---
title: "Canal drop structure, Q 5, FSL 110.5"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:canal-drop-q5-fsl1105"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-drop-q5-fsl1105
site: BCT Engineering
---

# Canal drop structure, Q 5, FSL 110.5

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2065 Shrawan (old course), 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/canal-drop-q5-fsl1105
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design a canal drop structure for the data given below: Q = 5 m³/s; FSL, upstream = 110.5 m; FSL, downstream = 109.5 m; Normal Depth, u/s and d/s = 1.5 m; Bed width = 3.0 m; Bligh's coefficient, C = 7.

## Short answer

A Sarda-type vertical fall with rectangular crest (length 3 m) yields crest level 109.59 m, head 0.910 m, top width 0.69 m. Cistern length 4.84 m, depth 0.24 m, floor level 107.76 m. Total impervious floor 9.07 m (u/s 1.0 m, d/s 6.40 m); cut-offs 1.10 m and 1.35 m; d/s floor thickness 1.11 m reducing to 0.52 m.

## Answer

#### Given data and assumptions
- $Q = 5$ m³/s, bed width $B = 3$ m, side slope 1:1, u/s FSL $= 110.50$ m, u/s bed $= 109.00$ m, d/s bed $= 108.00$ m.
- Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 109.50$ m); drop in water level $H_L = 1$ m.
- Bligh's coefficient $C = 7$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- Bed levels: u/s $= 110.5 - 1.5 = 109.0$ m, d/s $= 109.5 - 1.5 = 108.0$ m. Side slope 1:1 is assumed. The drop is designed as a Sarda-type vertical fall.

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (3 + 1\times1.5)(1.5) = 6.750\ \text{m}^2 \\ V_a &= Q/A = 0.741\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.028\ \text{m} \end{aligned}
$$

#### 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:
$$
H = 0.910\ \text{m}, \qquad \text{crest level} = 110.50 - 0.910 = 109.59\ \text{m}, \qquad d = 109.59 - 108.00 = 1.59\ \text{m}
$$
Top width $B_c = 0.55\sqrt{1.59} = 0.69$ m. Base width of the crest wall:
$$
B_w = \frac{H + d}{\sqrt{G}} = \frac{0.910 + 1.59}{\sqrt{2.24}} = 1.67\ \text{m}
$$

#### 3. Cistern
Energy head over the crest $E = H + h_a = 0.910 + 0.028 = 0.938$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.938\times1} = 4.84\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.938)^{2/3} = 0.240\ \text{m} \end{aligned}
$$
Cistern floor level $= 108.00 - 0.240 = 107.76$ 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: $H_s = d = 1.59$ m. Required creep length $L = C\,H_s = 7\times1.59 = 11.13$ m.
- Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.10$ m; d/s $d_2 = D/2 + 0.6 = 1.35$ m.
- Horizontal floor length $b = L - 2(d_1 + d_2) = 11.13 - 4.90 = 6.23$ m.
- Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.40$ m, which is more than $L_c$; adopt $L_d = 6.40$ m.
- U/s floor $= b - L_d - B_w = 6.23 - 6.40 - 1.67 = -1.84$ m, so a minimum u/s floor of **1.0 m** is provided. Actual creep $= 13.97$ m $\ge 11.13$ m, safe.
- Total floor length $= 1.00 + 1.67 + 6.40 = 9.07$ m.

#### 5. Floor thickness
Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 4.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 4.87}{L} = 1.59\times\dfrac{13.97 - 4.87}{13.97} = 1.036$ m (with $L = 13.97$ m, the creep provided).
$$
t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.036}{1.24} = 1.11\ \text{m}
$$
At the end of the cistern the creep is 9.71 m, $h_2 = 0.485$ m and $t_2 = 0.52$ m. Adopt a d/s floor **1.11 m** thick near the crest wall, reducing to 0.52 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 109.59 m, head $H$ = 0.910 m |
| Crest width (top / base) | 0.69 m / 1.67 m |
| Cistern (length / depth) | 4.84 m / 0.24 m, floor level 107.76 m |
| Floor (u/s / wall base / d/s) | 1.00 / 1.67 / 6.40 m, total 9.07 m |
| Cut-offs (u/s / d/s) | 1.10 m / 1.35 m |
| D/s floor thickness | 1.11 m, reducing to 0.52 m |

**Answer:** crest level 109.59 m (rectangular crest 3 m long, top width 0.69 m, base 1.67 m); cistern 4.84 m long and 0.24 m deep; impervious floor 9.07 m long (u/s 1.00 m, d/s 6.40 m) with cut-offs of 1.10 m (u/s) and 1.35 m (d/s); d/s floor thickness 1.11 m, reducing to 0.52 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 of the drop obtained?
   A. FSL u/s + h_a + H
   B. FSL u/s + h_a - H, where H is the total head over the crest from Q = 1.838 L H^(3/2)
   C. d/s bed + H
   D. FSL d/s - H
   Answer: B. H is the total head including the velocity head h_a, so crest level = 110.500 + 0.028 - 0.937 = 109.59 m. (Key point: Crest head and level)

2. What is the total creep length required by Bligh's theory here?
   A. L = C x H_L = 7 x 1.0 = 7.0 m
   B. L = C x D_u = 7 x 1.5 = 10.5 m
   C. L = H_L / C
   D. L = C x E_1 = 7 x 2.528
   Answer: A. Bligh's creep length is the coefficient times the head causing seepage, here the 1.0 m drop in water level. (Key point: Total creep length)

3. What condition fixes the cistern depth X?
   A. The conjugate depth y2 of the jump must match the tail water over the cistern, D_d + X
   B. The cistern depth equals the crest height
   C. The cistern depth equals the u/s flow depth
   D. The cistern depth equals the top width of the wall
   Answer: A. The jump is held in the cistern when y2 equals the tail water depth over the cistern floor; here the jump is already drowned so a nominal 0.25 m is used. (Key point: Cistern and jump)

4. How is the floor thickness near the wall found?
   A. t = 4/3 x h/(G - 1), using the residual uplift head h at that point
   B. t = h x G
   C. t = H_L / C
   D. t = Bw / 2
   Answer: A. Thickness resists uplift pressure head h with a 4/3 safety factor: 4/3 x 0.648/1.24 = 0.70 m. (Key point: Floor thickness)

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