14-35
14-78 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of
heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire
process (m a1  m a 2  m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are
negligible.
Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of
the air at various states are determined from the psychrometric chart (Figure A-31) to be
h1  23.5 kJ/kg dry air
1  0.0053 kg H 2 O/kg dry air (  2 )
v 1  0.809 m /kg dry air
3
Heating
coils
h3  42.3 kJ/kg dry air
 3  0.0087 kg H 2 O/kg dry air
Analysis (a) The amount of moisture in the air remains
constant it flows through the heating section ( 1 =  2),
but increases in the humidifying section ( 3 >  2). The
mass flow rate of dry air is
m a 
Sat. vapor
100C
Humidifier
AIR
10C
70%
35 m3/min
20C
60%
1 atm
1
2
3
V1
35 m3 / min

 43.3 kg/min
v1 0.809 m3 / kg
Noting that Q = W =0, the energy balance on the humidifying section can be expressed as
E in  E out  E system0 (steady)  0
E in  E out
 m i hi   m e he


m w hw  m a 2 h2  m a h3
( 3   2 )hw  h2  h3
Solving for h2,
h2  h3  ( 3   2 )h g @ 100C  42.3  (0.0087  0.0053)(2675.6)  33.2 kJ/kg dry air
Thus at the exit of the heating section we have 2 = 0.0053 kg H2O dry air and h2 = 33.2 kJ/kg dry air, which completely
fixes the state. Then from the psychrometric chart we read
T2  19.5C
 2  37.8%
(b) The rate of heat transfer to the air in the heating section is
Q in  m a (h2  h1 )  (43.3 kg/min)(33.2  23.5) kJ/kg  420 kJ/min
(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation
of water in the humidifying section,
m w  m a ( 3   2 )  (43.3 kg/min)(0.0087  0.0053)  0.15 kg/min
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