CHEMICAL ENGINEERING

290 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Thus, putting D
υt/υ , the heat flow becomes:

QDusTsCpυ

[∫x

0

2. 25 

y/υ^2 /  0. 75 

y/υ^4

1 / C 1 /^3 C 0. 25 

y/υ^6 /^3 d

y/υ

C

∫l

x

1. 5 

y/υ 0. 5 

y/υ^3 d

y/υ

]

DusTsCpυ[ 

0. 75 3 /  0. 15 5

1 / C 1 /^3 C 0. 036 7 /^3

C 

0. 75

1  2  0. 125

1  4 ]

DusTsCpυ

 0. 625  0. 15 2 C 0. 0107 4

The heat flow fromυtoυŁt in the absence of boundary layersD
υυŁt usTsCp.

∴
υυŁt Dυ
0. 625  0. 15 2 C 0. 0107 4

and:
υŁt/υ D 0. 375 C 0. 15 2  0. 0107 4

When <1, then DPr^0.^33
and neglecting the 4 term, an approximate value is:

υŁt/υD 0. 375 C 0. 15 Pr^0.^67

WhenPrD 0 .7,Pr^0.^67 D 0. 788

and:
υŁt/υ D
0. 375 C 0. 15 / 0. 788 D 0. 185

(Since this is much less than 1, neglecting the 4 term is justified.)

PROBLEM 11.8

Explain why it is necessary to use concepts, such as the displacement thickness and the
momentum thickness, for a boundary layer in order to obtain a boundary layer thickness
which is largely independent of the approximation used for the velocity profile in the
neighbourhood of the surface.
It is found that the velocityuat a distanceyfrom the surface can be expressed as a
simple power function (u/yn) for the turbulent boundary layer at a plane surface. What
is the value ofnif the ratio of the momentum thickness to the displacement thickness
is 1.78?

Solution

The first part of this problem is discussed in Section 11.1. If the displacement and the
momentum thicknesses areυŁandυmrespectively, then:

the momentum fluxD

∫υ

0

uydyuyD

∫υ

υm

usdyus

and: the mass fluxD

∫υ

0

uydyD

∫υ

υŁ

usdy