Modern Engineering Thermodynamics

426 CHAPTER 12: Mixtures of Gases and Vapors
12.8 PSYCHROMETRIC ENTHALPIES
The psychrometric chart also contains enthalpy information that is useful in energy balance calculations. Though
water vapor may be added or removed from the mixture by an air conditioning system, the mass flow rate of
the dry air component is usually constant throughout the system. This makes it convenient to define the
mixture’s specific enthalpy on a per unit mass of dry air basis, rather than on a per unit mass of mixture basis.
Specific enthalpies so constructed are referred to as psychrometric enthalpies, denotedbyh # to distinguish them
from the ordinary form of the specific enthalpy. Thus, the specific psychrometric enthalpy is defined as
h # = H m /m a
whereas the ordinary mixture specific enthalpy is defined as
h m = H m /m m
Note that, since H m = m a h # = m m h m = (m a + m w )h m , we have
Further, since
then,
h # = ð1 + ωÞh m
H m = H a + H w = m a h a + m w h w
H m /m a = h # = h a + ðm w /m a Þh w = h a + ωh w (12.32)
Values of h # are given on the psychrometric charts in Charts D.5 and D.6. 7
EXAMPLE 12.10
Air has a dry bulb temperature of 50.0°C and a relative humidity of 40.0% at a total pressure of 0.101 MPa. Calculate the
value of h # from its definition and compare this value with that found in Chart D.6 for these conditions.
Solution
The basic definition of h # is given by Eq. (12.32) as h # = h a + ωh w , where h a = c p (T DB − T ref ). When the psychrometric Charts D.5
and D.6 were developed, the reference temperature used in the h a ideal gas equation was chosen to be 0°C. From Table C.13b,
we find for air that c p = 1.004 kJ/(kg·K). Then,
h a = ½1:004 kJ=ðkg.KÞŠ½ð50:0 + 273:15 KÞ − ð0 + 273:15 KÞŠ = 50:2kJ=ðkg dry airÞ
Equation (12.26b) gives the humidity ratio ω for this mixture as
ϕp sat
ω = 0:622
p m − ϕp sat
where, from Table C.1b, p sat = p sat (50.0°C) = 0.01235 MPa. Then,
0:400 × 0:01235
ω = 0:622
0:101 − 0:400 × 0:01235
kg water vapor
= 0:0319
kg dry air
Since the water vapor in the mixture is superheated, its specific enthalpy h w is determined from Table C.3b at the temperature
of the mixture and the partial pressure of the water vapor. From Eq. (12.24), we have p w = ϕp sat = 0.400 × 0.01235 = 5.00 ×
10 −3 MPa. Then, from Table C.3b at 50.0°C and 5.00 × 10 −3 MPa, we find that h w = 2593.6 kJ/kg water vapor. Using the definition
of h # given in Eq. (12.32), we now have
h # kJ
= h a + ωh w = 50:2
kg dry air + kg water vapor
kJ
0:0319 × 2593:6
kg dry air
kg water vapor
kJ
= 133
kg dry air
Chart D.6 also gives this value for ω at T DB = 50.0°C and ϕ = 40.0%.
7 Note that h a has a zero reference state at 0°F (not 0 R) in Chart D.6a and 0°C in Chart D.6b with the h w value coming from the
appropriate steam table in each case. Recall that the choice of a reference state is arbitrary, so long as property differences are used in
the calculations.