Thermodynamics An Engineering Approach Yunus A Cengel Michael A BolesParte17

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696 | Thermodynamics
Also,
Therefore,
(c) The reduced temperatures and pressures for both N2 and O2 at the initial
and final states and the corresponding enthalpy departure factors are, from
Fig. A–29,
N2:
O2:
From Eq. 12–58,
 TR1,O2 �
Tm,2
Tcr,O2
�
160 K
154.8 K
� 1.03
 PR,O2 �
Pm
Pcr,O2
�
10 MPa
5.08 MPa
� 1.97
 TR1,O2 �
Tm,1
Tcr,O2
�
220 K
154.8 K
� 1.42
 TR2,N2 �
Tm,2
Tcr,N2
�
160 K
126.2 K
� 1.27
 PR,N2 �
Pm
Pcr,N2
�
10 MPa
3.39 MPa
� 2.95
 TR1,N2 �
Tm,1
Tcr,N2
�
220 K
126.2 K
� 1.74
 � 3503 kJ/kmol
 � 3 16394 � 4650 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1132.2 K 2 11.0 � 2.6 2
 q�out � 1hm1,ideal � hm2,ideal 2 � RuTcr 1Zh1 � Zh2 2m
 � 4650 kJ>kmol
 � 10.79 2 14648 kJ>kmol 2 � 10.21 2 14657 kJ>kmol 2
 hm2,ideal � yN2h2,ideal,N2 � yO2h2,ideal,O2
 � 6394 kJ>kmol
 � 10.79 2 16391 kJ>kmol 2 � 10.21 2 16404 kJ>kmol 2
 hm1,ideal � yN2h1,ideal,N2 � yO2h1,ideal,O2
f Zh1,N2 � 0.9
f Zh2,N2 � 2.4
f Zh1,O2 � 1.3
f Zh2,O2 � 4.0
 � 5222 kJ>kmol
 � 3 16404 � 4657 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1154.8 K 2 11.3 � 4.0 2
 1h1 � h2 2O2 � 1h1,ideal � h2,ideal 2O2 � RuTcr 1Zh1 � Zh2 2O2
 � 3317 kJ>kmol
 � 3 16391 � 4648 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1126.2 K 2 10.9 � 2.4 2
 1h1 � h2 2N2 � 1h1,ideal � h2,ideal 2N2 � RuTcr 1Zh1 � Zh2 2N2
Therefore,
 � 3717 kJ/kmol
 � 10.79 2 13317 kJ>kmol 2 � 10.21 2 15222 kJ>kmol 2
 q�out � yN2 1h1 � h2 2N2 � yO2 1h1 � h2 2O2
cen84959_ch13.qxd 4/20/05 2:22 PM Page 696
Chapter 13 | 697
Discussion This result is about 6 percent greater than the result obtained in
part (b) by using Kay’s rule. But it is more than twice the result obtained by
assuming the mixture to be an ideal gas.
When two gases or two miscible liquids are brought into contact, they mix
and form a homogeneous mixture or solution without requiring any work
input. That is, the natural tendency of miscible substances brought into con-
tact is to mix with each other. As such, these are irreversible processes, and
thus it is impossible for the reverse process of separation to occur sponta-
neously. For example, pure nitrogen and oxygen gases readily mix when
brought into contact, but a mixture of nitrogen and oxygen (such as air)
never separates into pure nitrogen and oxygen when left unattended.
Mixing and separation processes are commonly used in practice. Separa-
tion processes require a work (or, more generally, exergy) input, and mini-
mizing this required work input is an important part of the design process of
separation plants. The presence of dissimilar molecules in a mixture affect
each other, and therefore the influence of composition on the properties must
be taken into consideration in any thermodynamic analysis. In this section
we analyze the general mixing processes, with particular emphasis on ideal
solutions, and determine the entropy generation and exergy destruction. We
then consider the reverse process of separation, and determine the minimum
(or reversible) work input needed for separation.
The specific Gibbs function (or Gibbs free energy) g is defined as the
combination property g � h � Ts. Using the relation dh � v dP � T ds, the
differential change of the Gibbs function of a pure substance is obtained by
differentiation to be
(13–27)
For a mixture, the total Gibbs function is a function of two independent
intensive properties as well as the composition, and thus it can be expressed
as G � G(P, T, N1, N2, . . . , Ni). Its differential is
(13–28)
where the subscript Nj indicates that the mole numbers of all components in
the mixture other than component i are to be held constant during 
differentiation. For a pure substance, the last term drops out since the com-
position is fixed, and the equation above must reduce to the one for a pure
substance. Comparing Eqs. 13–27 and 13–28 gives
(13–29)dG � V dP � S dT �a
i
m i dNi or dg� � v� dP � s� dT �a
i
m i dyi
dG � a 0G
0P
b
T,N
 dP � a 0G
0T
b
P,N
 dT �a
i
a 0G
0Ni
b
P,T,Nj
 dNi 1mixture 2
dg � v dP � s dT or dG � V dP � S dT 1pure substance 2
TOPIC OF SPECIAL INTEREST* Chemical Potential and the Separation Work of Mixtures
*This section can be skipped without a loss in continuity.
cen84959_ch13.qxd 4/15/05 12:07 PM Page 697
698 | Thermodynamics
i
 �
 � �
� h � Ts � g
 � gi,mixture
� hi,mixture �Tsi,mixture
Mixture:
 
Pure substance:
 
FIGURE 13–18
For a pure substance, the chemical
potential is equivalent to the Gibbs
function.
where yi � Ni /Nm is the mole fraction of component i (Nm is the total num-
ber of moles of the mixture) and
(13–30)
is the chemical potential, which is the change in the Gibbs function of the
mixture in a specified phase when a unit amount of component i in the same
phase is added as pressure, temperature, and the amounts of all other com-
ponents are held constant. The symbol tilde (as in v�, h
�
, and s�) is used to
denote the partial molar properties of the components. Note that the sum-
mation term in Eq. 13–29 is zero for a single component system and thus the
chemical potential of a pure system in a given phase is equivalent to the
molar Gibbs function (Fig. 13–18) since G � Ng � Nm, where
(13–31)
Therefore, the difference between the chemical potential and the Gibbs
function is due to the effect of dissimilar molecules in a mixture on each
other. It is because of this molecular effect that the volume of the mixture of
two miscible liquids may be more or less than the sum of the initial volumes
of the individual liquids. Likewise, the total enthalpy of the mixture of two
components at the same pressure and temperature, in general, is not equal to
the sum of the total enthalpies of the individual components before mixing,
the difference being the enthalpy (or heat) of mixing, which is the heat
released or absorbed as two or more components are mixed isothermally. For
example, the volume of an ethyl alcohol–water mixture is a few percent less
than the sum of the volumes of the individual liquids before mixing. Also,
when water and flour are mixed to make dough, the temperature of the dough
rises noticeably due to the enthalpy of mixing released.
For reasons explained above, the partial molar properties of the compo-
nents (denoted by an tilde) should be used in the evaluation of the extensive
properties of a mixture instead of the specific properties of the pure compo-
nents. For example, the total volume, enthalpy, and entropy of a mixture
should be determined from, respectively,
(13–32)
instead of
(13–33)
Then the changes in these extensive properties during mixing become
(13–34)
where �Hmixing is the enthalpy of mixing and �Smixing is the entropy of
mixing (Fig. 13–19). The enthalpy of mixing is positive for exothermic mix-
¢Vmixing �a
i
Ni 1v�i � v�i 2 , ¢Hmixing �a
i
Ni 1h�i � hi 2 , ¢Smixing �a
i
Ni 1 s�i � s�i 2
V * �a
i
Niv�i H* �a
i
Ni hi and S* �a
i
Ni s�i 
V �a
i
Niv�i H �a
i
Ni h
�
i and S �a
i
Nis
�
i 1mixture 2
m � a 0G
0N
b
P,T
� g� � h � Ts� 1pure substance 2
mi � a 0G0Ni b P,T,Nj � g
�
i � h
�
i � T s
�
i 1for component i of a mixture 2
Mixing
chamber
A
B
yB
yA A + B
mixture
FIGURE 13–19
The amount of heat released or
absorbed during a mixing process is
called the enthalpy (or heat) of
mixing, which is zero for ideal
solutions.
cen84959_ch13.qxd 4/7/05 10:51 AM Page 698
Chapter 13 | 699
ing processes, negative for endothermic mixing processes, and zero for
isothermal mixing processes during which no heat is absorbed or released.
Note that mixing is an irreversible process, and thus the entropy of mixing
must be a positive quantity during an adiabatic process. The specific volume,
enthalpy, and entropy of a mixture are determined from
(13–35)
where yi is the mole fraction of component i in the mixture.
Reconsider Eq. 13–29 for dG. Recall that properties are point functions, and
they have exact differentials. Therefore, the test of exactness can be applied to
the right-hand side of Eq. 13–29 to obtain some important relations. For the dif-
ferential dz �M dx � N dy of a function z(x, y), the test of exactness is
expressed as (�M/�y)x � (�N/�x)y. When the amount of component i in a mix-
ture is varied at constant pressure or temperature while other components (indi-
cated by j ) are held constant, Eq. 13–29 simplifies to
(13–36)
(13–37)
Applying the test of exactness to both of these relations gives
(13–38)
where the subscript N indicates that the mole numbers of all components
(and thus the composition of the mixture) is to remain constant. Taking the
chemical potential of a component to be a function of temperature, pressure,
and composition and thus mi � mi (P, T, y1, y2, . . . , yj . . .), its total differ-
ential can be expressed as
(13–39)
where the subscript y indicates that the mole fractions of all components
(and thus the composition of the mixture) is to remain constant. Substituting
Eqs. 13–38 into the above relation gives
(13–40)
For a mixture of fixed composition undergoing an isothermal process, it sim-
plifies to
(13–41)
Ideal-Gas Mixtures and Ideal Solutions
When the effect of dissimilar molecules in a mixture on each other is negli-
gible, the mixture is said to be an ideal mixture or ideal solution and the
chemical potential of a component in such a mixture equals the Gibbs function
dm i � v�i dP 1T � constant, yi � constant 2
dmi � v�i dP � s�i dT �a
i
a 0mi
0yi
b
P,T,yj
 dyi
dmi � dg
�
i � a 0mi0P b T,y dP � a
0mi
0T
b
P,y
 dT �a
i
a 0mi
0yi
b
P,T,yj
 dyi
a 0mi
0T
b
P,N
� � a 0S
0Ni
b
T,P,Nj
� � s�i and a 0mi0P b T,N � a
0V
0Ni
b
T,P,Nj
� v�i
dG � V dP � m i dNi 1for T � constant and Nj � constant 2
dG � �S dT � m i dNi 1for P � constant and Nj � constant 2
v� �a
i
yiv�i h �a
i
yih
�
i and s� �a
i
yis
�
i
cen84959_ch13.qxd 4/20/05 2:22 PM Page 699
700 | Thermodynamics
of the pure component. Many liquid solutions encountered in practice, espe-
cially dilute ones, satisfy this condition very closely and can be considered to
be ideal solutions with negligible error. As expected, the ideal solution
approximation greatly simplifies the thermodynamic analysis of mixtures. In
an ideal solution, a molecule treats the molecules of all components in the
mixture the same way—no extra attraction or repulsion for the molecules of
other components. This is usually the case for mixtures of similar substances
such as those of petroleum products. Very dissimilar substances such as
water and oil won’t even mix at all to form a solution.
For an ideal-gas mixture at temperature T and total pressure P, the partial
molar volume of a component i is v�i � vi � RuT/P. Substituting this relation
into Eq. 13–41 gives
(13–42)
since, from Dalton’s law of additive pressures, Pi � yi P for an ideal gas
mixture and
(13–43)
for constant yi. Integrating Eq. 13–42 at constant temperature from the total
mixture pressure P to the component pressure Pi of component i gives
(13–44)
For yi � 1 (i.e., a pure substance of component i alone), the last term in the
above equation drops out and we end up with mi(T, Pi) � mi(T, P), which is
the value for the pure substance i. Therefore, the term mi(T, P) is simply the
chemical potential of the pure substance i when it exists alone at total mix-
ture pressure and temperature, which is equivalent to the Gibbs function
since the chemical potential and the Gibbs function are identical for pure
substances. The term mi(T, P) is independent of mixture composition and
mole fractions, and its value can be determined from the property tables of
pure substances. Then Eq. 13–44 can be rewritten more explicitly as
(13–45)
Note that the chemical potential of a component of an ideal gas mixture
depends on the mole fraction of the components as well as the mixture tem-
perature and pressure, and is independent of the identity of the other con-
stituent gases. This is not surprising since the molecules of an ideal gas
behave like they exist alone and are not influenced by the presence of other
molecules.
Eq. 13–45 is developed for an ideal-gas mixture, but it is also applicable to
mixtures or solutions that behave the same way—that is, mixtures or solutions
in which the effects of molecules of different components on each other are
negligible. The class of such mixtures is called ideal solutions (or ideal mix-
tures), as discussed before. The ideal-gas mixture described is just one cate-
mi,mixture,ideal 1T, Pi 2 � mi,pure 1T, P 2 � RuT ln yi
m i 1T, Pi 2 � m i 1T, P 2 � RuT ln PiP � m i 1T, P 2 � RuT ln yi 1ideal gas 2
d ln Pi � d ln 1yiP 2 � d 1ln yi � ln P 2 � d ln P 1yi � constant 2
dm i �
RuT
P
 dP � RuTd ln P � RuTd ln Pi 1T � constant, yi � constant, ideal gas 2
cen84959_ch13.qxd 4/20/05 2:22 PM Page 700
Chapter 13 | 701
gory of ideal solutions. Another major category of ideal solutions is the dilute
liquid solutions, such as the saline water. It can be shown that the enthalpy of
mixing and the volume change due to mixing are zero for ideal solutions (see
Wark, 1995). That is,
(13–46)
Then it follows that v�i � v
–
i and h
�
i � h
–
i. That is, the partial molar volume
and the partial molar enthalpy of a component in a solution equal the specific
volume and enthalpy of that component when it existed alone as a pure sub-
stance at the mixture temperature and pressure. Therefore, the specific volume
and enthalpy of individual components do not change during mixing if they
form an ideal solution. Then the specific volume and enthalpy of an ideal
solution can be expressed as (Fig. 13–20)
(13–47)
Note that this is not the case for entropy and the properties that involve
entropy such as the Gibbs function, even for ideal solutions. To obtain a rela-
tion for the entropy of a mixture, we differentiate Eq. 13–45 with respect to
temperature at constant pressure and mole fraction,
(13–48)
We note from Eq. 13–38 that the two partial derivatives above are simply the
negative of the partial molar entropies. Substituting,
(13–49)
Note that ln yi is a negative quantity since yi � 1, and thus �Ru ln yi is always
positive. Therefore, the entropy of a component in a mixture is always greater
than the entropy of that component when it exists alone at the mixture temper-
ature and pressure. Then the entropy of mixing of an ideal solution is deter-
mined by substituting Eq. 13–49 into Eq. 13–34 to be
(13–50a)
or, dividing by the total number of moles of the mixture Nm,
(13–50b)
Minimum Work of Separation of Mixtures
The entropy balance for a steady-flow system simplifies to Sin � Sout � Sgen
� 0. Noting that entropy can be transferred by heat and mass only, the
¢ s�mixing,ideal �a
i
yi 1 s�i � s�i 2 � �Ru a
i
yi ln yi 1per unit mole of mixture 2
¢Smixing,ideal �a
i
Ni 1 s�i � s�i 2 � �Ru a
i
Ni ln yi 1ideal solution 2
s�i,mixture,ideal 1T, Pi 2 � s�i,pure 1T, P 2 � Ru ln y1 1ideal solution 2
a 0m i,mixing 1T, Pi 2
0T
b
P,y
� a 0m i,pure 1T, P 2
0T
b
P,y
� Ru ln yi
v�mixing,ideal �a
i
yiv
�
i �a
i
yiv�i,pure and hmixture,ideal �a
i
yih
�
i �a
i
yihi,pure
¢Vmixing,ideal �a
i
Ni 1v�i � v�i 2 � 0 and ¢Hmixing,ideal �a
i
Ni 1h�i � hi 2 � 0
� �
�V mixing,ideal � 0
�H mixing,ideal � 0
v mixture � �yivi,pure
� �vi,mixture � vi,pure
� �
h mixture � �yihi,pure
� �hi,mixing � hi,pure
i
i
FIGURE 13–20
The specific volume and enthalpy of
individual components do not change
during mixing if they form an ideal
solution (this is not the case for
entropy).
cen84959_ch13.qxd 4/20/05 2:22 PM Page 701
702 | Thermodynamics
entropy generation during an adiabatic mixing process that forms an ideal
solution becomes
(13–51a)
or
(13–51b)
Also noting that Xdestroyed � T0 Sgen, the exergy destroyed during this (and any
other) process is obtained by multiplying the entropy generation by the tem-
perature of the environment T0. It gives
(13–52a)
or
(13–52b)
Exergy destroyed represents the wasted work potential—the work that would
be produced if the mixing process occurred reversibly. For a reversible or
“thermodynamically perfect” process,the entropy generation and thus the
exergy destroyed is zero. Also, for reversible processes, the work output is a
maximum (or, the work input is a minimum if the process does not occur
naturally and requires input). The difference between the reversible work and
the actual useful work is due to irreversibilities and is equal to the exergy
destruction. Therefore, Xdestroyed � Wrev � Wactual. Then it follows that for a
naturally occurring process during which no work is produced, the reversible
work is equal to the exergy destruction (Fig. 13–21). Therefore, for the adia-
batic mixing process that forms an ideal solution, the reversible work (total
and per unit mole of mixture) is, from Eq. 13–52,
(13–53)
A reversible process, by definition, is a process that can be reversed without
leaving a net effect on the surroundings. This requires that the direction of all
interactions be reversed while their magnitudes remain the same when the
process is reversed. Therefore, the work input during a reversible separation
process must be equal to the work output during the reverse process of mix-
ing. A violation of this requirement will be a violation of the second law of
thermodynamics. The required work input for a reversible separation process
is the minimum work input required to accomplish that separation since the
work input for reversible processes is always less than the work input of cor-
responding irreversible processes. Then the minimum work input required for
the separation process can be expressed as
(13–54)Wmin,in � �RuT0 a
i
Ni ln yi and w�min,in � �RuT0 a
i
yi ln yi
Wrev � �RuT0 a
i
Ni ln yi and w�rev � �RuT0 a
i
yi ln yi
x�destroyed � T0 s�gen � �RuT0 a
i
yi ln yi 1per unit mole of mixture 2
X destroyed � T0 Sgen � �RuT0 a
i
Ni ln yi 1ideal soluton 2
s�gen � s�out � s�in � ¢smixing � �Ru a
i
yi ln yi 1per unit mole of mixture 2
Sgen � Sout � Sin � ¢Smixing � �Ru a
i
Ni ln yi 1ideal solution 2
Mixing
chamber
 T0
Wrev � Xdestruction � T0Sgen 
A
B
A + B
mixture
FIGURE 13–21
For a naturally occurring process
during which no work is produced or
consumed, the reversible work is equal
to the exergy destruction.
cen84959_ch13.qxd 4/6/05 9:35 AM Page 702
Chapter 13 | 703
It can also be expressed in the rate form as
(13–55)
where W
.
min,in is the minimum power input required to separate a solution that
approaches at a rate of (or ) into its compo-
nents. The work of separation per unit mass of mixture can be determined
from , where Mm is the apparent molar mass of the
mixture.
