1. Reference problem#

1.1. Geometry#

The field studied is a \(3m\) bar.

_images/1000000000000168000000A7B0D6772D8EDCB228.gif

Coordinates of points \((m)\):

\(A(0;0)\) \(C(3;0.1)\)

\(B(3;0)\) \(D(0;0.1)\)

1.2. Material properties#

Here we take data that leads to an almost unitary problem. The units then no longer have any physical meaning.

Gravity is taken in the positive \(x\) direction (which therefore corresponds to the vertical axis).

Liquid water

Density \(({\mathrm{kg.m}}^{-3})\) Viscosity \((\mathrm{Pa.}s)\)

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Gas

Viscosity \((\mathrm{Pa.}s)\)

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Homogenized parameters

Gravity \((\mathrm{m.}{s}^{-2})\) Permeability \(K({m}^{2})\) Porosity Sorption isotherm Relative permeability

\(g=(\mathrm{9,81 };\mathrm{0 };0)\) 1 0.5 \({S}_{\text{we}}=\frac{1}{{\left[1+{(\frac{{P}_{c}}{1})}^{\mathrm{1,5}}\right]}^{1/3}}\) \(\begin{array}{}{\text{kr}}_{w}(S)=1\\ {\text{kr}}_{\text{gz}}(S)=1\end{array}\)

1.3. Boundary and initial conditions#

There is no flow everywhere. Initially, the medium is desaturated with a saturation of \(S=\mathrm{0,5}\) over the entire domain, which corresponds to capillary pressure:

\(\mathrm{Pc}=\mathrm{3,6}\mathrm{Pa}\).

An initial gas pressure of \(1\mathrm{Pa}\) is taken.

1.4. No time#

We model \(1s\) in the following way:

  • from \(0\) to \(\mathrm{0,1}s\): 5 steps of time

  • from \(\mathrm{0,1}\) to \(1s\): 9 steps of time

1.5. Benchmark solution#

In the steady state, hydrostatic equilibrium must be achieved.

So we have to \({\mathrm{\Delta P}}_{\text{lq}}=\rho \text{.}g\text{.}\mathrm{\Delta x}\)

With the data we have, it therefore gives us \({\mathrm{\Delta P}}_{\text{lq}}=\mathrm{9,}\text{81}\ast 3=\mathrm{29,43}\mathrm{Pa}\)