---
title: "Periodic settling basin with and without turbulence"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-periodic-5cumec-with-without-turbulence"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-periodic-5cumec-with-without-turbulence
site: BCT Engineering
---

# Periodic settling basin with and without turbulence

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

- Asked in: 2069 Chaitra, 8 marks
- 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-periodic-5cumec-with-without-turbulence
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Compute the dimension of a periodic type settling basin considering and without considering the turbulence effect for a hydropower plant through settling theory. Take settling velocity = 6 cm/sec; discharge = 5 m³/sec; particle size to be removed = 0.2 mm; depth of basin = 2.4 m.

## Short answer

Using Camp velocity 0.1968 m/s and depth 2.4 m, total width is 10.59 m, split into two chambers of about 5.5 m. Without turbulence length is 7.9 m; with turbulence effective fall velocity drops to 0.0432 m/s, giving 10.9 m, so 15 m is adopted.

## Answer

**Given**: design discharge $Q=5.0$ m³/s; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=6.00$ cm/s $=0.0600$ m/s (given). Periodic (intermittent) flushing is used, so no extra flushing discharge is added; the basin is designed both without and with turbulence.

#### 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):
$$
v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}
$$
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
Take the flow depth $H=2.4$ m in the settling zone (given; sediment storage is extra).
$$
B=\frac{Q}{vH}=\frac{5.0}{0.1968\times2.4}=10.59\ \text{m}
$$
Provide $n=2$ chambers so that one can be flushed while the other works: $b=B/n=5.29$ m each, adopt 5.5 m.

#### Step 3: Length without turbulence
$$
L_0=\frac{vH}{w}=\frac{0.1968\times2.4}{0.0600}=7.9\ \text{m}
$$

#### Step 4: Effect of turbulence
Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.
$$
\begin{aligned}
\alpha&=\frac{0.132\times0.1968}{\sqrt{2.4}}=0.0168\ \text{m/s}\\
w'&=0.0600-0.0168=0.0432\ \text{m/s}\\
L&=\frac{vH}{w'}=\frac{0.1968\times2.4}{0.0432}=10.9\ \text{m}
\end{aligned}
$$

#### Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity $v$ | 0.197 m/s |
| Settling depth $H$ | 2.4 m (plus about 1.0 m sediment storage in the hopper) |
| Number of chambers | 2 |
| Width of each chamber | 5.5 m |
| Length of settling zone | 15 m |
| Length/width ratio | 2.7 |
| Inlet transition | gradual, with guide vanes (side slope not steeper than about 1:5) |

Total volume of water in the basin $\approx 396$ 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 maintained 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.

#### Comparison
| Case | Effective fall velocity | Length of settling zone |
|---|---|---|
| Without turbulence | $w=0.060$ m/s | $L_0=7.87$ m |
| With turbulence | $w'=0.0432$ m/s | $L=10.92$ m |

Turbulence increases the length by 39%; the width (10.59 m in total) and the depth are unchanged. For periodic flushing, a sediment storage depth is added below the settling zone, and each chamber is flushed in turn.

**Answer**: without turbulence, 2 chambers of 5.5 m x about 8 m x 2.4 m; with turbulence, 2 chambers of 5.5 m x about 11 m x 2.4 m (15 m adopted for safety), with $v=0.197$ m/s.

## Check yourself

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

1. Why is the design flow kept at 5.0 m^3/s with no extra water added?
   A. Periodic flushing is done by stopping flow, so no continuous flushing discharge is drawn
   B. Turbulence already supplies the extra water
   C. The particle is too small to need extra water
   D. Flushing water is taken from the tailrace
   Answer: A. In a periodic basin sediment is removed during separate flushing operations, so unlike continuous flushing no extra discharge is added to Q. (Key point: No flushing discharge for periodic type)

2. What total width is found and how many chambers?
   A. 10.59 m total, 2 chambers of about 5.5 m
   B. 5.29 m total, one chamber
   C. 21.2 m total, 4 chambers
   D. 2.4 m total, 2 chambers of 1.2 m
   Answer: A. B = 5.0/(0.1968 x 2.4) = 10.59 m; two chambers (one flushed while the other works) give 5.29 m, adopted 5.5 m. (Key point: Width and chambers)

3. What is the length without considering turbulence?
   A. L0 = vH/w = 0.1968 x 2.4 / 0.06 = 7.9 m
   B. L0 = Q/(vH) = 10.6 m
   C. L0 = wH/v = 0.73 m
   D. L0 = vH/w' = 10.9 m
   Answer: A. Ideal settling gives L0 = vH/w = 7.87 m. (Key point: Length without turbulence)

4. By about how much does considering turbulence increase the length, and why?
   A. About 39% (7.87 m to 10.92 m), because w' = 0.0432 < w
   B. About 10%, because velocity rises
   C. About 100%, because width is halved
   D. It does not change because width and depth are unchanged
   Answer: A. Turbulence reduces fall velocity by alpha = 0.0168 m/s, so L = vH/w' = 10.92 m, 39% more; width and depth stay the same. (Key point: Turbulence effect comparison)

## Related in Hydropower Engineering

- [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 60 m³/s, 0.2 mm, w = 1.5 cm/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-02mm-w15) — Hydropower Engineering, chapter 5, asked 4×
- [Settling basin 50 m³/s, 0.2 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-50cumec-02mm-w15) — Hydropower Engineering, chapter 5, asked 4×
- [Settling basin 25 m³/s, 0.2 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-25cumec-02mm-w15) — Hydropower Engineering, chapter 5, asked 4×
- [Settling basin 60 m³/s, 0.15 mm, head 100 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-head-100m-015mm) — Hydropower Engineering, chapter 5
- [Settling basin 30 m³/s with Velikanov check](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-30cumec-02mm-velikanov) — Hydropower Engineering, chapter 5

## Related in other subjects

- [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 velocity of silica particle 0.02 cm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c69) — Water Supply Engineering, chapter 6
- [Settling velocity of silica particle 0.05 cm at 10°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as69a) — Water Supply Engineering, chapter 6