The minimum work relations above are for complete separation of the
components in the mixture. The required work input will be less if the exit-
ing streams are not pure. The reversible work for incomplete separation can
be determined by calculating the minimum separation work for the incoming
mixture and the minimum separation works for the outgoing mixtures, and
then taking their difference.
Reversible Mixing Processes
The mixing processes that occur naturally are irreversible, and all the work
potential is wasted during such processes. For example, when the fresh water
from a river mixes with the saline water in an ocean, an opportunity to pro-
duce work is lost. If this mixing is done reversibly (through the use of semi-
permeable membranes, for example) some work can be produced. The
maximum amount of work that can be produced during a mixing process is
equal to the minimum amount of work input needed for the corresponding
separation process (Fig. 13–22). That is,
(13–56)
Therefore, the minimum work input relations given above for separation can
also be used to determine the maximum work output for mixing.
The minimum work input relations are independent of any hardware or
process. Therefore, the relations developed above are applicable to any sepa-
ration process regardless of actual hardware, system, or process, and can be
used for a wide range of separation processes including the desalination of
sea or brackish water.
Second-Law Efficiency
The second-law efficiency is a measure of how closely a process approxi-
mates a corresponding reversible process, and it indicates the range available
for potential improvements. Noting that the second-law efficiency ranges
from 0 for a totally irreversible process to 100 percent for a totally reversible
process, the second-law efficiency for separation and mixing processes can
be defined as
(13–57)
where W
.
act,in is the actual power input (or exergy consumption) of the separa-
tion plant and W
.
act,out is the actual power produced during mixing. Note that
hII,separation �
W
#
min,in
W
#
act,in
�
wmin,in
wact,in
and hII,mixing �
W
#
act,out
W
#
max,out
�
wact,out
wmax,out
Wmax,out,mixing � Wmin,in,separation
wmin,in � w
�
min,in>Mm
m
#
m � N
#
mMm kg>sN# m kmol>s
W
#
min,in � �RuT0 a
i
N
#
i ln yi � �N
#
mRuT0 a
i
yi ln yi 1kW 2
Mixing
chamber
A
B
(a) Mixing
(b) Separation
A + B
mixture
A + B
mixture
yA
yB
A
B
yA
yB
Separation
unit
Wmax,out = 5 kJ/kg mixture
Wmax,in = 5 kJ/kg mixture
FIGURE 13–22
Under reversible conditions, the work
consumed during separation is equal
to the work produced during the
reverse process of mixing.
cen84959_ch13.qxd 4/6/05 9:35 AM Page 703
704 | Thermodynamics
the second-law efficiency is always less than 1 since the actual separation
process requires a greater amount of work input because of irreversibilities.
Therefore, the minimum work input and the second-law efficiency provide a
basis for comparison of actual separation processes to the “idealized” ones
and for assessing the thermodynamic performance of separation plants.
A second-law efficiency for mixing processes can also be defined as the
actual work produced during mixing divided by the maximum work potential
available. This definition does not have much practical value, however, since
no effort is done to produce work during most mixing processes and thus the
second-law efficiency is zero.
Special Case: Separation of a 
Two Component Mixture
Consider a mixture of two components A and B whose mole fractions are yA
and yB. Noting that yB � 1 � yA, the minimum work input required to sepa-
rate 1 kmol of this mixture at temperature T0 completely into pure A and
pure B is, from Eq. 13–54,
(13–58a)
or
(13–58b)
or, from Eq. 13–55,
(13–58c)
Some separation processes involve the extraction of just one of the compo-
nents from a large amount of mixture so that the composition of the remaining
mixture remains practically the same. Consider a mixture of two components
A and B whose mole fractions are yA and yB, respectively. The minimum work
required to separate 1 kmol of pure component A from the mixture of Nm � NA
� NB kmol (with NA �� 1) is determined by subtracting the minimum work
required to separate the remaining mixture �RuT0[(NA � 1)ln yA � NB ln yB]
from the minimum work required to separate the initial mixture Wmin,in �
�RuT0(NA ln yA � NB ln yB). It gives (Fig. 13–23)
(13–59)
The minimum work needed to separate a unit mass (1 kg) of component A is
determined from the above relation by replacing Ru by RA (or by dividing the
relation above by the molar mass of component A) since RA � Ru/MA. Eq.
13–59 also gives the maximum amount of work that can be done as one unit
of pure component A mixes with a large amount of A � B mixture.
An Application: Desalination Processes
The potable water needs of the world is increasing steadily due to population
growth, rising living standards, industrialization, and irrigation in agriculture.
There are over 10,000 desalination plants in the world, with a total desalted
w�min,in � �RuT0 ln yA � RuT0 ln 11>yA 2 1kJ>kmol A 2
 � �m
#
mRmT0 1yA ln yA � yB ln yB 2 1kW 2
W
#
min,in � �N
#
mRuT0 1yA ln yA � yB ln yB 2 
Wmin,in � �RuT0 1NA ln yA � NB ln yB 2 1kJ 2
w�min,in � �RuT0 1yA ln yA � yB ln yB 2 1kJ>kmol mixture 2
�
Separation
unit
A + B
Separation
unit
 A + B
yA, yB
yA, yB
1 kmol
pure A
(1 kmol)
A + B
pure A
pure B(a) Separating 1 kmol of A from 
 a large body of mixture
(b) Complete separation of 
 1 kmol mixture into its 
 components A and B
wmin,in = –RuT0 ln yA (kJ/kmol A)
�wmin,in = –RuT0 ( yA ln yA + yB ln yB) 
(kJ/kmol mixture)
FIGURE 13–23
The minimum work required to
separate a two-component mixture for
the two limiting cases.
cen84959_ch13.qxd 4/20/05 2:22 PM Page 704
Chapter 13 | 705
water capacity of over 5 billion gallons a day. Saudi Arabia is the largest user
of desalination with about 25 percent of the world capacity, and the United
States is the second largest user with 10 percent. The major desalination
methods are distillation and reverse osmosis. The relations can be used
directly for desalination processes, by taking the water (the solvent) to be
component A and the dissolved salts (the solute) to be component B. Then
the minimum work needed to produce 1 kg of pure water from a large reser-
voir of brackish or seawater at temperature T0 in an environment at T0 is,
from Eq. 13–59,
Desalination: (13–60)
where Rw � 0.4615 kJ/kg � K is the gas constant of water and yw is the mole
fraction of water in brackish or seawater. The relation above also gives the
maximum amount of work that can be produced as 1 kg of fresh water (from
a river, for example) mixes with seawater whose water mole fraction is yw.
The reversible work associated with liquid flow can also be expressed in
terms of pressure difference �P and elevation difference �z (potential
energy) as wmin,in � �P/r � g �z where r is the density of the liquid. Com-
bining these relations with Eq. 13–60 gives
(13–61)
and
(13–62)
where �Pmin is the osmotic pressure, which represents the pressure differ-
ence across a semipermeable membrane that separates fresh water from the
saline water under equilibrium conditions, r is the density of saline water,
and �zmin is the osmotic rise, which represents the vertical distance the
saline water would rise when separated from the fresh water by a membrane
that is permeable to water molecules alone (again at equilibrium). For desali-
nation processes, �Pmin represents the minimum pressure that the saline
water must be compressed in order to force the water molecules in saline
water through the membrane to the fresh water side during a reverse osmosis
desalination process. Alternately, �zmin represents the minimum height above
the fresh water level that the saline water must be raised to produce the
required osmotic pressure difference across the membrane to produce fresh
water. The �zmin also represents the height that the water with dissolved
organic matter inside the roots will rise through a tree when the roots are
surrounded by fresh water with the roots acting as semipermeable mem-
branes. The reverse osmosis process with semipermeable membranes is also
used in dialysis machines to purify the blood of patients with failed kidneys.
¢zmin � wmin,in>g � RwT0 ln 11>yw 2 >g 1m 2
¢Pmin � rwmin,in � rRwT0 ln 11>yw 2 1kPa 2
wmin,in � �RwT0 ln 11>yw 2 1kJ>kg pure water 2
EXAMPLE 13–6 Obtaining Fresh Water from Seawater
Fresh water is to be obtained from seawater at 15°C with a salinity of 3.48
percent on mass basis (or TDS � 34,800 ppm). Determine (a) the mole
fractions of the water and the salts in the seawater, (b) the minimum work
input required to separate 1 kg of seawater completely into pure water and
pure salts, (c) the minimum work input required to obtain 1 kg of fresh
cen84959_ch13.qxd 4/20/05 2:22 PM Page 705
706 | Thermodynamics
water from the sea, and (d ) the minimum gauge pressure that the seawater
must be raised if fresh water is to be obtained by reverse osmosis using
semipermeable membranes.
Solution Fresh water is to be obtained from seawater. The mole fractions of
seawater, the minimum works of separation needed for two limiting cases,
and the required pressurization of seawater for reverse osmosis are to be
determined.
Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The
total dissolved solids in water can be treated as table salt (NaCl). 3 The envi-
ronment temperature is also 15°C.
Properties The molar masses of water and salt are Mw � 18.0 kg/kmol
and Ms � 58.44 kg/kmol. The gas constant of pure water is Rw � 0.4615
kJ/kg · K (Table A–1). The density of seawater is 1028 kg/m3.
Analysis (a) Noting that the mass fractions of salts and water in seawater
are mfs � 0.0348 and mfw � 1 � mfs � 0.9652, the mole fractions are
determined from Eqs. 13–4 and 13–5 to be
(b) The minimum work input required to separate 1 kg of seawater com-
pletely into pure water and pure salts is
Therefore, it takes a minimum of 7.98 kJ of work input to separate 1 kg of
seawater into 0.0348 kg of salt and 0.9652 kg (nearly 1 kg) of fresh water.
(c) The minimum work input required to produce 1 kg of fresh water from
seawater is
Note that it takes about 5 times more work to separate 1 kg of seawater
completely into fresh water and salt than it does to produce 1 kg of fresh
water from a large amount of seawater.
 � 1.50 kJ/kg fresh water
 � 10.4615 kJ>kg # K 2 1288.15 K 2 ln 11>0.9888 2
 wmin,in � RwT0 ln 11>yw 2
 wmin,in �
w�min,in
Mm
�
147.2 kJ>kmol
18.44 kg>kmol � 7.98 kJ/kg seawater
 � 147.2 kJ>kmol
 � � 18.314 kJ>kmol # K2 1288.15 K2 10.9888 ln 0.9888 � 0.0112 ln 0.01122
 w�min,in � �RuT0 1yA ln yA � yB ln yB 2 � �RuT0 1yw ln yw � ys ln ys 2
 ys � 1 � yw � 1 � 0.9888 � 0.0112 � 1.12%
 yw � mfw
Mm
Mw
� 0.9652 
18.44 kg>kmol
18.0 kg>kmol � 0.9888
 Mm �
1
a
mfi
Mi
�
1
mfs
Ms
�
mfw
Mw
�
1
0.0348
58.44
�
0.9652
18.0
� 18.44 kg>kmol
cen84959_ch13.qxd 4/20/05 2:22 PM Page 706
Chapter 13 | 707
(d ) The osmotic pressure in this case is
which is equal to the minimum gauge pressure to which seawater must be
compressed if the fresh water is to be discharged at the local atmospheric
pressure. As an alternative to pressurizing, the minimum height above the
fresh water level that the seawater must be raised to produce fresh water is
(Fig. 13–24)
Discussion The minimum separation works determined above also represent
the maximum works that can be produced during the reverse process of mix-
ing. Therefore, 7.98 kJ of work can be produced when 0.0348 kg of salt is
mixed with 0.9652 kg of water reversibly to produce 1 kg of saline water,
and 1.50 kJ of work can be produced as 1 kg of fresh water is mixed with
seawater reversibly. Therefore, the power that can be generated as a river
with a flow rate of 106 m3/s mixes reversibly with seawater through semiper-
meable membranes is (Fig. 13–25)
which shows the tremendous amount of power potential wasted as the rivers
discharge into the seas.
 � 1.5 	 106 MW
 W
#
max,out � rV
#
wmax,out � 11000 kg>m3 2 1106 m3>s 2 11.50 kJ>kg 2 a 1 MW103 kJ>s b
¢zmin �
wmin,in
g
�
1.50 kJ>kg
9.81 m>s2 a
1 kg # m>s2
1 N
b a 1000 N # m
1 kJ
b � 153 m
 � 1540 kPa
 � 11028 kg>m3 2 10.4615 kPa # m3>kg # K 2 1288.15 K 2 ln 11>0.9888 2
 ¢Pmin � rm Rw T0 ln 11>yw 2
Saline
water
∆z
Membrane
P2 P1
 Pure
water
∆P � P2 � P1
FIGURE 13–24
The osmotic pressure and the osmotic
rise of saline water.
Fresh and saline water
 mixing irreversibly
Fresh river
 water
Sea water salinity � 3.48%
�z � 153 m
Fresh and saline water mixing reversibly
through semi-permeable membranes, and 
producing power
FIGURE 13–25
Power can be produced by mixing solutions of different concentrations reversibly.
cen84959_ch13.qxd 4/20/05 2:22 PM Page 707
708 | Thermodynamics
A mixture of two or more gases of fixed chemical composi-
tion is called a nonreacting gas mixture The composition of a
gas mixture is described by specifying either the mole frac-
tion or the mass fraction of each component, defined as
where
The apparent (or average) molar mass and gas constant of a
mixture are expressed as
Also,
Dalton’s law of additive pressures states that the pressure
of a gas mixture is equal to the sum of the pressures each gas
would exert if it existed alone at the mixture temperature and
volume. Amagat’s law of additivevolumes states that the vol-
ume of a gas mixture is equal to the sum of the volumes each
gas would occupy if it existed alone at the mixture tempera-
ture and pressure. Dalton’s and Amagat’s laws hold exactly
for ideal-gas mixtures, but only approximately for real-gas
mixtures. They can be expressed as
Dalton’s law:
Amagat’s law:
Here Pi is called the component pressure and Vi is called
the component volume. Also, the ratio Pi /Pm is called the
pressure fraction and the ratio Vi /Vm is called the volume
fraction of component i. For ideal gases, Pi and Vi can be
related to yi by
The quantity yiPm is called the partial pressure and the quan-
tity yiVm is called the partial volume. The P-v-T behavior
of real-gas mixtures can be predicted by using generalized
Pi
Pm
�
Vi
Vm
�
Ni
Nm
� yi
Vm �a
k
i�1
Vi 1Tm, Pm 2
Pm �a
k
i�1
Pi 1Tm, Vm 2
mfi � yi 
Mi
Mm
and Mm �
1
a
k
i�1
mfi
Mi
Mm �
mm
Nm
�a
k
i�1
yiMi and Rm �
Ru
Mm
mm �a
k
i�1
mi and Nm �a
k
i�1
Ni
mfi �
mi
mm
and yi �
Ni
Nm
SUMMARY
compressibility charts. The compressibility factor of the mix-
ture can be expressed in terms of the compressibility factors
of the individual gases as
where Zi is determined either at Tm and Vm (Dalton’s law) or at
Tm and Pm (Amagat’s law) for each individual gas. The P-v-T
behavior of a gas mixture can also be predicted approximately
by Kay’s rule, which involves treating a gas mixture as a pure
substance with pseudocritical properties determined from
The extensive properties of a gas mixture, in general, can
be determined by summing the contributions of each compo-
nent of the mixture. The evaluation of intensive properties of
a gas mixture, however, involves averaging in terms of mass
or mole fractions:
and
These relations are exact for ideal-gas mixtures and approxi-
mate for real-gas mixtures. The properties or property changes
of individual components can be determined by using ideal-
gas or real-gas relations developed in earlier chapters.
 cp,m �a
k
i�1
mficp,i and c�p,m �a
k
i�1
yic�p,i
 cv,m �a
k
i�1
mficv,i and c�v,m �a
k
i�1
yic�v,i
 sm �a
k
i�1
mfisi and s�m �a
k
i�1
yi s�i
 hm �a
k
i�1
mfihi and hm �a
k
i�1
yihi
 um �a
k
i�1
mfiui and u�m �a
k
i�1
yiu�i
 Sm �a
k
i�1
Si �a
k
i�1
mi si �a
k
i�1
Ni s�i
 Hm �a
k
i�1
Hi �a
k
i�1
mi hi �a
k
i�1
Ni hi
 Um �a
k
i�1
Ui �a
k
i�1
miui �a
k
i�1
Niu�i
P ¿cr,m �a
k
i�1
yi Pcr,i and T ¿cr,m � a
k
i�1
yi Tcr,i
Zm �a
k
i�1
yi Zi
cen84959_ch13.qxd 4/6/05 9:35 AM Page 708
Chapter 13 | 709
1. A. Bejan. Advanced Engineering Thermodynamics. 2nd
ed. New York: Wiley Interscience, 1997.
