2. Benchmark solution#
2.1. Calculation method#
The reference solution is one-dimensional because it only depends on the vertical coordinate. We will therefore reason on a linear axis along z.
If we neglect gravity and consider that there is no external heat source, the thermal equation is given by the following expression:
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{h} _ {l} ^ {m}frac {d {m} _ {l}} {mathrm {dt}} +frac {delta Qtext {“}} {mathrm {dt}} {mathrm {dt}}}} D {m} _ {l} +mathrm {divq} =0
Where \({h}_{l}^{m}\) is the mass enthalpy of water, \({m}_{l}\) its mass, its mass, \({M}_{l}\) its flow, q the heat flow, and \(Q\text{'}\) the non-convected heat.
The mass conservation equation is as follows:
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frac {d {m} _ {l}}} {mathrm {dt}} +D {M} _ {l} =0
In the absence of water flow, equation (1) is simplified to become:
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frac {delta Qtext {“}}} {mathrm {dt}} +mathrm {divq} =0
In this equation, quantity \(Q\text{'}\) represents the heat received by the system in a transformation for which there is no heat input per fluid inlet. A low pressure variation and parameters are selected so that the temperature varies little and therefore that:
\(\frac{dQ\text{'}}{T}=\frac{dQ\text{'}}{{T}_{0}}\)
So \(\mathrm{dQ}\text{'}\) has this expression:
\(\mathrm{dQ}\text{'}=3{\alpha }_{0}{K}_{0}{T}_{0}d{\varepsilon }_{v}+{C}_{0}^{\varepsilon }\mathrm{dT}-3{\alpha }_{l}^{m}{T}_{0}{\mathrm{dp}}_{l}\)
with:
\({\alpha }_{0}\) the homogenized thermal expansion coefficient equivalent to that of solid \({\alpha }_{s}\).
\({K}_{0}\) the drained elasticity coefficient.
\({C}_{0}^{\varepsilon }\) heat has constant deformation which has the expression \({C}_{0}^{\varepsilon }={C}_{0}^{\sigma }-9{T}_{0}{K}_{0}{\alpha }_{0}^{2}\) and \({C}_{0}^{\sigma }\) the heat at constant stress.
\({\alpha }_{l}^{m}\) The coefficient of relative thermal expansion of the liquid with respect to the skeleton, it has for expression: \({\alpha }_{l}^{m}=(b-\phi ){\alpha }_{s}+\phi {\alpha }_{l}\) with b the Biot coefficient, \(\phi\) the porosity and \({\alpha }_{l}\) the coefficient of expansion of the liquid.
\(d{\varepsilon }_{v}\) The variation in volume deformation.
\({\mathrm{dp}}_{l}\) The change in liquid pressure.
Moreover, the heat flow has the following expression: \(q=-{\lambda }_{T}\partial \frac{T}{\partial z}\) where \({\lambda }_{T}\) is the thermal conductivity coefficient.
By replacing \(\mathrm{dQ}\text{'}\) and q by their value in equation (3) and limiting ourselves to only time step \(\Delta t\), we obtain:
\(3{\alpha }_{0}{K}_{0}{T}_{0}\Delta {\varepsilon }_{v}-3{\alpha }_{l}^{m}{T}_{0}\Delta {p}_{l}+{C}_{0}^{\varepsilon }\Delta T=-\Delta tdq={\lambda }_{l}\Delta t\frac{{\partial }^{2}T}{\partial {z}^{2}}\)
Considering temperatures and displacements that are initially zero, we can write:
\(\Delta T=T(t)-T(0)=T\) and \(\Delta \varepsilon =\varepsilon (t)-\varepsilon (0)=\varepsilon\) and \(\Delta {p}_{l}={p}_{l}(t)-{p}_{l}(0)={p}_{l}\)
So at the first step of time, we will have:
\(T-a\frac{{\partial }^{2}T}{\partial {z}^{2}}=b\)
with \(a=\frac{{\lambda }_{l}\Delta T}{{C}_{0}^{\varepsilon }}\) and \(b=-\frac{3{\alpha }_{0}{K}_{0}{T}_{0}\varepsilon -3{\alpha }_{l}^{m}{T}_{0}{p}_{l}}{{C}_{0}^{\varepsilon }}\)
The variational formulation of this expression (in the one-dimensional case) is then as follows:
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{int} _ {omega} Tmathrm {.} stackrel {} {T}mathrm {dz} +a {int} +a {int} _ {omega}frac {partialstackrel {} {T} {T}} {T}} {partial z}} {partial z}} {partial z}mathrm {dz} -a {int} _ {partialOmega}frac {partial T} {partial T} {partial T} {partial T} {partial T} {partial z}}stackrel {} {T}mathrm {dz} = {int} _ {omega} bmathrm {.} stackrel {} {T}mathrm {dz}
To establish the analytical solution, we consider a single element of degree 1 since in modeling THM the thermo-hydraulic part is treated by linear elements.
Let it be a linear element:
We recall the boundary conditions: \(\frac{\partial T}{\partial z}=0\) at both ends (z=0.5 and z = -0.5)
The temperature based on form functions is written as follows:
\(T(z,t)=\sum _{i=1}^{i=2}{T}^{i}(t){\lambda }_{i}(z)\)
with
\({\lambda }_{1}(z)=\mathrm{0,5}(1+\mathrm{2z})\)
and
\({\lambda }_{2}(z)=\mathrm{0,5}(1-\mathrm{2z})\)
The following matrices are then introduced:
\(\left[A\right]=\left[{A}_{\mathrm{ij}}\right];{A}_{\mathrm{ij}}={\int }_{-\mathrm{0,5}}^{\mathrm{0,5}}{\lambda }_{i}{\lambda }_{j}\mathrm{dz}\)
\(\left[B\right]=\left[{B}_{\mathrm{ij}}\right];{B}_{\mathrm{ij}}={\int }_{-\mathrm{0,5}}^{\mathrm{0,5}}\frac{\partial {\lambda }_{i}}{\partial z}\frac{\partial {\lambda }_{j}}{\partial z}\mathrm{dz}\)
which leads to:
\(\left[A\right]=\frac{1}{6}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\) and \(\left[B\right]=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\)
We then note classically:
\(\{T\}=\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}\)
and \(\{\frac{\partial T}{\partial t}\}=\left\{\begin{array}{c}\frac{\partial {T}^{1}}{\partial t}\\ \frac{\partial {T}^{2}}{\partial t}\end{array}\right\}\)
Equation (4) then becomes
\([A]\{T\}+a[B]\{T\}-a[A]\{\frac{\partial T}{\partial t}\}=[A]\{b\}\)
with the imposed boundary conditions (zero displacement at the bottom and zero temperature flow at both ends), we have:
\(\{b\}=\left\{\begin{array}{c}0\\ b\end{array}\right\}\mathrm{et}\{\frac{\partial T}{\partial t}\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}\)
Which in the end gives us
\(\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}+a{[A]}^{-1}[B]\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}=\left\{\begin{array}{c}0\\ b\end{array}\right\}\)
In the end we get:
\(\left[\begin{array}{cc}1+\mathrm{6a}& -\mathrm{6a}\\ -\mathrm{6a}& 1+\mathrm{6a}\end{array}\right]\{T\}=\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}\)
2.2. Reference quantities and results#
For a short time of 100s, we will have \(\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}=\left\{\begin{array}{c}{2.10}^{-14}\\ \mathrm{1,045}{.10}^{-7}\end{array}\right\}\)
We will consider \({T}^{1}\) to be null.
2.3. Uncertainty about the solution#
None