The Mathematics of Fluid Flow Through Porous Media - Myron B. Allen, III

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theta represents the azimuth about the axis. In this coordinate system, the Laplace operator has the form

      (B.7)nabla squared equals nabla dot nabla equals StartFraction partial-differential squared Over partial-differential z squared EndFraction plus StartFraction 1 Over r EndFraction StartFraction partial-differential Over partial-differential r EndFraction left-parenthesis r StartFraction partial-differential Over partial-differential r EndFraction right-parenthesis plus StartFraction 1 Over r squared EndFraction StartFraction partial-differential squared Over partial-differential theta squared EndFraction period

      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

      Exercise 2.10 Show that, under these conditions, the nonlinear term left-parenthesis bold v dot nabla right-parenthesis bold v vanishes. Work in Cartesian coordinates.

Geometric representation of the profile of flow through a thin circular cylinder having radius R.
.

bold 0 equals minus nabla p plus StartFraction 1 Over Re EndFraction nabla squared bold v period

      (Here we use the dimensionless form.)

StartFraction 1 Over r EndFraction StartFraction d Over d r EndFraction left-parenthesis r StartFraction d v Over d r EndFraction right-parenthesis equals p Superscript prime Baseline Re period

      Exercise 2.11 Verify that the general solution to this equation has the form

v equals StartFraction p prime Re Over 4 EndFraction left-parenthesis r squared plus upper C 1 log r plus upper C 2 right-parenthesis comma

      where log stands for the natural logarithm and upper C 1 and upper C 2 denote arbitrary constants.

      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:

      The condition (2.20) requires that upper C 1 equals 0.

      Exercise 2.12 Impose the no‐slip boundary condition (2.19) to show that

      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 upper R, as drawn in Figure 2.9. In the case of steady flow when the Reynolds number is much smaller than 1, we neglect the inertial terms, arriving at the following mass and momentum balance equations:

StartLayout 1st Row 1st Column nabla dot bold v 2nd Column equals 0 comma 2nd Row 1st Column mu nabla squared bold v 2nd Column equals nabla p period EndLayout

      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:

StartLayout <p style= Скачать книгу