Benchmark solution ===================== 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: .. math:: : label: EQ-None {h} _ {l} ^ {m}\ frac {d {m} _ {l}} {\ mathrm {dt}} +\ frac {\ delta Q\ text {'}} {\ mathrm {dt}} {\ mathrm {dt}}}} D {m} _ {l} +\ mathrm {divq} =0 Where :math:`{h}_{l}^{m}` is the mass enthalpy of water, :math:`{m}_{l}` its mass, its mass, :math:`{M}_{l}` its flow, q the heat flow, and :math:`Q\text{'}` the non-convected heat. The mass conservation equation is as follows: .. math:: : label: EQ-None \ frac {d {m} _ {l}}} {\ mathrm {dt}} +D {M} _ {l} =0 In the absence of water flow, equation (1) is simplified to become: .. math:: : label: EQ-None \ frac {\ delta Q\ text {'}}} {\ mathrm {dt}} +\ mathrm {divq} =0 In this equation, quantity :math:`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: :math:`\frac{dQ\text{'}}{T}=\frac{dQ\text{'}}{{T}_{0}}` So :math:`\mathrm{dQ}\text{'}` has this expression: :math:`\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: * :math:`{\alpha }_{0}` the homogenized thermal expansion coefficient equivalent to that of solid :math:`{\alpha }_{s}`. * :math:`{K}_{0}` the drained elasticity coefficient. * :math:`{C}_{0}^{\varepsilon }` heat has constant deformation which has the expression :math:`{C}_{0}^{\varepsilon }={C}_{0}^{\sigma }-9{T}_{0}{K}_{0}{\alpha }_{0}^{2}` and :math:`{C}_{0}^{\sigma }` the heat at constant stress. * :math:`{\alpha }_{l}^{m}` The coefficient of relative thermal expansion of the liquid with respect to the skeleton, it has for expression: :math:`{\alpha }_{l}^{m}=(b-\phi ){\alpha }_{s}+\phi {\alpha }_{l}` with b the Biot coefficient, :math:`\phi` the porosity and :math:`{\alpha }_{l}` the coefficient of expansion of the liquid. * :math:`d{\varepsilon }_{v}` The variation in volume deformation. * :math:`{\mathrm{dp}}_{l}` The change in liquid pressure. Moreover, the heat flow has the following expression: :math:`q=-{\lambda }_{T}\partial \frac{T}{\partial z}` where :math:`{\lambda }_{T}` is the thermal conductivity coefficient. By replacing :math:`\mathrm{dQ}\text{'}` and q by their value in equation (3) and limiting ourselves to only time step :math:`\Delta t`, we obtain: :math:`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: :math:`\Delta T=T(t)-T(0)=T` and :math:`\Delta \varepsilon =\varepsilon (t)-\varepsilon (0)=\varepsilon` and :math:`\Delta {p}_{l}={p}_{l}(t)-{p}_{l}(0)={p}_{l}` So at the first step of time, we will have: :math:`T-a\frac{{\partial }^{2}T}{\partial {z}^{2}}=b` with :math:`a=\frac{{\lambda }_{l}\Delta T}{{C}_{0}^{\varepsilon }}` and :math:`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: .. math:: : label: EQ-None {\ int} _ {\ omega} T\ mathrm {.} \ stackrel {} {T}\ mathrm {dz} +a {\ int} +a {\ int} _ {\ omega}\ frac {\ partial\ stackrel {} {T} {T}} {T}} {\ partial z}} {\ partial z}} {\ partial z}\ mathrm {dz} -a {\ int} _ {\ partial\ Omega}\ frac {\ partial T} {\ partial T} {\ partial T} {\ partial T} {\ partial T} {\ partial z}}\ stackrel {} {T}\ mathrm {dz} = {\ int} _ {\ omega} b\ mathrm {.} \ 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: .. image:: images/Object_23.svg :width: 489 :height: 206 .. _RefImage_Object_23.svg: We recall the boundary conditions: :math:`\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: :math:`T(z,t)=\sum _{i=1}^{i=2}{T}^{i}(t){\lambda }_{i}(z)` with :math:`{\lambda }_{1}(z)=\mathrm{0,5}(1+\mathrm{2z})` and :math:`{\lambda }_{2}(z)=\mathrm{0,5}(1-\mathrm{2z})` The following matrices are then introduced: :math:`\left[A\right]=\left[{A}_{\mathrm{ij}}\right];{A}_{\mathrm{ij}}={\int }_{-\mathrm{0,5}}^{\mathrm{0,5}}{\lambda }_{i}{\lambda }_{j}\mathrm{dz}` :math:`\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: :math:`\left[A\right]=\frac{1}{6}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]` and :math:`\left[B\right]=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]` We then note classically: :math:`\{T\}=\left\{\begin{array}{c}{T}^{1}\\ {T}^{2}\end{array}\right\}` and :math:`\{\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 :math:`[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: :math:`\{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 :math:`\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: :math:`\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\}` Reference quantities and results ------------------------ For a short time of 100s, we will have :math:`\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 :math:`{T}^{1}` to be null. Uncertainty about the solution --------------------------- None