Flow through a porous medium, with pipes!

Posted in math on Wednesday, July 16 2014

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)

porous tube

Suppose fluid is coming in through the walls of a pipe, say a ceramic tube with a pressure at the outside of P2 and a pressure inside the tube of P1. What can we expect for a mass flow-rate coming out of our tube, in the steady state?

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 v0 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.