Three reservoirs with pressurised closed tank

Data. A: water level 200 m. B: closed tank, water level 170 m, surface pressure 667 kN/m². C: outlet at 140 m discharging to the atmosphere. $f = 0.02$ for all pipes. Friction losses only (velocity heads and minor losses neglected). $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$.

Piezometric head of tank B

$$H_B = 170 + \frac{667}{9.81} = 170 + 67.99 = 237.99\ \text{m}$$

This is higher than A (200 m). So the effective heads are: B = 238.0 m, A = 200 m, C = 140 m. Tank B is the highest source, A is the middle one, and C is the lowest (free outlet).

Direction of flows

At first assume $H_J = 200$ m (A just balanced): $Q_2 = \sqrt{(237.99 - 200)/2582.1} = 0.1213$ m³/s into J, $Q_3 = \sqrt{60/408.0} = 0.3835$ m³/s out. The outflow exceeds the inflow, so A must also supply J, and $H_J < 200$ m. Hence: B and A both feed J; J feeds C: $Q_3 = Q_1 + Q_2$.

Trial of $H_J$

The balance is reached at $H_J = 160.05$ m.

Discharges

$$Q_1 = \sqrt{\frac{200 - 160.05}{17409.4}} = 0.0479\ \text{m}^3/\text{s},\quad Q_2 = \sqrt{\frac{237.99 - 160.05}{2582.1}} = 0.1737\ \text{m}^3/\text{s},\quad Q_3 = \sqrt{\frac{160.05 - 140}{408.0}} = 0.2216\ \text{m}^3/\text{s}$$

Check: $0.0479 + 0.1737 = 0.2216 = Q_3$.

Answer: Pipe 1 (A to J): 47.9 l/s; pipe 2 (B to J): 173.7 l/s; pipe 3 (J to C): 221.6 l/s. Head at J = 160.0 m.