CHEMICAL ENGINEERING

THE BOUNDARY LAYER 293

Solution

The derivation of the momentum equation is given in Section 11.2 to give:



∂

∂x

∫l

0

ux

usux dyDR 0 D dux/dyyD 0 (equation 11.9) (i)

If the velocity profile is a sine function, then:

ux/us Dsin[

/ 2 

y/υ]

Differentiating:

1 /us ∂ux/∂yD
/ 2 υ cos[
/ 2
y/υ ]

WhenyDυ,uxDusand

∂ux/∂yD 0

Differentiating again:

1 /us∂^2 ux/∂y^2 D ^2 / 4 υ^2 sin[

/ 2 

y/υ]

WhenyD0,uxD0and∂^2 ux/∂y^2 D 0
Substituting, noting thatuxDuswheny>υ:
∫l

0

ux

uxus dyDus^2

∫υ

0

sin

y/υ

/ 2 [1sin

y/υ

/ 2 ]dy

Du^2 sυ

∫ 1

0

[sin

y/υ

/ 2  0. 5

1 cos

y/υ]d

y/υ

Du^2 sυ

([

 

2 /cos

y/υ

/ 2

] 1

0 ^0.^5

[

y/υ

] 1

0 C^0.^5

[

1 /sin

y/υ

] 1

0

)

Du^2 sυ



 2 / 0. 5 C 0 D 0. 1366 u^2 sυ (as in Problem 11.8)

Substituting in equation (i):

∂

 0. 1366 us^2 υ/∂xD    

/ 2 us/υ

∴ υdυD

/ 2 / 0. 1366
/us dx

IfυD0whenxD0:

υ^2 /xD 11. 5 

    x/u^2 s

and:
υ/x ^2 D 11. 5
/usx and
υ/x D 3. 39 Rex^0.^5

The displacement thickness,υŁis given by:

us

υυŁ D

∫υ

0

ussin[

y/υ

/ 2 ]dyDus 

2 υ/

[

cos[

y/υ

/ 2 ]

] 1

0

∴ υυŁD 2 υ/D 0. 637 υ

and: υŁD 0. 363 υ

∴
υŁ/x D
0. 363 ð 3. 39 Rex^0.^5 D 1. 23 Rex^0.^5