---
title: "Intermittent vs continuous flushing basin"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-intermittent-vs-continuous-5cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-intermittent-vs-continuous-5cumec
site: BCT Engineering
---

# Intermittent vs continuous flushing basin

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

- Asked in: 2078 Bhadra, 10 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-intermittent-vs-continuous-5cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Design a settling basin (i) with intermittent flushing (ii) continuous flushing for a hydroelectric plant by using the simple settling theory. The design discharge of the plant is 5 m³/s and depth of the basin is 3.20 m. Take w = 2.5 cm/s and λ = 1.5. Compare and justify the result. Assume 15% flushing discharge and efficiency = 90%.

## Short answer

Intermittent: area 300 m², total width 6.25 m, length 48 m, two chambers of 3.5 m x 48 m. Continuous with 15% flushing flow: area 345 m², total width 7.19 m, same length 48 m, two chambers of 3.6 m x 48 m; continuous flushing avoids storage and interruption but wastes 0.75 m³/s.

## Answer

**Given**: $Q=5$ m³/s, depth $H=3.2$ m, $w=0.025$ m/s, $\lambda=1.5$ (this coefficient gives the required trap efficiency of about 90% with turbulence), flushing discharge 15% in the continuous case. Assume a horizontal velocity of $v=0.25$ m/s (within 0.2-0.4 m/s). Simple settling theory: $A=\lambda Q/w$.

#### (i) Intermittent flushing ($Q=5$ m³/s)
$$
A=\frac{\lambda Q}{w}=\frac{1.5\times5}{0.025}=300\ \text{m}^2,\qquad B=\frac{Q}{vH}=\frac{5}{0.25\times3.2}=6.25\ \text{m},\qquad L=\frac{A}{B}=48.0\ \text{m}
$$
Provide two chambers of 3.5 m x 48 m so that one works while the other is being flushed. Sediment storage is also added (about 1 m) because the sediment stays in the basin between flushes.

#### (ii) Continuous flushing ($Q'=1.15\times5=5.75$ m³/s)
$$
A=\frac{1.5\times5.75}{0.025}=345\ \text{m}^2,\qquad B=\frac{5.75}{0.25\times3.2}=7.19\ \text{m},\qquad L=\frac{A}{B}=48.0\ \text{m}
$$
Two chambers of 3.6 m x 48 m. The settled sediment drops continuously into a hopper and is removed by the flushing water of 0.75 m³/s.

#### Comparison and justification
| Item | Intermittent | Continuous |
|---|---|---|
| Design flow | 5.00 m³/s | 5.75 m³/s |
| Surface area | 300 m² | 345 m² |
| Width (total) | 6.25 m | 7.19 m |
| Length | 48.0 m | 48.0 m |
| Sediment storage | needed (1 m) | not needed |
| Water loss | none during working | 15% of flow |
| Operation | basin must be taken out and flushed | uninterrupted |

The length is the same ($L=\lambda vH/w$) while the continuous basin is 15% wider because it carries 15% more water. The continuous basin needs no storage volume, no gate operation and no interruption, but it consumes 0.75 m³/s water all the time; the intermittent basin needs no flushing water except when flushing, but it needs two chambers and storage. For a small plant with little spare water, intermittent flushing is justified; for a plant with silt-laden water and automatic operation, continuous flushing is better.

**Answer**: (i) 2 chambers of about 3.5 m x 48 m; (ii) 2 chambers of about 3.6 m x 48 m, at a depth of 3.2 m.

## Check yourself

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

1. Which formula gives the intermittent basin surface area?
   A. A = Q/(vH) = 6.25 m^2
   B. A = lambda Q / w = 1.5 x 5 / 0.025 = 300 m^2
   C. A = w Q / lambda = 0.083 m^2
   D. A = vH/w = 32 m^2
   Answer: B. A = lambda Q / w = 300 m^2; width B = Q/(vH) = 6.25 m then gives L = A/B = 48 m. (Key point: Simple settling theory formula)

2. What discharge is used for the continuous flushing basin?
   A. 5.00 m^3/s
   B. 5.75 m^3/s (5 + 15%)
   C. 4.25 m^3/s (5 - 15%)
   D. 6.25 m^3/s
   Answer: B. 15% flushing water is added: Q' = 1.15 x 5 = 5.75 m^3/s, giving A = 345 m^2 and B = 7.19 m. (Key point: Design flow for continuous case)

3. Why do both basins come out 48 m long?
   A. Both have the same sediment storage
   B. L = lambda v H / w does not contain Q, so extra discharge only widens the basin
   C. Continuous flushing shortens L by 15% but intermittent adds 15% storage
   D. The length is simply assumed equal
   Answer: B. Since L = A/B = lambda v H / w, it depends on v, H and w only; the extra 15% flow increases width (7.19 vs 6.25 m), not length. (Key point: Why length is the same)

4. Which statement correctly compares the two flushing types?
   A. Continuous needs sediment storage, intermittent does not
   B. Intermittent consumes 0.75 m^3/s of water all the time
   C. Continuous works uninterrupted but wastes flushing water constantly; intermittent needs storage and two chambers but loses water only when flushing
   D. Both lose the same water and need the same storage
   Answer: C. Intermittent suits plants with little spare water; continuous suits silt-laden water and automatic operation, at the cost of 15% water loss. (Key point: Choosing between the two)

## Related in Hydropower Engineering

- [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
- [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
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- [Continuous settling basin (10 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-continuous-10cumec-015mm) — Hydropower Engineering, chapter 5
- [Single chamber settling basin design (15 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-single-chamber-15cumec) — 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×

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