CHEMICAL ENGINEERING

292 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

If the velocity within the boundary layer may be represented by a sine function:

ux/us Dsin[

/ 2 

y/υ] ii

the integral:

∫l

0

ux

usux dyDu^2 sυ

∫ 1

0

ux/us 

1 

ux/us d

y/υ

and, substituting from equation (ii):

∫l

0

ux

usux dyDu^2 sυ

∫ 1

0

[sin

/ 2 

y/υsin^2 

/ 2 

y/υ]d

y/υ

Du^2 sυ

{

[

 

2 /cos[

/ 2 

y/υ]

] 1

0 ^0.^5

∫ 1

0

1 cosy/υd

y/υ

}

Du^2 sυf 

2 / 0. 5 C

[

1 /sin

y/υ

] 1

0 g

Du^2 sυ

 2 / 0. 5 D 0. 1366 u^2 sυ

From equation (ii): uxDussin[
/ 2 
y/υ]

∴ dux/dyDus
/ 2 υcos[
/ 2 
y/υ]

and whenyD0:

dux/dyyD 0 D

/ 2 

us/υ

But: R 0 D dux/dyyD 0 D
/ 2 
us/υ

Therefore, substituting in equation (i):

∂/∂x

 0. 1366 u^2 sυD

/ 2 

    us/υ

υdυD

    /us

/ 0. 2732 dx

υ^2 /xD

    x/us

/ 0. 2732

υ^2 /x^2 D

    /usx

/ 0. 1366

and:
υ/xD 4. 80 Rex^0.^5

PROBLEM 11.10

Derive the momentum equation for the flow of a viscous fluid over a small plane surface.
Show that the velocity profile in the neighbourhood of the surface can be expressed
as a sine function which satisfies the boundary conditions at the surface and at the outer
edge of the boundary layer.
Obtain the boundary layer thickness and its displacement thickness as a function of the
distance from the leading edge of the surface, when the velocity profile is expressed as a
sine function.