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MoistAirTab™ Example 7 - Solids Drying

Example 7

A continuous hot air drying system is employed to remove 3,500 lb of moisture per hour from the solid stock being dried. The system consists of a low-head fan, a steam-heated air heater, and a rotary drum dryer. The ambient air, under the design condition of 68°F, 760 mmHg and a relative humidity of 40%, is pressurized by an axial fan unit to 775 mmHg. The fan unit has an adiabatic efficicency of 70%. The steam heater would preheat the to 175°F with a pressure drop of 10 mm Hg; The air emerges from the drier at 95°F, a pressure of 755 mm Hg and a relative humidity of 90%. Seek to define all process units involved in terms of energy requirement, sizing, and flow rates.

Solution

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

State PointUnit1234Overall
Location Ambient AirFan OutletHeater OutletDryer OutletSummary
Pressuremm Hg760775765755 
 psia14.7014.9914.7914.60 
Dry bulb temperatureoF68.0072.22175.0095.00 
Relative humidity%40.00%  90.00% 
Humidity ratiolbw/lbdry air0.005820.005820.005820.03311 
       
Total moisture removedlbw/hr    3500
Moisture removal capacitylbw/lbdry air    0.02729
Air flowrate (dry basis)lbdry air/hr    128,250
       
Entropy of moist airBtu/(lbdry air.°F)0.03020.03080.07470.0994 
Enthalpy of moist airBtu/lbdry air15.0016.0241.0351.64 
       
Adiabtic efficiency% 70%   
Entropy (Ideal) of moist airBtu/(lbdry air.°F) 0.03024   
Enthalpy (Ideal) of moist airBtu/lbdry air 15.71   
Enthalpy Difference (Ideal)Btu/lbdry air 0.72   
Enthalpy Difference (Real)Btu/lbdry air 1.02   
Enthalpy of moist air (calc.)Btu/lbdry air 16.02   
       
Energy InputBtu/lbdry air 1.02  131,077
MotorkW/lbdry air 0.0003  38
       
Specific volume of moist airft3/lbdry air13.4213.2716.0414.82 
Volumetric flow rateACFH1,721,3001,701,4962,057,6501,900,480 
 ACFM28,68828,35834,29431,675