---
title: "Vertical fall components, Q 1.0 m3/s"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:vertical-fall-q10-bligh-floor"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-fall-q10-bligh-floor
site: BCT Engineering
---

# Vertical fall components, Q 1.0 m3/s

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2062 Baisakh (old course), 16 marks
- Video: [DESIGN OF SARDA FALL NUMERICAL WITH DESIGN STEPS (PART 1) IRRIGATION/ Water resources engineering](https://www.youtube.com/watch?v=I3A3mTWh6xk) — Study with Civil buddy
- 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/vertical-fall-q10-bligh-floor
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design following components of a vertical fall: a) crest level b) crest width c) cistern elements d) floor length and floor thickness using Bligh's creep theory. With the data given below: a) full supply discharge = 1.0 m³/s b) drop = 1.0 m c) full supply depth = 0.75 m d) bed level U/S and D/S = 100 m / 99 m e) bed width = 3.0 m f) Bligh's coefficient = 6.0.

## Short answer

Crest level 100.41 m; rectangular crest top width 0.65 m, base 1.17 m. Cistern length 2.98 m, depth 0.125 m (floor 98.87 m). Floor lengths: u/s 1.0 m, d/s 4.9 m, total 7.07 m; d/s floor thickness 0.97 m reducing to 0.55 m.

## Answer

#### Given data and assumptions
- $Q = 1$ m³/s, bed width $B = 3$ m, rectangular canal, u/s FSL $= 100.75$ m, u/s bed $= 100.00$ m, d/s bed $= 99.00$ m.
- Full supply depth u/s = d/s $D = 0.75$ m (d/s FSL $= 99.75$ m); drop in water level $H_L = 1$ m.
- Bligh's coefficient $C = 6$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- No side slope is given, so a rectangular canal ($m = 0$) is assumed. U/s FSL $= 100 + 0.75 = 100.75$ m, d/s FSL $= 99 + 0.75 = 99.75$ m, so $H_L = 1.0$ m. The vertical fall is designed as a Sarda-type fall.

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (3 + 0\times0.75)(0.75) = 2.250\ \text{m}^2 \\ V_a &= Q/A = 0.444\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.010\ \text{m} \end{aligned}
$$

#### 2. Crest
Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 3$ m, and
$$
Q = 1.835\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{d}
$$
where $H$ is the head over the crest (measured from the u/s FSL), $B_c$ is the top width of the crest and $d$ is the height of the crest above the d/s bed. Solving by trial:
$$
H = 0.344\ \text{m}, \qquad \text{crest level} = 100.75 - 0.344 = 100.41\ \text{m}, \qquad d = 100.41 - 99.00 = 1.41\ \text{m}
$$
Top width $B_c = 0.55\sqrt{1.41} = 0.65$ m. Base width of the crest wall:
$$
B_w = \frac{H + d}{\sqrt{G}} = \frac{0.344 + 1.41}{\sqrt{2.24}} = 1.17\ \text{m}
$$

#### 3. Cistern
Energy head over the crest $E = H + h_a = 0.344 + 0.010 = 0.354$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.354\times1} = 2.98\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.354)^{2/3} = 0.125\ \text{m} \end{aligned}
$$
Cistern floor level $= 99.00 - 0.125 = 98.87$ m.

#### 4. Impervious floor (Bligh's theory)
Maximum seepage head occurs with no flow, water up to the crest on the u/s side and the d/s side dry: $H_s = d = 1.41$ m. Required creep length $L = C\,H_s = 6\times1.41 = 8.43$ m.
- Cut-offs: u/s $d_1 = D/3 + 0.6 = 0.85$ m; d/s $d_2 = D/2 + 0.6 = 0.97$ m.
- Horizontal floor length $b = L - 2(d_1 + d_2) = 8.43 - 3.65 = 4.78$ m.
- Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 4.90$ m, which is more than $L_c$; adopt $L_d = 4.90$ m.
- U/s floor $= b - L_d - B_w = 4.78 - 4.90 - 1.17 = -1.29$ m, so a minimum u/s floor of **1.0 m** is provided. Actual creep $= 10.72$ m $\ge 8.43$ m, safe.
- Total floor length $= 1.00 + 1.17 + 4.90 = 7.07$ m.

#### 5. Floor thickness
Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 3.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 3.87}{L} = 1.41\times\dfrac{10.72 - 3.87}{10.72} = 0.898$ m (with $L = 10.72$ m, the creep provided).
$$
t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{0.898}{1.24} = 0.97\ \text{m}
$$
At the end of the cistern the creep is 6.85 m, $h_2 = 0.508$ m and $t_2 = 0.55$ m. Adopt a d/s floor **0.97 m** thick near the crest wall, reducing to 0.55 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

#### Summary
| Component | Value |
|---|---|
| Crest | rectangular, length 3 m, level 100.41 m, head $H$ = 0.344 m |
| Crest width (top / base) | 0.65 m / 1.17 m |
| Cistern (length / depth) | 2.98 m / 0.13 m, floor level 98.87 m |
| Floor (u/s / wall base / d/s) | 1.00 / 1.17 / 4.90 m, total 7.07 m |
| Cut-offs (u/s / d/s) | 0.85 m / 0.97 m |
| D/s floor thickness | 0.97 m, reducing to 0.55 m |

**Answer:** crest level 100.41 m (rectangular crest 3 m long, top width 0.65 m, base 1.17 m); cistern 2.98 m long and 0.13 m deep; impervious floor 7.07 m long (u/s 1.00 m, d/s 4.90 m) with cut-offs of 0.85 m (u/s) and 0.97 m (d/s); d/s floor thickness 0.97 m, reducing to 0.55 m.

Sketch:
```text
 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|
```

## Check yourself

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

1. How is the crest level found in this vertical fall design?
   A. Crest level = u/s FSL + velocity head of approach - total head H over the crest
   B. Crest level = u/s bed level + H
   C. Crest level = d/s FSL - H
   D. Crest level = u/s FSL + H
   Answer: A. H is the total head including approach velocity head, so the crest sits at FSL + h_a - H below the energy level, giving 100.44 m here. (Key point: Crest level from head over crest)

2. Why is only a nominal cistern depth of 0.25 m provided?
   A. The tail water depth already exceeds the conjugate depth, so the jump is drowned and no real depression is needed; 0.25 m just holds a water cushion
   B. The cistern must always be exactly 0.25 m by code
   C. The tail water is too low, so the cistern must be deep to force the jump
   D. The cistern depth equals the drop of 1.0 m divided by four
   Answer: A. Without depression the conjugate depth is below the d/s water depth, so the jump is already submerged; a small depression is kept only as a cushion. (Key point: Cistern need when the jump is drowned)

3. What total creep length does Bligh's theory require here, and how is it obtained?
   A. L = H_L / C = 0.17 m
   B. L = C x H_L = 6 x 1.0 = 6.0 m
   C. L = C x (u/s depth) = 6 x 0.75 = 4.5 m
   D. L = 2 x C x H_L = 12 m
   Answer: B. Bligh's safe creep length is the coefficient C times the head loss H_L, which is 6 x 1.0 = 6.0 m. (Key point: Required creep length (Bligh))

4. How is the floor thickness near the wall toe obtained?
   A. t = h1 x (G - 1), using the full head
   B. t = (4/3) x h1 / (G - 1), where h1 is the residual uplift head at that point from Bligh's creep
   C. t = the crest wall top width
   D. t = 4/3 x drop height
   Answer: B. The floor must resist uplift pressure head h1, so t = (4/3) h1/(G-1) = 0.64 m, with h1 found from the creep to that point. (Key point: Floor thickness from residual head)

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