One of the fluids wets the porous medium more than the other; we refer to this as the wetting phase fluid and we refer to the other as the non-wetting phase fluid. In an oil-gas system, oil is the wetting phase. Let us consider a bounded connex open Ω of \(R^{d}\) (\(d=2\mbox{ or }3\)), describing the porous medium (the reservoir), with a Lipchitz boundary Γ, and let *t* be the time variable *t* in \([0,T[\), \(T <\infty\). Let \(C_{Gg}\) be the mass fraction of the gas component in the gas phase, \(C_{Og}\) the mass fraction of the oil component (the light component) in the gas phase, and \(C_{Oo}\) the mass fraction of the oil component (the heavy component) in the oil phase which is equal to 1. While this distribution of the hydrocarbon components between the oil and gas phases plays an important role in a steam drive process, we cannot say that the mass of each phase is conserved because of the possibility of transfer of the oil component between the two phases. Instead, we observe that the total mass of each component must be conserved.

Then the mass flux of the oil and the gas components are

$$\begin{aligned}& C_{Og}\rho_{g}U_{g}+\rho_{o}U_{o}, \\& C_{Gg}\rho_{g}U_{g}. \end{aligned}$$

The oil and gas mass components are

$$\begin{aligned}& \phi ( C_{Og}\rho_{g}S_{g}+ \rho_{o}S_{o} ) , \\& \phi C_{Gg}\rho_{g}S_{g}. \end{aligned}$$

Then, for each fluid, we can write the conservation equations as

$$ \begin{aligned} &\frac{\partial}{\partial t} \bigl[ \phi ( C_{Og} \rho_{g}S_{g}+ \rho _{o}S_{o} ) \bigr] +\nabla\cdot ( C_{Og} \rho_{g}U_{g}+\rho _{o}U_{o} ) =0, \\ &\frac{\partial}{\partial t} \bigl[ \phi ( C_{Gg}\rho _{g}S_{g}) \bigr] +\nabla\cdot ( C_{Gg}\rho_{g}U_{g}) =0, \end{aligned} $$

(1)

where (\(i=o,g \)) \(S_{i}\), \(U_{i}\), \(\rho_{i}\) represent the saturation, the velocity, and the density of the phase *i*, respectively. The parameter *ϕ* is the porosity of the medium. We have

$$\begin{aligned} &C_{Og}+C_{Gg} =1, \\ &\rho_{g} =f_{1} ( P_{g},C_{Gg} ) , \\ &\rho_{o} =f_{2} ( P_{o} ), \end{aligned}$$

where \(P_{o}\) and \(P_{g}\) are the oil and gaz pressure respectively.

We consider compressible fluids, with constant dynamic viscosities and where the gravity effect is neglected. Under these hypotheses, Darcy’s law combined with the mass conservation equations for each one of the component leads to the following system of partial differential equations of parabolic convection-diffusion type:

$$\begin{aligned}& \phi(x)\frac{\partial}{\partial t} \bigl(\rho_{g}S_{g}+ \rho_{o}\omega _{o}^{l}S_{o} \bigr)+ \nabla\cdot \bigl(\rho_{g}U_{g}+\rho_{o} \omega_{o}^{l}U_{o} \bigr) =0, \end{aligned}$$

(2)

$$\begin{aligned}& \phi(x)\frac{\partial}{\partial t} \bigl(\rho_{o}\omega _{o}^{h}S_{o} \bigr)+\nabla \cdot \bigl(\rho_{o}\omega_{o}^{h}U_{o} \bigr) =0, \end{aligned}$$

(3)

$$\begin{aligned}& U_{o} =-K(x)\frac{k_{ro}}{\mu_{o}}\nabla P_{o}, \end{aligned}$$

(4)

$$\begin{aligned}& U_{g} =-K(x)\frac{k_{rg}}{\mu_{g}}\nabla P_{g}, \end{aligned}$$

(5)

where (\(i=o,g\)) \(P_{i}\), \(\mu_{i}\), \(k_{ri}\) represent the pressure, the viscosity, and the relative permeability of the phase *i*, respectively. The parameter *K* is the absolute permeability of the medium and \(\omega_{o}^{c}\), \(c=h,l\) is the mass fraction of the component *c*, denoted by *h* for the heavy component and by *l* for the light component in the oil phase. We have

$$\begin{aligned}& \mu_{g} =f_{3} ( P_{g},C_{Gg} ) , \\& \mu_{o} =f_{4} ( P_{o} ) , \\& k_{rg} =f_{5} ( S_{g},S_{o} ) , \\& k_{ro} =f_{6} ( S_{g},S_{o} ) . \end{aligned}$$

Furthermore, we shall use the subscript *S* to indicate standard conditions, *i.e.* appropriate conditions for the temperature and the pressure of medium.

Let \(\rho_{OS}\), \(\rho_{GS}\) be the density (measured at standard conditions) of the oil and the gas components, respectively. The gas formation volume factor, denoted by \(B_{G}\), is the ratio of the volume of free gas (all of which is gas component), measured at the reservoir conditions, to the volume of the same gas measured at standard conditions. Thus

$$B_{G} ( P,T ) =\frac{V_{G} ( P,T ) }{V_{GS}}. $$

Let \(W_{G}\) be the weight of free gas, since \(V_{G}=\frac{W_{G}}{\rho_{g}}\) and \(V_{GS}=\frac{W_{G}}{\rho_{gS}}\), then

$$B_{G}=\frac{\rho_{gS}}{\rho_{g}}, $$

so that the volatility of the oil in the gas is expressed by the ratio

$$R_{V}=\frac{V_{OS}}{V_{GS}}. $$

The mass fractions of the two components in the gas phase are

$$\begin{aligned}[b] &C_{Og} =\frac{R_{V}\rho_{oS}}{B_{G}\rho_{g}}, \\ &C_{Gg} =\frac{\rho_{gS}}{B_{G}\rho_{g}}. \end{aligned} $$

