---
title: "Minimum impervious floor length, 11000 m3/s"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:barrage-min-floor-length-q11000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/barrage-min-floor-length-q11000
site: BCT Engineering
---

# Minimum impervious floor length, 11000 m3/s

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2078 Baisakh, 8 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-q11000
- 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 11000 m³/s, through alluvium media (median diameter of particles = 0.34 mm). The flood level, pond level and d/s floor level are 208 m, 205 m and 199 m respectively. If the safe exit gradient is 1/7, compute minimum total impervious floor length required to safeguard the structure from piping. Prepare a conceptual section of the designed structure.

## Short answer

Downstream cutoff 6.8 m deep; required floor length b≈45.3 m satisfies exit gradient 1/7.

## Answer

#### Data and assumptions

$Q = 11000$ m³/s, $d_{50} = 0.34$ mm, flood level $= 208$ m, pond level $= 205$ m, d/s floor level $= 199$ m, safe exit gradient $= 1/7$. The design head for piping is the static head with no tail water, $H = 205 - 199 = 6$ m. The d/s cutoff is taken to a depth of $1.5R$ below the HFL.

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

$$
\begin{aligned}
f &= 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.34} = 1.026 \\
P &= 4.75\sqrt{Q} = 498.2\ \text{m},\qquad q = Q/P = 22.08\ \text{m}^3/\text{s/m} \\
R &= 1.35\left(\frac{q^2}{f}\right)^{1/3} = 10.53\ \text{m}
\end{aligned}
$$

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

Scour level $= 208 - 1.5R = 192.20$ m. Depth of the d/s pile below the floor:

$$
d = 199 - 192.20 = 6.80\ \text{m}
$$

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

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = 1/7\ \Rightarrow\ \frac{1}{\pi\sqrt{\lambda}} = 0.1429\times\frac{6.80}{6} = 0.1619
$$

$$
\lambda = \left(\frac{1}{\pi\times 0.1619}\right)^2 = 3.864,\qquad \lambda = \frac{1+\sqrt{1+\alpha^2}}{2} \Rightarrow \alpha = \sqrt{(2\lambda-1)^2-1} = 6.654
$$

$$
b = \alpha d = 6.654\times 6.80 = 45.3\ \text{m}
$$

Check: $G_E$ with $b = 45.3$ m is 0.1429 $= 1/7.0$.

**Answer:** Minimum total impervious floor length (between the end cutoffs) $\approx 45.3$ m, with the d/s cutoff 6.8 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 208 ~~~~~~~~~~~~~~~~~~~~~~~
  Pond level  205 ~~~~~~~~~|gate|
 u/s block apron  ___crest__|    |__ glacis __ d/s jump floor
 ====================================================== 199
   u/s pile (1.0-1.5R)  <--- b = 45.3 m ---> d/s pile 6.8 m
                                   inverted filter + block apron
```

## Check yourself

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

1. For d50 = 0.34 mm the silt factor is about:
   A. 1.03 (1.76 sqrt(0.34))
   B. 0.34
   C. 5.9
   D. 0.6
   Answer: A. f = 1.76 sqrt(0.34) = 1.026. (Key point: Silt factor)

2. The head used for the piping check is:
   A. 205 - 199 = 6 m (no tail water)
   B. 208 - 199 = 9 m
   C. 208 - 205 = 3 m
   D. 7 m
   Answer: A. Static head with no tail water is pond level minus d/s floor level. (Key point: Design head)

3. The d/s cutoff depth is 6.80 m since:
   A. Scour level = 208 - 1.5R (R = 10.53) = 192.20, and 199 - 192.20 = 6.8
   B. R - H
   C. 1.5 x H
   D. 2R
   Answer: A. The cutoff goes to 1.5R below HFL, measured down from the floor level. (Key point: Cutoff depth)

4. With G_E = 1/7 and d = 6.8 m, 1/(pi sqrt(lambda)) equals:
   A. (1/7)(6.8/6) = 0.1619
   B. (1/7)(6/6.8)
   C. 7 x 6.8
   D. 0.5
   Answer: A. G_E = (H/d)/(pi sqrt(lambda)) so 1/(pi sqrt(lambda)) = G_E d/H = 0.1619. (Key point: Solving for floor length)

5. The minimum floor length that results is about:
   A. 45.3 m
   B. 20 m
   C. 6.8 m
   D. 100 m
   Answer: A. lambda = 3.864, alpha = 6.654, b = 6.654 x 6.8 = 45.3 m. (Key point: Result)

## Close variants set in other exams

- [Minimum impervious floor length, 10000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abarrage-min-floor-length-q10000) — Irrigation and Drainage Engineering, chapter 5, asked 2×

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