How should I decide which principal units to use in dimensional analysis based on resulting pi's

Question

Consider an unknown equation involving six different variables:

• period $T$, in seconds
• velocity $v$, in meters per second
• pressure difference $\Delta p$, in Pascal
• length $L$, in meters
• volume $V$, in cubic meters
• density $\rho$, in kilograms per cubic meter

We want to find an expression for $T$. Applying dimensional analysis and Buckhingham's Pi Theorem, we need a basis of $3$ fundamental units and $3$ dimensionless $\pi$'s.

Consider the basis: $v$, $\Delta p$, $L$ ($T, M L^{-1} T^{-1}, L$)

The $\pi$ which includes $\rho$ will be

$$\pi_1 = \frac{\rho v^2}{\Delta p}$$

Now consider the basis: $v$, $\rho$, $L$ ($T, M L^{-3}, L$)

The $\pi$ which includes $\Delta p$ will be

$$\pi_1 = \frac{\Delta p}{\rho v^2}$$

which is the inverse of the $\pi$ found in the previous base. However the remaining two $\pi$'s remain exactly equal!

$$\pi_2 = \frac{Tv}{L}$$ $$\pi_3 = \frac{V}{L^3}$$

Which leads to the following contradicting result

$$\frac{Tv}{L} = \phi(\frac{\rho v^2}{\Delta p}, \frac{V}{L^3})$$

$$\frac{Tv}{L} = \phi(\frac{\Delta p}{\rho v^2}, \frac{V}{L^3})$$

So does $T$ depend on $v^2$ or on $v^{-2}$? How do I know which one it is? Did I make a mistake or is the answer to this question beyond the abilities of dimensional analysis?

Answer 1

Assuming your dimensional analysis is correct, the functional dependence is of the form: $$\frac{Tv}{L} = k_1\left(\frac{\rho v^2}{\Delta p}\right)^{\alpha_1}\left(\frac{V}{L^3}\right)^{\beta_1}$$ or: $$\frac{Tv}{L} = k_2\left(\frac{\Delta p}{\rho v^2}\right)^{\alpha_2}\left(\frac{V}{L^3}\right)^{\beta_2}$$

The only way to determine the numerical value of the coefficients is either through an analytical result or through an experiment.

In general you should find that $k_1=k_2=k$, $\alpha_1=-\alpha_2=\alpha$ and $\beta_1=\beta_2=\beta$ so it does not matter what the functional dependence is.