Answer:
Complete Question
[tex]u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ][/tex]
[tex]a.2.73\times 10^{-4} \\b.5.5\times 10^{-4} \\c.0.03\\d.2.3\times 10^{-5}[/tex]
Answer
τ at y = 0 = [tex]2.73 \times 10^{-4} \frac{lb}{ft^{2} }[/tex]
Explanation:
The work function is given as
[tex]u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ][/tex]
where
Umax = maximum velocity = 5 fps(foot per second)
h= depth (h) of the storm water = [tex]3^{''} = 0.25 feet[/tex]
Shear stress on the ground surface = ? τ=
[tex]u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ][/tex]
considering
[tex]u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ][/tex]
Considering
τ = [tex]\mu\frac{du}{dy}[/tex]
Taking derivatives
[tex]\frac{du}{dy}=\mu_{max} [\frac{0.5}{h} -\frac{3}{h^{2} } \times2y][/tex]
At y = 0
[tex]\frac{du}{dy}=5[\frac{0.5}{0.25} ] =10[/tex]
therefore
τ at y = 0 = [tex]\mu\frac{du}{dy}[/tex]
= [tex]2.73\times 10^{-5} \times 10[/tex]
τ at y = 0 = [tex]2.73 \times 10^{-4} \frac{lb}{ft^{2} }[/tex]
