Answer:
δ₂ = 15.07 mm
Explanation:
Given :
When the leading edge, [tex]x_{1}[/tex] is 2.2 m, then boundary layer thickness,δ₁ = 10 mm = 0.01 m
[tex]x_{2}[/tex] = 5 m
Now we know that for a laminar flow, the boundary layer thickness is
δ = [tex]\frac{5.x}{\sqrt{Re_{x}}}[/tex] -------(1)
and Reyonlds number, Re is
[tex]Re = \frac{\rho .v.x}{\mu }[/tex]------(2)
where ρ is density
v is velocity
x is distance from the leading edge
μ is dynamic viscosity
from (1) and (2), we get
δ∝[tex]x^{1/2}[/tex]
Therefore,
[tex]\frac{\delta _{1}}{x_{1}^{1/2}}= \frac{\delta _{2}}{x_{2}^{1/2}}[/tex]
[tex]\frac{10}{2.2^{1/2}}= \frac{\delta _{2}}{5^{1/2}}[/tex]
δ₂ = 15.07 mm
Therefore, boundary layer thickness is 15.07 mm when the leading edge is 5 m.