CHEMICAL ENGINEERING

176 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

and for a mean specific heat of 1.0 kJ/kg K, the heat load isQD 2. 455 ð 1. 0
665 
495 D 417 .4kW
For gas to gas heat transfer, an overall coefficient of 1/
1 / 60 C 1 / 60 D30 W/m^2 K
will be taken using the data in Table 9.17.

(a) Co-current flow

 1 D 

845  495 D350 deg K, 2 D 

675  665 D10 deg K

and in equation 9.9:mD
350  10 /ln
350 / 10 D 95 .6degK.

Therefore in equation 9.1:AD
417. 4 ð 103 /
30 ð 95. 6 D 145 .5m^2.
For 25 mm i.d. tubes an o.d. of 32 mm will be assumed for which the outside areaD

$ð 0. 032 ð 1. 0 D 0 .1005 m^2 /m

and total length of tubingD
145. 5 / 0. 1005 D1447 m.

At a mean air temperature of 580 K:D
29 / 22. 4
273 / 580 D 0 .609 kg/m^3.
∴ volume flow of airD
2. 445 / 0. 609 D 4 .03 m^3 /s.

For a reasonable gas velocity of say 15 m/s: area for flowD 

4. 03 / 15 D 0 .268 m^2.

Cross-sectional area of one tubeD

$/ 40. 0252 D 0 .00050 m^2.

∴number of tubes/passD
0. 268 / 0. 00050 D 545 ,each of lengthD
1447 / 545 D 2 .65 m.

In practice, the standard length of 2.44 m would be adoptedwith 

1447 / 2. 44 D 594

tubes in a single pass.

(b) Countercurrent flow

In this case, 1 D
845  665 D180 deg K,  2 D
675  495 D180 deg K, and mD
180 deg K

In equation 9.1:AD
417. 4 ð 103 /
30 ð 180 D 77 .3m^2

and total length of tubingD
77. 3 / 0. 1005 D769 m.

With a velocity of 15 m/s, each tube would be 

769 / 545 D 1 .41 m long.

A better arrangement would be the use of 

769 / 2. 44 D315 tubes, 2.44 mlong, though

this would give a higher velocity and hence an increased air side pressure drop. With such
an arrangement, 315ð32 mm o.d. tubes could be accommodated in a 838 mm i.d. shell
on 40 mm triangular pitch.

(c) Cross flow

As in (b),mD180 deg K.

From equation 9.213:XD
t 2 t 1 /
T 1 t 1 D
665  495 /
845  495 D 0. 486

and: YD
T 1 T 2 /
t 2 t 1 D
845  675 /
665  495 D 1. 0