2. Y. A. Çengel, Y. Cerci, and B. Wood, “Second Law
Analysis of Separation Processes of Mixtures,” ASME
International Mechanical Engineering Congress and
Exposition, Nashville, Tennessee, 1999.
3. Y. Cerci, Y. A. Çengel, and B. Wood, “The Minimum
Separation Work for Desalination Processes,” ASME
REFERENCES AND SUGGESTED READING
International Mechanical Engineering Congress and
Exposition, Nashville, Tennessee, 1999.
4. J. P. Holman. Thermodynamics. 3rd ed. New York:
McGraw-Hill, 1980.
5. K. Wark, Jr. Advanced Thermodynamics for Engineers.
New York: McGraw-Hill, 1995.
Composition of Gas Mixtures
13–1C What is the apparent gas constant for a gas mixture?
Can it be larger than the largest gas constant in the mixture?
13–2C Consider a mixture of two gases. Can the apparent
molar mass of this mixture be determined by simply taking
the arithmetic average of the molar masses of the individual
gases? When will this be the case?
13–3C What is the apparent molar mass for a gas mixture?
Does the mass of every molecule in the mixture equal the
apparent molar mass?
13–4C Consider a mixture of several gases of identical
masses. Will all the mass fractions be identical? How about
the mole fractions?
13–5C The sum of the mole fractions for an ideal-gas mix-
ture is equal to 1. Is this also true for a real-gas mixture?
13–6C What are mass and mole fractions?
13–7C Using the definitions of mass and mole fractions,
derive a relation between them.
13–8C Somebody claims that the mass and mole fractions
for a mixture of CO2 and N2O gases are identical. Is this
true? Why?
13–9C Consider a mixture of two gases A and B. Show that
when the mass fractions mfA and mfB are known, the mole
fractions can be determined from
where MA and MB are the molar masses of A and B.
yA �
MB
MA 11>mfA � 1 2 � MB and yB � 1 � yA
PROBLEMS*
13–10 The composition of moist air is given on a molar
basis to be 78 percent N2, 20 percent O2, and 2 percent water
vapor. Determine the mass fractions of the constituents of air.
13–11 A gas mixture has the following composition on a
mole basis: 60 percent N2 and 40 percent CO2. Determine the
gravimetric analysis of the mixture, its molar mass, and gas
constant.
13–12 Repeat Prob. 13–11 by replacing N2 by O2.
13–13 A gas mixture consists of 5 kg of O2, 8 kg of N2,
and 10 kg of CO2. Determine (a) the mass fraction of each
component, (b) the mole fraction of each component, and
(c) the average molar mass and gas constant of the mixture.
13–14 Determine the mole fractions of a gas mixture that
consists of 75 percent CH4 and 25 percent CO2 by mass.
Also, determine the gas constant of the mixture.
13–15 A gas mixture consists of 8 kmol of H2 and 2 kmol of
N2. Determine the mass of each gas and the apparent gas con-
stant of the mixture. Answers: 16 kg, 56 kg, 1.155 kJ/kg · K
13–16E A gas mixture consists of 5 lbmol of H2 and
4 lbmol of N2. Determine the mass of each gas and the appar-
ent gas constant of the mixture.
13–17 A gas mixture consists of 20 percent O2, 30 percent
N2, and 50 percent CO2 on mass basis. Determine the volu-
metric analysis of the mixture and the apparent gas constant.
P-v-T Behavior of Gas Mixtures
13–18C Is a mixture of ideal gases also an ideal gas? Give
an example.
13–19C Express Dalton’s law of additive pressures. Does this
law hold exactly for ideal-gas mixtures? How about nonideal-
gas mixtures?
13–20C Express Amagat’s law of additive volumes. Does this
law hold exactly for ideal-gas mixtures? How about nonideal-
gas mixtures?
*Problems designated by a “C” are concept questions, and students
are encouraged to answer them all. Problems designated by an “E”
are in English units, and the SI users can ignore them. Problems
with a CD-EES icon are solved using EES, and complete solutions
together with parametric studies are included on the enclosed DVD.
Problems with a computer-EES icon are comprehensive in nature,
and are intended to be solved with a computer, preferably using the
EES software that accompanies this text.
cen84959_ch13.qxd 4/6/05 9:35 AM Page 709
13–21C How is the P-v-T behavior of a component in an
ideal-gas mixture expressed? How is the P-v-T behavior of a
component in a real-gas mixture expressed?
13–22C What is the difference between the component pres-
sure and the partial pressure? When are these two equivalent?
13–23C What is the difference between the component vol-
ume and the partial volume? When are these two equivalent?
13–24C In a gas mixture, which component will have the
higher partial pressure—the one with the higher mole number
or the one with the larger molar mass?
13–25C Consider a rigid tank that contains a mixture of
two ideal gases. A valve is opened and some gas escapes. As
a result, the pressure in the tank drops. Will the partial pres-
sure of each component change? How about the pressure
fraction of each component?
13–26C Consider a rigid tank that contains a mixture of
two ideal gases. The gas mixture is heated, and the pressure
and temperature in the tank rise. Will the partial pressure of
each component change? How about the pressure fraction of
each component?
13–27C Is this statement correct? The volume of an ideal-
gas mixture is equal to the sum of the volumes of each indi-
vidual gas in the mixture. If not, how would you correct it?
13–28C Is this statement correct? The temperature of an
ideal-gas mixtureis equal to the sum of the temperatures of
each individual gas in the mixture. If not, how would you
correct it?
13–29C Is this statement correct? The pressure of an ideal-
gas mixture is equal to the sum of the partial pressures of
each individual gas in the mixture. If not, how would you
correct it?
13–30C Explain how a real-gas mixture can be treated as a
pseudopure substance using Kay’s rule.
13–31 A rigid tank contains 8 kmol of O2 and 10 kmol of
CO2 gases at 290 K and 150 kPa. Estimate the volume of the
tank. Answer: 289 m3
13–32 Repeat Prob. 13–31 for a temperature of 400 K.
13–33 A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2
at 250 kPa and 280 K. The mixture is now heated to 400 K.
Determine the volume of the tank and the final pressure of the
mixture.
13–34 A gas mixture at 300 K and 200 kPa consists of 1 kg
of CO2 and 3 kg of CH4. Determine the partial pressure of
each gas and the apparent molar mass of the gas mixture.
13–35E A gas mixture at 600 R and 20 psia consists of
1 lbm of CO2 and 3 lbm of CH4. Determine the partial pressure
of each gas and the apparent molar mass of the gas mixture.
13–36 A 0.3-m3 rigid tank contains 0.6 kg of N2 and 0.4 kg
of O2 at 300 K. Determine the partial pressure of each gas
710 | Thermodynamics
and the total pressure of the mixture. Answers: 178.1 kPa,
103.9 kPa, 282.0 kPa
13–37 A gas mixture at 350 K and 300 kPa has the follow-
ing volumetric analysis: 65 percent N2, 20 percent O2, and
15 percent CO2. Determine the mass fraction and partial pres-
sure of each gas.
13–38 A rigid tank that contains 1 kg of N2 at 25°C and
300 kPa is connected to another rigid tank that contains 3 kg
of O2 at 25°C and 500 kPa. The valve connecting the two
tanks is opened, and the two gases are allowed to mix. If the
final mixture temperature is 25°C, determine the volume of
each tank and the final mixture pressure. Answers: 0.295 m3,
0.465 m3, 422 kPa
O2
3 kg
25°C
500 kPa
N2
1 kg
25°C
300 kPa
FIGURE P13–38
Ar
1 kmol
220 K
5 MPa
N2
3 kmol
190 K
8 MPa
FIGURE P13–40
13–39 A volume of 0.3 m3 of O2 at 200 K and 8 MPa is
mixed with 0.5 m3 of N2 at the same temperature and pres-
sure, forming a mixture at 200 K and 8 MPa. Determine the
volume of the mixture, using (a) the ideal-gas equation of
state, (b) Kay’s rule, and (c) the compressibility chart and
Amagat’s law. Answers: (a) 0.8 m3, (b) 0.79 m3, (c) 0.80 m3
13–40 A rigid tank contains 1 kmol of Ar gas at 220 K
and 5 MPa. A valve is now opened, and 3 kmol
of N2 gas is allowed to enter the tank at 190 K and 8 MPa.
The final mixture temperature is 200 K. Determine the pres-
sure of the mixture, using (a) the ideal-gas equation of state
and (b) the compressibility chart and Dalton’s law.
13–41 Reconsider Prob. 13–40. Using EES (or other)
software, study the effect of varying the moles
of nitrogen supplied to the tank over the range of 1 to 10
kmol of N2. Plot the final pressure of the mixture as a func-
tion of the amount of nitrogen supplied using the ideal-gas
equation of state and the compressibility chart with Dalton’s
law.
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Chapter 13 | 711
13–42E A rigid tank contains 1 lbmol of argon gas at 400 R
and 750 psia. A valve is now opened, and 3 lbmol of N2 gas
is allowed to enter the tank at 340 R and 1200 psia. The final
mixture temperature is 360 R. Determine the pressure of the
mixture, using (a) the ideal-gas equation of state and (b) the
compressibility chart and Dalton’s law. Answers: (a) 2700
psia, (b) 2507 psia
Properties of Gas Mixtures
13–43C Is the total internal energy of an ideal-gas mixture
equal to the sum of the internal energies of each individual gas
in the mixture? Answer the same question for a real-gas mixture.
13–44C Is the specific internal energy of a gas mixture
equal to the sum of the specific internal energies of each indi-
vidual gas in the mixture?
13–45C Answer Prob. 13–43C and 13–44C for entropy.
13–46C Is the total internal energy change of an ideal-gas
mixture equal to the sum of the internal energy changes of
each individual gas in the mixture? Answer the same question
for a real-gas mixture.
13–47C When evaluating the entropy change of the compo-
nents of an ideal-gas mixture, do we have to use the partial
pressure of each component or the total pressure of the mixture?
13–48C Suppose we want to determine the enthalpy change
of a real-gas mixture undergoing a process. The enthalpy
change of each individual gas is determined by using the gen-
eralized enthalpy chart, and the enthalpy change of the mixture
is determined by summing them. Is this an exact approach?
Explain.
13–49 A process requires a mixture that is 21 percent 
oxygen, 78 percent nitrogen, and 1 percent argon by volume.
All three gases are supplied from separate tanks to an adia-
batic, constant-pressure mixing chamber at 200 kPa but at dif-
ferent temperatures. The oxygen enters at 10°C, the nitrogen at
60°C, and the argon at 200°C. Determine the total entropy
change for the mixing process per unit mass of mixture.
13–50 A mixture that is 15 percent carbon dioxide, 5 percent
carbon monoxide, 10 percent oxygen, and 70 percent nitrogen
by volume undergoes an adiabatic compression process having
a compression ratio of 8:1. If the initial state of the mixture is
300 K and 100 kPa, determine the makeup of the mixture on a
mass basis and the internal energy change per unit mass of
mixture.
13–51 Propane and air are supplied to an internal combus-
tion engine such that the air-fuel ratio is 16:1 when the pres-
sure is 95 kPa and the temperature is 30°C. The compression
ratio of the engine is 9.5:1. If the compression process is
isentropic, determine the required work input for this com-
pression process, in kJ/kg of mixture.
13–52 An insulated rigid tank is divided into two compart-
ments by a partition. One compartment contains 2.5 kmol of
CO2 at 27°C and 200 kPa, and the other compartment con-
tains 7.5 kmol of H2 gas at 40°C and 400 kPa. Now the parti-
tion is removed, and the two gases are allowed to mix.
Determine (a) the mixture temperature and (b) the mixture
pressure after equilibrium has been established. Assume con-
stant specific heats at room temperature for both gases.
13–53 A 0.9-m3 rigid tank is divided into two equal com-
partments by a partition. One compartment contains Ne at
20°C and 100 kPa, and the other compartment contains Ar at
50°C and 200 kPa. Now the partition is removed, and the two
gases are allowed to mix. Heat is lost to the surrounding air
during this process in the amount of 15 kJ. Determine (a) the
final mixture temperature and (b) the final mixture pressure.
Answers: (a) 16.2°C, (b) 138.9 kPa
13–54 Repeat Prob. 13–53 for a heat loss of 8 kJ.
13–55 Ethane (C2H6) at 20°C and 200 kPa and meth-
ane (CH4) at 45°C and 200 kPa enter an adia-
batic mixing chamber. The mass flow rate of ethane is 9 kg/s,
which is twice the mass flow rate of methane. Determine
(a) the mixture temperature and (b) the rate of entropy gener-
ation during this process, in kW/K. Take T0 � 25°C.
13–56 Reconsider Prob. 13–55. Using EES (or other)
software, determine the effect of the mass frac-
tion of methane in the mixture on the mixture temperature
and the rate of exergy destruction. The total mass flow rate is
maintained constant at 13.5 kg/s, and the mass fraction of
methane is varied from 0 to 1. Plot the mixture temperature
and the rate of exergy destruction against the mass fraction,
and discuss the results.
13–57 An equimolar mixture of helium and argon gases is to
be used as the working fluid in a closed-loop gas-turbine cycle.
H2
7.5 kmol
40°C
400 kPa
CO2
2.5 kmol
27°C
200 kPa
FIGURE P13–52
2.5 MPa
1300 K
200 kPa
He - Ar
turbine 
FIGURE P13–57
cen84959_ch13.qxd 4/20/05 2:22 PM Page 711
The mixture enters the turbine at 2.5 MPa and 1300 K and
expands isentropically to a pressure of 200 kPa. Determine the
work output of the turbine per unit mass of the mixture.
13–58E A mixture of 80 percent N2 and 20 percentCO2 gases (on a mass basis) enters the nozzle
of a turbojet engine at 90 psia and 1800 R with a low veloc-
ity, and it expands to a pressure of 12 psia. If the isentropic
efficiency of the nozzle is 92 percent, determine (a) the exit
temperature and (b) the exit velocity of the mixture. Assume
constant specific heats at room temperature.
13–59E Reconsider Prob. 13–58E. Using EES (or
other) software, first solve the stated problem
and then, for all other conditions being the same, resolve the
problem to determine the composition of the nitrogen and
carbon dioxide that is required to have an exit velocity of
2600 ft /s at the nozzle exit.
13–60 A piston–cylinder device contains a mixture of
0.5 kg of H2 and 1.6 kg of N2 at 100 kPa and 300 K. Heat is
now transferred to the mixture at constant pressure until the
volume is doubled. Assuming constant specific heats at the
average temperature, determine (a) the heat transfer and
(b) the entropy change of the mixture.
13–61 An insulated tank that contains 1 kg of O2 at 15°C
and 300 kPa is connected to a 2-m3 uninsulated tank that
contains N2 at 50°C and 500 kPa. The valve connecting the
two tanks is opened, and the two gases form a homogeneous
mixture at 25°C. Determine (a) the final pressure in the tank,
(b) the heat transfer, and (c) the entropy generated during this
process. Assume T0 � 25°C.
Answers: (a) 444.6 kPa, (b) 187.2 kJ, (c) 0.962 kJ/K
712 | Thermodynamics
during this process by treating the mixture (a) as an ideal gas
and (b) as a nonideal gas and using Amagat’s law.
