---
title: "Distance to within 5% of normal depth"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:wide-channel-5m3s-depth-3-5m-5-percent-normal"
source: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-5m3s-depth-3-5m-5-percent-normal
site: BCT Engineering
---

# Distance to within 5% of normal depth

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

- Asked in: 2068 Magh, 7+1 marks
- Video: [Backwater calculations in open channel. (2) Standard step methods](https://www.youtube.com/watch?v=nzhI2uJJH0M) — Hubert Chanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-5m3s-depth-3-5m-5-percent-normal
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A wide rectangular channel conveys a discharge of 5 m³/sec with a bed slope of 1 in 3600 with Manning's coefficient (n) = 0.02. If the depth at a section is 3.5 m, determine how far upstream or downstream of the section, the depth would vary within 5% of the normal depth. Find the nature of profile and make calculation with direct step method and take only 2 steps for calculation.

## Short answer

M1 profile. Normal depth 2.93 m, critical 1.37 m. The depth reaches 5% of normal depth (3.08 m) about 4.85 km upstream of the 3.5 m section using direct step method with two steps.

## Answer

**Given:** wide rectangular channel, discharge taken as $q = 5\ \text{m}^2/\text{s}$ per metre width (a wide channel is analysed per unit width), $S_0 = 1/3600$, $n = 0.02$, $y = 3.5$ m. Wide channel: $R \approx y$.

**Depths**

$$
y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5} = \left(\frac{0.02 \times 5}{1/60}\right)^{0.6} = 6^{0.6} = 2.930\ \text{m},\qquad y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m}
$$

**Profile type:** $y_n > y_c$ (mild) and $y = 3.5 > y_n$ (zone 1), so the profile is **M1** (backwater). The depth falls going upstream towards $y_n$. The target depth is within 5% of $y_n$: $1.05\,y_n = 3.077$ m, which lies **upstream** of the 3.5 m section.

**Direct step method** (2 equal steps, $\Delta y = 0.2117$ m), with $S_f = n^2q^2/y^{10/3}$:

| $y$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ ($\times 10^{-4}$) | $\bar S_f$ ($\times 10^{-4}$) | $S_0 - \bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 3.5000 | 1.429 | 3.6040 | 1.536 | | | |
| 3.2883 | 1.520 | 3.4062 | 1.891 | 1.714 | 0.0001064 | -1859 |
| 3.0767 | 1.625 | 3.2113 | 2.361 | 2.126 | 0.0000652 | -2990 |

$$
L = 1859 + 2990 \approx 4850\ \text{m upstream}
$$

(A finer calculation gives about 4.9 km.)

**Answer: M1 profile; the depth comes within 5% of the normal depth about 4.85 km upstream of the 3.5 m section.**

## Check yourself

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

1. How is yn found for a wide channel?
   A. yn = (q^2/g)^(1/3)
   B. yn = (nq/sqrt(S0))^(3/5), using R = y
   C. yn = q/n
   D. yn = S0 q
   Answer: B. With R approx y, Manning gives yn = (nq/sqrt(S0))^(3/5) = 2.93 m. (Key point: Normal depth wide channel)

2. Why is the profile M1?
   A. y < yc
   B. Mild slope with y = 3.5 m above yn = 2.93 m
   C. y between yn and yc
   D. Steep slope
   Answer: B. y > yn > yc on a mild slope is M1 backwater. (Key point: Profile)

3. What end depth is used for 'within 5% of normal depth'?
   A. 0.95 yn
   B. yn
   C. 1.05 yn = 3.077 m
   D. yc
   Answer: C. M1 lies above yn, so 1.05 yn is the limit, upstream of the section. (Key point: Target depth)

4. Which friction slope expression is used for the wide channel?
   A. Sf = V^2/gy
   B. Sf = n^2 q^2 / y^(10/3)
   C. Sf = S0
   D. Sf = nq/y
   Answer: B. For R = y, Manning gives Sf = n^2 V^2/y^(4/3) = n^2 q^2/y^(10/3). (Key point: Friction slope)

## Related in Hydraulics

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- [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6

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