Reference solution (A, B and C models)
================================================


Validation of the water exchange term from the heat exchange term 
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A certain number of hypotheses are put forward in order to make the analogy with the thermal case:


* the temperature is constant (no thermal calculation);


* the gas pressure is constant;


* Vapor pressure is negligible;


* We are neglecting the broadcast of Fick;


* The material is non-deformable (no mechanical calculations);


* Gravity is zero;


* Water is incompressible :math:`{\rho }_{w}=\mathit{cste}`.

Recall that Darcy's law (see R7.01.11) for the gas phase is written as:

.. math::
 :label: eq-2

    {M} _ {\ mathit {gz}} =- {\ rho}} =- {\ rho} _ {\ mathit {gz}} {\ mathit {gz}}} ^ {H}\nabla {p} _ {\ mathit {gz}}

with:


* :math:`{M}_{\mathit{gz}}` the gas flow;


* :math:`{\rho }_{\mathit{gz}}` the density of the gas;


* :math:`{\mathrm{\lambda }}_{\mathit{gz}}^{H}` the hydraulic conductivity of the gas;


* :math:`{p}_{\mathit{gz}}` gas pressure.

Gold:

.. math::
 :label: eq-3

    {p} _ {\ mathit {gz}} =\ mathit {cste}\ Rightarrow\nabla {p} _ {\ mathit {gz}} =0

So:

.. math::
: label: eq-4

    {M} _ {\ mathit {gz}} = {M}} = {M} _ {\ mathit {as}} + {\ mathit {gz}} = 0

Since we're neglecting the Fick broadcast, then:

.. math::
: label: eq-5

    {M} _ {\ mathit {ad}} = {M} _ {\ mathit {VP}} =0

By () and ():

.. math::
 :label: eq-6

    {M} _ {\ mathit {as}} =0

Thus we can simplify the problem, which boils down to the equation for the conservation of the mass of liquid water:

.. math::
 :label: eq-7

    \ dot {{m} _ {w}}} +\ text {Div} ({M} _ {w}) =0

Or :math:`{m}_{w}={\rho }_{w}\mathrm{\varphi }S(1+\mathit{Tr}(\epsilon ))-{\rho }_{w}^{0}{\mathrm{\varphi }}^{0}{S}^{0}`, with:


* :math:`{\rho }_{w}` the density of water, :math:`{\rho }_{w}^{0}` the density in the initial state;


* :math:`\mathrm{\varphi }=\frac{{V}_{\mathit{vide}}}{V}` the porosity of the material, :math:`{V}_{\mathit{vide}}` corresponding to the unfilled pores, :math:`V` to the porosity of the material in the initial state;


* :math:`S=\frac{{V}_{\mathit{lq}}}{{V}_{\mathit{vide}}}` water saturation, :math:`{S}^{0}` initial saturation;


* :math:`\epsilon` the deformation of the material;

The material is considered to be non-deformable: :math:`\epsilon =0` and :math:`\mathrm{\varphi }=\mathit{cste}={\mathrm{\varphi }}^{0}` and that water is incompressible, :math:`{\rho }_{w}={\rho }_{w}^{0}`. So:

.. math::
 :label: eq-8

    {m} _ {w} (t) = {\ rho} _ {w} _ {w} ^ {0}\ mathrm {\ varphi} (S (t) - {S} ^ {0})

This involves:

.. math::
 :label: eq-9

    \ dot {{m} _ {w}} (t) = {\ rho} _ {w} _ {w}\ mathrm {\ varphi} {\ partial} _ {t} S ({p} _ {c} (t)) = {\ rho} (t)) = {\ rho} _ {t)))) = {\ rho} _ {t}))) = {\ rho} _ {t}))) = {\ rho} _ {w} _ {w}\ mathrm {\ varphi} {S} ^ {\ text {'}}} ({p} _ {c} (t))) = {\ rho} _ {t}))) = {\ rho} _ {t})) = {\ rho} _ {t})\ dot {{p} _ {c}} (t)

Since the effects of gravity are overlooked, Darcy's law for the liquid phase is written:

.. math::
 :label: eq-10

    {M} _ {\ mathit {aq}} = {M} _ {w}} =- {\ rho} _ {w} {\ lambda} _ {\ mathit {q}}} ^ {H} (S) (S)\nabla {p} _ {w}

Gold :math:`{p}_{c}={p}_{\mathit{gz}}-{p}_{\mathit{lq}}={p}_{\mathit{atm}}-{p}_{w}`. So :math:`\nabla {p}_{c}=-\nabla {p}_{w}` and therefore :math:`{M}_{w}={\mathrm{\rho }}_{w}{\mathrm{\lambda }}_{w}^{H}(S)\nabla {p}_{c}`.

