The Mathematics of Fluid Flow Through Porous Media. Myron B. Allen, III
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Appendix B reviews the derivation of this expression.
In view of the symmetry of the problem about the axis of the tube, we seek solutions of the form
that is, the axial component depends only on distance from the axis of the tube, and the radial and azimuthal coordinates of the velocity vanish, as drawn in Figure 2.10. We allow the pressure to vary with
Exercise 2.10 Show that, under these conditions, the nonlinear term
Figure 2.10 Profile of flow through a thin circular cylinder having radius
.Since the flow is steady, the Navier–Stokes equation therefore reduces to
(Here we use the dimensionless form.)
For a fluid velocity of the form (2.17), we need to only solve the first coordinate equation,
Since the second term on the right side of Eq. (2.18) is independent of
Exercise 2.11 Verify that the general solution to this equation has the form
where
For boundary conditions, we assume no slip at the wall of the tube and insist that the velocity along the axis of the tube remain finite:
(2.19) (2.20)
The condition (2.20) requires that
Exercise 2.12 Impose the no‐slip boundary condition (2.19) to show that
Equation (2.21) indicates that the fluid velocity has a parabolic profile, with the velocity reaching its maximum magnitude along the axis of the tube and vanishing at the walls. We encounter this profile again in Section 5.1.2, in discussing why solutes spread as they move through the microscopic channels of a porous medium.
2.4.2 The Stokes Problem
Another classic problem derived from the Navier–Stokes equation examines the slow, incompressible, viscous flow of a fluid around a solid sphere of radius
On the surface of the solid sphere, the velocity vanishes, while as one moves far away from the sphere the velocity approaches a uniform far‐field value: