u-a+by-cy”, where y is the perpendicolar distance measured from the surface of the flat plate and a, b and c are constants. Determine the expression of velocity distribution in dimensionless form as $\frac{\square}{\square}=(\frac{y}{δ})$ where, U is main stream velocity at boundary layer thickness δ. Further also find boundary layer thickness in terms of Reynolds number.

Velocity distribution: u=$a+by-cy^2$

The following condition must be satisfied:

i)At y=0 , u=0

∴u=$a+by-cy^2$

0=a+0-0 ∴a=0

ii)At y=δ, u=U

∴U=$bδ-cδ^2$………………………(i)

iii)At y=$δ, \frac{du}{dy}$=0

∴$(\frac{du}{dy})_(y=δ)=\frac{d}{dy} (a+by-cy^2 )=b-2cy=b-2cδ=0$…………(ii)

Substituting values of b=2cδ from (ii) in (i), we get

U=$2cδ^2-cδ^2$

$U=cδ^2$

$c=\frac{U}{δ^2}$

Hence from the velocity distrubution is:

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

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

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

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

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

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

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

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

$(\frac{du}{dy})_(y=0)=U(\frac{2}{δ}-0)=\frac{2U}{δ} ∴τ_0=\frac{2μU}{δ}$

Equating the value of $τ_0 $

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

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

$δ∙dδ=\frac{15μ}{ρU}$ dx

Integrating both the sides we get,

$\frac{δ^2}{2}=\frac{15μ}{ρU}x+C$

At x=0,δ=0 ∴C=0

$\frac{δ^2}{2}=\frac{15μ}{ρU} x$

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