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Deriving Physical Relationships

Consider a viscous fluid flowing through a circular pipe. The volume rate of fluid flow through the pipe, $Q$, might plausibly depend on the radius of the pipe, $a$, the viscosity of the fluid, $\eta$, and the pressure gradient along the pipe, ${\mit\Delta} p/l$. Here, ${\mit\Delta} p$ is the pressure difference between the two ends of the pipe, and $l$ is the length of the pipe. Let us guess that

$\displaystyle Q = A\,a^x \,\eta^{\,y}\left(\frac{{\mit\Delta}p}{l}\right)^z,$ (1.5)

where $x$, $y$, and $z$ are, as yet, unknown exponents, and $A$ is a dimensionless constant. Now, the dimensions of $Q$ are $[L]^3/[T]$, the dimensions of $a$ are $[L]$, the dimensions of $\eta$ are $[M]/([L]\,[T])$, and the dimensions of ${\mit\Delta} p/l$ are $\{([M]\,[L]/[T]^2)/[L]^2\}/[L] = [M]/([L]^2\,[T]^2)$. Thus, equating the dimensions of the left- and right-hand sides of the previous equation, we obtain

$\displaystyle \frac{[L]^3}{[T]} = [L]^x\left(\frac{[M]}{[L]\,[T]}\right)^y\left(\frac{[M]}{[L]^2\,[T]^2}\right)^z.$ (1.6)

Now, if Equation (1.5) is to be dimensionally consistent then we can separately equate the exponents of length, mass, and time in the previous expression. Equating the exponents of $[L]$, we obtain

$\displaystyle 3 = x -y-2\,z.$ (1.7)

Equating the exponents of $[M]$, we get

$\displaystyle 0= y+z.$ (1.8)

Finally, equating the exponents of $[T]$, we obtain

$\displaystyle -1=-y-2\,z.$ (1.9)

It is easily seen that $x=4$, $y=-1$, and $z=1$. Hence, we deduce that

$\displaystyle Q = A\,\frac{a^4}{\eta}\left(\frac{{\mit\Delta}p}{l}\right).$ (1.10)

next up previous
Next: Scaling Laws Up: Units Previous: Dimensional Analysis