---
title: "Backwater distance within 100 mm of normal depth"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:dam-backwater-distance-within-100mm-normal"
source: https://www.bctengineer.com/q/semester_4/hydraulics/dam-backwater-distance-within-100mm-normal
site: BCT Engineering
---

# Backwater distance within 100 mm of normal depth

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

- Asked in: 2076 Baisakh, 8 marks
- Video: [Backwater profiles in gradually-varied open channel flows](https://www.youtube.com/watch?v=sJghflcmCek) — Hubert Chanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/dam-backwater-distance-within-100mm-normal
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A dam is built across a channel of rectangular cross section which carries water at the rates of 8.75 m³/s. As a result the depth just upstream of the dam is increased to 2.5 m. The channel is 5 m wide and the slope of the bed is 1 in 5000. The channel is lined with concrete (Manning's n = 0.015). How far upstream is the depth within 100 mm of the normal depth?

## Short answer

The dam creates an M1 backwater curve, with normal depth 1.80 m and critical depth 0.678 m. Direct stepping from 2.5 m to 1.90 m gives the depth within 100 mm of normal depth about 8.15 km upstream of the dam.

## Answer

**Given:** $Q = 8.75\ \text{m}^3/\text{s}$, $b = 5\ \text{m}$, $S_0 = 1/5000 = 0.0002$, $n = 0.015$, depth at the dam $y = 2.5$ m.

**Depths**

$$
q = 1.75\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.75^2}{9.81}\right)^{1/3} = 0.678\ \text{m}
$$

By Manning's equation (trial, $R = A/P$): $y_n = 1.80\ \text{m}$.

Since $y = 2.5 > y_n > y_c$ on a mild slope, the profile is **M1** (backwater). Going upstream, the depth falls towards $y_n$. We need the distance upstream to the point where $y = y_n + 0.1 = 1.90$ m.

**Direct step method**, 4 equal steps from 2.5 m to 1.90 m ($\Delta y = 0.15$ 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) | $S_f$ ($\times 10^{-4}$) | $\bar S_f$ ($\times 10^{-4}$) | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 2.500 | 1.250 | 0.700 | 2.5250 | 0.82 | | |
| 2.350 | 1.211 | 0.745 | 2.3784 | 0.97 | 0.89 | -1323 |
| 2.200 | 1.170 | 0.795 | 2.2325 | 1.15 | 1.06 | -1552 |
| 2.050 | 1.126 | 0.853 | 2.0875 | 1.40 | 1.28 | -2003 |
| 1.900 | 1.080 | 0.921 | 1.9438 | 1.72 | 1.56 | -3270 |

$$
L = 1323 + 1552 + 2003 + 3270 \approx 8150\ \text{m}
$$

The negative sign means upstream. A finer integration gives about 8.26 km.

**Answer: The depth is within 100 mm of the normal depth about 8.15 km (roughly 8.2 km) upstream of the dam** (M1 backwater curve).

## Check yourself

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

1. Why is the profile upstream of the dam M1?
   A. 2.5 m > yn = 1.80 m > yc = 0.678 m on a mild slope
   B. Depth is below yc
   C. Depth is between yn and yc
   D. The slope is steep
   Answer: A. Depth above normal on a mild slope gives M1 backwater. (Key point: Profile type)

2. What end depth is used for the calculation?
   A. yc + 0.1 = 0.778 m
   B. The normal depth 1.80 m exactly
   C. yn + 0.1 = 1.90 m
   D. yn - 0.1 = 1.70 m
   Answer: C. The question asks where depth is within 100 mm of yn, i.e. 1.90 m, since M1 approaches yn from above. (Key point: Target depth)

3. Why can the calculation not run to exactly yn?
   A. The formula fails at y = yc
   B. M1 approaches yn asymptotically, giving infinite length
   C. Friction becomes zero
   D. The channel ends
   Answer: B. At y = yn, S0 = Sf and dx is infinite, so a depth slightly above yn is used. (Key point: Why not reach yn)

4. About how far upstream is the depth 1.90 m?
   A. About 815 m
   B. About 81.5 km
   C. About 8.1 km
   D. About 0.8 km
   Answer: C. Summing the four steps gives about 8150 m upstream of the dam. (Key point: Result)

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