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Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by:

Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by: uU=2(yδ)(yδ)2

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Solution :

Given :

The velocity distribution in boundary layer given by,

uU=2(yδ)(yδ)2

(i) Displacement thickness δ

δ=δ0(1uU)dy

Substituting the value of uU=2(yδ)(yδ)2, we have

δ=δ0{1[2(yδ)(yδ)2]}dy=δ0{12(yδ)+(yδ)2}dy=[y2y22δ+y33δ2]δ0=δδ2δ+δ33δ2=δδ+δ3=δ3

(ii) Momentum thickness θ

θ=δ0uU{1uU}dy=δ0(2yδy2δ2)[1(2yδy2δ2)]dy=δ0[2yδy2δ2][12yδ+y2δ2]dy=δ0[2yδ4y2δ2+2y3δ3y2δ2+2y3δ3y4δ4]dy=δ0[2yδ5y2δ2+4y3δ3y4δ4]dy=[2y22δ5y33δ2+4y44δ3y55δ4]δ0=[δ2δ5δ33δ2+δ4δ3δ55δ4]=δ5δ3+δδ5=15δ25δ+15δ3δ15=30δ28δ15=2δ15

(iii) Energy thickness δ

δ=δ0uU[1u2U2]dy=δ0(2yδy2δ2)(1[2yδy2δ2]2)dy=δ0(2yδy2δ2)(1[4y2δ2+y4δ44y3δ3])dy=δ0(2yδy2δ2)(14y2δ2y4δ4+4y3δ3)dy=δ0(2yδ8y3δ32y5δ5+8y4δ4y2δ2+4y4δ4+y6δ64y5δ5)dy=00[2yδy2δ28y3δ3+12y4δ46y5δ3+y6δ6]dy=[2y22δy33δ28y44δ3+12y55δ46y66δ5+y77δ6]δ0=δ2δδ33δ22δ4δ3+12δ55δ4δ6δ5+δ77δ6=δδ32δ+125δδ+δ7=2δδ3+125δ+δ7=210δ35δ+252δ+15δ105=245δ+267δ105=22δ105

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