๐๏ธ Manning Equation โ Open Channel
Uniform flow in open channels using Manning\
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Manning Uniform Flow
Gravity flows in open channels are governed by boundary friction. Uniform flow conditions establish when gravitational accelerating force matches frictional drag forces exactly, aligning water surface and bed slopes.
โข Normal Depth ($y_n$): Iterative depth solution corresponding to equilibrium uniform flow.
โข Hydraulic Radius ($R_h$): Geometric ratio indicating channel flow cross-section efficiency.
โข Manning Presets: Roughness coefficients ($n$) representing wall friction effects.
โข Hydraulic Radius ($R_h$): Geometric ratio indicating channel flow cross-section efficiency.
โข Manning Presets: Roughness coefficients ($n$) representing wall friction effects.
๐ Configuration
Manning's Equation:
$Q = \frac{1}{n} A R_h^{2/3} S_0^{1/2}$
$V = Q / A$
$Fr = V / \sqrt{g A / T}$
$Q = \frac{1}{n} A R_h^{2/3} S_0^{1/2}$
$V = Q / A$
$Fr = V / \sqrt{g A / T}$
๐ Results & Visualization
Configure inputs and click Calculate Hydraulics to view results.
๐ Calculation Methodology
Mathematical Model & Theory
Open channel flow under gravity is modeled by Manning's Equation. It relates flow velocity ($V$) and volumetric flow ($Q$) to wetted perimeter roughness and hydraulic gradient:
$$Q = \frac{1}{n} A R_h^{2/3} S_0^{1/2} \quad \text{(SI Units)}$$
$$R_h = \frac{A}{P}$$
Variable Definitions & Units:
- $Q$: Volumetric discharge [mยณ/s]
- $n$: Manning's roughness coefficient [s/m^{1/3}]
- $A$: Cross-sectional flow area [mยฒ]
- $P$: Wetted perimeter [m], $R_h$: Hydraulic radius [m]
- $S_0$: Bed slope [m/m]
Assumptions
- Steady, uniform flow (constant depth, area, and velocity downstream).
- Prismatic channel geometry (cross-section shape is constant along length).
- Incompressible liquid (constant density).
- Frictional losses are represented purely by Manning's empirical roughness ($n$).
Academic References
- Chow, V. T.: Open-Channel Hydraulics, McGraw-Hill.
- Manning, R.: On the Flow of Water in Open Channels and Pipes, Transactions of the Institution of Civil Engineers of Ireland.
Worked Engineering Example
Problem Statement:
A rectangular concrete channel ($n = 0.013$) has a bottom width of 3 m and a bed slope of 0.001. Find the discharge rate $Q$ when the flow depth is 1.5 m.
Step-by-step Solution:
1. Calculate cross-sectional area $A$ and wetted perimeter $P$:
$$A = b \times y = 3 \times 1.5 = 4.5 \text{ m}^2$$ $$P = b + 2y = 3 + 2(1.5) = 6 \text{ m}$$ 2. Calculate hydraulic radius $R_h$:
$$R_h = A / P = 4.5 / 6 = 0.75 \text{ m}$$ 3. Apply Manning's Equation:
$$Q = \frac{1}{0.013} \times 4.5 \times (0.75)^{2/3} \times (0.001)^{1/2}$$ $$Q = 76.923 \times 4.5 \times 0.8255 \times 0.03162 = 9.04 \text{ m}^3/\text{s}$$
Final Result:
The channel discharge is 9.04 mยณ/s.
A rectangular concrete channel ($n = 0.013$) has a bottom width of 3 m and a bed slope of 0.001. Find the discharge rate $Q$ when the flow depth is 1.5 m.
Step-by-step Solution:
1. Calculate cross-sectional area $A$ and wetted perimeter $P$:
$$A = b \times y = 3 \times 1.5 = 4.5 \text{ m}^2$$ $$P = b + 2y = 3 + 2(1.5) = 6 \text{ m}$$ 2. Calculate hydraulic radius $R_h$:
$$R_h = A / P = 4.5 / 6 = 0.75 \text{ m}$$ 3. Apply Manning's Equation:
$$Q = \frac{1}{0.013} \times 4.5 \times (0.75)^{2/3} \times (0.001)^{1/2}$$ $$Q = 76.923 \times 4.5 \times 0.8255 \times 0.03162 = 9.04 \text{ m}^3/\text{s}$$
Final Result:
The channel discharge is 9.04 mยณ/s.