Second parallel pipe to increase flow

Data. $H = 100$ m, $L = 10$ km, $f = 0.03$, $Q = 200$ l/s $= 0.2$ m³/s. Only friction loss is considered.

Diameter of the single pipe

$$H = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D^5 = \frac{8(0.03)(10\,000)(0.2)^2}{\pi^2(9.81)(100)} = 0.00992 \;\Rightarrow\; D = 0.397\ \text{m}$$

($D \approx 397$ mm; check $V = 1.61$ m/s.) For this pipe the head loss over a length $L$ at discharge $Q$ is $h_f = kLQ^2$, where

$$k = \frac{8f}{\pi^2gD^5} = 0.2500\ \text{s}^2/\text{m}^6 \text{ per metre of pipe}$$

(so that $100 = k(10\,000)(0.2)^2$).

Second pipe

The extra flow is $5\times10^6$ l/day $= 5000$ m³/day $= 0.05787$ m³/s, so the new total discharge is

$$Q' = 0.2 + 0.05787 = 0.25787\ \text{m}^3/\text{s}$$

A second pipe of the same diameter ($D$, same $f$) is laid in parallel along a length $x$, measured from one end. Over this length the flow divides equally ($Q'/2$ in each pipe); over the remaining length $(L - x)$ the single pipe carries $Q'$.

$$H = k(L - x)Q'^2 + k\,x\left(\frac{Q'}{2}\right)^2 = kQ'^2\left(L - 0.75x\right)$$ $$100 = 0.2500(0.25787)^2(10\,000 - 0.75x) \;\Rightarrow\; 10\,000 - 0.75x = 6015.3$$ $$x = \frac{10\,000 - 6015.3}{0.75} = 5313\ \text{m}$$

Answer: Diameter of the pipe $D = 0.397$ m (about 397 mm). The second pipe, of the same diameter, must be laid in parallel for a length of about $5313$ m ($5.31$ km, about 53% of the line).