Question

Stormwater is running over a surface at a maximum velocity (Umax) of 5 fps. The velocity (u) profile can be modeled by the following equation. If the depth (h) of the stormwater is 3", then what is the shear stress on the ground surface?

Answer 1

Answer:

Complete Question

$u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ]$

$a.2.73\times 10^{-4} \\b.5.5\times 10^{-4} \\c.0.03\\d.2.3\times 10^{-5}$

Answer

τ at y = 0 = $2.73 \times 10^{-4} \frac{lb}{ft^{2} }$

Explanation:

The work function is given as

$u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ]$

where

Umax = maximum velocity = 5 fps(foot per second)

h= depth (h) of the storm water = $3^{''} = 0.25 feet$

Shear stress on the ground surface = ? τ=

$u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ]$

considering

$u=U_{max} [0.5\frac{y}{h}-3\frac{y^{2} }{h} ]$

Considering

τ =  $\mu\frac{du}{dy}$

Taking derivatives

$\frac{du}{dy}=\mu_{max} [\frac{0.5}{h} -\frac{3}{h^{2} } \times2y]$

At   y = 0

$\frac{du}{dy}=5[\frac{0.5}{0.25} ] =10$

therefore

τ at y = 0 = $\mu\frac{du}{dy}$

                = $2.73\times 10^{-5} \times 10$

τ at y = 0 = $2.73 \times 10^{-4} \frac{lb}{ft^{2} }$