---
title: "Hydraulic jump stilling basin (25 m³/s/m)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:stilling-basin-25cumec-per-m-2074"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-25cumec-per-m-2074
site: BCT Engineering
---

# Hydraulic jump stilling basin (25 m³/s/m)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2074 Ashwin, 6 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-25cumec-per-m-2074
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Design a hydraulic jump stilling basin for the maximum discharge of 25 m³s⁻¹m⁻¹ flowing from an overflow spillway, with the spillway crest 50 m above the downstream gravel river bed with a slope $S_0 = 0.001$ and $n = 0.028$.

## Short answer

y1=0.724 m, y2=12.91 m, Fr1=12.96; depress floor 6.5 m; basin length about 65 m.

## Answer

**Assumptions** (not given in the question): $C_d=0.75$ in $q=\tfrac23C_d\sqrt{2g}H^{3/2}$; losses on the spillway face neglected; the tail water depth in the river is the normal depth (wide channel, Manning); basin length $L=5\,y_2$.

#### Step 1: Head over crest
$$
25=2.2147\,H^{3/2}\Rightarrow H=5.03\ \text{m}
$$

#### Step 2: Tail water (Manning, wide channel)
$$
y_n=\left(\frac{qn}{\sqrt{S_0}}\right)^{3/5}=\left(\frac{25\times0.028}{\sqrt{0.001}}\right)^{0.6}=6.41\ \text{m}
$$

#### Step 3: Conjugate depth at the toe (floor at bed level)
$E=50+5.03=55.03$ m, so $y_1=0.766$ m, $V_1=32.6$ m/s, $Fr_1=11.90$ and
$$
y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=12.52\ \text{m}
$$
Since $y_n=6.41$ m is less than $y_2$, the jump would be swept away and the floor must be depressed.

#### Step 4: Depression of floor
Require $y_n+z=y_2(z)$ where $E=50+H+z$:

| $z$ (m) | $E$ (m) | $y_1$ (m) | $Fr_1$ | $y_2$ (m) | $y_n+z$ (m) |
|---|---|---|---|---|---|
| 0 | 55.03 | 0.766 | 11.90 | 12.52 | 6.41 |
| 6.50 | 61.53 | 0.724 | 12.96 | 12.91 | 12.91 |

Depression of the floor: **$z=6.50$ m** (say 6.5 m).

#### Step 5: Length and type
$$
L=5\,y_2=5\times12.91=64.6\ \text{m}\ \Rightarrow\ \text{adopt }65\ \text{m}
$$
$Fr_1=12.96>4.5$ and $V_1=34.5$ m/s $>18$ m/s, hence a USBR Type II basin: chute blocks of height $y_1=0.724$ m and a dentated end sill $0.2y_2=2.58$ m high. Protect the d/s bed with riprap.

**Answer**: $y_1=0.724$ m, $y_2=12.91$ m, $Fr_1=12.96$; depress the floor by 6.5 m; basin length about 65 m.

## Check yourself

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

1. How is H found from q = 25?
   A. From q = (2/3) Cd sqrt(2g) H^(3/2) with an assumed Cd = 0.75, giving H = 5.03 m
   B. H = q/g
   C. H = 25 m
   D. H = q^2
   Answer: A. Cd was assumed because it is not given; solving gives H = 5.03 m. (Key point: Head over crest)

2. Why is the floor depressed by 6.5 m?
   A. Tail water yn = 6.41 m is below y2 = 12.52 m, so yn + z must equal y2
   B. yn is greater than y2
   C. Fr1 is less than 1
   D. To reduce q
   Answer: A. Trial shows yn + z = y2(z) at z = 6.50 m. (Key point: Depression of floor)

3. Which basin arrangement is chosen and why?
   A. USBR Type II since Fr1 = 12.96 > 4.5 and V1 = 34.5 m/s > 18 m/s
   B. Type III with baffle blocks since V1 is low
   C. Type IV since Fr1 is 3
   D. No blocks at all
   Answer: A. High velocity rules out baffle blocks, so chute blocks and a dentated end sill are used. (Key point: Basin type)

## Related in Hydropower Engineering

- [Hydraulic jump stilling basin at spillway toe](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-hydraulic-jump-80cumec-2070) — Hydropower Engineering, chapter 4
- [Hydraulic jump stilling basin (28 m³/s/m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-28cumec-per-m-2075) — Hydropower Engineering, chapter 4
- [Is a hydraulic jump stilling basin required?](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-required-check-100cumec) — Hydropower Engineering, chapter 4
- [USBR stilling basin design (80 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-usbr-80cumec-2082) — Hydropower Engineering, chapter 4
- [Discharge from hydraulic jump depths 1.5 m to 4 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-hydraulic-jump-depth-change-120m) — Hydropower Engineering, chapter 4
- [Stilling basin invert level and USBR type](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astilling-basin-usbr-300cumec-12m) — Hydropower Engineering, chapter 4

## Related in other subjects

- [Discharge over overflow spillway](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aoverflow-spillway-discharge) — Hydro Power, chapter 6
- [Hydraulic jump definition and downstream Froude number](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahydraulic-jump-what-is-and-froude-after-jump) — Hydraulics, chapter 9
- [Maximum permissible slope in 60 mm gravel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agravel-channel-max-permissible-slope) — Hydraulics, chapter 10
- [Jump at spillway toe; repelled and submerged jump](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ajump-at-toe-of-spillway-repelled-submerged) — Hydraulics, chapter 9