The 3 equations are:

Mass conservation laws for water and for CO2:

(1)

$\displaystyle

\phi\frac{\partial s} {\partial t} + \frac{\partial u}{\partial x} \( F(s) \) =0 $

(2)

$\displaystyle

\phi \frac{\partial \rho} {\partial t}(P)(1-s)+\frac{\partial \rho}{\partial x}P(u)

(1-F(s))=0

$

The Darcy Law for both phases, water and gas is

(3)

$\displaystyle

u = -k (\frac{s(k_(rw))}{\mu_w}+\frac{s(k_(rg))}{\mu_g})

(\frac{\partial P}{\partial x}) $

Initial and boundary conditions

(x is distance and t is time,P is pressure,s is saturation):

At t=0, s=1

At x=0. s=s_wi

At x=0, P=P_1

At x=L, P=P_2

Three variables to be found:

$\displaystyle s(x,t); u(x,t); P(x,t); $

F is a function of s and F is ratio of relative premeability and viscosity.

All others are known....rw and rg are relative premeabilities for water and gas...

mu_w and mu_g are viscosities for water and gas.

Numerical solution adopted by me:

Consider finite steps,

(\Delta x) and (\Delta t).

$\displaystyle {s_i} ^ k =s (i \Delta x,k \Delta t) $ and the same for P and u.

Then (for the explicit method), we can write approximately using discretization as

$\displaystyle \frac{\partial s}{\partial t}= \frac{({s_i}^(k+1)-{s_i}^(k))}{\Delta t}$

and

$\displaystyle

\frac{\partial u}{\partial x}F(s)=\frac{uF_(i+1)^k-uF_(i)^k}{\Delta x} $

On substitution in (1), we get an equation for s at (i,k+1) .

Now , i tried to do the same for the other 2 equations but could not separate the

variables u and p.Also did not know how to use the initial and boundary conditions.

But i think the procedure could be like:

The solution at the layer k=0 (t=0) is known from initial conditions.

Assume that the solution at layer k has been calculated. In order to find the solution at the layer k+1,

1) Find the values of saturation s_i^k+1, for each i, from Eq. (s);

2) Find the values of rho_i^k+1= rho(P_i^k+1) from Eq. (r);

3) Re-calculate P_i^k+1 based on the known values of rho_i^k+1;

4) Find the values of u_i^k+1 from Eq. (u).

So , please help me as its urgent...thank you so much...[/quote]