Answer:
1.12 m
0.08291 m
Explanation:
u = Upstream velocity = 0.4 m/s
Re = Reynold's number = [tex]5\times 10^5[/tex] (turbulent)
[tex]\nu[/tex] = Viscosity of water = [tex]1.12\times 10^{-6}\ Pas[/tex]
Here the flow is turbulent so we have the relation
[tex]Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m[/tex]
The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m
Boundary layer thickness relation is given by
[tex]\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m[/tex]
The boundary layer thickness is 0.08291 m
