Question

The Chezy's coefficient C is related to Darcy-Weisbach friction factor 'f' as

• A. $C=\sqrt{\left(\frac{g}{8f}\right)}$
• B. $C=\sqrt{\left(\frac{8g}{f^{\frac{1}{4}}}\right)}$
• C. $C=\sqrt{\left(\frac{8g}{f}\right)}$
• D. $C=\sqrt{\left(\frac{f}{8g}\right)}$

Answer 1

The correct option is C $C=\sqrt{\left(\frac{8g}{f}\right)}$

$h_f=f\frac{L}{4R}\frac{V^2}{2g}\text{(Darcy-Weisbach equation for non-circular section)}$

$V=C\sqrt{RS}\text{(Chezy's equation)}$

$\text{Slope,}S=\frac{h_f}{L}$

$\text{Equating both equations, we get}C=\sqrt{\frac{8g}{f}}$