Question

A very wide rectangular channel carries a discharge $Q=70~\mathrm{m^3/s}$ per meter width. Its bed slope changes from $S_0=0.0001$ to $S_0=0.0009$ at a point $P$ (not to scale). The Manning's roughness coefficient is $n=0.01$. What water-surface profile(s) exist(s) near the point $P$?

• A. M$_2$ and S$_2$
• B. M$_2$ only
• C. S$_2$ only
• D. S$_2$ and hydraulic jump

Hint:

For very wide channels: use $q=\tfrac{1}{n}y^{5/3}S_0^{1/2}$ and $y_c=(q^2/g)^{1/3}$. $y_n>y_c \Rightarrow$ mild; $y_n<y_c \Rightarrow$ steep. Drawdown on mild $\Rightarrow$ M$_2$; approach to $y_n$ on steep from critical $\Rightarrow$ S$_2$.

Answer 1

The Correct Option is A

Step 1: Critical depth for a very wide rectangular channel.
For unit width, critical depth is \[ y_c=\left(\frac{q^2}{g}\right)^{1/3}, q=Q=70~\mathrm{m^2/s}. \] \[ y_c=\left(\frac{70^2}{9.81}\right)^{1/3}\approx 7.93~\mathrm{m}. \]

Step 2: Normal depth on each reach (Manning, very wide).
For a very wide rectangle, $A=y$, $R\approx y$, and \[ q=\frac{1}{n}\,y^{5/3}\,S_0^{1/2}\;\Rightarrow\; y_n=\left(q\,n\,S_0^{-1/2}\right)^{3/5}. \] Upstream ( $S_0=0.0001$ ): \[ y_{n1}=\left(\frac{70\cdot 0.01}{\sqrt{0.0001}}\right)^{3/5}\approx 12.80~\mathrm{m}. \] Downstream ( $S_0=0.0009$ ): \[ y_{n2}=\left(\frac{70\cdot 0.01}{\sqrt{0.0009}}\right)^{3/5}\approx 6.62~\mathrm{m}. \]

Step 3: Classify slopes.
- Upstream: $y_{n1}>y_c$ $\Rightarrow$ mild slope (M).
- Downstream: $y_{n2}<y_c$ $\Rightarrow$ steep slope (S).

Step 4: Profiles near $P$.
Approaching the steeper reach, depth must decrease from $y_{n1}$ toward the control; on a mild slope such a drawdown with $y_c<y<y_n$ is profile M$_2$. Immediately downstream on the steep slope, the flow adjusts from near-critical toward $y_{n2}(<y_c)$; the curve with $y_n<y<y_c$ is profile S$_2$. \[ \boxed{\text{Profiles near }P:\; \text{M}_2 \text{ (upstream)} \;\text{and}\; \text{S}_2 \text{ (downstream).}} \]