---
title: "Settling basin 60 m³/s, 0.15 mm, head 300 m"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-60cumec-300m-015mm-turbulence"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-60cumec-300m-015mm-turbulence
site: BCT Engineering
---

# Settling basin 60 m³/s, 0.15 mm, head 300 m

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

- Asked in: 2076 Ashwin, 8+4 marks; 2075 Ashwin, 7 marks (asked 2 times)
- Video: [Design Of Settling Basin | Hazen's and Vetter's Equation |Hydropower Engineering numerical](https://www.youtube.com/watch?v=BKq6dZonTQU) — OPengineer
- 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-60cumec-300m-015mm-turbulence
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Design a settling basin for a high head project in a river which utilizes 60 m³/s discharge and gross head of 300 m. The sediment particles larger than 0.15 mm (fall velocity = 1.5 cm/s) need to be trapped in the basin. Consider the effect of turbulence as well. Also draw the plan and section of the basin showing major components.

## Short answer

Use Camp velocity v=44√0.15=0.170 m/s, depth 6 m, total width 58.7 m split into 3 chambers of 20 m. Turbulence reduces fall velocity to 0.0058 m/s, giving length 175.8 m, adopted 180 m. Basin: 3 chambers, each 20 m wide x 180 m long x 6 m deep.

## Answer

**Given**: design discharge $Q=60.0$ m³/s and gross head 300 m (a high-head Pelton plant, so particles above 0.15 mm must be removed); particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). 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):
$$
v=44\sqrt{0.15}=17.04\ \text{cm/s}=0.1704\ \text{m/s}
$$
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 $H=6.0$ m in the settling zone (sediment storage is extra).
$$
B=\frac{Q}{vH}=\frac{60.0}{0.1704\times6.0}=58.68\ \text{m}
$$
Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=19.56$ m each, adopt 20.0 m.

#### Step 3: Length without turbulence
$$
L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}
$$

#### Step 4: Effect of turbulence
Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.
$$
\begin{aligned}
\alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\
w'&=0.0150-0.0092=0.0058\ \text{m/s}\\
L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m}
\end{aligned}
$$

#### Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity $v$ | 0.170 m/s |
| Settling depth $H$ | 6.0 m (plus about 1.0 m sediment storage in the hopper) |
| Number of chambers | 3 |
| Width of each chamber | 20.0 m |
| Length of settling zone | 180 m |
| Length/width ratio | 9.0 |
| Inlet transition | gradual, with guide vanes (side slope not steeper than about 1:5) |

Total volume of water in the basin $\approx 3\times20\times180\times6=64800$ 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      | \         /
         /         \ +-----------------+ /         \
                     |  chamber 3      |
                     +=================+
  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 chambers continue to supply the plant.

**Answer**: 3 chambers, each about 20.0 m wide x 180 m long x 6.0 m deep (settling zone), with $v=0.170$ m/s.

## Check yourself

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

1. How is the horizontal velocity chosen for the 0.15 mm particle?
   A. v = Q / H
   B. v = w x L
   C. Camp: v = 44 sqrt(d) = 0.170 m/s
   D. v = 3 m/s flushing velocity
   Answer: C. For 0.1<d<1 mm Camp's relation gives 17 cm/s, within 0.1-0.4 m/s so deposits are not re-suspended. (Key point: Horizontal velocity)

2. How is the total width found?
   A. B = Q / (v H) = 58.7 m, split into 3 chambers
   B. B = Q x v x H
   C. B = vH / w
   D. B = Q / w
   Answer: A. Area Q/v divided by depth H gives width; chambers allow flushing one at a time. (Key point: Basin width)

3. What does turbulence do to the design?
   A. Increases fall velocity and shortens the basin
   B. Reduces effective fall velocity w' = w - alpha, so length increases from 68 m to about 176 m
   C. Has no effect on length
   D. Reduces the required width only
   Answer: B. alpha = 0.132 v/sqrt(H) = 0.0092 m/s; w' = 0.0058 m/s and L = vH/w'. (Key point: Turbulence effect)

4. How is one chamber flushed while the plant keeps running?
   A. Stop the plant completely
   B. Open the inlet fully to raise velocity
   C. Dredge it from above
   D. Close its inlet, drain it and flush to the river while others operate
   Answer: D. Multiple chambers with a sloping floor and flushing channel allow continuous supply. (Key point: Flushing)

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