Answers: (a) 4273 kJ, (b) 4745 kJ
13–64 Determine the total entropy change and exergy
destruction associated with the process described in Prob.
13–63 by treating the mixture (a) as an ideal gas and (b) as a
nonideal gas and using Amagat’s law. Assume constant spe-
cific heats at room temperature and take T0 � 30°C.
13–65 Air, which may be considered as a mixture of 79 per-
cent N2 and 21 percent O2 by mole numbers, is compressed
isothermally at 200 K from 4 to 8 MPa in a steady-flow
device. The compression process is internally reversible, and
the mass flow rate of air is 2.9 kg/s. Determine the power
input to the compressor and the rate of heat rejection by treat-
ing the mixture (a) as an ideal gas and (b) as a nonideal gas
and using Amagat’s law. Answers: (a) 115.3 kW, 115.3 kW,
(b) 143.6 kW, 94.2 kW
O2
1 kg
15°C
300 kPa
N2
2 m3
50°C
500 kPa
FIGURE P13–61
Heat
6 kg H2
21 kg N2
160 K
5 MPa
FIGURE P13–63
200 K
8 MPa
200 K
4 MPa
79% N2
21% O2
W
·
FIGURE P13–65
13–62 Reconsider Prob. 13–61. Using EES (or other)
software, compare the results obtained assum-
ing ideal-gas behavior with constant specific heats at the
average temperature, and using real-gas data obtained from
EES by assuming variable specific heats over the temperature
range.
13–63 A piston–cylinder device contains 6 kg of H2 and
21 kg of N2 at 160 K and 5 MPa. Heat is now transferred to
the device, and the mixture expands at constant pressure until
the temperature rises to 200 K. Determine the heat transfer
13–66 Reconsider Prob. 13–65. Using EES (or other)
software, compare the results obtained by
assuming ideal behavior, real gas behavior with Amagat’s
law, and real gas behavior with EES data.
13–67 The combustion of a hydrocarbon fuel with air results
in a mixture of products of combustion having the composition
on a volume basis as follows: 4.89 percent carbon dioxide,
cen84959_ch13.qxd 4/20/05 2:22 PM Page 712
Chapter 13 | 713
6.50 percent water vapor, 12.20 percent oxygen, and 76.41 per-
cent nitrogen. Determine the average molar mass of the mix-
ture, the average specific heat at constant pressure of the
mixture at 600 K, in kJ/kmol � K, and the partial pressure of
the water vapor in the mixture for a mixture pressure of
200 kPa.
13–68 A mixture that is 20 percent carbon dioxide, 10 per-
cent oxygen, and 70 percent nitrogen by volume undergoes a
process from 300 K and 100 kPa to 500 K and 400 kPa.
Determine the makeup of the mixture on a mass basis and the
enthalpy change per unit mass of mixture.
Special Topic: Chemical Potential 
and the Separation Work of Mixtures
13–69C It is common experience that two gases brought into
contact mix by themselves. In the future, could it be possible
to invent a process that will enable a mixture to separate into
its components by itself without any work (or exergy) input?
13–70C A 2-L liquid is mixed with 3 L of another liquid,
forming a homogeneous liquid solution at the same tempera-
ture and pressure. Can the volume of the solution be more or
less than the 5 L? Explain.
13–71C A 2-L liquid at 20°C is mixed with 3 L of another
liquid at the same temperature and pressure in an adiabatic
container, forming a homogeneous liquid solution. Someone
claims that the temperature of the mixture rose to 22°C after
mixing. Another person refutes the claim, saying that this
would be a violation of the first law of thermodynamics. Who
do you think is right?
13–72C What is an ideal solution? Comment on the vol-
ume change, enthalpy change, entropy change, and chemical
potential change during the formation of ideal and nonideal
solutions.
13–73 Brackish water at 12°C with total dissolved solid con-
tent of TDS � 780 ppm (a salinity of 0.078 percent on mass
basis) is to be used to produce fresh water with negligible salt
content at a rate of 280 L/s. Determine the minimum power
input required. Also, determine the minimum height to which
the brackish water must be pumped if fresh water is to be
obtained by reverse osmosis using semipermeable membranes.
13–74 A river is discharging into the ocean at a rate of
400,000 m3/s. Determine the amount of power that can be gen-
erated if the river water mixes with the ocean water reversibly.
Take the salinity of the ocean to be 3.5 percent on mass basis,
and assume both the river and the ocean are at 15°C.
13–75 Reconsider Prob. 13–74. Using EES (or other)
software, investigate the effect of the salinity of
the ocean on the maximum power generated. Let the salinity
vary from 0 to 5 percent. Plot the power produced versus the
salinity of the ocean, and discuss the results.
13–76E Fresh water is to be obtained from brackish water
at 65°F with a salinity of 0.12 percent on mass basis (or
TDS � 1200 ppm). Determine (a) the mole fractions of the
water and the salts in the brackish water, (b) the minimum
work input required to separate 1 lbm of brackish water com-
pletely into pure water and pure salts, and (c) the minimum
work input required to obtain 1 lbm of fresh water.
13–77 A desalination plant produces fresh water from sea-
water at 10°C with a salinity of 3.2 percent on mass basis at a
rate of 1.4 m3/s while consuming 8.5 MW of power. The salt
content of the fresh water is negligible, and the amount of
fresh water produced is a small fraction of the seawater used.
Determine the second-law efficiency of this plant.
13–78 Fresh water is obtained from seawater at a rate of
0.5 m3/s by a desalination plant that consumes 3.3 MW of
power and has a second-law efficiency of 18 percent. Deter-
mine the power that can be produced if the fresh water pro-
duced is mixed with the seawater reversibly.
Review Problems
13–79 Air has the following composition on a mole basis:
21 percent O2, 78 percent N2, and 1 percent Ar. Determine
the gravimetric analysis of air and its molar mass. Answers:
23.2 percent O2, 75.4 percent N2, 1.4 percent Ar, 28.96 kg/kmol
13–80 Using Amagat’s law, show that
for a real-gas mixture of k gases, where Z is the compressibil-
ity factor.
13–81 Using Dalton’s law, show that
for a real-gas mixture of k gases, where Z is the compressibil-
ity factor.
13–82 A mixture of carbon dioxide and nitrogen flows
through a converging nozzle. The mixture leaves the nozzle
at a temperature of 500 K with a velocity of 360 m/s. If the
velocity is equal to the speed of sound at the exit tempera-
ture, determine the required makeup of the mixture on a mass
basis.
13–83 A piston–cylinder device containsproducts of com-
bustion from the combustion of a hydrocarbon fuel with air.
The combustion process results in a mixture that has the com-
position on a volume basis as follows: 4.89 percent carbon
dioxide, 6.50 percent water vapor, 12.20 percent oxygen, and
76.41 percent nitrogen. This mixture is initially at 1800 K and
1 MPa and expands in an adiabatic, reversible process to
200 kPa. Determine the work done on the piston by the gas,
in kJ/kg of mixture. Treat the water vapor as an ideal gas.
Zm �a
k
i�1
yi Zi
Zm �a
k
i�1
yi Zi
cen84959_ch13.qxd 4/6/05 9:35 AM Page 713
13–84 A rigid tank contains 2 kmol of N2 and 6 kmol of
CH4 gases at 200 K and 12 MPa. Estimate the volume of the
tank, using (a) the ideal-gas equation of state, (b) Kay’s rule,
and (c) the compressibility chart and Amagat’s law.
13–85 A steady stream of equimolar N2 and CO2 mixture at
100 kPa and 18°C is to be separated into N2 and CO2 gases at
100 kPa and 18°C. Determine the minimum work required
per unit mass of mixture to accomplish this separation
process. Assume T0 � 18°C.
13–86 A gas mixture consists of O2 and N2. The ratio of the
mole numbers of N2 to O2 is 3:1. This mixture is heated dur-
ing a steady-flow process from 180 to 210 K at a constant
pressure of 8 MPa. Determine the heat transfer during this
process per mole of the mixture, using (a) the ideal-gas
approximation and (b) Kay’s rule.
13–87 Reconsider Prob. 13–86. Using EES (or other)
software, investigate the effect of the mole
fraction of oxygen in the mixture on heat transfer using real-
gas behavior with EES data. Let the mole fraction of oxygen
vary from 0 to 1. Plot the heat transfer against the mole frac-
tion, and discuss the results.
13–88 Determine the total entropy change and exergy
destruction associated with the process described in Prob.
13–86, using (a) the ideal-gas approximation and (b) Kay’s
rule. Assume constant specific heats and T0 � 30°C.
13–89 A rigid tank contains a mixture of 4 kg of He and
8 kg of O2 at 170 K and 7 MPa. Heat is now transferred to the
tank, and the mixture temperature rises to 220 K. Treating the
He as an ideal gas and the O2 as a nonideal gas, determine
(a) the final pressure of the mixture and (b) the heat transfer.
13–90 A mixture of 60 percent carbon dioxide and 40 per-
cent methane on a mole basis expands through a turbine from
1600 K and 800 kPa to 100 kPa. The volume flow rate at the
turbine entrance is 10 L/s. Determine the rate of work done by
the mixture using (a) ideal-gas approximation and (b) Kay’s
rule.
13–91 A pipe fitted with a closed valve connects two tanks.
One tank contains a 5-kg mixture of 62.5 percent CO2 and
37.5 percent O2 on a mole basis at 30°C and 125 kPa. The
second tank contains 10 kg of N2 at 15°C and 200 kPa. The
valve in the pipe is opened and the gases are allowed to mix.
During the mixing process 100 kJ of heat energy is supplied
to the combined tanks. Determine the final pressure and tem-
perature of the mixture and the total volume of the mixture.
13–92 Using EES (or other) software, write a program
to determine the mole fractions of the compo-
nents of a mixture of three gases with known molar masses
when the mass fractions are given, and to determine the mass
fractions of the components when the mole fractions are given.
Run the program for a sample case, and give the results.
714 | Thermodynamics
13–93 Using EES (or other) software, write a program
to determine the apparent gas constant, con-
stant volume specific heat, and internal energy of a mixture
of three ideal gases when the mass fractions and other prop-
erties of the constituent gases are given. Run the program for
a sample case, and give the results.
13–94 Using EES (or other) software, write a program
to determine the entropy change of a mixture
of three ideal gases when the mass fractions and other prop-
erties of the constituent gases are given. Run the program for
a sample case, and give the results.
Fundamentals of Engineering (FE) Exam Problems
13–95 An ideal-gas mixture whose apparent molar mass is
36 kg/kmol consists of N2 and three other gases. If the mole
fraction of nitrogen is 0.30, its mass fraction is
(a) 0.15 (b) 0.23 (c) 0.30 (d ) 0.39 (e) 0.70
13–96 An ideal-gas mixture consists of 2 kmol of N2 and
6 kmol of CO2. The mass fraction of CO2 in the mixture is
(a) 0.175 (b) 0.250 (c) 0.500 (d ) 0.750 (e) 0.875
13–97 An ideal-gas mixture consists of 2 kmol of N2 and
4 kmol of CO2. The apparent gas constant of the mixture is
(a) 0.215 kJ/kg � K (b) 0.225 kJ/kg � K (c) 0.243 kJ/kg � K
(d) 0.875 kJ/kg � K (e) 1.24 kJ/kg � K
13–98 A rigid tank is divided into two compartments by a
partition. One compartment contains 3 kmol of N2 at 600 kPa
and the other compartment contains 7 kmol of CO2 at 200
kPa. Now the partition is removed, and the two gases form a
homogeneous mixture at 300 kPa. The partial pressure of N2
in the mixture is
(a) 75 kPa (b) 90 kPa (c) 150 kPa (d ) 175 kPa (e) 225 kPa
13–99 An 80-L rigid tank contains an ideal-gas mixture of
5 g of N2 and 5 g of CO2 at a specified pressure and temper-
ature. If N2 were separated from the mixture and stored at
mixture temperature and pressure, its volume would be
(a) 32 L (b) 36 L (c) 40 L (d ) 49 L (e) 80 L
13–100 An ideal-gas mixture consists of 3 kg of Ar and
6 kg of CO2 gases. The mixture is now heated at constant
volume from 250 K to 350 K. The amount of heat transfer is
(a) 374 kJ (b) 436 kJ (c) 488 kJ
(d ) 525 kJ (e) 664 kJ
13–101 An ideal-gas mixture consists of 30 percent helium
and 70 percent argon gases by mass. The mixture is now
expanded isentropically in a turbine from 400°C and 1.2 MPa
to a pressure of 200 kPa. The mixture temperature at turbine
exit is
(a) 195°C (b) 56°C (c) 112°C
(d ) 130°C (e) 400°C
cen84959_ch13.qxd 4/20/05 2:22 PM Page 714
Chapter 13 | 715
13–102 One compartment of an insulated rigid tank con-
tains 2 kmol of CO2 at 20°C and 150 kPa while the other
compartment contains 5 kmol of H2 gas at 35°C and 300 kPa.
Now the partition between the two gases is removed, and the
two gases form a homogeneous ideal-gas mixture. The tem-
perature of the mixture is
(a) 25°C (b) 29°C (c) 22°C (d ) 32°C (e) 34°C
13–103 A piston–cylinder device contains an ideal-gas mix-
ture of 3 kmol of He gas and 7 kmol of Ar gas at 50°C and
400 kPa. Now the gas expands at constant pressure until its
volume doubles. The amount of heat transfer to the gas mix-
ture is
(a) 6.2 MJ (b) 4.2 MJ (c) 27 MJ
(d ) 10 MJ (e) 67 MJ
13–104 An ideal-gas mixture of helium and argon gases
with identical mass fractions enters a turbine at 1200 K and
1 MPa at a rate of 0.3 kg/s, and expands isentropically to
100 kPa. The power output of the turbine is
(a) 478 kW (b) 619 kW (c) 926 KW
(d ) 729 kW (e) 564 kW
Design and Essay Problem
13–105 Prolonged exposure to mercury even at relatively
low but toxic concentrations in the air is known to cause per-
manent mental disorders, insomnia, and pain and numbness
in the hands and the feet, among other things. Therefore, the
maximum allowable concentration of mercury vapor in the
air at work places is regulated by federal agencies. These reg-
ulations require that the average level of mercury concentra-
tion in the air does not exceed 0.1 mg/m3.
Consider a mercury spill that occurs in an airtight storage
room at 20°C in San Francisco during an earthquake. Calcu-
late the highest level of mercury concentration in the air that
can occur in the storage room, in mg/m3, and determine if it
is within the safe level. The vapor pressure of mercury at
20°C is 0.173 Pa. Propose some guidelines to safeguard
against the formation of toxic concentrations of mercury
vapor in air in storage rooms and laboratories.
cen84959_ch13.qxd 4/6/05 9:35 AM Page 715
cen84959_ch13.qxd 4/6/05 9:35 AM Page 716
Chapter 14
GAS–VAPOR MIXTURES AND AIR-CONDITIONING
| 717
At temperatures below the critical temperature, the gasphase of a substance is frequently referred to as avapor. The term vapor impliesa gaseous state that is
close to the saturation region of the substance, raising the
possibility of condensation during a process.
In Chap. 13, we discussed mixtures of gases that are usu-
ally above their critical temperatures. Therefore, we were not
concerned about any of the gases condensing during a
process. Not having to deal with two phases greatly simplified
the analysis. When we are dealing with a gas–vapor mixture,
however, the vapor may condense out of the mixture during a
process, forming a two-phase mixture. This may complicate
the analysis considerably. Therefore, a gas–vapor mixture
needs to be treated differently from an ordinary gas mixture.
Several gas–vapor mixtures are encountered in engineer-
ing. In this chapter, we consider the air–water-vapor mixture,
which is the most commonly encountered gas–vapor mixture
in practice. We also discuss air-conditioning, which is the pri-
mary application area of air–water-vapor mixtures.
Objectives
The objectives of Chapter 14 are to:
• Differentiate between dry air and atmospheric air.
• Define and calculate the specific and relative humidity of
atmospheric air.
• Calculate the dew-point temperature of atmospheric air.
• Relate the adiabatic saturation temperature and wet-bulb
temperatures of atmospheric air.
• Use the psychrometric chart as a tool to determine the
properties of atmospheric air.
• Apply the principles of the conservation of mass and energy
to various air-conditioning processes.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 717
14–1 � DRY AND ATMOSPHERIC AIR
Air is a mixture of nitrogen, oxygen, and small amounts of some other
gases. Air in the atmosphere normally contains some water vapor (or mois-
ture) and is referred to as atmospheric air. By contrast, air that contains no
water vapor is called dry air. It is often convenient to treat air as a mixture
of water vapor and dry air since the composition of dry air remains rela-
tively constant, but the amount of water vapor changes as a result of con-
densation and evaporation from oceans, lakes, rivers, showers, and even the
human body. Although the amount of water vapor in the air is small, it plays
a major role in human comfort. Therefore, it is an important consideration
in air-conditioning applications.
