CHEMICAL ENGINEERING

THE BOUNDARY LAYER 291

∴

∫υ

0

u^2 ydyDu^2 s

υυm

and:

∫υ

0

uydyDus

υυŁ

∫ 1

0

uy/us 2 d

y/υD 1 

υmυ

and:

∫ 1

0

uy/us d

y/υD 1 

υŁ/υ

∴
υm/υŁ D

[

1 

∫ 1

0

uy/us 2 d

y/υ

]/[

1 

∫ 1

0

uy/us d

y/υ

]

If
uy/us D
y/υ n, then:

∫ 1

0

uy/us d

y/υD

∫ 1

0

y/υnd

y/υD 1 /

nC 1

and:

∫ 1

0

uy/us d

y/υD

∫ 1

0

y/υ^2 nd

y/υD 1 /

 2 nC 1

∴
υm/υŁ D[1 1 /
2 nC 1 ]/[1 1 /
nC 1 ]D 2
nC 1 /
2 nC 1

When

υm/υŁ D 1 .78:

1. 78 D 2 nC 2 /

 2 nC 1 D 3. 56 nC 1. 78 D 2 nC 2

and: nD
0. 22 / 1. 56 D 0 .141 or approximately 1/ 7.

PROBLEM 11.9

Derive the momentum equation for the flow of a fluid over a plane surface for conditions
where the pressure gradient along the surface is negligible. By assuming a sine function
for the variation of velocity with distance from the surface (within the boundary layer)
for streamline flow, obtain an expression for the boundary layer thickness as a function
of distance from the leading edge of the surface.

Solution

Using the nomenclature of Fig. 11.5, the argument presented in Section 11.2 results in
the expression known as the momentum equation, given in equation 11.9, which may be
expressed as:

∂/∂x

∫l

0

ux

usux dyDR 0 i