---
title: "Discharge in 20 km pipe between reservoirs"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:long-pipeline-discharge-reservoirs"
source: https://www.bctengineer.com/q/semester_4/hydraulics/long-pipeline-discharge-reservoirs
site: BCT Engineering
---

# Discharge in 20 km pipe between reservoirs

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2068 Magh, 8 marks
- Video: [Solving Flow Velocity Between Two Reservoirs Using Energy Equation and Healdloss Equation](https://www.youtube.com/watch?v=lYpKPl2zP24) — High Water Studio
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/long-pipeline-discharge-reservoirs
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

A 2 cm diameter 20 km long pipeline connects two reservoirs filled with water open to the atmosphere. What is the discharge in the pipeline if the surface elevation difference of the reservoirs level is 5 m? $\nu_{water} = 1.02 \times 10^{-6}$ m²/s.

## Short answer

Low velocity suggests laminar flow. Hagen-Poiseuille gives V=0.0301 m/s, Re=589 (<2000). Discharge Q = 9.44 × 10⁻⁶ m³/s (0.0094 L/s, 34 L/h).

## Answer

**Data.** $D = 0.02$ m, $L = 20\,000$ m, $H = 5$ m, $\nu = 1.02\times10^{-6}$ m²/s. The pipe is very long and thin, so the velocity will be small; try **laminar** flow first.

#### Laminar flow assumption
The available head is used up in friction (velocity head and minor losses are negligible against 5 m over 20 km). Hagen-Poiseuille:
$$
h_f = \frac{32\nu LV}{gD^2} = H \;\Rightarrow\; V = \frac{gD^2H}{32\nu L} = \frac{9.81(0.02)^2(5)}{32(1.02\times10^{-6})(20\,000)} = 0.0301\ \text{m/s}
$$

#### Check
$$
Re = \frac{VD}{\nu} = \frac{0.0301(0.02)}{1.02\times10^{-6}} = 589 < 2000
$$
The flow is laminar, so the assumption is right. (Velocity head $V^2/2g = 4.6\times10^{-5}$ m is negligible against 5 m.)

#### Discharge
$$
Q = \frac{\pi}{4}D^2V = \frac{\pi}{4}(0.02)^2(0.0301) = 0.0000094\ \text{m}^3/\text{s} = 0.00944\ \text{l/s} \approx 34.0\ \text{l/hour}
$$

**Answer:** $Q \approx 9.44\times10^{-6}$ m³/s (about 0.0094 l/s, 34 l/h); laminar flow with $Re \approx 589$.

## Check yourself

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

1. Why try laminar flow first?
   A. The pipe is very long and thin so velocity is small
   B. The pipe is smooth
   C. The reservoirs are open
   D. The head is high
   Answer: A. 5 m over 20 km in a 2 cm pipe means a very small velocity. (Key point: Laminar assumption)

2. Which equation gives V?
   A. H = f L V^2/(2 g D) with f = 0.02
   B. H = 32 nu L V/(g D^2) (Hagen-Poiseuille)
   C. V = sqrt(2 g H)
   D. H = V^2/2g
   Answer: B. Friction head in laminar flow is h_f = 32 nu L V/(g D^2), set equal to H. (Key point: Governing equation)

3. How is the assumption verified?
   A. Compute Re = VD/nu and check it is below 2000
   B. Compare V with sqrt(2gH)
   C. Check the head is 5 m
   D. No check is needed
   Answer: A. Re of about 589 is below 2000 so laminar flow is correct. (Key point: Verification)

4. Why can the velocity head and minor losses be neglected?
   A. They are zero
   B. They are tiny (about 5x10^-5 m) compared with 5 m
   C. They are included in f
   D. They cancel each other
   Answer: B. V^2/2g is about 4.6x10^-5 m, negligible against the 5 m head. (Key point: Neglected terms)

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