The temperature of air in air-conditioning applications ranges from about
�10 to about 50°C. In this range, dry air can be treated as an ideal gas with
a constant cp value of 1.005 kJ/kg · K [0.240 Btu/lbm · R] with negligible
error (under 0.2 percent), as illustrated in Fig. 14–1. Taking 0°C as the ref-
erence temperature, the enthalpy and enthalpy change of dry air can be
determined from
(14–1a)
and
(14–1b)
where T is the air temperature in °C and �T is the change in temperature. In
air-conditioning processes we are concerned with the changes in enthalpy
�h, which is independent of the reference point selected.
It certainly would be very convenient to also treat the water vapor in the
air as an ideal gas and you would probably be willing to sacrifice some
accuracy for such convenience. Well, it turns out that we can have the con-
venience without much sacrifice. At 50°C, the saturation pressure of water
is 12.3 kPa. At pressures below this value, water vapor can be treated as an
ideal gas with negligible error (under 0.2 percent), even when it is a satu-
rated vapor. Therefore, water vapor in air behaves as if it existed alone and
obeys the ideal-gas relation Pv � RT. Then the atmospheric air can be
treated as an ideal-gas mixture whose pressure is the sum of the partial pres-
sure of dry air* Pa and that of water vapor Pv:
(14–2)
The partial pressure of water vapor is usually referred to as the vapor pres-
sure. It is the pressure water vapor would exert if it existed alone at the
temperature and volume of atmospheric air.
Since water vapor is an ideal gas, the enthalpy of water vapor is a function
of temperature only, that is, h � h(T ). This can also be observed from the 
T-s diagram of water given in Fig. A–9 and Fig. 14–2 where the constant-
enthalpy lines coincide with constant-temperature lines at temperatures
P � Pa � Pv 1kPa 2
¢hdry air � cp¢T � 11.005 kJ>kg # °C 2 ¢T 1kJ>kg 2
hdry air � cpT � 11.005 kJ>kg # °C 2T 1kJ>kg 2
718 | Thermodynamics
T, °C
s
h = const.
50
FIGURE 14–2
At temperatures below 50°C, the 
h � constant lines coincide with the
T � constant lines in the superheated
vapor region of water.
DRY AIR
T,°C cp,kJ/kg ·°C
–10
0
10
20
30
40
50
1.0038
1.0041
1.0045
1.0049
1.0054
1.0059
1.0065
FIGURE 14–1
The cp of air can be assumed to be
constant at 1.005 kJ/kg · °C in the
temperature range �10 to 50°C with
an error under 0.2 percent.
*Throughout this chapter, the subscript a denotes dry air and the subscript v denotes
water vapor.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 718
below 50°C. Therefore, the enthalpy of water vapor in air can be taken to be
equal to the enthalpy of saturated vapor at the same temperature. That is,
(14–3)
The enthalpy of water vapor at 0°C is 2500.9 kJ/kg. The average cp value of
water vapor in the temperature range �10 to 50°C can be taken to be 1.82
kJ/kg · °C. Then the enthalpy of water vapor can be determined approxi-
mately from
(14–4)
or
(14–5)
in the temperature range �10 to 50°C (or 15 to 120°F), with negligible
error, as shown in Fig. 14–3.
14–2 � SPECIFIC AND RELATIVE HUMIDITY OF AIR
The amount of water vapor in the air can be specified in various ways.
Probably the most logical way is to specify directly the mass of water vapor
present in a unit mass of dry air. This is called absolute or specific humid-
ity (also called humidity ratio) and is denoted by v:
(14–6)
The specific humidity can also be expressed as
(14–7)
or
(14–8)
where P is the total pressure.
Consider 1 kg of dry air. By definition, dry air contains no water vapor,
and thus its specific humidity is zero. Now let us add some water vapor to
this dry air. The specific humidity will increase. As more vapor or moisture
is added, the specific humidity will keep increasing until the air can hold no
more moisture. At this point, the air is said to be saturated with moisture,
and it is called saturated air. Any moisture introduced into saturated air
will condense. The amount of water vapor in saturated air at a specified
temperature and pressure can be determined from Eq. 14–8 by replacing Pv
by Pg, the saturation pressure of water at that temperature (Fig. 14–4).
The amount of moisture in the air has a definite effect on how comfort-
able we feel in an environment. However, the comfort level depends more
on the amount of moisture the air holds (mv) relative to the maximum
amount of moisture the air can hold at the same temperature (mg). The ratio
of these two quantities is called the relative humidity f (Fig. 14–5)
(14–9)f �
mv
mg
�
PvV>RvT
PgV>RvT �
Pv
Pg
v �
0.622Pv
P � Pv
1kg water vapor>kg dry air 2
v �
mv
ma
�
PvV>RvT
PaV>RaT �
Pv >Rv
Pa >Ra � 0.622 
Pv
Pa
v �
mv
ma
1kg water vapor>kg dry air 2
hg 1T 2 � 1060.9 � 0.435T 1Btu>lbm 2 T in °F
hg 1T 2 � 2500.9 � 1.82T 1kJ>kg 2 T in °C
hv 1T, low P 2 � hg 1T 2
Chapter 14 | 719
WATER VAPOR
–10
0
10
20
30
40
50
2482.1
2500.9
2519.2
2537.4
2555.6
2573.5
2591.3
2482.7
2500.9
2519.1
2537.3
2555.5
2573.7
2591.9
–0.6
0.0
0.1
0.1
0.1
–0.2
–0.6
hg,kJ/kg
T,°C Table A-4 Eq. 14-4 
Difference,
kJ/kg 
FIGURE 14–3
In the temperature range �10 to 50°C,
the hg of water can be determined
from Eq. 14–4 with negligible error.
AIRAIR
 2525°C,100 kC,100 kPaPa
(Psat,Hsat,H 2O @ 25O @ 25°C = 3.1698 k = 3.1698 kPa)Pa)
Pv = = 0
Pv < 3.1698 k < 3.1698 kPaPa
Pv = 3.1698 k = 3.1698 kPaPa
dry airdry air
unsaturated airunsaturated air
saturated airsaturated air
FIGURE 14–4
For saturated air, the vapor pressure is
equal to the saturation pressure of
water.
cen84959_ch14.qxd 4/29/05 11:33 AM Page 719
where
(14–10)
Combining Eqs. 14–8 and 14–9, we can also express the relative humidity as
(14–11a, b)
The relative humidity ranges from0 for dry air to 1 for saturated air. Note
that the amount of moisture air can hold depends on its temperature. There-
fore, the relative humidity of air changes with temperature even when its
specific humidity remains constant.
Atmospheric air is a mixture of dry air and water vapor, and thus
the enthalpy of air is expressed in terms of the enthalpies of the dry air and
the water vapor. In most practical applications, the amount of dry air in the
air–water-vapor mixture remains constant, but the amount of water vapor
changes. Therefore, the enthalpy of atmospheric air is expressed per unit
mass of dry air instead of per unit mass of the air–water vapor mixture.
The total enthalpy (an extensive property) of atmospheric air is the sum of
the enthalpies of dry air and the water vapor:
Dividing by ma gives
or
(14–12)
since hv � hg (Fig. 14–6).
Also note that the ordinary temperature of atmospheric air is frequently
referred to as the dry-bulb temperature to differentiate it from other forms
of temperatures that shall be discussed.
h � ha � vhg 1kJ>kg dry air 2
h �
H
ma
� ha �
mv
ma
 hv � ha � vhv
H � Ha � Hv � maha � mv hv
f �
vP
10.622 � v 2Pg and v �
0.622fPg
P � fPg
Pg � Psat @ T
720 | Thermodynamics
AIRAIR
2525°C,1 atmC,1 atm
ma = = 1 kg1 kg
mv = =
mv, , maxmax = = 
0.01 kg0.01 kg
0.02 kg0.02 kg
Specific humidity: Specific humidity: ω = 0.01 = 0.01
Relative humidity: Relative humidity: φ = 50% = 50%
kg Hkg H2O
kg dry air kg dry air 
FIGURE 14–5
Specific humidity is the actual amount
of water vapor in 1 kg of dry air,
whereas relative humidity is the ratio
of the actual amount of moisture in 
the air at a given temperature to the
maximum amount of moisture air can
hold at the same temperature.
moisture
ω kg
hg
 Dry air
1 kg
ha
(1 + ω) kg of
moist air
h = ha + ωhg,kJ/kg dry air
FIGURE 14–6
The enthalpy of moist (atmospheric)
air is expressed per unit mass of dry
air, not per unit mass of moist air.
T = 25°C
P = 100 kPa
φ = 75%
ROOM
5 m × 5 m × 3 m
FIGURE 14–7
Schematic for Example 14–1.
EXAMPLE 14–1 The Amount of Water Vapor in Room Air
A 5-m � 5-m � 3-m room shown in Fig. 14–7 contains air at 25°C and 100
kPa at a relative humidity of 75 percent. Determine (a) the partial pressure
of dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dry
air, and (d ) the masses of the dry air and water vapor in the room.
Solution The relative humidity of air in a room is given. The dry air pres-
sure, specific humidity, enthalpy, and the masses of dry air and water vapor
in the room are to be determined.
Assumptions The dry air and the water vapor in the room are ideal gases.
Properties The constant-pressure specific heat of air at room temperature is
cp�1.005 kJ/kg · K (Table A–2a). For water at 25°C, we have Tsat � 3.1698
kPa and hg � 2546.5 kJ/kg (Table A–4).
cen84959_ch14.qxd 4/26/05 4:00 PM Page 720
14–3 � DEW-POINT TEMPERATURE
If you live in a humid area, you are probably used to waking up most summer
mornings and finding the grass wet. You know it did not rain the night before.
So what happened? Well, the excess moisture in the air simply condensed on
the cool surfaces, forming what we call dew. In summer, a considerable
amount of water vaporizes during the day. As the temperature falls during the
Chapter 14 | 721
Analysis (a) The partial pressure of dry air can be determined from Eq. 14–2:
where
Thus,
(b) The specific humidity of air is determined from Eq. 14–8:
(c) The enthalpy of air per unit mass of dry air is determined from
Eq. 14–12:
The enthalpy of water vapor (2546.5 kJ/kg) could also be determined from
the approximation given by Eq. 14–4:
which is almost identical to the value obtained from Table A–4.
(d ) Both the dry air and the water vapor fill the entire room completely.
Therefore, the volume of each gas is equal to the volume of the room:
The masses of the dry air and the water vapor are determined from the ideal-
gas relation applied to each gas separately:
The mass of the water vapor in the air could also be determined from
Eq. 14–6:
mv � vma � 10.0152 2 185.61 kg 2 � 1.30 kg
 mv �
PvVv
RvT
�
12.38 kPa 2 175 m3 2
10.4615 kPa # m3>kg # K 2 1298 K 2 � 1.30 kg
 ma �
PaVa
RaT
�
197.62 kPa 2 175 m3 2
10.287 kPa # m3>kg # K 2 1298 K 2 � 85.61 kg
Va � Vv � Vroom � 15 m 2 15 m 2 13 m 2 � 75 m3
hg @ 25°C � 2500.9 � 1.82 125 2 � 2546.4 kJ>kg
 � 63.8 kJ/kg dry air
 � 11.005 kJ>kg # °C 2 125°C 2 � 10.0152 2 12546.5 kJ>kg 2
 h � ha � vhv � cpT � vhg
v �
0.622Pv
P � Pv
�
10.622 2 12.38 kPa 2
1100 � 2.38 2 kPa � 0.0152 kg H2O/kg dry air
Pa � 1100 � 2.38 2 kPa � 97.62 kPa
Pv � fPg � fPsat @ 25°C � 10.75 2 13.1698 kPa 2 � 2.38 kPa
Pa � P � Pv
cen84959_ch14.qxd 4/26/05 4:00 PM Page 721
night, so does the “moisture capacity” of air, which is the maximum amount
of moisture air can hold. (What happens to the relative humidity during this
process?) After a while, the moisture capacity of air equals its moisture con-
tent. At this point, air is saturated, and its relative humidity is 100 percent.
Any further drop in temperature results in the condensation of some of the
moisture, and this is the beginning of dew formation.
The dew-point temperature Tdp is defined as the temperature at which
condensation begins when the air is cooled at constant pressure. In other
words, Tdp is the saturation temperature of water corresponding to the vapor
pressure:
(14–13)
This is also illustrated in Fig. 14–8. As the air cools at constant pressure, the
vapor pressure Pv remains constant. Therefore, the vapor in the air (state 1)
undergoes a constant-pressure cooling process until it strikes the saturated
vapor line (state 2). The temperature at this point is Tdp, and if the tempera-
ture drops any further, some vapor condenses out. As a result, the amount of
vapor in the air decreases, which results in a decrease in Pv. The air remains
saturated during the condensation process and thus follows a path of
100 percent relative humidity (the saturated vapor line). The ordinary 
temperature and the dew-point temperature of saturated air are identical.
You have probably noticed that when you buy a cold canned drink from a
vending machine on a hot and humid day, dew forms on the can. The for-
mation of dew on the can indicates that the temperature of the drink is
below the dew-point temperature of the surrounding air (Fig. 14–9).
The dew-point temperature of room air can be determined easily by cool-
ing some water in a metal cup by adding small amounts of ice and stirring.
The temperature of the outer surface of the cup when dew starts to form on
the surface is the dew-point temperature of the air.
Tdp � Tsat @ Pv
722 | Thermodynamics
T
s
T1
Tdp
2
1
P v
 =
 c
on
st
.
FIGURE 14–8
Constant-presssure cooling of moist
air and the dew-point temperature on
the T-s diagram of water.
MOIST
AIR
Liquid water
droplets
(dew)
T < Tdp
FIGURE 14–9
When the temperature of a cold drink
is below the dew-point temperature of
the surrounding air, it “sweats.”
COLD
OUTDOORS
10°C
AIR
20°C, 75%
Typical temperature
distribution
16°C
18°C 20°C 20°C 20°C 18°C
16°C
FIGURE 14–10
Schematic for Example 14–2.
EXAMPLE 14–2 Fogging of the Windows in a House
In cold weather, condensation frequently occurs on the inner surfaces of the
windows due to the lower air temperatures near the window surface. Consider
a house, shown in Fig. 14–10, that contains air at 20°C and 75 percent rela-
tive humidity. At what window temperature will the moisture in the air start
condensing on the inner surfaces of the windows?
Solution The interior of a house is maintained at a specified temperature
and humidity. The window temperature at which fogging starts is to be
determined.
Properties The saturation pressure of water at 20°C is Psat � 2.3392 kPa
(Table A–4).
Analysis The temperature distribution in a house, in general, is not uniform.
When the outdoor temperature drops in winter, so does the indoor tempera-
ture near the walls and thewindows. Therefore, the air near the walls and
the windows remains at a lower temperature than at the inner parts of a
house even though the total pressure and the vapor pressure remain constant 
throughout the house. As a result, the air near the walls and the windows
undergoes a Pv � constant cooling process until the moisture in the air
cen84959_ch14.qxd 4/26/05 4:00 PM Page 722
14–4 � ADIABATIC SATURATION AND 
WET-BULB TEMPERATURES
Relative humidity and specific humidity are frequently used in engineering
and atmospheric sciences, and it is desirable to relate them to easily measur-
able quantities such as temperature and pressure. One way of determining
the relative humidity is to determine the dew-point temperature of air,
as discussed in the last section. Knowing the dew-point temperature, we
can determine the vapor pressure Pv and thus the relative humidity. This
approach is simple, but not quite practical.
Another way of determining the absolute or relative humidity is related to
an adiabatic saturation process, shown schematically and on a T-s diagram
in Fig. 14–11. The system consists of a long insulated channel that contains
a pool of water. A steady stream of unsaturated air that has a specific
humidity of v1 (unknown) and a temperature of T1 is passed through this
channel. As the air flows over the water, some water evaporates and mixes
with the airstream. The moisture content of air increases during this process,
and its temperature decreases, since part of the latent heat of vaporization of
the water that evaporates comes from the air. If the channel is long enough,
the airstream exits as saturated air (f � 100 percent) at temperature T2,
which is called the adiabatic saturation temperature.
If makeup water is supplied to the channel at the rate of evaporation at
temperature T2, the adiabatic saturation process described above can be ana-
lyzed as a steady-flow process. The process involves no heat or work inter-
actions, and the kinetic and potential energy changes can be neglected. Then
the conservation of mass and conservation of energy relations for this two-
inlet, one-exit steady-flow system reduces to the following:
Mass balance:
or
m
#
av1 � m
#
f � m
#
av2
 m
#
w1 � m
#
f � m
#
w2
 m
#
a1 � m
#
a2 � m
#
a
Chapter 14 | 723
starts condensing. This happens when the air reaches its dew-point tempera-
ture Tdp, which is determined from Eq. 14–13 to be
where
Thus,
Discussion Note that the inner surface of the window should be maintained
above 15.4°C if condensation on the window surfaces is to be avoided.
Tdp � Tsat @ 1.754 kPa � 15.4 °C
Pv � fPg @ 20°C � 10.75 2 12.3392 kPa 2 � 1.754 kPa
Tdp � Tsat @ Pv
T
s
2
1
Adiabatic
saturation
temperature
Dew-point
temperature
Unsaturated air 
T1, ω1
f1
Saturated air 
T2, ω2
f2 � 100%
1 2
Liquid water
at T2
Liquid water
P v
1
FIGURE 14–11
The adiabatic saturation process and
its representation on a T-s diagram of
water.
