Roughness height from velocity ratio

Data. $D = 100$ mm so $R = 50$ mm. Distances from the wall: $y_1 = 10$ mm, $y_2 = 30$ mm. $u_2 = 1.90$ m/s at 30 mm; $u_1/u_2 = 0.838$. Water, $\rho = 1000$ kg/m³.

Velocity distribution in a hydrodynamically rough pipe:

$$\frac{u}{u_*} = 5.75\log_{10}\frac{y}{k} + 8.5$$

Shear velocity

$u_1 = 0.838(1.90) = 1.5922$ m/s. Subtracting the equation at $y_1$ from the one at $y_2$ (the constants and $k$ cancel):

$$\frac{u_2 - u_1}{u_*} = 5.75\log_{10}\frac{y_2}{y_1} = 5.75\log_{10}3 = 2.7434$$ $$u_* = \frac{1.90 - 1.5922}{2.7434} = \frac{0.3078}{2.7434} = 0.1122\ \text{m/s}$$

(i) Average roughness height

At $y_2 = 30$ mm: $\dfrac{1.90}{0.1122} = 16.935 = 5.75\log_{10}\dfrac{30}{k} + 8.5$

$$\log_{10}\frac{30}{k} = \frac{16.935 - 8.5}{5.75} = 1.4669 \;\Rightarrow\; k = \frac{30}{10^{1.4669}} = 1.024\ \text{mm}$$

Check: $\dfrac{u_*k}{\nu} = \dfrac{0.1122(1.024\times10^{-3})}{10^{-6}} = 115 > 100$, so the pipe is hydrodynamically rough, as given.

(ii) Wall shear stress

$$\tau_0 = \rho u_*^2 = 1000(0.1122)^2 = 12.59\ \text{Pa}$$

(iii) Mean velocity

For rough pipes $\dfrac{V}{u_*} = 5.75\log_{10}\dfrac{R}{k} + 4.75$, with $R/k = \dfrac{50}{1.024} = 48.84$:

$$V = 0.1122\left[5.75\log_{10}(48.84) + 4.75\right] = 1.622\ \text{m/s}$$

Answer: $k = 1.024$ mm (about 1.0 mm); $\tau_0 = 12.59$ Pa (about 12.6 Pa); $V = 1.622$ m/s.