---
title: "Profile for abrupt slope change 0.0016 to 0.0150"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:sketch-profile-abrupt-slope-change-0016-to-015"
source: https://www.bctengineer.com/q/semester_4/hydraulics/sketch-profile-abrupt-slope-change-0016-to-015
site: BCT Engineering
---

# Profile for abrupt slope change 0.0016 to 0.0150

Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF)

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/sketch-profile-abrupt-slope-change-0016-to-015
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

Sketch the water-surface profile along rectangular channel (n = 0.014), if the channel is 3 m wide; the flow rate is 9.6 m³/s; and there is an abrupt change in slope from 0.0016 to 0.0150.

## Short answer

Upstream of the break, the mild slope has normal depth 1.392 m and the surface falls through an M2 curve to critical depth 1.014 m at the break. Downstream, an S2 curve falls to the steep-slope normal depth 0.629 m; no hydraulic jump occurs.

## Answer

**Given:** $b = 3\ \text{m}$, $Q = 9.6\ \text{m}^3/\text{s}$ ($q = 3.2\ \text{m}^2/\text{s}$), $n = 0.014$, $S_{01} = 0.0016$, $S_{02} = 0.0150$.

**Depths**

$$
y_c = \left(\frac{3.2^2}{9.81}\right)^{1/3} = 1.014\ \text{m}
$$

From Manning's equation, $R = A/P$:

| Reach | $S_0$ | $y_n$ (m) | Class |
|---|---|---|---|
| Upstream | 0.0016 | 1.392 | mild ($y_n > y_c$) |
| Downstream | 0.0150 | 0.629 | steep ($y_n < y_c$) |

**Profile**

1. **Upstream (mild) reach:** the flow is subcritical and uniform at $y_n = 1.392$ m far upstream. The abrupt change to the steep slope forces critical depth at the break. The surface falls from the NDL to $y_c = 1.014$ m as an **M2** curve, asymptotic to the NDL far upstream and meeting the CDL at the break.
2. **Downstream (steep) reach:** the flow starts at $y_c$ at the break and continues as an **S2** curve, falling and becoming asymptotic to the uniform supercritical depth $y_n = 0.629$ m.
3. No hydraulic jump occurs because the flow goes from subcritical to supercritical.

```text
 MILD  (yn=1.392)       |  STEEP (yn=0.629)
 ---- NDL ......___      |
              M2   ''--._|
 CDL - - - - - - - - - - *  (yc = 1.014, break)
                         |'-._  S2
 NDL(steep) - - - - - - -|- - -''--- yn = 0.629
```

**Answer: M2 on the upstream mild slope (1.392 m falling to 1.014 m at the break); S2 on the downstream steep slope (1.014 m falling to 0.629 m).**

## Check yourself

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

1. With yc = 1.014 m, how are the reaches classed?
   A. Upstream mild (yn 1.392 m), downstream steep (yn 0.629 m)
   B. Both mild
   C. Upstream steep, downstream mild
   D. Both steep
   Answer: A. yn > yc upstream gives mild; yn < yc downstream gives steep. (Key point: Slope classes)

2. What happens at the break from mild to steep slope?
   A. A hydraulic jump forms
   B. Flow passes through critical depth at the break
   C. Flow stays at normal depth
   D. Depth jumps to yn2
   Answer: B. The control is critical depth at the break, which links subcritical upstream flow to supercritical downstream flow. (Key point: Critical depth at break)

3. What is the upstream profile?
   A. M1 rising to the break
   B. M3 rising
   C. S2 falling
   D. M2 falling from 1.392 m to 1.014 m
   Answer: D. Depth between yn and yc on a mild slope is M2 drawdown, meeting the CDL at the break. (Key point: Upstream profile)

4. Why is there no hydraulic jump?
   A. Flow changes from subcritical to supercritical, which is gradual through yc
   B. Slope is too small
   C. Depth is below normal
   D. Froude number is one throughout
   Answer: A. A jump needs supercritical to subcritical flow; here the change is the opposite, so S2 follows smoothly. (Key point: Jump)

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