Steel pipe size for given energy gradient

Data. $Q = 0.03$ m³/s, permissible energy gradient $S = h_f/L = 0.05$. Commercial steel pipe: $k = 0.046$ mm (Moody table). Water at 20 °C, $\nu = 1.0\times10^{-6}$ m²/s. This is a "find the diameter" problem, solved by trial.

Working equation

$$S = \frac{f}{D}\frac{V^2}{2g},\quad V = \frac{4Q}{\pi D^2} \;\Rightarrow\; S = \frac{8fQ^2}{\pi^2 g D^5} \;\Rightarrow\; D = \left(\frac{8fQ^2}{\pi^2 g S}\right)^{1/5}$$

Trials (f from the Colebrook-White equation at each new D)

The values converge to $D = 0.1210$ m. Final check: $V = 2.610$ m/s, $Re = 315734$, $k/D = 0.000380$, $f = 0.01742$ (Colebrook), and $S = fV^2/(2gD) = 0.0500$, which matches the required 0.05.

Diameter required: $D \approx 121$ mm. In practice a 125 mm commercial size would be adopted (it gives $S$ = 0.0424 < 0.05 at $V$ = 2.44 m/s).

Smooth or transition?

$$u_* = V\sqrt{\frac{f}{8}} = 2.610\sqrt{\frac{0.01742}{8}} = 0.1218\ \text{m/s},\qquad Re_* = \frac{u_*k}{\nu} = \frac{0.1218(0.046\times10^{-3})}{10^{-6}} = 5.60$$

Since $4 < Re_* < 100$, the boundary acts in the transition zone (neither smooth nor fully rough), as the Colebrook-White equation assumes.

Answer: $D \approx 121$ mm (use 125 mm); the boundary is in the transition regime.