---
title: "Diversion weir: Bligh and Khosla, 4000 m3/s"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:diversion-weir-bligh-khosla-q4000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/diversion-weir-bligh-khosla-q4000
site: BCT Engineering
---

# Diversion weir: Bligh and Khosla, 4000 m3/s

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2071 Bhadra, 12 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/diversion-weir-bligh-khosla-q4000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

A diversion weir with a vertical drop to be designed for an irrigation system has the following data: Design flood = 4000 m³/s; Natural width of the source river = 300 m; Bed material = Coarse sand, Bligh's C = 12; Lacey's f = 1.2; Height of weir above low water = 3.0 m; Top width of the crest = 2.0 m. Fix the length of the floor according to Bligh's principle and design the length of floor and depth of cutoffs using Khosla's seepage theory. Compute the thickness of the floor at key points. Make suitable assumptions if necessary. Draw a neat sketch of the designed weir.

## Short answer

Bligh creep length 36 m. Khosla design: floor length between piles 12 m, upstream cutoff 3.8 m, downstream cutoff 8.3 m; provided floor thickness 1.0 m (upstream) and 1.5 m (downstream).

## Answer

#### Data and assumptions

$Q = 4000$ m³/s, river width $300$ m, Bligh's $C = 12$, Lacey's $f = 1.2$, weir height above low water $3.0$ m, crest width $2.0$ m. Assumed: river bed (low water) level $= 100.0$ m, broad-crested weir with $C_d = 1.7$ ($Q = 1.7LH^{3/2}$), afflux $= 1.0$ m, specific gravity of concrete $G = 2.4$.

#### 1. Waterway and flood levels

$$
P = 4.75\sqrt{4000} = 300.4\ \text{m}\approx 300\ \text{m}\ (\text{equal to the river width})
$$

Crest level $= 100.0 + 3.0 = 103.0$ m. Discharge per metre $q = 4000/300 = 13.33$ m³/s/m.

$$
H = \left(\frac{4000}{1.7\times 300}\right)^{2/3} = 3.95\ \text{m},\quad \text{HFL}_{u/s} = 103.0 + 3.95 = 106.95\ \text{m},\quad \text{HFL}_{d/s} = 105.95\ \text{m}
$$

#### 2. Scour depth and cutoffs

$$
R = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{13.33^2}{1.2}\right)^{1/3} = 7.14\ \text{m}
$$

- U/s pile bottom $= 106.95 - 1.5R = 96.23$ m, so $d_1 = 100 - 96.23 = 3.77$ m; adopt 3.8 m.
- D/s pile bottom $= 105.95 - 2R = 91.66$ m, so $d_2 = 8.34$ m; adopt 8.3 m.

#### 3. Floor length by Bligh's principle

Static head $H = 103.0 - 100.0 = 3.0$ m. Creep length $L_c = CH = 12\times 3 = 36$ m.

$$
b = L_c - 2(d_1 + d_2) = 36 - 2(3.8 + 8.3) = 11.8\ \text{m}
$$

#### 4. Khosla's check and design

Adopt a floor length $b = 12$ m between the piles: the 2 m wide crest body occupies the first 2 m, and the d/s floor below the vertical drop the remaining 10 m. Pile depths 3.8 m (u/s) and 8.3 m (d/s); floor thickness 1.0 m (u/s, under the crest) and 1.5 m (d/s) provisionally.

#### Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 12$ m.

| Pile | d (m) | α | λ | φE (%) | φD (%) |
|---|---|---|---|---|---|
| U/s pile | 3.80 | 3.158 | 2.156 | 47.69 | 31.98 |
| D/s pile | 8.30 | 1.446 | 1.379 | 64.87 | 41.14 |

#### Pressures at the key points

Pressure is expressed as a percentage of the head $H = 3.00$ m (no flow, pond at crest, d/s dry).

Mutual interference ($b' = b = 12$ m):

- at $C_1$ (due to the d/s pile): $C = 19\sqrt{\dfrac{8.3}{12}}\times\dfrac{3.8 + 8.3}{12} = +15.93\%$
- at $E_2$ (due to the u/s pile): $C = 19\sqrt{\dfrac{3.8}{12}}\times\dfrac{8.3 + 3.8}{12} = -10.78\%$

Floor thickness: at $C_1$, $+\dfrac{68.02 - 52.31}{3.8}\times 1.0 = +4.13\%$; at $E_2$, $-\dfrac{64.87 - 41.14}{8.3}\times 1.5 = -4.29\%$.

| Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) |
|---|---|---|---|---|
| E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 3.00 |
| D1 | 68.02 | - | 68.02 | 2.04 |
| C1 | 52.31 | int +15.93, thk +4.13 | 72.37 | 2.17 |
| E2 | 64.87 | int -10.78, thk -4.29 | 49.80 | 1.49 |
| D2 | 41.14 | - | 41.14 | 1.23 |
| C2 (d/s end) | 0 | 0 | 0 | 0 |

Corrections: int = mutual interference of piles, thk = floor thickness (no slope correction, since the floor is horizontal).

