Chapter 15 Ideal Gas Mixtures and
Combustion
15.1 Introduction
In the United States, homes and commercial buildings consume about 36% of the total energy
used; buildings also use two-thirds of all electricity generated nationally. Maintaining a relatively constant
temperature and humidity in buildings, houses, factories, and commercial establishments consumes a large
fraction of this energy use. The systems used to accomplish this task may involve either the cooling or
heating of an airflow and, possibly, humidification or dehumidification (respectively, the addition or
removal of water vapor from an air stream) so that people are comfortable in the conditioned environment.
Note that some equipment (e.g., computers and other electronic devices) and processes (e.g., textiles and
printing industries) also require temperature and humidity levels within specified ranges to ensure proper
operation or quality. In a cooling application, when we couple a refrigeration cycle to an air handling
system, we obtain the traditional “air conditioning” system.
In Chapter 8 we studied the refrigeration cycle. In Chapters 2, 5, and 7 we studied properties of
non-reacting pure fluids, such as water, refrigerants, and ideal gases. Note that even though air is
composed of several different pure gases (oxygen, nitrogen, argon, carbon dioxide, etc.) we treated it as if it
were a pure single-component gas.
There are many applications in which the thermodynamic properties of mixtures of gases must be
evaluated. Anytime we use air properties we use the results of such an evaluation. When we deal with air
conditioning, we must account for water vapor contained in the air. Other situations in which gas mixtures
are encountered are in cooling towers, chemical process plants, and in the products of combustion of fossil
fuels. In previous chapters we dealt with mixtures of the same substance but with different phases (e.g.,
liquid water and water vapor in a two-phase mixture). In this chapter we examine the thermodynamic
properties of mixtures of different substances. We focus exclusively on mixtures of ideal gases, including
applications in which water vapor is one of the ideal gases.
While non-reacting fluids (e.g., water, air) and processes are used in man applications, situations
in which reactions occur also are common. For example, heat input to Rankine, Brayton, Otto, and Diesel
cycles (discussed in Chapter 8) often is accomplished through the combustion of a fossil fuel, either refined
oil (e.g., kerosene, fuel oil, jet fuel, gasoline), coal, or natural gas. In previous chapters we ignored the
actual combustion process—the chemical reaction between a fuel and an oxidant that releases energy stored
in the hydrocarbon fuel. In this chapter we develop the thermodynamic methods needed to analyze reacting
mixtures.
15.2 Ideal Gas Mixtures
An ideal gas mixture is a mixture of two or more ideal gases that do not react. In typical
applications (e.g., the heat transfer from the products of combustion of a fuel to the water in a boiler), we
need the thermodynamic properties of the gas mixture for use with conservation of mass and energy.
Because a nearly infinite number of mixture compositions is possible, it is not feasible to tabulate mixture
properties. Instead, we apply to the ideal gas mixture all the ideal gas equations we have already derived.
For example, by appropriate definition and use, the ideal gas equation, PV = mRT M , can be used for a
pure gas or for a mixture of pure gases (as we have assumed whenever we have dealt with air). We use the
properties of the individual gases in the mixture to evaluate the properties of the mixture.
Several relationships can be developed for use in evaluating the thermodynamic properties of ideal
gas mixtures. To begin the development of mixture properties, consider a mixture of several gases. Each
gas, i, has a mass, mi, and a molecular weight, Mi. We add the masses of each gas to obtain the total mass
in the mixture:
k
mmix = m1 + m2 + m3 + " = ∑ mi
Eq. S15-1
i =1
We divide this expression by the total mixture mass, mmix:
1=
k
k
m
m
m1
m
+ 2 + 3 +" = ∑ i = ∑ Xi
mmix mmix mmix
i =1 mmix
i =1
Eq. S15-2
The quantity on the right hand side of the equation is defined as the mass fraction, X i = mi mmix , and the
sum of all the individual mass fractions is unity.
In addition to mass, we often describe mixtures in terms of the number of moles of each
component present, ni, which is defined in terms of the mass, mi, and molecular weight, Mi, of the
component i:
ni =
mi
Mi
Eq. S15-3
The total number of moles in a mixture, nmix, is simply the contribution from each gas:
k
nmix = n1 + n2 + n3 + " = ∑ ni
Eq. S15-4
i =1
We divide this expression by the total number of moles in the mixture, nmix:
1=
k
k
n
n
n1
n
+ 2 + 3 + " = ∑ i = ∑ Yi
nmix nmix nmix
i =1 nmix
i =1
Eq. S15-5
The quantity of the right hand side of the equation is defined as the mole fraction, Yi = ni nmix , and the sum
of all the individual mole fractions is unity.
Using Eq. S15-3 we can develop an expression for the apparent molecular weight, Mmix, of a
mixture:
nmix =
mmix
M mix
Eq. S15-6
Solving Eq. S15-6 for the apparent molecular weight and incorporating Eq. S15-3, Eq. S15-4, and Eq.
S15-5:
M mix =
mmix
1
1
=
( n1 M 1 + n2 M 2 + n3 M 3 + ") =
nmix nmix
nmix
k
k
∑n M = ∑
i =1
i
i
i =1
k
ni M i
= ∑ Yi M i
nmix
i =1
Eq. S15-7
This equation states that the apparent molecular weight of the mixture is calculated using a mole-weighted
average of the molecular weights of each gas. The mass fraction also can be used to calculate the mixture
apparent molecular weight. With a derivation similar to that which produced Eq. S15-7, we obtain:
2
k
1
1
= ∑ Xi
M mix i =1
Mi
Eq. S15-8
The relationships developed above apply, in general, to all mixtures. We now want to use them
with ideal gas mixtures. Two models—Dalton’s law of additive pressures and Amagat’s law of additive
volumes—are idealizations that give identical results when applied to mixtures of ideal gases. The basic
idea behind the evaluation of mixture properties (in the absence of chemical reaction) is the principle of
superposition: For a mixture of ideal gases contained in a given volume at a given (common) temperature,
each gas behaves as if the other gases were not present. The additive contributions of each gas results in
the mixture properties and thermodynamic condition. Why does superposition work with ideal gases?
Because of the large spaces between molecules, molecules of one gas rarely interact with molecules of
another gas. Consequently, the molecules of one gas behave as if all the other gases were absent.
Consider a mixture of gases at a common temperature, T, contained in a volume, V, with a total
pressure P, as illustrated on the left hand side of Figure S15-1. According to Dalton’s law, each gas obeys
the ideal gas equation:
PV
i i = ni RTi
Eq. S15-9
We solve for the individual number of moles, ni and substitute this into Eq. S15-4 for each component to
obtain:
k
PmixVmix PV
PV
P V PV
= 1 1 + 2 2 + 3 3 +" = ∑ i i
RTmix
RT1 RT2 RT3
i =1 RTi
Eq. S15-10
But Dalton’s law states that each gas occupies the same volume and has the same temperature as the
mixture so that
V = Vmix = V1 = V2 = V3 = "
and
T = Tmix = T1 = T2 = T3 = "
Canceling common terms in Eq. S15-10 results in:
k
P = Pmix = P1 + P2 + P3 + " = ∑ Pi
Eq. S15-11
i =1
Figure S15-1 Superposition principle for a mixture of ideal gases
3
where P is the total pressure of the mixture, and Pi represents the partial pressures of each gas, which is the
pressure an individual gas i would exert in a volume V at temperature T if all the other gases were removed
from the volume. The partial pressure of each gas is less than the total pressure of the mixture, as
illustrated on the right hand side of Figure S15-1.
If we divide all terms in Eq. S15-11 with the total pressure P, we obtain a series of terms in the
form Pi P :
k
P
P1 P2 P3
+ + +" = ∑ i
P P P
i =1 P
1=
Using the ideal gas law, P = nRT V , to express the pressures, and canceling common terms we obtain:
k
k
k
n RT V
P
n
n1 RT V
n RT V
+ 2
+ 3
+ " = ∑ i = ∑ i = ∑ Yi
nmix RT V nmix RT V nmix RT V
i =1 P
i =1 nmix
i =1
1=
Eq. S15-12
This shows that the ratio of partial pressure to total pressure is equal to the mole fraction:
Pi
n
= i = Yi
P nmix
Eq. S15-13
The partial volume of a gas, Vi, based on Amagat’s Law, is obtained when we assume that each
component in the mixture behaves as an ideal gas if it existed separately at the mixture pressure, P, and
temperature, T. Using the ideal gas law:
ni RT
P
Vi =
Dividing the partial volume by the total volume:
Vi
n RT P
n
= i
= i = Yi
V nmix RT P nmix
k
Because we know that the sum of the mole fractions is unity ( ∑ Yi = 1 ), then
i =1
k
Vi
∑V
=1
Eq. S15-14
i =1
With the quantities defined above, we can develop expressions for the thermodynamic properties
ρ, u, h, and s. For a single gas, U = mu , and the total internal energy contained within the volume shown
in Figure S15-1 is the sum of the contributions from each gas:
U = U1 + U 2 + U 3 +"
Eq. S15-15
Describing each contribution in terms of individual masses and internal energies and dividing by the total
mass:
u=
k
m
m1
m
u1 + 2 u2 + 3 u3 + " = ∑ X i ui
mmix
mmix
mmix
i =1
Eq. S15-16
4
which states that the specific internal energy of a mixture is composed of the mass-weighted contributions
from each gas.
Gas properties are expressed in molar units also. For example, the specific molar internal energy
is u = Mu , and the total internal energy is U = nu . Using the latter expression in Eq. S15-15 and
simplifying, we obtain the mixture molar specific internal energy, u :
k
u = ∑ Yi ui
Eq. S15-17
i =1
which states that the molar specific internal energy of a mixture is composed of the mole-weighted
contributions from each gas. Because these expressions are for ideal gases, the specific internal energies
are all evaluated at the common mixture temperature.
Expressions for mixture enthalpy and specific heats are developed in a comparable manner with
individual contributions evaluated at the common mixture temperature:
k
h = ∑ X i hi
Eq. S15-18
i =1
k
h = ∑ Yi hi
Eq. S15-19
i =1
k
cv = ∑ X i cv ,i
Eq. S15-20
i =1
k
cv = ∑ Yi cv ,i
Eq. S15-21
i =1
k
c p = ∑ X i c p ,i
Eq. S15-22
i =1
k
c p = ∑ Yi c p ,i
Eq. S15-23
i =1
Density and entropy are handled in a slightly different manner. On a mass basis, density is mass
per unit volume:
ρ=
k
mmix m1 + m2 + m3 + "
=
= ρ1 + ρ 2 + ρ3 + " = ∑ ρi
V
V
i =1
Eq. S15-24
On a molar basis, we obtain a comparable expression:
ρ=
k
nmix
= ρ1 + ρ 2 + ρ3 + " ∑ ρi
V
i =1
Eq. S15-25
The individual gas densities, ρi, are evaluated at the common T and V, and at the partial pressures of each
gas, not at the total pressure of the mixture; for example, ρ i = PV
RT .
i
To find the entropy of the mixture:
s = ∑ X i si
Eq. S15-26
s = ∑ Yi si
Eq. S15-27
The equations given in Chapter 7 for the entropy differences of ideal gases are used as before:
5
s2,i − s1,i = ∫ c p ,i (T )
dT R ⎛ P2,i
−
ln ⎜
T M i ⎜⎝ P1,i
s2,i − s1,i = ∫ c p ,i (T )
⎛P
dT
− R ln ⎜ 2,i
⎜P
T
⎝ 1,i
⎞
dT R ⎛ v2,i
+
ln ⎜
⎟⎟ = ∫ cv ,i (T )
T M i ⎜⎝ v1,i
⎠
⎞
⎛v
dT
+ R ln ⎜ 2,i
⎟⎟ = ∫ cv ,i (T )
⎜v
T
⎠
⎝ 1,i
⎞
⎟⎟
⎠
Eq. S15-28
⎞
⎟⎟
⎠
Eq. S15-29
Note, however, that the mixture temperatures and the partial pressures of the components are used. These
equations can be simplified if the specific heats are assumed to be constant:
⎛ T ⎞ R ⎛ P2,i ⎞
⎛ T2 ⎞ R ⎛ v2,i ⎞
s2,i − s1,i = c p ,i (T ) ln ⎜ 2 ⎟ −
ln ⎜
ln ⎜
⎟⎟ = cv ,i (T ) ln ⎜ ⎟ +
⎟⎟
⎜
⎜
⎝ T1 ⎠ M i ⎝ P1,i ⎠
⎝ T1 ⎠ M i ⎝ v1,i ⎠
⎛T
s2,i − s1,i = c p ,i (T ) ln ⎜ 2
⎝ T1
⎛ P2,i
⎞
⎟ − R ln ⎜⎜
⎠
⎝ P1,i
⎞
⎛ v2,i
⎛ T2 ⎞
⎟⎟ = cv ,i (T ) ln ⎜ ⎟ + R ln ⎜⎜
⎝ T1 ⎠
⎠
⎝ v1,i
⎞
⎟⎟
⎠
Eq. S15-30
Eq. S15-31
We can also incorporate so into the first parts of Eq. S15-28 and Eq. S15-29:
s2,i − s1,i = s2,o i − s1,o i −
R ⎛ P2,i ⎞
ln ⎜
⎟
M i ⎜⎝ P1,i ⎟⎠
⎛P
s2,i − s1,i = s2,oi − s1,oi − R ln ⎜ 2,i
⎜P
⎝ 1,i
⎞
⎟⎟
⎠
Eq. S15-32
Eq. S15-33
Note that with a constant composition, Y2,i = Y1,i so that:
P2,i
P1,i
=
Y2,i P2
Y1,i P1
=
P2
P1
Eq. S15-34
Whenever we are presented with a problem or application involving a mixture of ideal gases,
conservation of mass, conservation of energy, the entropy balance equation, equations for cycle
performance, isentropic efficiency, etc. are used as they have always been. The only difference for a fixed
composition mixture is that the evaluation of the fluid properties is handled as given in this section of the
textbook.
Example S15-1 Mixture of ideal gases
A 10 m3 tank contains a gas mixture at 25 °C, 100 kPa and has the following molar composition:
N2, 55%; CO2, 25%; O2, 10%; H2O, 10%. Determine:
a) the mass fraction of each gas
b) the apparent molecular weight of the mixture
c) the partial pressure of each component (in kPa)
d) the mass of the mixture (in kg).
Approach:
We assume the gases are ideal and that each gas in the mixture behaves as if it alone fills the
volume at the mixture temperature. The ideal gas mixture relations developed above are applied.
6
Solution:
Assumptions:
a) The mass fraction of each gas can be evaluated by considering 1 kmol of
mixture [A1], [A2] and applying Eq. S15-2. Molecular weight of each gas is obtained
from Table A-1.
Mi
Gas
ni
mi = ni M i
X i = mi mmix
N2
CO2
O2
H2O
0.55
0.25
0.10
0.10
28.01
44.01
32.00
18.02
15.406
11.003
3.200
1.802
∑ = mmix = 31.41kg
0.491
0.350
0.102
0.057
∑ = 1.000
A1. Each gas is ideal.
A2. The mixture
behaves as an ideal gas
and follows Dalton’s
Law.
b) The apparent molecular weight can be determined from either Eq. S15-7 or
Eq. S15-8. Using Eq. S15-7, we have M mix = ∑ Yi M i . Because we used 1 kmol to
create the table above, the column labeled ni is the mole fraction, so that the entry
labeled mmix is also Mmix = 31.41kg/kmol.
c) The partial pressures of each gas is given by Eq. S15-13, Pi P = ni nmix = Yi
or Pi = Yi P :
PN2 = 0.55 (100 kPa ) =55 kPa
PCO2 = 0.25 (100 kPa ) =25 kPa
PO2 = 0.10 (100 kPa ) =10 kPa
PH 2O = 0.10 (100 kPa ) =10 kPa
d) The mass of the mixture is determined with the ideal gas equation:
kN ⎞
kg ⎞ ⎛ 1kJ ⎞
⎛
3 ⎛
⎟
⎜100 2 ⎟ (10 m ) ⎜ 31.41
⎟⎜
kmol ⎠ ⎝ kN m ⎠
m ⎠
PVM mix ⎝
⎝
=
= 12.68 kg
mmix =
kJ ⎞
RT
⎛
⎜ 8.314
⎟ ( 25+273) K
kmolK ⎠
⎝
Comments:
The mixture molecular weight depends both on the mole fraction and the molecular weight of each
gas. Note that even though CO2 only comprises 25% mole fraction of the mixture, it added
11.003/31.41×100% = 35% to the molecular weight of the mixture because the CO2 molecular weight is
larger than that of the other gases.
7
Example S15-2 Compression of a mixture of ideal gases
A mixture of ideal gases enters a piston-cylinder assembly at 0.1 MPa, 27 °C and is compressed
adiabatically to 0.6 MPa, 197 °C. The mixture has a mass of 0.67 kg, and the mass fractions of the
constituent gases are: N2, 40%; CO2, 30%, O2, 10%; H2O, 20%. Determine:
a) the work required (in kJ)
b) the final volume of the mixture (in m3).
Approach:
Conservation of energy is used to calculate the work required. Properties of the mixture must be
evaluated. The final volume requires the use of the ideal gas equation; the apparent molecular weight will
needed.
Solution:
Assumptions:
a) For a closed system, conservation of energy is:
Q − W = ∆E
Assuming [A1] and [A2], we solve this equation for work:
W = ∆U
k
Eq. S15-16 shows that u = ∑ X i ui ([A3], [A4]). With constant mass fractions, Xi, we
i =1
k
can write ∆u = ∑ X i ∆ui so that:
i =1
k
k
W
= ∑ X i ∆ui
mmix i =1
i =1
The ideal gas properties are determined from Table SA-1 at the appropriate
temperatures. Note that in the table the molar internal energy is given with units of
kJ/kmol, so we must divide by the molecular weight to put it into mass units:
k
k
∆u
W
= ∑ X i ∆ui = ∑ X i i
mmix i =1
Mi
i =1
W = ∆U = mmix ∑ X i ∆ui →
A1. The system is
adiabatic.
A2. Potential and
kinetic energy effects
are negligible.
A3. Each gas is ideal.
A4. The mixture
behaves as an ideal gas
and follows Dalton’s
Law.
