Question 3.10: Radial Flow Through a Porous-Walled Tube Consider pressure-d......

Writing Eq. (3.22) in reduced form (see Assumptions and Concept):

\rm\nabla p=-\frac{\mu}{æ}\vec{\bf u}-\underbrace{C_F\frac{\rho\vec{\bf u}|\vec{\bf u}|}{\sqrt æ} }_{\underset{term}{Forchheimer} } +\underbrace{\hat\mu\nabla^2\vec{\bf u}}_{\underset{term}{Brinkman} }                 (3.22)

\rm \vec{\bf {v}}=-\frac{æ}{\mu} \nabla p                         (E.3.10.1)

and taking the divergence of Eq. (E.3.10.1) we have with \rm\nabla \cdot \vec v=0 (continuity equation for incompressible fluids)

0=-\frac{æ}{\mu} \nabla^2 p                        (E.3.10.2)

where ∇² is the Laplace operator (App. A). For our 1-D case in cylindrical coordinates, Eq. (3.10.2) is simply

\rm \frac{1}{r} \frac{d}{dr} \left\lgroup r\frac{dp}{dr} \right\rgroup =0                         (E.3.10.3)

subject to \rm p(r=R_1)=p_i\,\text{and}\,p(r=R_2)=p_o. Double integration leads to

And finally for \rm p_o\gt p_i:

\rm \frac{{ p}-{ p_{i}}}{{ p}_{o}-{ p_{i}}}=\frac{\mathrm{ln}({ r}/{ R}_{1})}{\mathrm{ln}({ R}_{2}/\,{ R}_{1})}                           (E.3.10.4)

Now, with Darcy’s law in the r-direction, i.e.,

\rm v=-\frac{æ}{\mu} \left\lgroup \frac{dp}{dr} \right\rgroup                  (E.3.10.5a)

we obtain with the given p(r) and \rm \Delta p\equiv  p_o-p_i

\rm v(r)=-\frac{æ}{\mu} \frac{\Delta p}{\ln (R_2/R_1)} \frac{1}{r}                      (E.3.10.5b)

A radial mass balance provides the added mass flow rate

\rm \Delta \dot m =-\rho v\Big|_{r=R_1}A_{surface}                     (E.3.10.6a)

where \rm A_{surface}=2\pi R_1L, so that

\Delta\rm\dot m=\frac{2\pi æ\rho \Delta p}{\mu\ln(R_2/R_1)} L                 (E.3.10.6b)

Graph:

Comments: The v(r) -function (E.3.10.5b) is hyperbolic, i.e., for a given Δp, æ , μ and geometry, v decreases inversely with r. As expected radial mass influx increases with wall permeability, tube length, and pressure difference.