Modern Engineering Thermodynamics

422 CHAPTER 12: Mixtures of Gases and Vapors
CRITICAL THINKING
A sling psychrometer can be used to determine the humidity of the surrounding air because its wet bulb temperature is
nearly the same as the adiabatic saturation temperature, and Eq. (12.31) can then be used to find ω 1 = ω air in Figure 12.2.
The combined heat and mass transfer rate analysis of a wet bulb thermometer yields the following result:
ω adiabatic − ω air
= ω
WB − ω air Pr 2/3
T air − T adiabatic T air − T WB Sc
where ω adiabatic = ω 3 in Figure 12.2, ω WB is the humidity ratio in the vicinity of the wet bulb, T air = T DB = T 1 in Figure 12.2,
Pr = c p μ/k is the Prandtl number, and Sc = μ/(ρD) is the Schmidt number. Pr and Sc are traditional dimensionless numbers
composed of viscosity μ, constant pressure specific heat c p , specific heat ratio k, density ρ, and mass diffusivity D. From the
equation, we see that T WB = T adiabatic only if (Pr/Sc) 2/3 = 1. It turns out that, for water vapor in air, the Prandtl to Schmidt
number ratio is about 1.0, so the wet bulb and adiabatic saturation temperatures are about the same. However, for any
other chemical vapor–air mixture, (Pr/Sc) 2/3 generally is not equal to 1, and the wet bulb and adiabatic saturation temperatures
are not equal.
Suppose you had a serious leak or spill of a dangerous liquid chemical, like a refrigerant that subsequently evaporated into
the air of a closed room, and you wanted to use a sling psychrometer to estimate the resulting concentration of the chemical
in the air. Is it possible to use this equation to correct the wet bulb temperature (with the bulb wetted with the spilled
liquid) so that it could be used with a psychrometric chart (or Eq. (12.31)) to give an estimate of the concentration of the
spilled chemical in the air?
Thermodynamics. Note that the dry bulb temperature is just the temperature registered on any ordinary thermometer
and the psychrometric chart is just part of the p-T diagram for saturated and superheated water vapor in
the low-pressure region. When the mixture is saturated with water vapor (ϕ = 100%), no water can evaporate
from the wet bulb wick and T WB = T DB = T DP . Note also that a psychrometric chart is drawn for a fixed total
pressure, thus Charts D.5 and D.6 are valid only for mixtures at 1 atm total pressure.
EXAMPLE 12.7
Wet and dry bulb temperature measurements made outside on a cold day reveal that T DB = 5.0°C andT WB = 4.0°C. Using
the psychrometric chart, determine
a. ϕ, ω, T DP , and p w for the outside air
b. The values of ϕ, ω, T WB , and p w if this mixture is heated at constant pressure to 25.0°C.
Solution
a. From Chart D.6 at T DB1 = 5.0°C and T WB1 = 4.0°C, we read ϕ 1 = 80.%, ω 1 = 0.004 kg of water vapor per kg of dry air,
T DP1 = 20°C, and p w1 = 700. N/m 2 .
b. Now the mixture is heated at constant pressure until its dry bulb temperature increases to 25.0°C. Note that, when the
temperature is stated without a modifier (i.e., “wet” or “dry”), we presume it is the ordinary, or dry bulb, temperature.
Then, Chart D.6 gives ϕ 2 ≈ 20.%, ω 2 = ω 1 , T WB2 = 13°C, T DP2 = T DP1 , and p w2 = p w1 .
This is shown in Figure 12.5.
Notice that, under these conditions, the relative humidity and the wet and dry bulb temperatures change, but none of the
other characteristics change. This is because the amount of water vapor and the amount of air present do not change.
Exercises
15. Determine the values of ϕ, ω, T WB , and p w when the mixture in Example 12.7 is reheated to 20.0°C rather than 25.0°C
and all the other variables remain the same. Answer: ϕ = 28%, ω = 0.004 kg H 2 O per kg of dry air, T WB = 11°C, and
p w = 700. N/m 2 .
16. If the dry bulb temperature in Example 12.7 is 8.0°C rather than 5.0°C and all the other variables remain the same,
determine ϕ, ω, and p w for the outside air. Answer: ϕ = 50%, ω = 0.0035 kg H 2 O per kg of dry air, and p w = 550 N/m 2 .
17. Rework Example 12.7 for a dry bulb temperature of 10.0°C and a wet bulb temperature of 8.0°C. Answer: (a)ϕ =
75%, ω = 0.006 kg H 2 O per kg of dry air, T DP = 6.0°C, and p w = 900 N/m 2 ;and(b)ϕ = 30%, ω = 0.006 kg H 2 Oper
kg of dry air, T WB = 14°C, and p w = 900 N/m 2 .