---
title: "Three reservoirs 120 m, 70 m and 50 m"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:three-reservoir-a120-b70-c50"
source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-a120-b70-c50
site: BCT Engineering
---

# Three reservoirs 120 m, 70 m and 50 m

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2068 Bhadra, 10 marks
- Video: [Hydraulics | Three Reservoir Problems CALTECH](https://www.youtube.com/watch?v=euaVThq7OW8) — Engr. Clidez
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-a120-b70-c50
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Reservoir A, water surface elevation 120 m is connected to reservoir B and C having surface elevation 70 m and 50 m respectively. A pipe line 150 mm diameter and 400 m long connects reservoir A to Junction D. Reservoir B and C are connected to Junction D by 75 mm diameter 100 m long and 100 mm diameter 250 m long pipeline respectively. Assuming friction factor f = 0.04 for all pipes, estimate the rate of flow for each pipe, neglecting minor head losses.

## Short answer

Resistances r=8fL/(π²gD⁵): AD=17409, DB=139276, DC=82627. Assume B receives; trial H_D balances Q_AD=Q_DB+Q_DC. Converges at H_D=95.9 m: Q_AD=37.2 L/s (A→D), Q_DB=13.6 L/s (D→B), Q_DC=23.6 L/s (D→C).

## Answer

**Data.** $Z_A = 120$ m, $Z_B = 70$ m, $Z_C = 50$ m. Pipes: AD 400 m, 150 mm; DB 100 m, 75 mm; DC 250 m, 100 mm; $f = 0.04$ for all. Friction only: $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$.

| Pipe | L (m) | D (m) | r (s²/m⁵) |
|---|---|---|---|
| AD | 400 | 0.150 | 17409 |
| DB | 100 | 0.075 | 139276 |
| DC | 250 | 0.100 | 82627 |

#### Method
A is the highest reservoir and supplies water; C is the lowest and receives. Test whether B receives or supplies by assuming $H_D = 70$ m (B balanced): $Q_{AD} = \sqrt{50/17409} = 0.0536$ m³/s in, $Q_{DC} = \sqrt{20/82627} = 0.0155$ m³/s out. Inflow exceeds outflow, so the surplus goes to B: $H_D > 70$ m and $Q_{AD} = Q_{DB} + Q_{DC}$.

| $H_D$ (m) | $Q_{AD}$ | $Q_{DB}$ | $Q_{DC}$ | $Q_{AD} - Q_{DB} - Q_{DC}$ |
|---|---|---|---|---|
| 70.5 | 0.05332 | 0.00189 | 0.01575 | +0.03568 |
| 85 | 0.04484 | 0.01038 | 0.02058 | +0.01388 |
| 90 | 0.04151 | 0.01198 | 0.02200 | +0.00753 |
| 95.9 | 0.03721 | 0.01364 | 0.02357 | +0.00000 |

The balance is at $H_D = 95.90$ m.

#### Discharges
$$
Q_{AD} = \sqrt{\frac{120 - 95.90}{17409}} = 0.03721\ \text{m}^3/\text{s} = 37.21\ \text{l/s}
$$
$$
Q_{DB} = \sqrt{\frac{95.90 - 70}{139276}} = 0.01364\ \text{m}^3/\text{s} = 13.64\ \text{l/s}
$$
$$
Q_{DC} = \sqrt{\frac{95.90 - 50}{82627}} = 0.02357\ \text{m}^3/\text{s} = 23.57\ \text{l/s}
$$
Check: $13.64 + 23.57 = 37.21$ l/s $= Q_{AD}$.

**Answer:** $Q_{AD} = 37.2$ l/s (A to D), $Q_{DB} = 13.6$ l/s (D to B), $Q_{DC} = 23.6$ l/s (D to C); $H_D = 95.9$ m.

## Check yourself

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

1. With only friction losses considered, how is the head loss in each pipe written in this solution?
   A. h_f = rQ^2 with r = 8fL/(pi^2 g D^5)
   B. h_f = rQ with r = 8fL/(pi^2 g D^5)
   C. h_f = rQ^2 with r = 8fL/(pi^2 g D^2)
   D. h_f = rQ^0.5 with r = fL/D
   Answer: A. Darcy-Weisbach in terms of discharge gives h_f = 8fLQ^2/(pi^2 g D^5), so loss is proportional to Q^2. (Key point: Friction-only head loss form)

2. Why is it first tested whether B receives or supplies water by assuming H_D = 70 m?
   A. Because 70 m is the average of the three reservoir levels
   B. Because with H_D = 70 m, pipe DB carries no flow, so comparing inflow from A with outflow to C shows whether B must take the surplus or supply the deficit
   C. Because the head at the junction must always equal the middle reservoir level
   D. Because B has the smallest pipe so it always supplies
   Answer: B. At H_D = 70 m no flow goes to or from B; if A's inflow exceeds C's outflow the surplus must go to B (H_D > 70 m). (Key point: Direction of flow in B)

3. Inflow (0.0536) exceeds outflow to C (0.0155) when H_D = 70 m. What condition is then used to find the true H_D?
   A. Q_AD = Q_DB - Q_DC
   B. Q_AD + Q_DB = Q_DC
   C. Q_AD = Q_DB + Q_DC, with H_D above 70 m
   D. Q_DB = Q_DC
   Answer: C. Surplus flows to B, so the flow from A equals the sum of flows to B and C, which fixes H_D by trial. (Key point: Continuity at junction D)

4. At the balanced junction head H_D = 95.9 m, which set of discharges is correct?
   A. Q_AD = 37.2, Q_DB = 23.6, Q_DC = 13.6 l/s
   B. Q_AD = 37.2, Q_DB = 13.6, Q_DC = 23.6 l/s
   C. Q_AD = 23.6, Q_DB = 13.6, Q_DC = 37.2 l/s
   D. Q_AD = 50.0, Q_DB = 13.6, Q_DC = 36.4 l/s
   Answer: B. Q_DB = sqrt(25.9/139276) = 13.6 l/s, Q_DC = sqrt(45.9/82627) = 23.6 l/s, and their sum 37.2 l/s equals Q_AD. (Key point: Final discharges)

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