Cast iron pipe size for given head loss

Data. $Q = 0.4$ m³/s, $L = 1000$ m, $h_f = 2$ m, $e = 0.26$ mm, $\mu = 0.0014$ Pa s, so $\nu = \mu/\rho = 1.40\times10^{-6}$ m²/s. This is a "find the diameter" problem, so iterate.

Working equation

From Darcy-Weisbach with $V = \dfrac{4Q}{\pi D^2}$:

$$h_f = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5} = \left(\frac{8f(1000)(0.4)^2}{\pi^2(9.81)(2)}\right)^{1/5}$$

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

The diameter converges to $D = 0.6452$ m. Check: $V = 1.223$ m/s, $Re = 563815$, $e/D = 0.00040$, $f = 0.01692$, and

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.01692\left(\frac{1000}{0.6452}\right)\frac{(1.223)^2}{19.62} = 2.000\ \text{m}$$

Answer: $D \approx 645$ mm; a 650 mm cast iron pipe would be used.