Example Problems On Dehumidification | Mass Transfer - Chemical Engineering PDF Download

EXAMPLE PROBLRMS ON DEHUMIDIFICATION 
Example Problem 6.4: 
Moist warm air is to be cooled and dehumidified to a wetbulb temperature of 30ºC using water of 25ºC in a dehumidification column. The air enters the tower with a dry-bulb temperature of 85ºC and wet-bulb temperature of 45ºC at 4800 kg/m²h rate. The overall gas phase mass transfer coefficient is estimated to be 2150 kg/m³h. The flow rate of water is 1.4 times the minimum. Calculate height of the tower.

Solution 6.4: 

Given,
T_G1=85ºC; T_W1=45ºC
From psychrometric chart, Y^/₁ = 0.065 kg moisture/kg dry air and H^/₁ = 210 kJ/kg dry air.
or, Y^/ =0.047 and
H^/₁ = [(1.005 + 1.88 X 0.047) X 85 + 2500 X 0.047] kJ/kg
=210.4 kJ/kg
Wet-bulb temperature of outlet air is 30ºC. Y^/₂ = 0.027 kg moisture/kg dry air and
H^/₂ = 98 kJ/kg.
We have, inlet water temperature, T_L2=25ºC
Locate top point Q(T_L2,H^/₂) (lower terminal of operating line) at Q(25, 98) on T_L-H^/ plane.
The enthalpy of feed moist air at the bottom point P (upper terminal) of the dehumidification tower is 210 kJ/kg. But the temperature of the exit water is unknown.
 

Calculation of minimum water rate: 
The equilibrium line is convexed downward. The “pinch point” is obtained by drawing the horizontal line through H^/1 = 210 kJ/kg to meet equilibrium line at P^/ . Points Q and P^/ are joined. Line QP^/ is the operating line for minimum water flow rate.
The water temperature at P^/ is T_L1,max=45ºC (T_W1).

= 4507.04 kg/h.m² 
Now, the minimum water rate L_s,min can be obtained from the following enthalpy balance equation:

4507.04×(210-98)=L_s,min(45-25)×4.187
L_s,min=6028.05 kg/h.m2
Actual water flow rate=1.4×G_s,min=1.4×6028.05 kg/h.m² =8439.27 kg/h.m²
Outlet temperature of water TL1 is obtained from Equationas follows:
4507.04×(210-98)=8439.27×(T_L1-25)×4.187
T_L1=39.29ºC
Moisture removed from air=water condensed per hour
=Gs(Y^/₁ - Y^/₂)
=4507.04×(0.065-0.027) kg/h.m² .
=171.27 kg/h.m² . 

Height of dehumidification tower, Z=H_toG×N_toG=2.096×2.912 m=6.104 m.