The document Example Problem On Humidification Chemical Engineering Notes | EduRev is a part of the Chemical Engineering Course Mass Transfer.

All you need of Chemical Engineering at this link: Chemical Engineering

**HUMIDIFICATION AND AIR CONDITIONING****Example Problem 6.2: **

It is planned to cool water from 43.3^{0}C to 29.4^{0}C in a packed countercurrent water-cooling tower using entering air at 29.4^{0}C with a wet bulb temperature of 23.9^{0}C. The water flow is 9764.9 kg/h.m^{2} and the air flow is 6835.4 kg/h.m^{2} . The overall mass transfer coefficient is =2500 kg/m^{3}h (Î”Y^{/} ). 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: 0^{0}C)**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 43.3^{0}C

(a) G_{s, min}=4546.4 kg/h.m^{2} .

N_{toG}=Area under the curve= (154.8-71.7)Ã—0.036148=3.004

H_{toG}=2.734 m

(b)Tower height= 2.734Ã—3.004 m=8.213 m (Ans.)**Example Problem 6.3: **

A cooling tower of 50 m^{2} cross-sectional area is required to cool the warm water from 42^{0}C to 29^{0}C at a rate of 425250 kg/h. The ambient air at 32^{0}C has a wet-bulb temperature of 22^{0}C and air rate (moist) is 6000 kg/h.m^{2} . The overall mass transfer coefficient, =740.375 kg/m^{3}h (Î”Y^{/} ) 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 25.5^{0}C, 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}=32ÂºC, T_{w1}=22ÂºC, 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.m^{2} =8505 kg/h.m^{2}

c_{wL}=4.187 kJ/kgÂºC

Gs_{min}=3800.47 kg/h.m^{2}

Gs=6000/(1+0.013) kg/h.m^{2} =5923 kg/h.m^{2}

Slope of the operating line for actual air rate:

Draw operating line with slope 6.006 through point Q. T_{L2}=42ÂºC. 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=8Ã—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 29ÂºC. Assume T_{L1} as 32^{Âº}C.

Get height of the cooling tower. If height is 22.56 m, 32^{Âº}C is the answer. Otherwise guess another T_{L1}.

Or,

T_{G1}=32^{Âº}C, T_{w1}=25.5^{Âº}C, 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^{Âº}C

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

29 videos|46 docs|44 tests

### Example Problems On Dehumidification

- Doc | 1 pages
### Special Module - Mass Transfer

- Doc | 5 pages

- Evaporation Loss Of Water In Cooling Tower
- Doc | 3 pages
- Cooling Tower Thermal Design
- Video | 04:31 min
- Key Points In The Design Of Cooling Tower And Step By Step Design Procedure Of Cooling Tower
- Doc | 2 pages
- Design Calculations Of Cooling Tower
- Doc | 6 pages