⎛ 9786 − 6229 ⎞
⎛ 12444 − 6939 ⎞
= 0.40 ⎜
⎟ + 0.30 ⎜
⎟
28.01
44.01
⎝
⎠
⎝
⎠
kJ
⎛ 9935 − 6242 ⎞
⎛ 11869 − 7472 ⎞
+ 0.10 ⎜
⎟ + 0.20 ⎜
⎟ = 148.7
18.02
kg
⎝ 32.00 ⎠
⎝
⎠
k
⎛ W ⎞
⎛
kJ ⎞
Finally, W = mmix ⎜
⎟ = ∆U = mmix ∑ X i ∆ui = ( 0.67 kg ) ⎜ 148.7 ⎟ =99.6 kJ
kg ⎠
i =1
⎝
⎝ mmix ⎠
The final volume is determined with the ideal gas equation
PV = mmix RT M mix . The apparent molecular weight is calculated with Eq. S15-8:
8
1
1
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
= ∑ Xi
= 0.40 ⎜
⎟ + 0.30 ⎜
⎟ + 0.10 ⎜
⎟ + 0.20 ⎜
⎟
M mix
Mi
⎝ 28.01 ⎠
⎝ 44.01 ⎠
⎝ 32.00 ⎠
⎝ 18.02 ⎠
M mix = 28.31kg kmol
b) The final volume is:
⎛
V =
mmix RT
=
PM mix
( 0.67 kg ) ⎜ 8.314
kJ ⎞
⎟ (197+273) K
kmol K ⎠
⎝
kN
kg ⎞ ⎛ 1kJ ⎞
⎛
⎞⎛
⎟
⎜ 600 2 ⎟ ⎜ 28.31
⎟⎜
m ⎠⎝
kmol ⎠ ⎝ kN m ⎠
⎝
= 0.154 m3
Comments:
This problem could have been solved using specific heats of each compound evaluated at the
average process temperature. The same result would have been obtained.
Example S15-3 Turbine with products of combustion
In a Brayton cycle, hot gas leaves the combustion chamber and enters the turbine at 1127 °C, 800
kPa with a volumetric flow rate of 400 m3/min. The mole fractions of the constituent gases are: CO2, 2.2%;
H2O, 4.4%; O2, 16.1%; N2, 77.3%. The gas leaves the well-insulated turbine at 627 °C, 96 kPa.
Determine:
a) the mass flow rate (in kg/s)
b) the power developed (in kW)
c) the isentropic efficiency.
Approach:
The mass flow rate is determined from the definition of mass flow in terms of density and volume
flow rate. We use the ideal gas equation to determine the density. Power is calculated by applying
conservation of mass and energy to a control volume surrounding the turbine and using the given inlet and
exit temperatures. The isentropic efficiency is obtained by application of its definition and determining
what the outlet state of the turbine would be for an isentropic process. Throughout the solution, the mixture
equations are used to evaluate properties.
Solution:
Assumptions:
The schematic of the turbine is shown below along with the given information:
9
a) Mass flow rate is m = ρ V A = ρV . The mixture density is determined with
Eq. S15-24 [A1], [A2]:
k
ρ = ∑ ρi
i =1
using the partial pressures of each gas in the ideal gas equation. Because the mole
fraction of each gas is given, the partial pressures are given by Eq. S15-13
PCO2 = 0.022 ( 800 kPa ) =17.6 kPa
Pi P = ni nmix = Yi or Pi = Yi P :
PH 2O = 0.044 ( 800 kPa ) =35.2 kPa ,
PN2 = 0.773 ( 800 kPa ) =618.4 kPa .
A1. Each gas is ideal.
A2. The mixture
behaves as an ideal gas
and follows Dalton’s
Law.
PO2 = 0.161( 800 kPa ) =128.8 kPa
For the density of N2, we use the ideal gas equation:
kN ⎞ ⎛
kg ⎞ ⎛ 1kJ ⎞
⎛
618.4 2 ⎟ ⎜ 28.01
⎟
⎜
⎟⎜
PM N 2 ⎝
kmol ⎠ ⎝ kN m ⎠
m ⎠⎝
kg
ρ N2 =
=
= 1.488 3
RT
m
⎛
kJ ⎞
⎜ 8.314
⎟ (1127+273) K
kmol
K
⎝
⎠
In a similar manner, the densities of the three other gases are calculated:
ρCO2 = 0.0665 kg m3 , ρ H 2O = 0.0545kg m3 , ρO2 = 0.354 kg m3 . The mixture density is
ρ = 1.488 + 0.0665 + 0.0545 + 0.354 = 1.963 kg m 3
and the total mass flow rate is:
kg ⎞ ⎛
m3 ⎞ ⎛ 1min ⎞
kg
⎛
m = ⎜ 1.963 3 ⎟ ⎜ 400
⎟⎜
⎟ =13.1
m
min
60s
s
⎝
⎠⎝
⎠
⎠⎝
b) The power developed is obtained by applying conservation of energy and
mass to the control volume around the turbine. Assuming [A1], [A2], [A3], [A4], and
[A5], conservation of mass gives us m i , A = m i , B = m i , and conservation of energy gives
us:
W = ∑ m i ∆hi
Using Eq. S15-6 and recognizing that it can be recast in terms of flow rates, m i = ni M i .
The enthalpies obtained from Table SA-1 are in molar units and must be divided by
molecular weight to be put into mass units, ∆hi = ∆hi M i . Incorporating both of these
into the energy equation:
W = ∑ m i ∆hi = ∑ ( ni M i ) ( ∆hi M i ) = ∑ ni ∆hi
A3. The system is
steady.
A4. Potential and
kinetic energy effects
are negligible.
A5. The turbine is
adiabatic.
Using the definition of mole fraction (Eq. S15-13) and recognizing that it can be recast in
terms of flow rates, ni = Yi nmix , and again using Eq. S15-6 but for the total flow,
nmix = m mix M mix , we incorporate these into the energy equation to obtain:
⎛ m ⎞
m
W = ∑ (Yi nmix ) ∆hi = ∑ ⎜ Yi mix ⎟ ∆hi = mix ∑ Yi ∆hi
M
M
mix ⎠
mix
⎝
The mole fraction is given, we calculated the mass flow rate, so the quantity we still need
is the apparent molecular weight of the mixture, which can be calculated with Eq. S15-7:
M mix = ( 0.022 )( 44.01) + ( 0.044 )(18.02 ) + ( 0.161)( 32.00 ) + ( 0.773 )( 28.01)
= 28.56 kg kmol
With the enthalpies from Table SA-1 at the appropriate temperatures:
⎛ 13.1kg s ⎞
W = ⎜
⎟ [0.022 ( 65271-37405 ) +0.044 ( 53351-32828 )
⎝ 28.56 kg kmol ⎠
+0.161( 45648-27928 ) +0.773 ( 43605-26890 ) ]
kJ
=7930 kW
kmol
c) Isentropic efficiency is defined as:
10
W
W
m
h −h
= 1 2
ηs = act = act
Wideal Wideal m h1 − h2 s
We need to evaluate the enthalpy at the exit of the turbine assuming it is isentropic, or
from the entropy balance equation assuming [A3], [A5], [A6]:
S2 = S1 or S2 − S1 = 0
The total entropy must remain constant, so multiplying Eq. S15-33 by the
individual molar flow rates, recognizing that the ratio of partial pressures is the same as
the ratio of total pressures (Eq. S15-34), and summing:
A6. Entropy generation
rate is zero.
⎡
⎛ P ⎞⎤
S 2 − S1 = ∑ ni ( s2,i − s1,i ) = ∑ ni ⎢ s2,oi − s1,oi − R ln ⎜ 2 ⎟ ⎥ = 0
⎝ P1 ⎠ ⎦
⎣
Expressing the component flow rate in terms of the mole fraction and total molar flow
rate, ni = Yi nmix , the total molar flow rate cancels from the equation, and we are left with:
⎡
⎛ P ⎞⎤
− s1,i ) = ∑ Yi ⎢ s2,oi − s1,oi − R ln ⎜ 2 ⎟ ⎥ = 0
⎝ P1 ⎠ ⎦
⎣
We know the inlet conditions, the mole fractions, and the pressures. The
unknown outlet temperature must be determined with an iterative solution. With
properties from Table SA-1, the equation to solve is:
⎡
⎡ o
⎛ 96 ⎞ ⎤
⎛ 96 ⎞ ⎤
0.022 ⎢ s2,oCO2 − 288.11 − 8.314 ln ⎜
⎟ ⎥ + 0.044 ⎢ s2, H 2O − 247.24 − 8.314 ln ⎜
⎟⎥
⎝ 800 ⎠ ⎦
⎝ 800 ⎠ ⎦
⎣
⎣
∑Y ( s
i
2, i
⎡
⎡ o
⎛ 96 ⎞ ⎤
⎛ 96 ⎞ ⎤
+0.161 ⎢ s2,oO2 − 255.45 − 8.314 ln ⎜
⎟ ⎥ + 0.773 ⎢ s2, N 2 − 239.38 − 8.314 ln ⎜
⎟⎥ = 0
⎝ 800 ⎠ ⎦
⎝ 800 ⎠ ⎦
⎣
⎣
The procedure is to guess a temperature, determine each s2,oi at that temperature, and
evaluate the equation. Once the left hand side equals zero, we have a converged solution.
Performing the iteration, the outlet temperature is about 840 K = 567 °C, and the
corresponding enthalpies are: hCO2 = 34251 kJ/kmol, hH2O = 29454 kJ/kmol, hO2 = 25877
kJ/kmol, hN2 = 24974 kJ/kmol. We can use the energy equation we developed above to
calculate the maximum power:
⎛ 13.1kg s ⎞
W = ⎜
⎟ [0.022 ( 65271-34251) +0.044 ( 53351-29454 )
⎝ 28.56 kg kmol ⎠
+0.161( 45648-25877 ) +0.773 ( 43605-24974 ) ]
kJ
=8861kW
kmol
Finally, the isentropic efficiency is
7930 kW
ηs =
= 0.895
8861kW
Comments:
Because the composition remains constant, in the evaluation of the entropy change the ratio of
partial pressures equals the ratio of total pressures. Iterative solutions can be solved easily with available
software, if the properties of the ideal gases are included.
11
15.3 Psychrometrics
The previous section discussed mixtures of ideal gases that did not change composition. In the
present section, we discuss a particular mixture of two ideal gases: air and water vapor. Air by itself is
called dry air, an air-water vapor mixture is called moist air, and the study of this special mixture is called
psychrometrics. Processes that involve moist air mixtures must be handled differently than those described
in Section 15.2, because the amount of water vapor in the mixture often changes. Hence, the mixture
composition and its fluid properties change.
The “sweating” of a glass of ice water is a simple example of what happens when a moist air
mixture is cooled; some of the water vapor in the mixture in contact with the cold glass condenses on the
glass, and the amount of water vapor in the mixture decreases. Other examples of what happens when
moist air mixtures are cooled include the formation of a plume above a cooling tower on a cold day, water
dripping down poorly insulated windows on winter days, mirrors fogging in steamy bathrooms, fog
forming over cold lakes on humid days, ice forming in freezer compartments of refrigerators, and frost
formation on car windows. Water vapor is added to moist air when liquid water evaporates into the air.
Examples of this process include sweating of humans, the slow decrease in the water level in a container
sitting on the counter in a kitchen, and the water vapor rising from a pot of boiling water and being
absorbed by the air.
Air conditioning systems are the most common application in which properties of air-water vapor
mixtures must be analyzed. Consider the refrigeration cycle shown in Figure S15-2. If this system is to be
used to cool air and to make the environment inside a building more comfortable, then the air handling
system must blow air through ductwork and the evaporator (cooling coil) of the refrigeration system. If the
building were located in a humid region (e.g., Louisiana) and the air flow cooled sufficiently, water vapor
would be removed from the air stream. This is called dehumidification. The condensate flow would need
to be drained from the cooling coil. Likewise, in a dry region (e.g., Las Vegas), water vapor is added to the
air stream to make the environment inside a building more comfortable. This is called humidification.
We often talk about humidity, even though we may not have a working definition. Meteorologists
measure and report this quantity daily on the news, usually in terms of a percent. The amount of moisture
in air has an effect on how comfortable we feel. However, instead of the absolute amount of moisture in
the air affecting our comfort level, comfort depends more on the amount of moisture in the air relative to
the maximum amount of moisture the air can hold at the same temperature. At a relative humidity of
φ = 100% fog forms, and in the desert, the relative humidity is often very low, in the order of φ = 0 − 15% .
Humidity is simply a convenient way to describe the amount of water vapor in an air-water vapor mixture.
Figure S15-2 Schematic of refrigeration cycle and air handling system
12
In a humid climate, dehumidification is used to set the relative humidity to a comfortable range
(usually 45-55% in buildings). However, when the air-water vapor mixture is cooled enough to condense
some of the water vapor, in addition to the temperature dropping to a low (and possibly uncomfortable)
temperature, the relative humidity goes to one hundred percent. This combination of low temperature and
high humidity would result in a very clammy, damp environment. Mold could grow easily. To set both the
temperature and humidity to appropriate levels, we first cool the air/water vapor mixture to a temperature
lower than a target value to decrease the water content in the air, and then we heat the air/water vapor
mixture to reach the target temperature. The heating is performed in a “reheat” section, as shown on Figure
S15-2. On the other hand, if the relative humidity is low (such as in the winter or in a desert environment),
water vapor must be added (humidification) to the air stream. This can be accomplished by the injection of
a very fine mist or fog of water into the air stream. The water drops evaporate by absorbing heat from the
air and, thus, cool and humidify the air simultaneously. To analyze these processes, we apply conservation
of mass and energy. The main issue with solving the equations is the evaluation of the air-water vapor
mixture properties.
Before we present definitions for use with air-water vapor mixtures, we want to emphasize two
important points. The first is that the mass of water vapor in the mixture is quite small compared to the
mass of dry air. For example, at 20 °C and 50% relative humidity (RH), there is only ∼0.0075 kg water per
kg dry air; at 35 °C and 80% RH, the ratio is ∼0.029 kg water per kg dry air. The second point is that to
add or remove some of this small amount of water vapor requires large amounts of energy compared to just
heating or cooling the dry air.
We describe the amount of water vapor in an air-water vapor mixture in two ways. The first
quantity is the specific humidity, ω (also called humidity ratio and absolute humidity). It is the ratio of the
mass of water vapor to the mass of dry air in the mixture and is used whenever we apply conservation of
mass and/or energy to an air-water vapor mixture. It is defined as:
ω=
mass of water vapor mv
=
mass of dry air
ma
Eq. S15-35
which has units of kg of water vapor/kg of dry air (or lbm water vapor/lbm dry air). The denominator is
not the total mass of the mixture, mmix; it is the mass of the dry air, ma, so that mmix = ma + mv . In an airwater vapor mixture, only by adding water vapor to or removing water vapor from dry air can we change
the specific humidity. By definition, the specific humidity of dry air is zero.
The expression for specific humidity can be expressed also in terms of the partial pressure of the
water vapor, Pv, and the partial pressure of the dry air, Pa. Substitute the ideal gas equation for the mass of
the air and of the water vapor into Eq. S15-35, and insert the molecular weights of air and water:
ω=
M v PV
RT M v Pv 18.02 Pv
P
v
=
=
= 0.622 v
M a PaV RT M a Pa 28.97 Pa
Pa
Eq. S15-36
The total pressure of the air-water vapor mixture is found by the adding the partial pressures of
both gases, P = Pa + Pv and is equal to the barometric pressure. Solve this for Pa and substitute into Eq.
S15-36 to get an expression for the specific humidity in terms of the partial pressure of the water vapor and
the barometric pressure.
ω = 0.622
Pv
P − Pv
Eq. S15-37
Why is this expression useful? The total pressure P is easily measured with a barometer. (This is the
pressure the meteorologists report each day on the local weather.) Likewise, the partial pressure of the
water vapor, Pv, is also easily obtained, as discussed below.
Relative humidity, φ, describes the actual amount (mv) of moisture in the air relative to the
maximum amount (mg) of moisture air can hold at the same temperature. A relative humidity of φ = 0%
13
represents dry air. A relative humidity of φ = 100% represents saturated air, that is, the situation when air
holds the maximum possible amount of water vapor at a given temperature. Attempts to add more water
vapor to an already saturated mixture would result in the formation of fog and liquid water. We define the
relative amount of water vapor in dry air (i.e., the relative humidity) as:
φ=
mv
mg
Eq. S15-38
Using the ideal gas equation to describe the mass of the water vapor in a mixture, Eq. S15-38 is recast as:
φ=
M H 2O PV
RT
v
M H 2O PgV RT
=
Pv
Pg
Eq. S15-39
The quantity Pg is the partial pressure of water vapor corresponding to the saturation pressure of water at
the mixture temperature, and Pv is the actual partial pressure of the water vapor in the mixture. Note that
from this relationship, relative humidity is seen to be independent of the pressure and density of the dry air
and of the barometric pressure.
Consider Figure S15-3 for pure water. The relationship between the partial pressure of the water
vapor, the saturated air pressure, and various temperatures is shown. We are interested in the superheated
vapor region and have already seen that water vapor in this region can be treated as an ideal gas. One
constant pressure line represents the actual partial pressure of the water vapor in the mixture (Pv) and the
other constant pressure lines represent pressures of the water vapor (Pg1 and Pg2) for the saturated air at
different temperatures. For a given relative humidity, the actual partial pressure of the water vapor, Pv, is
obtained using Eq. S15-39.
Figure S15-3 T-v diagram for water
The pressure at which air becomes saturated with water vapor, Pg, is found from the saturated
steam tables, Table A-10 or B-10. This pressure is the saturation pressure evaluated at the mixture
temperature. That is:
14
Pg = Psat , water ( evaluated at the mixture temperature )
Eq. S15-40
The mixture temperature is called the dry-bulb temperature. It is simply the temperature of the mixture as
measured by any of several types of ordinary thermometers placed in the mixture. The term "dry bulb" is
used to distinguish the temperature of the mixture from the temperature reading obtained from a
thermometer which has its temperature sensitive element wrapped in gauze and soaked in water (wet-bulb
temperature, which is discussed below).
As the mixture temperature decreases from T1 to T2 (see Figure S15-3), the saturation pressure of
the water decreases from Pg1 to Pg2. From Eq. S15-39, if Pv remains constant, then the relative humidity
increases from φ1 to φ2 . We can further decrease the mixture temperature to T3, at which point Pg3 = Pv,
φ3 = 100% , and the air is saturated. The temperature at point 3 is called the dew-point temperature.
