2. Analytical solution#

2.1. Calculation method#

The monophasic unsteady and one-dimensional problem can be written in a general form such as:

\(N\frac{\partial P}{\partial t}-{K}_{i}\Delta P=0\)

\(P(t=0)={P}_{0}\)

\(P(t,x=0)=0\)

\(\frac{\partial P}{\partial x}(t,x,L)=0\)

This problem admits an analytical solution obtained by development in Fourier series.

\(P={\sum }_{k=0}^{K}\frac{{\mathrm{4P}}_{0}}{(\mathrm{2k}+1)\pi }\mathrm{exp}(\frac{-{K}_{i}}{N}{\omega }_{k}^{2}t)\mathrm{sin}({\omega }_{k}x)\) with \({\omega }_{k}=(k+\frac{1}{2})\frac{\pi }{L}\)

The number of terms \(K\) in this series is determined as follows:

Let \({n}_{x}\) be the number of \({x}_{i}\) points where the solution is evaluated at an instant \(t\).

We ask:

\({a}_{k}^{i}=\frac{4}{(\mathrm{2k}+1)\pi }\mathrm{exp}(\frac{-{K}_{i}}{N}{\omega }_{k}^{2}t)\mathrm{sin}({\omega }_{k}{x}_{i})\)

So the solution can be written as: \(P({x}_{i})={\sum }_{k=0}^{K}{P}_{0}\mathrm{.}{a}_{k}^{i}\)

We choose \(K\) such as: \(\frac{1}{{n}_{x}}\sqrt{{\sum }_{i=1}^{\mathrm{nx}}{({a}_{k}^{i})}^{2}}<\epsilon\)

In practice we took \(\varepsilon ={10}^{-10}\).

The shapes of the analytical solution at times 1, 10, 100, 1000 are shown in the figure below:

_images/100000000000025800000190C21A5A0074EC48A0.png

The following table shows the number of terms by time:

Time

Number of serial terms

1

194

10

64

100

22

1000

8

Table 2.1-1: Representation of term number as a function of time

2.2. Simplifying hypotheses#

It is considered that the medium is completely saturated with gas and a zero liquid pressure is imposed on all nodes in aster. An initial gas pressure \({P}_{g}^{\mathit{ref}}\) is imposed and boundary conditions are given which correspond to a variation in this reference pressure, so \(\delta {P}_{g}\) this pressure variation is given. The gas mass conservation equation will be written as:

\(\frac{\partial (\varphi \delta {P}_{g})}{\partial t}+d(\frac{{K}_{i}{k}_{g}}{{\mu }_{g}}({P}_{g}^{r}+\delta {P}_{g})d({P}_{g}^{r}+\delta {P}_{g}))=0\)

Assuming \(\delta {P}_{g}\) small in front of \({P}_{g}^{\mathit{ref}}\), this equation becomes:

\(\frac{\partial (\varphi \delta {P}_{g})}{\partial t}+\frac{{K}_{i}{k}_{g}}{{\mu }_{g}}{P}_{g}^{r}\Delta (\delta {P}_{g})=0\)

It is therefore \(\delta {P}_{g}\) that we will identify with the solution of the model analytical equation.

In order to find the coefficients of the model problem, we will take:

\(\varphi \mathrm{=}1\)

\({k}_{g}\mathrm{=}{\mu }_{g}\mathrm{=}1\)

and we will make sure that

\({K}_{\mathit{int}}{P}_{\mathit{ref}}\mathrm{=}{10}^{\mathrm{-}3}\).