---
title: "Minimum impervious floor length, 10000 m3/s"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:barrage-min-floor-length-q10000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/barrage-min-floor-length-q10000
site: BCT Engineering
---

# Minimum impervious floor length, 10000 m3/s

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2072 Asoj, 7+1 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/barrage-min-floor-length-q10000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

An irrigation barrage has to be designed to pass a flood of 10,000 m³/s, through alluvium media (median diameter of particles = 0.33 mm). The flood level, pond level and downstream floor level are 207.0 m, 204 m and 198.0 m respectively. If the safe exit gradient is 1/6, compute minimum total impervious floor length required to safeguard the structure from piping. Prepare a conceptual section of the designed structure.

## Short answer

From safe exit gradient 1/6, downstream cutoff depth 6.4 m below floor (198 m), minimum impervious floor length ≈34.2 m between cutoffs.

## Answer

#### Data and assumptions

$Q = 10000$ m³/s, $d_{50} = 0.33$ mm, flood level $= 207$ m, pond level $= 204$ m, d/s floor level $= 198$ m, safe exit gradient $= 1/6$. The design head for piping is the static head with no tail water, $H = 204 - 198 = 6$ m. The d/s cutoff is taken to a depth of $1.5R$ below the HFL.

#### 1. Silt factor, waterway and scour depth

$$
\begin{aligned}
f &= 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.33} = 1.011 \\
P &= 4.75\sqrt{Q} = 475.0\ \text{m},\qquad q = Q/P = 21.05\ \text{m}^3/\text{s/m} \\
R &= 1.35\left(\frac{q^2}{f}\right)^{1/3} = 10.26\ \text{m}
\end{aligned}
$$

#### 2. Depth of d/s cutoff

Scour level $= 207 - 1.5R = 191.62$ m. Depth of the d/s pile below the floor:

$$
d = 198 - 191.62 = 6.38\ \text{m}
$$

#### 3. Floor length from the exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = 1/6\ \Rightarrow\ \frac{1}{\pi\sqrt{\lambda}} = 0.1667\times\frac{6.38}{6} = 0.1773
$$

$$
\lambda = \left(\frac{1}{\pi\times 0.1773}\right)^2 = 3.223,\qquad \lambda = \frac{1+\sqrt{1+\alpha^2}}{2} \Rightarrow \alpha = \sqrt{(2\lambda-1)^2-1} = 5.353
$$

$$
b = \alpha d = 5.353\times 6.38 = 34.2\ \text{m}
$$

Check: $G_E$ with $b = 34.2$ m is 0.1667 $= 1/6.0$.

**Answer:** Minimum total impervious floor length (between the end cutoffs) $\approx 34.2$ m, with the d/s cutoff 6.4 m deep. If a deeper d/s cutoff (2R) is used, the exit gradient is satisfied for any floor length, and the length is then fixed by the uplift thickness and the jump.

#### Conceptual section

```text
  Flood level 207 ~~~~~~~~~~~~~~~~~~~~~~~
  Pond level  204 ~~~~~~~~~|gate|
 u/s block apron  ___crest__|    |__ glacis __ d/s jump floor
 ====================================================== 198
   u/s pile (1.0-1.5R)  <--- b = 34.2 m ---> d/s pile 6.4 m
                                   inverted filter + block apron
```

## Check yourself

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

1. Silt factor from median particle diameter is:
   A. f = 1.76 sqrt(d in mm)
   B. f = 1.76 d^2
   C. f = 4.75 sqrt(d)
   D. f = 1.35 d
   Answer: A. Lacey's f = 1.76 sqrt(d_mm) = 1.011 for 0.33 mm. (Key point: Silt factor)

2. Head used for piping design here is:
   A. Static head with no tail water: pond - d/s floor = 6 m
   B. Flood - pond = 3 m
   C. Flood - d/s floor = 9 m
   D. Pond level
   Answer: A. The worst case for exit gradient is no tail water, so H = 204 - 198 = 6 m. (Key point: Design head)

3. The d/s cutoff depth 6.38 m comes from:
   A. Floor level - (flood level - 1.5R) with R = 10.26 m
   B. R/2
   C. H + R
   D. Pond - 1.5R
   Answer: A. Scour level 207 - 1.5R = 191.62 m; depth below floor 198 - 191.62 = 6.38 m. (Key point: Cutoff depth)

4. How is the minimum floor length found?
   A. Set G_E = 1/6, solve lambda, then alpha, then b = alpha d
   B. b = C H
   C. b = 2d
   D. b = H/G_E
   Answer: A. From G_E = (H/d)(1/(pi sqrt(lambda))), lambda = 3.223, alpha = 5.353, b = 34.2 m. (Key point: Floor length from exit gradient)

## Close variants set in other exams

- [Minimum impervious floor length, 11000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abarrage-min-floor-length-q11000) — Irrigation and Drainage Engineering, chapter 5, asked 2×

## Related in Irrigation and Drainage Engineering

- [Khosla: barrage HGL and floor thickness](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akhosla-barrage-hgl-thickness-pond1135) — Irrigation and Drainage Engineering, chapter 5