Three reservoirs with pump in pipe 2

Data. Reservoir levels: 100 m (pipe 1), 90 m (pipe 2, with pump), 30 m (pipe 3). Pump head $H_p = 30$ m in pipe 2 (flow from the reservoir to J). Only friction loss, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$.

Method

The pump adds 30 m of head, so on the pipe-2 side the effective head at the reservoir end is $90 + 30 = 120$ m. This is higher than the 100 m level of reservoir 1, so the junction head must lie between 30 m and 120 m; reservoir 1 supplies J if $H_J < 100$ m and receives water if $H_J > 100$ m. Let $H_J$ be the energy (piezometric) head at J.

• Pipe 1 (reservoir 100 m to J): $Q_1 = \sqrt{(100 - H_J)/r_1}$ (if $H_J < 100$)
• Pipe 2 (reservoir 90 m plus pump to J): $Q_2 = \sqrt{(90 + 30 - H_J)/r_2}$
• Pipe 3 (J to reservoir 30 m): $Q_3 = \sqrt{(H_J - 30)/r_3}$
• Continuity: $Q_1 + Q_2 = Q_3$

Trial of $H_J$

Result

$$H_J = 78.65\ \text{m}$$ $$Q_1 = 0.661\ \text{m}^3/\text{s},\qquad Q_2 = 4.387\ \text{m}^3/\text{s},\qquad Q_3 = 5.048\ \text{m}^3/\text{s}$$

Check: $Q_1 + Q_2 = 0.661 + 4.387 = 5.048 = Q_3$. The directions are: reservoir 1 to J, pump-reservoir to J, and J to reservoir 3. Velocities: $V_1 = 2.34$, $V_2 = 5.59$, $V_3 = 4.46$ m/s.

Answer: $Q_1 = 0.661$, $Q_2 = 4.387$, $Q_3 = 5.048$ m³/s; energy head at J $= 78.65$ m.