---
title: "Sarda fall: Q 9, drop 1.2 m"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-q9-drop12"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-q9-drop12
site: BCT Engineering
---

# Sarda fall: Q 9, drop 1.2 m

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2071 Bhadra, 8 marks
- Video: [Design of Sarda Fall water Resources engineering](https://www.youtube.com/watch?v=Rncmf2oZ16o) — 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-q9-drop12
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design the crest and cistern of a drop structure (Sarda type) for a discharge of 9 cumecs and a drop height of 1.2 m: FSL u/s and d/s = 105.7 m and 104.5 m; Bed Level u/s and d/s = 104.2 m and 103.0 m; Bed Width u/s and d/s = 8 m; Side Slope of Channel = 1:1.

## Short answer

Rectangular crest at 104.97 m (top width 0.77 m, base 1.80 m); cistern 4.73 m long and 0.23 m deep, floor level 102.77 m.

## Answer

#### Given data and assumptions
- $Q = 9$ m³/s, bed width $B = 8$ m, side slope 1:1, u/s FSL $= 105.70$ m, u/s bed $= 104.20$ m, d/s bed $= 103.00$ m.
- Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 104.50$ m); drop in water level $H_L = 1.2$ m.
- Specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- Only the crest and the cistern are asked for, so the floor is not designed.

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (8 + 1\times1.5)(1.5) = 14.250\ \text{m}^2 \\ V_a &= Q/A = 0.632\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.020\ \text{m} \end{aligned}
$$

#### 2. Crest
Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 8$ 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:
$$
H = 0.727\ \text{m}, \qquad \text{crest level} = 105.70 - 0.727 = 104.97\ \text{m}, \qquad d = 104.97 - 103.00 = 1.97\ \text{m}
$$
Top width $B_c = 0.55\sqrt{1.97} = 0.77$ m. Base width of the crest wall:
$$
B_w = \frac{H + d}{\sqrt{G}} = \frac{0.727 + 1.97}{\sqrt{2.24}} = 1.80\ \text{m}
$$

#### 3. Cistern
Energy head over the crest $E = H + h_a = 0.727 + 0.020 = 0.747$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.747\times1.2} = 4.73\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.896)^{2/3} = 0.232\ \text{m} \end{aligned}
$$
Cistern floor level $= 103.00 - 0.232 = 102.77$ m.

#### Summary
| Component | Value |
|---|---|
| Crest | rectangular, length 8 m, level 104.97 m, head $H$ = 0.727 m |
| Crest width (top / base) | 0.77 m / 1.80 m |
| Cistern (length / depth) | 4.73 m / 0.23 m, floor level 102.77 m |

**Answer:** crest level 104.97 m (rectangular crest 8 m long, top width 0.77 m, base 1.80 m); cistern 4.73 m long and 0.23 m deep.

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. What crest length is used for this Sarda fall with B = 8 m and D = 1.5 m?
   A. 8.0 m
   B. 8.9 m
   C. 9.5 m
   D. 12.0 m
   Answer: B. L = B + 0.6D = 8 + 0.9 = 8.90 m. (Key point: Sarda crest length)

2. How is the top width of the crest wall obtained?
   A. T = 0.55 sqrt(H + d_c), about 0.90 m
   B. T = H + d_c
   C. T = 0.55 (H + d_c)^2
   D. T = d_c / 0.55
   Answer: A. The empirical formula T = 0.55 sqrt(H + d_c) = 0.55 sqrt(0.672 + 2.03) = 0.90 m. (Key point: Crest wall top width)

3. Why is the cistern depth only a nominal 0.25 m?
   A. The conjugate depth is greater than tail water, so a deep cistern is needed
   B. The tail water depth already exceeds the conjugate depth, so the jump is drowned and only a nominal cushion is kept
   C. No tail water exists
   D. The cistern depth must equal the crest head H
   Answer: B. Since the jump is already submerged without depression, a small depression only holds a water cushion. (Key point: Cistern depth logic)

4. Which formula is used for the bottom width of the crest wall?
   A. B_w = (H_w + H)/sqrt(G)
   B. B_w = H_w x sqrt(G)
   C. B_w = T/2
   D. B_w = H_w + H x G
   Answer: A. A gravity wall has B_w = (H_w + H)/sqrt(G) = 2.95/1.497 = 1.97 m. (Key point: Crest wall bottom width)

## Related in Irrigation and Drainage Engineering

- [Sarda fall: Q 12, bed level 103 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q12-bed103) — Irrigation and Drainage Engineering, chapter 7
- [Crest width, cistern length of vertical drop, Q 1.55](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q155-drop075) — Irrigation and Drainage Engineering, chapter 7, asked 2×
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