Modern Engineering Thermodynamics

428 CHAPTER 12: Mixtures of Gases and Vapors
EXAMPLE 12.12
Moist air at atmospheric pressure and 25.0°C with a relative humidity of 80.0% is to be cooled and dehumidified to 20.0°C
with a relative humidity of 40.0%. On a per unit mass of dry air basis, determine (a) the amount of water to be removed,
(b) the cooling heat transfer rate, and (c) the reheating heat transfer rate.
Solution
The schematic for this process is the same as Figure 12.7. From the psychrometric chart (Chart D.6), we can find the following
information: 8 State 1 State 2 State 3
T DB1 = 25:0°C
ϕ 1 = 80:0%
h # 1
= 67 kg/ ð kg da Þ
ω 1 = 0:016 kg H 2 O/ ðkg daÞ
T WB2 = 6:0°C
ϕ 2 = 100:%
h # 1
= 21 kg/ðkg daÞ
ω 2 = 0:0056 kg H 2 O/ ðkg daÞ
T DB3 = 20:0°C
ϕ 3 = 40:0%
h # 3
= 35 kg/ðkg daÞ
ω 3 = ω 2
a. The amount of water removed per unit mass of dry air is
ω 1 − ω 2 = 0:016 − 0:0056 = 0:010 kg H 2 O/ðkg dry airÞ
b. The amount of cooling required per unit mass of dry air is given by an energy rate balance on the cooling section as
_Q cooling / _m dry air = h # 2 − h# 1 + ðω 1 − ω 2 Þh f ðT 2 Þ
= 21 − 67 + 0:010ð25:2Þ = −45:7 kJ/ðkg dry airÞ
c. The reheating heat transfer rate is given by an energy rate balance on the reheating section:
_Q reheating / _m dry air = h # 3 − h# 2 = 35 − 21 = 14 kJ/ðkg dry airÞ
Exercises
30. Suppose the inlet air in Example 12.12 has a relative humidity of 100.% instead of 80.0% and all the other variables
remain the same. How much water then has to be removed per unit mass of dry air to achieve the outlet state of
T DB2 = 20.0°C and ϕ 2 = 40.0%? Answer: ω 1 − ω 2 = 0.0144 kg H 2 O per kg of dry air
31. If the final dry bulb temperature in Example 12.12 is 25.0°C (and ϕ 3 ≈ 30.0%) rather than 20.0°C and all the other
parameters remain the same (i.e., the mixture is dehumidified but has no net cooling), determine the cooling and
reheating heat transfer rates per unit mass flow rate of dry air. Answer: _Q cooling / _m dry air = −45:7 kJ/ ðkg dry airÞ and
_Q reheating / _m dry air = 24:0 kJ/ ðkg dry airÞ
32. Rework Example 12.12 for an inlet condition of T DB1 = 100.°F and ϕ 1 = 80.0% and an outlet condition of T DB2 = 60.0°F
and ϕ 2 = 40.0%. Answer: (a) ω 1 − ω 2 = 0.0247 lbm H 2 O/(lbm dry air), (b) _Q cooling / _m dry air = −41:9 Btu/ ðlbm dry airÞ,
and (c) _Q reheating / _m dry air = 5:5 Btu/ ðlbm dry airÞ
8 Values taken from reading a chart are typically accurate to only two significant figures.
Another common air conditioning design problem where the psychrometric chart is put to good use is in the
mixing of two or more wet airstreams. This normally involves determining how to mix the inlet airstreams to
produce a desired output conditional airstream or predicting the outlet airstream properties when all the inlet
airstream properties are known.
The conservation of mass equation for water when wet airstreams 1 and 2 are adiabatically and aergonically
mixed to form wet airstream 3 is
or
_m w3 = _m w1 + _m w2
or
_m w3 = _m a3 ω 3 = _m a1 ω 1 + _m a2 ω 2
ω 3 = ð _m a1 / _m a3 Þω 1 + ð _m a2 / _m a3 Þω 2 (12.33)
From the energy rate balance applied to this process, we get
_Q − _W + _m a1 h # 1 + _m a2h # 2 − _m a3h # 3 = 0
⎵
0