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

For a fluid velocity of the form (2.17), we need to only solve the first coordinate equation,

(2.18)

Since the second term on the right side of Eq. (2.18)is independent of by Eq. (2.17), so is the pressure gradient . It follows that is constant, and hence varies linearly along the tube. Equation (2.18)therefore reduces to the following ordinary differential equation:

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

where stands for the natural logarithm and and 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:

(2.19) (2.20)

The condition (2.20)requires that .

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

(2.21)

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.

Another classic problem derived from the Navier–Stokes equation examines the slow, incompressible, viscous flow of a fluid around a solid sphere of radius , 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:

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:

In 1851, in a tour de force of vector calculus, Stokes [139] published the solution to this boundary‐value problem, along with an expression for the total viscous force exerted on the sphere: . This force is called the Stokes drag.

For our purposes, we need not examine the calculation of in detail. Instead, we use a simpler dimensional analysis, exploiting concepts from elementary linear algebra, to deduce the functional form of the drag force. Since the only parameters in the boundary‐value problem are , , and , any solution to the problem of calculating defines a relationship of the form

(2.22)

for some function . By a theorem widely attributed to American physicist Edgar Buckingham [31], this relationship, involving variables that have physical dimensions, implies the existence of an equivalent relationship

involving only dimensionless variables Appendix Creviews this theorem.

Thus, we seek relationship equivalent to Eq. (2.22), involving only dimensionless variables formed using powers of the dimensional variables . For any such variable, denoted generically by ,

(2.23)

for exponents to be determined. Equation (2.23)implies that the exponents of , , and must vanish, yielding the following homogeneous linear system for , , , and :