---
title: "Settling basin design for 0.15 mm particles"
subject: "Hydro Power"
chapter: 5
key: "semester_6/hydro_power:settling-basin-design-0-15mm"
source: https://www.bctengineer.com/q/semester_6/hydro_power/settling-basin-design-0-15mm
site: BCT Engineering
---

# Settling basin design for 0.15 mm particles

Hydro Power · Chapter 5: Components of Hydropower System

- Asked in: 2081 Asoj, 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-0-15mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5

## Question

Design a settling basin for a ROR hydropower plant considering the turbulent effect by using the settling theory. The basin should serve to remove particles greater than 0.15 mm diameter from the water conveying mainly sand. The design discharge is 10 m³/s and the depth of the basin is 4.0 m. Take w = 6 cm/s and λ = 1.50.

## Short answer

Using Camp's velocity (v =0.170 m/s), width from continuity (B ≈14.7 m, adopt 15 m), and Velikanov's turbulent length (7.83 m). The greater ideal settling length 11.36 m governs; adopted basin 12 m long × 15 m wide ×4 m deep in two chambers.

## Answer

**Given:** $Q = 10$ m³/s, $h = 4.0$ m, particle size $d = 0.15$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, $\lambda = 1.50$.

**Method:** Camp's formula for flow velocity, continuity for width, settling theory with Velikanov's turbulence correction for length.

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

$$
v = 44\sqrt{0.15} = 17.04\ \text{cm/s} = 0.1704\ \text{m/s}
$$

#### 2. Width
$$
B = \frac{Q}{v\,h} = \frac{10}{0.1704 \times 4.0} = 14.67\ \text{m}
$$

#### 3. Ideal settling length
$$
\begin{aligned}
t &= \frac{h}{w} = \frac{4.0}{0.06} = 66.7\ \text{s} \\
L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.1704}{0.06} = 11.36\ \text{m}
\end{aligned}
$$

#### 4. Length considering turbulence (Velikanov)
$$
\begin{aligned}
L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\
&= \frac{1.5^2 \times 0.1704^2 \times (2.0 - 0.2)^2}{7.51 \times 0.06^2} \\
&= \frac{2.25 \times 0.02904 \times 3.24}{0.02704} = 7.83\ \text{m}
\end{aligned}
$$

#### 5. Adopted dimensions
Since the length cannot be less than the ideal settling length, $L_0 = 11.36$ m governs; adopt $L = 12$ m.

| Item | Value |
|---|---|
| Flow velocity | 0.170 m/s |
| Width | 14.67 m, say 15 m (2 chambers × 7.5 m) |
| Depth | 4.0 m + sediment storage depth |
| Length (Velikanov) | 7.83 m |
| Length (ideal) | 11.36 m |
| Adopted length | 12 m + inlet/outlet transitions |

**Answer:** $v = 0.170$ m/s, $B \approx 14.7$ m, Velikanov length $= 7.83$ m; adopt a basin about **12 m long × 15 m wide × 4 m deep**, divided into two chambers so that one can be flushed while the other operates.

Note: the given $w = 6$ cm/s is much larger than the real fall velocity of 0.15 mm sand (about 1.5 cm/s), which is why the computed length is short; with the real value the basin would be several times longer.

## Check yourself

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

1. For d = 0.15 mm, v = 44 sqrt(d) is about:
   A. 11 cm/s
   B. 17.0 cm/s
   C. 44 cm/s
   D. 6.6 cm/s
   Answer: B. 44 x 0.3873 = 17.04 cm/s. (Key point: Camp velocity)

2. With Q = 10, v = 0.1704 m/s and h = 4 m, B is about:
   A. 7.3 m
   B. 2.5 m
   C. 58.7 m
   D. 14.7 m
   Answer: D. B = 10/(0.1704 x 4) = 14.67 m. (Key point: Width)

3. Ideal length is 11.36 m, Velikanov 7.83 m. Adopted length is based on:
   A. Width
   B. The ideal length (rounded to 12 m)
   C. The average
   D. 7.83 m
   Answer: B. Length cannot be less than ideal settling length. (Key point: Governing length)

4. Why is the computed basin short in this problem?
   A. Sand is heavy
   B. Q is small
   C. h is small
   D. The given w = 6 cm/s is much larger than real fall velocity of 0.15 mm sand
   Answer: D. Real w is about 1.5 cm/s, so a real basin would be several times longer. (Key point: Note on w)

## Related in Hydro Power

- [Settling basin design for 0.2 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-2mm) — 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
- [Settling basin design principle and components](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-principle-components) — Hydro Power, chapter 5

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