```14-73
14-114 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required
makeup water are to be determined.
Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire
process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling
Analysis (a) The mass flow rate of dry air through the tower remains constant (m a1  m a 2  m a ) , but the mass flow rate of
liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process.
The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass
and energy balances yields
Dry Air Mass Balance:
 m a ,i   m a ,e


m a1  m a 2  m a
Water Mass Balance:
AIR
32C
2
EXIT
95%
 m w,i   m w,e  m 3  m a11  m 4  m a 2  2
m 3  m 4  m a ( 2  1 )  m makeup
Energy Balance:
WARM
WATER
E in  E out  E system 0 (steady)  0
40C
40 kg/s
E in  E out
 i hi   m
 e he (since Q = W = 0)
m
 e he   m
 i hi
0  m
 a 2 h2  m
 4 h4  m
 a1h1  m
 3h3
0m
 a ( h2  h1 )  ( m
3  m
 makeup )h4  m
 3h3
0m
Solving for m a ,
m a 
m 3 (h3  h4 )
(h2  h1 )  ( 2   1 )h4
3
4
COOL
WATER
From the psychrometric chart (Fig. A-31 or EES),
1 AIR
INLET
1 atm
Tdb = 22C
Twb = 16C
30C
Makeup water
h1  44.7 kJ/kg dry air
1  0.008875 kg H 2 O/kg dry air
v 1  0.848 m 3 /kg dry air
and
h2  106.6 kJ/kg dry air
 2  0.02905 kg H 2 O/kg dry air
From Table A-4,
h3  h f @ 40C  167.53 kJ/kg H 2 O
h4  h f @ 30C  125.74 kJ/kg H 2 O
Substituting,
m a 
(40 kg/s)(167.53  125.74)kJ/kg
 28.17 kg/s
(106.6  44.7) kJ/kg  (0.02905  0.008875)(125.74) kJ/kg
Then the volume flow rate of air into the cooling tower becomes
V1  m av 1  (28.17 kg/s)(0.848 m 3 / kg )  23.9 m 3 /s
(b) The mass flow rate of the required makeup water is determined from
m makeup  m a ( 2  1 )  (28.17 kg/s)(0.02905  0.008875)  0.568 kg/s
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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