Mean velocity law for rough pipes

Start. For a hydrodynamically rough pipe the velocity at distance $y$ from the wall, from Prandtl's mixing length theory and Nikuradse's experiments, is

$$\frac{u}{u_*} = 5.75\log_{10}\frac{y}{K} + 8.5 \quad\cdots(1)$$

where $u_* = V_* = \sqrt{\tau_0/\rho}$ is the shear velocity and $K$ is the roughness height. It is assumed to hold over the whole section. Mean velocity is

$$V = \frac{1}{\pi R^2}\int_0^R u\,2\pi r\,dr$$

Put $y = R - r$, so $r = R - y$ and $dr = -dy$:

$$V = \frac{2}{R^2}\int_0^R u\,(R - y)\,dy$$

Split (1) as $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{R}{K} + 8.5 + 5.75\log_{10}\dfrac{y}{R}$. The first two terms are constant, and for them $\dfrac{2}{R^2}\displaystyle\int_0^R (R-y)\,dy = 1$. For the last term use $\log_{10}x = 0.4343\ln x$ and let $\eta = y/R$:

$$\frac{2}{R^2}\int_0^R (R - y)\ln\frac{y}{R}\,dy = 2\int_0^1 (1 - \eta)\ln\eta\,d\eta$$

Using $\int_0^1 \ln\eta\,d\eta = -1$ and $\int_0^1 \eta\ln\eta\,d\eta = -\dfrac14$:

$$2\left[-1 - \left(-\frac14\right)\right] = 2\left(-\frac34\right) = -\frac32$$

Hence the area-weighted average of $5.75\log_{10}(y/R)$ is

$$5.75(0.4343)\left(-\frac32\right) = 2.5\left(-\frac32\right) = -3.75$$

Result.

$$\frac{V}{V_*} = 5.75\log_{10}\frac{R}{K} + 8.5 - 3.75$$ $$\boxed{\frac{V}{V_*} = 5.75\log_{10}\left(\frac{R}{K}\right) + 4.75}$$

(The same integral applied to the smooth-pipe law gives $\dfrac{V}{V_*} = 5.75\log_{10}\dfrac{V_*R}{\nu} + 1.75$.) The equation also gives $f$ for rough pipes: since $V_* = V\sqrt{f/8}$, $\dfrac{1}{\sqrt f} = 2.03\log_{10}\dfrac{R}{K} + 1.68$, close to the von Karman-Prandtl form.