---
title: "Backwater distance into reservoir, 6 m channel"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:clean-earth-channel-entering-reservoir-distance"
source: https://www.bctengineer.com/q/semester_4/hydraulics/clean-earth-channel-entering-reservoir-distance
site: BCT Engineering
---

# Backwater distance into reservoir, 6 m channel

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

- Asked in: 2070 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/clean-earth-channel-entering-reservoir-distance
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

The clean earth (n = 0.020) channel in figure below is 6 m wide and laid on a slope of 0.005236. Water flows at 30 m³/s in the channel and enters a reservoir so that the channel depth is 3 m just before the entry. Assuming gradually varied flow, calculate the distance L. [Figure: channel with a 2 m depth marked at the upstream end of the reach and a 3 m depth at the reservoir entry; L is the length of the reach between them]

## Short answer

The profile is M1 (backwater). Using the direct step method, the distance L from the 2 m depth upstream to the 3 m depth at the reservoir entry is approximately 202 m.

## Answer

**Given:** $b = 6$ m, $n = 0.020$, $S_0 = 0.005236$, $Q = 30\ \text{m}^3/\text{s}$. The depth is 3 m at the reservoir entry and 2 m at the upstream end of the reach. Find the length $L$ between them. (Assumed reading of the figure: the 2 m depth is upstream, at distance $L$ from the entry.)

**Depths** ($q = 5\ \text{m}^2/\text{s}$):

$$
y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m},\qquad y_n = 1.417\ \text{m}\ (\text{Manning, } R = A/P)
$$

$y_n > y_c$, so the slope is (just) mild. Depths of 2 m to 3 m are above $y_n$, so the profile is **M1** (backwater from the reservoir). The depth falls going upstream.

**Direct step method**, 4 steps of 0.25 m, from $y = 3.0$ m to $y = 2.0$ m:

$$
E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}
$$

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 2.00 | 1.200 | 2.500 | 2.3186 | 0.00196 | | |
| 2.25 | 1.286 | 2.222 | 2.5017 | 0.00141 | 0.00169 | 51.6 |
| 2.50 | 1.364 | 2.000 | 2.7039 | 0.00106 | 0.00124 | 50.5 |
| 2.75 | 1.435 | 1.818 | 2.9185 | 0.00082 | 0.00094 | 49.9 |
| 3.00 | 1.500 | 1.667 | 3.1416 | 0.00065 | 0.00073 | 49.5 |

(Steps are taken from the upstream depth 2 m towards the reservoir, so $\Delta x$ is measured downstream.)

$$
L = 51.6 + 50.5 + 49.9 + 49.5 \approx 201.6\ \text{m}
$$

A finer integration gives 201.0 m, which agrees.

**Answer: $L \approx 202\ \text{m}$** (M1 backwater profile).

## Check yourself

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

1. What is the slope class (yc = 1.366, yn = 1.417)?
   A. Steep
   B. Critical
   C. Horizontal
   D. Mild, just
   Answer: D. yn is slightly above yc, so the slope is mild. (Key point: Slope class)

2. Why is the profile M1?
   A. Depths 2 to 3 m exceed yn on a mild slope due to reservoir backwater
   B. Depths are below yc
   C. Depths are between yn and yc
   D. The slope is adverse
   Answer: A. y > yn > yc is zone 1, an M1 backwater curve. (Key point: Profile)

3. How does depth vary on this profile going upstream?
   A. Rises
   B. Stays the same
   C. Falls toward yn
   D. Oscillates
   Answer: C. M1 depth decreases upstream toward normal depth, so the 2 m depth lies upstream of the 3 m. (Key point: Direction of depth)

4. What is L?
   A. About 20 m
   B. About 202 m
   C. About 2 km
   D. About 50 m
   Answer: B. The four step lengths (about 50 m each) sum to about 202 m. (Key point: Total length)

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