#### Floor thickness

Unbalanced head $h$ = hydraulic-gradient level $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$. The HGL is taken as a straight line between the corrected values at $C_1$ ($x = 0$) and $E_2$ ($x = 12$ m).

| Location | x (m) | φ (%) | HG level (m) | Unbalanced head (m) | Required t (m) | Provided t (m) | Check |
|---|---|---|---|---|---|---|---|
| C1 (under the crest) | 0 | 72.37 | 102.17 | balanced by the weir body and the pond water above | nominal | 1.00 | safe |
| Toe of the drop | 2 | 68.61 | 102.06 | 2.06 | 1.47 | 1.50 | safe |
| E2 (d/s pile) | 12 | 49.80 | 101.49 | 1.49 | 1.07 | 1.50 | safe |

The u/s part of the floor lies under the masonry crest and the pond water, whose weight balances the uplift, so a nominal 1.0 m is enough there. The critical point is just downstream of the vertical drop; provide **1.5 m** there. It may be tapered towards the d/s end, but not below about 1.1 m.

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{3.00}{8.30}\times\frac{1}{\pi\sqrt{1.379}} = 0.0980\ (\approx 1/10.2)
$$

Permissible $= 0.1667$, so the floor is safe against piping.

Add u/s and d/s aprons (block apron with filter, length $\approx 1.5$-$2$ times the scour depth, say about 15 m).

#### Sketch

```text
  HFL u/s ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  Crest 103.0  _____ 2.0 m
 u/s apron    |     |   vertical drop     d/s apron
 ============ |_____|=========================
 100.0 bed    |<------ b = 12 m ------>|
   pile 3.8 m      floor 1.0 m | 1.5 m   pile 8.3 m
```

**Answer:** Waterway 300 m; crest 103.0 m; u/s cutoff 3.8 m; d/s cutoff 8.3 m; Bligh floor length $\approx 11.8$ m, adopted 12 m (Khosla exit gradient 0.098 $< 1/6$); floor thickness 1.0 m (nominal) under the crest and 1.5 m on the d/s floor below the drop.

## Check yourself

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

1. Why is the waterway taken as 300 m for Q = 4000 m3/s?
   A. Lacey's P = 4.75 sqrt(Q) = 300.4 m, equal to river width
   B. River width is divided by two
   C. It is the weir height times 100
   D. It is fixed from Bligh's coefficient
   Answer: A. The Lacey perimeter of about 300 m matches the natural river width. (Key point: Waterway)

2. Bligh floor length b is found from:
   A. b = C H - 2(d1 + d2) = 36 - 24.2 = 11.8 m
   B. b = C H + 2(d1 + d2)
   C. b = H/C
   D. b = 5R
   Answer: A. Creep length 12 x 3 = 36 m minus vertical creep 2(3.8 + 8.3) leaves 11.8 m. (Key point: Bligh floor length)

3. The d/s pile depth 8.3 m results from:
   A. d/s HFL - 2R, minus bed level, with R = 7.14 m
   B. u/s HFL - 1.5R
   C. Weir height x 2
   D. Crest width x 4
   Answer: A. D/s pile bottom is 105.95 - 2(7.14) = 91.66 m, i.e. 8.34 m below the 100 m bed. (Key point: Cutoff depths)

4. How is the required floor thickness found at a key point?
   A. t = h/(G-1), h = unbalanced head (HG level minus floor top)
   B. t = h (G-1)
   C. t = G/h
   D. t = H/12
   Answer: A. The floor weight (G-1)t must balance the unbalanced uplift head h. (Key point: Required thickness)

5. At the u/s pile (C1) required thickness is 1.55 m against 1.0 m provided. What is concluded?
   A. Safe
   B. Not safe; thickness there must be increased
   C. Exit gradient fails
   D. Pile is too deep
   Answer: B. Provided thickness is less than required, so it is not safe and the larger thickness must be adopted. (Key point: Thickness check result)

6. Exit gradient 0.098 against permissible 1/6 means:
   A. Floor safe against piping
   B. Floor unsafe
   C. Pile must be shorter
   D. Floor length must be doubled
   Answer: A. 0.098 is below 0.167, so piping is not a danger. (Key point: Exit gradient)

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