Permissible roughness for minimum energy loss

Idea. A pipe has the minimum energy loss when it is hydrodynamically smooth, since then $f$ depends only on $Re$ and the roughness adds nothing. The pipe stays smooth as long as the roughness is buried in the laminar sub-layer:

$$\frac{u_*k}{\nu} \le 4 \quad\Rightarrow\quad k \le \frac{4\nu}{u_*}$$

Data. $D = 1$ m, $Q = 1.0$ m³/s, $\nu = 1.0\times10^{-6}$ m²/s (water at about 20 °C).

Step 1: velocity and Reynolds number

$$V = \frac{Q}{\pi D^2/4} = 1.2732\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 1273240$$

Step 2: smooth-pipe friction factor (Prandtl: $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$)

$$f = 0.01118$$

Step 3: shear velocity

$$u_* = V\sqrt{\frac{f}{8}} = 1.2732\sqrt{\frac{0.01118}{8}} = 0.04759\ \text{m/s}$$

Step 4: permissible roughness

$$k \le \frac{4\nu}{u_*} = \frac{4(1.0\times10^{-6})}{0.04759} = 0.0840\ \text{mm}$$

Answer: The surface roughness height must not exceed about $0.084$ mm (0.08 mm) for minimum loss. With the looser limit $u_*k/\nu \le 5$, the value is 0.105 mm.