2. Benchmark solution

2.1. Calculation method

2.1.1. Calculation of the conservation of air mass

The conservation of the mass of gas is written as:

\(\frac{{\mathrm{dm}}_{\mathrm{air}}}{\mathrm{dt}}+d({M}_{\mathrm{as}}+{M}_{\mathrm{ad}})\) 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:

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

therefore

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

\(d{\rho }_{\mathrm{as}}=\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}{\mathrm{dp}}_{\mathrm{as}}\) and \(d{\rho }_{\mathrm{ad}}=\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{{K}_{H}}{\mathrm{dp}}_{\mathrm{as}}\)

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

Speed calculation:

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

since \({F}_{\mathrm{vp}}=0\) and \(\nabla {p}_{\mathrm{vp}}=0\)

and

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

Like \(\nabla {p}_{\mathrm{lq}}=\nabla {p}_{w}+\nabla {p}_{\mathrm{ad}}=\nabla {p}_{\mathrm{ad}}=\frac{\mathrm{RT}}{{K}_{H}}\nabla {p}_{\mathrm{as}}\)

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

[éq 2.1.1-1] can then be simplified to the following form:

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

with

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

and

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

Heat equation whose result is known.

2.2. Benchmark results

With the previous numerical values, we find:

\({p}_{\mathrm{as}}={10}^{5}\Rightarrow {p}_{\mathrm{ad}}^{0}=\frac{\mathrm{RT}}{{K}_{H}}{p}_{\mathrm{as}}^{0}=4992\)

\({\rho }_{\mathrm{as}}^{0}=\frac{{M}_{\mathrm{as}}^{\mathrm{ol}}}{\mathrm{RT}}{p}_{\mathrm{as}}^{0}=0.4\) and \({\rho }_{\mathrm{ad}}^{0}=\frac{{M}_{\mathrm{ad}}^{\mathrm{ol}}}{\mathrm{RT}}{p}_{\mathrm{ad}}^{0}=0.02\)

\({\rho }_{\mathrm{vp}}^{0}={\rho }_{\mathrm{vp}}={4.10}^{-3}\)

The constants of the heat equation are then:

\(C=2.48{10}^{-6}\)

\(L=1.4{10}^{-16}\)

2.3. Uncertainties

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