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Question 3.9: Creeping Flow in a Channel Filled with a Porous Medium A sla......

Creeping Flow in a Channel Filled with a Porous Medium

A slab of homogeneous porous medium of thickness h, bound by impermeable channel walls, represents a good base case for axial flow through porous insulation, a filter, a catalytic converter, tissue, i.e., extracellular matrix, etc.

Concepts Assumptions Sketch
Use of the Darcy- Brinkman equation to invoke noslip at channel wall • Creeping flow, i.e., 〈u〉 ≡ u , where u(y) only
• ∂p / ∂x = ¢
• Constant properties
• Symmetry
Step-by-Step
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The 1-D form of Eq. (3.22) without the “high-speed” Forchheimer term reads:

\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)

{\frac{\mathrm{d}\mathrm{p}}{\mathrm{d}\mathrm{x}}}=-{\frac{\mathrm{\mu}}{æ}}\mathrm{u}+\mathrm{\mu}{\frac{\mathrm{d}^{2}\mathrm{u}}{\mathrm{d}\mathrm{y}^{2}}}={¢}                     (E.3.9.1a)

or

\rm \frac{\mathrm{d}^{2}\mathrm{u}}{\mathrm{d}y^{2}}\!-\!\frac{1}{\mathrm{æ}}\!\mathrm{u}\!=\!\frac{1}{\mu\mathrm{}}\left\lgroup \frac{dp}{dx} \right\rgroup=¢                      (E.3.9.1b)

subject to \rm u\left\lgroup y=\frac{h}{2} \right\rgroup =0 and \rm \frac{du}{dy} =0 \,\text{at}\,y=0.

The homogeneous solution to Eq. (E.3.9.1b) can be found on page 130 of Polyanin & Zaitsev (1995) or other math handbooks:

\rm \mathrm{u}(y)=-\frac{\mathrm{æ}\mathrm{h}}\mu\left\lgroup\frac{\mathrm{d}{ p}}{\mathrm{d}{x}}\right\rgroup\left[1-\frac{\cosh\!\left({ y}/\sqrt{æ}\right)}{\cosh\!\left(\mathrm{h/}{\sqrt{æ}}\right)}\right]                       (E.3.9.2)

An effective hydraulic conductivity can be estimated, given the channel flow rate per unit depth, i.e.,

\hat Q=2\int_0^{h/2}u\mathrm{d}\mathbf{y}\,:=-{\frac{\mathrm{æh}}{\mathrm{{\mu}}}}\left\lgroup {\frac{\mathrm{d}\mathrm{p}}{\mathrm{d}\mathrm{x}}}\right\rgroup \left[1-{\frac{2{\sqrt{æ}}}{\mathrm{{h}}}}\operatorname{tanh}({\frac{\mathrm{{h}}}{2{\sqrt{æ}}}})\right]                      (E.3.9.3a)

Defining K_{eff} as \overline{{{\mathrm{u}}}}/(\mathrm{d}{\mathrm{p}}/\mathrm{d}\mathrm{x})=-{\frac{\hat{\mathrm{Q}}/\mathrm{h}}{\mathrm{d}{\mathrm{p}}/\mathrm{d}\mathrm{x}}} we obtain:

\frac{\mathrm{K}_{\mathrm{eff}}}{\mathrm{æ}/\mathrm{\mu}}\equiv\frac{\mathrm{K}_{\mathrm{eff}}}{\mathrm{K}}=1-\frac{2\sqrt{æ}}{\mathrm{h}}{\tanh}\left\lgroup\frac{\mathrm{h}}{2\sqrt{æ}}\right\rgroup                      (E.3.9.3b)

Graphs:

Comments: The first graph depicts a u(y)-profile determined by the (1− cosh y) function (i.e., compare Eq. (E.3.9.1b) with Eq. (E.3.4.5) for Poiseuille flow). The second graph shows the expected nonlinear increase of the nondimensionalized \rm K_{eff} with channel height h.

{\rm u(r)=\frac{r_0^2}{4\mu}} \left\lgroup\frac{\Delta {\rm p}}{l} \right\rgroup\rm \left[1-\left\lgroup\frac{r}{r_0} \right\rgroup^2 \right] =u_{max}\left[1-\left\lgroup\frac{r}{r_0} \right\rgroup^2 \right]                        (E.3.4.5a, b)
example 3.9

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