Wall shear stress as dimensionless function

Variables: $\tau_w, U, \delta, u', \rho, dp/dx$. So $n = 6$, $m = 3$ (M, L, T), giving $n - m = 3$ $\pi$ terms. Repeating variables: $\rho, U, \delta$ (as given).

Dimensions

$\tau_w = ML^{-1}T^{-2}$, $U = u' = LT^{-1}$, $\delta = L$, $\rho = ML^{-3}$, $dp/dx = ML^{-2}T^{-2}$.

$\pi_1$ (with $\tau_w$)

$$\pi_1 = \tau_w\,\rho^{a}U^{b}\delta^{c}$$

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

$$\pi_1 = \frac{\tau_w}{\rho U^2}$$

$\pi_2$ (with $u'$)

It has the same dimensions as $U$, so:

$$\pi_2 = \frac{u'}{U}$$

$\pi_3$ (with $dp/dx$)

$$\pi_3 = \frac{dp}{dx}\rho^{a}U^{b}\delta^{c}$$

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

$$\pi_3 = \frac{\delta}{\rho U^2}\,\frac{dp}{dx}$$

Result

$$\frac{\tau_w}{\rho U^2} = \phi\left(\frac{u'}{U},\ \frac{\delta}{\rho U^2}\frac{dp}{dx}\right)$$

The left side is half the skin friction coefficient ($C_f/2$), and the second argument is a dimensionless pressure gradient.