---
title: "Slope break and distance to depth 0.4 m"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:wide-channel-slope-break-direct-step-3-steps"
source: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-slope-break-direct-step-3-steps
site: BCT Engineering
---

# Slope break and distance to depth 0.4 m

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

- Asked in: 2081 Chaitra, 2+6 marks
- Video: [GRADUALLY VARIED FLOW M1PROFILES |SumamMiss| FLUID MECHANICS Lecture Videos:M5 –36](https://www.youtube.com/watch?v=0WQ6YBLP0Js) — Sumam Miss
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-slope-break-direct-step-3-steps
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A wide rectangular channel has an average bed slope of 0.0002 and discharge per unit width of 3.12 m³/s/m. Assume a Manning's value n = 0.02. If a break in slope occurs and the slope becomes 0.001, classify the flow profiles in both the slopes and use direct step method (3 steps only) to determine the distance downstream of the slope change at which the water surface level reaches 0.4 m.

## Short answer

Both slopes are mild (y_n1=2.437 m, y_n2=1.504 m > y_c=0.997 m); upstream M2 drawdown, downstream uniform flow. The depth 0.4 m is below critical and cannot occur. The M2 profile from the break upstream to 2.388 m depth extends about 5.1 km, computed with three direct steps.

## Answer

**Given:** wide rectangular channel, $q = 3.12\ \text{m}^2/\text{s}$, $n = 0.02$, $S_{01} = 0.0002$, $S_{02} = 0.001$. Wide channel: $R \approx y$.

**Critical and normal depths**

$$
y_c = \left(\frac{3.12^2}{9.81}\right)^{1/3} = 0.997\ \text{m}
$$

$$
y_{n1} = \left(\frac{0.02 \times 3.12}{\sqrt{0.0002}}\right)^{3/5} = 2.437\ \text{m},\qquad y_{n2} = \left(\frac{0.02 \times 3.12}{\sqrt{0.001}}\right)^{3/5} = 1.504\ \text{m}
$$

Both $y_{n1}$ and $y_{n2}$ exceed $y_c$, so **both slopes are mild**, the second being steeper. Flow is subcritical throughout, so the control is downstream.

**Classification**

- Downstream slope ($S_{02}$): uniform flow at $y_{n2} = 1.504$ m.
- Upstream slope ($S_{01}$): the depth must fall from $y_{n1} = 2.437$ m (far upstream) to $y_{n2} = 1.504$ m at the break. This lies between $y_n$ and $y_c$, so the profile is an **M2** (drawdown) curve.

**Reading of the 0.4 m depth.** A depth of 0.4 m is below $y_c = 0.997$ m, which cannot occur on a mild slope for this discharge (it would be an M3 curve and needs a gate or other supercritical control). The calculation below is therefore done for the actual M2 curve: the distance upstream of the break at which the depth has recovered to $0.98\,y_{n1} = 2.388$ m.

#### Direct step (3 steps, $S_0 = 0.0002$)

$$
E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2 q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}
$$

| $y$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $S_0 - \bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 1.5035 | 2.075 | 1.7230 | 0.001000 | | | |
| 1.7983 | 1.735 | 1.9518 | 0.000551 | 0.000775 | -0.000575 | -397.6 |
| 2.0932 | 1.491 | 2.2064 | 0.000332 | 0.000441 | -0.000241 | -1055.6 |
| 2.3880 | 1.307 | 2.4750 | 0.000214 | 0.000273 | -0.0000729 | -3682.9 |

The negative sign means the distances are measured **upstream** of the slope change.

$$
\sum \Delta x = -397.6 - 1055.6 - 3682.9 = -5136\ \text{m}
$$

**Answer:** Both reaches are mild (M2 profile on the flatter upstream slope, uniform flow on the steeper one). The depth rises from 1.50 m at the break to 2.39 m at about **5.1 km upstream**. A depth of 0.4 m is below critical depth (0.997 m) and cannot occur in this flow without a control.

## Check yourself

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

1. Why are both slopes mild here?
   A. yn1 and yn2 are both below yc
   B. Both normal depths (2.437 m, 1.504 m) exceed yc = 0.997 m
   C. Both slopes are smaller than 0.01
   D. Flow is supercritical on both
   Answer: B. Mild means yn > yc. Both normal depths exceed critical depth. (Key point: Classification)

2. Why is the profile on the upstream slope an M2 curve?
   A. Depth falls from yn1 to yn2 at the break, lying between yn and yc
   B. Depth rises above yn
   C. Depth is below yc
   D. Depth is constant at yn1
   Answer: A. Slope steepens, so normal depth falls; depth drops from yn1 toward yn2 in zone 2, between yn and yc: M2. (Key point: Profile on upstream slope)

3. Why can the 0.4 m depth not be reached by the profile in this problem?
   A. 0.4 m is above normal depth
   B. 0.4 m is below yc = 0.997 m, which would need a supercritical control on a mild slope
   C. Direct step cannot handle small depths
   D. 0.4 m equals the sequent depth
   Answer: B. On a mild slope a depth below yc is an M3 profile and needs a gate or similar control, which is absent. (Key point: Depth of 0.4 m)

4. What does the negative sign of the computed dx mean?
   A. The calculation failed
   B. Flow is supercritical
   C. Distances are measured upstream of the slope change
   D. The channel slope is adverse
   Answer: C. Negative dx means the points lie upstream of the break, about 5.1 km. (Key point: Sign of distance)

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