Answer:
specific energy = 2.65 ft
y2 = 1.48 ft
Explanation:
given data
average speed v = 6.5 ft/s
width = 5 ft
depth of the water y = 2 ft
solution
we get here specific energy that is express as
specific energy = y + [tex]\frac{v^2}{2g}[/tex] ...............1
put here value and we get
specific energy = [tex]2 + \frac{6.5^2}{2\times 9.8\times 3.281}[/tex]
specific energy = 2.65 ft
and
alternate depth is
y2 = [tex]\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})[/tex]
and
here Fr² = [tex]\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}[/tex]
Fr² = 0.8025
put here value and we get
y2 = [tex]\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})[/tex]
y2 = 1.48 ft
