---
title: "Sarda fall: Q 12, bed level 103 m"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-q12-bed103"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-q12-bed103
site: BCT Engineering
---

# Sarda fall: Q 12, bed level 103 m

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2068 Chaitra (old course), 12 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/sarda-fall-q12-bed103
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design a 1.5 Sarda type fall for a canal having a discharge of 12 cumecs with the following data: Bed level u/s = 103.0 m; Side slopes of canal = 1:1; Bed level d/s = 101.5 m; Full supply level u/s = 104.5 m; Bed width u/s and d/s = 10 m; Soil = Good loam; Bligh's Coefficient = 6.

## Short answer

Rectangular crest at 103.74 m, 10 m long, top width 0.82 m, base 2.00 m. Cistern 5.42 m long, 0.28 m deep, floor level 101.22 m. Total floor 9.90 m; d/s floor thickness 1.56 m reducing to 0.68 m.

## Answer

#### Given data and assumptions
- $Q = 12$ m³/s, bed width $B = 10$ m, side slope 1:1, u/s FSL $= 104.50$ m, u/s bed $= 103.00$ m, d/s bed $= 101.50$ m.
- Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 103.00$ m); drop in water level $H_L = 1.5$ m.
- Bligh's coefficient $C = 6$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- D/s depth = u/s depth = 1.5 m, so d/s FSL $= 103.0$ m and the drop is 1.5 m. For good loam soil, Bligh's $C = 6$ (given).

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (10 + 1\times1.5)(1.5) = 17.250\ \text{m}^2 \\ V_a &= Q/A = 0.696\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.025\ \text{m} \end{aligned}
$$

#### 2. Crest
Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 10$ 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:
$$
H = 0.760\ \text{m}, \qquad \text{crest level} = 104.50 - 0.760 = 103.74\ \text{m}, \qquad d = 103.74 - 101.50 = 2.24\ \text{m}
$$
Top width $B_c = 0.55\sqrt{2.24} = 0.82$ m. Base width of the crest wall:
$$
B_w = \frac{H + d}{\sqrt{G}} = \frac{0.760 + 2.24}{\sqrt{2.24}} = 2.00\ \text{m}
$$

#### 3. Cistern
Energy head over the crest $E = H + h_a = 0.760 + 0.025 = 0.785$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.785\times1.5} = 5.42\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(1.177)^{2/3} = 0.279\ \text{m} \end{aligned}
$$
Cistern floor level $= 101.50 - 0.279 = 101.22$ 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: $H_s = d = 2.24$ m. Required creep length $L = C\,H_s = 6\times2.24 = 13.44$ 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 $b = L - 2(d_1 + d_2) = 13.44 - 4.90 = 8.54$ m.
- Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.90$ m, which is more than $L_c$; adopt $L_d = 6.90$ m.
- U/s floor $= b - L_d - B_w = 8.54 - 6.90 - 2.00 = -0.37$ m, so a minimum u/s floor of **1.0 m** is provided. Actual creep $= 14.80$ m $\ge 13.44$ m, safe.
- Total floor length $= 1.00 + 2.00 + 6.90 = 9.90$ m.

#### 5. Floor thickness
Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 5.20$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 5.20}{L} = 2.24\times\dfrac{14.80 - 5.20}{14.80} = 1.452$ m (with $L = 14.80$ m, the creep provided).
$$
t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.452}{1.24} = 1.56\ \text{m}
$$
At the end of the cistern the creep is 10.63 m, $h_2 = 0.632$ m and $t_2 = 0.68$ m. Adopt a d/s floor **1.56 m** thick near the crest wall, reducing to 0.68 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 10 m, level 103.74 m, head $H$ = 0.760 m |
| Crest width (top / base) | 0.82 m / 2.00 m |
| Cistern (length / depth) | 5.42 m / 0.28 m, floor level 101.22 m |
| Floor (u/s / wall base / d/s) | 1.00 / 2.00 / 6.90 m, total 9.90 m |
| Cut-offs (u/s / d/s) | 1.10 m / 1.35 m |
| D/s floor thickness | 1.56 m, reducing to 0.68 m |

**Answer:** crest level 103.74 m (rectangular crest 10 m long, top width 0.82 m, base 2.00 m); cistern 5.42 m long and 0.28 m deep; impervious floor 9.90 m long (u/s 1.00 m, d/s 6.90 m) with cut-offs of 1.10 m (u/s) and 1.35 m (d/s); d/s floor thickness 1.56 m, reducing to 0.68 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 are the d/s conditions fixed for this 1.5 m Sarda fall?
   A. D/s depth equals u/s depth 1.5 m, so d/s FSL = 101.5 + 1.5 = 103.0 m
   B. D/s depth is zero
   C. D/s FSL equals u/s FSL
   D. D/s depth equals the drop, 1.5 m, at bed 103 m
   Answer: A. The canal depth is unchanged, so the d/s FSL is the d/s bed plus 1.5 m, which is 103.0 m, a head loss of 1.5 m. (Key point: Downstream FSL for 1.5 m fall)

2. Which results are right for the crest?
   A. L = 10 m, crest level 104.5 m
   B. L = 10.90 m, H = 0.711 m, crest level 103.79 m
   C. L = 10.90 m, crest level 105.21 m
   D. L = 11.5 m, H = 1.5 m
   Answer: B. L = 10 + 0.6x1.5 = 10.9 m; H = 0.711 m, so crest level = 104.5 - 0.711 = 103.79 m. (Key point: Crest head and level)

3. How is the d/s floor length obtained?
   A. Bligh's formula L_d = 2.21 C sqrt(H_L sqrt(q/75)), which must also contain the cistern
   B. L_d = C x H_L
   C. L_d = 2 x crest length
   D. L_d = cut-off depth x 6
   Answer: A. The solution uses Bligh's d/s floor formula giving 5.65 m, long enough to hold the 5.57 m cistern. (Key point: Downstream floor length)

4. Why is the floor thickest near the wall toe?
   A. Because the residual uplift head h1 is greatest there, since less head has been lost along the creep path
   B. Because the water is deepest there
   C. Because the cut-off is deeper there
   D. Because Bligh's C is largest there
   Answer: A. Bligh's head loss is proportional to creep; near the toe little has been lost, so h1 (0.933 m) and t = 4/3 h1/(G-1) = 1.00 m are largest. (Key point: Floor thickness)

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