140 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
and in equation 9.9:
mD[89
372 T ]/ln[89/
372 T ]D
T 283 /ln[89/
372 T ]Area
For 230 mm o.d. tubes, outside area per unit lengthD
$ð 0. 230 ð 1. 0 D 0 .723 m^2 /m.
Total length of tubesD
30 ð 40 ð 2 D2400 m and hence heat transfer area,AD
2400 ð 0. 723 D 1735 .2m^2.
Heat load
The cross-sectional area for flow/tubeD
$/ 4
0. 230 2 D 0 .0416 m^2 /tube.
Assuming a single-pass arrangement, there are 1200 tubes per pass and hence area for
flowD
1200 ð 0. 0416 D 49 .86 m^2.
For a velocity of 1.0 m/s, the volumetric flowD
0. 1 ð 49. 86 m^3 /sandthemass
flowD
1000 ð 4. 986 D4986 kg/s.
Thus the heat load,QD 4986 ð 4. 18
T 283 D 20 , 840
T 283 kW or 2. 084 ð
107
T 283 W.
Substituting forQ,U,A,andmin equation 9.1:
2. 084 ð 107
T 283 D
1923 ð 1735. 2
T 283 /ln[89/
372 T ]or: ln[89/
372 T ]D 0 .1601 andTD296 K
The total heat load is, therefore,QD 20 , 840
296 283 D 2. 71 ð 105 kWand the mass of steam condensedD
2. 71 ð 105 / 2250 D 120 .4 kg/s.
PROBLEM 9.12
In an oil cooler, water flows at the rate of 360 kg/h per tube through metal tubes of
outer diameter 19 mm and thickness 1.3 mm, along the outside of which oil flows in the
opposite direction at the rate of 6.675 kg/s per tube. If the tubes are 2 m long and the
inlettemperatures of the oil and water are 370 K and 280 K respectively, what will be the
outlet oil temperature? The coefficient of heat transfer on the oil side is 1.7kw/m^2 K
and on the water side 2.5kW/m^2 K and the specific heat of the oil is 1.9 kJ/kg K.
Solution
In the absence of information as to the geometry of the unit, the solution will be worked
on the basis ofone tube— a valid approach as the number of tubes effectively appears
on both sides of equation 9.1.
IfTwandToare the outlet temperature of the water and the oil respectively, then: