---
title: "Three reservoirs with pump in pipe 2"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:pump-three-reservoir-flow-and-head-j"
source: https://www.bctengineer.com/q/semester_4/hydraulics/pump-three-reservoir-flow-and-head-j
site: BCT Engineering
---

# Three reservoirs with pump in pipe 2

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2079 Chaitra, 10 marks
- Video: [THREE RESERVOIR PROBLEM PART 2 | HYDRAULICS   | DE LA CRUZ TUTORIALS](https://www.youtube.com/watch?v=67IVvAwwa9E) — Engr. Jobert S. De La Cruz
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/pump-three-reservoir-flow-and-head-j
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

For a three-reservoir system as shown in figure, due to installation of pump at pipe 2, the flow is from the reservoir to the junction J in pipe 2. The pipes data are: $L_1$ = 2300 m, $L_2$ = 2000 m, $L_3$ = 2500 m, $D_1$ = 0.6 m, $D_2$ = 1 m, $D_3$ = 1.2 m, $f_1$ = 0.02, $f_2$ = 0.013, $f_3$ = 0.023, where L = length, D = diameter, f = Darcy friction factor. If the pump head is 30 m, determine the flow rates in the pipes and energy head at J. [Figure: reservoir at 100 m connected by pipe 1 to junction J; pipe 2 with a pump connects J to a reservoir at 90 m; pipe 3 connects J to a reservoir at 30 m; datum below]

## Short answer

By guessing junction head H_J and solving continuity Q₁+Q₂=Q₃ with the pump adding 30 m to the reservoir head, the flow rates are Q₁=0.661, Q₂=4.387, Q₃=5.048 m³/s and the energy head at J is 78.65 m.

## Answer

**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 $r = \dfrac{8fL}{\pi^2gD^5}$.

| Pipe | L (m) | D (m) | f | r (s²/m⁵) |
|---|---|---|---|---|
| 1 | 2300 | 0.6 | 0.020 | 48.879 |
| 2 | 2000 | 1.0 | 0.013 | 2.1483 |
| 3 | 2500 | 1.2 | 0.023 | 1.9093 |

#### 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 $H_J$
| $H_J$ (m) | $Q_1$ | $Q_2$ | $Q_3$ | $Q_1 + Q_2 - Q_3$ |
|---|---|---|---|---|
| 70 | 0.783 | 4.824 | 4.577 | +1.031 |
| 80 | 0.640 | 4.315 | 5.117 | -0.163 |
| 78 | 0.671 | 4.422 | 5.014 | +0.079 |
| 78.65 | 0.661 | 4.387 | 5.048 | 0.00 |

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the pump in pipe 2 accounted for in the solution?
   A. The pump head is ignored and only r2 is changed
   B. Effective head at the pipe-2 end becomes 90 + 30 = 120 m
   C. Pipe 2 head becomes 90 - 30 = 60 m
   D. The pump head is added to the junction head H_J
   Answer: B. The pump adds 30 m of head to reservoir 2's 90 m, so Q2 = sqrt((120 - H_J)/r2). (Key point: Effect of the pump)

2. Which continuity equation applies at the junction J here?
   A. Q1 + Q2 = Q3
   B. Q1 = Q2 + Q3
   C. Q1 + Q3 = Q2
   D. Q1 = Q2 = Q3
   Answer: A. Reservoirs 1 (100 m) and 2 (120 m with pump) feed J, and J feeds the 30 m reservoir 3. (Key point: Continuity at J)

3. At the trial H_J = 80 m the table gives Q1 + Q2 - Q3 = -0.163. What should be done next?
   A. Raise H_J, because inflow is too small
   B. Keep H_J, since the sign is negative
   C. Reduce the pump head
   D. Lower H_J, because outflow exceeds inflow
   Answer: D. A negative value means Q3 exceeds Q1 + Q2; lowering H_J increases Q1 and Q2 and decreases Q3, moving toward balance at about 78.65 m. (Key point: Adjusting the trial H_J)

4. At the balanced H_J = 78.65 m, which set of flows is correct?
   A. Q1 = 0.661, Q2 = 5.048, Q3 = 4.387 m^3/s
   B. Q1 = 0.661, Q2 = 4.387, Q3 = 3.726 m^3/s
   C. Q1 = 0.661, Q2 = 4.387, Q3 = 5.048 m^3/s
   D. Q1 = 4.387, Q2 = 0.661, Q3 = 5.048 m^3/s
   Answer: C. Q1 + Q2 = 0.661 + 4.387 = 5.048 = Q3, satisfying continuity. (Key point: Final result)

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