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Cooling/Dehumidification of an air stream

A 1000 m3/minute air stream at atmospheric pressure is at a temperature of 82.2oC and a humidity of 60%. We want to cool the air + water vapor to 8.9oC to condense the water vapor. Determine the sensible cooling and latent cooling required to drop the temperature of the air mass from 82.2oC to 8.9oC and condense the water vapor. Determine both the sensible and latent cooling requirements in units of KW and BTU/sec?

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Please see the attached Word file for complete solution. The following is simply a copy of the content in the file.

General concepts:
(1) Sensible cooling would be the energy needed to reduce the temperature of the air mass from 82.2oC to 8.9oC
(2) Latent cooling would incorporate the heat of vaporization and would be the energy needed to create the phase change from water vapor to liquid water (condensate)
(3) Total cooling req't = Sensible + Latent

Given:
Air flow = 1000 m3/min = 16.67 m3/sec
Inlet temperature = 82.2oC = 355.4oK
Outlet temperature = 8.9oC = 282oK
Humidity = 60%
Pressure = 1.00 atm = 1.013 bar

Inlet Conditions:

(a) Since the air mass is not saturated, we must determine the partial pressure of the water vapor:

From steam (saturated) tables @ temp = 82oC

Pressuresat = .5177 bar
Enthalpy (saturated liquid) = hl = 2303.2 kj/kg
Enthalpy (saturated vapor) = hv = 2647.3 kj/kg
Density (saturated liquid) = l = 970.3 kg/m3
Density (saturated vapor) = v = .3188 kg/m3

(b) Therefore, partial pressure of the water vapor can be calculated by:

Ppw =  (Humidity) x Pressure

Ppw = .60 x .5177 bar = .3106 bar = partial pressure of the water ...

Solution Summary

The solution provides very detailed theoretical calculations within a 4-page Word file.

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