Economical Rectangular Channel Width Calculation

The problem asks for the width of a rectangular channel designed for the most economical section. We are given the discharge ($Q$), bed slope ($S$), and Manning's roughness coefficient ($n$).

Key Concepts for Economical Section

• For a rectangular channel, the most economical section occurs when the hydraulic radius ($R$) is maximized for a given cross-sectional area ($A$). This condition is met when the width ($B$) is twice the flow depth ($y$), i.e., $B = 2y$.
• Manning's equation relates discharge to channel geometry and slope: $ Q = \frac{1}{n} A R^{2/3} S^{1/2} $

Derivation and Calculation

1. Define Geometric Properties:
   • Given: $Q = 5 \, m^3 s^{-1}$, $n = 0.01$, $S = 0.05\% = 0.0005$.
   • For the most economical rectangular section: $B = 2y$.
   • Cross-sectional Area: $A = B \times y = (2y) \times y = 2y^2$.
   • Wetted Perimeter: $P = B + 2y = (2y) + 2y = 4y$.
   • Hydraulic Radius: $R = \frac{A}{P} = \frac{2y^2}{4y} = \frac{y}{2}$.
2. Apply Manning's Equation:
   • Substitute the derived properties into Manning's equation: $ 5 = \frac{1}{0.01} (2y^2) \left(\frac{y}{2}\right)^{2/3} (0.0005)^{1/2} $
   • Simplify the equation: $ 5 = 100 \times 2y^2 \times \frac{y^{2/3}}{2^{2/3}} \times \sqrt{0.0005} $ $ 5 = 200 \times y^{2 + 2/3} \times \frac{1}{\sqrt[3]{4}} \times 0.02236 $ $ 5 = 200 \times y^{8/3} \times \frac{1}{1.5874} \times 0.02236 $ $ 5 \approx 125.99 \times y^{8/3} \times 0.02236 $ $ 5 \approx 2.817 y^{8/3} $
3. Solve for Depth ($y$):
   • Rearrange to find $y^{8/3}$: $ y^{8/3} = \frac{5}{2.817} \approx 1.7749 $
   • Calculate $y$: $ y = (1.7749)^{3/8} = (1.7749)^{0.375} $ $ y \approx 1.256 \, m $
4. Calculate Width ($B$):
   • Use the condition for the economical section: $B = 2y$. $ B = 2 \times 1.256 $ $ B \approx 2.512 \, m $

The calculated width is approximately 2.512 m. This value lies close to the provided answer range of 2.45 m to 2.5 m.