---
title: "Bed slope for critical normal depth"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:bed-slope-for-critical-flow-at-normal-depth"
source: https://www.bctengineer.com/q/semester_4/hydraulics/bed-slope-for-critical-flow-at-normal-depth
site: BCT Engineering
---

# Bed slope for critical normal depth

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2068 Magh, 6 marks
- Video: [LGCIV2051 OC5 - Calculation of the critical depth and critical slope](https://www.youtube.com/watch?v=k5Ey5W0e2Ec) — Hydraulics and beyond - Sandra Soares-Frazao
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/bed-slope-for-critical-flow-at-normal-depth
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Find at what bed slope a 4 m wide rectangular channel be laid so that the flow is critical at a normal depth of 1.25 m, with Manning's coefficient (n) = 0.015.

## Short answer

To have normal depth equal to critical depth, the bed slope must be the critical slope. For y_n = 1.25 m, V = sqrt(g y) = 3.50 m/s. With Manning's equation, R = 0.769 m, S0 = (0.015×3.50 / 0.769^(2/3))^2 = 0.0039 (≈1 in 255).

## Answer

**Given:** rectangular channel, $b = 4\ \text{m}$, normal depth $y_n = 1.25\ \text{m}$ equals the critical depth, $n = 0.015$.

**Step 1: Critical flow velocity**

$$
V = \sqrt{g\,y} = \sqrt{9.81 \times 1.25} = 3.502\ \text{m/s}
$$

**Step 2: Geometry**

$$
A = 4 \times 1.25 = 5\ \text{m}^2,\quad P = 4 + 2(1.25) = 6.5\ \text{m},\quad R = \frac{5}{6.5} = 0.769\ \text{m}
$$

The discharge is $Q = AV = 17.51\ \text{m}^3/\text{s}$.

**Step 3: Manning's equation**, $V = \dfrac{1}{n}R^{2/3}S_0^{1/2}$:

$$
S_0 = \left(\frac{nV}{R^{2/3}}\right)^2 = \left(\frac{0.015 \times 3.502}{0.769^{2/3}}\right)^2 = 0.003915
$$

**Answer: Bed slope $S_0 \approx 0.0039$ (about 1 in 255).** This is the critical slope for this channel and discharge.

## Check yourself

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

1. For critical flow at depth 1.25 m in a rectangular channel, the velocity is:
   A. V = sqrt(g y) = 3.50 m/s
   B. V = 2 sqrt(g y)
   C. V = q/y with q = 4 m^2/s
   D. V = g y = 12.3 m/s
   Answer: A. Fr = 1 gives V = sqrt(g y) = sqrt(9.81 x 1.25) = 3.502 m/s. (Key point: Critical velocity)

2. What hydraulic radius is used?
   A. R = y = 1.25 m
   B. R = A/P = 5/6.5 = 0.769 m
   C. R = A/b = 1.25 m
   D. R = P/A = 1.3 m
   Answer: B. A = 4 x 1.25 = 5 m^2, wetted perimeter P = 4 + 2(1.25) = 6.5 m, so R = 0.769 m. (Key point: Hydraulic radius)

3. Because normal depth equals critical depth, how is the slope found?
   A. S0 = Sc from Manning with the critical velocity, S0 = (nV/R^(2/3))^2
   B. S0 = V^2/(g y)
   C. S0 = y/b
   D. S0 = n V R^(2/3)
   Answer: A. Uniform flow at critical velocity needs S0 = (nV/R^(2/3))^2, about 0.0039. (Key point: Use of Manning)

4. The calculated slope 0.0039 is called:
   A. Mild slope
   B. Adverse slope
   C. Critical slope for this channel and discharge
   D. Steep slope for all discharges
   Answer: C. It is the slope at which uniform flow is exactly critical for Q = 17.5 m^3/s. (Key point: Meaning of result)

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