By adding the last two equations and noting that \(C_{Gg}+C_{Og}=1\), we obtain

$$\rho_{g}=\frac{ ( \rho_{GS}+R_{V}\rho_{OS} ) }{B_{G}}. $$

The substitution of these mass fractions and densities into (1) gives, for the gas and the oil components,

$$ \begin{aligned} &\frac{\partial}{\partial t} \biggl[ \phi \biggl( \frac{R_{V}}{B_{G}}\rho _{oS}S_{g}+\rho_{o}S_{o} \biggr) \biggr] +\nabla\cdot \biggl( \frac {R_{V}}{B_{G}}\rho_{oS}U_{g}+ \rho_{o}U_{o} \biggr) =0, \\ &\frac{\partial}{\partial t} \biggl[ \phi \biggl( \frac{\rho _{gS}S_{g}}{B_{G}} \biggr) \biggr] +\nabla\cdot \biggl( \frac{\rho_{gS}U_{g}}{B_{G}} \biggr) =0. \end{aligned} $$

(6)

We suppose that it is a saturated regime and is expressed by

The capillary pressure is given by

$$ P_{g}-P_{o}=P_{c}(S_{o})=p_{c}(S_{o})p_{cM}, $$

(8)

where

$$ p_{cM}=\sup \bigl\vert P_{c}(S_{o}) \bigr\vert \quad\mbox{and}\quad0\leq p_{c}(S_{o})\leq1. $$

(9)

We define the mobility of each phase by the formula

$$ \lambda_{i}=\frac{k_{ri}}{\mu_{i}},\quad i=o,g, $$

(10)

and the total mobility *λ* by

$$ \lambda=\lambda_{o}+\lambda_{g}. $$

(11)

To simplify, we set

$$\begin{aligned}& \rho_{o}^{h} =\rho_{o}\omega_{o}^{h}, \end{aligned}$$

(12)

$$\begin{aligned}& \rho=\rho_{g}+\rho_{o}, \end{aligned}$$

(13)

$$\begin{aligned}& b =\rho_{g}\lambda_{g}+\rho_{o} \lambda_{o}, \end{aligned}$$

(14)

$$\begin{aligned}& d =\rho_{g}-\rho_{o}. \end{aligned}$$

(15)

Let us now introduce the new unknowns, namely the reduced saturation and the global pressure in the following way: if we denote by \(S_{i,m}\) and \(S_{i,M}\), the residual and the maximum saturations of the fluid \(i=o,g\), respectively; the reduced saturation is given by

$$\begin{aligned}& S =\frac{S_{o}-S_{o,m}}{1-S_{g,m}-S_{o,m}}, \end{aligned}$$

(16)

$$\begin{aligned}& 0 \leq S\leq1. \end{aligned}$$

(17)

The ‘global pressure’ was first introduced by Chavent and Jaffre [1] in the following form:

$$ P=\frac{1}{2} ( P_{g}+P_{o} ) +\gamma ( S ) , $$

(18)

with

$$ \gamma ( S ) =\frac{1}{2}\int_{S_{o,m}}^{S} \frac {\lambda _{g}-\lambda_{o}}{\lambda}p_{c}^{\prime} ( \xi ) p_{cM}\,d\xi. $$

(19)

Hence

$$ \gamma(S)=\frac{1}{2}\int_{0}^{S}\alpha ( \xi )\,d\xi, $$

(20)

where

$$ \alpha ( S ) =\frac{\lambda_{g} ( S ) -\lambda _{o} ( S ) }{\lambda ( S ) }p_{c}^{\prime} ( S ) P_{cM} $$

(21)

is the capillary diffusion. Therefore, our model is given by the following simplified system:

$$\begin{aligned}& \begin{aligned}[b] &\Phi(x)\frac{\partial}{\partial t} \bigl( \rho_{o}^{h}S \bigr) +\phi (x)S_{o,m}\frac{\partial}{\partial t} \bigl( \rho_{o}^{h} \bigr) \\ &\quad{}-\nabla \cdot \bigl( K(x)\rho_{o}^{h}\lambda_{o}(S) \nabla P \bigr) +\nabla\cdot \bigl( K(x)\rho _{o}^{h}\alpha (S) \nabla(S) \bigr) =0, \end{aligned} \end{aligned}$$

(22)

$$\begin{aligned}& \begin{aligned}[b] &\Phi(x)\frac{\partial}{\partial t} ( \rho S ) +\phi(x)\frac{ \partial}{\partial t} ( \rho S_{o,m}+\rho_{g} ) \\ &\quad{}-\nabla \cdot \bigl(K(x)b(S,P)\nabla P \bigr)+ \nabla\cdot \bigl( K(x)d ( P ) \alpha(S)\nabla S \bigr) =0, \end{aligned} \end{aligned}$$

(23)

$$\begin{aligned}& \lambda ( S ) \nabla P\cdot\eta=0, \qquad\alpha(S)\nabla S=0, \quad\mbox{on }\Gamma\times ( 0,T ) , \end{aligned}$$

(24)

$$\begin{aligned}& S ( x,0 ) =S^{0} ( x ),\qquad P ( x,0 ) =P^{0} ( x ) ,\quad\mbox{in }\Omega, \end{aligned}$$

(25)

where \(\phi(x)=\varphi(x)(1-S_{o,m}-S_{o,g})\).