CHEMICAL ENGINEERING

THE BOUNDARY LAYER 287

Solution

The shear stress in the fluid at the surface:RD ux/y

From the equation given: RD 0. 03 u^2 s
/usx^0.^2

∴ uxD
0. 03 u^2 sy/ 
/usx^0.^2

If the velocity at the edge of the laminar sub-layer isub,uxDubwhenyDυb.

∴ ubD
0. 03 u^2 sυb/ 
/usυ^0.^2

and:
υb/υD 33. 3
ub/us
/usυ^0.^8

From equation 11.24, the velocity distribution is given by:

υb/υ^1 /^7 D

ub/us

and hence:
ub/us 7 D 33. 3
ub/us
/usυ^0.^8

or:
ub/us D 1. 65
/usυ^0.^115

D 1. 65 Reυ^0.^115

Substituting 0. 376 x^0.^8 

    /us^0.^2 forυfrom equation 11.29:

ub/us D 1 .65[0. 376 usx^0.^8  0.^2 /

    u^0 s.^2 ^0.^2 ]^0.^115

D 

1. 65 / 0. 3760.^115 

us^0.^8 x^0.^8 ^0.^8 / ^0.^8 ^0.^115

D 1. 85 Rex^0.^09

Now:
υb/υD
ub/us 7 D 74. 2 Re^0 x.^63

From equation 11.31:
υ/xD 0. 376 Rex^0.^2

∴
υb/xD
74. 2 ð 0. 376 /
Re^0 x.^63 Re^0 x.^2

D 27. 9 Rex^0.^83

In this case: RexD
1 ð 5 ð 1. 3 / 17 ð 10 ^6

D 3. 82 ð 105

υbD 1. 0 ð 27. 9

3. 82 ð 105 ^0.^83

D 6. 50 ð 10 ^4 mor 0.65 mm

PROBLEM 11.6

Obtain the momentum equation for an element of the boundary layer. If the velocity profile
in the laminar region can be represented approximately by a sine function, calculate the
boundary layer thickness in terms of distance from the leading edge of the surface.