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For the velocity profile for laminar boundary flow

For the velocity profile for laminar boundary flow υU=sin(πy2δ)

where u is the velocity at the 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 the

boundary layer thickness and the average drag coefficient in terms of the Reynolds Number.

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Solution :

The velocity profile is vU=sin(π2yδ)

τ0ρU2=x[δ0vU(1vU)dy]=x[δ0sin(π2yδ)[1sin(π2yδ)]dy]=x[δ0[sin(π2yδ)sin2(π2yδ)]dy]=[x[cosπy2δπ2δ][πy2δ×12π2δsin2(π2yδ)4×π2δ]]δ0=x[(cosπ2δδπ2δ+cosπ2×0δπ2δ)[π2δδ×12π2δ0]]=x[(0+1π2δ)(π4)π2δ]=x[2δππ4×2δπ]=x[2δπδ2]=x[4π2π]δ=(4π2π)δxτ0=(4π2π)pU2δx.....(i)

τ0 is also equal =μ(dvdy)aty=0

But

v=Usin(π2yδ)(dvdy)=Ucos(π2yδ)×π2δ(dvdy)y=0=U×π2δcos(π2×0δ)=Uπ28

τ0=μ(vy)y=0=μUπ2δ.....(ii)

Equating the two values τ0 given by equations (i) and (ii)

(4π2π)ρU2δx=μUπ2δδδ=μUπ2×2π4π×1ρU2xδδ=π2(4π)μUρU2.x=11.4975μρUx

Integrating, we get δ22=11.4975μρUx+C

At x=0,δ=0 and hence C=0

δ22=11.4975μρUxδ=2×11.4975μρUx=4.795μρU×=4.795μρUx=4.795μρUx×x=4.795xRe.....(iii)

(ii) Shear Stress (τ0)

From equation (ii)

τ0=μUπ2δ=μUπ2×4.795xRe=μUπRe2×4.795x=π2×4.795μUxRe=0.327μUxRe

(iii) Drag force (Fb)

on one side of the plate is given by

FD=L0τ0×b×dx=L00.327μUxRe×b×dx=0.327μU×bL01xρUxμdx=0.327μU×b×ρUμL0x12dx=0.327μU×b×ρUμ[x1/212]L0=0.327×2×μU×bρUμ×L=0.655×μU×b×ρULμ

(iv) Co-efficient of drag, CD

CD=FD12ρAU2, where A=b×LCD=0.655×μU×b×ρULμ12ρU2×b×L=0.655×2×μρUL×ρULμ=1.31×1ρUL=1.31Re

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