Consider dew forming on grass. In the summer, a considerable amount of water vaporizes during
the day. As the temperature falls during the night (pressure is held constant), so does the ability of the air
to hold water vapor. If the temperature decreases enough (such as if the mixture temperature falls from T2
to T3), then the moisture carrying capacity of the air will equal the moisture content of the air. At this point
Pg3 = Pv, and the air is saturated (φ = 100%). Any further drop in mixture temperature (such as from T3 to
T4 on Figure S15-3) results in condensation of water vapor from the air-water vapor mixture, and the partial
pressure of the vapor (Pv) must decrease; the air remains saturated during the condensation process. The
same process governs the formation of fog on a mirror in a steamy bathroom and causes water to drip from
cold water pipes. The excess moisture in the air simply condenses on a cool surface forming dew.
Likewise, when a low-pressure weather system passes through a location, clouds form and often it rains.
The dew point temperature, TDP, is equal to the saturation temperature of water at the partial pressure of the
water vapor:
TDP = Tsat , water ( evaluated at Pv )
Eq. S15-41
Note that Pv = φ Pg . Combining Eq. S15-37 and Eq. S15-39 we obtain a relationship between the
specific humidity and the relative humidity:
φ=
ωP
( 0.622 + ω ) Pg
or
ω=
0.622φ Pg
P − φ Pg
Eq. S15-42
For a given specific humidity, as temperature increases relative humidity decreases, and vice versa. The
amount of water vapor in saturated air at a specified temperature and barometric pressure can be
determined with Eq. S15-42. We do this by setting φ = 100%.
Example S15-4 Compression of a moist air mixture
In a piston-cylinder assembly, moist air is compressed isothermally from 20 °C, 100 kPa, 68%
relative humidity to 250 kPa. The initial volume is 0.75 m3. Determine:
a) the mass of the mixture (in lbm)
b) the amount of water condensed during the process if condensation occurs or the final relative
humidity if condensation does not occur.
Approach:
15
The mass is determined with the ideal gas equation, which is used to calculate the initial density of
both gases (air and water vapor); the partial pressures and the mixture temperature must be used. With the
relative humidity, dry-bulb temperature, and total pressure known, the specific humidity can be calculated
at both states. If the specific humidity decreases, then water vapor has condensed from the mixture. The
amount can be calculated from the specific humidity change and the mass of dry air.
Solution:
Assumptions:
a) Using assumptions [A1] and [A2] for both the air and water vapor, the mass
of the mixture is m = ρ mixV = ρ aV + ρ vV . The densities are evaluated with the ideal gas
equation and the partial pressures of each gas. Using Eq. S15-39, Pv1 = φ1 Pg1 . From
Table A-10, Pg1(20 °C) = Psat(20 °C) = 2.239 kPa, and Pa1 = 100 kPa –2.239 kPa =
97.76 kPa. The density of air is determined from:
kN ⎞⎛
kg ⎞ ⎛ 1kJ ⎞
⎛
⎟
⎜ 97.76 2 ⎟⎜ 28.97
⎟⎜
m
kmol
PM ⎝
kg
⎠⎝
⎠ ⎝ kN m ⎠
=
= 1.163 3
ρa =
kJ ⎞
RT
m
⎛
⎜ 8.314
⎟ ( 20+273) K
kmolK ⎠
⎝
Similarly, the water vapor density is determined to be ρv = 0.017 kg/m3, so the total
mass is:
mmix = mv + ma = ( 0.017 kg m3 )( 0.75 m3 ) + (1.163kg m3 )( 0.75 m3 )
A1. Each gas is ideal.
A2. The mixture
behaves as an ideal gas
and follows Dalton’s
Law.
=0.013+0.872=0.885 kg
b) If no condensation occurs, then ω1 = ω 2 , and Pv 2 ≤ Pg1 = Pg 2 ( 20 o C ) . We
can check this using Eq. S15-37:
P
Pv 2
P
P
P
ω1 = ω 2 = 0.622 v1 = 0.622
→ 2 = 1 → Pv 2 = 2 Pv1
P1 − Pv1
P2 − Pv 2
Pv 2 Pv1
P1
Using Eq. S15-39, Pv1 = φ1 Pg1 , and when this is substituted into the above expression:
Pv 2 =
P2
250
φ1 Pg1 =
( 0.68 ) Pg1 = 1.70 Pg1
100
P1
For no condensation, Pv 2 ≤ Pg1 = Pg 2 ( 20 o C ) , so because Pv 2 > Pg1 , condensation does
occur, and the amount can be determined from the specific humidities and mass of dry
air.
The specific humidity at the initial and final states can be determined from Eq.
S15-42, where φ1 = 0.68 , φ2 = 1.00 , P1 = 100 kPa , P2 = 250 kPa , and Pg1 = Pg 2 :
16
ω1 =
0.622φ1 Pg1
ω2 =
P1 − φ Pg1
=
0.622 ( 0.68)( 2.239 )
100 − ( 0.68 )( 2.239 )
0.622 (1.00 )( 2.239 )
250 − (1.00 )( 2.239 )
= 0.00562
= 0.00962
kg vapor
kg dry air
kg vapor
kg dry air
The amount of water condensed is:
mliquid = mv1 − mv 2 = ma (ω1 − ω 2 )
= ( 0.872 kg )( 0.00962-0.00562 )
kg vapor
=0.0035 kg
kg dry air
Comments:
The mass of the water vapor present in the mixture is quite small compared to that of the air. Note
that during compression, almost 27% of the water vapor condensed to liquid. This liquid must be removed
from the compressor so that equipment damage can be avoided.
Relative humidity and specific humidity are frequently used in engineering and atmospheric
sciences, but neither is easy to measure directly. If one of these quantities is known along with the dry bulb
temperature and the barometric pressure, then the other one can be calculated with Eq. S15-42. One way to
indirectly measure specific humidity is through the use of an adiabatic saturation process, a schematic of
which is shown on Figure S15-4. In this device, moist air, which enters at a dry bulb temperature T1, is
forced through a very long insulated chamber. Liquid water at temperature T3 is added to the chamber at a
rate just sufficient to balance that which is lost to evaporation into the moist air stream. The chamber
length is designed so that at its exit the moist air is saturated (φ2 = 100%) and the measured dry bulb
temperature at the exit is T2, which is known as the adiabatic-saturation temperature. We analyze this
steady flow process by application of conservation of mass and energy.
For mass, we track the water and dry air separately. Thus, for steady flow
Air:
m a1 = m a 2 = m a
Water:
m v1 + m L 3 = m v 2
Figure S15-4 Adiabatic saturation process
The amount of dry air remains constant. The amount of water vapor in the moist air between the inlet and
exit increases by the amount of make-up liquid water added at 3. Specific humidity is given in Eq. S15-35
as the ratio of two masses. We can also write this equation in terms of two mass flow rates:
17
ω=
m v
m a
Eq. S15-43
We use Eq. S15-43 and the constancy of the dry airflow rate to rewrite the conservation of mass equation
for the water:
ω1m a + m L 3 = ω 2 m a
or
m L 3 = (ω 2 − ω1 ) m a
For energy, assuming the process is adiabatic, steady, with no work, and neglecting changes in
potential and kinetic energy, we obtain:
m a1ha1 + m v1hv1 + m L 3 h3 − m a 2 ha 2 − m v 2 hv 2 = 0
Now incorporating conservation of mass for the air and water, dividing through by the dry air mass flow
rate, using the definition of specific humidity, and solving for ω1:
ω1 =
( ha 2 − ha1 ) + ω 2 ( hv 2 − h3 )
hv1 − h3
Eq. S15-44
All the quantities on the right hand side of the equation can be evaluated from measured
temperatures and barometric pressure. This is more easily seen if we further simplify this expression. The
first term in the numerator on the right hand side of Eq. S15-44, ( ha 2 − ha1 ) , is the enthalpy change of the
dry air between the inlet and exit of the device. Assuming the air specific heat is constant, we can replace
this enthalpy difference with ∆h = c p ∆T . Evaporation from the liquid water decreases both the pool and
air temperatures (heat from the air is absorbed by the water to evaporate some of it), and the outlet air
becomes saturated; this causes T3 = T2. The make-up liquid water at 3 is subcooled, so we can evaluate its
enthalpy using the subcooled liquid approximation, h3 ≈ h f (T3 ) = h f (T2 ) . The water vapor enthalpy can
be approximated as the saturated vapor enthalpy, hv = hg , evaluated at the appropriate temperature (T1 or
T2).
Thus, the enthalpy difference in the second term in the numerator becomes
hv 2 − h3 = hg 2 − h f 2 = h fg (T2 ) . Incorporating all of these simplifications into Eq. S15-44, we obtain:
ω1 =
c p , a (T2 − T1 ) + ω 2 h fg ,2
hg1 − h f 3
Eq. S15-45
The temperatures T1 and T2 are the measured dry bulb temperatures. The saturated steam table (Table A10) is used to evaluate the enthalpy of vaporization (at T2), the saturated vapor enthalpy (at T1), and the
saturated liquid enthalpy (at T3). The specific humidity ω2 is evaluated using Eq. S15-42 with φ2 = 100%
and Pg evaluated at T2.
There are some difficulties involved with the use of an adiabatic saturator to determine humidity,
particularly the long channel length required to ensure saturated air at the exit. A more practical approach
is to use a thermometer whose bulb is covered with a wicking material saturated with water. When air is
blown over the wet material the temperature of the water decreases because of evaporation of the water into
the air. (Think of how a wet washcloth becomes cold if it sets for a while on the edge of the bathtub after
you have taken a shower.) Evaporation will continue until the wick material reaches an equilibrium state.
The measured temperature is the wet-bulb temperature, and for air-water vapor mixtures at "normal"
pressures and temperatures, the wet-bulb temperature is a very good approximation to the adiabaticsaturation temperature obtained at the exit of an adiabatic saturator (T2 in the above derivation). Thus, if
18
the dry-bulb temperature, the wet-bulb temperature, and the barometric pressure are measured, then both
relative humidity and specific humidity can be determined.
The traditional and simple approach for determining humidity is to measure the wet-bulb
temperature, TWB, and the dry-bulb temperature, TDB, with a sling psychrometer. (See Figure S15-5.) The
psychrometer is spun in the air so that water evaporates from the wick material surrounding the
thermometer bulb, and the water temperature falls below the dry-bulb temperature. At steady state, the
measured wet-bulb temperature, dry-bulb temperature, and barometric pressure are used in Eq. S15-42 with
φ = 100% to determine ω2. Then Eq. S15-44 or Eq. S15-45 is used to determine ω1. Rewriting these
equations in terms of the sling psychrometer measured quantities and indicating at what temperature each
property must be evaluated:
ω=
ha (TWB ) − ha (TDB ) + ω ∗ h fg (TWB )
where
ω∗ =
hg (TDB ) − h f (TWB )
0.622 Pg (TWB )
=
c p , a (TWB − TDB ) + ω ∗ h fg (TWB )
hg (TDB ) − h f (TWB )
Eq. S15-46
P − Pg (TWB )
Figure S15-5 Measurement of wet- and dry-bulb temperatures with a sling psychrometer (M. J. Moran, H.
N. Shapiro, Fundamentals of Engineering Thermodynamics, 3rd edition, John Wiley & Sons,
1996. Used with permission.)
19
Example S15-5 Use of wet- and dry-bulb temperatures
With a sling psychrometer, you measure a dry-bulb temperature of 35 °C and a wet-bulb
temperature of 30 °C. The barometric pressure is 96 kPa. Determine:
a) the specific humidity
b) the relative humidity
c) If the barometric pressure rises to 104 kPa and the wet- and dry-bulb temperatures remain the
same, what are the new relative and specific humidities?
Approach:
Because we have the barometric pressure and the dry- and wet-bulb temperatures, we assume that
the wet-bulb temperature is a good approximation to the adiabatic saturation temperature, so that we can
use Eq. S15-42 and Eq. S15-46, respectively, to determine relative humidity and specific humidity.
Solution:
Assumptions:
a) We begin with Eq. S15-46 to calculate the specific humidity with
assumptions [A1], [A2], and [A3]. The water enthalpies are obtained from Table A-10,
and the air enthalpies are obtained (by interpolation) from Table A-9.
0.622 Pg (TWB ) 0.622 ( 4.246 kPa )
kg vapor
=0.0288
ω∗ =
=
96 kPa-4.246 kPa
kg dry air
P − Pg (TWB )
ω=
ha (TWB ) − ha (TDB ) + ω ∗ h fg (TWB )
hg (TDB ) − h f (TWB )
303.21 − 308.23 + ( 0.0288 )( 2430.5 )
kg vapor
= 0.0266
2565.3 − 125.79
kg dry air
b) Using Eq. S15-42, we can determine the relative humidity:
( 0.0266 )( 96 )
ωP
=
= 0.70 = 70%
φ=
( 0.622 + ω ) Pg (TDB ) ( 0.622 + 0.0266 )( 5.628 )
=
A1. The air water vapor
mixture is an ideal gas
mixture.
A2. A sling
psychrometer is a good
approximation to an
adiabatic saturator.
A3. The system is
steady.
c) For the higher barometric pressure of 104 kPa:
0.622 ( 4.246 kPa )
kg vapor
ω∗ =
=0.0265
104 kPa-4.246 kPa
kg dry air
ω=
φ=
303.21 − 308.23 + ( 0.0265 )( 2430.5 )
2565.3 − 125.79
= 0.0243
kg vapor
kg dry air
( 0.0243)(104 )
ωP
=
= 0.69 = 69%
( 0.622 + ω ) Pg (TDB ) ( 0.622 + 0.0243)( 5.628)
Comments:
Changes in absolute total pressure affect the humidity when the temperatures remain constant.
However, the differences are not large, and as an approximation, the psychrometric chart at one atmosphere
is often used for different pressures.
20
To determine the internal energy, enthalpy, or entropy of an air-water vapor mixture, the
contributions of each component are added, as was done earlier for mixtures of other ideal gases. For
example, if we define Hmix ≡ mahmix = Ha + Hv = ma ha + mv hv, and then divide by ma we get:
H mix
m
≡ hmix ≡ ha + v hv = ha + ω hv
ma
ma
Eq. S15-47
where ha is the enthalpy per unit mass of dry air from ideal gas tables at Tmix, hv is evaluated from the
saturation table of water using hv ≅ hg at Tmix, and hmix is the mixture enthalpy per unit mass of dry air. This
is also called the moist air enthalpy. Internal energy and entropy are expressed in a similar manner as:
S mix
m
= smix = sa + v sv = sa + ω sv
ma
ma
Eq. S15-48
U mix
m
= umix = ua + v uv = ua + ω uv
ma
ma
Eq. S15-49
As can be seen, superposition of the properties of both gases is used to give a mass-weighted
average property. The contribution of the water vapor is weighted by the specific humidity. Note that we
can describe hmix in another manner. Recall that for the air in the air-water vapor mixture we assumed
∆ha = c p , a ∆T = c p , a (T2 − T1 ) . We can carry out the multiplication to obtain ∆ha = c p , aT2 − c p , aT1 , and
define the air enthalpy as ha = c p , aT . Now we can rewrite the mixture enthalpy as:
hmix = ha + ω hv = c p , aT + ω hv
Eq. S15-50
Because hv is only a function of temperature, the mixture enthalpy is often given on psychrometric charts,
which graphically show the properties of air-water vapor mixtures. As shown on Figure S15-6, the specific
humidity, dry-bulb temperature, wet-bulb temperature, mixture specific volume (per unit mass of dry air),
mixture enthalpy (per unit mass of dry air), and relative humidity all can be displayed on the same figure.
Note that this figure is for a total pressure of one atmosphere (101.325 kPa). For other barometric
pressures, either a chart prepared specifically for that pressure or the equations/approach given above must
be used to determine the mixture properties. This is a very “busy” graph. Figure S15-7 is a simplified
version of the full psychrometric chart that more clearly shows the six properties given in the chart.
The psychrometric chart is useful for showing various processes when heating, cooling,
dehumidifying, reheating, or humidifying air. Figure S15-8 illustrates simple heating or cooling of a moist
air stream; the dry-bulb temperature and the relative humidity is changed but the specific humidity remains
constant. When cooling of an air stream is sufficient to reduce the dry-bulb temperature to its dewpoint
(see Figure S15-9) the relative humidity reaches 100%. Further removal of heat causes water vapor to
condense; the specific humidity and dry-bulb temperature decrease while the relative humidity remains at
100%. When a target value of specific humidity is reached, the air flow is reheated; specific humidity
remains constant, and dry-bulb temperature and relative humidity increase. Heating and humidifying an air
stream is illustrated in Figure S15-10. When heating, the dry-bulb temperature is increased at constant
specific humidity; adding water vapor increases the specific humidity and the resulting dry-bulb
temperature will depend on the temperature of the water spray.
21
Figure S15-6 Psychrometric chart for 101.3 kPa (M. J. Moran, H. N. Shapiro, Fundamentals of Engineering Thermodynamics, 3rd edition, John Wiley
& Sons, 1996. Used with permission.)
22
Figure S15-7 Simplified representation of a psychrometric chart.
Figure S15-8 Simple heating and cooling processes illustrated on psychrometric chart
23
Figure S15-9 Dehumidification and reheat process illustrated on psychrometric chart
Figure S15-10 Humidification and heating process illustrated on psychrometric chart
24
Example S15-6 Dehumidification
Air enters an air-conditioning system at 101.3 kPa, 34 °C, and 70% relative
humidity with a volumetric flow rate of 8 m3/min. The air flows over a cooling coil, where the air is
cooled and dehumidified, and then over resistance heating wires where the air is heated such that the exit
air condition is 22 °C and 50% relative humidity. The condensed water vapor is drained from the system.
Use the psychrometric chart to evaluate properties. Determine:
a) the temperature of the air before it enters the heating section (in °C)
b) the condensate flow rate (in kg/s
c) the heat transfer rate in the cooling section (in kW)
d) the heat transfer rate in the heating section (in kW).
Approach:
A schematic of the system and the given information are shown below:
Because water vapor is condensed from the air flow, the air is saturated after it leaves the cooling coil. No
water vapor is added or removed in the heating coil, so the specific humidity at locations B and C are equal.
The exit state, C, is known, so we can work our way back from there to determine the state at B. The heat
transfer rates in the cooling and heating sections can be determined by applying conservation of energy.
The mass flow rate is needed and can be evaluated from the given volumetric flow and density. All the
properties will be obtained from the psychrometric chart since the total pressure is one atmosphere.
Solution:
a) For the overall system we assume [A1], [A2], [A3], and [A4]. No water is
added or removed in the heating section so that ω 2 = ω3 . At location 3 we are given the
dry-bulb temperature and relative humidity. From the psychrometric chart, Figure
S15-6, we obtain: ω 3 ≈ 0.0084 kg vapor kg dry air , h3 ≈ 43kJ kg dry air .
