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Transient 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 $\alpha_1$, $\alpha_2$. Initially T1=T2 then T1 suddenly drops. Calculate the transient response of the slab for the surface coefficients being functions of the wall temperatures Tw1, Tw2 and compare these results with the steady-state solution III.1.

\begin{figure}
\centering\hspace{0pt}
\epsfclipon\epsfxsize=8cm\epsffile{slabt2.ps}
\end{figure}

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, $\rho\!\cdot\! c=3\times 10^6$ J/m3K.
Boundary conditions: $T_1=-20^\circ$C, $T_2=25^\circ$C at time t>0.

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

\begin{displaymath}\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.
\end{displaymath}

Initial conditions: $T(x)\equiv 25^\circ$C at time t=0.
Solution: The surface coefficients $\alpha_1$, $\alpha_2$ and the initial temperature field are processed in the same way as in section III.1. Numerical analysis is carried out in the time interval (0,200h) with the initial time step set to STEP=10h. The step length is controlled by the parameter TOL which is matched with the convergence tolerance TOL= EDIF=0.1$^\circ$C.


 
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Next: Quadrilaterals, BEAM6T2 Up: Heat Transfer Previous: OUTPUT