$\frac{u}{U}=2(\frac{y}{δ})-2(\frac{y}{δ})^2+(\frac{y}{δ})^3$ where u is the velocity at distance y from the surface of the flat plate and U be the free stream velocity at the boundary layer thickness $δ$ Obtain an expression for boundary layer thickness, shear stress and force on one side of the plate in terms of Reynolds number.

Consider the Laminar boundary layer velocity distribution over a flat plate,

$\frac{u}{U} =2(\frac{Y}{δ})-2(\frac{Y}{δ})^2+(\frac{y}{δ})^3$

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,

$θ =∫_0^δ[2(\frac{Y}{δ})-2(\frac{Y}{δ})^2+(\frac{y}{δ})^3]{1-[2(\frac{Y}{δ})-2(\frac{Y}{δ})^2+(\frac{y}{δ})^3]} dY$

$θ =∫_0^δ[\frac{2Y}{δ}-\frac{4Y^2}{δ^2} +\frac{4Y^3}{δ^3} -\frac{2Y^4}{δ^4} -\frac{2Y^2}{δ^2} +\frac{4Y^3}{δ^3} -\frac{4Y^4}{δ^4} +\frac{2Y^5}{δ^5} +\frac{Y^3}{δ^3} -\frac{2Y^4}{δ^4} +\frac{2Y^5}{δ^5} -\frac{Y^6}{δ^6}] dY$

$θ =[\frac{2Y^2}{2δ}-\frac{4Y^3}{3δ^2}+\frac{4Y^4}{4δ^3}-\frac{2Y^5}{5δ^4}-\frac{2Y^3}{3δ^2} +\frac{4Y^4}{4δ^3}-\frac{4Y^5}{5δ}^4 +\frac{2Y^6}{6δ^5}+\frac{Y^4}{4δ^3} -\frac{2Y^5}{5δ}^4 +\frac{2Y^6}{6δ^5 }-\frac{Y^7}{7δ^6 }]_0^δ$

$θ =[\frac{73δ}{420}]$

$∴\frac{∂θ}{∂x}=\frac{73}{420} \frac{∂δ}{∂x}$

Hence,

$\frac{\tau_0}{ρU^2}=\frac{73}{420} \frac{∂δ}{∂x}$

$∴\tau_0=\frac{73ρU^2}{420}\frac{∂δ}{∂x}$

Now, By Newton’s Law of Viscosity for Newtonian fluids,

∴$\tau_0=μ\frac{du}{dY}$

where, $\frac{du}{dY}=\frac{2}{δ}-\frac{4Y}{δ^2} +\frac{3Y^2}{δ^3}$

Hence, equating it to the above equation,

$\frac{73ρU^2}{420} \frac{∂δ}{∂x}=μ \frac{du}{dY}$

$\frac{73ρU^2}{420}\frac{∂δ}{∂x}=μU[\frac{2}{δ}-\frac{4y}{δ^2} +\frac{3y^2}{δ^3}]_(Y=0)$

$\frac{73ρU^2}{420} \frac{ ∂δ}{∂x}=μU(\frac{2}{δ})$

$\frac{73ρU}{840} \frac{∂δ}{∂x}=(\frac{μ}{δ})$

$\frac{73}{840} \frac{∂δ}{∂x}=(\frac{μ}{ρU})$

Integrating the above differential equation by variable separable method,

$\frac{73δ^2}{1680}=(\frac{μ}{ρU})x$

$\frac{δ^2}{x^2} =\frac{1680}{73} (\frac{μ}{ρUx})$

We know that, Local Reynolds number is given by the equation,

$Re_x=\frac{ρUx}{μ}$

∴$\frac{δ^2}{x^2} =\frac{1680}{73} (\frac{1}{Re_x})$

∴$\frac{δ}{x}=\sqrt{\frac{1680}{73Re_x}}$

∴$\frac{δ}{x}=\frac{4.797}{\sqrt{Re_x}}$

… This is the boundary layer thickness equation.

Shear Stress on one side of the plate is given by,

$\tau_0=μ \frac{du}{dY}$

∴$τ_0=μU[\frac{2}{δ}-\frac{4y}{δ^2} +\frac{3y^2}{δ^3}]_(Y=0)$

∴$\tau_0=μU[\frac{2}{δ}-\frac{4}{δ}+\frac{3}{δ}]$

∴$\tau_0=\frac{μU}{δ}$

Substituting the value of δ

∴$\tau_0=\frac{μU}{x\sqrt{\frac{1680}{73Re_x}}}$

∴$\tau_0=\frac{μU}{x} \sqrt{\frac{73Re_x}{1680}}$

∴$\tau_0=\frac{μU}{x} \sqrt{\frac{73ρUx}{1680μ}}$

∴$\tau_0=ρU^2 \sqrt{\frac{73μ}{1680ρUx}}$

∴$\tau_0=ρU^2 \sqrt{\frac{73}{1680Re_x}}$

∴$\tau_0=\frac{0.2084ρU^2}{\sqrt{Re_x }}$

… This is the Shear Stress equation.

Force per unit width on one side of the plate due to drag from leading to the trailing edge (length=l) is given by,

$F_D=∫_0^l \tau_0 dx$

$F_D=∫_0^l ρU^2 \sqrt{\frac{73\mu}{1680Re_x}}dx$

$F_D=∫_0^l ρU^2 \sqrt{\frac{73μ}{1680ρUx}}dx$

$F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} ∫_0^l\frac{1}{\sqrt{x}}dx$

$F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} [2\sqrt{x}]_0^l$

$F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} 2\sqrt{l}$

$F_D=ρU^2 \sqrt{\frac{292μl}{1680ρU}}$

$F_D=ρU^2 l\sqrt{\frac{292μ}{1680ρUl}}$

We know that the Reynolds number for the flat plate is given by,

Re=$\frac{ρUl}{μ}$

∴$F_D=ρU^2 l\sqrt{\frac{292}{1680Re}}$

∴$F_D=\frac{0.4169ρU^2 l}{\sqrt{Re}}$

… This is the Drag force per unit width on one side of the plate