---
title: "Hydraulic jump stilling basin (28 m³/s/m)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:stilling-basin-28cumec-per-m-2075"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-28cumec-per-m-2075
site: BCT Engineering
---

# Hydraulic jump stilling basin (28 m³/s/m)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2075 Ashwin, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-28cumec-per-m-2075
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Design a hydraulic jump stilling basin for the flood discharge 28 m³/s/m flowing from an ogee spillway with the spillway crest 55 m above the downstream gravel river bed with a slope 1:1000 and Manning's roughness coefficient 0.028. Assume coefficients of discharge, depth and length are 0.75, 1.2 and 4.5 respectively. Also assume sp.gr of sediment as 2.65.

## Short answer

y1=0.757 m, y2=14.16 m, Fr1=13.58; depress floor 10.13 m; basin length 64 m; riprap approx 0.7 m.

## Answer

**Given**: $q=28$ m³/s/m, crest 55 m above the d/s river bed, bed slope 1:1000, $n=0.028$ (gravel bed), $C_d=0.75$, coefficient of tail-water depth 1.2 (the tail water over the floor must be at least $1.2y_2$), coefficient of length 4.5 ($L=4.5y_2$), relative density of sediment 2.65 (used for the riprap). Energy losses on the spillway face are neglected.

#### Step 1: Head over the crest
$$
28=\tfrac23(0.75)(4.429)H^{3/2}=2.2147\,H^{3/2}\Rightarrow H=5.43\ \text{m}
$$

#### Step 2: Tail water in the river (wide channel, Manning)
$$
y_n=\left(\frac{q\,n}{\sqrt{S_0}}\right)^{3/5}=\left(\frac{28\times0.028}{\sqrt{0.001}}\right)^{0.6}=6.86\ \text{m}
$$

#### Step 3: Jump at the toe (floor at bed level)
$E=55+5.43=60.43$ m:
$$
y_1=0.819\ \text{m},\quad V_1=34.2\ \text{m/s},\quad Fr_1=12.07,\quad y_2=13.57\ \text{m}
$$
Required tail water $1.2y_2=16.28$ m $\gg y_n=6.86$ m, so the floor must be depressed by $z$.

#### Step 4: Depression of the floor
The condition is $y_n+z\ge1.2\,y_2(z)$, with $E=55+H+z$:

| $z$ (m) | $E$ (m) | $y_1$ (m) | $Fr_1$ | $y_2$ (m) | $1.2y_2$ (m) | $y_n+z$ (m) |
|---|---|---|---|---|---|---|
| 0 | 60.43 | 0.819 | 12.07 | 13.57 | 16.28 | 6.86 |
| 10.13 | 70.56 | 0.757 | 13.58 | 14.16 | 16.99 | 16.99 |

So depress the floor by **$z\approx10.13$ m** below the river bed (say 10.5 m).

#### Step 5: Length and details
$$
L=4.5\,y_2=4.5\times14.16=63.7\ \text{m}\ \Rightarrow\ \text{adopt }64\ \text{m}
$$
- $Fr_1=13.58>4.5$ and $V_1=37.0$ m/s: a Type II (chute blocks, dentated sill) arrangement is used; chute-block height $\approx y_1=0.757$ m; end sill $0.2y_2=2.83$ m.
- Downstream protection: velocity over the sill $V=q/y_n=4.08$ m/s. By Isbash's formula with $C=0.86$ and $S_s=2.65$, the stone size is $d=\dfrac{V^2}{2gC^2(S_s-1)}=0.69$ m, so use riprap of about 0.7-0.8 m size on a filter layer.

**Answer**: $y_1=0.757$ m, $y_2=14.16$ m, $Fr_1=13.58$; floor depressed by 10.13 m; basin length 64 m; riprap about 0.7 m stones.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is tail water depth found for a wide gravel channel?
   A. yn = (q n / sqrt(S0))^(3/5) = 6.86 m
   B. yn = q / sqrt(g)
   C. yn = H
   D. yn = q n S0
   Answer: A. For a wide rectangular channel, Manning gives yn = (qn/sqrt(S0))^0.6. (Key point: Tail water Manning)

2. What does the coefficient 1.2 mean in this design?
   A. Tail water over the floor should be at least 1.2 y2
   B. Basin length is 1.2 y2
   C. Sill height is 1.2 y1
   D. Head is 1.2 times crest height
   Answer: A. The condition is yn + z >= 1.2 y2, a safety margin, giving z about 10.13 m. (Key point: Tail water coefficient 1.2)

3. How is the length found?
   A. L = 4.5 y2 = 4.5 x 14.16 = 64 m
   B. L = 4.5 y1
   C. L = 1.2 y2
   D. L = 4.5 H
   Answer: A. The question gives a length coefficient of 4.5 applied to y2. (Key point: Basin length)

4. What is the sediment specific gravity 2.65 used for?
   A. Sizing riprap by Isbash's formula for d/s protection
   B. Finding the head over crest
   C. Computing Froude number
   D. Computing tail water
   Answer: A. Isbash's formula d = V^2/(2gC^2(Ss-1)) gives about 0.69 m stone. (Key point: Riprap)

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