---
title: "Settling basin 8 m³/s, 0.3 mm, 90% trap"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-8cumec-03mm-90-percent"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-8cumec-03mm-90-percent
site: BCT Engineering
---

# Settling basin 8 m³/s, 0.3 mm, 90% trap

Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants

- Asked in: 2074 Ashwin, 8 marks; 2073 Shrawan, 6 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-8cumec-03mm-90-percent
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

With considering turbulent effect, design a settling basin to remove the sediment size greater than 0.3 mm diameter. Assume the design discharge of the basin is 8 m³/s and trap efficiency as 90%.

## Short answer

Use Camp velocity v=44√0.30=0.241 m/s, depth 3.5 m, width 9.48 m split into 2 chambers of 5 m. Turbulence gives w'=0.023 m/s and length 36.7 m; for 90% trap efficiency multiply by 2.303, giving 84.5 m, adopted 85 m. Basin: 2 chambers, each 5 m x 85 m x 3.5 m.

## Answer

**Given**: design discharge $Q=8.0$ m³/s; particles of size $d\ge 0.30$ mm to be trapped, fall velocity $w=4.00$ cm/s $=0.0400$ m/s (assumed for 0.3 mm quartz at about 20 °C, after Rouse). The required trap efficiency is 90%, and turbulence is considered.

#### Step 1: Horizontal velocity (Camp)
For $0.1<d<1$ mm, $v=a\sqrt d$ with $a=44$ (v in cm/s, d in mm):
$$
v=44\sqrt{0.30}=24.10\ \text{cm/s}=0.2410\ \text{m/s}
$$
This is within the usual 0.1-0.4 m/s range, so deposited sediment is not re-suspended.

#### Step 2: Cross-section and width
Assume a flow depth $H=3.5$ m in the settling zone (sediment storage is extra).
$$
B=\frac{Q}{vH}=\frac{8.0}{0.2410\times3.5}=9.48\ \text{m}
$$
Provide $n=2$ chambers so that one can be flushed while the other works: $b=B/n=4.74$ m each, adopt 5.0 m.

#### Step 3: Length without turbulence
$$
L_0=\frac{vH}{w}=\frac{0.2410\times3.5}{0.0400}=21.1\ \text{m}
$$

#### Step 4: Effect of turbulence
Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.
$$
\begin{aligned}
\alpha&=\frac{0.132\times0.2410}{\sqrt{3.5}}=0.0170\ \text{m/s}\\
w'&=0.0400-0.0170=0.0230\ \text{m/s}\\
L&=\frac{vH}{w'}=\frac{0.2410\times3.5}{0.0230}=36.7\ \text{m}
\end{aligned}
$$

For a trap efficiency of 90% (Vetter's relation $\eta=1-e^{-w'L/(vH)}$), the length is multiplied by $\ln\frac{1}{1-\eta}=2.303$:
$$
L=2.303\times36.7=84.5\ \text{m}
$$
(This is conservative; Vetter's relation with the still-water fall velocity $w$ alone would give $2.303\times21.1=48.6$ m.)

#### Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity $v$ | 0.241 m/s |
| Settling depth $H$ | 3.5 m (plus about 1.0 m sediment storage in the hopper) |
| Number of chambers | 2 |
| Width of each chamber | 5.0 m |
| Length of settling zone | 85 m |
| Length/width ratio | 17.0 |
| Inlet transition | gradual, with guide vanes (side slope not steeper than about 1:5) |

Total volume of water in the basin $\approx 2\times5\times85\times3.5=2975$ m³.

```text
PLAN (flushing gates at the downstream end)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +=================+ /         \
  trash-                  |  |  flush gates
  rack                    v  v  --> flushing channel --> river
```

```text
LONGITUDINAL SECTION
 inlet          settling zone (depth H)      outlet
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  WL
 \    --->   --->   --->   --->   --->   --->  |
  \  ______________________________________   | weir
   \/   sediment storage / hopper slope 1:30 \__|__ to headrace
          flush gate -> flushing channel -> river
```

**Flushing arrangement**: the floor of each chamber slopes at about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. A flushing velocity of at least 3 m/s is kept in the channel. The chamber to be flushed is closed at the inlet, drained and flushed to the river, while the other chamber continues to supply the plant.

**Answer**: 2 chambers, each about 5.0 m wide x 85 m long x 3.5 m deep (settling zone), with $v=0.241$ m/s.

## Check yourself

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

1. What is the horizontal velocity for 0.3 mm particles?
   A. 0.044 m/s
   B. 2.41 m/s
   C. 0.30 m/s
   D. v = 44 sqrt(0.30) = 0.241 m/s
   Answer: D. Camp's relation for 0.1<d<1 mm with d in mm, v in cm/s. (Key point: Camp velocity)

2. What is the turbulence-reduced fall velocity?
   A. w' = w - 0.132 v/sqrt(H) = 0.023 m/s
   B. w' = w + alpha = 0.057 m/s
   C. w' = w = 0.040 m/s
   D. w' = w x alpha
   Answer: A. Turbulence reduces effective settling, lengthening the basin to 36.7 m. (Key point: Turbulence correction)

3. How is the 90% trap efficiency accounted for?
   A. Divide the length by 0.9
   B. Add 10% to the length
   C. Multiply the length by ln(1/(1-eta)) = 2.303 (Vetter)
   D. Double the width
   Answer: C. From eta = 1 - exp(-w'L/(vH)), L = 2.303 x 36.7 = 84.5 m. (Key point: Trap efficiency)

4. Why provide two chambers?
   A. To halve the velocity
   B. One can be flushed while the other works
   C. To double the trap efficiency
   D. To avoid turbulence
   Answer: B. Multiple chambers keep supply continuous during flushing. (Key point: Chambers)

## Related in Hydropower Engineering

- [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 with Hazen coefficient and sediment depth](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-hazen-sediment-depth-5000ppm) — 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
- [Settling basin 40 m³/s, 0.15 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-40cumec-015mm-w15) — Hydropower Engineering, chapter 5, asked 4×
- [Settling basin 60 m³/s, 0.15 mm, head 300 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-300m-015mm-turbulence) — 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

## Related in other subjects

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- [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
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