---
title: "Settling basin 30 m³/s with Velikanov check"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-30cumec-02mm-velikanov"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-30cumec-02mm-velikanov
site: BCT Engineering
---

# Settling basin 30 m³/s with Velikanov check

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

- Asked in: 2075 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-30cumec-02mm-velikanov
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Find the dimensions of the settling basin for a high head project of a Himalayan river which carries a discharge of 30 m³/s and a gross head of 100 m. The sediment size to be removed is up to 0.20 mm and fall velocity ω = 2 cm/sec. If the turbulence is considered, what will be the dimension of the basin? Check the length of the settling basin using Velikanov's method given correction factor λ = 1.5.

## Short answer

For 30 m³/s trapping >0.2 mm (w=0.02 m/s), v=0.197 m/s, depth 5 m. Two chambers 15.5 m wide; turbulence length 117 m, Velikanov 120 m, adopt 125 m. Dimensions: 2 chambers 15.5 m × 125 m × 5 m.

## Answer

**Given**: design discharge $Q=30.0$ m³/s and gross head 100 m; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=2.00$ cm/s $=0.0200$ m/s (given). The length is first found without turbulence, then with turbulence, and finally checked by Velikanov's method ($\lambda=1.5$).

#### 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
Assume a flow depth $H=5.0$ m in the settling zone (sediment storage is extra).
$$
B=\frac{Q}{vH}=\frac{30.0}{0.1968\times5.0}=30.49\ \text{m}
$$
Provide $n=2$ chambers so that one can be flushed while the other works: $b=B/n=15.25$ m each, adopt 15.5 m.

#### Step 3: Length without turbulence
$$
L_0=\frac{vH}{w}=\frac{0.1968\times5.0}{0.0200}=49.2\ \text{m}
$$

#### Step 4: Effect of turbulence
Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.
$$
\begin{aligned}
\alpha&=\frac{0.132\times0.1968}{\sqrt{5.0}}=0.0116\ \text{m/s}\\
w'&=0.0200-0.0116=0.0084\ \text{m/s}\\
L&=\frac{vH}{w'}=\frac{0.1968\times5.0}{0.0084}=117.4\ \text{m}
\end{aligned}
$$

#### Step 5: Check by Velikanov's method
$$
L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w^2}=\frac{1.5^2\times0.1968^2\times(\sqrt{5.0}-0.2)^2}{7.51\times0.0200^2}=120.2\ \text{m}
$$

The Velikanov length (120.2 m) is 2% greater than the turbulence length (117.4 m). The larger of the two is adopted and rounded up to 125 m.

#### Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity $v$ | 0.197 m/s |
| Settling depth $H$ | 5.0 m (plus about 1.0 m sediment storage in the hopper) |
| Number of chambers | 2 |
| Width of each chamber | 15.5 m |
| Length of settling zone | 125 m |
| Length/width ratio | 8.1 |
| Inlet transition | gradual, with guide vanes (side slope not steeper than about 1:5) |

Total volume of water in the basin $\approx 19375$ 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.

**Answer**: 2 chambers, each about 15.5 m wide x 125 m long x 5.0 m deep (settling zone), with $v=0.197$ m/s.

## Check yourself

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

1. What horizontal velocity is used for d = 0.20 mm (Camp: v = a sqrt(d), a = 44, v in cm/s, d in mm)?
   A. v = 44 sqrt(0.20) = 19.68 cm/s = 0.1968 m/s
   B. v = 44 x 0.20 = 8.8 cm/s = 0.088 m/s
   C. v = 44 sqrt(0.20) = 19.68 m/s
   D. v = 44/sqrt(0.20) = 98 cm/s = 0.98 m/s
   Answer: A. For 0.1 < d < 1 mm Camp gives v = a sqrt(d); the result must be converted from cm/s to m/s and lies in the 0.1-0.4 m/s range so deposits are not resuspended. (Key point: Camp horizontal velocity)

2. With H = 5.0 m and flow 30.0 m^3/s, what is the total width B and why are chambers provided?
   A. B = Q/v = 152 m; chambers are provided to reduce the length
   B. B = Q/(vH) = 30.49 m; 2 chambers so that one can be flushed while the others work
   C. B = Q/(wH) = 300 m; chambers are provided to increase velocity
   D. B = Q/(vH) = 30.49 m; chambers are provided only to reduce construction cost
   Answer: B. Continuity Q = vBH gives the width, and splitting the basin into chambers allows flushing one chamber without stopping the plant. (Key point: Width of basin)

3. What is the effective fall velocity w' with turbulence and the resulting length?
   A. w' = w + alpha = 0.0316 m/s, L = 31.1 m
   B. w' = alpha = 0.0116 m/s, L = 84.8 m
   C. alpha = 0.132 v/sqrt(H) = 0.0116 m/s, w' = 0.0084 m/s, L = vH/w' = 117.1 m
   D. w' = w = 0.0200 m/s, L = 49.2 m
   Answer: C. Turbulence is subtracted from the fall velocity (w' = w - alpha) and the length is vH/w'. (Key point: Turbulence correction)

4. How is the final length decided after the Velikanov check (lambda = 1.5)?
   A. The smaller value 117.1 m is adopted because it is more economical
   B. L_V = 120.2 m is adopted without rounding because Velikanov replaces the turbulence method
   C. The two values are added: 237 m
   D. L_V = 120.2 m is greater than the turbulence length 117.1 m, so the larger is adopted and rounded up to 125 m
   Answer: D. Velikanov's relation L = lambda^2 v^2 (sqrt(H) - 0.2)^2 /(7.51 w^2) is a check; the longer length is on the safe side. (Key point: Velikanov check)

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