written 7.2 years ago by | modified 3.0 years ago by |
Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by: uU=2(yδ)−(yδ)2
written 7.2 years ago by | modified 3.0 years ago by |
Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by: uU=2(yδ)−(yδ)2
written 3.0 years ago by | • modified 3.0 years ago |
Solution :
Given :
The velocity distribution in boundary layer given by,
uU=2(yδ)−(yδ)2
(i) Displacement thickness δ∗
δ∗=∫δ0(1−uU)dy
Substituting the value of uU=2(yδ)−(yδ)2, we have
δ∗=∫δ0{1−[2(yδ)−(yδ)2]}dy=∫δ0{1−2(yδ)+(yδ)2}dy=[y−2y22δ+y33δ2]δ0=δ−δ2δ+δ33δ2=δ−δ+δ3=δ3
(ii) Momentum thickness θ
θ=∫δ0uU{1−uU}dy=∫δ0(2yδ−y2δ2)[1−(2yδ−y2δ2)]dy=∫δ0[2yδ−y2δ2][1−2yδ+y2δ2]dy=∫δ0[2yδ−4y2δ2+2y3δ3−y2δ2+2y3δ3−y4δ4]dy=∫δ0[2yδ−5y2δ2+4y3δ3−y4δ4]dy=[2y22δ−5y33δ2+4y44δ3−y55δ4]δ0=[δ2δ−5δ33δ2+δ4δ3−δ55δ4]=δ−5δ3+δ−δ5=15δ−25δ+15δ−3δ15=30δ−28δ15=2δ15
(iii) Energy thickness δ∗∗
δ∗∗=∫δ0uU[1−u2U2]dy=∫δ0(2yδ−y2δ2)(1−[2yδ−y2δ2]2)dy=∫δ0(2yδ−y2δ2)(1−[4y2δ2+y4δ4−4y3δ3])dy=∫δ0(2yδ−y2δ2)(1−4y2δ2−y4δ4+4y3δ3)dy=∫δ0(2yδ−8y3δ3−2y5δ5+8y4δ4−y2δ2+4y4δ4+y6δ6−4y5δ5)dy=∫00[2yδ−y2δ2−8y3δ3+12y4δ4−6y5δ3+y6δ6]dy=[2y22δ−y33δ2−8y44δ3+12y55δ4−6y66δ5+y77δ6]δ0=δ2δ−δ33δ2−2δ4δ3+12δ55δ4−δ6δ5+δ77δ6=δ−δ3−2δ+125δ−δ+δ7=−2δ−δ3+125δ+δ7=−210δ−35δ+252δ+15δ105=−245δ+267δ105=22δ105