Calculating Rectangular Channel Flow Depth

This solution calculates the flow depth ($y$) in a rectangular channel for maximum hydraulic efficiency using Manning's equation.

Applying Hydraulic Efficiency Condition

For a rectangular channel, maximum hydraulic efficiency (minimum wetted perimeter for a given area) occurs when the width ($b$) is twice the flow depth ($y$). Therefore, we set:

$ b = 2y $

Manning's Equation Setup

Manning's equation is:

$ Q = \frac{1}{n} A R^{2/3} S^{1/2} $

Where:

• $Q$ = Discharge = 64 m$^3$/s
• $n$ = Manning's roughness coefficient = 1/80
• $S$ = Bed slope = 0.01
• $A$ = Cross-sectional area
• $R$ = Hydraulic radius

For a rectangular channel with $b = 2y$:

• Area, $A = b \times y = (2y) \times y = 2y^2$
• Wetted Perimeter, $P = b + 2y = 2y + 2y = 4y$
• Hydraulic Radius, $R = \frac{A}{P} = \frac{2y^2}{4y} = \frac{y}{2}$

Solving for Flow Depth (y)

Substitute the expressions for $A$ and $R$ into Manning's equation:

$ 64 = \frac{1}{(1/80)} (2y^2) \left(\frac{y}{2}\right)^{2/3} (0.01)^{1/2} $

Simplify the equation:

$ 64 = 80 \times 2y^2 \times \frac{y^{2/3}}{2^{2/3}} \times 0.1 $

$ 64 = 16 \times y^{2 + 2/3} \times \frac{1}{\sqrt[3]{4}} $

$ 64 = 16 \times y^{8/3} \times \frac{1}{\sqrt[3]{4}} $

Isolate $y^{8/3}$:

$ y^{8/3} = \frac{64 \times \sqrt[3]{4}}{16} $

$ y^{8/3} = 4 \times 4^{1/3} = 4^{4/3} $

Solve for $y$ by raising both sides to the power of $3/8$:

$ y = \left(4^{4/3}\right)^{3/8} = 4^{(4/3) \times (3/8)} = 4^{1/2} $

$ y = 2 $

The calculated flow depth is 2.00 m.