---
title: "Vertical drop crest and cistern, Q 4.5"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q45-fsl1065"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q45-fsl1065
site: BCT Engineering
---

# Vertical drop crest and cistern, Q 4.5

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2072 Asoj, 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/vertical-drop-q45-fsl1065
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design the crest and cistern of a vertical drop structure for the data given below. Discharge = 4.5 cumec, Bed level u/s = 105.00, side slope of channel = 1:1, bed level d/s = 103.5, FSL u/s = 106.5, Bed width u/s and d/s = 3.0 m, Top width of crest = 0.75 m (for initial assumption), Cd = 0.41.

## Short answer

Crest level 105.64 m, crest length 3 m, head 0.88 m; crest wall top width 0.80 m, bottom 2.19 m. Cistern depth 0.25 m (floor 103.25 m) and length 6.28 m.

## Answer

#### Given data and assumptions
- $Q = 4.5$ m³/s, bed width $B = 3$ m, side slope 1:1, u/s FSL $= 106.50$ m, u/s bed $= 105.00$ m, d/s bed $= 103.50$ m.
- Full supply depth u/s $D_u = 1.500$ m; the d/s depth is taken equal to the u/s depth, $D_d = 1.500$ m (d/s FSL $= 103.5 + 1.5 = 105.000$ m); drop in water level $H_L = 1.500$ m.
- Specific gravity of masonry $G = 2.24$ (assumed, not given).

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD_u)D_u = (3 + 1\times1.500)(1.500) = 6.750\ \text{m}^2 \\ v &= Q/A = 0.667\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.023\ \text{m} \end{aligned}
$$

#### 2. Crest length, head and crest level
Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 0.41\times4.429 = 1.816$, where $H$ is the total head over the crest (including $h_a$).
Take crest length equal to the bed width, $L = B$:
$$
H = \left(\frac{Q}{1.816\,L}\right)^{2/3} = \left(\frac{4.5}{1.816\times3}\right)^{2/3} = 0.880\ \text{m}
$$
$$
\text{Crest level} = FSL_u + h_a - H = 106.500 + 0.023 - 0.880 = 105.64\ \text{m}
$$
Height of crest above d/s bed $d_c = 105.64 - 103.50 = 2.14$ m.

#### 3. Cistern (energy dissipation)
Discharge per metre length of crest $q = Q/L = 1.500$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 3.023$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:
- Without any depression the conjugate depth (1.41 m) is less than the tail water depth $D_d = 1.500$ m, so the jump is already drowned and hydraulically no depression is needed. A nominal cistern depth is still provided to hold a water cushion under the jet: **$X = 0.25$ m** (cistern floor level $= 103.25$ m).
- With this $X$: $y_1 = 0.193$ m, $F_1 = 5.65$, $y_2 = 1.448$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.448 - 0.193) = 6.28$ m.

#### 4. Crest (drop) wall
Check the assumed top width of 0.75 m: for a rectangular crest $T = 0.55\sqrt{d_c} = 0.55\sqrt{2.14} = 0.80$ m $> 0.75$ m, so adopt **$T = 0.80$ m**. Height of wall from cistern floor to crest $H_w = 105.64 - 103.25 = 2.39$ m. Bottom width for a gravity masonry wall:
$$
B_w = \frac{H_w + H}{\sqrt{G}} = \frac{2.39 + 0.88}{\sqrt{2.24}} = 2.19\ \text{m}
$$
Adopt bottom width $B_w = 2.19$ m (not less than the top width), with the d/s face vertical or slightly battered.

#### Summary
| Component | Value |
|---|---|
| Crest level | 105.64 m |
| Crest length / head $H$ | 3.00 m / 0.88 m |
| Crest wall top / bottom width | 0.80 m / 2.19 m |
| Cistern depth / length | 0.25 m / 6.28 m |

**Answer:** crest level 105.64 m, crest length 3.00 m, crest wall 0.80 m top and 2.19 m bottom width, cistern depth 0.25 m (level 103.25 m) and length 6.28 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. With Cd = 0.41, the constant in Q = Cd*sqrt(2g)*L*H^(3/2) is
   A. 0.41
   B. 1.838
   C. 1.816
   D. 4.43
   Answer: C. 0.41 x sqrt(2 x 9.81) = 0.41 x 4.43 = 1.816, used instead of 1.838 (which belongs to Cd = 0.415). (Key point: Weir coefficient)

2. Why is only a nominal cistern depth of 0.25 m provided here?
   A. Because the jump is already drowned: the conjugate depth is less than the tail water depth, so no depression is hydraulically needed
   B. Because the discharge is too large for a cistern
   C. Because the cistern must equal the crest top width
   D. Because the crest is below the d/s bed
   Answer: A. Tail water (1.5 m) exceeds the conjugate depth, so the jump is submerged; a small cistern is kept only to hold a water cushion. (Key point: Drowned jump and nominal cistern)

3. With Hw = 105.64 - 103.25 = 2.39 m, H = 0.88 m and G = 2.24, what is Bw = (Hw + H)/sqrt(G)?
   A. 2.19 m
   B. 1.46 m
   C. 3.27 m
   D. 1.63 m
   Answer: A. Bw = (2.39 + 0.88)/1.497 = 3.27/1.497 = 2.19 m. (Key point: Crest wall bottom width)

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