text{ $^\circ$C},$$ verifying the FEM results.