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

The Mathematics of Fluid Flow Through Porous Media

1st Column bold v left-parenthesis bold x right-parenthesis 2nd Column equals bold 0 comma 3rd Column double-vertical-bar bold x double-vertical-bar 4th Column equals upper R semicolon 2nd Row 1st Column bold v left-parenthesis bold x right-parenthesis 2nd Column right-arrow v Subscript infinity Baseline bold e 1 comma 3rd Column double-vertical-bar bold x double-vertical-bar 4th Column right-arrow infinity period EndLayout"/>

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: bold upper F equals upper F bold e 1 equals 6 pi mu upper R v Subscript infinity Baseline bold e 1 . This force is called the Stokes drag.

For our purposes, we need not examine the calculation of upper F 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 , v Subscript infinity , and upper R , any solution to the problem of calculating upper F defines a relationship of the form

(2.22) phi left-parenthesis upper F comma mu comma v Subscript infinity Baseline comma upper R right-parenthesis equals 0 comma

for some function phi . 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

normal upper Phi left-parenthesis normal upper Pi 1 comma normal upper Pi 2 comma ellipsis right-parenthesis equals 0

involving only dimensionless variables normal upper Pi 1 comma normal upper Pi 2 comma ellipsis Appendix C reviews this theorem.

Thus, we seek relationship equivalent to Eq. (2.22), involving only dimensionless variables formed using powers of the dimensional variables upper F comma mu comma v Subscript infinity Baseline comma and upper R . For any such variable, denoted generically by normal upper Pi ,

(2.23)

for exponents n 1 comma n 2 comma n 3 comma n 4 to be determined. Equation (2.23) implies that the exponents of normal upper M , normal upper L , and normal upper T must vanish, yielding the following homogeneous linear system for n 1 , n 2 , n 3 , and n 4 :

(2.24) Start 3 By 4 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 0 4th Column 0 2nd Row 1st Column 1 2nd Column negative 1 3rd Column 1 4th Column 1 3rd Row 1st Column negative 2 2nd Column negative 1 3rd Column negative 1 4th Column 0 EndMatrix Start 4 By 1 Matrix 1st Row n 1 2nd Row n 2 3rd Row n 3 4th Row n 4 EndMatrix equals StartBinomialOrMatrix 0 Choose 0 EndBinomialOrMatrix period

Exercise 2.13 Row‐reduce Eq. (2.24) to deduce that n 4 is a free variable, for which one may choose any value, and n 3 equals n 2 equals minus n 1 equals n 4 .

Arbitrarily picking n 4 equals negative 1 yields the single dimensionless variable normal upper Pi equals upper F mu Superscript negative 1 Baseline v Subscript infinity Superscript negative 1 Baseline upper R Superscript negative 1 ; all other dimensionless variables for this problem must be multiples of this product.

The calculation in Exercise 2.13 shows that any relationship equivalent to Eq. (2.22) but involving only dimensionless variables has the form normal upper Phi left-parenthesis normal upper Pi right-parenthesis equals 0 . Solutions to such an equation are constant values of . Setting normal upper Pi equals upper C for a generic constant upper C , we conclude that Stokes drag has the form

(2.25) upper F equals upper C mu upper R v Subscript infinity Baseline period

This result is consistent with that of Stokes's original calculation, except that we have an undetermined constant upper C instead of Скачать книгу