(The mass flow rate of dry air
remains constant)
(The mass flow rate of vapor in the
air increases by an amount equal
to the rate of evaporation m. f )
cen84959_ch14.qxd 4/26/05 4:00 PM Page 723
Thus,
Energy balance:
or
Dividing by m
.
a gives
or
which yields
(14–14)
where, from Eq. 14–11b,
(14–15)
since f2 � 100 percent. Thus we conclude that the specific humidity (and
relative humidity) of air can be determined from Eqs. 14–14 and 14–15 by
measuring the pressure and temperature of air at the inlet and the exit of an
adiabatic saturator.
If the air entering the channel is already saturated, then the adiabatic satu-
ration temperature T2 will be identical to the inlet temperature T1, in which
case Eq. 14–14 yields v1 � v2. In general, the adiabatic saturation tempera-
ture is between the inlet and dew-point temperatures.
The adiabatic saturation process discussed above provides a means of
determining the absolute or relative humidity of air, but it requires a long
channel or a spray mechanism to achieve saturation conditions at the exit. A
more practical approach is to use a thermometer whose bulb is covered with
a cotton wick saturated with water and to blow air over the wick, as shown in
Fig. 14–12. The temperature measured in this manner is called the wet-bulb
temperature Twb, and it is commonly used in air-conditioning applications.
The basic principle involved is similar to that in adiabatic saturation.
When unsaturated air passes over the wet wick, some of the water in the
wick evaporates. As a result, the temperature of the water drops, creating a
temperature difference (which is the driving force for heat transfer) between
the air and the water. After a while, the heat loss from the water by evapora-
tion equals the heat gain from the air, and the water temperature stabilizes.
The thermometer reading at this point is the wet-bulb temperature. The wet-
bulb temperature can also be measured by placing the wet-wicked ther-
mometer in a holder attached to a handle and rotating the holder rapidly,
that is, by moving the thermometer instead of the air. A device that works
v2 �
0.622Pg2
P2 � Pg2
v1 �
cp 1T2 � T1 2 � v2hfg2
hg1 � hf2
1cpT1 � v1hg1 2 � 1v2 � v1 2hf2 � 1cpT2 � v2hg2 2
h1 � 1v2 � v1 2hf2 � h2
m
#
a h1 � m
#
a 1v2 � v1 2hf2 � m# a h2
 m
#
a h1 � m
#
f hf2 � m
#
a h2
 E
#
in � E
#
out 1since Q# � 0 and W# � 0 2
m
#
f � m
#
a 1v2 � v1 2
724 | Thermodynamics
Ordinary
thermometer
Wet-bulb
thermometer
Air
flow
Liquid
water
Wick
FIGURE 14–12
A simple arrangement to measure the
wet-bulb temperature.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 724
on this principle is called a sling psychrometer and is shown in Fig. 14–13.
Usually a dry-bulb thermometer is also mounted on the frame of this device
so that both the wet- and dry-bulb temperatures can be read simultaneously.
Advances in electronics made it possible to measure humidity directly in a
fast and reliable way. It appears that sling psychrometers and wet-wicked ther-
mometers are about to become things of the past. Today, hand-held electronic
humidity measurement devices based on the capacitance change in a thin poly-
mer film as it absorbs water vapor are capable of sensing and digitally display-
ing the relative humidity within 1 percent accuracy in a matter of seconds.
In general, the adiabatic saturation temperature and the wet-bulb tempera-
ture are not the same. However, for air–water vapor mixtures at atmospheric
pressure, the wet-bulb temperature happens to be approximately equal to the
adiabatic saturation temperature. Therefore, the wet-bulb temperature Twb
can be used in Eq. 14–14 in place of T2 to determine the specific humidity
of air.
Chapter 14 | 725
Dry-bulb
thermometer
Wet-bulb
thermometer
wick
Wet-bulb
thermometer
FIGURE 14–13
Sling psychrometer.
EXAMPLE 14–3 The Specific and Relative Humidity of Air
The dry- and the wet-bulb temperatures of atmospheric air at 1 atm (101.325
kPa) pressure are measured with a sling psychrometer and determined to be
25 and 15°C, respectively. Determine (a) the specific humidity, (b) the rela-
tive humidity, and (c) the enthalpy of the air.
Solution Dry- and wet-bulb temperatures are given. The specific humidity,
relative humidity, and enthalpy are to be determined.
Properties The saturation pressure of water is 1.7057 kPa at 15°C, and
3.1698 kPa at 25°C (Table A–4). The constant-pressure specific heat of air
at room temperature is cp � 1.005 kJ/kg · K (Table A–2a).
Analysis (a) The specific humidity v1 is determined from Eq. 14–14,
where T2 is the wet-bulb temperature and v2 is
Thus,
(b) The relative humidity f1 is determined from Eq. 14–11a to be
f1 �
v1P2
10.622 � v1 2Pg1 �
10.00653 2 1101.325 kPa 2
10.622 � 0.00653 2 13.1698 kPa 2 � 0.332 or 33.2%
 � 0.00653 kg H2O/kg dry air
 v1 �
11.005 kJ>kg # °C 2 3 115 � 25 2°C 4 � 10.01065 2 12465.4 kJ>kg 2
12546.5 � 62.982 2 kJ>kg
 � 0.01065 kg H2O>kg dry air
 v2 �
0.622Pg2
P2 � Pg2
�
10.622 2 11.7057 kPa 2
1101.325 � 1.7057 2 kPa
v1 �
cp 1T2 � T1 2 �v2hfg2
hg1 � hf2
cen84959_ch14.qxd 4/26/05 4:00 PM Page 725
14–5 � THE PSYCHROMETRIC CHART
The state of the atmospheric air at a specified pressure is completely speci-
fied by two independent intensive properties. The rest of the properties can
be calculated easily from the previous relations. The sizing of a typical air-
conditioning system involves numerous such calculations, which may even-
tually get on the nerves of even the most patient engineers. Therefore, there
is clear motivation to computerize calculations or to do these calculations
once and to present the data in the form of easily readable charts. Such
charts are called psychrometric charts, and they are used extensively in
air-conditioning applications. A psychrometric chart for a pressure of 1 atm
(101.325 kPa or 14.696 psia) is given in Fig. A–31 in SI units and in Fig.
A–31E in English units. Psychrometric charts at other pressures (for use at
considerably higher elevations than sea level) are also available.
The basic features of the psychrometric chart are illustrated in Fig. 14–14.
The dry-bulb temperatures are shown on the horizontal axis, and the spe-
cific humidity is shown on the vertical axis. (Some charts also show the
vapor pressure on the vertical axis since at a fixed total pressure P there is a
one-to-one correspondence between the specific humidity v and the vapor
pressure Pv, as can be seen from Eq. 14–8.) On the left end of the chart,
there is a curve (called the saturation line) instead of a straight line. All the
saturated air states are located on this curve. Therefore, it is also the curve
of 100 percent relative humidity. Other constant relative-humidity curves
have the same general shape.
Lines of constant wet-bulb temperature have a downhill appearance to the
right. Lines of constant specific volume (in m3/kg dry air) look similar, except
they are steeper. Lines of constant enthalpy (in kJ/kg dry air) lie very nearly
parallel to the lines of constant wet-bulb temperature. Therefore, the constant-
wet-bulb-temperature lines are used as constant-enthalpy lines in some charts.
For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures are
identical (Fig. 14–15). Therefore, the dew-point temperature of atmospheric
air at any point on the chart can be determined by drawing a horizontal line (a
line of v � constant or Pv � constant) from the point to the saturated curve.
The temperature value at the intersection point is the dew-point temperature.
The psychrometric chart also serves as a valuable aid in visualizing the air-
conditioning processes. An ordinary heating or cooling process, for example,
appears as a horizontal line on this chart if no humidification or dehumidifica-
tion is involved (that is, v � constant). Any deviation from a horizontal line
indicates that moisture is added or removed from the air during the process.
726 | Thermodynamics
(c) The enthalpy of air per unit mass of dry air is determined from Eq. 14–12:
Discussion The previous property calculations can be performed easily using
EES or other programs with built-in psychrometric functions.
 � 41.8 kJ/kg dry air
 � 11.005 kJ>kg # °C 2 125°C 2 � 10.00653 2 12546.5 kJ>kg 2
 h1 � ha1 � v1hv1 � cpT1 � v1hg1
Dry-bulb temperature
Sp
ec
if
ic
 h
um
id
ity
,ω
Sa
tu
ra
tio
n 
lin
e,
φ
= 
10
0%
φ
= 
co
ns
t.
T
wb = const.
h = const.
v = const.
FIGURE 14–14
Schematic for a psychrometric chart.
Saturation line
Tdp = 15°C
T d
b 
=
 1
5°
C
T
w
b = 15°C
15°C
15°C
FIGURE 14–15
For saturated air, the dry-bulb,
wet-bulb, and dew-point 
temperatures are identical.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 726
14–6 � HUMAN COMFORT AND AIR-CONDITIONING
Human beings have an inherent weakness—they want to feel comfortable.
They want to live in an environment that is neither hot nor cold, neither
humid nor dry. However, comfort does not come easily since the desires of
the human body and the weather usually are not quite compatible. Achieving
comfort requires a constant struggle against the factors that cause discomfort,
such as high or low temperatures and high or low humidity. As engineers, it
is our duty to help people feel comfortable. (Besides, it keeps us employed.)
Chapter 14 | 727
EXAMPLE 14–4 The Use of the Psychrometric Chart
Consider a room that contains air at 1 atm, 35°C, and 40 percent relative
humidity. Using the psychrometric chart, determine (a) the specific humidity,
(b) the enthalpy, (c) the wet-bulb temperature, (d ) the dew-point tempera-
ture, and (e) the specific volume of the air.
Solution The relative humidity of air in a room is given. The specific humid-
ity, enthalpy, wet-bulb temperature, dew-point temperature, and specific vol-
ume of the air are to be determined using the psychrometric chart.
Analysis At a given total pressure, the state of atmospheric air is completely
specified by two independent properties such as the dry-bulb temperature
and the relative humidity. Other properties are determined by directly read-
ing their values at the specified state.
(a) The specific humidity is determined by drawing a horizontal line from the
specified state to the right until it intersects with the v axis, as shown in
Fig. 14–16. At the intersection point we read
(b) The enthalpy of air per unit mass of dry air is determined by drawing a
line parallel to the h � constant lines from the specific state until it inter-
sects the enthalpy scale, giving
(c) The wet-bulb temperature is determined by drawing a line parallel to the
Twb � constant lines from the specified state until it intersects the satura-
tion line, giving
(d ) The dew-point temperature is determined by drawing a horizontal line from
the specified state to the left until it intersects the saturation line, giving
(e) The specific volume per unit mass of dry air is determined by noting the
distances between the specified state and the v � constant lines on both sides
of the point. The specific volume is determined by visual interpolation to be
Discussion Values read from the psychrometric chart inevitably involve read-
ing errors, and thus are of limited accuracy.
v � 0.893 m3/kg dry air
Tdp � 19.4°C
Twb � 24°C
h � 71.5 kJ/kg dry air
v � 0.0142 kg H2O/kg dry air
T = 35°C
Tdp
Twb
h
φ
= 
40
%
ω
v
FIGURE 14–16
Schematic for Example 14–4.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 727
It did not take long for people to realize that they could not change the
weather in an area. All they can do is change it in a confined space such as a
house or a workplace (Fig. 14–17). In the past, this was partially accomplished
by fire and simple indoor heating systems. Today, modern air-conditioning
systems can heat, cool, humidify, dehumidify, clean, and even deodorize the
air–in other words, condition the air to peoples’ desires. Air-conditioning sys-
tems are designed to satisfy the needs of the human body; therefore, it is
essential that we understand the thermodynamic aspects of the body.
The human body can be viewed as a heat engine whose energy input is
food. As with any other heat engine, the human body generates waste heat
that must be rejected to the environment if the body is to continue operat-
ing. The rate of heat generation depends on the level of the activity. For an
average adult male, it is about 87 W when sleeping, 115 W when resting or
doing office work, 230 W when bowling, and 440 W when doing heavy
physical work. The corresponding numbers for an adult female are about
15 percent less. (This difference is due to the body size, not the body 
temperature. The deep-body temperature of a healthy person is maintained
constant at about 37°C.) A body will feel comfortable in environments in
which it can dissipate this waste heat comfortably (Fig. 14–18).
Heat transfer is proportional to the temperature difference. Therefore in
cold environments, a body loses more heat than it normally generates,
which results in a feeling of discomfort. The body tries to minimize the
energy deficit by cuttingdown the blood circulation near the skin (causing a
pale look). This lowers the skin temperature, which is about 34°C for an
average person, and thus the heat transfer rate. A low skin temperature
causes discomfort. The hands, for example, feel painfully cold when the
skin temperature reaches 10°C (50°F). We can also reduce the heat loss
from the body either by putting barriers (additional clothes, blankets, etc.)
in the path of heat or by increasing the rate of heat generation within the
body by exercising. For example, the comfort level of a resting person
dressed in warm winter clothing in a room at 10°C (50°F) is roughly equal
to the comfort level of an identical person doing moderate work in a room
at about �23°C (�10°F). Or we can just cuddle up and put our hands
between our legs to reduce the surface area through which heat flows.
In hot environments, we have the opposite problem—we do not seem to
be dissipating enough heat from our bodies, and we feel as if we are going
to burst. We dress lightly to make it easier for heat to get away from our
bodies, and we reduce the level of activity to minimize the rate of waste
heat generation in the body. We also turn on the fan to continuously replace
the warmer air layer that forms around our bodies as a result of body heat
by the cooler air in other parts of the room. When doing light work or walk-
ing slowly, about half of the rejected body heat is dissipated through perspi-
ration as latent heat while the other half is dissipated through convection and
radiation as sensible heat. When resting or doing office work, most of the
heat (about 70 percent) is dissipated in the form of sensible heat whereas
when doing heavy physical work, most of the heat (about 60 percent) is dis-
sipated in the form of latent heat. The body helps out by perspiring or sweat-
ing more. As this sweat evaporates, it absorbs latent heat from the body and
cools it. Perspiration is not much help, however, if the relative humidity of
728 | Thermodynamics
FIGURE 14–17
We cannot change the weather, but we
can change the climate in a confined
space by air-conditioning.
© Vol. 77/PhotoDisc
23°C
Waste
 heat
37°C
FIGURE 14–18
A body feels comfortable when it can
freely dissipate its waste heat, and no
more.
cen84959_ch14.qxd 4/27/05 10:46 AM Page 728
the environment is close to 100 percent. Prolonged sweating without any
fluid intake causes dehydration and reduced sweating, which may lead to a
rise in body temperature and a heat stroke.
Another important factor that affects human comfort is heat transfer by
radiation between the body and the surrounding surfaces such as walls and
windows. The sun’s rays travel through space by radiation. You warm up in
front of a fire even if the air between you and the fire is quite cold. Likewise,
in a warm room you feel chilly if the ceiling or the wall surfaces are at a
considerably lower temperature. This is due to direct heat transfer between
your body and the surrounding surfaces by radiation. Radiant heaters are
commonly used for heating hard-to-heat places such as car repair shops.
The comfort of the human body depends primarily on three factors: the
(dry-bulb) temperature, relative humidity, and air motion (Fig. 14–19). The
temperature of the environment is the single most important index of com-
fort. Most people feel comfortable when the environment temperature is
between 22 and 27°C (72 and 80°F). The relative humidity also has a con-
siderable effect on comfort since it affects the amount of heat a body can
dissipate through evaporation. Relative humidity is a measure of air’s ability
to absorb more moisture. High relative humidity slows down heat rejection
by evaporation, and low relative humidity speeds it up. Most people prefer a
relative humidity of 40 to 60 percent.
Air motion also plays an important role in human comfort. It removes the
warm, moist air that builds up around the body and replaces it with fresh
air. Therefore, air motion improves heat rejection by both convection and
evaporation. Air motion should be strong enough to remove heat and mois-
ture from the vicinity of the body, but gentle enough to be unnoticed. Most
people feel comfortable at an airspeed of about 15 m/min. Very-high-speed
air motion causes discomfort instead of comfort. For example, an environ-
ment at 10°C (50°F) with 48 km/h winds feels as cold as an environment at
�7°C (20°F) with 3 km/h winds as a result of the body-chilling effect of the
air motion (the wind-chill factor). Other factors that affect comfort are air
cleanliness, odor, noise, and radiation effect.
14–7 � AIR-CONDITIONING PROCESSES
Maintaining a living space or an industrial facility at the desired temperature
and humidity requires some processes called air-conditioning processes.
These processes include simple heating (raising the temperature), simple cool-
ing (lowering the temperature), humidifying (adding moisture), and dehumidi-
fying (removing moisture). Sometimes two or more of these processes are
needed to bring the air to a desired temperature and humidity level.
Various air-conditioning processes are illustrated on the psychrometric
chart in Fig. 14–20. Notice that simple heating and cooling processes appear
as horizontal lines on this chart since the moisture content of the air remains
constant (v � constant) during these processes. Air is commonly heated
and humidified in winter and cooled and dehumidified in summer. Notice
how these processes appear on the psychrometric chart.