The cooling coil cools the mixture such that water vapor condenses from the
mixture, which results in the mixture exiting the coil as saturated air and φ2 = 100% .
Because ω 2 = ω3 , we now have two properties at 2, so from the psychrometric chart:
T2 ≈ 11.2 o C , h2 ≈ 32 kJ kg dry air .
b) From the psychrometric chart, at location 1 conditions: ρ = 0.904 kg m 3 ,
h1 ≈ 96 kJ kg dry air , ω1 ≈ 0.0242 kg vapor kg dry air . The mass flow rate is:
Assumptions:
A1. The system is
steady.
A2. Potential and
kinetic energy effects
are negligible.
A3. No work occurs in
the system.
A4. The moist air
behaves as an ideal gas
mixture.
kg ⎞ ⎛ m3 ⎞
kg
kg
⎛
m a = ρV = ⎜ 0.904 3 ⎟ ⎜ 8
= 0.121
⎟ = 7.23
m ⎠ ⎝ min ⎠
min
s
⎝
25
The condensate flow at location 4 is obtained from conservation of mass with
[A1] applied only to the water flow:
m v1 − m v 2 − m 4 = 0
From Eq. S15-43, m v = ω m a . Substituting this into the above equation and solving for
the condensate flow rate:
m 4 = m v1 − m v 2 = ω1m a − ω 2 m a = (ω1 − ω 2 ) m a
kg ⎞
kg
⎛
= ( 0.0242 − 0.0084 ) ⎜ 0.121 ⎟ =0.00191
s ⎠
s
⎝
c) The heat transfer rate in the cooling coil is obtained by conservation of
energy. Define the control volume to include the flows in the duct from 1 to 2, and
exclude the coil itself. With assumptions [A1], [A2], [A3], and [A4], conservation of
energy gives:
Q + m 1h1 − m 2 h2 − m 4 h4 = 0
Evaluating the enthalpy at locations 1 and 2 from the psychrometric chart gives the
mixture enthalpy, hmix = ha + ω hv , which is per unit mass of dry air. For consistency,
the flow rates at 1 and 2 must be the dry air flows. From conservation of mass for a
steady system, the dry air flow rate is constant, so m a ,1 = m a ,2 = m a ,3 = m a = ρV and the
energy equation is rewritten:
Q = m a ( h2 − h1 ) − m 4 h4
The temperature of the liquid water exiting at 4 is not given. We assume that
temperature is the same as at location 2, and the enthalpy at 4 is the saturated liquid
enthalpy. From Table A-10 at 11.2 °C (by interpolation), h4 = h f ,4 = 47.0 kJ kg .
Therefore, the heat transfer rate in the cooling coil is:
kg ⎞
kJ ⎛
kg ⎞ ⎛
kJ ⎞ ⎛ 1kW ⎞
⎛
Q = ⎜ 0.121 ⎟ ( 32-96 ) - ⎜ 0.00191 ⎟ ⎜ 47.0 ⎟ ⎜
⎟ =-7.83kW
s ⎠
kg ⎝
s ⎠⎝
kg ⎠ ⎝ 1kJ s ⎠
⎝
The negative sign indicates that heat is being removed from the air stream.
d) The heat transfer rate in the reheat section (between locations 2 and 3) is
also determined from conservation of energy. No water vapor is added or removed in
this section, so with the same assumptions as used previously:
kg ⎞
kJ ⎛ 1kW ⎞
⎛
Q = m a ( h3 − h2 ) = ⎜ 0.121 ⎟ ( 43-32 ) ⎜
⎟ =1.33kW
s ⎠
kg ⎝ 1kJ s ⎠
⎝
The heat transfer is positive, which indicates we are adding heat to the air stream (the
temperature rises from 11.2 °C to 22 °C).
Comments:
To cool the dry air from T1 to T2 (ignoring the condensing water vapor and the condensate flow
rate), Q = m a ( ha ,2 − ha ,1 ) . From Table A-9, ha,1(34 °C) = 307.23 kJ/kg, ha,2(11.2 °C) = 284.33 kJ/kg, so
that Q = -2.77 kW. As can be seen by comparing this value with that which takes into account the
condensation (-7.83 kW), ignoring the condensation would result in a very large error. For the reheat
section, ignoring the flow of the water vapor, Q = m a ( ha ,3 − ha ,2 ) , ha,3(22 °C) = 295.17 kJ/kg, and Q =
1.31 kW. The difference between this value and that given above in the problem solution (1.33 kW) shows
that because of the small amount of water vapor in the mixture, ignoring the water vapor has little effect on
the sensible heating. Note also that there are uncertainties in the results using the charts because of reading
accuracy.
26
Example S15-7 Cooling tower
Cooling water circulated through the tubes of the condenser in a Rankine cycle must be
maintained at a relatively constant inlet temperature. Where cooling water is not available from a lake,
ocean, or river, cooling towers are used to remove heat from the circulating water so that the water can be
recycled nearly indefinitely; some water is lost during the cooling process due to evaporation, and
additional water (make-up) must be added to maintain a fixed amount of water in the system.
Hot water at a flow rate of 500,000 gal/min enters the cooling tower shown below at a temperature
of 120 °F and must be cooled to 80 °F. Air enters the tower at 70 °F with a relative humidity of 50% and
leaves at 115 °F with a relative humidity of 95% (to avoid fog formation). The barometric pressure is 14.1
psia. The make-up water is at 60 °F. Determine:
a) the required air flow rate (in lbm/min)
b) the rate of water evaporation (in lbm/min and gal/min).
Approach:
A schematic of the system and the given information are shown on the figure below. We are given
all the inlet and exit moist air conditions and the inlet liquid water flow rate; we also know the exit liquid
water flow rate because make-up water is added to maintain the liquid flow constant. The dry air mass
flow rate is constant. The inlet water vapor flow rate can be determined from its specific humidity, as can
the exit water vapor flow rate, if we can calculate the dry air flow rate. We can determine the air flow rate
by application of conservation of mass to the water and air individually, and conservation of energy to the
whole system. Note that the psychrometric chart cannot be used because this system is not operating at one
standard atmosphere.
27
Solution:
a) For the control volume drawn around the complete cooling tower, we
assume [A1], [A2], [A3], [A4], [A5], and [A6]. Conservation of mass applied to the air
and water individually gives:
Dry air: m a1 − m a 2 = 0 → m a1 = m a 2 = m a
Water: m v1 + m L 3 + m L 4 − m v 2 − m L 5 = 0
The total liquid flows at 3 and 5 are equal: m L 3 = m L 5 , so the water mass balance
equation is:
m v1 + m L 4 − m v 2 = 0 → m L 4 = m v 2 − m v1
where m L 4 is the make-up water flow rate (or water evaporation rate). From Eq.
S15-43, m v = ω m a , so substituting this into the water mass balance equation:
m L 4 = m v 2 − m v1 = ω 2 m a − ω1m a = (ω 2 − ω1 ) m a
The make-up water flow rate can be calculated once the dry air mass flow rate is known.
Applying conservation of energy to the control volume:
( m a ha1 + m v1hv1 ) + m L3 hL3 + m L 4 hL 4 − ( m a ha 2 + m v 2 hv 2 ) − m L5 hL5 = 0
Assumptions:
A1. The system is
steady.
A2. Potential and
kinetic energy effects
are negligible.
A3. No work occurs in
the system.
A4. The cooling tower is
adiabatic.
A5. Each gas is ideal.
A6. The mixture
behaves as an ideal gas
mixture and follows
Dalton’s Law.
The enthalpy of liquid water and the enthalpy of vapor can be approximated with their
respective saturation enthalpies evaluated at the given temperatures. Likewise, the air
enthalpies can be evaluated at their given temperatures. Examination of this equation
shows that the only unknown is the dry air mass flow rate. Solving for that quantity:
28
m L 3 ( hL 5 − hL 3 )
m a =
ha1 − ha 2 + ω1hv1 − ω 2 hv 2 + (ω 2 − ω1 ) hL 4
To calculate the specific humidities we use Eq. S15-42, where φ1 = 0.50 ,
φ2 = 0.95 , P1 = P2 = 14.1 psia , T1 = 70 o F , T2 = 115 o F
ω1 =
ω2 =
0.622φ1 Pg1
P1 − φ Pg1
=
0.622 ( 0.50 )( 0.3632 )
14.1 − ( 0.50 )( 0.3632 )
0.622 ( 0.95 )(1.486 )
14.1 − ( 0.95 )(1.486 )
= 0.0692
= 0.00812
lbm vapor
lbm dry air
lbm vapor
lbm dry air
The water enthalpies and density are determined from Table B-11, and the air enthalpies
are from Table B-9. The liquid water mass flow is:
lbm ⎞ ⎛
gal ⎞ ⎛
ft 3 ⎞
⎛
6 lbm
0.1337
m L 3 = ρ LV = ⎜ 61.69 3 ⎟ ⎜ 500,000
⎟ =4.12×10
⎟⎜
ft ⎠ ⎝
min ⎠ ⎝
gal ⎠
min
⎝
⎛
6 lbm ⎞
⎜ 4.12×10
⎟ ( 48.09 − 88.00 )
min ⎠
⎝
m a =
126.7 − 137.5 + ( 0.00812 )(1092.0 ) − ( 0.0692 )(1111.4 ) + ( 0.0692 − 0.00812 )( 28.08 )
= 2.13×106
lbm
min
b) The make-up water flow rate (evaporation mass flow rate) is:
m L 4 = ( 0.0692 − 0.00812 ) ( 2.13×106 lbm min ) = 130,100 lbm min
or in terms of volume flow VL 4 = m L 4 ρ = 15,800 gal/min.
Comments:
The heat transfer rate from the cooling water is approximately:
Q out ≈ m L 3 c p , w (T3 − T5 ) = 1.65 × 108 Btu min ≈ 2900 MW .
For a heat power cycle, η = W net Q in = W net
( Q
out
)
+ W net . Assuming a cycle thermal efficiency of 40%, the
net power output would be ≈ 1933 MW. This indicates that this cooling tower is for a large power plant.
Example S15-8 Air compressor with inlet fogging
Power requirements increase in the summer because of the demand for air conditioning. However,
as the air temperature rises at the inlet to the compressor of a Brayton cycle, the net power produced by the
cycle decreases. A technique to reduce the inlet air temperature is to humidify the air. This is particularly
effective in dry climates. In Example 8-12 we analyzed a simple Brayton cycle using an air-standard
analysis; the air was dry. We now want to analyze the compressor with inlet humidification (fogging) and
incorporate the effects of water vapor in the analysis. A fogging unit consists of numerous spray nozzles
that atomize liquid water into very small drops that evaporate quickly in the dry air.
In a high desert environment, air at 14 lbf/in.2, 100 °F, 10% relative humidity with a volumetric
flow rate of 10,000 ft3/min enters a fogging assembly ahead of the compressor inlet in a Brayton cycle that
has a pressure ratio of 14. Sufficient water at 66 °F is added to the air to raise its relative humidity to 100%
and to lower its temperature before it enters the compressor; the volume flow rate remains the same. The
isentropic efficiency of the compressor is 83%. Determine:
a) the required water flow rate (in lbm/min)
b) the air temperature at the inlet to the compressor (°F)
29
c) the compressor power (in kW).
Compare to the results of Example 8-12.
Approach:
A schematic of the system and the given information is shown below:
The liquid water added is determined from the difference between the water vapor flows at locations 1 and
2. The specific humidity is needed at those two locations, as well as the dry air mass flow rate. Because
we have two properties at location 1, the specific humidity there can be determined. At location 2, we
know only that the air is saturated. However, the fogging unit works similar to an adiabatic saturation
device, and the outlet temperature is equal to the inlet water temperature. To determine the compressor
power, we apply conservation of mass and energy in the same way as we have done previously when we
have analyzed a compressor. The main difference is the evaluation of the properties, which now must
include the effects of water vapor, instead of only dry air.
Solution:
a,b) We begin by applying conservation of mass to the control volume around
the fogging unit. Assuming [A1], [A2], [A3], [A4], and [A5], we obtain:
m a1 − m a 2 = 0 → m a1 = m a 2 = m a
Air:
Water: m v1 + m L 4 − m v 2 = 0 → m L 4 = m v 2 − m v1
Using the definition of specific humidity, the liquid mass flow rate can be rewritten as:
m L 4 = (ω 2 − ω1 ) m a
At location 1, we know T1 = 100 o F, φ1 = 10% , so we can determine ω1 . Using Table
A-10 to evaluate the saturation pressure of water at 100 ºF:
0.622φ1 Pg1 0.622 ( 0.10 )( 0.9503)
lbm vapor
=
= 0.00425
ω1 =
14.0 − ( 0.10 )( 0.9503)
lbm dry air
P1 − φ Pg1
Assumptions:
A1. The system is steady.
A2. Potential and kinetic
energy effects are
negligible.
A3. The system is
adiabatic.
A4. Each gas is ideal.
A5. The mixture behaves
as an ideal gas and
follows Dalton’s law.
We do not have enough given information to calculate ω 2 immediately. We know
φ2 = 100% , and we can determine a second property by considering the fogging
process. Air enters the device and flows through it while water vapor is evaporated into
30
it; at the exit, the air is saturated. This is the same process that occurs in an adiabatic
saturation device as described above. Because of this, we know that the outlet air
temperature equals the inlet liquid water temperature; that is, T2 = T4 = 66 ºF. With this,
we can calculate the specific humidity at 2:
0.622 (1.00 )( 0.3165 )
lbm vapor
ω2 =
= 0.0144
14.0 − (1.00 )( 0.3165 )
lbm dry air
Once we have the dry air mass flow rate we can calculate the amount of water
needed. The density of the dry air is:
lbf ⎞ ⎛
lbm ⎞ ⎛ 144in.2 ⎞
⎛
⎟
⎜14.0 2 ⎟ ⎜ 28.97
⎟⎜
in. ⎠ ⎝
lbmol ⎠ ⎝ ft 2 ⎠
lbm
PM ⎝
ρa =
=
= 0.0675 3
ft
RT
⎛
⎞
ft lbf
⎜ 1545
⎟ (100+460 ) R
lbmol R ⎠
⎝
and the dry air mass flow rate is:
lbm ⎞ ⎛
ft 3 ⎞
lbm
⎛
m a = ρV = ⎜ 0.0675 3 ⎟ ⎜ 10000
⎟ = 675
ft ⎠ ⎝
min ⎠
min
⎝
The liquid water flow rate is:
lbm ⎞
lbm
⎛
m L 4 = (ω 2 − ω1 ) m a = ( 0.0144 − 0.00425 ) ⎜ 675
⎟ =6.85
min ⎠
min
⎝
c) Using the compressor control volume defined in the schematic, and
assuming [A1], [A2], [A3], [A4], [A5], and [A6], conservation of mass and energy
gives us:
WC , s = ( m a ha 3 + m v 3 hv 3 ) s − ( m a ha 2 + m v 2 hv 2 ) = m a ⎡⎣( ha 3 + ω 3 hv 3 ) s − ( ha 2 + ω 2 hv 2 ) ⎤⎦
A6. The compressor is
isentropic.
with the subscript s added to emphasize the isentropic process. Using the definition of
isentropic efficiency, we have:
W
WC = C ,ideal
ηC
Our task is to evaluate the exit enthalpies for an isentropic process. We know the exit
pressure; we need the (common) exit temperature.
Application of the entropy balance equation to the control volume with
assumptions [A1], [A3], and [A7], we obtain:
S 2 = S3 → m a sa 2 + m v sv 2 = m a sa 3 + m v sv 3 → sa 2 + ω 2 sv 2 = sa 3 + ω3 sv 3
The dry air mass flow rate cancels out of the equation. Because no water is added or
removed in the compressor, ω 2 = ω3 . Rearranging the equation and bringing in the
expression for the entropy change of an ideal gas:
⎡
⎛P ⎞
R
R ⎛ P3 ⎞ ⎤
sa 2 − sa 3 = ω3 ( sv 3 − sv 2 ) → sao2 − sao3 −
ln ⎜ 2 ⎟ = ω 3 ⎢ svo3 − svo2 −
ln ⎜ ⎟ ⎥
M a ⎝ P3 ⎠
M v ⎝ P2 ⎠ ⎦
⎣
A7. Entropy generation is
zero.
The inlet entropies are known, and the ratio of the partial pressures is equal to the ratio
of the total pressures because the composition is fixed. Once the exit temperature is
known, the exit entropies can be determined and this equation evaluated. We can do
this iteratively. From Table A-9 (by interpolation), sao2 = 0.59444 Btu lbm ⋅ R and
sv02 = ( 44.91 18.02 ) = 2.4923Btu lbm ⋅ R , so that the equation to solve is:
1.986 ⎛ 14 ⎞
⎡ o
1.986 ⎛ 196 ⎞ ⎤
ln ⎜
ln ⎜
⎟ = ( 0.0144 ) ⎢ sa 3 − 2.4923 −
⎟
28.97 ⎝ 196 ⎠
18.02 ⎝ 14 ⎠ ⎥⎦
⎣
Consistent units are used and are not shown for brevity. Rearranging the equation:
0.59444 − sao3 −
31
0.59444 −
1.986 ⎛ 14 ⎞
⎡
1.986 ⎛ 14 ⎞ ⎤
ln ⎜
ln ⎜
⎟ + ( 0.0144 ) ⎢ 2.4923 +
⎟
28.97 ⎝ 196 ⎠
18.02 ⎝ 196 ⎠ ⎥⎦
⎣
= 0.8154 = ( 0.0144 ) svo3 + sao3
We guess T3, evaluate sao3 and svo3 , and then evaluate the right hand side of the equation.
Note that the water vapor properties are in molar units and must be divided by the
molecular weight of water (18.02).
