written 2.9 years ago by
RakeshBhuse
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modified 2.9 years ago
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Solution :
The velocity profile is vU=sin(π2yδ)
τ0ρU2=∂∂x[∫δ0vU(1−vU)dy]=∂∂x[∫δ0sin(π2yδ)[1−sin(π2yδ)]dy]=∂∂x[∫δ0[sin(π2yδ)−sin2(π2yδ)]dy]=[∂∂x[−cosπy2δπ2δ]−[πy2δ×12π2δ−sin2(π2yδ)4×π2δ]]δ0=∂∂x[(−cosπ2δδπ2δ+cosπ2×0δπ2δ)−[π2δδ×12π2δ−0]]=∂∂x[(0+1π2δ)−(π4)π2δ]=∂∂x[2δπ−π4×2δπ]=∂∂x[2δπ−δ2]=∂∂x[4−π2π]δ=(4−π2π)∂δ∂x∴τ0=(4−π2π)pU2∂δ∂x.....(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(∂v∂y)y=0=μUπ2δ.....(ii)
Equating the two values τ0 given by equations (i) and (ii)
(4−π2π)ρU2∂δ∂x=μUπ2δ⇒δ∂δ=μUπ2×2π4−π×1ρU2∂x∴δ∂δ=π2(4−π)μUρU2.∂x=11.4975μρU∂x
Integrating, we get δ22=11.4975μρUx+C
At x=0,δ=0 and hence C=0
δ22=11.4975μρUxδ=√2×11.4975μρUx=4.795√μρU×=4.795√μρUx=4.795√μρUx×x=4.795x√Re.....(iii)
(ii) Shear Stress (τ0)
From equation (ii)
τ0=μUπ2δ=μUπ2×4.795x√Re=μUπ√Re2×4.795x=π2×4.795μUx√Re=0.327μUx√Re
(iii) Drag force (Fb)
on one side of the plate is given by
FD=∫L0τ0×b×dx=∫L00.327μUx√Re×b×dx=0.327μU×b∫L01x√ρUxμdx=0.327μU×b×√ρUμ∫L0x−12dx=0.327μU×b×√ρUμ[x1/212]L0=0.327×2×μU×b√ρUμ×√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=1.31×1√ρUL=1.31√Re