0
5.1kviews
Derive Momentum thickness and energy thickness for the given velocity profile

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

2 Answers
1
276views

A) Momentum thickness, θ is,

θ=δouU(1uU)dy

velocity profile

uU=2(yδ)(yδ)2

θ=δo{2(yδ)(yδ)2} {12(yδ)(yδ)2} dy

= δo[2yδy2δ2][12yδ+y2δ2] dy

= δo[2yδ4y2δ2+2y3δ3y2δ2+2y3δ3y4δ4] dy

=δo(2yδ5y2δ2+4y3δ2y4δ4)dy

= [2y22δ5y33δ2+4y44δ3y45δ4]δ

=(δ2δ5δ33δ2+δ4δ3δ55δ4)

= δ5δ3=0δ5=2δ15

B) Energy thickness δ

δ=δouu(1u2u2)dy=δo(2yδy2δ2)(1(2yδy2δ2)2)dy

= = δo(2yδy2δ2)(1[4y2/δ2+y4/δ44y3/δ3])dy

= δo(2yδy2/δ2)(14y2/δ2y4/δ4+4y3/δ3)dy

= δo(2y/δ8y2/δ32y5/δ5+8y4/δ4y2/δ2+4y4/δ4+y6/δ64y5/δ5)dy

= δo(2y/δy2/δ28y3/δ3+12y4/δ46y5/δ5+y6/δ6)dy

= [2y22δy33δ28y4δ3+12y55δ46y66δ5+y7/7δ6]δo

=δ2δδ33δ22δ4δ3+12δ55δ4δ6δ5+δ77δ6

= δδ/32δ+12δδ+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)

Please log in to add an answer.