Right hand side of
T3
sao3
svo3
equation
(Btu/lbm·R)
(R)
(Btu/lbm·R)
(Btu/lbm·R)
1000
0.75042
2.7853
0.7905
1120
0.77880
2.8397
0.8197
1100
0.77422
2.8309
0.8150
This is close enough, so T3 ≈ 1100 R = 640 ºF, and the enthalpies evaluated at the
appropriate temperatures are ha2 = 125.7 Btu/lbm, ha3 = 266.0 Btu/lbm, hv2 = 231.4
Btu/lbm, and hv3 = 496.2 Btu/lbm. Finally,
675lbm min
Btu
WC =
⎡ 266.0+ ( 0.0144 )( 496.2 ) ⎦⎤ - ⎣⎡125.7+ ( 0.0144 )( 231.4 ) ⎦⎤
⎣
0.83
lbm
⎛ 1min ⎞⎛ 1.0551kW ⎞
×⎜
⎟⎜
⎟ =2060 kW
⎝ 60s ⎠⎝ 1Btu s ⎠
{
}
Comments:
The compressor in Example 8-12 (without inlet fogging) required 2150 kW and the outlet
temperature was 745 ºF. With inlet fogging, the power required is 2060 kW and the outlet temperature was
640 ºF; note also that the mass flow rate through the compressor is about 1% greater than that without
fogging. The reduced compressor power would translate directly into additional net power produced by the
cycle. However, more heat must be added in the combustor to raise the mixture temperature from 640 ºF
with fogging to 745 ºF without fogging.
15.4 Combustion
Until now we have focused on nonreacting fluids and mixtures. However, chemically reacting
systems are common in industrial practice, such as in power production (e.g., heat input into Rankine and
Otto cycles through the combustion of fossil fuels), chemical processing, fuel cells, pollution control (e.g.,
catalytic converters), and other applications. We focus on one particular type of chemically reacting
system, the combustion of a hydrocarbon fuel (i.e., a combustible material) and air. During combustion of
a fuel and air, a large quantity of energy is released and different chemical compounds are created. The
same conservation equations developed and used for nonreacting systems are applicable to chemically
reacting systems as well. The primary difference is that the properties used in the conservation equations
must be modified; in addition, we must take into account the differences in chemical composition before
and after the combustion process.
Hydrocarbon fuels contain hydrogen and carbon, and often sulfur, nitrogen, and other chemical
substances. We will ignore everything except the hydrogen and carbon. The fuels exist in solid, liquid,
and gaseous form. Typical gases include methane, butane, and propane (natural gas is composed mostly of
these three components); liquids include gasoline, kerosene, and diesel fuel; and a solid hydrocarbon fuel is
coal, which can have a wide range of properties.
The combustion of a hydrocarbon fuel requires oxygen, which is supplied by atmospheric air. Dry
air is composed of oxygen (O2), nitrogen (N2), and many other elements at small concentrations; its
chemical composition is shown in Table S15-1. Atmospheric air is dry air plus water vapor. For simplicity
we will ignore everything except the oxygen and nitrogen. That is, we will assume that the mole fraction of
oxygen is 21% and will account for all the other gases with a mole fraction of nitrogen of 79%. Stated
another way, there are 79/21 = 3.76 moles of nitrogen for each mole of oxygen in the air. The actual
combustion process is quite complex. We idealize the process by assuming we have complete combustion,
32
which implies that sufficient oxygen is present such that the products of combustion will consist of carbon
dioxide (CO2), oxygen (O2), nitrogen (N2), and water (H2O). No unoxidized carbon (C), carbon monoxide
(CO), hydrogen (H2), or hydroxyl (OH) will exist in the products of combustion. Nitrogen is a relatively
stable element and does not react readily with other gases. It will be considered to be inert and nonreactive
with any of the other gases at the temperatures we use. An idealized schematic of the combustion process
is given in Figure S15-11.
Figure S15-11
Table S15-1 Approximate chemical composition of dry air
Component
Mole Fraction (%)
Mass Fraction (%)
75.5
78.08
Nitrogen (N2)
23.1
20.95
Oxygen (O2)
1.3
0.93
Argon (Ar)
0.05
0.03
Carbon dioxide (CO2)
<0.02
<0.01
Neon, helium, methane,
and others
Combustion is governed by a chemical equation of the form:
Reactants ⇒ Products
or
Fuel + Oxidizer ⇒ Products
Total mass is conserved (there are no nuclear reactions) in this process. In addition, the mass of each
element must be conserved, although the elements will be combined in different compounds on the two
sides of the equation (except for nitrogen, N2, as noted above). We ensure this conservation through the
use of a chemical equation, as described in introductory chemistry courses.
Consider one of the simplest reactions, the complete combustion of hydrogen and oxygen.
A(H 2 ) + B ( O2 ) ⇒ C ( H 2 O )
Eq. S15-51
Figure S15-11 Idealized combustion process in a device
In this reaction, “A” moles of hydrogen and “B” moles of oxygen combine to form “C” moles of water. We
need to find the stoichiometric coefficients A, B, and C to ensure conservation of each element. Balances
on the number of atoms of hydrogen and oxygen give
Hydrogen:
2 A = 2C
Oxygen:
2B = C
33
There are three unknowns and only two equations. We arbitrarily choose A = 1, so that we always deal
with one mole of the fuel. Solving the two balance equation, we obtain B = 0.5 and C = 1. This results in
the balanced equation:
H 2 +0.5O 2 ⇒ H 2 O
Eq. S15-52
This equation states that one mole (or one kmol or one lbmol) of hydrogen plus a half mole (or one-half
kmol or one-half lbmol) of oxygen combine to form one mole (or one kmol or one lbmol) of water. The
important point is that mass is conserved but different numbers of moles of compounds may be present on
the two sides of the equation. Another way to look at this equation is to remember that the molecular
weight of hydrogen is about 2 kg/kmol, for oxygen it is 32 kg/kmol, and for water it is 18 kg/kmol. Thus,
the equation can be interpreted as 2 kg of hydrogen plus 16 kg of oxygen combine to form 18 kg of water.
(Similar results are obtained with English units.)
The air-fuel ratio ( AFR ) is often used to quantify the amount of air per unit mass of fuel in a
combustion process; the ratio can also be described in terms of per unit mole of fuel ( AFR ).
AFR =
ma
M
= AFR a
mF
MF
Eq. S15-53
The minimum amount of air required for complete combustion is called the theoretical air or
stoichiometric air. If additional air is present, the amount can be expressed in terms of percent of
theoretical air or in terms of percent excess air (which is theoretical air percent minus 100%). Percent of
theoretical air can be determined by dividing the actual mass air-fuel ratio by the mass air-fuel ratio for
stoichiometric combustion:
Percent theoretical air =
AFRact
AFR act
=
AFRstoichiometric AFR stoichiometric
Eq. S15-54
Example S15-9 Combustion of methane
Methane (CH4) is burned with the stoichiometric amount of air. Determine the air-fuel ratio on a
mass and a molar basis.
Approach:
The chemical equation for combustion of methane and air must be written and the number of
moles needed for conservation of each element determined. Once those are found, the mass of air and fuel
can be calculated from the stoichiometric coefficients and molecular weights of each substance. Finally,
Eq. S15-53 can be used to convert the mass AFR to the molar AFR .
Solution:
For complete combustion of methane and air, we assume [A1], [A2], [A3], and
[A4]. The reaction equation for burning methane and air is:
A(CH 4 ) + B ( O2 +3.76N 2 ) ⇒ C ( N 2 ) + D ( H 2 O ) + E ( CO2 )
Balances on the elements gives:
Assumptions:
A1. Air is composed only
of 3.76 moles of N2 for
each mole of O2.
A2. The products of
combustion behave as an
ideal gas mixture.
A3. N2 is inert.
34
Carbon:
Hydrogen:
Oxygen:
A=E
4 A = 2D
2 B =D + 2 E
Nitrogen ( N 2 ): 3.76 B = C
Note that nitrogen is balance as N2, because we have assumed that it never reacts with
any other substance. We arbitrarily set A =1 so that we deal with one mole of fuel.
From the carbon balance, E = 1, the hydrogen balance, D = 2, the oxygen balance, B =
2, and the nitrogen balance, C = 7.52, so that the chemical equation is:
CH 4 + 2 ( O2 +3.76N 2 ) ⇒ 7.52 ( N 2 ) + 2 ( H 2 O ) + CO2
A4. Complete
combustion occurs with
combustion products only
CO2, H2O, and N2.
The molar air-fuel ratio is
AFR =
2 (1 + 3.76 )
nair
moles air
=
= 9.52
n fuel
1
mole fuel
The mass air-fuel ratio is obtained with Eq. S15-53, and the molecular weights
are from Table A-1:
AFR = AFR
M air
kg air
lbm air
⎛ 28.97 ⎞
= 9.52 ⎜
= 17.2
⎟ = 17.2
16.043
kg
fuel
lbm
fuel
M fuel
⎝
⎠
Comment:
With stoichiometric combustion, no excess oxygen is present in the products of the reaction.
Example S15-10 Excess air in combustion
The methane in Example S15-9 is burned with air at 14.7 psia in a constant pressure process. If
the wall temperature of the duct in which the products of combustion flow (e.g., an exhaust stack) is lower
than the dew-point temperature, water vapor will condense out of the mixture. This can cause, among other
problems, corrosion and damage to the duct. Determine:
a) the percentage of theoretical air if the mass air-fuel ratio is 35
b) the dew-point temperature of the products with the excess air found in part (a) (in °F).
Approach:
The definition of percent theoretical air (Eq. S15-54) is used with the given information. The
dew-point temperature is the saturation temperature of the water vapor at its partial pressure. The water
vapor partial pressure is determined with its mole fraction, and the mole fraction requires that the chemical
reaction equation be balanced for the excess air situation.
Solution:
a) The percent theoretical air can be calculated with Eq. S15-54. The actual
AFR is given in the problem statement; the stoichiometric AFR was calculated in
Example S15-9:
AFRact
35lbm air lbm fuel
Percent theoretical air =
=
= 2.03 = 203%
AFRstoichiometric 17.2 lbm air lbm fuel
Note that we can express this number in terms of excess air:
Assumptions:
Same as in Example
S15-9
35
Percent excess air = Percent theoretical air − 100% = 203% − 100% = 103%
b) To determine the dew-point temperature, we first need to determine the
mole fraction of the water vapor in the products of combustion, so we need to rebalance
the reaction equation, taking into account the additional air:
A(CH 4 ) + 2.03B ( O2 +3.76N 2 ) ⇒ C ( N 2 ) + D ( H 2 O ) + E ( CO2 ) + F ( O2 )
where the factor 2.03 represents the theoretical air.
Following the approach given in Example S15-9, the reaction equation is again
balanced, and balances on the elements give:
A=E
Carbon:
Hydrogen:
4 A = 2D
Oxygen:
Nitrogen ( N 2 ):
2..03B ( 2 ) = D + 2 E + 2 F
2.03B ( 3.76 ) = C
Again, we arbitrarily set A = 1. We also know the value of B from the stoichiometric
balance equation, B = 2. From the equations, we determine C = 15.3, D = 2, E = 1, and
F = 2.06, so that the final equation is:
CH 4 + 4.06 ( O 2 +3.76N 2 ) ⇒ 15.3 ( N 2 ) + 2 ( H 2 O ) + CO2 + 2.06 ( O2 )
The vapor pressure is the water mole fraction times the total pressure:
2
lbf
lbf
⎡
⎤
14.7 2 =1.44 2
Pv = YH 2O Ptot = ⎢
⎥
in.
in.
⎣15.3 + 2 + 1 + 2.06 ⎦
From Table B-11, the saturation temperature corresponding to this pressure is 113.9 °F.
Comments:
With excess air used in the combustion process, oxygen is present in the products of the reaction.
Note that the stoichiometric coefficient on the oxygen in the products can be calculated directly if the
excess air is known: F = B×(% excess air) = 2(1.03) = 2.06.
The reaction between hydrogen and oxygen shown by the mass balance in Eq. S15-52 is
accompanied by the liberation of energy. Thus, in addition to applying conservation of mass to a reaction,
we also must apply conservation of energy.
Application of conservation of energy to pure and nonreacting substances involves evaluation of
differences in properties (enthalpy or internal energy) at two different states. Because of this characteristic,
properties of water or ideal gases listed in the appendices at the back of the book are given and used
without regard to the arbitrary reference states used to create the tables. We can do this because the
substance does not change its chemical composition; that is, it does not combine with other substances, it
only changes state. During combustion, the reactants (individual substances) combine to form new
products (or compounds). Application of conservation of energy to a reaction must account for the energy
involved in the formation of the new compound from its constituent parts. Thus, an appropriate and
common reference state for the enthalpy or internal energy of the reactants before the reaction and of the
products after the reaction must be defined.
The arbitrarily chosen standard reference state is defined at Tref = 298.15 K (25 ºC or 77 ºF) and
Pref = 1 atm. The enthalpies of stable elements (e.g., H2, O2, N2, C) are given a value of zero at the standard
reference state; monotonic H, O, and N are not stable and do not have an enthalpy of zero at the reference
state. The symbol for the reference enthalpy is hº = 0, with the superscript º indicating the reference state;
on a molar basis, h o = 0 . With this chosen datum, changes in properties caused by changes in chemical
composition can be found.
We define the enthalpy of formation, ∆h fo , of a compound as the difference between the enthalpy
of the compound per unit mole at the reference state and the enthalpies of its constituent stable elements
also at the reference state. It is the energy released (an exothermic reaction) or absorbed (an endothermic
reaction) when the compound is formed.
36
k
o
∆h fo = hcompound
− ∑ ni hi o
Eq. S15-55
i =1
Because each stable element by definition has an enthalpy of zero at the reference state, the second term on
the right hand side of Eq. S15-55 is zero and the enthalpy of the compound at the reference state is equal to
the enthalpy of formation
o
∆h fo = hcompound
Eq. S15-56
which can be measured experimentally or through use of other thermodynamic techniques. Values of this
quantity are given in Tables SA-3 and SB-3 for common compounds. Note that both positive and negative
values are listed, which are consistent with the sign convention adopted for heat transfer in Chapter 2; that
is, heat transfer to a system is positive and from a system is negative. If a reaction is exothermic, then the
sign on the enthalpy of formation is negative; for an endothermic reaction, the sign is positive. Because
many of these substances can exist in either a gaseous or liquid state, sometimes the state is indicated by
either a “g” or an “ A .” For example, liquid propane can be designated as C3 H 8 A .
Combustion systems are designed for the fuel and air to enter at temperatures and pressures
similar to the standard reference state (e.g., the gasoline and air used in an automobile engine) or at
elevated temperatures and pressures (e.g., the high pressure air used in a Brayton cycle is preheated in a
regenerator to a high temperature before combustion occurs). In both cases, the products of combustion
typically leave at a high temperature. The specific enthalpy of a compound at a state different than the
standard reference state can be determined with:
h (T , P ) = ∆h fo + ⎡⎣ h (T , P ) − h (Tref , Pref ) ⎤⎦ = ∆h fo + ∆h
Eq. S15-57
The term in brackets on the right hand side of Eq. S15-57 represents the enthalpy change, ∆h , of the
elements or compounds (constant composition) on either side of a chemical equation.
Consider the application of conservation of energy to a steady, open system that undergoes a
chemical reaction; kinetic and potential energies are ignored:
Q − W + ∑ m i hi − ∑ m i hi = 0
in
out
It is convenient to use molar flow rates and enthalpies because we obtain the molar quantities from the
chemical reaction equation. From Eq. S15-6, we can show that ni = m i M i is the molar flow rate of each
constituent. Solving for mass flow rate and substituting this into the above equation results in terms
involving ni ( hi M i ) = ni hi , so that the energy equation can be written as:
Q − W + ∑ ni hi − ∑ ni hi = 0
in
Eq. S15-58
out
The constituents flowing into the system are the reactants; the constituents flowing out of the system are the
products. (See Figure S15-11.) We substitute Eq. S15-57 into Eq. S15-58 and rearrange to obtain:
⎡
Q − W = ⎢ ∑ ni ∆h fo
⎣ out
(
) − ∑ n ( ∆h ) ⎤⎥ + ⎡⎢∑ n ( ∆h ) − ∑ n ( ∆h ) ⎤⎥
i
i
in
o
f i
⎦ ⎣ out
i
i
i
in
i
⎦
Eq. S15-59
We usually write the chemical balance equation based on one mole of the fuel (such as shown in Eq.
S15-52); the stoichiometric coefficients for the other compounds are found accordingly. We divide Eq.
S15-59 by the molar flow rate of fuel, nF , so that everything is described per mole of fuel. The ratio
37
ni n F is simply equal to the stoichiometric coefficient in the chemical reaction equation if one mole of fuel
is used. Thus, the energy equation is:
Q W ⎡
⎤ ⎡
⎤
−
= ∑ n ∆h fo − ∑ ni ∆h fo ⎥ + ⎢ ∑ ni ∆h − ∑ ni ∆h ⎥
i
i
i
i
nF nF ⎢⎣ out i
in
in
⎦ ⎣ out
⎦
(
)
(
)
( )
( )
Eq. S15-60
The first bracketed term on the right hand side of Eq. S15-60 is composed of only the enthalpy of
formation and associated stoichiometric coefficients. For complete combustion of fuel with air as the
oxidizer, we define this term as the enthalpy of combustion, ∆hCo (or heat of combustion). The enthalpy of
combustion can be determined experimentally and is often used because the enthalpy of formation of some
fuels is not known. Thus, we rewrite Eq. S15-60 as:
Q W
⎡
⎤
−
= ∆hCo + ⎢ ∑ ni ∆h − ∑ ni ∆h ⎥
i
i
n F n F
in
⎣ out
⎦
( )
( )
Eq. S15-61
One of the compounds produced in the combustion of a hydrocarbon fuel is water. Water can
exist in a liquid or a vapor state. The value of the enthalpy of combustion when liquid water is produced is
called the higher heating value, HHV, and when water vapor is produced the enthalpy of combustion is
called the lower heating value, LHV. The difference between these two quantities is the energy required to
condense the water vapor in the products of combustion:
HHV = LHV + (nh fg ) H 2 O
Eq. S15-62
where n is the moles of water in the products and h fg is the enthalpy of vaporization at standard
temperature. Note that the HHV and LHV are expressed as positive numbers. Values are given in Tables
SA-3 and SB-3.
Example S15-11 Heat transfer from a combustion process
Liquid propane ( C3 H 8 A ) at 25 °C is mixed and burned with 250% theoretical air (150% excess
air) that enters the combustion chamber at 17 °C. The propane flow rate is 0.25 kg/s. The combustion
process is complete, and the products of combustion leave the chamber at 727 °C. Determine:
a) the air mass flow rate (in kg/s)
b) the heat transfer rate from the combustion chamber (in kW).
Approach:
A schematic of the system and given information are shown on the schematic below:
38
Because we know the fuel flow rate, if we can determine the AFR, then the air mass flow rate can be
calculated. The air-fuel ratio can be determined by balancing the chemical equation and using Eq. S15-53.
