Figure shows an anchored barge that extends across a canal by distance d=30 mand into the water by distance b=12 m. The canal has a widthD=55 m, a water depth H=14 m , and a uniform water flow speed ${\mathit{v}}_{\mathbf{i}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. Assume that the flow around the barge is uniform. As the water passes the bow, the water level undergoes a dramatic dip known as the canal effect. If the dip has depth h=0.80 m , what is the water speed alongside the boat through the vertical cross sections at (a) point a and if the dip has depth h=0.80 m , what is the water speed alongside the boat through the vertical cross sections at (b) point b? The erosion due to the speed increase is a common concern to hydraulic engineers.

1. The width of canal, D=55 m
2. The width of barge, d=30 m
3. The depth of canal, H=14 m
4. The depth of barge in water, h=12 m
5. The uniform water speed of canal, v=1.5 m/s

We can find the area of a cross-section of the canal and the area at point a. Then inserting these values in the continuity equation, we can find the speed of water alongside the boat at point a. Similarly, we can find the speed of water alongside the boat at point b.

Formula:

The continuity equation at two ends of a liquid flowing,$A_1{v}_1=A_2{v}_2$ (i)

Where,

A₁, v₁ and A₂, v₂ are the cross-sectional area and velocity of two ends 1& 2

From equation(i),we get

$v_2=\frac{A{v}_1}{A_2}$

The area of the cross-section of the canal is given as:

$\begin{array}{l}{A}_1=HD\\ =55\times 14\\ =770m^2\end{array}$

From the figure, we can write for the area of the cross-section at point a as:

$\begin{array}{ll}{A}_2=A_a=(H-h)D-(b-h)d& \\ =\left(14\mathrm{m}-0.80m\right)(55& m)-(12m-0.8m\left)\right(30m)\\ =390{\mathrm{m}}^2& \end{array}$

So, the speed at point a is

$$\begin{gathered} v_2=\frac{A_1{v}_1}{A_2} \\ {v}_{2}or{v}_a=\frac{770m^2\times 1.5m/s}{3.90m^2} \\ =2.96m/s \end{gathered}$$

Therefore, the speed of water alongside the boat at point a is 2.96 m/s

The area of the cross-section at point b is given as:

$\begin{array}{l}{A}_2=H\times D-b\times d\\ =\left(14\mathrm{m}\right)(55m)-(12m)(30m)\\ =410m^2\end{array}$

So, the speed at point b is given as:

$$\begin{gathered} v_{2}or{v}_b=\frac{A_1{v}_1}{A_2} \\ =2.82m/s \end{gathered}$$

Therefore, the speed of water alongside the boat at point b is 2.82 m/s