2. Reference solution

2.1. Calculation method

It is an analytical solution. In this configuration the problem is one-dimensional and the heat equation is reduced to the following differential equation:

\(\frac{{d}^{2}T}{{\mathit{dy}}^{2}}=0\) with conditions \(T(y\mathrm{=}0)\mathrm{=}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}}\) and \(T(y\mathrm{=}\mathit{LY})\mathrm{=}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{sup}}\)

This equation has the following solution:

\(T(y)\mathrm{=}(\text{}\stackrel{ˉ}{T}{\text{}}^{\text{sup}}\mathrm{-}\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}})\frac{y}{\mathit{LY}}+\text{}\stackrel{ˉ}{T}{\text{}}^{\text{inf}}\)

2.2. Reference quantities and results

The temperature is tested at points \({P}^{\text{+}}\), \({P}^{\text{-}}\) and \(Q\) (see Figure):

\(T(\frac{\mathit{LY}}{2})=\frac{(\text{}\stackrel{̄}{T}{\text{}}^{\text{sup}}+\text{}\stackrel{̄}{T}{\text{}}^{\text{inf}})}{2}\), digital application: \(T(\frac{\mathit{LY}}{2})=15°C\)

The two components (space constants) of the temperature gradient are also tested:

\({\partial }_{x}T=0\), digital application: \({\partial }_{x}T=0°{\mathit{C.m}}^{-1}\)

\({\partial }_{y}T=\frac{(\text{}\stackrel{̄}{T}{\text{}}^{\text{sup}}-\text{}\stackrel{̄}{T}{\text{}}^{\text{inf}})}{\mathit{LY}}\), digital application: \({\partial }_{y}T=10°{\mathit{C.m}}^{-1}\)

Identification	Reference type	Reference value
Points \({P}^{\text{+}}\), \({P}^{\text{-}}\), and \(Q\) - \(\mathit{TEMP}\)	“ANALYTIQUE”	\(15°C\)
In every way, \(\mathit{DTX}\)	“ANALYTIQUE”	\(0°{\mathit{C.m}}^{-1}\)
In every way, \(\mathit{DTY}\)	“ANALYTIQUE”	\(10°{\mathit{C.m}}^{-1}\)