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%.
- To what temperature must the air stream be cooled in the condenser?
- 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: | Units | Ambient Air | Compressor Outlet | Condenser Outlet | Desiccant Dryer Outlet |
Flow rate (dry air) | lbdry air/hr | 250 | 250 | 250 | 250 |
Moisture removal | % | | | 90% | |
Pressure | mm Hg | 760 | | | |
| atm | | 11 | | |
| psia | 14.69 | 161.59 | 156.59 | 151.59 |
Releative humidity | % | 55% | | 100% | |
Dry bulb temperature | °F | 85 | | 68.66 | |
Dew point temperature | °F | 67.04 | | | -10 |
Humidity ratio | lbw/lbdry air | 0.0143 | 0.0143 | 0.0014 | 4.64949E-05 |
Moisture flow | lbw/hr | 3.568 | | 0.357 | 0.012 |
Moisture removal | lbw/hr | | | 3.211 | 0.345 |