$u/U= 2(y/δ) - (y/δ)^{2}$Where U is the mean stream velocity and δ is the boundary layer thickness . Determine the displacement thickness and momentum thickness.

1)Determine momentum thickness:

θ=$∫_0^δ\frac{u}{U}(1-\frac{u}{U})dy$

Substitute value of $\frac{u}{U}$ in above equation we get,

$θ=[∫_0^δ{\frac{2y}{δ}-\frac{y^2}{δ^2}}{1-(\frac{2y}{δ}-\frac{y^2}{δ^2})} dy]$

Making substitution

Let $\frac{y}{δ}=$t

∴$y=δt$

$dy=δ$dt

Changing limit at $y=0, t=0$

At $y=δ,t=0$

∴$θ=∫_0^1[2t-t^2 ][1-2t+t^2 ]δdt$

=$δ[∫_0^1[2t-t^2 ][1-2t+t^2 ]$

$θ=0.1333δ ……… (ans)(1)$

Boundary layer thickness:

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

We know, $\frac{\tau_0}{ρU^2}=\frac{d}{dx} [∫_0^δ \frac{u}{U}(1-\frac{u}{U})dy]$

Substituting the value of $\frac{u}{U}$, we get

$\frac{\tau_0}{ρU^2}=\frac{d}{dx} [∫_0^δ{\frac{2y}{δ}-\frac{y^2}{δ^2}}{1-(\frac{2y}{δ}-\frac{y^2}{δ^2})} dy]$

$\frac{\tau_0}{ρU^2}=\frac{d}{dx} [∫_0^δ{\frac{2y}{δ}-\frac{y^2}{δ^2}}{1-\frac{2y}{δ}+\frac{y^2}{δ^2}}dy]$

=$\frac{d}{dx} [∫_0^δ{\frac{2y}{δ}-\frac{4y^2}{δ^2} +\frac{2y^3}{δ^3} -\frac{y^2}{δ^2} +\frac{(2y^3}{δ^3} -\frac{y^4}{δ^4}} dy]$

=$\frac{d}{dx} [∫_0^δ\frac{2y}{δ}-\frac{5y^2}{δ^2} +\frac{4y^3}{δ^3}-\frac{y^4}{δ^4}]dy$

=$\frac{d}{dx} [(\frac{2y^2}{2δ}-\frac{5y^3}{3δ^2}+\frac{4y^4}{4δ^3}-\frac{y^5}{5δ^4 }]_0^δ=\frac{d}{dx} [δ-\frac{5}{3} δ+δ-\frac{1}{5} δ]=\frac{d}{dx} (\frac{2}{15δ})$

$\tau_0=ρU^2×\frac{d}{dx} (\frac{2}{15 δ})=\frac{2}{15} ρU^2 \frac{dδ}{dx}$

Also, according to Newton’s law of viscosity,

$\tau_0=μ(\frac{du}{dy})_{y=0}$

$u=U(\frac{2y}{δ}-(\frac{y}{δ})^2)$

$\frac{du}{dy}=U(\frac{2}{δ}-\frac{2y}{δ^2})$,U being constant

$(\frac{du}{dy})_(y=0)=U(\frac{2}{δ}-0)=\frac{2U}{δ} ∴\tau_0=\frac{2μU}{δ}$

Equating the value of $\tau_0$

$\frac{2}{15} ρU^2 dδ/dx=2μU/δ$

$δ\frac{dδ}{dx}=\frac{15μU}{ρU^2}=\frac{15μ}{ρU}$

$δ∙dδ=\frac{15μ}{ρU}dx$

Integrating both the sides we get,

$\frac{δ^2}{2}=\frac{15μ}{ρU}$ x+C

At x=0,δ=0 ∴C=0

$\frac{δ^2}{2}=\frac{15μ}{ρU}$ x

$δ=\sqrt{\frac{2×15μx}{ρU}}=5.48\sqrt{\frac{μx}{ρU}}=5.48\sqrt{\frac{μx×x}{ρU×x}}=5.48\sqrt{\frac{x^2}{Re_x}}=5.48 \sqrt{\frac{x}{Re_x}}$……(ans)

Displacement thickness(δ^*):-

$δ^*=∫_0^δ(1-\frac{u}{U})dy$

$δ^*=∫_0^δ(1-\frac{2y}{δ}+(\frac{y}{δ})^2 )dy$

$=∫_0^1 (1-2t+t^2)δdt$

=0.333δ

=$5.48 \frac{x}{\sqrt{Re_x}}$×0.3333….(ans)