2. Benchmark solution#
2.1. Calculation method used for the reference solution#
The bar is subjected to a \(r(T)\mathrm{=}{r}_{0}\mathrm{-}{r}_{1}T\) heat source, where \({r}_{1}>0\) for reasons of thermal stability. Its initial temperature is \({T}_{0}(x)\) and the ends of the bar are maintained at zero temperature. The evolution of temperature obeys the heat equation:
\(\rho c\dot{T}\mathrm{=}\lambda {\mathrm{\nabla }}^{2}T+r(T)\); \(T(x\mathrm{,0})\mathrm{=}{T}_{0}(x)\); \(T(\mathrm{-}L,t)\mathrm{=}T(L,t)\mathrm{=}0\)
By normalization, we can reduce ourselves without loss of generality to the following equation:
\(\frac{\mathrm{\partial }u}{\mathrm{\partial }t}\mathrm{=}\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^{2}}+1\mathrm{-}{\omega }^{2}u\); \(u(x\mathrm{,0})\mathrm{=}{u}_{0}(x)\); \(u(\mathrm{-}\mathrm{1,}t)\mathrm{=}u(\mathrm{1,}t)\mathrm{=}0\)
To solve this equation, we first look at the asymptotic solution \({u}_{\mathrm{\infty }}(x)\) which verifies:
\(0\mathrm{=}\frac{{\mathrm{\partial }}^{2}{u}_{\mathrm{\infty }}}{\mathrm{\partial }{x}^{2}}+1\mathrm{-}{\omega }^{2}{u}_{\mathrm{\infty }}\); \({u}_{\mathrm{\infty }}(\mathrm{-}1)\mathrm{=}{u}_{\mathrm{\infty }}(1)\mathrm{=}0\)
The solution of this linear differential equation of the second order is equal to:
\({u}_{\mathrm{\infty }}(x)\mathrm{=}\frac{1}{{\omega }^{2}}(1\mathrm{-}\frac{\mathrm{cosh}\omega x}{\mathrm{cosh}\omega })\)
The solution of the transient equation is then obtained by projecting \(v\mathrm{=}u\mathrm{-}{u}_{\mathrm{\infty }}\) onto the Laplacian eigenfunctions onto \(\left\}\mathrm{-}\mathrm{1,}1\right\{\). To simplify the analysis, we adopt an initial condition \({u}_{0}\) equal to the first eigenmode, namely:
\({u}_{0}(x)\mathrm{=}{u}_{\mathrm{\infty }}(x)\mathrm{-}\mathrm{cos}\frac{\pi x}{2}\)
Only the first mode being activated, we are brought back to the resolution of a first-order time differential equation, to finally obtain the solution:
\(u(x,t)\mathrm{=}{u}_{\mathrm{\infty }}(x)\mathrm{-}\mathrm{exp}(\mathrm{-}{\omega }^{2}t\mathrm{-}\frac{{\pi }^{2}}{4}t)\mathrm{cos}\frac{\pi x}{2}\)
Finally, as before, we go back from \(u\) to \(T\) by adopting a specific set of parameters, without taking into account the units, so that \(T\mathrm{=}u\). To do this, we take \(\lambda \mathrm{=}{r}_{0}\mathrm{=}\rho c\), \({r}_{1}\mathrm{=}{\omega }^{2}{r}_{0}\),, \(L\mathrm{=}1\) and \({T}_{0}(x)\mathrm{=}{u}_{0}(x)\).
2.2. Benchmark results#
The test case is conducted with \(\omega \mathrm{=}\sqrt{2}\) and the temperature at \(t=1\) is examined at a node in the plane of symmetry \((x=0)\). The data is as follows:
Thermal conductivity |
LAMBDA |
|
Volume heat capacity |
RHO_CP |
|
Initial temperature |
\({T}_{0}\) |
|
Heat source |
\({r}_{0}\) \({r}_{1}\) |
2. 4. |
Size tested |
\(T\) (\(x\mathrm{=}\mathrm{0,}t\mathrm{=}1\)) |
Reference value |
\(0.258974\) |