# 2.4: How to Construct Dimensionless Variables

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

You may be wondering about how you could have constructed the dimensionless variable $$\rho U D / \mu$$ on your own instead of having it presented to you. Start with a very general product $$\rho^{a} U^{b} D^{c} \mu^{d}$$. The exponents $$a$$ through $$d$$ have to be adjusted so that the $$\mathrm{M}$$, $$\mathrm{L}$$, and $$\mathrm{T}$$ dimensions of the product cancel out. One of the exponents can be chosen arbitrarily, say $$d=1$$, but then $$a$$, $$b$$, and $$c$$ have to be adjusted by solving three equations, one each for $$\mathrm{M}$$, $$\mathrm{L}$$, and $$\mathrm{T}$$, expressing the condition that the product be dimensionless. Using length as an example, you can see from the list of dimensions above that length enters into $$\rho$$ to the power $$-3$$, into $$U$$ to the power $$+1$$, into $$D$$ to the power $$+1$$, and into $$\mu$$ to the power $$-1$$. So for the length dimension to cancel out of $$\rho^{a} U^{b} D^{c} \mu$$, the following condition must be met: $$-3 a+b+c-1=0$$. (Keep in mind that we have already chosen $$d$$ to be $$1$$. Two more conditions, one for $$\mathrm{M}$$ and one for $$T$$, give three linear equations in the three unknowns $$a$$, $$b$$, and $$c$$:

$\begin{array}{rlrl}{-3 a} & {+b} & {+c} & {-1} & {=0} & {(\text { for } L)} \\ {+a} & {} & {}& {+1} & {=0} & {(\text { for } M)} \\ {} & {-b} & {} & {-1} & {=0} & {(\text { for } T)}\end{array} \label{lineq}$

The solution is $$a = -1$$, $$b = -1$$, $$c = -1$$, so the product takes the form $$\mu / \rho U D$$. This is the inverse of the Reynolds number introduced above. If $$d$$ had been taken as $$-1$$ at the outset, the result would have been the Reynolds number itself.

This page titled 2.4: How to Construct Dimensionless Variables is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by John Southard (MIT OpenCourseware) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.