Question

A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000. It carries a discharge of $4m^3/s$ and its Manning's roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach, the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance, $\frac{dy}{dx}$, at this section is _________ .

$(useg=10m/s^2$)

• A. 0.0028
• B. 0.0035
• C. 0.0032
• D. 0.0056

Answer 1

The correct option is C 0.0032

$Adverseslope=-\frac{1}{10000}$

$Q=4m^3/s$

n = 0.01
y = 0.5 m

$\frac{dy}{dx}=\frac{S_0-S_1}{1-F_t^2}$

$F_r=\frac{V}{\sqrt{gy}}=\frac{Q}{2y\sqrt{gy}}$

$=\frac{4}{2\times 0.5\sqrt{10\times 0.5}}=1.789$

$R=\frac{A}{P}=\frac{2\times 0.5}{2+2\times 0.5}=\frac{1}{3}$

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

$Onsolvingfor{S}_f$

$S_f^{1/2}=6.923\times {10}^{-3}$

$\frac{dy}{dx}=\frac{-\frac{1}{10000}-6.923\times {10}^{-3}}{1-(1.789{)}^2}$

$=3.191\times {10}^{-3}\simeq 0.0032$