---
title: "Desander design, two basins (16 m³/s)"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:desander-16cumec-two-basins"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/desander-16cumec-two-basins
site: BCT Engineering
---

# Desander design, two basins (16 m³/s)

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

- Asked in: 2081 Bhadra, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/desander-16cumec-two-basins
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Design a desander using the following data: discharge = 16 m³/s, particle size to be settled ≥ 0.2 mm, particle fall velocity = 0.022 m/s, horizontal flow velocity = 0.2 m/s. Consider two basins with a factor of safety for basin area of 1.5. Use length-to-breadth ratio of 6. Assume a sediment concentration of 1 kg/m³, a sediment particle density of 2650 kg/m³, a settling time of 1 day, a sediment packing factor of 0.5, and an additional 10% discharge for flushing.

## Short answer

Design flow is 17.6 m3/s. Total area is 1.5 x 17.6 / 0.022 = 1200 m2, split into two basins, so each is 600 m2; with L/B=6 adopt 10 m x 60 m. Depth 4.5 m. Store about 1.0 m of sediment in each hopper.

## Answer

**Given**: $Q=16$ m³/s, particle $\ge0.2$ mm with $w=0.022$ m/s, horizontal velocity $v=0.2$ m/s, two basins, factor of safety 1.5, $L/B=6$, $C=1$ kg/m³, $\rho_s=2650$ kg/m³, settling (flushing) time 1 day, packing factor 0.5, extra discharge for flushing 10%.

#### Step 1: Design flow
$$
Q'=1.10\times16=17.6\ \text{m}^3/\text{s}
$$

#### Step 2: Surface area
$$
A_{total}=\frac{FS\,Q'}{w}=\frac{1.5\times17.6}{0.022}=1200.0\ \text{m}^2\quad\Rightarrow\quad A_{basin}=\frac{1200.0}{2}=600.0\ \text{m}^2
$$

#### Step 3: Plan size of one basin
$$
B=\sqrt{\frac{A_{basin}}{6}}=\sqrt{\frac{600.0}{6}}=10.00\ \text{m},\qquad L=6B=60.0\ \text{m}
$$
Adopt $B=10$ m, $L=60$ m.

#### Step 4: Depth of flow
Flow in one basin $=Q'/2=8.8$ m³/s; with $v=0.2$ m/s:
$$
H=\frac{Q'/2}{B\,v}=\frac{8.8}{10\times0.2}=4.40\ \text{m}\ \Rightarrow\ \text{adopt }4.5\ \text{m}
$$

#### Step 5: Sediment storage (1 day)
$$
\begin{aligned}
\text{Mass}&=C\,Q'\,t=1\times17.6\times86400=1520640\ \text{kg}\\
\text{Volume}&=\frac{1520640}{2650\times0.5}=1148\ \text{m}^3\ \text{(both basins)}\\
\text{Depth per basin}&=\frac{1148/2}{600.0}=0.96\ \text{m}
\end{aligned}
$$
Provide about 1.0 m of sediment storage in the hopper below the flow depth (or flush more often, as in practice).

| Item | Value |
|---|---|
| Number of basins | 2 |
| Size of each basin | 10 m wide x 60 m long |
| Flow depth | 4.5 m |
| Sediment storage depth | about 1.0 m |
| Total depth | about 5.5 m + 0.5 m free board |

**Answer**: 2 basins, each 10 m x 60 m in plan with a flow depth of 4.5 m and 1.0 m storage depth.

## Check yourself

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

1. What flow is used to design the desander?
   A. Q' = 16 m^3/s
   B. Q' = 1.10 x 16 = 17.6 m^3/s, including 10% for flushing
   C. Q' = 16/2 = 8 m^3/s
   D. Q' = 16 x 1.5 = 24 m^3/s
   Answer: B. The flushing allowance of 10% is added to the discharge. (Key point: Design flow)

2. How is the area of one basin found with FS = 1.5 and two basins?
   A. A = 17.6/0.022 = 800 m^2 per basin
   B. A = 1200 m^2 per basin
   C. A_total = 1.5 x 17.6/0.022 = 1200 m^2, so each basin is 600 m^2
   D. A = 1.5 x 8.8/0.2 = 66 m^2 per basin
   Answer: C. Total area is FS x Q'/w and is shared by the two basins. (Key point: Area per basin)

3. How is the flow depth of each basin obtained (B = 10 m, v = 0.2 m/s)?
   A. H = Q'/(B v) = 8.8 m using the full flow
   B. H = Q/(B w) = 72.7 m
   C. H = w/v = 0.11 m
   D. H = (Q'/2)/(B v) = 8.8/(10 x 0.2) = 4.4 m, adopt 4.5 m
   Answer: D. Each basin carries half the flow, so continuity with the given horizontal velocity gives the depth. (Key point: Depth of flow)

4. What is the sediment volume in 1 day (C = 1 kg/m^3, 2650 kg/m^3, packing factor 0.5)?
   A. Mass = 17.6 x 86400 = 1,520,640 kg, volume = mass/(2650 x 0.5) = 1148 m^3 for both basins
   B. Volume = 1,520,640/2650 = 574 m^3
   C. Volume = 1,520,640 x 0.5/2650 = 287 m^3
   D. Volume = 17.6 x 86400 = 1,520,640 m^3
   Answer: A. Dividing the mass by density and packing factor gives the deposit volume, then half goes to each basin (about 0.96 m depth). (Key point: Sediment storage)

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