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Given the velocity distribution in a laminar boundary layer on a flat plate as

uU=2(yδ)2(yδ)2+(yδ)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.

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

uU=2(Yδ)2(Yδ)2+(yδ)3

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

τ0ρU2=θx

Where,

τ0→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,

enter image description here

θ=δ0[2(Yδ)2(Yδ)2+(yδ)3]1[2(Yδ)2(Yδ)2+(yδ)3]dY

θ=δ0[2Yδ4Y2δ2+4Y3δ32Y4δ42Y2δ2+4Y3δ34Y4δ4+2Y5δ5+Y3δ32Y4δ4+2Y5δ5Y6δ6]dY

θ=[2Y22δ4Y33δ2+4Y44δ32Y55δ42Y33δ2+4Y44δ34Y55δ4+2Y66δ5+Y44δ32Y55δ4+2Y66δ5Y77δ6]δ0

enter image description here

θ=[73δ420]

∴\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,

enter image description here

\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

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