---
title: "Three reservoirs: levels 100, 70 and 30 m"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:three-reservoirs-fixed-levels-100-70-30"
source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoirs-fixed-levels-100-70-30
site: BCT Engineering
---

# Three reservoirs: levels 100, 70 and 30 m

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2074 Bhadra, 10 marks; 2071 Magh, 8 marks (asked 2 times)
- Video: [Unit:2 | Pipe Networking | Numerical from Three Reservoirs| Types:1 | Prashant YT | BE Civil |](https://www.youtube.com/watch?v=pIXbQd4ZiiM) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoirs-fixed-levels-100-70-30
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Determine the discharge rate in each pipeline and the piezometric head at D for the following three-reservoir problem. [Figure: reservoir A with water surface EL 100 m connected to junction D (EL 50.0 m) by pipe L = 1000 m, d = 254 mm, f = 0.020; reservoir B at EL 70.0 m connected to D by pipe L = 600 m, d = 305 mm, f = 0.018; reservoir C at EL 30.0 m connected to D by pipe L = 500 m, d = 152 mm, f = 0.025]

## Short answer

With Darcy–Weisbach resistance r = 8fL/(π²gD⁵), trial for H_D satisfying continuity gives: Q_AD = 133.9 L/s (A→D), Q_DB = 76.5 L/s (D→B), Q_DC = 57.4 L/s (D→C); H_D = 71.98 m.

## Answer

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

| Pipe | L (m) | d (m) | f | r (s²/m⁵) |
|---|---|---|---|---|
| AD | 1000 | 0.254 | 0.020 | 1563.1 |
| BD | 600 | 0.305 | 0.018 | 338.1 |
| CD | 500 | 0.152 | 0.025 | 12729.6 |

#### 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 $Q_A = Q_B + Q_C$):

| $H_D$ (m) | $Q_A$ | $Q_B$ | $Q_C$ | $Q_A - Q_B - Q_C$ |
|---|---|---|---|---|
| 70.0 | 0.1385 | 0 | 0.0561 | +0.0824 |
| 72.0 | 0.1338 | 0.0769 | 0.0574 | -0.0005 |
| 75.0 | 0.1265 | 0.1216 | 0.0595 | -0.0546 |

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.

## Check yourself

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

1. How is flow in each pipe computed for a trial H_D?
   A. Q = sqrt(|H_i - H_D|/r_i), from higher to lower head
   B. Q = (H_i - H_D) r_i
   C. Q = r_i / H
   D. Q is the same in all pipes
   Answer: A. h_f = rQ^2, so Q = sqrt(h_f/r), directed from higher head to lower. (Key point: Basic relation)

2. At the trial H_D = 70 m, why must H_D be raised?
   A. Inflow from A exceeds outflow to C, so extra water must leave via B, requiring H_D above 70
   B. Q_B is already positive
   C. C is higher than D
   D. Outflow exceeds inflow
   Answer: A. 0.1385 in against 0.0561 out leaves a surplus that must go to B, which needs H_D above 70 m. (Key point: Deciding flow direction at B)

3. What condition fixes H_D?
   A. Q_A = Q_B + Q_C at the junction
   B. Q_A = Q_B
   C. Q_B = Q_C
   D. H_D = average of levels
   Answer: A. Flow into the junction from A must equal the flow out to B and C. (Key point: Continuity condition)

4. H_D = 71.98 m and junction elevation is 50 m. What is the pressure head at D?
   A. 71.98 m
   B. 21.98 m
   C. 50 m
   D. 121.98 m
   Answer: B. Pressure head = piezometric head - elevation = 21.98 m. (Key point: Piezometric head vs pressure)

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