---
title: "Settling tank 20 m x 2.5 m; velocity for 0.02 mm"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-b66"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-b66
site: BCT Engineering
---

# Settling tank 20 m x 2.5 m; velocity for 0.02 mm

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2066 Bhadra (old course)
- Video: [Sedimentation tank, design of sedimentation tank, detention time, overflow rate, flow velocity, TSS](https://www.youtube.com/watch?v=43wzunPfwmA) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-b66
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

In a continuous flow settling tank 20 m long and 2.5 m deep, what velocity of water would you recommend for effective removal of 0.02 mm particles at 25°C? Assume specific gravity of particles = 2.65 and kinematic viscosity of water = 0.01 cm$^2$/sec.

## Short answer

For complete removal, horizontal velocity v_h = v_s × L/H. Stokes' settling velocity is 0.036 cm/s (Re<1). With L=20 m, H=2.5 m, v_h = 0.288 cm/s (17.3 cm/min).

## Answer

For complete removal in an ideal horizontal-flow tank, the particle must settle the depth $H$ in the time the water takes to flow the length $L$:
$$
\frac{H}{v_s}=\frac{L}{v_h}\ \Rightarrow\ v_h=v_s\,\frac{L}{H}
$$

#### Step 1: Settling velocity of the 0.02 mm particle
$d=0.002$ cm, $S_s=2.65$, $\nu=0.01$ cm²/s (given):
$$
v_s=\frac{981\times1.65\times0.002^2}{18\times0.01}=0.03597\ \text{cm/s}\quad(Re=0.007<1)
$$

#### Step 2: Horizontal velocity
$L=20$ m, $H=2.5$ m:
$$
v_h=0.03597\times\frac{20}{2.5}=0.2878\ \text{cm/s}=17.27\ \text{cm/min}=0.173\ \text{m/min}
$$
This is below the 0.3 m/min limit for scour, so it is acceptable.

**Answer: the velocity of flow should not exceed 0.288 cm/s (17.3 cm/min).**

## Check yourself

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

1. What condition gives complete removal of a particle in an ideal horizontal-flow tank?
   A. Settling velocity equals horizontal velocity
   B. Settling time H/vs equals flow-through time L/vh
   C. Tank depth equals tank length
   D. Detention time is exactly 1 hour
   Answer: B. The particle entering at the top must reach the bottom by the time the water travels the length L, so H/vs = L/vh. (Key point: Ideal tank condition)

2. Why is Stokes' law acceptable for the 0.02 mm particle?
   A. The particle is heavier than water
   B. Re is greater than 1000
   C. Re = 0.007, which is less than 1
   D. Temperature is 25 °C
   Answer: C. Stokes' law applies for Re < 1, and the check gave Re = 0.007. (Key point: Stokes validity)

3. How is the recommended flow velocity obtained from vs?
   A. vh = vs × L/H = 0.03597 × 20/2.5 = 0.288 cm/s
   B. vh = vs × H/L = 0.0045 cm/s
   C. vh = vs = 0.036 cm/s
   D. vh = vs × (L + H) = 0.81 cm/s
   Answer: A. Rearranging H/vs = L/vh gives vh = vs L/H, and L/H = 8. (Key point: Horizontal velocity)

4. Why is 0.173 m/min checked against 0.3 m/min?
   A. To make sure water does not freeze
   B. To fix the particle size
   C. To raise the viscosity
   D. To make sure the flow does not scour settled sludge
   Answer: D. Above about 0.3 m/min the flow can resuspend deposited solids, so the velocity must stay below it. (Key point: Scour limit)

## Close variants set in other exams

- [Settling tank 30 m x 3 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk68) — Water Supply Engineering, chapter 6, asked 2×

## Related in Water Supply Engineering

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## Related in other subjects

- [Continuous flow sedimentation tank particle size](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-continuous-4m-60m) — Sanitary Engineering, chapter 7