With both the air and fuel flow rates known, conservation of energy can be applied to the control volume
shown in the schematic, with the properties evaluated from the appropriate tables.
Solution:
a) We begin the solution by assuming [A1], [A2], [A3], and [A4]. When the
fuel is burned with stoichiometric air, the combustion products are only CO2, H2O, and
N2. When excess air is used, the combustion products again will be CO2, H2O, and N2,
but now there also will be free O2 in the products. The amounts of CO2 and H2O in the
products will be the same for both situations, because all of the fuel is oxidized, so we
first balance the reaction equation for propane and stoichiometric air:
A(C3 H8 A) + B ( O2 +3.76N 2 ) ⇒ C ( N 2 ) + D ( H 2 O ) + E ( CO2 )
Balancing the element equations:
Carbon:
3A = E
Hydrogen:
8 A = 2D
Oxygen:
2.B = D + 2 E
Assumptions:
A1. Air is composed only
of 3.76 moles of N2 for
each mole of O2.
A2. The products of
combustion behave as an
ideal gas mixture.
A3. N2 is inert.
A4. Complete
combustion occurs with
combustion products only
CO2, H2O, N2, and O2.
Nitrogen ( N 2 ): 3.76B = C
We arbitrarily set A =1 so that we deal with one mole of fuel. Solving for the
stoichiometric coefficients, we obtain B = 5, C = 18.8, D = 4, and E = 3:
C3 H8 A + 5 ( O2 +3.76N 2 ) ⇒ 18.8 ( N 2 ) + 4 ( H 2 O ) + 3 ( CO2 )
Now we write the reaction equation for burning liquid propane with 150%
excess air using the stoichiometric coefficients for the fuel, CO2, and H2O:
C3 H8 A + (1.5 + 1) B ( O 2 +3.76N 2 )
⇒ (1.5 + 1) B ( 3.76 )( N 2 ) + 4 ( H 2 O ) + 3 ( CO 2 ) + 1.5 B ( O 2 )
The quantity (1.5+1) represents the theoretical air used in this reaction, and the
coefficient of 1.5 on O2 represents the excess air. Using the balance on the O2:
(1.5 + 1) 2B = 4 + 6 + (1.5) 2B
Solving for B = 5, and the chemical equation becomes:
C3 H8 A + 12.5 ( O 2 +3.76N 2 ) ⇒ 47.0 ( N 2 ) + 4 ( H 2 O ) + 3 ( CO2 ) + 7.5 ( O2 )
The molar air-fuel ratio is
12.5 (1 + 3.76 )
n
moles air
AFR = a =
= 59.5
1
mole fuel
nF
and the mass air-fuel ratio (with molecular weights from Table A-1) is:
M
kg air
⎛ 28.97 ⎞
AFR = AFR a = 59.5 ⎜
⎟ = 39.1
kg fuel
MF
⎝ 44.09 ⎠
Finally, the air mass flow rate is:
⎛
kg air ⎞ ⎛
kg fuel ⎞
kg air
m a = ( AFR ) m F = ⎜ 39.1
⎟⎜ 0.25
⎟ =9.78
kg fuel ⎠ ⎝
s ⎠
s
⎝
b) The heat transfer rate from the combustion process is calculated with
conservation of energy. Assuming [A5], [A6], and [A7], Eq. S15-60 is applicable:
Q ⎡
⎤ ⎡
⎤
= ∑ ni ( ∆h fo ) − ∑ ni ( ∆h fo ) ⎥ + ⎢ ∑ ni ( ∆h ) − ∑ ni ( ∆h ) ⎥
i
i
i
i
n F ⎢⎣ out
in
in
⎦ ⎣ out
⎦
Rearranging this equation to group the inflow and outflow terms:
Q
= ∑ ni ( ∆h fo + ∆h ) − ∑ ni ( ∆h fo + ∆h )
i
i
n F out
in
Expanding each term, and letting the subscript “R” (for reactant) indicate that the
A5. The system is steady.
A6. Potential and kinetic
energy effects are
negligible.
A7. No work occurs in
the system.
39
property must be evaluated at the inlet state, “P” (for product) for the property to be
evaluated at the exit state, and “ref” for the property to be evaluated at the reference
state:
Q
= nN2 , P ∆h fo + hP − href
+ nH 2 O , P ∆h fo + hP − href
+ nCO2 , P ∆h fo + hP − href
N2
H 2O
CO2
nF
(
)
(
( ∆h
)
)
+ nO2 , P ∆h fo + hP − href
− nN 2 , R
o
f
+ hR − href
(
O2
)
(
− nC3 H8 , R ∆h fo + hR − href
)
C3 H 8
(
)
(
)
− nO2 , R ∆h fo + hR − href
O2
N2
Note that the inlet temperature of the propane is 25 °C = 298 K, the air is at 17 °C = 290
K, and the reference temperature is 25 °C = 298 K. The properties are obtained from
Tables SA-1 and SA-3. Note that for the fuel, because it is liquid, we must add the
enthalpy
of
vaporization
to
the
enthalpy
of
formation,
and
o
o
∆h f = −103,850 − 15, 060 = −118, 640 kJ kmol . For both N2 and O2, ∆h f = 0 , because
those two compounds already exist. Entering the enthalpies (not showing units for
brevity), the equation is:
Q
= ( 47 )( 0 + 30129 − 8669 ) N + ( 4 )( −241820 + 35882 − 9904 ) H O
2
2
n F
+ ( 3)( −393520 + 42769 − 9364 )CO + ( 7.5 )( 0 + 31389 − 8682 )O
(
− (1) −118640 + h298 − h298
)
− ( 47.0 )( 0 + 8432 − 8669 ) N
2
C3 H 8
2
− (12.5 ) ( 0 + 8443 − 8682 )O
2
2
Q
kJ
= −632, 000
nF
kmol fuel
The molar flow rate is
nF = m F M F = ( 0.25 kg s ) ( 44.09 kg kmol ) =0.00567 kmol fuel s
and the heat transfer rate is:
⎛
⎞⎛
kJ
kmol fuel ⎞
Q = ⎜ −632, 000
⎟ ⎜ 0.00567
⎟ = −3,584 kW
kmol
fuel
s
⎠
⎝
⎠⎝
Comments:
The minus sign on the heat transfer rate indicates that the heat transfer is out of the system. If the
products of reaction were cooled to a sufficiently low temperature, then some of the water vapor in the
products would condense out of the mixture, and the remaining gas would be saturated ( φ = 100% ).
There can be two objectives when a combustion process is used in an industrial application. The
first is the desire for a large energy release for heat transfer to a different substance (e.g., vaporization of
water in the boiler of a Rankine cycle power plant). The second is to achieve a given products temperature
for some purpose (e.g., setting the gas temperature at the inlet to the turbine in a Brayton cycle). Regarding
the second objective, excess air decreases the temperature the products of combustion can attain. After the
minimum quantity of air needed for complete combustion is used, additional air contributes nothing to the
combustion process, and part of the heat released by the combustion goes to raise the temperature of the
excess air. Hence, the temperature obtainable in a combustion process where there is no heat transfer or
work from the products of combustion is called the adiabatic flame temperature. The maximum possible
combustion temperature would be obtained when the minimum amount of air is used.
Consider Eq. S15-60. For a combustion system with no heat transfer and no work, this equation
states that the enthalpy of the reactants and the enthalpy of the products must be equal. With a little
rearrangement we obtain:
40
∑ n ( ∆h
i
in
o
f
+ ∆h
) = ∑ n ( ∆h
i
i
out
o
f
+ ∆h
)
i
Eq. S15-63
The left hand side of the equation (the reactants) can be evaluated once we have specified the reactants and
their states. The right hand side of the equation (the products) can be evaluated for the temperature once
we have established the products of combustion. However, an iterative solution must be used because the
terms ∆hi on the left hand side represent the enthalpy change of each product from Tref to the temperature
sought. This temperature will be the highest possible temperature that could be obtained by combustion of
the fuel and air used. Note that this temperature can be varied depending on the temperature of the
reactants. One temperature would be obtained if the reactants entered at the standard reference state. A
higher temperature would be realized if the reactants were preheated, such as in a Brayton cycle that uses a
regenerator. Example S15-12 illustrates the iteration required.
Example S15-12 Adiabatic flame temperature
For the conditions given in Example S15-11 assume that the combustion chamber is insulated so
that no heat is lost from the combustion process. The enthalpy of vaporization plus the enthalpy of
formation of propane is –118,640 kJ/kmol. Determine:
a) the adiabatic flame temperature when the liquid propane is burned in stoichiometric air
b) the adiabatic flame temperature when the propane is burned in 250% theoretical air.
Approach:
The adiabatic flame temperature is determined with Eq. S15-63. The inlet states are known. The
outlet temperature is what is sought. Referring to Example S15-11, we used the energy equation to obtain
the heat transfer from the combusting fuel. If we take the energy balance equation developed in that
problem, but set Q = 0 , then we would have Eq. S15-63. For stoichiometric air, no excess oxygen would
exist in the products of combustion; with excess air, oxygen would exist. In both cases, the stoichiometric
coefficients would be different. Because the outlet properties are not linear functions of temperature, an
iterative solution is required.
Solution:
a) The stoichiometric chemical equation for the combustion of liquid propane
with air is (from Example S15-11):
C3 H8 A + 5 ( O2 +3.76N 2 ) ⇒ 18.8 ( N 2 ) + 4 ( H 2 O ) + 3 ( CO2 )
To calculate the adiabatic flame temperature, we use conservation of energy with [A1A8] to obtain Eq. S15-63. Expanding each term, as we did in Example S15-11, and
letting the subscript “R” (for reactant) indicate that the property must be evaluated at the
initial state, “P” (for product) for the property to be evaluated at the final state, and “ref”
for the property to be evaluated at the reference state, we obtain:
Assumptions:
A1. Air is composed only of
3.76 moles of N2 for each mole
of O2.
A2. The products of
combustion behave as an ideal
gas mixture.
A3. N2 is inert.
A4. Complete combustion
occurs with combustion
41
(
nC3 H8 , R ∆h fo + hR − href
)
(
(
)
+ nO2 , R ∆h fo + hR − href
C3 H 8
= nN2 , P ∆h fo + hP − href
)
N2
(
(
+ nN 2 , R ∆h fo + hR − href
O2
+ nH 2O , P ∆h fo + hP − href
)
H 2O
)
(
N2
+ nCO2 , P ∆h fo + hP − href
)
CO2
From Tables SA-1 and SA-3, those properties that are known are substituted
into the equation:
+ ( 5 )( 0 + 8443 − 8682 )O + (18.8 )( 0 + 8432 − 8669 ) N
(1) −118640 + h298 − h298
(
)
(
C3 H 8
= (18.8 ) 0 + hP − 8669
)
(
N2
2
2
(
+ ( 4 ) −241820 + hP − 9904
+ ( 3) −393520 + hP − 9364
)
)
products only CO2, H2O, N2,
and O2.
A5. The system is steady.
A6. Potential and kinetic
energy effects are negligible.
A7. No work occurs in the
system.
A8. The system is adiabatic.
H2O
CO2
(Units are omitted for brevity.) The number –118,640 kJ/kmol represents the sum of the
enthalpy of formation plus the enthalpy of vaporization, since liquid propane is being
burned. Note that the left hand side of the equation is all known; the right hand side can
be evaluated once we have a temperature. The procedure is to estimate an adiabatic
flame temperature, evaluate the enthalpies of the products at that temperature, and
compare the value of the right and left hand sides of the equation. When they are equal,
the correct temperature has been determined.
To start the iteration, an estimate of the adiabatic flame temperature can be
made by assuming that all the products are nitrogen, which comprises 18.8 moles of the
25.8 moles total in the products. Rewriting the above equation (and leaving many terms
uncalculated to show where they came from), we have:
(1)( −118640 )C H + ( 5)(8443 − 8682 )O + (18.8)( 8432 − 8669 ) N + (18.8)(8669 ) N
3
8
2
2
2
+ ( 4 )( 241820 + 9904 ) H O + ( 3)( 393520 + 9364 )CO
( )
= (18.8 ) hP
2
N2
( )
+ ( 4 ) hP
H2O
( )
+ ( 3) hP
2
CO2
( )
≈ (18.8 + 4 + 3) hP
( )
N2
( )
kmol N 2
kJ
kJ
≈ 27.8
hP
→ hP
≈ 81090
N2
N2
kmol fuel
kmol fuel
kmol N 2
From Table SA-1, this enthalpy corresponds to a temperature of about TP ≈ 2450 K. We
can now use this to start our iteration using the complete equation in which the three
unknown enthalpies, hP , are on the right hand side of the equation:
or 2.254 × 106
( )
2.254 × 106 = 18.8 hP
N2
( )
+ 4 hP
H2O
( )
+ 3 hP
CO2
Estimated TP
(K)
Right hand side of
H2O: hP
CO2: hP
N2: hP
equation (kJ/kmol)
(kJ/kmol)
(kJ/kmol)
(kJ/kmol)
2450
81149
106183
128219
2.335 × 106
2400
79320
103508
125152
2.281 × 106
2350
77496
100846
122091
2.227 × 106
By interpolation, we determine TP ≈ 2375 K.
b) For the situation with 150% excess air, the chemical balance equation (again
from the previous example) is:
C3 H8 A + 12.5 ( O 2 +3.76N 2 ) ⇒ 47.0 ( N 2 ) + 4 ( H 2 O ) + 3 ( CO2 ) + 7.5 ( O2 )
Now the excess O2 and N2 in the products must be taken into account.
nC3 H8 , R ∆h fo + hR − href
− nO2 , R ∆h fo + hR − href
− nN2 , R ∆h fo + hR − href
(
)
(
= nN2 , P ∆h + hP − href
o
f
(
)
N2
+ nCO2 , P ∆h + hP − href
o
f
(
C3 H 8
)
(
O2
+ nH 2O , P ∆h + hP − href
)
CO2
o
f
(
(
)
N2
H2O
+ nO2 , P ∆h + hP − href
o
f
)
)
O2
Substituting in values:
42
(1)( −118640 )C H
3
8
+ (12.5 )( 8443 − 8682 )O + ( 47.0 )( 8432 − 8669 ) N
(
)
+ ( 3) ( −393520 + h
= ( 47.0 ) hP − 8669
N2
P
2
(
+ ( 4 ) −241820 + hP − 9904
− 9364
)
CO2
(
)
2
H 2O
+ ( 7.5 ) hP − 8682
)
O2
Rearranging the equation so all the known quantities are on the left hand side, and the
unknown quantities are on the right hand side:
(1)( −118640 )C H + (12.5)( 8443 − 8682 )O + ( 47.0 )( 8432 − 8669 ) N
3
8
2
2
+ ( 47.0 )( 8669 ) N + ( 4 )( 241820 + 9904 ) H O + ( 3)( 393520 + 9364 )CO + ( 7.5 )( 8682 )O
2
( )
= ( 47.0 ) hP
N2
( )
+ ( 4 ) hP
2
H 2O
( )
+ ( 3) hP
CO2
( )
+ ( 7.5 ) hP
2
2
O2
Again, we obtain our first estimate of the adiabatic flame temperature by assuming that
all of the products are N2. Doing this, we obtain:
kmol N 2
kJ
≈ ( 47 + 4 + 3 + 7.5 ) hP
= 61.5
hP
2.527 × 106
N2
N2
kmol fuel
kmol fuel
( )
(h )
( )
kJ
kmol N 2
Therefore, using this enthalpy value and Table SA-1 for N2, our first estimate of the
adiabatic flame temperature with 150% excess air is TP ≈ 1330 K. Iterating on the main
equation with excess air as in part (a) for stoichiometric air, we obtain TP ≈ 1277 K.
P N
2
≈ 41090
Comments:
The adiabatic flame temperature is significantly lower when excess air is used compared to when
stoichiometric air is used (1277 K versus 2375 K). The excess air absorbs some of the energy that
previously would have been used to raise the temperature of the products of combustion to a higher level.
In Chapter 8, we discussed how the limiting factor in the Brayton cycle efficiency is the inlet temperature
to the turbine; this temperature is set by using an appropriate amount of excess air during the combustion
process. Note also that the initial estimates of the adiabatic flame temperature were good. This shows how
approximations can be used to make relatively quick estimates.
15.5 Summary
In non-reacting ideal gas mixtures, properties are determined through the use of the ideal gas
equation and ideal gas properties applied to individual gases in the mixture. Several relationships are
independent of the mixture model used. The composition of the mixture is described by two quantities:
k
m
Xi = 1
mass fraction, X i = i
and
∑
mmix
i =1
k
ni
Yi = 1
and
∑
nmix
i =1
To use the ideal gas equation, we need the apparent molecular weight, Mmix, of the mixture, which
is defined as:
k
k
m
1
1
M mix = mix = ∑ Yi M i
or
= ∑ Xi
M mix i =1
Mi
nmix
i =1
For ideal gas mixtures, two idealizations of the mixture behavior give identical results: Dalton’s
law of additive pressures and Amagat’s law of additive volumes. The basic idea behind the evaluation of
mixture properties (in the absence of chemical reaction) is the principle of superposition: For a mixture of
ideal gases contained in a given volume at a given (common) temperature, each gas behaves as if the other
gases were not present. The additive contributions of each gas results in the mixture properties and
thermodynamic condition.
mole fraction, Yi =
43
We let P equal the total pressure of a mixture. The pressure an individual gas i would exert in a
volume V at temperature T if all the other gases were removed from the volume is called Pi, the partial
pressures of each gas in a mixture and is lower than the total pressure. The partial pressures is related to
the mole fraction by:
k
k
k
Pi
n
= ∑ i = ∑ Yi = 1
∑
i =1 P
i =1 nmix
i =1
The partial volume is related to the mole fraction by:
k
k
Vi
= ∑ Yi = 1
∑
i =1 V
i =1
The thermodynamic properties of a mixture can be described in mass and molar units. With mass
units, the mixture properties are composed of the mass-weighted contributions from each gas:
k
u = ∑ X i ui
i =1
k
h = ∑ X i hi
i =1
k
cv = ∑ X i cv ,i
i =1
k
c p = ∑ X i c p ,i
i =1
s = ∑ X i si
With molar units, the mixture properties are composed of the mole-weighted contributions from
each gas:
k
u = ∑ Yi ui
i =1
k
h = ∑ Yi hi
i =1
k
cv = ∑ Yi cv ,i
i =1
k
c p = ∑ Yi c p ,i
i =1
s = ∑ Yi si
To convert from molar to mass units, the molar quantity is divided by the molecular weight (for example,
ui = ui M i ).
