Pipe diameter to halve pumping cost

Data. $\nu = 0.6\times10^{-6}$ m²/s, $Q = 0.8$ m³/s, $D = 0.5$ m, same absolute roughness $k$ for the new pipe.

$$V = \frac{4Q}{\pi D^2} = 4.074\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 3395305 > 10^6$$

So $f \propto (k/D)^{1/3}$ may be used.

Part 1: Diameter for half the pumping cost

Pumping cost is proportional to the power $P = \rho g Q h_f$. With $Q$ fixed, cost $\propto h_f$ (neglecting all losses except friction).

$$h_f = \frac{fLV^2}{2gD}, \qquad V = \frac{4Q}{\pi D^2} \;\Rightarrow\; h_f \propto \frac{f}{D^5}$$

For constant $k$: $f \propto \left(\dfrac{k}{D}\right)^{1/3} \propto D^{-1/3}$, so

$$h_f \propto D^{-1/3}\,D^{-5} = D^{-16/3}$$

Cost halved: $\left(\dfrac{D'}{D}\right)^{-16/3} = \dfrac{1}{2}$ gives

$$\frac{D'}{D} = 2^{3/16} = 1.1388 \;\Rightarrow\; D' = 0.5(1.1388) = 0.569\ \text{m} \approx 569\ \text{mm}$$

Check: new velocity $V' = 3.142$ m/s and $Re' = 2981506 > 10^6$, so the assumption holds.

Part 2: n equal pipes in parallel at the same Reynolds number

The same $Re = VD/\nu$ for the same fluid means $V'D' = VD$. Each of the $n$ pipes carries $Q/n$:

$$\frac{Q}{n} = \frac{\pi}{4}D'^2V' \;\Rightarrow\; \frac{\pi}{4}D^2V = n\frac{\pi}{4}D'^2V' = n\frac{\pi}{4}D'\,(VD)\;\Rightarrow\; D' = \frac{D}{n},\quad V' = nV$$

Relative roughness rises to $k/D' = n(k/D)$, so $f' = n^{1/3}f$. Head loss of each pipe (all pipes have the same $h_f$ in parallel):

$$h_f' = \frac{f'LV'^2}{2gD'} = \frac{n^{1/3}f\,L\,n^2V^2}{2g(D/n)} = n^{3+1/3}h_f = n^{10/3}h_f$$

The discharge is the same, so the power and the running cost increase by the factor

$$\boxed{\frac{\text{cost}'}{\text{cost}} = n^{10/3}}$$

For example, two pipes ($n = 2$) cost $2^{10/3} = 10.1$ times as much. Using one large pipe is much cheaper than several small ones.

Answer: $D' \approx 569$ mm (about 570 mm); running cost rises by $n^{10/3}$ for $n$ parallel pipes.