Three reservoirs with two parallel pipes

Layout assumed from the figure. A feeds junction D by pipe 1; B joins D by pipe 2; from D two pipes, 3 and 4, run in parallel to a junction E; pipe 5 then runs from E to reservoir C. Minor losses are included through $\sum K$: for each pipe the head loss is $h = \left(\dfrac{fL}{D} + \sum K\right)\dfrac{V^2}{2g} = rQ^2$, with

$$r = \frac{8}{\pi^2gD^4}\left(\frac{fL}{D} + \sum K\right)$$

(i) Relation between the flows in the two parallel pipes

They have the same head loss, so $r_3Q_3^2 = r_4Q_4^2$:

$$\frac{Q_3}{Q_4} = \sqrt{\frac{r_4}{r_3}} = \sqrt{\frac{15.578}{7.536}} = 1.438 \;\Rightarrow\; Q_3 = 1.438\,Q_4$$

The equivalent resistance of the pair is given by $\dfrac{1}{\sqrt{r_{par}}} = \dfrac{1}{\sqrt{r_3}} + \dfrac{1}{\sqrt{r_4}}$, so $r_{par} = 2.621$ s²/m⁵, and $Q_5 = Q_3 + Q_4$.

(ii) HGL at junction D

Let $H_D$ be the HGL at D. Pipe 5 is in series with the parallel pair (3, 4), so $H_D - Z_C = (r_{par} + r_5)Q_5^2$.

• Pipe 1: $Q_1 = \sqrt{(185 - H_D)/r_1}$ (A to D)
• Pipe 2: $Q_2 = \sqrt{|105 - H_D|/r_2}$, towards B if $H_D > 105$, from B if $H_D < 105$
• Pipe 5: $Q_5 = \sqrt{(H_D - 70)/(r_{par} + r_5)}$
• Continuity at D: $Q_1 = Q_2 + Q_5$ (when B is receiving), or $Q_1 + Q_2 = Q_5$ (when B is supplying).

Trials ($Q_2 > 0$ means B supplies D, $Q_2 < 0$ means B receives):

The sum changes sign between 108 and 110 m, and the balance is at

$$H_D = 108.95\ \text{m}$$

Since $H_D > 105$ m, pipe 2 carries water to reservoir B.

(iii) Flow rates

Pipe 5: $Q_5 = \sqrt{(108.95 - 70)/(2.621 + 8.539)} = 1.868$ m³/s. The HGL at junction E is $70 + r_5Q_5^2 = 99.80$ m, so $Q_3 = \sqrt{(108.95 - 99.80)/7.536} = 1.102$ and $Q_4 = \sqrt{(108.95 - 99.80)/15.578} = 0.766$ m³/s.

Continuity check at D: $Q_1 = 1.942 = Q_2 + Q_5 = 0.073 + 1.868$.

Answer: (i) $Q_3 = 1.438\,Q_4$; (ii) HGL at D $\approx 109.0$ m; (iii) $Q_1 = 1.942$, $Q_2 = 0.073$ (to B), $Q_3 = 1.102$, $Q_4 = 0.766$, $Q_5 = 1.868$ m³/s.