Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

1 atm, either other charts for that pressure or the relations developed earlier

should be used. In our case, the choice is clear:

since v 2 v 1. Then the rate of heat transfer to air in the heating section

becomes

(b) The mass balance for water in the humidifying section can be expressed as

or

where

Thus,

Discussion The result 0.539 kg/min corresponds to a water requirement of

close to one ton a day, which is significant.

Cooling with Dehumidification

The specific humidity of air remains constant during a simple cooling

process, but its relative humidity increases. If the relative humidity reaches

undesirably high levels, it may be necessary to remove some moisture from

the air, that is, to dehumidify it. This requires cooling the air below its dew-

point temperature.

0.539 kg/min

m

#

w^1 55.2 kg>min^21 0.01206
0.0023^2

0.01206 kg H 2 O>kg dry air

v 3 

0.622f 3 Pg 3

P 3 
f 3 Pg 3



0.622 1 0.60 21 3.1698 kPa 2

3100 
 1 0.60 21 3.1698 24 kPa

m

#

wm

#

a^1 v 3 
v 22

m

#

a 2 v 2 m

#

wm

#

a 3 v 3

673 kJ/min

Q

#

inm

#

a^1 h 2 
h 12 ^1 55.2 kg>min^231 28.0
15.8^2 kJ>kg^4

28.0 kJ>kg dry air

h 2 cpT 2 v 2 hg 2  1 1.005 kJ>kg#°C 21 22°C 2  1 0.0023 21 2541.0 kJ>kg 2

15.8 kJ>kg dry air

h 1 cpT 1 v 1 hg 1  1 1.005 kJ>kg#°C 21 10°C 2  1 0.0023 21 2519.2 kJ>kg 2

v 1 

0.622Pv 1

P 1 
Pv 1



0.622 1 0.368 kPa 2

1100 
0.368 2 kPa

0.0023 kg H 2 O>kg dry air

m

#

a

V

#

1

v 1



45 m^3 >min

0.815 m^3 >kg

55.2 kg>min

v 1 

RaT 1

Pa



1 0.287 kPa#m^3 >kg#K 21 283 K 2

99.632 kPa

0.815 m^3 >kg dry air

Pa 1 P 1 
Pv 1  1100 
0.368 2 kPa99.632 kPa

Pv 1 f 1 Pg 1 fPsat @ 10°C 1 0.3 21 1.2281 kPa 2 0.368 kPa

732 | Thermodynamics