Because each gas behaves as if the other gases were not present, mixture density is obtained by
simply adding the individual gas densities ρi, evaluated at the common T and V, and at the partial pressures
of each gas, not at the total pressure of the mixture, so that:
k
k
m
n
ρ = mix = ∑ ρi and ρ = mix = ∑ ρi
V
V
i =1
i =1
Note that these properties are used in conservation of mass and energy and in the entropy balance
equation in the same manner as for a pure substance.
The ambient environment in offices, homes, and some industrial or commercial processes are
controlled with air conditioning and heating systems to ensure reasonable comfort for the inhabitants and to
ensure quality of some products. This special ideal gas mixture—air and water vapor—can be analyzed
with the previous equations, but because the amount of water vapor in a mixture often changes during a
process, other methods also must be used. Air by itself is called dry air, an air-water vapor mixture is
called moist air, and the study of this special mixture is called psychrometrics. The process in which water
vapor is added to an air stream is called humidification, and when water vapor is removed from an air
stream the process is called dehumidification.
The ratio of the mass of water vapor to the mass of dry air in a mixture is called the specific
humidity, ω, and is defined as:
m
ω= v
ma
Using the ideal gas equation, this ratio can be expressed in terms of the partial pressure of the water vapor,
Pv, and the total pressure of the mixture, P:
P
ω = 0.622 v
P − Pv
The relative humidity, φ, describes the actual amount (mv) of moisture air can hold relative to the
maximum amount (mg) of moisture air can hold at the same temperature and is defined:
m
P
φ= v = v
mg Pg
where Pg is the saturation pressure of water at the mixture temperature. A relative humidity of φ = 0%
represents dry air. A relative humidity of φ = 100% represents saturated air, that is, the situation when air
holds the maximum possible amount of water vapor. The quantity Pg is the partial pressure of water vapor
corresponding to the saturation pressure of water at the mixture temperature:
44
Pg = Psat , water ( evaluated at the mixture temperature )
The mixture temperature is called the dry-bulb temperature and is measured by an ordinary
thermometer placed in the mixture. The wet-bulb temperature is the temperature reading obtained from a
thermometer which has its temperature sensitive element wrapped in gauze and soaked in water. When we
cool a mixture sufficiently (lowering the mixture temperature at constant pressure) until Pg = Pv,
φ = 100% , and the air is saturated, the resulting temperature is called the dew-point temperature, TDP,
because any additional cooling will cause some of the water vapor to condense:
TDP = Tsat , water ( evaluated at Pv )
The relationship between specific humidity and relative humidity can be obtained by combining
equations:
0.622φ Pg
ωP
φ=
ω=
or
P − φ Pg
( 0.622 + ω ) Pg
The analysis of an adiabatic saturation process can be used to develop equations with which
specific humidity and relative humidity can be determined. The measurements required are the dry-bulb
temperature, wet-bulb temperature, and barometric pressure. A sling psychrometer provides an excellent
approximation to an adiabatic saturator and is much simpler to construct and operate. With the measured
quantities, the specific humidity is determined from:
h (T ) − ha (TDB ) + ω ∗ h fg (TWB ) c p , a (TWB − TDB ) + ω ∗ h fg (TWB )
=
ω = a WB
hg (TDB ) − h f (TWB )
hg (TDB ) − h f (TWB )
where
ω∗ =
0.622 Pg (TWB )
P − Pg (TWB )
For air-water vapor mixtures, the thermodynamic properties are described in terms of the air and
water vapor properties, with the specific humidity acting as a weighting function:
hmix = ha + ω hv
umix = ua + ω uv
smix = sa + ω sv
Psychrometric charts can be developed that show the relationship among specific humidity,
relative humidity, dry-bulb temperature, wet-bulb temperature, mixture specific volume (per unit mass of
dry air), and mixture enthalpy (per unit mass of dry air). These charts are developed for a given total
pressure.
Combustion of hydrocarbon fuels and air (the reactants) release large quantities of energy, and
different chemical compounds are created (the products). Several simplifications were made for analyzing
the combustion process. The hydrocarbon fuel is assumed to be composed only of hydrogen and carbon.
The air is assumed to be composed only of oxygen and nitrogen, with a mole fraction ratio of 1:3.76.
Nitrogen is assumed to be totally inert, and the products of combustion consist only of nitrogen, oxygen,
carbon dioxide, and water.
Total mass is conserved during a combustion process. In addition, the mass of each element is
conserved. To ensure this conservation, we use chemical equations to determine the stoichiometric
coefficients needed for the combustion of a fuel plus an arbitrary amount of air. A common quantity to
describe the amount of air present in a combustion process is the air-fuel ratio:
m
M
AFR = a = AFR a
mF
MF
If the amount of air is the minimum quantity needed for complete combustion of the fuel, then we call it the
stoichiometric air or theoretical air. If more air is present than what is actually needed for complete
combustion, the amount can be expressed in terms of percent of theoretical air or in terms of percent
excess air (which is theoretical air percent minus 100%):
AFRact
AFR act
=
Percent theoretical air =
AFRstoichiometric AFR stoichiometric
During combustion, the reactants (individual substances) combine to form new products (or
compounds). Application of the energy equation to a reaction must account for the energy involved in the
formation of the new compound from its constituent parts. Thus, an appropriate and common reference
45
state for the enthalpy or internal energy of the reactants before the reaction and of the products after the
reaction must be defined. The arbitrarily chosen standard reference state is defined at Tref = 298.15 K (25
ºC or 77 ºF) and Pref = 1 atm.
The enthalpy of formation, ∆h fo , of a compound is defined as the difference between the enthalpy
of the compound per unit mole at the reference state and the enthalpies of its constituent stable elements
also at the reference state. It is the energy released (an exothermic reaction) or absorbed (an endothermic
reaction) when the compound is formed.
The chemical balance equation is usually based on one mole of the fuel, and the stoichiometric
coefficients for the other compounds are found accordingly. Conservation of energy is often written in
terms of the fuel flow rate, with the stoichiometric coefficients are used to weight the terms:
Q W ⎡
⎤ ⎡
⎤
−
= ∑ ni ( ∆h fo ) − ∑ ni ( ∆h fo ) ⎥ + ⎢ ∑ ni ( ∆h ) − ∑ ni ( ∆h ) ⎥
i
i
i
i
n F n F ⎢⎣ out
in
in
⎦ ⎣ out
⎦
The first bracketed term on the right hand side of this equation is composed of only the enthalpy of
formation and associated stoichiometric coefficients. For complete combustion of fuel with air as the
oxidizer, we define this term as the enthalpy of combustion, ∆hCo (or heat of combustion). The second
bracketed term represents for each compound in the process the difference between its enthalpy at the
temperature at which it exists and its enthalpy at its reference state.
The value of the enthalpy of combustion when liquid water is produced is called the higher
heating value, HHV, and when water vapor is produced the enthalpy of combustion is called the lower
heating value, LHV. The difference between these two quantities is the energy required to condense the
water vapor in the products of combustion:
HHV = LHV + (nh fg ) H 2 O
where n is the moles of water in the products and h fg is the enthalpy of vaporization at standard
temperature.
The temperature that a combustion process will attain with no work or heat transfer from the
products of combustion is called the adiabatic flame temperature. The maximum possible temperature
obtainable occurs when the minimum amount of air is used. Excess air is used to decrease the temperature
of the products of combustion. After the minimum quantity of air needed for complete combustion is used,
additional air contributes nothing to the combustion process, and part of the heat released by the
combustion goes to raise the temperature of the excess air.
15.6 Selected References
Althouse, A.D., Turnquist, C.H., and Bracciano, A.F., Modern Refrigeration and Air Conditioning, 2nd
Edition, Goodheart and Willcox Co., 2001.
Barnard, J.A. and Bradley, J.N., Flame and Combustion, 2nd edition, Chapman and Hall, 1985.
Çengel, Y.A. and Boles, M.A., Thermodynamics An Engineering Approach, 3rd edition, McGraw-Hill, New
York, 1998.
Glassman, I., Combustion, 3rd edition, Academic Press, 1996.
Kuo, K. K.-Y., Principles of Combustion, John Wiley & Sons, 1986.
Moran, M.J. and Shapiro, H.N., Fundamentals of Engineering Thermodynamics, 5th edition, John Wiley &
Sons, New York, 2003.
Parsons, R., editor, Fundamentals: 2001 ASHRAE Handbook, American Society of Heating, Refrigerating,
and Air-conditioning Engineers, 2001.
46
Sonntag, R.E., Borgnakke, C., and Van Wylen, G.J., Fundamentals of Thermodynamics, 5th edition, John
Wiley & Sons, New York, 1998.
Turns, S.R., An Introduction to Combustion: Concepts and Applications w/Software, McGraw-Hill Science,
1999.
Williams, F.A., Combustion Theory, 2nd edition, Benjamin/Cummings Publishing Company, 1985.
15.7 Problems
Ideal gas mixtures
P 15-1 A gas mixture consists of 12 kg of CO2, 16 kg of N2, and 10 kg of O2. Determine:
a) the mass fraction of each component
b) the mole fraction of each component
c) the mixture average molecular weight.
P 15-2 An ideal gas mixture at 210 kPa, 50 °C has a molar analysis as follows: N2, 35%: CO2, 25%; O2,
40%. Determine:
a) the mass fractions
b) the partial pressures of each component (in kPa)
c) the volume occupied by 1 kg of the mixture (in m3).
P 15-3 A gas mixture at 20 ºC consists of 0.95 kg O2, 0.82 kg N2, 1.32 kg CO2, and 0.091 kg CO and is
contained in a 0.825-m3 tank. Determine the total pressure (in kPa).
P 15-4 The composition of natural gas varies depending on its source. One sample has the following
volume fraction analysis: methane (CH4), 84%; ethane (C2H6), 14.8%; carbon dioxide (CO2),
0.70%; nitrogen (N2), 0.50%. For the mixture, determine:
a) mass fractions
b) the mixture molecular weight
c) the mixture density if the total mixture pressure is 175 kPa at a temperature of 27 °C.
P 15-5 The pressure and temperature in a rigid tank are 375 kPa and 46 ºC, respectively. The ideal gas
mixture contained in the tank is composed of 0.75 kmol argon (Ar) and 1.35 kmol nitrogen (N2).
Determine:
a) the volume of the tank (in m3)
b) the final pressure if the mixture is heated to 127 ºC (in kPa).
P 15-6 An ideal gas mixture of 1 lbm air and 1 lbm water vapor is compressed isentropically in a closed
system from 50 lbf/in.2, 250 ºF to 100 lbf/in.2. Determine:
a) the final temperature (in ºF)
b) the work required (in Btu).
P 15-7 A pipe connects two tanks with a valve in between. One tank contains 2 kg methane (CH4) at 200
kPa, 10 °C; the other tank contains 4 kg oxygen (O2) at 600 kPa, -10 °C. The valve between the
tanks is opened, and the two gases mix adiabatically. Determine:
a) the final mixture temperature (in °C)
b) the final mixture pressure (in kPa).
P 15-8 Two gas streams enter a mixer and a single stream exits. The first stream, with a flow rate of 0.2
kg/s, is a mixture of 13% methane (CH4) and 87% air (mole fractions). The second stream is air. If
the mole fraction of methane must be 5% in the outlet stream, determine:
a) the mass flow rate of the second entering stream of air (in kg/s)
47
b) the mass flow rate of oxygen in the exiting stream (in kg/s).
P 15-9 An ideal gas mixture at 500 kPa, 300 K consisting of 20 kg O2 and 16 kg N2 is contained in a pistoncylinder assembly. Heat is added, and the mixture expands at constant pressure until it reaches a
temperature of 450 K. Determine the heat transfer during this process (in kJ).
P 15-10 Hydrogen at 0.2 MPa, 70 ºC and nitrogen at 0.2 MPa, 270 ºC enter an adiabatic mixing chamber in
separate flow streams. The molar flow ratio is 3:1, respectively. The mixture leaves at 0.19 MPa.
Using constant specific heats evaluated at 27 ºC, determine:
a) the exit temperature (in ºC)
b) the entropy generation rate per kg of mixture (in kJ/kg·K).
P 15-11 Natural gas with a molar analysis of methane (CH4), 70% and ethane (C2H6), 30% is compressed
isothermally from 600 kPa, 77 °C to 2000 kPa. No entropy is generated. Determine the power
input per kmol of mixture (in kJ/kmol).
P 15-12 Two gas streams are adiabatically mixed. The first stream is methane is at 4 bar, 80 ºC with a flow
rate of 8 kg/min; the second stream is air at 1.2 bar, 30 ºC. After mixing, the mixture is at 1 bar, 40
ºC. Determine:
a) the mass flow rate of the 30 ºC air (in kg/min)
b) the entropy generation rate (in kJ/min·K).
P 15-13 A mixture of 0.3 lbmol of N2 and 0.2 lbmol of O2 at 15 lbf/in.2, 500 °F is compressed isothermally
to 60 lbf/in2. Heat transfer during the process is to the surroundings that are at 40 °F. Determine:
a) the work (in Btu)
b) the heat transfer (in Btu)
c) the entropy produced (in Btu/R).
P 15-14 A mixture of N2 and CO2 (mole fractions 60 % and 40%, respectively) with a flow rate of 2.5 kg/s is
compressed adiabatically from 150 kPa, 77 ºC to 450 kPa, 207 ºC. Determine:
a) the power required (in kW)
b) the compressor isentropic efficiency.
P 15-15 Hydrogen (H2) at 14.7 psia, 70 °F with a flow rate of 0.1 lbm/s is mixed adiabatically with methane
(CH4) at 14.7 psia, 150 °F to form a mixture that is 50% hydrogen (by volume). Determine:
a) the methane flow rate (in lbm/s)
b) the exit temperature (in °F).
P 15-16 A rigid tank is divided into two compartments, each of which has a volume of 0.25 m3. One
compartment contains CO2 at 200 kPa, 27 ºC, and the other compartment contains O2 at 100 kPa, 50
ºC. The partition separating the gases is removed and the gases mix. Heat is lost to the
surroundings in the amount of 15 kJ. Determine:
a) the final mixture temperature (in ºC)
b) the final mixture pressure (in kPa).
P 15-17 A mixture with a molar analysis of N2, 83%, CO2, 13%, and O2, 4% with an inlet flow rate of 50
m3/min is expanded in an adiabatic turbine from 400 kPa, 227 °C to 105 kPa, 77 °C. Determine the
power output (in kW).
P 15-18 A mixture of 40% carbon dioxide and 60% nitrogen (by weight) is compressed isentropically from
100 kPa, 27 °C to 400 kPa. Assume the specific heats are constant at the inlet temperature.
Determine:
a) the outlet temperature (in °C)
b) the work required (in kJ/kg).
P 15-19 Two gas streams are mixed in an insulated device. The first stream is nitrogen (N2) at 20 psia, 120
ºF with a volume flow rate of 300 ft3/min. The second stream is oxygen (O2) at 20 psia, 200 ºF with
48
a mass flow rate of 50 lbm/min. The exiting stream is at 17 psia. Determine the exit temperature
(in ºF).
P 15-20 A mixture of N2, CO2, and H2O at 100 kPa, 27 ºC is compressed isentropically in a steady-flow
compressor to 600 kPa. The molar ratio is 4:1:1, respectively. Using constant specific heats
evaluated at the inlet temperature, determine:
a) the outlet temperature (in ºC)
b) the required work per kmol (in kJ/kmol).
P 15-21 A rigid tank contains 1 kg of CO2 at 100 kPa, 27 °C. A second tank contains 0.8 kg O2 at 500 kPa,
127 °C. The valve in the pipe connecting the two tanks is opened, and the gases are allowed to mix
until an equilibrium temperature of 77 °C is obtained. Determine:
a) the volume of each tank (in m3)
b) the final pressure (in kPa)
c) the heat transfer to or from the gases (in kJ).
P 15-22 A mixture of N2 and CO2 (mole fractions 70 % and 30%, respectively) with a flow rate of 1.5 kg/s is
compressed adiabatically from 125 kPa, 27 ºC to 450 kPa. The compressor has an isentropic
efficiency of 83%. Determine:
a) the exit temperature (in ºC)
b) the power required (in kW).
P 15-23 A 10-m3 tank, initially filled with O2 at 40 kPa, 27 ºC, is connected to a pipeline that contains N2 at
200 kPa, 27 ºC. Nitrogen is allowed to flow into the smaller tank until the pressure reaches 110 kPa.
Heat transfer to the surroundings results in an isothermal process. Determine:
a) the mass of nitrogen that enters the tank (in kg)
b) the heat transfer (in kJ).
P 15-24 A mixture of 1 kg methane (CH4) and 15 kg air initially at 27 ºC, 101.3 kPa is compressed
isentropically to one tenth of its initial volume. Determine the final temperature (in ºC) and pressure
(in kPa).
P 15-25 A mixture with a molar analysis of CO2, 50%, CO, 34%, and O2, 16% with a flow rate of 2 kg/s is
adiabatically compressed from 100 kPa, 27 °C to 550 kPa, 227 °C. The inlet velocity is 40 m/s and
the outlet velocity is 80 m/s. Determine the required power input (in kW). How does the power
change if velocity is not taken into account?
P 15-26 A mixture of oxygen (O2) and nitrogen (N2) (mass fractions 60% and 40%, respectively) is
expanded adiabatically from 500 kPa, 450 °C to 150 kPa. The flow rate is 2.5 kg/s, and the
isentropic efficiency is 80%. Determine:
a) the exit temperature (in °C)
b) the output power (in kW)
Psychrometrics
P 15-27 Sometimes on a winter day, eyeglasses fog up when a person enters a building from the outdoors.
Assuming an outdoor air temperature of 5 ºC and an indoor air temperature of 25 ºC, determine the
maximum relative humidity the building can have for a person’s eyeglasses not to become fogged.
P 15-28 Air at a pressure of 101.3 kPa with a flow rate of 150 m3/min enters a humidifier with a dry-bulb
temperature of 38 °C and a wet-bulb temperature of 20 °C. The air exits the cooler with a dry-bulb
temperature of 25 °C and a relative humidity of 65%. Without using the psychrometric chart,
determine the make-up water flow rate (in kg/min).
49
P 15-29 Outside air at 30 °F, 60% relative humidity is to be conditioned and delivered to a building at 70 °F,
50% relative humidity. The volume flow rate is 7500 ft3/min. Using the psychrometric chart,
determine the required water addition per lbm of dry air.
P 15-30 When feed corn is harvested, it must be dried before it is stored so that it does not mildew and rot.
