Page 130 - Modelling in Transport Phenomena A Conceptual Approach

110 CHAPTER 4. EVALUATION OF TRANSFER COEFFICIENTS

The friction factor can be calculated from the Chen correlation, Eq. (4.5-16).
Taking €ID M 0,
1.1098
7.1490

= (m) = 1.16 x
7.1490 '~3'~'

1
-- 5.0452
--410g ---
d7 : ! : ( Re
5
= - 41og [- 5.0452 log(1.16 x + f =0.0072
13,274
Hence Eq. (7) becomes

L (0.072)(2)(2.56)2/3
-= = 37.4
D 0.0072
The required length is then

L = (37.4)(2.5) = 93.5cm
Linton-Sherwood correlation

Substitution of Eq. (4.5-33) into Eq. (6) gives
L
- = 3.13 Sc2i3
D
= 3.13 (13, 274)0*17(2.56)2/3 29.4
=
The tube length is
L = (29.4)(2.5) = 73.5cm

4.5.4 Flow in Non-Circular Ducts
The correlations given for friction factor, heat transfer coefficient, and mass transfer
coefficient are only valid for ducts of circular cross-section. These correlations can
be used for flow in non-circular ducts by introducing the concept of hydraulic
equivalent diameter, Dh, defined by

Flow area
Dh =4 (4. $36)
Wetted perimeter
The Reynolds number based on the hydraulic equivalent diameter is

(4.5-37)