---
title: "Slope change 1 in 95 to 1 in 1420 with jump"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:slope-change-95-to-1420-jump-height-or-spillway"
source: https://www.bctengineer.com/q/semester_4/hydraulics/slope-change-95-to-1420-jump-height-or-spillway
site: BCT Engineering
---

# Slope change 1 in 95 to 1 in 1420 with jump

Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF)

- Asked in: 2068 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/slope-change-95-to-1420-jump-height-or-spillway
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A wide channel with uniform rectangular section has a change of slope from 1 in 95 to 1 in 1420 and the flow is 3.75 m³/s per m width. Determine the normal depth of flow corresponding to each slope and show that a hydraulic jump will occur in the region of the junction. Calculate the height of the jump and sketch the surface profiles between the upstream and downstream regions of uniform flow. Manning's coefficient n = 0.013 and it may be assumed that the channel is wide in comparison with the depth of flow, so that the hydraulic mean depth is approximately equal to the depth of flow.

## Short answer

Normal depths: yn1=0.64 m (steep, supercritical) and yn2=1.44 m (mild, subcritical). Jump on mild slope after M3 curve from 0.64 m to 0.86 m; jump height = 1.44 − 0.86 = 0.58 m, location ≈ 95 m downstream of junction.

## Answer

**Given:** wide rectangular channel, $q = 3.75\ \text{m}^2/\text{s}$, $n = 0.013$, slopes $1/95 = 0.01053$ and $1/1420 = 0.000704$; $R \approx y$.

#### Normal and critical depths

$$
y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.75^2}{9.81}\right)^{1/3} = 1.128\ \text{m}
$$

$$
y_{n1} = \left(\frac{0.013 \times 3.75}{\sqrt{0.01053}}\right)^{0.6} = 0.640\ \text{m}\quad (< y_c)
$$

$$
y_{n2} = \left(\frac{0.013 \times 3.75}{\sqrt{0.000704}}\right)^{0.6} = 1.440\ \text{m}\quad (> y_c)
$$

The upstream slope is steep with supercritical uniform flow, $V_1 = 5.86$ m/s and $Fr_1 = 2.34$; the downstream slope is mild with subcritical uniform flow. The transition must be a **hydraulic jump**.

#### Location and height

Sequent depth of $y_{n1}$:

$$
y_2 = \frac{0.640}{2}\left(\sqrt{1 + 8(2.34)^2} - 1\right) = 1.821\ \text{m}
$$

This is more than $y_{n2} = 1.440$ m, so the downstream depth is insufficient to force the jump at the junction. The jump forms **on the mild slope, downstream of the junction**, after an **M3** curve in which the depth rises from 0.640 m. The jump begins at the depth $y_j$ whose sequent depth equals 1.440 m:

$$
y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g\,y_j^3}} - 1\right) = 1.440 \;\Rightarrow\; y_j = 0.864\ \text{m}
$$

**Height of the jump** $= y_{n2} - y_j = 1.440 - 0.864 = 0.577$ m.

Direct step for the M3 length (3 steps from 0.640 to 0.864 m, $S_0 = 0.000704$, $S_f = n^2q^2/y^{10/3}$): $\Delta x = 33.2 + 32.1 + 30.1 \approx 95$ m. So the jump is about 95 m downstream of the junction.

#### Sketch

```text
   steep 1/95            |       mild 1/1420
 -- yn1 = 0.640 ---------+--.                      
                         |   '-.  M3     J    _____ yn2=1.440
                         |      '-..____/'|---'
                         |   0.640 -> 0.864 | 1.440
```

**Answer: $y_{n1} = 0.640$ m, $y_{n2} = 1.440$ m; a jump from 0.864 m to 1.440 m (height 0.58 m) forms about 95 m downstream of the junction, preceded by an M3 curve.**

## Check yourself

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

1. How is the normal depth found for a wide rectangular channel?
   A. yn = (n q/sqrt(S0))^(3/5), since R is about y
   B. yn = (q^2/g)^(1/3)
   C. yn = q/V critical
   D. yn = n q S0
   Answer: A. With R = y, Manning gives q = (1/n) y^(5/3) S0^(1/2). (Key point: Normal depth wide channel)

2. Why does a jump occur?
   A. yn1 = 0.640 m is below yc = 1.128 m and yn2 = 1.440 m is above it
   B. Both depths are below yc
   C. Both depths are above yc
   D. The slope is steeper downstream
   Answer: A. Supercritical to subcritical uniform flow requires a jump. (Key point: Why jump)

3. Why does the jump form downstream of the junction?
   A. The sequent depth 1.821 m is less than yn2
   B. The sequent depth of yn1 (1.821 m) is more than yn2 = 1.440 m, so the tailwater is too low for the jump at the junction
   C. The slope is steep downstream
   D. yn2 equals yc
   Answer: B. The supercritical flow continues as an M3 curve until its sequent depth equals 1.440 m. (Key point: Where the jump forms)

4. How is the height of the jump found?
   A. yn2 - yj, with yj = 0.864 m, giving 0.58 m
   B. y2 - yn1
   C. yc - yn1
   D. yn2 - yc
   Answer: A. The jump goes from yj to yn2. (Key point: Jump height)

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