Equivalent roughness from head loss test

Data. $D = 0.5$ m, $L = 100$ m, water with $\nu = 1.0\times10^{-6}$ m²/s. Area $A = 0.19635$ m².

Friction factor from the test

$f = \dfrac{h_f\,2gD}{LV^2}$ with $V = Q/A$:

At high flows $f$ levels off at about 0.0193, which means the pipe is fully rough there.

(i) Equivalent sand-grain roughness

Use the highest flow ($Q$ = 800 l/s, $Re = 2037183$, $f = 0.01932$) in the Colebrook-White equation:

$$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)$$

Solving, $\dfrac{k}{D} = 0.000903$, so

$$k = 0.000903(500) = 0.45\ \text{mm}$$

Check: Colebrook with this $k$ gives $f$ = 0.0218, 0.0198, 0.0195 and 0.0193 for the four flows, which agree with the test values.

(ii) Maximum discharge for smooth behaviour

The pipe is smooth while $\dfrac{u_*k}{\nu} \le 4$:

$$u_* = \frac{4\nu}{k} = \frac{4(10^{-6})}{0.45\times10^{-3}} = 0.0089\ \text{m/s}$$

With smooth-pipe $f$ (Prandtl law) at the limiting $Re$ (found by iteration, $Re \approx 92619$, $f = 0.0183$):

$$V = \frac{u_*}{\sqrt{f/8}} = 0.185\ \text{m/s},\qquad Q = AV = 36.4\ \text{l/s}$$

(iii) Discharge for fully rough behaviour

The pipe is fully rough when $\dfrac{u_*k}{\nu} \ge 100$, i.e. $u_* = \dfrac{100\nu}{k} = 0.2214$ m/s. For the rough pipe, $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$ gives $f = 0.01914$:

$$V = \frac{u_*}{\sqrt{f/8}} = 4.526\ \text{m/s}, \qquad Q = 889\ \text{l/s}$$

This is the limit of the rough regime: for $Q$ above about 889 l/s (it is a minimum, not a maximum, flow for rough behaviour) the pipe is fully rough and $f$ no longer depends on $Re$.

Answer: (i) $k \approx 0.45$ mm; (ii) smooth for $Q \lesssim 36$ l/s; (iii) fully rough for $Q \gtrsim 889$ l/s (transition in between).