Modern Engineering Thermodynamics

424 CHAPTER 12: Mixtures of Gases and Vapors
EXAMPLE 12.8 (Continued )
which, with T WB = 60.0°F and T DB = 70.0°F, the psychrometric chart, Chart D.5, gives approximately
ω 1 = 61 grains of water per lbm of dry air
which is essentially the same as that calculated from Eq. (12.31).
Exercises
18. If the dry bulb temperature in the room discussed in Example 12.8 is 80.0°F rather than 70.0°F, calculate the humidity
ratio in the room and compare your answer with that obtained from the psychrometric chart. Answer: ω 1 = 45.2 grains
of water per lbm of dry air.
19. Rework Example 12.8 for wet and dry bulb temperatures of 65.0°F and 85.0°F, respectively. Answer: ω 1 = 60.2 grains of
water per lbm of dry air.
20. Using the method of your choice, determine the relative humidity ratio in the room discussed in Example 12.8 when the
wet and dry bulb temperatures are 18.0°C and 22.0°C, respectively. Answer: ω = 0.0115 kg of water per kg of dry air.
6 The grain is the smallest of the ancient Egyptian measures of weight (see Chapter 1) and originally represented the average weight of a grain of barley
corn. Today, it is still used in some engineering fields (e.g., in the heating, ventilating, and air conditioning field) as a mass unit, with 7000 grains =
1 lbm.
Equation (12.31) gives essentially the same values as obtained from the psychrometric chart, but the chart is
much easier and quicker to use.
12.7 AIR CONDITIONING
Complete air conditioning involves producing an environment with desired pressure, temperature, humidity,
purity, and circulation characteristics. In this section, we are concerned only with altering the temperature and
the humidity in typical air conditioning applications.
In Example 12.8, we see how winter air is severely dehumidified if it is simply heated up to room temperature.
Water vapor must be added to bring its humidity up into the 40 to 50% relative humidity range. This can easily
be done by blowing the heated air across a moist surface, as shown in Figure 12.6. This is the technique used in
a common room humidifier.
The humidification process 2–3 shown in Figure 12.6 is also an example of evaporative cooling. When unsaturated
air is brought into contact with liquid water at the same (dry bulb) temperature, some of the water evaporates
(thus cooling the mixture) and the resulting air–water vapor mixture moves upward along the T WB =
constant line, as shown in Figure 12.6b. The minimum dry bulb temperature that can be produced by evaporative
cooling occurs when the outlet air becomes saturated with water vapor (ϕ = 100%), then T DB = T WB .This
concept is illustrated in the following example.
Electrical work
input
Liquid water
input
Cold
outside
air
1 2 3
Heater Fan Moist
screen
(a)
Warm
conditioned
air
p w
φ 1
T WB
φ 1 > φ 2
3 ω
1
2
φ 2
T DB
φ = 100%
(b)
Humidification
and cooling
2–3
Heating
1–2
FIGURE 12.6
Temperature and humidity conditioning of cold winter air.