Example Problem On Humidification | Mass Transfer - Chemical Engineering PDF Download

HUMIDIFICATION AND AIR CONDITIONING
Example Problem 6.2: 
It is planned to cool water from 43.3⁰C to 29.4⁰C in a packed countercurrent water-cooling tower using entering air at 29.4⁰C with a wet bulb temperature of 23.9⁰C. The water flow is 9764.9 kg/h.m² and the air flow is 6835.4 kg/h.m² . The overall mass transfer coefficient is  =2500 kg/m³h (ΔY^/ ). Calculate (a) minimum air rate that can be used and (b) tower height needed if air flow of 6835.4 kg/h.m2 is used.
Given: Height of transfer unit, H_toG=G_s/  a. Tie lines are vertical.
Enthalpies of saturated air-water vapor (Base temperature: 0⁰C)


Solution 6.2:
 

Y^/₁ =0.0165 kg/kg dry air [From Psychrometric chart]
H^/₁ =71.7 kJ/kg
H^/₂ =154.8 kJ/kg
From Graph, at G_s,min, H^/₂ =197 kJ/kg at 43.3⁰C 
(a) G_{s, min}=4546.4 kg/h.m² .

N_toG=Area under the curve= (154.8-71.7)×0.036148=3.004
H_toG=2.734 m

(b)Tower height= 2.734×3.004 m=8.213 m (Ans.)

Example Problem 6.3: 
A cooling tower of 50 m² cross-sectional area is required to cool the warm water from 42⁰C to 29⁰C at a rate of 425250 kg/h. The ambient air at 32⁰C has a wet-bulb temperature of 22⁰C and air rate (moist) is 6000 kg/h.m² . The overall mass transfer coefficient,  =740.375 kg/m³h (ΔY^/ ) where a is specific interfacial area of air-water contact. Determine (a) the minimum air rate and (b) overall gas-phase enthalpy transfer units. (c) Keeping other conditions unchanged, if the wet-bulb temperature is changed to 25.5⁰C, what will the cold water temperature?
Given: Antoine Equation: ln P^V_A (bar) =   temperature in K. Total pressure is 1 atm.

Solution 6.3: 

T_G1=32ºC, T_w1=22ºC, From psychrometric chart, Y^/₁ = 0.013 kg/ kg dry air
H^/₁ = 65.44 kJ / kg
(a) Draw tangent to equilibrium curve from point Q.
Slope of the operating line for minimum air rate:

L=425250/50 kg/h.m² =8505 kg/h.m²
c_wL=4.187 kJ/kgºC
Gs_min=3800.47 kg/h.m²
Gs=6000/(1+0.013) kg/h.m² =5923 kg/h.m²
Slope of the operating line for actual air rate:

Draw operating line with slope 6.006 through point Q. T_L2=42ºC. Locate point P. (Get H^/₂ from graph).
or,

8505 X 4.187(42-29) 5923(H^/₂ - 65.5)
H^/₂ =143.6 kJ/kg 

Height of the cooling tower=8×2.82 m=22.56 m
(b) Overall gas-phase enthalpy transfer unit (Nto_G)=2.82
(c) T_L1 is unknown. Height of the cooling tower is same, i.e., 22.56 m. Slope of the operating line is as before, i.e., 6.006. T_L1 should be greater than 29ºC. Assume T_L1 as 32^ºC.
Get height of the cooling tower. If height is 22.56 m, 32^ºC is the answer. Otherwise guess another T_L1.
Or,
T_G1=32^ºC, T_w1=25.5^ºC, From psychrometric chart, Y^/₁ 0.017 kg / kg dry air
H^/₁ = 75.682 KJ / kg
Lcw_L (T_L2 - T_L1) = Gs(H^/₂ - H^/₁)

T_L1=30.8^ºC