Darcy's flow

Porous Media Flow...

Darcy's law: $$u=-\frac{1}{\mu}(K\cdot \nabla p).$$ Here

• $u$ is the Darcy velocity (total discharge rate divided by the cross-sectional area),
• $K$ is the permeability tensor, its three components are the permeability constants for the three directions (unequal in the anisotropic case);
• $p$ is the pressure;
• $\mu$ is the viscosity of the fluid.
• $v$ is the pore velocity is the Darcy velocity divided by the porosity $\phi$.

Assumed:

• The permeability is proportional to the average pore diameter.
• Both the fluid and the rock are incompressible.
• Density and porosity are constant.

For the single fluid phase, the conservation of mass produces the steady state equation called the pressure equation: $$\nabla\Big( \frac{1}{\mu}(K\cdot \nabla p) + \frac{q}{\rho} \Big) =0.$$ Here

• $q$ is the mass flow rate per unit volume;
• $\rho$ is the density of the flow.

For a dual phase flow, we have two fluids in a porous media. Then we assign index $*_w$ for wet and $*_n$ for non-wet analogues of the quantities above. Then the respective Darcy's velocities satisfy: $$u_w=-\frac{k_{rw}}{\mu _w}\nabla p_w,$$ $$u_n=-\frac{k_{rn}}{\mu _n}\nabla p_n.$$ Here we assume the case is isotropic and, hence, $K=kI$, where $k$, the absolute permeability, is a scalar that may depend on location (but not direction). The the relative permeability constants are given by: $$k_{rw}=\frac{k_w}{k},$$ $$k_{rn}=\frac{k_n}{k}.$$ The capillary pressure: $p_c=p_n-p_w$.

We have wet $S_w$ and non-wet $S_n$ saturation. Then it follows $S_w+S_n=1$. The total mobility is $$\lambda (S_w) = \frac{k_{rn}(S_w)}{\mu _n} + \frac{k_{rw}(S_w)}{\mu _w}.$$ We have then a PDE: $$\nabla \cdot u = - \nabla \cdot \Big( k\lambda \nabla p \Big) = \frac{q_n}{\rho _n} + \frac{q_w}{\rho _w}.$$ Here:

• $p$ is the average pressure of the two fluids,
• $u$ is the total fluid velocity.