---
title: "Vertical drop in 10 m wide canal, Q 20"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-20cumec-10m-canal"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-20cumec-10m-canal
site: BCT Engineering
---

# Vertical drop in 10 m wide canal, Q 20

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2072 Magh, 1+6 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/vertical-drop-20cumec-10m-canal
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Drawing a definition sketch, design a vertical drop in a 10 m wide canal (side slope 1:1) discharging a flow as 20 m³/s. The canal bed level upstream and downstream is 102 m and 100 m respectively, whereas the FSL upstream and downstream is 105 m and 130 m [as printed, unclear] respectively. Determine design level, length of cistern and upstream floor length using Bligh's safe hydraulic gradient as 1/8.

## Short answer

Design crest level 103.98 m; trapezoidal crest 10 m long, top width 1.23 m, base 3.05 m. Cistern 7.20 m long, 0.41 m deep (floor 99.59 m). Upstream floor length 10.96 m; total impervious floor 24.41 m with cut-offs 1.60 m and 2.10 m.

## Answer

#### Given data and assumptions
- $Q = 20$ m³/s, bed width $B = 10$ m, side slope 1:1, u/s FSL $= 105.00$ m, u/s bed $= 102.00$ m, d/s bed $= 100.00$ m.
- Full supply depth u/s = d/s $D = 3$ m (d/s FSL $= 103.00$ m); drop in water level $H_L = 2$ m.
- Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- The printed d/s FSL (130 m) is clearly a misprint. The d/s depth is taken equal to the u/s depth (3 m), so d/s FSL $= 100 + 3 = 103$ m and $H_L = 105 - 103 = 2$ m.
- Bligh's safe hydraulic gradient of 1/8 means $C = 8$. The vertical drop is designed as a Sarda-type fall; the "design level" asked for is the crest level.

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (10 + 1\times3)(3) = 39.000\ \text{m}^2 \\ V_a &= Q/A = 0.513\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.013\ \text{m} \end{aligned}
$$

#### 2. Crest
Since $Q > 14$ m³/s, a **trapezoidal crest** is used, with crest length equal to the bed width, $L = 10$ m, and
$$
Q = 1.99\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{H + d}
$$
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:
$$
H = 1.024\ \text{m}, \qquad \text{crest level} = 105.00 - 1.024 = 103.98\ \text{m}, \qquad d = 103.98 - 100.00 = 3.98\ \text{m}
$$
Top width $B_c = 0.55\sqrt{1.024 + 3.98} = 1.23$ m. With a u/s batter of 1:3 and a d/s batter of 1:8, the base width $= 1.23 + 3.98\left(\tfrac13 + \tfrac18\right) = 3.05$ m.

#### 3. Cistern
Energy head over the crest $E = H + h_a = 1.024 + 0.013 = 1.037$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{1.037\times2} = 7.20\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(2.075)^{2/3} = 0.407\ \text{m} \end{aligned}
$$
Cistern floor level $= 100.00 - 0.407 = 99.59$ 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: $H_s = d = 3.98$ m. Required creep length $L = C\,H_s = 8\times3.98 = 31.81$ m.
- Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.60$ m; d/s $d_2 = D/2 + 0.6 = 2.10$ m.
- Horizontal floor length $b = L - 2(d_1 + d_2) = 31.81 - 7.40 = 24.41$ m.
- Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 10.40$ m, which is more than $L_c$; adopt $L_d = 10.40$ m.
- U/s floor $= b - L_d - B_w = 24.41 - 10.40 - 3.05 = $ **10.96 m**.
- Total floor length $= 10.96 + 3.05 + 10.40 = 24.41$ m.

#### Summary
| Component | Value |
|---|---|
| Crest | trapezoidal, length 10 m, level 103.98 m, head $H$ = 1.024 m |
| Crest width (top / base) | 1.23 m / 3.05 m |
| Cistern (length / depth) | 7.20 m / 0.41 m, floor level 99.59 m |
| Floor (u/s / wall base / d/s) | 10.96 / 3.05 / 10.40 m, total 24.41 m |
| Cut-offs (u/s / d/s) | 1.60 m / 2.10 m |

**Answer:** crest (design) level 103.98 m (trapezoidal crest 10 m long, top width 1.23 m); cistern 7.20 m long and 0.41 m deep (floor at 99.59 m); u/s floor length 10.96 m (total floor 24.41 m, cut-offs 1.60 m u/s and 2.10 m d/s).

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. A Bligh's safe hydraulic gradient of 1/8 means C = 8. With HL = 2.0 m, the required creep length is
   A. 0.25 m
   B. 10 m
   C. 16 m
   D. 22 m
   Answer: C. Required creep L = C x HL = 8 x 2.0 = 16 m. (Key point: Bligh's gradient and creep)

2. How is the printed d/s FSL of 130 m treated?
   A. It is used as printed, giving a huge drop
   B. It is treated as a typing error; d/s depth is taken equal to u/s depth (3 m), so d/s FSL = 103 m and HL = 2 m
   C. It is replaced by the u/s bed level
   D. It is ignored and HL is taken as zero
   Answer: B. A d/s FSL of 130 m is physically impossible, so the d/s depth is assumed equal to the u/s depth: 100 + 3 = 103 m, giving a drop of 2 m. (Key point: Handling the unclear d/s FSL)

3. Since no top width is given, how is it found?
   A. Taken as 0.5 m always
   B. T = 0.55 sqrt(H + dc), about 1.23 m
   C. Equal to the crest length
   D. Equal to the cistern depth
   Answer: B. The empirical relation T = 0.55 sqrt(H + dc) = 0.55 sqrt(1.058 + 3.96) gives 1.23 m. (Key point: Crest wall top width)

4. The computed u/s floor b_u is negative (-5.02 m). What does this mean?
   A. The design is unsafe and must be rejected
   B. The creep from cut-offs, wall and d/s floor already exceeds the required 16 m, so only a nominal u/s floor (1.0 m) is provided
   C. The u/s floor must be 5.02 m long
   D. The cistern should be removed
   Answer: B. Other elements already supply more creep than needed, so Bligh's requirement is satisfied (actual 22.02 m >= 16 m) and just a nominal floor is adopted. (Key point: Negative u/s floor requirement)

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