Flow in and around pipes is old hat at this point, so to keep things fresh what about flow through porous medium, into a pipe? (Transport Phenomena 4C.4)

Suppose fluid is coming in through the walls of a pipe, say a ceramic tube with a pressure at the outside of *P _{2}* and a pressure inside the tube of

*P*. What can we expect for a mass flow-rate coming out of our tube, in the steady state?

_{1}When dealing with porous media it is common to use a *smoothed* equation, where we average over the pores of the medium to get an effective flow. The smoothed continuity equation is:
$$ \varepsilon \frac{\partial \rho}{\partial t} = - \left( \nabla \cdot \rho \mathbf{v_0} \right) $$

Where $latex \varepsilon$ is the porosity and *v _{0}* is the superficial velocity.

Darcy's law is given by: $$ \mathbf{v_0} = -\frac{\kappa}{\mu} \nabla \mathcal{P} $$

Where *κ* is the permeability of the porous medium.

Substituting Darcy's law into the smoothed continuity equation, for an incompressible fluid, leads to:

$$ \nabla^2 \mathcal{P} = 0 $$

Returning to our pipe, suppose the pressure depends only on the radial direction, *r*, and not on the angle or length along the pipe. In cyclindrical coordinates:

$$ \nabla^2 \mathcal{P} = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial}{\partial r} \mathcal{P} \right) = 0 $$

Integrating twice we arrive at: $$ \mathcal{P} = A\cdot \ln r + B $$

Where *A* and *B* are constants to be determined.

The boundary conditions are: $$ \mathcal{P} = P_1 \mbox{ at } r = R_1 $$

and

$$ \mathcal{P} = P_2 \mbox{ at } r = R_2 $$

To solve for *A*:
$$ P_1 - P_2 = A \cdot \ln \left( \frac{R_1}{R_2} \right) $$
$$ A = \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} $$

Then: $$ \mathcal{P} - P_1 = A \cdot \ln \left( \frac{r}{R_1} \right) $$ $$ \mathcal{P} - P_1 = \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} \cdot \ln \left( \frac{r}{R_1} \right) $$ $$ \mathcal{P} = \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} \cdot \ln \left( \frac{r}{R_1} \right) + P_1 $$

Plugging this into Darcy's Law: $$ \mathbf{v_0} = -\frac{\kappa}{\mu} \nabla \mathcal{P} = -\frac{\kappa}{\mu} \frac{\partial}{\partial r} \mathcal{P} $$ $$ \mathbf{v_0} = -\frac{\kappa}{\mu} \frac{\partial}{\partial r} \left( \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} \cdot \ln \left( \frac{r}{R_1} \right) + P_1 \right) $$ $$ \mathbf{v_0} = - \frac{\kappa}{\mu} \cdot \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} \cdot \frac{1}{r} $$

We can easily obtain the mass flow rate out of that ceramic tube by integrating over the flow area: $$ w = \int \limits_{0}^{h} \int \limits_{0}^{2 \pi} \int \limits_{R_1}^{R_2} \rho \cdot \mathbf{v_0} \cdot r dr d\theta dz $$ $$ w = - \frac{2 \pi h \rho \kappa}{\mu} \cdot \frac{P_1 - P_2}{\ln \left( \frac{R_1}{R_2} \right)} $$

Which is what we sought to find out from the very beginning.