Benchmark solution
=====================


Calculation method
------------------


Calculation of the conservation of air mass
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The conservation of the mass of gas is written as:


.. _RefEquation 2.1.1-1:

 :math:`\frac{{\mathrm{dm}}_{\mathrm{air}}}{\mathrm{dt}}+d({M}_{\mathrm{as}}+{M}_{\mathrm{ad}})` eq 2.1.1-1

We write that the total mass of water and the total mass of air are conserved (because there is no water or gas flow at the edge) and we get:

:math:`{m}_{\mathrm{air}}={m}_{\mathrm{as}}+{m}_{\mathrm{ad}}={S}_{0}({\rho }_{\mathrm{ad}}-{\rho }_{\mathrm{ad}}^{0})+(1-{S}_{0})({\rho }_{\mathrm{as}}-{\rho }_{\mathrm{as}}^{0})`

therefore




.. _RefEquation 2.1.1-2:

 :math:`d({m}_{\mathrm{as}}+{m}_{\mathrm{ad}})={S}_{0}d{\rho }_{\mathrm{ad}}+(1-{S}_{0})d{\rho }_{\mathrm{as}}` eq 2.1.1-2

 :math:`d{\rho }_{\mathrm{as}}=\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}{\mathrm{dp}}_{\mathrm{as}}` and :math:`d{\rho }_{\mathrm{ad}}=\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{{K}_{H}}{\mathrm{dp}}_{\mathrm{as}}`

 :math:`\frac{{\mathrm{dm}}_{\mathrm{air}}}{\mathrm{dt}}=d(\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{{K}_{H}}{S}_{0}+(1-{S}_{0})\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}{\mathrm{dp}}_{\mathrm{as}})`

Speed calculation:


.. _RefEquation 2.1.1-3:

 :math:`\frac{{M}_{\mathrm{as}}}{{\rho }_{\mathrm{as}}}={\lambda }_{\mathrm{gz}}(-\nabla {p}_{\mathrm{as}})` eq 2.1.1-3

since :math:`{F}_{\mathrm{vp}}=0` and :math:`\nabla {p}_{\mathrm{vp}}=0`

and

:math:`{M}_{\mathrm{ad}}={\rho }_{\mathrm{ad}}{\lambda }_{\mathrm{lq}}(-\nabla {p}_{\mathrm{lq}})-{F}_{\mathrm{ad}}\nabla {C}_{\mathrm{ad}}` with :math:`{C}_{\mathrm{ad}}={\rho }_{\mathrm{ad}}`

Like :math:`\nabla {p}_{\mathrm{lq}}=\nabla {p}_{w}+\nabla {p}_{\mathrm{ad}}=\nabla {p}_{\mathrm{ad}}=\frac{\mathrm{RT}}{{K}_{H}}\nabla {p}_{\mathrm{as}}`

 :math:`{M}_{\mathrm{ad}}={\rho }_{\mathrm{ad}}{\lambda }_{\mathrm{lq}}\frac{\mathrm{RT}}{{K}_{H}}(-\nabla {p}_{\mathrm{as}})-\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{{K}_{H}}{F}_{\mathrm{ad}}\nabla {p}_{\mathrm{as}}`

[:ref:`éq 2.1.1-1 <éq 2.1.1-1>`] can then be simplified to the following form:

:math:`C\frac{{\mathrm{dp}}_{\mathrm{as}}}{\mathrm{dt}}=Ld(\nabla {p}_{\mathrm{as}})`

with

:math:`C=\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{{K}_{H}}{S}_{0}+(1-{S}_{0})\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}`

and

:math:`L={\rho }_{\mathrm{as}}^{0}{\lambda }_{\mathrm{gz}}+\frac{\mathrm{RT}}{{K}_{H}}{\rho }_{\mathrm{ad}}^{0}{\lambda }_{\mathrm{lq}}+\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{{K}_{H}}{F}_{\mathrm{ad}}`

Heat equation whose result is known.


Benchmark results
----------------------

With the previous numerical values, we find:

:math:`{p}_{\mathrm{as}}={10}^{5}\Rightarrow {p}_{\mathrm{ad}}^{0}=\frac{\mathrm{RT}}{{K}_{H}}{p}_{\mathrm{as}}^{0}=4992`

:math:`{\rho }_{\mathrm{as}}^{0}=\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}{p}_{\mathrm{as}}^{0}=0.4` and :math:`{\rho }_{\mathrm{ad}}^{0}=\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{\mathrm{RT}}{p}_{\mathrm{ad}}^{0}=0.02`

:math:`{\rho }_{\mathrm{vp}}^{0}={\rho }_{\mathrm{vp}}={4.10}^{-3}`

The constants of the heat equation are then:

 :math:`C=2.48{10}^{-6}`

 :math:`L=1.4{10}^{-16}`


Uncertainties
------------

The uncertainties are quite large because the analytical solution is an approximate solution due to the linearization of the equations.