Pressure drop in pipe by Buckingham-Pi

Use the Buckingham $\pi$ method with the variables $\Delta P, D, L, \rho, \mu, V$ and $\epsilon/D$.

Variables and dimensions

$\epsilon/D$ is already dimensionless, so it is a $\pi$ term by itself. For the other six variables: $n = 6$, fundamental dimensions $m = 3$ (M, L, T), so $n - m = 3$ more $\pi$ terms. Total $= 4$.

Repeating variables: $D$ (geometric), $V$ (kinematic), $\rho$ (fluid property).

$\pi_1$ (with $\Delta P$)

$$\pi_1 = \Delta P\,D^{a}V^{b}\rho^{c}$$

M: $1 + c = 0 \Rightarrow c = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-1 + a + b - 3c = 0 \Rightarrow a = 0$.

$$\pi_1 = \frac{\Delta P}{\rho V^2}$$

$\pi_2$ (with $L$)

Same dimension as $D$, so $\pi_2 = \dfrac{L}{D}$.

$\pi_3$ (with $\mu$)

$$\pi_3 = \mu\,D^{a}V^{b}\rho^{c}:\quad c = -1,\; b = -1,\; a = -1 \;\Rightarrow\; \pi_3 = \frac{\mu}{\rho V D} = \frac1{Re}$$

Result

$$\frac{\Delta P}{\rho V^2} = \phi\left(\frac{\rho V D}{\mu},\ \frac LD,\ \frac\epsilon D\right)$$

Experiments show $\Delta P \propto L/D$, so

$$\Delta P = \frac{L}{D}\,\rho V^2\,\phi_1\left(Re,\ \frac\epsilon D\right)$$

This is the Darcy-Weisbach form, with friction factor $f$ related to $\phi_1$ by $\phi_1 = f/2$.