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MoistAirTab™ Example 6 - Air Dehydration

Example 6

Ambient air at 85°F and 760 mm Hg and 55% relative humidity is compressed to 10 atm (gauge) to supply a requirement for 250 lb/hr of air (dry basis). The mixture is cooled in a condenser to remove 90% of the original moisture content. The air stream is further   dehydrated to -10°F dew point (at the line pressure) in a desiccant dryer. The pressure drop in condenser and desiccant dryer is 5 psi each. The isentropic efficiency of the air compressor is 85%.

  1. To what temperature must the air stream be cooled in the condenser?
  2. How much water needs to be removed in the desiccant dryer?

Solution

The solution is presented below as a table generated from Excel. The gray shaded areas represent the input data.

 

Solution:UnitsAmbient AirCompressor OutletCondenser OutletDesiccant Dryer Outlet
Flow rate (dry air)lbdry air/hr250250250250
Moisture removal%  90% 
Pressuremm Hg760   
 atm 11  
 psia14.69 161.59156.59151.59
Releative humidity%55% 100% 
Dry bulb temperature°F85 68.66 
Dew point temperature°F67.04  -10
Humidity ratiolbw/lbdry air0.01430.01430.00144.64949E-05
Moisture flowlbw/hr3.568  0.357 0.012
Moisture removallbw/hr  3.211 0.345