---
title: "Settling basin design for 0.25 mm sediment"
subject: "Hydro Power"
chapter: 5
key: "semester_6/hydro_power:settling-basin-design-0-25mm"
source: https://www.bctengineer.com/q/semester_6/hydro_power/settling-basin-design-0-25mm
site: BCT Engineering
---

# Settling basin design for 0.25 mm sediment

Hydro Power · Chapter 5: Components of Hydropower System

- Asked in: 2073 Bhadra, 8 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-0-25mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5

## Question

With considering turbulent effect, design a settling basin to remove the sediment size greater than 0.25 mm diameter. Assume design discharge of the basin is 10 m³/s and trap efficiency as 95%.

## Short answer

Assumeddepth 4.0 m, fall velocity 0.027 m/s for0.25 mm sand, λ =1.65 for95% efficiency. Camp's velocity v =0.22 m/s, width B ≈11.4 m (adopt 12 m as two 6-m chambers). Velikanov length 78 m governs over ideal 32.6 m; basin 78 m long × 12 m wide ×4 m deep.

## Answer

**Given:** $Q = 10$ m³/s, particle size $d = 0.25$ mm, trap efficiency $\eta = 95\%$.

**Assumptions** (not given): basin depth $h = 4.0$ m; water at about 20 °C; fall velocity of 0.25 mm quartz sand from the standard fall-velocity table, $w = 2.70$ cm/s $= 0.027$ m/s; Velikanov coefficient for 95% efficiency $\lambda = 1.65$ (Velikanov table: 90% → 1.28, 95% → 1.65, 98% → 2.05).

#### 1. Horizontal velocity (Camp)
$$
v = a\sqrt{d} = 44\sqrt{0.25} = 22\ \text{cm/s} = 0.22\ \text{m/s}
$$

#### 2. Width
$$
B = \frac{Q}{v\,h} = \frac{10}{0.22 \times 4.0} = 11.36\ \text{m}
$$

#### 3. Ideal settling length (no turbulence)
$$
\begin{aligned}
t &= \frac{h}{w} = \frac{4.0}{0.027} = 148.1\ \text{s} \\
L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.22}{0.027} = 32.6\ \text{m}
\end{aligned}
$$

#### 4. Length considering turbulence (Velikanov)
Turbulent eddies keep particles in suspension longer, so the length is increased:

$$
\begin{aligned}
L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\
&= \frac{1.65^2 \times 0.22^2 \times (2.0 - 0.2)^2}{7.51 \times 0.027^2} \\
&= \frac{2.7225 \times 0.0484 \times 3.24}{0.005475} = 77.98\ \text{m}
\end{aligned}
$$

#### 5. Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity | 0.22 m/s |
| Settling depth | 4.0 m (+ about 1 m for sediment storage) |
| Total width | 11.36 m, say 12 m = 2 chambers × 6 m |
| Ideal length | 32.6 m |
| Length with turbulence | 77.98 m, say 78 m |
| Detention time | $78/0.22 \approx 355$ s |

**Answer:** a two-chamber settling basin about **78 m long, 12 m wide (2 × 6 m) and 4 m deep** (plus storage depth, inlet and outlet transitions). The turbulence effect more than doubles the length compared with the ideal settling length of 32.6 m. Each chamber has $L/B = 78/6 = 13$, which gives uniform flow.

## Check yourself

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

1. How does 95% trap efficiency enter the Velikanov design?
   A. Through changing Q
   B. Through the coefficient lambda (1.65 for 95%)
   C. Through the depth
   D. It is ignored
   Answer: B. Higher efficiency means a larger lambda and so a longer basin. (Key point: Trap efficiency effect)

2. With v = 0.22 m/s, h = 4 m, Q = 10, B is about:
   A. 44 m
   B. 5.7 m
   C. 2.2 m
   D. 11.4 m
   Answer: D. B = 10/(0.22 x 4) = 11.36 m. (Key point: Width)

3. Ideal length L0 = h v / w with w = 0.027 m/s is:
   A. 148 m
   B. 32.6 m
   C. 11 m
   D. 78 m
   Answer: B. 4 x 0.22 / 0.027 = 32.6 m. (Key point: Ideal length)

4. Turbulence-corrected length (78 m) compared with 32.6 m shows:
   A. No change
   B. Length halves
   C. Turbulence reduces length
   D. Turbulence more than doubles the required length
   Answer: D. Eddies keep particles suspended, so a much longer basin is needed. (Key point: Effect of turbulence)

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

## Related in other subjects

- [Settling basin 8 m³/s, 0.3 mm, 90% trap](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-8cumec-03mm-90-percent) — Hydropower Engineering, chapter 5, asked 2×
- [Continuous flushing settling basin, 95% efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-continuous-60cumec-velikanov) — Hydropower Engineering, chapter 5
- [Settling basin trap efficiency (Camp's diagram)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-camp-diagram-80cumec) — Hydropower Engineering, chapter 5
- [Settling basin 6 m³/s, 0.25 mm, depth 3 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-6cumec-025mm-depth3m) — Hydropower Engineering, chapter 5