Page 197 - Chemical process engineering design and economics

Process Heat Transfer 179

By solving Equation 4.54 using Polymath, F = 0.9471. Because F = 0.9471
is greater than the minimum recommended value of 0.85, the design is acceptable.
Finally, using Equation 4.5.3, we find that the required surface area,

3
5
2
2
A = (3.669xl0 ) / 1.33xl0 (0.9471) (5.944) = 49.0 m (527.2 ft )
Because of the uncertainty in the overall heat-transfer coefficient, allow for
2
a safety factor of 20%, which results in an area of 58.80 m . Round off the area to
2
2
60 m (64.6 ft ).

RATING HEAT EXCHANGERS

The objective of a rating problem is to determine if an existing process unit will
satisfy process conditions. To arrive at an approximate calculation procedure for
rating a heat exchanger, first define a clean overall heat-transfer coefficient, i.e., in
= 0 in Equation 4.15.
the absence of any fouling. Therefore, R fi and R fo
1
U oC =——————— (4.16)
1
D n

For many situations, D 0 / D ; « 1.
Substitute Equation 4.16 into Equation 4.15 and let Rf,- and Rf 0 equal
(Rf ;)A and (Rf)A, the available fouling resistances. Then, the overall heat-transfer
0
coefficient,
U 0 = l/[(R fi) A + 1/Ucc + (Rfo) A] (4.17)

If the individual fouling resistances are added to obtain the total fouling re-
sistance, R,, A, then

U 0 =l/(R oA -l/U oC ) (4.18)

Rearranging Equation 4.18, the total available fouling resistance,

U oC -U 0
ROA=————— (4.19)
U oCU 0

Copyright © 2003 by Taylor & Francis Group LLC