🌊 Manning's Equation Calculator
Calculate uniform open channel flow hydraulics — find normal depth or flow rate for rectangular, trapezoidal, triangular, and circular sections.
📝 Configuration
Manning's Equation:
$Q = \frac{1}{n} A R_h^{2/3} S_0^{1/2}$
$V = Q / A$
$Fr = V / \sqrt{g A / T}$
$Q = \frac{1}{n} A R_h^{2/3} S_0^{1/2}$
$V = Q / A$
$Fr = V / \sqrt{g A / T}$
📊 Results & Visualization
Configure the inputs and click Calculate to see results.
ℹ️ About Manning's Equation
Manning's equation is an empirical formula estimating the average velocity of a liquid flowing in a conduit that does not completely enclose the liquid, i.e., open channel flow.
Key features:
• **Uniform Flow assumption:** The bed slope is parallel to the water surface, and water velocity remains constant along the channel.
• **Iterative normal depth:** For shapes other than simple rectangular, normal depth $y_n$ cannot be solved analytically and requires numerical solution.
Manning's equation is an empirical formula estimating the average velocity of a liquid flowing in a conduit that does not completely enclose the liquid, i.e., open channel flow.
Key features:
• **Uniform Flow assumption:** The bed slope is parallel to the water surface, and water velocity remains constant along the channel.
• **Iterative normal depth:** For shapes other than simple rectangular, normal depth $y_n$ cannot be solved analytically and requires numerical solution.
📘 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 [m³/s]
- $n$: Manning's roughness coefficient [s/m^{1/3}]
- $A$: Cross-sectional flow area [m²]
- $P$: Wetted perimeter [m], $R_h$: Hydraulic radius [m]
- $S_0$: Bed slope [m/m]
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 m³/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 m³/s.