CHEMICAL ENGINEERING

200 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Mass of liquid:mD
0. 147 ð 980 D 144 .3kg

and: mCpD
144. 3 ð 3000 D 432 ,957 J/K.

∴ 432 ,957 dT/dtD 100 Ac
383 T 21. 6
T 288

D 

38 , 300 AcC 6221  

100 AcC 21. 6 T

∴

∫ 368

288

dT/



 38 , 300 AcC 6221 /

 100 AcC 21. 6 T

D 



100 AcC 21. 6 / 432 , 957

∫ 1800

0

dt

∴ lnf[
38 , 300 AcC 6221 /
100 AcC 21. 6 288]/

[ 

38 , 300 AcC 6221 /

 100 AcC 21. 6  368 ]g

D 0. 00416

100 AcC 21. 6

This equation is solved by trial and error to give:AcD 5 .0m^2.

PROBLEM 9.69

Liquid oxygen is distributed by road in large spherical vessels, 1.82 m in internal diameter.
If the vessels were unlagged and the coefficient for heat transfer from the outside of the
vessel to the atmosphere were 5 W/m^2 K, what proportion of the contents would evaporate
during a journey lasting an hour? Initially the vessels are 80% full.
What thickness of lagging would be required to reduce the losses to one tenth? Atmo-
spheric temperatureD288 K. Boiling point of oxygenD90 K. Density of oxygenD
1140 kg/m^3. Latent heat of vaporisation of oxygenD214 kJ/kg. Thermal conductivity
of laggingD 0 .07 W/m K.

Solution

Volume of the vessel D$d^3 / 6 D
$ð 1. 823 / 6 D 3 .16 m^3
Volume of liquid oxygen D
80 / 1003. 16 D 2 .53 m^3
Mass of liquid oxygen D
2. 53 ð 1140 D2879 kg
Surface area of unlagged vesselD
$ð 1. 822 D 10 .41 m^2
Heat leakage into the vessel DhcA
T 1 T 2 D 5. 0 ð 10. 41
288  90
D 10 ,302 W or 10.3kW

∴ Evaporation rate of oxygenD
10. 3 / 214 D 0 .048 kg/s

∴ Evaporation taking place during 1 hD
0. 048 ð 3600 D 173 .3kg

which is 

100 ð 173. 3 / 2879 D 6 .02% of the contents

In order to reduce the losses to one tenth, the heat flow into the vessel must be 1.03 kW
and this will be achieved by reducing the temperature driving force to:

288  90 / 10 D 19 .8degK.