---
title: "Settling basin design for 0.2 mm particles"
subject: "Hydro Power"
chapter: 5
key: "semester_6/hydro_power:settling-basin-design-0-2mm"
source: https://www.bctengineer.com/q/semester_6/hydro_power/settling-basin-design-0-2mm
site: BCT Engineering
---

# Settling basin design for 0.2 mm particles

Hydro Power · Chapter 5: Components of Hydropower System

- Asked in: 2081 Chaitra, 6 marks
- Video: [Design Settling Basin | Theory + Numerical | Hydropower Engineering](https://www.youtube.com/watch?v=bgMZO7zXuaI) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/settling-basin-design-0-2mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5

## Question

Design a settling basin for a run-of-river hydropower station, considering turbulence effects using settling theory. The basin should remove particles larger than 0.2 mm in diameter from sand-laden water. Given a design discharge of 12 m³/s and a basin depth of 3.5 m, use w = 6 cm/s and λ = 1.50 for the design.

## Short answer

Horizontal velocity by Camp's formula (v = 0.197 m/s), width from continuity (B ≈17.4 m, adopt 17.5 m), length considering turbulence by Velikanov (9.0 m) but ideal settling length 11.48 m governs. Adopted basin dimensions: 12 m long × 17.5 m wide × 3.5 m deep (two chambers).

## Answer

**Given:** $Q = 12$ m³/s, $h = 3.5$ m, particle size $d = 0.2$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, turbulence coefficient $\lambda = 1.50$.

**Method:** horizontal velocity by Camp's formula, width from continuity, length by settling theory corrected for turbulence by Velikanov's formula.

#### 1. Horizontal (flow-through) velocity
Camp: $v = a\sqrt{d}$ cm/s with $a = 44$ for $0.1 < d < 1$ mm.

$$
\begin{aligned}
v &= 44\sqrt{0.2} = 19.68\ \text{cm/s} = 0.197\ \text{m/s}
\end{aligned}
$$

#### 2. Width of basin
$$
\begin{aligned}
B &= \frac{Q}{v\,h} = \frac{12}{0.1968 \times 3.5} = 17.42\ \text{m}
\end{aligned}
$$

#### 3. Length by ideal settling theory (no turbulence)
$$
\begin{aligned}
t &= \frac{h}{w} = \frac{3.5}{0.06} = 58.3\ \text{s} \\
L_0 &= v\,t = \frac{h\,v}{w} = \frac{3.5 \times 0.1968}{0.06} = 11.48\ \text{m}
\end{aligned}
$$

#### 4. Length considering turbulence (Velikanov)
$$
\begin{aligned}
L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\
&= \frac{1.5^2 \times 0.1968^2 \times (\sqrt{3.5} - 0.2)^2}{7.51 \times 0.06^2} \\
&= \frac{2.25 \times 0.03872 \times 2.7916}{0.02704} = 9.00\ \text{m}
\end{aligned}
$$

#### 5. Adopted dimensions
The length must not be less than the ideal settling length, so the larger value (11.48 m) governs; adopt $L = 12$ m.

| Item | Value |
|---|---|
| Flow velocity $v$ | 0.197 m/s |
| Width $B$ | 17.42 m, say 17.5 m (2 chambers of 8.75 m) |
| Settling depth $h$ | 3.5 m (+ about 0.5–1.0 m sediment storage) |
| Length (Velikanov) | 9.00 m |
| Length (ideal) | 11.48 m |
| Adopted length $L$ | 12 m (+ inlet and outlet transitions) |

**Answer:** $v = 0.197$ m/s, $B \approx 17.4$ m, turbulent (Velikanov) length $= 9.0$ m; adopted basin about **12 m long × 17.5 m wide × 3.5 m deep** (split into two chambers for flushing).

Note: the given $w = 6$ cm/s is high for 0.2 mm sand (real value about 2–2.5 cm/s); with a smaller $w$ the required length becomes much larger, so in practice the basin is made longer, with $L/B$ about 8–10 per chamber.

## Check yourself

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

1. For d = 0.2 mm, Camp's v = 44 sqrt(d) gives approximately:
   A. 2 cm/s
   B. 19.7 cm/s
   C. 8.8 cm/s
   D. 44 cm/s
   Answer: B. 44 x sqrt(0.2) = 44 x 0.447 = 19.68 cm/s. (Key point: Camp velocity)

2. With Q = 12, v = 0.1968 m/s, h = 3.5 m, the width B is about:
   A. 40 m
   B. 2.3 m
   C. 6.1 m
   D. 17.4 m
   Answer: D. B = Q/(v h) = 12/(0.1968 x 3.5) = 17.4 m. (Key point: Width)

3. Velikanov gave 9.0 m and ideal settling 11.48 m. Why is 12 m adopted?
   A. Velikanov is always adopted
   B. Length cannot be less than the ideal settling length, so the larger governs
   C. The width forces it
   D. The smaller value is safer
   Answer: B. The larger value (11.48 m, rounded to 12 m) governs. (Key point: Which length governs)

4. The ideal settling length is:
   A. L = v/h
   B. L = w/(h v)
   C. L = Q/w
   D. L = h v / w
   Answer: D. 3.5 x 0.1968 / 0.06 = 11.48 m. (Key point: Ideal length)

## Related in Hydro Power

- [Settling basin design for 0.15 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-15mm) — Hydro Power, chapter 5
- [Settling basin design for 0.25 mm sediment](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-25mm) — Hydro Power, chapter 5

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