Chapter 14 | 729
23°C
f = 50%
Air motion
15 m/min
FIGURE 14–19
A comfortable environment.
© Reprinted with special permission of King
Features Syndicate.
Cooling
Heating
H
um
id
if
yi
ng
D
eh
um
id
if
yi
ng
Co
ol
in
g 
an
d
de
hu
m
id
ify
in
g
He
ati
ng
 an
d
hu
m
id
ify
in
g
FIGURE 14–20
Various air-conditioning processes.
cen84959_ch14.qxd 4/26/05 4:00 PM Page 729
Most air-conditioning processes can be modeled as steady-flow processes,
and thus the mass balance relation m
.
in � m
.
out can be expressed for dry air
and water as
Mass balance for dry air: (14–16)
Mass balance for water: (14–17)
Disregarding the kinetic and potential energy changes, the steady-flow
energy balance relation E
.
in � E
.
out can be expressed in this case as
(14–18)
The work term usually consists of the fan work input, which is small rela-
tive to the other terms in the energy balance relation. Next we examine
some commonly encountered processes in air-conditioning.
Simple Heating and Cooling (V � constant)
Many residential heating systems consist of a stove, a heat pump, or an elec-
tric resistance heater. The air in these systems is heated by circulating it
through a duct that contains the tubing for the hot gases or the electric resis-
tance wires, as shown in Fig. 14–21. The amount of moisture in the air
remains constant during this process since no moisture is added to or
removed from the air. That is, the specific humidity of the air remains con-
stant (v � constant) during a heating (or cooling) process with no humidifi-
cation or dehumidification. Such a heating process proceeds in the direction
of increasing dry-bulb temperature following a line of constant specific
humidity on the psychrometric chart, which appears as a horizontal line.
Notice that the relative humidity of air decreases during a heating process
even if the specific humidity v remains constant. This is because the relative
humidity is the ratio of the moisture content to the moisture capacity of air
at the same temperature, and moisture capacity increases with temperature.
Therefore, the relative humidity of heated air may be well below comfort-
able levels, causing dry skin, respiratory difficulties, and an increase in 
static electricity.
A cooling process at constant specific humidity is similar to the heating
process discussed above, except the dry-bulb temperature decreases and the
relative humidity increases during such a process, as shown in Fig. 14–22.
Cooling can beaccomplished by passing the air over some coils through
which a refrigerant or chilled water flows.
The conservation of mass equations for a heating or cooling process that
involves no humidification or dehumidification reduce to m
.
a1
� m
.
a2
� m
.
a for
dry air and v1 � v2 for water. Neglecting any fan work that may be present,
the conservation of energy equation in this case reduces to
where h1 and h2 are enthalpies per unit mass of dry air at the inlet and the
exit of the heating or cooling section, respectively.
Q
#
� m
#
a 1h2 � h1 2 or q � h2 � h1
Q
#
in � W
#
in �a
in
m
#
h � Q
#
out � W
#
out �a
out
m
#
h
 a
in
m
#
w �a
out
m
#
w or a
in
m
#
a v �a
out
m
#
a v
 a
in
m
#
a �a
out
m
#
a 1kg>s 2
730 | Thermodynamics
ω2 = ω1
Heating coils
Heat
Air T2
T1, ω1, f1
f2 < f1
FIGURE 14–21
During simple heating, specific
humidity remains constant, but relative
humidity decreases.
1
2
12°C 30°C
v = constant
Cooling 
f2 = 80% f1 = 30%
FIGURE 14–22
During simple cooling, specific
humidity remains constant, but relative
humidity increases.
cen84959_ch14.qxd 4/27/05 10:46 AM Page 730
Heating with Humidification
Problems associated with the low relative humidity resulting from simple
heating can be eliminated by humidifying the heated air. This is accom-
plished by passing the air first through a heating section (process 1-2) and
then through a humidifying section (process 2-3), as shown in Fig. 14–23.
The location of state 3 depends on how the humidification is accom-
plished. If steam is introduced in the humidification section, this will result
in humidification with additional heating (T3 � T2). If humidification is
accomplished by spraying water into the airstream instead, part of the latent
heat of vaporization comes from the air, which results in the cooling of the
heated airstream (T3 � T2). Air should be heated to a higher temperature in
the heating section in this case to make up for the cooling effect during the
humidification process.
EXAMPLE 14–5 Heating and Humidification of Air
An air-conditioning system is to take in outdoor air at 10°C and 30 percent
relative humidity at a steady rate of 45 m3/min and to condition it to 25°C
and 60 percent relative humidity. The outdoor air is first heated to 22°C in
the heating section and then humidified by the injection of hot steam in the
humidifying section. Assuming the entire process takes place at a pressure
of 100 kPa, determine (a) the rate of heat supply in the heating section and
(b) the mass flow rate of the steam required in the humidifying section.
Solution Outdoor air is first heated and then humidified by steam injec-
tion. The rate of heat transfer and the mass flow rate of steam are to be
determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of
dry air remains constant during the entire process. 2 Dry air and water vapor
are ideal gases. 3 The kinetic and potential energy changes are negligible.
Properties The constant-pressure specific heat of air at room temperature is
cp � 1.005 kJ/kg · K, and its gas constant is Ra � 0.287 kJ/kg · K (Table
A–2a). The saturation pressure of water is 1.2281 kPa at 10°C, and 3.1698
kPa at 25°C. The enthalpy of saturated water vapor is 2519.2 kJ/kg at 10°C,
and 2541.0 kJ/kg at 22°C (Table A–4).
Analysis We take the system to be the heating or the humidifying section,
as appropriate. The schematic of the system and the psychrometric chart of
the process are shown in Fig. 14–24. We note that the amount of water
vapor in the air remains constant in the heating section (v1 � v2) but
increases in the humidifying section (v3 � v2).
(a) Applying the mass and energy balances on the heating section gives
Dry air mass balance:
Water mass balance:
Energy balance:
The psychrometric chart offers great convenience in determining the properties
of moist air. However, its use is limited to a specified pressure only, which is 1
atm (101.325 kPa) for the one given in the appendix. At pressures other than
 Q
#
in � m
#
a h1 � m
#
a h2 S Q
#
in � m
#
a 1h2 � h1 2
 m
#
a1v1 � m
#
a2v2 S v1 � v2
 m
#
a1 � m
#
a2 � m
#
a
Chapter 14 | 731
Air
Heating 
coils
ω2 = ω1
1 2 3
Heating
section
Humidifying
section
ω3 > ω2
Humidifier
FIGURE 14–23
Heating with humidification.
·
Air
1 2 3
V1 = 45 m
3/min
10°C 22°C 25°C
1 2
3
1
 = 
30
%
3
= 6
0%
T1 = 10°C
T2 = 22°C
T3 = 25°C
Humidifier
f1 = 30% f3 = 60%
Heating 
coils
f
f
FIGURE 14–24
Schematic and psychrometric chart for
Example 14–5.
cen84959_ch14.qxd 4/26/05 4:01 PM Page 731
1 atm, either other charts for that pressure or the relations developed earlier
should be used. In our case, the choice is clear:
since v2 � v1. Then the rate of heat transfer to air in the heating section
becomes
(b) The mass balance for water in the humidifying section can be expressed as
or
where
Thus,
Discussion The result 0.539 kg/min corresponds to a water requirement of
close to one ton a day, which is significant.
Cooling with Dehumidification
The specific humidity of air remains constant during a simple cooling
process, but its relative humidity increases. If the relative humidity reaches
undesirably high levels, it may be necessary to remove some moisture from
the air, that is, to dehumidify it. This requires cooling the air below its dew-
point temperature.
 � 0.539 kg/min
 m
#
w � 155.2 kg>min 2 10.01206 � 0.0023 2
 � 0.01206 kg H2O>kg dry air
 v3 �
0.622f3Pg3
P3 � f3Pg3
�
0.622 10.60 2 13.1698 kPa 2
3100 � 10.60 2 13.1698 2 4 kPa
m
#
w � m
#
a 1v3 � v2 2
m
#
a2v2 � m
#
w � m
#
a3v3
 � 673 kJ/min
 Q
#
in � m
#
a 1h2 � h1 2 � 155.2 kg>min 2 3 128.0 � 15.8 2 kJ>kg 4
 � 28.0 kJ>kg dry air
 h2 � cpT2 � v2hg2 � 11.005 kJ>kg # °C 2 122°C 2 � 10.0023 2 12541.0 kJ>kg 2
 � 15.8 kJ>kg dry air
 h1 � cpT1 � v1hg1 � 11.005 kJ>kg # °C 2 110°C 2 � 10.0023 2 12519.2 kJ>kg 2
 v1 �
0.622Pv1
P1 � Pv1
�
0.622 10.368 kPa 2
1100 � 0.368 2 kPa � 0.0023 kg H2O>kg dry air
 m
#
a �
V
#
1
v1
�
45 m3>min
0.815 m3>kg � 55.2 kg>min
 v1 �
RaT1
Pa
�
10.287 kPa # m3>kg # K 2 1283 K 2
99.632 kPa
� 0.815 m3>kg dry air
 Pa1 � P1 � Pv1 � 1100 � 0.368 2 kPa � 99.632 kPa
 Pv1 � f1Pg1 � fPsat @ 10°C � 10.3 2 11.2281 kPa 2 � 0.368 kPa
732 | Thermodynamics
cen84959_ch14.qxd 4/26/05 4:01 PM Page 732
The cooling process with dehumidifying is illustrated schematically and
on the psychrometric chart in Fig. 14–25 in conjunction with Example
14–6. Hot, moist air enters the cooling section at state 1. As it passes
through the cooling coils, its temperature decreases and its relative humidity
increases at constant specific humidity. If the cooling section is sufficiently
long, air reaches its dew point (state x, saturated air). Further cooling of air
results in the condensation of part of the moisture in the air. Air remains sat-
urated during the entire condensation process, which follows a line of 100
percent relative humidity until the final state (state 2) is reached. The water
vapor that condenses out of the air during this process is removed from the
cooling section through a separate channel. The condensate is usually
assumed to leave the cooling section at T2.
The cool, saturated air at state 2 is usually routed directly to the room,
where it mixes with the room air. In some cases, however, the air at state 2
may be at the right specific humidity but at a very low temperature. In such
cases, air is passed through a heating section where its temperature is raised
to a more comfortable level before it is routed to the room.
EXAMPLE 14–6 Cooling and Dehumidification of Air
Air enters a window air conditioner at 1 atm, 30°C, and 80 percent relative
humidity at a rate of 10 m3/min, and it leaves as saturated air at 14°C. Part
of the moisture in the air that condenses during the process is also removed
at 14°C. Determine the rates of heat and moisture removal from the air.
Solution Air is cooled and dehumidified by a window air conditioner. The
rates of heat andmoisture removal are to be determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry
air remains constant during the entire process. 2 Dry air and the water vapor
are ideal gases. 3 The kinetic and potential energy changes are negligible.
Properties The enthalpy of saturated liquid water at 14°C is 58.8 kJ/kg
(Table A–4). Also, the inlet and the exit states of the air are completely spec-
ified, and the total pressure is 1 atm. Therefore, we can determine the prop-
erties of the air at both states from the psychrometric chart to be
h1 � 85.4 kJ/kg dry air h2 � 39.3 kJ/kg dry air
v1 � 0.0216 kg H2O/kg dry air and v2 � 0.0100 kg H2O/kg dry air
v1 � 0.889 m3/kg dry air
Analysis We take the cooling section to be the system. The schematic of
the system and the psychrometric chart of the process are shown in Fig.
14–25. We note that the amount of water vapor in the air decreases during
the process (v2 � v1) due to dehumidification. Applying the mass and
energy balances on the cooling and dehumidification section gives
Dry air mass balance:
Water mass balance:
Energy balance: a
in
m
#
h � Q
#
out �a
out
m
#
h S Q
#
out � m
# 1h1 � h2 2 � m# w hw
 m
#
a1v1 � m
#
a2v2 � m
#
w S m
#
w � m
#
a 1v1 � v2 2
 m
#
a1 � m
#
a2 � m
#
a
Chapter 14 | 733
·
Air
Cooling coils
2 1
14°C 30°C
1
2
f1 = 80%
f2 = 100%
 
T2 = 14°C
f2 = 100%
T1 = 30°C
f1 = 80%
V1 = 10 m
3/minCondensate
removal
14°C
Condensate
x
FIGURE 14–25
Schematic and psychrometric chart for
Example 14–6.
cen84959_ch14.qxd 4/26/05 4:01 PM Page 733
Then,
Therefore, this air-conditioning unit removes moisture and heat from the air
at rates of 0.131 kg/min and 511 kJ/min, respectively.
Evaporative Cooling
Conventional cooling systems operate on a refrigeration cycle, and they can
be used in any part of the world. But they have a high initial and operating
cost. In desert (hot and dry) climates, we can avoid the high cost of cooling
by using evaporative coolers, also known as swamp coolers.
Evaporative cooling is based on a simple principle: As water evaporates,
the latent heat of vaporization is absorbed from the water body and the sur-
rounding air. As a result, both the water and the air are cooled during the
process. This approach has been used for thousands of years to cool water.
A porous jug or pitcher filled with water is left in an open, shaded area.
A small amount of water leaks out through the porous holes, and the pitcher
“sweats.” In a dry environment, this water evaporates and cools the remain-
ing water in the pitcher (Fig. 14–26).
You have probably noticed that on a hot, dry day the air feels a lot cooler
when the yard is watered. This is because water absorbs heat from the air as
it evaporates. An evaporative cooler works on the same principle. The evap-
orative cooling process is shown schematically and on a psychrometric chart
in Fig. 14–27. Hot, dry air at state 1 enters the evaporative cooler, where it
is sprayed with liquid water. Part of the water evaporates during this process
by absorbing heat from the airstream. As a result, the temperature of the
airstream decreases and its humidity increases (state 2). In the limiting case,
the air leaves the evaporative cooler saturated at state 2�. This is the lowest
temperature that can be achieved by this process.
The evaporative cooling process is essentially identical to the adiabatic satu-
ration process since the heat transfer between the airstream and the surround-
ings is usually negligible. Therefore, the evaporative cooling process follows a
line of constant wet-bulb temperature on the psychrometric chart. (Note that
this will not exactly be the case if the liquid water is supplied at a temperature
different from the exit temperature of the airstream.) Since the constant-wet-
bulb-temperature lines almost coincide with the constant-enthalpy lines, the
enthalpy of the airstream can also be assumed to remain constant. That is,
(14–19)
and
(14–20)
during an evaporative cooling process. This is a reasonably accurate approx-
imation, and it is commonly used in air-conditioning calculations.
h � constant
Twb � constant
 � 511 kJ/min
 Q
#
out � 111.25 kg>min 2 3 185.4 � 39.3 2 kJ>kg 4 � 10.131 kg>min 2 158.8 kJ>kg 2
 m
#
w � 111.25 kg>min 2 10.0216 � 0.0100 2 � 0.131 kg/min
 m
#
a �
V
#
1
v1
�
10 m3>min
0.889 m3>kg dry air � 11.25 kg>min
734 | Thermodynamics
Water that
leaks out
Hot, dry
air
FIGURE 14–26
Water in a porous jug left in an open,
breezy area cools as a result of
evaporative cooling.
HOT, 
DRY
AIR
2 1
1
2
2'
COOL, 
MOIST
AIR
Liquid
water
Twb = const.
 h = const.
~
~
FIGURE 14–27
Evaporative cooling.
cen84959_ch14.qxd 4/26/05 4:01 PM Page 734
EXAMPLE 14–7 Evaporative Cooling of Air by a Swamp Cooler
Air enters an evaporative (or swamp) cooler at 14.7 psi, 95°F, and 20 percent
relative humidity, and it exits at 80 percent relative humidity. Determine
(a) the exit temperature of the air and (b) the lowest temperature to which
the air can be cooled by this evaporative cooler.
Solution Air is cooled steadily by an evaporative cooler. The temperature
of discharged air and the lowest temperature to which the air can be cooled
are to be determined.
Analysis The schematic of the evaporative cooler and the psychrometric
chart of the process are shown in Fig. 14–28.
(a) If we assume the liquid water is supplied at a temperature not much dif-
ferent from the exit temperature of the airstream, the evaporative cooling
process follows a line of constant wet-bulb temperature on the psychrometric
chart. That is,
The wet-bulb temperature at 95°F and 20 percent relative humidity is deter-
mined from the psychrometric chart to be 66.0°F. The intersection point of
the Twb � 66.0°F and the f � 80 percent lines is the exit state of the air.
The temperature at this point is the exit temperature of the air, and it is
determined from the psychrometric chart to be
(b) In the limiting case, air leaves the evaporative cooler saturated (f � 100
percent), and the exit state of the air in this case is the state where the Twb
� 66.0°F line intersects the saturation line. For saturated air, the dry- and
the wet-bulb temperatures are identical. Therefore, the lowest temperature to
which air can be cooled is the wet-bulb temperature, which is
Discussion Note that the temperature of air drops by as much as 30°F in
this case by evaporative cooling.
Adiabatic Mixing of Airstreams
Many air-conditioning applications require the mixing of two airstreams.
