i. Check if boundary layer separation occurs.

ii. Determine boundary layer thickness (in terms of Re)

Consider the Laminar boundary layer velocity distribution,

$\frac{u}{U} =2(\frac{Y}{δ})-(\frac{Y}{δ})^2$

Now, it can be seen from the figure that the flow can be identified as,

Which is positive, hence the boundary layer does not separate.

For the Boundary layer thickness, we know the Von Karman’s Momentum Integral equation,

$\frac{\tau_0}{ρU^2}= \frac{∂θ}{∂x}$

where,

$\tau_{0}$→Shear stress in the fluid

$ρ$→Density of the fluid U→Free stream velocity of the fluid $θ$→Momentum thickness We know that the Momentum thickness is given by the equation, $θ =∫_0^{δ}\frac{u}{U} {1-\frac{u}{U}}$ dY $θ =∫_0^δ[2(\frac{Y}{δ})-(\frac{Y}{δ})^2 ]{1-[2(\frac{Y}{δ})-(\frac{Y}{δ})^2]}$ dY $θ =∫_0^δ[\frac{2Y}{δ}-\frac{4Y^2}{δ^2} +\frac{2Y^3}{δ^3} -\frac{Y^2}{δ^2} +\frac{2Y^3}{δ^3} -\frac{Y^4}{δ^4}]$ dY ![enter image description here][2] $θ =[δ-\frac{4δ}{3}+\frac{δ}{2}-\frac{δ}{3}+\frac{δ}{2}-\frac{δ}{5}]$ $θ =[\frac{2δ}{15}]$ $∴\frac{∂θ}{∂x}=\frac{2}{15} \frac{∂δ}{∂x}$ Hence, $\frac{\tau_0}{ρU^2}=\frac{2}{15} \frac{∂δ}{∂x}$ ∴$\tau_0=\frac{2ρU^2}{15} \frac{∂δ}{∂x}$ Now, By Newton’s Law of Viscosity for Newtonian fluids, ∴$\tau_0=μ \frac{du}{dY}$ Hence, equating it to the above equation, $\frac{2ρU^2}{15} \frac{∂δ}{∂x}=μ \frac{du}{dY}$ $\frac{2ρU^2}{15} \frac{∂δ}{∂x}=μU[(\frac{2}{δ})-(\frac{2Y}{δ^2})]_(Y=0)$ $\frac{2ρU^2}{15} \frac{∂δ}{∂x}=μU(\frac{2}{δ})$ $\frac{ρU}{15} \frac{∂δ}{∂x}=(\frac{μ}{δ})$ $\frac{1}{15} \frac{∂δ}{∂x}=(\frac{μ}{ρU})$ Integrating the above differential equation by variable separable method, $∫\frac{δ}{15}∂δ=(\frac{μ}{ρU}) ∫∂x$ $\frac{δ^2}{30}=(\frac{μ}{ρU})x$ $\frac{δ^2}{x^2} =30(\frac{μ}{ρUx})$ We know that, Local Reynolds number is given by the equation, $Re_x=\frac{ρUx}{μ}$ ∴$\frac{δ^2}{x^2} =30(\frac{1}{Re_x})$ ∴$\frac{δ}{x}=\sqrt{\frac{30}{Re}}_x$ ∴$\frac{δ}{x}=\frac{5.477}{\sqrt{{e_{x}}}}$

....Which is the required equation