148 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
pressure drop, anddiD
19. 0
2 ð 1. 65 D 15 .7 mm or 0.0157 m.
∴ hiD 4280
0. 00488 ð 299 1 1. 00.^8 / 0. 01570.^2
D
4280 ð 0. 459 ð 1. 0 / 0. 436 D4506 W/m^2 Kor, based on the outer diameter,hioD
4. 506 ð 0. 0157 / 0. 019 D 3 .72 kW/m^2 K
Wall:
For steel,kD45 W/m K andxD 0 .00163 m
and hence:x/kD
0. 00163 / 45 D 0 .0000362 m^2 K/W or 0.0362 m^2 K/kW
Outside:
hoD 1 .1kW/m^2 KIgnoring scale resistance: 1/UD 1 /hoCx/kC 1 /hio
D
0. 9091 C 0. 0362 C 0. 2686 and: UD 0 .823 kW/m^2 K
Area
QDUAmand hence: AD 1340 /
0. 823 ð 32. 8 D 49 .6m^2
Outer area of 0.019 m diameter tubeD
$ð 0. 019 ð 1. 0 D 0 .0597 m^2 /m and hence
total length of tubing requiredD
49. 6 / 0. 0597 D 830 .8m.
Thus with 2.5, 3.6, and 5.0 m tubes, the number of tubes will be 332, 231 or 166.
The total cooling water flowD 14 .57 kg/sand foruD1 m/s, the flow through 1 tube is 0.20 kg/s
∴ the number of tubes/passD
14. 57 / 0. 20 D 73
Clearly 3 passes are usually to be avoided, and hence 2 or 4 are suitable, that is 146
or 292 tubes, 5.0 or 2.5 m long.
The former is closer to a standard shell size and 166ð19 mm tubes on 25.4 mm square
pitch with two tube side passes can be fitted within a 438 mm i.d. shell. In this event,
the water velocity would be slightly less than 1 m/s in fact
1 ð 146 / 166 D 0 .88 m/s,
though this would not affect the overall coefficient to any significant extent.
The proposed unit is therefore 166ð19 mm o.d. tubes on 25.4 mm square pitch 5.0m
long with a 438 mm i.d. shell.
In making such calculations it is good practice to add an overload factor to the heat
load, say 10%, to allow for errors in predicting film coefficients, although this is often
taken into account in allowing for extra tubes within the shell. In this particular example,
the fact that the unit is to be installed 15 m above ground level is of significance in
limiting the pressure drop and it may be that in an actual situation space limitations
would immediately specify the tube length.