Buckingham pi theorem for pipe pressure drop

Buckingham's pi theorem

If a physical phenomenon involves $n$ variables and these contain $m$ fundamental dimensions (M, L, T), then the relation among them can be expressed in terms of $(n - m)$ independent dimensionless groups ($\pi$-terms).

Step 1: Variables

$\Delta p = f(D, L, V, \rho, \mu, \varepsilon)$, so $n = 7$.

Step 2: Number of pi terms

$m = 3$ (M, L, T), so the number of $\pi$ terms $= 7 - 3 = 4$.

Step 3: Repeating variables

Choose 3 repeating variables that together contain M, L and T and cannot form a dimensionless group among themselves: $D$ (geometric), $V$ (kinematic) and $\rho$ (dynamic).

Step 4: Form the pi terms

$\pi_1$ 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 $L$: length has the same dimension as $D$, so $\pi_2 = L/D$.

$\pi_3$ with $\mu$: $\pi_3 = \mu D^a V^b \rho^c$

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

$$\pi_3 = \frac{\mu}{\rho V D} = \frac{1}{Re}$$

$\pi_4$ with $\varepsilon$: $\pi_4 = \varepsilon/D$ (relative roughness).

Step 5: Final relation

$$\frac{\Delta p}{\rho V^2} = \phi\left(\frac{L}{D}, Re, \frac{\varepsilon}{D}\right)$$

Experiments show that $\Delta p$ is directly proportional to $L$. Therefore

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

where $f = 2\phi_1(Re, \varepsilon/D)$ is the Darcy friction factor. This is the Darcy-Weisbach equation, $h_f = f L V^2/(2 g D)$.