324 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
PROBLEM 13.8
Water is to be cooled in a packed tower from 330 to 295 K by means of air flowing
countercurrently. The liquid flows at the rate of 275 cm^3 /m^2 s and the air at 0.7m^3 /m^2 s.
The entering air has a temperature of 295 K and a humidity of 20%. Calculate the required
height of tower and the condition of the air leaving at the top.
The whole of the resistance to heat and mass transfer can be considered as being within
the gas phase and the product of the mass transfer coefficient and the transfer surface per
unit volume of columnhDamay be taken as 0.2s^1.
Solution
Assuming, the latent heat of water at 273 KD2495 kJ/kg
specific heat capacity of dry air D 1 .003 kJ/kg K
specific heat capacity of water vapour D 2 .006 kJ/kg K
then the enthalpy of the inlet air stream is:
HG 1 D 1. 003 295 273 CH 2495 C 2. 006 295 273
From Fig. 13.4, whenD295 K, at 20% humidity,HD 0 .003 kg/kg, and:
HG 1 D 1. 003 ð 22 C 0. 003 2495 C 2. 006 ð 22 D 29 .68 kJ/kg
In the inlet air, the humidity is 0.003 kg/kg dry air or 0. 003 / 18 / 1 / 29 D
0 .005 kmol/kmol dry air.
Hence the flow of dry airD 1 0. 005 0. 70 D 0 .697 m^3 /m^2 s.
Density of air at 295 KD 29 / 22. 4 273 / 295 D 1 .198 kg/m^3.and hence the mass flow of dry airD 0. 697 ð 1. 198 D 0 .835 kg/m^2 s
and the mass flow of waterD 275 ð 10 ^6 m^3 /m^2 sor 275 ð 10 ^6 ð 1000 D
0 .275 kg/m^2 s.
The slope of the operating line, given by equation 13.37 is:
LCL/GD 0. 275 ð 4. 18 / 0. 835 D 1. 38
The coordinates of the bottom of the operating line are:
L 1 D295 K andHG 1 D 29 .7kJ/kg
Hence, on an enthalpy – temperature diagram (Fig. 13a), the operating line of slope 1.38
is drawn through the point (29.7, 295).
The top point of the operating line is given byL 2 D330 K, and from Fig. 13a,HG 2 D
78 .5kJ/kg.
From Figs 13.4 and 13.5 the curve representing the enthalpy of saturated air as a
function of temperature is obtained and drawn in. This plot may also be obtained by
calculation using equation 13.60.
The integral: ∫
dHG/HfHG