Parallel pipe added in second half

Data. $D = 0.6$ m, $L = 1500$ m, $f = 0.01$, head $H = 0.3$ m (as given). Minor losses are neglected, so the head is used up in friction only. Here $Q_{before}$ and $Q_{after}$ are the total discharges in the single pipe and in the modified line (they are not the discharges $Q_1$, $Q_2$ in the two parallel pipes of the figure).

Before: single pipe of length 1500 m

$$H = \frac{fLV_b^2}{2gD} \;\Rightarrow\; \frac{V_b^2}{2g} = \frac{HD}{fL} = \frac{0.3(0.6)}{0.01(1500)} = 0.01200\ \text{m}$$ $$V_b = 0.4852\ \text{m/s},\qquad Q_{before} = \frac{\pi}{4}(0.6)^2V_b = 0.13719\ \text{m}^3/\text{s}$$

After: first 750 m single, second 750 m has two identical pipes in parallel

Let $V_a$ be the velocity in the first half. In the second half the flow divides equally between two pipes of the same size, so the velocity in each is $V_a/2$.

$$H = \frac{f(750)V_a^2}{2gD} + \frac{f(750)(V_a/2)^2}{2gD} = \frac{f(750)}{D}\,\frac{V_a^2}{2g}\,(1 + 0.25)$$ $$\frac{V_a^2}{2g} = \frac{HD}{f(750)(1.25)} = \frac{0.3(0.6)}{0.01(750)(1.25)} = 0.01920\ \text{m} \;\Rightarrow\; V_a = 0.6138\ \text{m/s}$$ $$Q_{after} = \frac{\pi}{4}(0.6)^2V_a = 0.17354\ \text{m}^3/\text{s}$$

(each parallel pipe carries 0.08677 m³/s).

Increase in discharge

$$\Delta Q = Q_{after} - Q_{before} = 0.17354 - 0.13719 = 0.03635\ \text{m}^3/\text{s}, \text{ about } 26.5\%\ \text{increase}$$

(In general $\dfrac{Q_{after}}{Q_{before}} = \sqrt{\dfrac{1500}{937.5}} = \sqrt{1.6} = 1.265$.)

Answer: The discharge increases by $0.0364$ m³/s, i.e. by about 26.5%.