Minimum hydrodynamically smooth pipe size

Criterion. A pipe is hydrodynamically smooth when the roughness is buried in the laminar sub-layer, i.e. when the roughness Reynolds number (Nikuradse) satisfies

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

with $u_* = V\sqrt{f/8}$ and $k = 0.15$ mm.

Step 1: friction factor of a smooth pipe at $Re = 2\times10^5$. Prandtl's smooth-pipe law $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$ gives (by iteration)

$$f = 0.01564, \qquad \sqrt{f/8} = 0.04421$$

Step 2: express the criterion in terms of $D$. Since $Re = VD/\nu$,

$$\frac{u_* k}{\nu} = \frac{V\sqrt{f/8}\,k}{\nu} = Re\,\sqrt{\frac{f}{8}}\,\frac{k}{D} \le 4$$ $$D \ge \frac{Re\,k\sqrt{f/8}}{4} = \frac{(2\times10^5)(0.00015)(0.04421)}{4} = 0.332\ \text{m}$$

Answer: Minimum diameter $\approx 0.332$ m (about 330 mm). Any smaller pipe at this Reynolds number is not smooth. (Using the limit $u_*k/\nu = 5$ instead of 4 gives 0.265 m.)