SECTION 13Humidification and Water Cooling
PROBLEM 13.1
In a process in which benzene is used as a solvent, it is evaporated into dry nitrogen.
The resulting mixture at a temperature of 297 K and a pressure of 101.3kN/m^2 has a
relative humidity of 60%. It is required to recover 80% of the benzene present by cooling
to 283 K and compressing to a suitable pressure. What must this pressure be? Vapour
pressures of benzene: at 297 KD 12 .2kN/m^2 :at283KD 6 .0kN/m^2.
Solution
See Volume 1, Example 13.1
PROBLEM 13.2
0 .6m^3 /s of gas is to be dried from a dew point of 294 K to a dew point of 277.5 K.
How much water must be removed and what will be the volume of the gas after drying?
Vapour pressure of water at 294 KD 2 .5kN/m^2. Vapour pressure of water at 277.5KD
0 .85 kN/m^2.
Solution
When the gas is cooled to 294 K, it will be saturated andPw 0 D 2 .5kN/m^2.
From Section 13.2:
mass of vapourDPw 0 Mw/RTD 2. 5 ð 18 / 8. 314 ð 294 D 0 .0184 kg/m^3 gas.When water has been removed, the gas will be saturated at 277.5 K, and
PwD 0 .85 kN/m^2.
At this stage, mass of vapourD 0. 85 ð 18 / 8. 314 ð 277. 5 D 0 .0066 kg/m^3 gas
Hence, water to be removedD 0. 0184 0. 0066 D 0 .0118 kg/m^3 gasor: 0. 0118 ð 0. 6 D 0 .00708 kg/s
Assuming the gas flow, 0.6m^3 /s, is referred to 273 K and 101.3kN/m^2 , 0.00708 kg/s
of water is equivalent to 0. 00708 / 18 D 3. 933 ð 10 ^4 kmol/s.
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