This is particularly true for large buildings, most production and process
plants, and hospitals, which require that the conditioned air be mixed with a
certain fraction of fresh outside air before it is routed into the living space.
The mixing is accomplished by simply merging the two airstreams, as
shown in Fig. 14–29.
The heat transfer with the surroundings is usually small, and thus the mix-
ing processes can be assumed to be adiabatic. Mixing processes normally
involve no work interactions, and the changes in kinetic and potential ener-
gies, if any, are negligible. Then the mass and energy balances for the adia-
batic mixing of two airstreams reduce to
Mass of dry air: (14–21)
Mass of water vapor: (14–22)
Energy: (14–23) m
#
a1h1 � m
#
a2h2 � m
#
a3h3
 v1m
#
a1 � v2m
#
a2 � v3m
#
a3
 m
#
a1 � m
#
a2 � m
#
a3
Tmin � T2¿ � 66.0°F
T2 � 70.4°F
Twb � constant
Chapter 14 | 735
2' 1
1
2
2'
AIR
Tmin T2 95°F
2
T1 = 95°F
f = 20%
P = 14.7 psia
f 2
 =
 8
0%
f 1
 =
 2
0%
FIGURE 14–28
Schematic and psychrometric chart for
Example 14–7.
cen84959_ch14.qxd 4/26/05 4:01 PM Page 735
Eliminating m
.
a3
from the relations above, we obtain
(14–24)
This equation has an instructive geometric interpretation on the psychro-
metric chart. It shows that the ratio of v2 � v3 to v3 � v1 is equal to the
ratio of m
.
a1
to m
.
a2. The states that satisfy this condition are indicated by the
dashed line AB. The ratio of h2 � h3 toh3 � h1 is also equal to the ratio of
m
.
a1
to m
.
a2
, and the states that satisfy this condition are indicated by the dashed
line CD. The only state that satisfies both conditions is the intersection point
of these two dashed lines, which is located on the straight line connecting
states 1 and 2. Thus we conclude that when two airstreams at two different
states (states 1 and 2) are mixed adiabatically, the state of the mixture (state 3)
lies on the straight line connecting states 1 and 2 on the psychrometric chart,
and the ratio of the distances 2-3 and 3-1 is equal to the ratio of mass flow
rates m
.
a1
and m
.
a2
.
The concave nature of the saturation curve and the conclusion above lead
to an interesting possibility. When states 1 and 2 are located close to the sat-
uration curve, the straight line connecting the two states will cross the satu-
ration curve, and state 3 may lie to the left of the saturation curve. In this
case, some water will inevitably condense during the mixing process.
EXAMPLE 14–8 Mixing of Conditioned Air with Outdoor Air
Saturated air leaving the cooling section of an air-conditioning system at
14°C at a rate of 50 m3/min is mixed adiabatically with the outside air at
32°C and 60 percent relative humidity at a rate of 20 m3/min. Assuming that
the mixing process occurs at a pressure of 1 atm, determine the specific
humidity, the relative humidity, the dry-bulb temperature, and the volume
flow rate of the mixture.
Solution Conditioned air is mixed with outside air at specified rates. The
specific and relative humidities, dry-bulb temperature, and the flow rate of
the mixture are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor
are ideal gases. 3 The kinetic and potential energy changes are negligible.
4 The mixing section is adiabatic.
Properties The properties of each inlet stream are determined from the psy-
chrometric chart to be
and
 v2 � 0.889 m
3>kg dry air
 v2 � 0.0182 kg H2O>kg dry air
 h2 � 79.0 kJ>kg dry air
 v1 � 0.826 m
3>kg dry air
 v1 � 0.010 kg H2O>kg dry air
 h1 � 39.4 kJ>kg dry air
m
#
a1
m
#
a2
�
v2 � v3
v3 � v1
�
h2 � h3
h3 � h1
736 | Thermodynamics
1
2
3
A
C
h2
h3
h1
D
ω2 – ω3
ω3 – ω1
ω2
ω3
ω1
B
2
1
3Mixing
section
ω1
 h1
ω2
 h2
ω 3
 h3
h2 – h3
h3 – h1
FIGURE 14–29
When two airstreams at states 1 and 2
are mixed adiabatically, the state of
the mixture lies on the straight line
connecting the two states.
cen84959_ch14.qxd 4/26/05 4:01 PM Page 736
Chapter 14 | 737
Analysis We take the mixing section of the streams as the system. The
schematic of the system and the psychrometric chart of the process are
shown in Fig. 14–30. We note that this is a steady-flow mixing process.
The mass flow rates of dry air in each stream are
From the mass balance of dry air,
The specific humidity and the enthalpy of the mixture can be determined
from Eq. 14–24,
which yield
These two properties fix the state of the mixture. Other properties of the mix-
ture are determined from the psychrometric chart:
Finally, the volume flow rate of the mixture is determined from
Discussion Notice that the volume flow rate of the mixture is approximately
equal to the sum of the volume flow rates of the two incoming streams. This
is typical in air-conditioning applications.
Wet Cooling Towers
Power plants, large air-conditioning systems, and some industries generate
large quantities of waste heat that is often rejected to cooling water from
nearby lakes or rivers. In some cases, however, the cooling water supply is
limited or thermal pollution is a serious concern. In such cases, the waste
heat must be rejected to the atmosphere, with cooling water recirculating
and serving as a transport medium for heat transfer between the source and
the sink (the atmosphere). One way of achieving this is through the use of
wet cooling towers.
A wet cooling tower is essentially a semienclosed evaporative cooler. An
induced-draft counterflow wet cooling tower is shown schematically in
V
#
3 � m
#
a3v3 � 183 kg>min 2 10.844 m3>kg 2 � 70.1 m3/min
 v3 � 0.844 m
3>kg dry air
 f3 � 89%
 T3 � 19.0°C
 h3 � 50.1 kJ>kg dry air
 v3 � 0.0122 kg H2O/kg dry air
 
60.5
22.5
�
0.0182 � v3
v3 � 0.010
�
79.0 � h3
h3 � 39.4
 
m
#
a1
m
#
a2
�
v2 � v3
v3 � v1
�
h2 � h3
h3 � h1
m
#
a3 � m
#
a1 � m
#
a2 � 160.5 � 22.5 2 kg>min � 83 kg>min
 m
#
a2 �
V
#
2
v2
�
20 m3>min
0.889 m3>kg dry air � 22.5 kg>min
 m
#
a1 �
V
#
1
v1
�
50 m3>min
0.826 m3>kg dry air � 60.5 kg>min
·
·
T2 = 32°C
f2 = 60%
V2 = 20 m
3/min
Saturated air
T1 = 14°C
V1 = 50 m
3/min
2
1
3Mixing
section
P = 1 atm
V3
v3
f3
T3
23
1
14°C 32°C
f 1
 =
 1
00
%
f
2 
= 
60
%
·
FIGURE 14–30
Schematic and psychrometric chart for
Example 14–8.
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Fig. 14–31. Air is drawn into the tower from the bottom and leaves through
the top. Warm water from the condenser is pumped to the top of the tower
and is sprayed into this airstream. The purpose of spraying is to expose a
large surface area of water to the air. As the water droplets fall under the
influence of gravity, a small fraction of water (usually a few percent) evapo-
rates and cools the remaining water. The temperature and the moisture content
of the air increase during this process. The cooled water collects at the bottom
of the tower and is pumped back to the condenser to absorb additional waste
heat. Makeup water must be added to the cycle to replace the water lost by
evaporation and air draft. To minimize water carried away by the air, drift
eliminators are installed in the wet cooling towers above the spray section.
The air circulation in the cooling tower described is provided by a fan,
and therefore it is classified as a forced-draft cooling tower. Another popular
type of cooling tower is the natural-draft cooling tower, which looks like a
large chimney and works like an ordinary chimney. The air in the tower has
a high water-vapor content, and thus it is lighter than the outside air. Conse-
quently, the light air in the tower rises, and the heavier outside air fills the
vacant space, creating an airflow from the bottom of the tower to the top.
The flow rate of air is controlled by the conditions of the atmospheric air.
Natural-draft cooling towers do not require any external power to induce the
air, but they cost a lot more to build than forced-draft cooling towers. The
natural-draft cooling towers are hyperbolic in profile, as shown in Fig.
14–32, and some are over 100 m high. The hyperbolic profile is for greater
structural strength, not for any thermodynamic reason.
The idea of a cooling tower started with the spray pond, where the warm
water is sprayed into the air and is cooled by the air as it falls into the pond,
as shown in Fig. 14–33. Some spray ponds are still in use today. However,
they require 25 to 50 times the area of a cooling tower, water loss due to air
drift is high, and they are unprotected against dust and dirt.
We could also dump the waste heat into a still cooling pond, which is
basically a large artificial lake open to the atmosphere. Heat transfer from
the pond surface to the atmosphere is very slow, however, and we would
need about 20 times the area of a spray pond in this case to achieve the
same cooling.
EXAMPLE 14–9 Cooling of a Power Plant by a Cooling Tower
Cooling water leaves the condenser of a power plant and enters a wet cooling
tower at 35°C at a rate of 100 kg/s. Water is cooled to 22°C in the cooling
tower by air that enters the tower at 1 atm, 20°C, and 60 percent relative
humidity and leaves saturated at 30°C. Neglecting the power input to the
fan, determine (a) the volume flow rate of air into the cooling tower and (b)
the mass flow rate of the required makeup water.
Solution Warm cooling water from a power plant is cooled in a wet cooling
tower. The flow rates of makeup water and air are to be determined.
Assumptions 1 Steady operating conditions exist and thus the mass flowrate of dry air remains constant during the entire process. 2 Dry air and the
water vapor are ideal gases. 3 The kinetic and potential energy changes are
negligible. 4 The cooling tower is adiabatic.
738 | Thermodynamics
WARM
WATER
COOL
WATER
AIR
INLET
FIGURE 14–32
A natural-draft cooling tower.
FIGURE 14–33
A spray pond.
Photo by Yunus Çengel.
COOL
WATER
AIR EXIT
WARM
WATER
AIR
INLET
FAN
FIGURE 14–31
An induced-draft counterflow cooling
tower.
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Chapter 14 | 739
COOL
WATER
WARM
WATER
AIR
35°C
100 kg/s
100 kg/s
System
boundary
4
1
3
2
1 atm
20°C
f1 = 60%
30°C
f2 = 100%
V1
Makeup 
water
22°C
·
FIGURE 14–34
Schematic for Example 14–9.
Properties The enthalpy of saturated liquid water is 92.28 kJ/kg at 22°C
and 146.64 kJ/kg at 35°C (Table A–4). From the psychrometric chart,
h1 � 42.2 kJ/kg dry air h2 � 100.0 kJ/kg dry air
v1 � 0.0087 kg H2O/kg dry air v2 � 0.0273 kg H2O/kg dry air
v1 � 0.842 m3/kg dry air
Analysis We take the entire cooling tower to be the system, which is shown
schematically in Fig. 14–34. We note that the mass flow rate of liquid water
decreases by an amount equal to the amount of water that vaporizes in the
tower during the cooling process. The water lost through evaporation must be
made up later in the cycle to maintain steady operation.
(a) Applying the mass and energy balances on the cooling tower gives
Dry air mass balance:
Water mass balance:
or
Energy balance:
or
Solving for m·a gives
Substituting,
Then the volume flow rate of air into the cooling tower becomes
(b) The mass flow rate of the required makeup water is determined from
Discussion Note that over 98 percent of the cooling water is saved and
recirculated in this case.
m
#
makeup � m
#
 a 1v2 � v1 2 � 196.9 kg>s 2 10.0273 � 0.0087 2 � 1.80 kg/s
V
#
1 � m
#
av1 � 196.9 kg>s 2 10.842 m3>kg 2 � 81.6 m3/s
m
#
a �
1100 kg>s 2 3 1146.64 � 92.28 2 kJ>kg 4
3 1100.0 � 42.2 2 kJ>kg 4 � 3 10.0273 � 0.0087 2 192.28 2 kJ>kg 4 � 96.9 kg>s
m
#
a �
m
#
3 1h3 � h4 2
1h2 � h1 2 � 1v2 � v1 2h4
m
#
3 h3 � m
#
a 1h2 � h1 2 � 1m# 3 � m# makeup 2h4
a
in
m
#
h �a
out
m
#
h S m# a1h1 � m
#
3 h3 � m
#
a2h2 � m
#
4 h4
m
#
3 � m
#
4 � m
#
a 1v 2 � v1 2 � m# makeup 
 m
#
3 � m
#
a1v1 � m
#
4 � m
#
a2v2 
 m
#
a1 � m
#
a2 � m
#
a 
SUMMARY
In this chapter we discussed the air–water-vapor mixture,
which is the most commonly encountered gas–vapor mixture
in practice. The air in the atmosphere normally contains some
water vapor, and it is referred to as atmospheric air. By con-
trast, air that contains no water vapor is called dry air. In the
temperature range encountered in air-conditioning applica-
tions, both the dry air and the water vapor can be treated as
ideal gases. The enthalpy change of dry air during a process
can be determined from
The atmospheric air can be treated as an ideal-gas mixture
whose pressure is the sum of the partial pressure of dry air Pa
and that of the water vapor Pv,
P � Pa � Pv
¢hdry air � cp ¢T � 11.005 kJ>kg # °C 2 ¢T
cen84959_ch14.qxd 4/26/05 4:01 PM Page 739
740 | Thermodynamics
The enthalpy of water vapor in the air can be taken to be equal
to the enthalpy of the saturated vapor at the same temperature:
hv(T, low P) � hg(T) � 2500.9 � 1.82T (kJ/kg) T in °C
� 1060.9 � 0.435T (Btu/lbm) T in °F
in the temperature range �10 to 50°C (15 to 120°F).
The mass of water vapor present per unit mass of dry air is
called the specific or absolute humidity v,
where P is the total pressure of air and Pv is the vapor pres-
sure. There is a limit on the amount of vapor the air can hold
at a given temperature. Air that is holding as much moisture
as it can at a given temperature is called saturated air. The
ratio of the amount of moisture air holds (mv) to the maxi-
mum amount of moisture air can hold at the same tempera-
ture (mg) is called the relative humidity f,
where Pg � Psat @ T. The relative and specific humidities can
also be expressed as
Relative humidity ranges from 0 for dry air to 1 for saturated
air.
The enthalpy of atmospheric air is expressed per unit mass
of dry air, instead of per unit mass of the air–water-vapor
mixture, as
The ordinary temperature of atmospheric air is referred to
as the dry-bulb temperature to differentiate it from other forms
of temperatures. The temperature at which condensation begins
if the air is cooled at constant pressure is called the dew-point
temperature Tdp:
Relative humidity and specific humidity of air can be deter-
mined by measuring the adiabatic saturation temperature of
air, which is the temperature air attains after flowing over
water in a long adiabatic channel until it is saturated,
v1 �
cp 1T2 � T1 2 � v2hfg2
hg1 � hf2
Tdp � Tsat @ Pv
h � ha � vhg 1kJ>kg dry air 2
f �
vP
10.622 � v 2Pg and v �
0.622fPg
P � fPg
f �
mv
mg
�
Pv
Pg
v �
mv
ma
�
0.622Pv
P � Pv
1kg H2O>kg dry air 2
where
and T2 is the adiabatic saturation temperature. A more practi-
cal approach in air-conditioning applications is to use a ther-
mometer whose bulb is covered with a cotton wick saturated
with water and to blow air over the wick. The temperature
measured in this manner is called the wet-bulb temperature
Twb, and it is used in place of the adiabatic saturation temper-
ature. The properties of atmospheric air at a specified total
pressure are presented in the form of easily readable charts,
called psychrometric charts. The lines of constant enthalpy
and the lines of constant wet-bulb temperature are very
nearly parallel on these charts.
The needs of the human body and the conditions of the
environment are not quite compatible. Therefore, it often
becomes necessary to change the conditions of a living space
to make it more comfortable. Maintaining a living space or
an industrial facility at the desired temperature and humidity
may require simple heating (raising the temperature), simple
cooling (lowering the temperature), humidifying (adding
moisture), or dehumidifying (removing moisture). Sometimes
two or more of these processes are needed to bring the air to
the desired temperature and humidity level.
Most air-conditioning processes can be modeled as steady-
flow processes, and therefore they can be analyzed by apply-
ing the steady-flow mass (for both dry air and water) and
energy balances,
Dry air mass:
Water mass:
Energy:
The changes in kinetic and potential energies are assumed to
be negligible.
During a simple heating or cooling process, the specific
humidity remains constant, but the temperature and the rela-
tive humidity change. Sometimes air is humidified after it is
heated, and some cooling processes include dehumidification.
In dry climates, air can be cooled via evaporative cooling by
passing it through a section where it is sprayed with water. In
locations with limited cooling water supply, large amounts of
waste heat can be rejected to the atmosphere with minimum
water loss through the use of cooling towers.
 Q
#
in � W
#
in �a
in
m
#
h � Q
#
out � W
#
out �a
out
m
#
h 
 a
in
m
#
w �a
out
m
#
w or a
in
m
#
av �a
out
m
#
av 
 a
in
m
#
a �a
out
m
#
a 
v2 �
0.622Pg2
P2 � Pg2
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