# 2.4: How to Construct Dimensionless Variables

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- 4158

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.