---
title: "Pool drained by hose, find water depth"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:swimming-pool-hose-drain-depth"
source: https://www.bctengineer.com/q/semester_4/hydraulics/swimming-pool-hose-drain-depth
site: BCT Engineering
---

# Pool drained by hose, find water depth

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2071 Bhadra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/swimming-pool-hose-drain-depth
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Small swimming pool is drained with velocity of 1.2 m/sec using a pipe with hose diameter 20 mm, length 30 m, and absolute roughness e = 0.2 mm. Find the water depth d at instant shown in figure below considering minor head loss coefficient at entrance K = 0.5. [Figure: pool with water depth d above the hose inlet; the hose runs over the pool rim and discharges 3 m below the inlet level]

## Short answer

Colebrook friction factor f=0.0403. Energy equation including entrance loss, friction, and velocity head gives d + 3 = 4.543 m, so water depth above hose inlet d = 1.54 m.

## Answer

**Data.** $V = 1.2$ m/s, $D = 0.02$ m, $L = 30$ m, $e = 0.2$ mm, entrance loss $K = 0.5$, water $\nu = 1.0\times10^{-6}$ m²/s. The hose outlet (open to the atmosphere) is 3 m below the hose inlet level. Let $d$ be the depth of water above the hose inlet.

#### Friction factor
$$
Re = \frac{VD}{\nu} = \frac{1.2(0.02)}{10^{-6}} = 24000, \qquad \frac{e}{D} = 0.010
$$
Colebrook-White (iteration) gives $f = 0.0403$.

#### Energy equation
Between the pool surface (atmospheric, velocity zero, elevation $d + 3$ above the outlet) and the free outlet (velocity $V$):
$$
d + 3 = \frac{V^2}{2g}\left(1 + K + f\frac{L}{D}\right), \qquad \frac{V^2}{2g} = \frac{1.2^2}{2(9.81)} = 0.0734\ \text{m}
$$
$$
f\frac{L}{D} = 0.0403\left(\frac{30}{0.02}\right) = 60.40
$$
$$
d + 3 = 0.0734\left(1 + 0.5 + 60.40\right) = 0.0734(61.90) = 4.543\ \text{m}
$$
$$
d = 4.543 - 3 = 1.54\ \text{m}
$$

**Answer:** The water depth above the hose inlet is about $1.54$ m.

## Check yourself

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

1. How is f found for the hose?
   A. f = 64/Re since the velocity is low
   B. Colebrook-White with Re = 24000 and e/D = 0.01
   C. f = 0.02 from experience
   D. From the relative roughness alone with no Re
   Answer: B. Re = 24000 is turbulent and e/D = 0.01 is large, so Colebrook-White gives f of about 0.040. (Key point: Friction factor method)

2. Which energy equation gives the depth d?
   A. d = (V^2/2g)(1 + K + fL/D)
   B. d + 3 = (V^2/2g)(1 + K + fL/D)
   C. d + 3 = (V^2/2g) fL/D only
   D. d - 3 = (V^2/2g)(K + fL/D)
   Answer: B. The pool surface is d + 3 above the outlet; it supplies exit velocity head, entrance loss and friction. (Key point: Energy equation)

3. What does the '1' in (1 + K + fL/D) represent?
   A. Exit kinetic energy V^2/2g at the free outlet
   B. Exit loss coefficient of the pool
   C. Velocity head in the pool
   D. Bend loss
   Answer: A. The hose discharges freely to atmosphere with velocity V, so one velocity head leaves as kinetic energy. (Key point: Terms in the bracket)

4. Given d + 3 = 4.543 m, what is d?
   A. 4.54 m
   B. 7.54 m
   C. 1.54 m
   D. 3.0 m
   Answer: C. d = 4.543 - 3 = 1.54 m above the hose inlet. (Key point: Final depth)

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