CHEMICAL ENGINEERING

THE BOUNDARY LAYER 295

Du^2 sυ

{[



2



cos



2

y

υ

]l

0



1

2

∫ 1

0

(

1 cos

y

υ

)

d

(y

υ

)

}

Du^2 sυ

{

2





1

2

C

[

1



sin

y

υ

]l

0

}

Du^2 sυ

{

2





1

2

}

D 0. 1366 u^2 sυ.

Differentiating:

dux

dy

DusÐ

(

2 υ

cos



2

y

υ

)

(

dux

dy

)

yD 0

D

(

2

)

Ð

(u

s

υ

)

and: R 0 D

(

dux

dy

)

yD 0

D



2

    us

υ

.

Thus: 

∂

∂x

0. 1366 u^2 sυD



2

    us

υ

∴ υdυD

(

us

)(



0. 2732

)

dx.

∴

υ^2

2

D

(

(^) x
us

)(



0. 2732

)

∴

υ^2

x^2

D

(

usx

)(



0. 1366

)

and:

υ

x

D 4. 796 Rex^1 /^2

PROBLEM 11.12

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 may 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.

Solution

The momentum flux across 1 – 2 is:

∫L

0 u

2

xdy

The change from 1–2 to 3–4 is:

∂

∂x

{∫

L

0 u

2

xdy

}

dx