CHEMICAL ENGINEERING

288 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

The derivation of the momentum equation for an element of the boundary layer is
presented in detail in Section 11.2 and the final expression is:

RoD

∂/∂x

∫l

0

usuxuxdy

A sine function may be developed as follows. WhenyD0,uxD0andwhenyDυ,
uxDus.

Thus: uxDussin
ay

and whenyDυ,sinayD/2oraυD/2andaD/ 2 υ.

∴ uxDussin
y/ 2 d

and over the range 0<y<υ,

ux/us Dsin[

/ 2 

y/υ]

The integral in the momentum equation may now be evaluated for the laminar boundary
layer considering the ranges 0<y<υandυ<y<lseparately.

∴

∫l

0

usuxuxdyD

∫υ

0

u^2 sf 1 sin[

/ 2 

y/υ]gfsin[

/ 2 

y/υ]gdy

C

∫l

υ

usususdy

Du^2 s

∫υ

0

[sin

y/ 2 υsin^2 

y/ 2 υ]dy

Du^2 s[[cos

y/ 2 υ]/

/ 2 υy/ 2 Csin

y/υ/

 2 /υ]υ 0

Du^2 sυ[ 

2 / 

1 / 2 ]

R 0 D  

∂ux/∂yyD 0 D  us/ 2 υ

and substituting in the momentum equation:

∂

[

u^2 sυ

(

2





1

2

)]/

∂xD us/ 2 υ

∴ υdυD ^2 dx/us
4 

∴ υ^2 / 2 D[^2 /
4 ]
x/us

and: υD 4. 80
x/us 0.^5

∴
υ/xD 4. 80
/xus 0.^5 D 4. 80 Rex^0.^5