Modern Engineering Thermodynamics

420 CHAPTER 12: Mixtures of Gases and Vapors
In these formulae, h a is found in the gas tables (Table C.16), h w is found in the superheated steam tables, and w a
and w w are the mass fractions of the dry air and water vapor. However, since psychrometrics involves only a twocomponent
mixture, there is no particular advantage to using the complicated premixed mixture formula. Therefore,
we confine our attention to the simpler unmixed component form illustrated in Eqs. (12.28) and (12.29).
12.5 THE ADIABATIC SATURATOR
Evaporative humidification processes normally occur without external heat transfer and are therefore adiabatic. If
the outlet of an evaporative humidifier is saturated with water vapor (ϕ = 100%), then the device is known as
an adiabatic saturator. A simple adiabatic saturator is shown in Figure 12.2. It consists of an inlet air–water vapor
flow stream at temperature T 1 , a liquid makeup water flow stream at temperature T 2 , and an outlet air–water
vapor flow stream. If the unit is insulated and made long enough, the outlet flow stream will be saturated with
water vapor, and the temperature T 3 of the outlet flow stream is then called the adiabatic saturation temperature.
Since this device is adiabatic and aergonic, its steady state, steady flow energy rate balance (ERB) reduces to
(neglecting changes in flow stream kinetic and potential energy)
_m a1 h a1 + _m w1 h w1 + _m w2 h w2 − _m a3 h a3 − _m w3 h w3 = 0
wherewehavechosentoseparatethecontributions from the air and water components according to
Eq. (12.28). From the conservation of mass, _m a1 = _m a3 = _m a and _m w2 = _m w3 − _m w1 .
Then, the ERB becomes
or
_m a ðh a1 − h a3 Þ + ð _m w3 − _m w1 Þh w2 + _m w1 h w1 − _m w3 h w3 = 0
_m a ðh a1 − h a3 Þ + _m w1 ðh w1 − h w2 Þ + _m w3 ðh w2 − h w3 Þ = 0
Dividing by _m a and introducing the humidity ratios ω 1 = _m w1 / _m a and ω 3 = _m w3 / _m a , and solving for ω 1 gives
ω 1 = ðh a3 − h a1 Þ + ω 3 ðh w3 − h w2 Þ
(12.30)
h w1 − h w2
Since we can treat the air here as an ideal gas and assuming T 3 = T 2 ,
h a3 − h a1 = c pa ðT 3 − T 1 Þ = c pa ðT 2 − T 1 Þ
and since the liquid makeup water is only a slightly compressed liquid, we can write
h w2 ≈ h f ðT 2 Þ = h f 2
Finally, since the outlet state contains saturated water vapor at the adiabatic saturation temperature, T 3 = T 2 ,
h w3 = h g ðT 3 Þ = h g ðT 2 Þ = h g2
The water vapor in the inlet region is superheated. A quick check of the Mollier diagram (Figure 7.15) reveals
that the isotherms in the low-pressure superheated region are very nearly horizontal. Therefore, the enthalpy of
water vapor in this region depends only on temperature, so we can take
h w1 = h g ðT 1 Þ = h g1
Insulation
Air−water
vapor
mixture
1
Evaporation
3
Water
Saturated
air−water
vapor mixture
2
Liquid makeup water
FIGURE 12.2
An adiabatic saturator.