written 6.2 years ago by | modified 3.1 years ago by |
Derive Momentum thickness and energy thickness for the given velocity profile
u/U=2(y/δ)- (y/δ)^2
(10 Marks) May-2018
Subject Fluid Mechanics 2
Topic Boundary layer theory
Difficulty Medium
written 6.2 years ago by | modified 3.1 years ago by |
Derive Momentum thickness and energy thickness for the given velocity profile
u/U=2(y/δ)- (y/δ)^2
(10 Marks) May-2018
Subject Fluid Mechanics 2
Topic Boundary layer theory
Difficulty Medium
written 6.0 years ago by | • modified 6.0 years ago |
A) Momentum thickness, θ is,
θ=∫δouU(1−uU)dy
velocity profile
uU=2(yδ)−(yδ)2
∴θ=∫δo{2(yδ)−(yδ)2} {1−2(yδ)−(yδ)2} dy
= ∫δo[2yδ−y2δ2][1−2yδ+y2δ2] dy
= ∫δo[2yδ−4y2δ2+2y3δ3−y2δ2+2y3δ3−y4δ4] dy
=∫δo(2yδ−5y2δ2+4y3δ2−y4δ4)dy
= [2y22δ−5y33δ2+4y44δ3−y45δ4]δ
=(δ2δ−5δ33δ2+δ4δ3−δ55δ4)
= δ−5δ3=0−δ5=2δ15
B) Energy thickness δ∗∗
δ∗∗=∫δouu(1−u2u2)dy=∫δo(2yδ−y2δ2)(1−(2yδ−y2δ2)2)dy
= = ∫δo(2yδ−y2δ2)(1−[4y2/δ2+y4/δ4−4y3/δ3])dy
= ∫δo(2yδ−y2/δ2)(1−4y2/δ2−y4/δ4+4y3/δ3)dy
= ∫δo(2y/δ−8y2/δ3−2y5/δ5+8y4/δ4−y2/δ2+4y4/δ4+y6/δ6−4y5/δ5)dy
= ∫δo(2y/δ−y2/δ2−8y3/δ3+12y4/δ4−6y5/δ5+y6/δ6)dy
= [2y22δ−y33δ2−8y4δ3+12y55δ4−6y66δ5+y7/7δ6]δo
=δ2δ−δ33δ2−2δ4δ3+12δ55δ4−δ6δ5+δ77δ6
= δ−δ/3−2δ+12δ−δ+δ/7=22δ105
(Now, always put δ=1 in above equation to get 22105, which is then multiplied by δ i.e. 22105×δ=22105δ as ans)