Three reservoirs: levels 100, 70 and 30 m

Data. Reservoir levels: A = 100 m, B = 70 m, C = 30 m; junction D at elevation 50 m. Only friction losses are considered (velocity heads and minor losses neglected). For each pipe $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$:

Method (trial of the piezometric head $H_D$)

Let $H_D$ be the piezometric head at the junction. Flow in each pipe: $Q_i = \sqrt{|H_i - H_D|/r_i}$, directed from the higher head to the lower. Continuity at D: $\sum Q_{in} = \sum Q_{out}$.

$H_D$ must lie between the levels of the highest and lowest reservoirs (100 m and 30 m). First test whether B supplies or receives water:

• Trial $H_D = 70$ m (no flow in BD): $Q_A = \sqrt{30/1563.1} = 0.1385$ m³/s in, $Q_C = \sqrt{40/12729.6} = 0.0561$ m³/s out. Inflow exceeds outflow, so the surplus must go to B, and $H_D$ must rise above 70 m. So A feeds both B and C, with $Q_A = Q_B + Q_C$.

Trials (with $Q_A = Q_B + Q_C$):

The balance is reached at

$$H_D = 71.98\ \text{m}$$

Discharges

$$Q_{AD} = \sqrt{\frac{100 - 71.98}{1563.1}} = 0.1339\ \text{m}^3/\text{s} = 133.9\ \text{l/s}\quad(\text{A to D})$$ $$Q_{DB} = \sqrt{\frac{71.98 - 70}{338.1}} = 0.0765\ \text{m}^3/\text{s} = 76.5\ \text{l/s}\quad(\text{D to B})$$ $$Q_{DC} = \sqrt{\frac{71.98 - 30}{12729.6}} = 0.0574\ \text{m}^3/\text{s} = 57.4\ \text{l/s}\quad(\text{D to C})$$

Check: $76.5 + 57.4 = 133.9$ l/s $= Q_{AD}$.

Piezometric head and pressure at D

Piezometric head at D $= 71.98$ m. As the junction is at elevation 50 m, the pressure head there is $71.98 - 50 = 21.98$ m of water ($p_D \approx 216$ kPa).

Answer: $Q_{AD} = 133.9$ l/s (in), $Q_{DB} = 76.5$ l/s and $Q_{DC} = 57.4$ l/s (out); $H_D = 71.98$ m.