By replacing :math:`{M}_{w}` and :math:`\dot{{m}_{w}}` with their last expressions in () and assuming :math:`{\lambda }_{w}^{H}(S)=\frac{K\times {K}_{\mathit{rl}}(S)}{\mu }` to be a constant:

.. math::
 :label: eq-11

    {\ mathrm {\ rho}} _ {w}\ mathrm {\ varphi} {S} {S} ^ {\ text {'}} ({p} _ {c})\ times\ dot {{p} _ {c} _ {c}} _ {c}}} (t) + {\ mathrm {\ lambda}}} _ {w} ^ {H} (S)\ text {Div} _ {c} _ {c})\ times\ dot {p} _ {p} _ {c} _ {c} _ {c} _ {c}} (t) + {\ mathrm {\ lambda}}} _ {w} ^ {H} (S)\ text {Div} _ {c}} (s)\ text {Div} _ {c}\ rho}} _ {w}\ mathrm {.} \nabla {p} _ {c}) =0

Water is assumed to be incompressible so:

.. math::
 :label: eq-12

    {\ mathrm {\ rho}} _ {w}\ mathrm {\ varphi} {S}} ^ {\ text {'}} ({p} _ {c})\ times\ dot {{p} _ {c} _ {c}} (t) + {\ mathrm {\ rho}} (t) + {\ mathrm {\ rho}}} _ {w} {\ lambda}} _ {w} ^ {H}} _ {w} {\ lambda}} _ {w} ^ {H} (S)\ text {Div} (\nabla {p} _ {c}) =0

:math:`{\rho }_{w}\mathrm{\varphi }{S}^{\text{'}}({p}_{c})\times \dot{{p}_{c}}(t)+{\rho }_{w}{\lambda }_{w}^{H}(S)\text{Div}(\text{}\nabla {p}_{c})=0`

With the Laplacian operator:

.. math::
 :label: eq-13

    {\ mathrm {\ rho}} _ {w}\ mathrm {\ varphi} {S}} ^ {\ text {'}} ({p} _ {c})\ times\ dot {{p} _ {c} _ {c}} (t) + {\ mathrm {\ rho}} (t) + {\ mathrm {\ rho}}} _ {w} {\ lambda}} _ {w} ^ {H}} _ {w} {\ lambda}} _ {w} ^ {H} (S)\ mathrm {\ Delta} {p} {p} _ {c} =0

Finally:

.. math::
 :label: eq-14

    {\ partial} _ {t} {p} _ {c} +\ frac {{\ mathrm {\ lambda}} _ {w} ^ {H} (S)} {\ mathrm {\ varphi} {S} {S} {S}} {S}} ^ {\ text {c}} ^ {c}} ^ {c}} ^ {c})}\ mathrm {\ Delta} {p} {p} {p}} _ {c} =0

To close the system, we add:


* the initial condition: :math:`{p}_{c}(x\mathrm{,0})\mathrm{:}=10000\mathit{Pa}`;


* the mixed condition on [BC] (or [BCFG] depending on the geometry): :math:`{\mathrm{\lambda }}^{H}\nabla ({p}_{c}\mathrm{.}n)={h}^{H}({p}_{\mathit{ext}}-{p}_{c})`.

This formulation is similar to a heat equation without a source term, of the form:

.. math::
 :label: eq-15

    \ mathrm {\ rho} {C} _ {p} {p} {\ partial} _ {t} T+\ text {Div} (q) =0

By taking Fourier's law that relates heat flow :math:`q` to the temperature gradient, namely:

.. math::
 :label: eq-16

    q=-\ mathrm {\ lambda}\nabla T

Assuming that :math:`\mathrm{\lambda }` is a constant, characteristic of the material:

.. math::
 :label: eq-17

    {\ partial} _ {t} T-\ frac {\ mathrm {\ lambda}} {\ mathrm {\ rho} {C} _ {p}}\ mathrm {\ Delta} T=0

Where :math:`\mathrm{\rho }` is the density of concrete, :math:`{C}_{p}` is the specific heat of the material, and :math:`\mathrm{\lambda }` is the thermal conductivity.

The thermal problem is thus already treated in THER_NON_LINE. We can then compare the results of the thermal problem with those of problem THM, by taking closure conditions of the same type as in the case THM, i.e. by asking:

:math:`\begin{array}{c}T(x\mathrm{,0})=\mathit{cste}\\ \mathrm{\lambda }(\text{}\nabla T\mathrm{.}n)={h}_{T}({T}_{\mathit{ext}}-T)\mathit{au}\mathit{bord}\end{array}`

In addition, ensuring that:


* :math:`T(x\mathrm{,0})=\mathit{cste}\mathrm{:}={p}_{c}(x\mathrm{,0})=10000`;


* :math:`-\rho {C}_{p}≔\mathrm{\varphi }{S}^{\text{'}}({p}_{c})`


* :math:`\lambda \mathrm{:}={\lambda }^{H}`


* :math:`{h}_{T}\mathrm{:}={h}^{H}`


* :math:`{T}_{\mathit{ext}}\mathrm{:}={p}_{\mathit{ext}}`

Subsequently, we then rely on thermal models AXIS, PLAN_DIAG, and 3D, and 3D, which are compared with models AXIS_THH2MS, D_ PLAN_THH2MS and 3D_ THH2MS, respectively.