Consider corn that enters a dryer at 40% moisture by mass (solid plus moisture). Dry air enters the
dryer at 80 ºC at a rate of 16 kg/kg of corn entering. Moist air exits the dryer at 100 kPa, 35 ºC, and
55% relative humidity. Determine the moisture content by mass of the corn leaving the dryer.
P 15-31 Dry air (0% relative humidity) at 14.7 lbf/in.2, 150 ºF is used to dry a damp fabric that enters the
dryer with a 45% moisture content by mass (solid plus moisture) and leaves with a 5% moisture
content. The mass flow rate of the exiting fabric (solid plus moisture) is 10 lbm/min. Moist air
exits at 14.7 lbf/in.2, 130 ºF, and 65% relative humidity. Determine the required volumetric flow
rate of the entering air (in ft3/min).
P 15-32 The design specifications for a new room humidifier are: air volume flow rate 0.015 m3/s, inlet
relative humidity 40%, outlet relative humidity 60%, dry bulb temperature fixed at 21 ºC, and
barometric pressure of 101.3 kPa. Water is to be added only once per day. If the humidifier is to
run continuously for 24 hrs before the water supply (assume it is at 21 ºC) must be refilled,
determine the required supply tank volume (in cm3).
P 15-33 Air at 14 lbf/in.2, 75 ºF, and 75% relative humidity is compressed to 70 lbf/in.2, 280 ºF. The air then
passes through a water-cooled heat exchanger and exits as saturated air at 68 lbf/in.2, 100 ºF.
Determine:
a) the specific humidity and relative humidity at the compressor discharge
b) the amount of water vapor condensed in the heat exchanger for each lbm of dry air flowing
through it.
P 15-34 Air at 101.3 kPa, 20 °C, and 55% relative humidity flows from a humidifier into a room. From
equipment and people in the room, 6 kW of heat is added to the air; assume no water vapor is added.
If the exit temperature of the air cannot exceed 26 °C, determine:
a) the inlet volumetric flow rate of air required (in m3/hr)
b) the percent error if the water vapor in the air is ignored.
P 15-35 Shown in the figure below is a heating/humidifier unit. Air enters at 101.3 kPa, -5 ºC, and 90%
relative humidity. The preheating coil raises the temperature to 2 ºC. The air is then humidified
with water whose temperature is held at the dew-point temperature of the air exiting from the
complete unit. Reheating is used to set the outlet to 22 ºC and 35% relative humidity. The make-up
water enters at 4 ºC. Using the psychrometric chart, determine:
a) the water added per kg of dry air
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b) the total heat transfer (preheater plus spray water heater plus reheat) to the air per kg of dry
air (kJ/kg dry air).
P 15-36 For energy conservation and indoor air quality reasons, fresh air and recirculation air from a
building are mixed before being used again in the building. Consider the following mixing
situation. Air exits from the cooling coil in an air conditioning system at 12 ºC and 100% relative
humidity and is adiabatically mixed with fresh air at 36 ºC and 30% relative humidity. The fresh air
has a mass flow rate 30% of that exiting from the cooling coils. Determine the final temperature
and relative humidity of the mixture.
P 15-37 Air with a volume flow rate of 600 m3/min enters the cooling/dehumidifying coils of an air
conditioner at 101.3 kPa, 40 °C, 80% relative humidity. The design specification is for the air to
leave the coil as saturated air at 25 °C. Determine:
a) the water removal rate (in kg/min)
b) the heat transfer rate (in kW).
P 15-38 A warm-air furnace/humidifier unit is used to heat 1000 ft3/min of air from 14.5 lbf/in.2, 60 ºF, and
60% relative humidity to 120 ºF and 12% relative humidity. Liquid water is supplied at 65 ºF.
Determine:
a) the liquid water flow rate (in lbm/hr)
b) the required heat transfer rate (in Btu/hr).
P 15-39 Products of combustion with a molar analysis of N2, 74%, H2O, 14%, CO2, 7%, and O2, 5% at 14.7
psia, 800 ºF with a volumetric flow rate of 300 ft3/min enter a counterflow heat exchanger and exit
at 200 º F. On the other side of the heat exchanger moist air enters at 14.7 psia, 60 ºF, and 30%
relative humidity and exits at 100 ºF. Determine the mass flow rate of the moist air (in lbm/min).
P 15-40 To remove dust particles and odors from air, an air washer is often used. This device is composed
of a chamber in which water is sprayed and liquid water is removed from the bottom, taking the dust
and odors with it. Air with a flow rate of 5000 ft3/min enters the device shown below at 14.7
lbf/in.2, 65 ºF, and 70% relative humidity and leaves as saturated air at 60 ºF. Water at 70 ºF with a
flow of 30 lbm/min is sprayed into the air. For the water exiting the chamber, determine:
a) the flow rate (in lbm/min)
b) its temperature (in ºF).
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P 15-41 Moist air enters an adiabatic compressor at 95 kPa, 20 ºC, and 60% relative humidity with a
volumetric flow rate of 20 m3/min and leaves at 200 kPa, 100 ºC. Determine:
a) the relative humidity at the exit
b) the power input (in kW).
P 15-42 An air duct is routed through an uninsulated space in a building. The air enters at 14.7 psia with a
dry-bulb temperature of 82 ºF, a wet-bulb temperature of 68 ºF, and a flow rate of 10 lbm/min. The
air exits the duct at 62 ºF. Using the psychrometric chart, determine:
a) the relative humidity at the duct inlet
b) the relative humidity at the duct outlet
c) the heat transfer rate (in Btu/hr).
P 15-43 Air enters an industrial dryer at 100 kPa, 25 ºC, and 65% relative humidity and is heated to 140 ºC
at constant pressure. The hot air is passed over the material to be dried and exits the dryer at 60 ºC
and 30% relative humidity. Water must be removed from the material at a rate of 20 kg/min.
Determine:
a) the dry air mass flow rate (in kg/min)
b) the heat transfer rate (in kW).
P 15-44 Air at 14.7 lbf/in.2, 50 ºF, and 40% relative humidity enters a heater/humidifier device and exits at
70 ºF and 50% relative humidity. This is a constant pressure process, and liquid water at 45 ºF is
used. Determine per lbm of dry air:
a) the amount of water injected (in lbm water/lbm dry air)
b) the heat transfer rate (in Btu/lbm dry air).
P 15-45 For energy conservation and air quality purposes, designers of building air handling systems mix
fresh air from outside with recycled air from inside. Consider two air streams, both at 14.7 lbf/in.2,
that are mixed in a steady flow adiabatic mixer. The first stream has a flow of 650 ft3/min and is at
55 °F and 10% relative humidity. The second stream has a flow of 900 ft3/min and is at 85 °F and
80% relative humidity. Determine the outlet relative humidity, temperature, and specific humidity.
P 15-46 In an air conditioning system, air with a volume flow rate of 25 m3/min enters the dehumidifying
section at 101.3 kPa, 28 ºC, and 70% relative humidity and leaves as saturated moist air; some water
vapor condenses, and it leaves the system at the same temperature as the saturated moist air. The air
then is reheated to 24 ºC and 40% relative humidity. Using the psychrometric chart, determine:
a) the temperature of the moist air leaving the dehumidifier (in ºC)
b) the water condensation rate (in kg/min)
c) the heat transfer rate in the cooling coil (in kW)
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d) the heat transfer rate in the reheat section (in kW).
P 15-47 Air at 101.3 kPa, 12 °C, and 70% relative humidity flows into a heater/humidifier and leaves at 38
°C, and 60% relative humidity. Determine per kg of dry air:
a) the amount of water to be added (in kg water/kg dry air)
b) the heat transfer required if the water is supplied at 18 °C (in kJ/kg dry air)
c) the percent error if only the heat transfer required by the dry air (water vapor and make-up
water ignored) is taken into account (in kJ/kg dry air).
P 15-48 The air conditioning system for a large building that has a cooling load of 1.5 MW is shown in the
figure below. Refrigerant R-134a enters the compressor as a saturated vapor at 360 kPa and leaves
at 1.2 MPa; the flow rate is 6 kg/s. The refrigerant leaves the condenser as a saturated liquid. The
water flowing through the condenser can have an 8 ºC temperature rise. The pressure rise across the
pump is 75 kPa. The compressor isentropic efficiency is 70% and that of the pump is 60%. Using
appropriate assumptions and approximations, determine:
a) the compressor power (in kW)
b) the condenser heat transfer rate (in kW)
c) the water flow rate through the condenser (in kg/min)
d) the required make-up water flow rate (in kg/min)
e) the pump power (in kW).
P 15-49 In an air conditioning system, air at 14.7 psia, 84 °F, and 70% relative humidity flows over the
cooling coil with a flow of 15,000 ft3/min. The air is cooled and dehumidified, and the liquid water
leaves the system at 50 °F. The air then flows through the reheat section and exits at 75 °F and 40%
relative humidity. Using the psychrometric chart, determine:
a) the condensate flow rate (in lbm/hr)
b) the heat transfer rate in the cooling coil (in Btu/hr)
c) the heat transfer rate in the reheat section (in Btu/hr).
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P 15-50 Cooling towers often are operated so that the air leaving the tower is not saturated and no fog forms.
Consider a tower in which warm water enters at 38 ºC with a flow rate of 900 kg/min and leaves at
24 ºC. Air enters the cooling tower at 101.3 kPa, 20 ºC, and 40% relative humidity with a flow rate
of 600 m3/min. Make-up water at 22 ºC is added at a rate of 15 kg/min. Does fog form? That is,
determine the dry-bulb temperature (in ºC) and the relative humidity of the air leaving the cooling
tower. Use the psychrometric chart where appropriate.
P 15-51 In a cooling tower, 400 gal/min of liquid water enters at 120 ºF. Air enters at 1 atm, 80 ºF dry-bulb
temperature, and 45% relative humidity, and exits at 100 ºF and 95% relative humidity. Make-up
water is available at 70 ºF. Determine the mass flow rates of the dry air and the make-up water (in
lbm/min).
P 15-52 Warm water at a flow rate of 100,000 kg/hr enters a cooling tower at 40 °C and leaves at 25 °C. Air
enters the cooling tower at 98 kPa, 20 °C, 40% relative humidity and leaves saturated at 35 °C.
Determine the required 20 °C make-up water flow rate (in kg/hr) and the air flow rate (in kg/hr).
P 15-53 The heat rejected to the cooling water in the condenser of a Rankine cycle power plant is 850 MW.
The cooling water enters the cooling tower at 40 ºC and exits at 25 ºC. Air enters the cooling tower
at 101.3 kPa, 10 ºC, and 35% relative humidity and exits at 30 ºC and 95% relative humidity.
Make-up water is supplied at 10 ºC. Determine the mass flow rate of the air, cooling water, and
make-up water (in kg/s).
Combustion
P 15-54 For the combustion of octane (C8H18) with air, determine the air-fuel ratio for this reaction on both a
molar and a mass basis assuming:
a) stoichiometric air
b) 150% excess air.
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P 15-55 Assume gasoline can be treated as octane (C8H18) with a specific gravity of 0.68, and it is to be
burned with air at 101.3 kPa, 27 °C. Determine the theoretical volume of air required to burn a liter
of gasoline (in m3).
P 15-56 Dodecane (C12H26) is burned with 50% excess air. Determine:
a) the molar and mass air-fuel ratios
b) the dew-point temperature of the combustion products at 101.3 kPa (in °C).
P 15-57 Assume gasoline can be treated as octane (C8H18), and it is to be burned with 100% theoretical air at
101.3 kPa. Determine:
a) the partial pressure of the water vapor (in kPa) and the dew-point temperature of the products
(in °C) for stoichiometric combustion
b) the partial pressure (in kPa) and dew-point temperature (in °C) for 65% excess air.
P 15-58 An Orsat analysis of the products from the combustion of propane (C3H8) and air yields the
following mole fractions: CO2, 11.5%; O2, 2.7%, CO, 0.7%. Water vapor and nitrogen are also
present. Determine the reaction equation and the percent theoretical air for the reaction.
P 15-59 When a fuel is burned with atmospheric air, water vapor is present in addition to the dry air.
Consider a gaseous fuel mixture that has the following molar analysis: CH4, 72%; N2, 14%; H2, 9%;
CO2, 3%; O2, 2%. The fuel is burned with moist air to form gaseous products at one atmosphere;
only CO2, H2O, and N2 are present in the products. If the dew-point temperature of the products is
60 °C, determine the amount of water vapor present in the combustion air (in kmol/kmol fuel).
P 15-60 At an oil refinery, an impure gas mixture is combusted for use in process heating. The gas at 20
lbf/in.2, 300 ºF has the following volumetric analysis: CH4, 20%; C2H6, 40%; N2, 30%, H2O, 10%.
Air is supplied at 15 lbf/in.2, 100 ºF at a rate of 115% theoretical air. Determine:
a) the volume flow of air supplied per volume flow of fuel
b) the dew-point temperature of the products of combustion if the products exit at 15 lbf/in2.
P 15-61 Decane (C10H22) is burned with 25% excess air, which is at 98 kPa, 30 °C, and 75% relative
humidity. The combustion products are cooled to 98 kPa, 30 °C. Determine the amount of water
vapor that condenses (in kg water/kg fuel).
P 15-62 In an oil-fired furnace, the fuel is burned with 30% excess air. The fuel, which is composed of 83%
carbon, 14.2% hydrogen, and 2.8% inert material on a weight basis, has a flow rate of 9500 kg/hr.
Determine the volumetric flow rate of the products at 101.3 kPa, 400 °C (in m3/hr).
P 15-63 Methane (CH4) is burned completely with 200% theoretical air, which is at 103 kPa, 15 °C, and
45% relative humidity. Determine:
a) the balanced reaction equation
b) the dew-point temperature of the combustion products at 103 kPa (in °C).
P 15-64 Ethane (C2H6) burns with stoichiometric air; both gases enter the combustion chamber at 14.7
lbf/in.2, 77 °F. The products of combustion are cooled to 14.7 lbf/in.2, 77 °F. Determine the heat
transfer rate (in Btu/lbmol fuel).
P 15-65 Methane (CH4) at 101.3 kPa, 34 °C with a volumetric flow rate of 0.98 m3/min is burned with 20%
excess air. If the products are to be cooled from 600 °C to 200 °C, determine the heat transfer rate
(in kW).
P 15-66 A heat transfer rate of 10 MW is to be obtained by burning benzene (C6H6) at 101.3 kPa, 25 °C with
stoichiometric air at the same conditions. The products of combustion exit at 477 °C. Determine
the required fuel mass flow rate (in kg/s).
P 15-67 The boiler in a simple Rankine cycle uses methane (CH4) at 101.3 kPa, 25 ºC which combusts with
110% theoretical air at 101.3 kPa, 25 ºC. The products of combustion exit the smoke stack at 101.3
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kPa, 150 ºC. The cycle thermal efficiency is 40%. Determine the required fuel mass flow rate per
MW of net power output (in kg/hr MW).
P 15-68 Gaseous pentane (C5H12) at 25ºC is burned with 22% excess air that is at 37 ºC. The products of
combustion are cooled to 400 °C. Determine the heat transfer per kg mole of fuel (in kJ/kgmol).
P 15-69 Propane (C3H8) and pure oxygen (O2) in stoichiometric proportions are combusted in a water-cooled
burner. The reactants both enter at 90 °F and 2 atm and exit at 850 °F. The products (CO2 and H2O
only) have a flow rate of 250 lbm/hr. Determine the heat transfer rate (in Btu/hr).
P 15-70 Overall plant efficiency in a Rankine cycle is improved by using the exhaust gases from the boiler to
preheat the incoming combustion air. Consider a system in which methane (CH4) at 101.3 kPa, 25
°C is burned in the boiler with 150% theoretical air. After the products of combustion are used to
boil the water in the boiler, the gases leave the boiler at 727 °C, enter a heat exchanger to preheat
the combustion air, and leave the heat exchanger at 587 °C. The air enters the heat exchanger at 25
°C. Determine the combustion air temperature as it leaves the heat exchanger and enters the boiler
(in °C).
P 15-71 In the boiler of a Rankine cycle, methane gas (CH4) at 101.3 kPa, 25 °C is burned with 40% excess
air at 101.3 kPa, 127 °C. The products of combustion exit at 101.3 kPa, 427 °C. The heat transfer
is used to vaporize and superheat water, which enters as a saturated liquid at 8 MPa and leaves at
480 °C; the water flow rate is 100 kg/s. Determine the required volumetric flow rate of the methane
(in m3/s).
P 15-72 Determine the adiabatic flame temperature for hydrogen (H2) burning with 200% theoretical air.
The hydrogen and air both enter at 101.3 kPa, 25 °C (in °C).
P 15-73 A hot gas at 1140 ºF is needed for a process. Gaseous propane (C3H8) is to be burned with air in an
adiabatic steady flow combustion chamber; both gases enter at 77 ºF. Determine the required airfuel mass ratio and the percent excess air.
P 15-74 Determine the enthalpy of combustion at 25 °C for ethane (C2H6) when gaseous ethane reacts with
air producing liquid water in the products (in kJ/kmol). Do the same for acetylene (C2H2).
P 15-75 Butane (C4H10) is to be burned with sufficient excess air so that the products of combustion enter a
gas turbine at 1600 K. Fuel and air enter at 25 °C. Determine the required percent excess air.
P 15-76 In a regenerative Brayton cycle, the maximum operating temperature of a gas turbine is not
exceeded by using excess air during the combustion of the fuel. Consider a turbine whose
maximum temperature is 1600 °F. Methane (CH4) at 14.7 lbf/in.2, 77 °F is burned completely with
air that is preheated in the regenerator to 400 °F. Determine the required percent excess air.
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P 15-77 Gaseous propane (C3H8) at 25 ºC is burned in an adiabatic chamber with 110% theoretical air at 25
ºC to ensure complete combustion. Determine the exit temperature of the products of combustion
(in ºC).
P 15-78 In a simple Brayton cycle power plant, air enters the adiabatic compressor (ηC = 83%) at 100 kPa,
22 °C and leaves at 450 kPa. The high-pressure air enters the combustion chamber where it burns
with methane (CH4) that enters at 450 kPa, 25 °C with a molar flow rate of 0.15 kmol/s. The
products of combustion at 450 kPa, 727 °C enter the adiabatic turbine (ηT = 90%) and expand to 100
kPa. Determine:
a) the percent theoretical air
b) the net power output (in kW).
P 15-79 Methane (CH4) is burned in an insulated vessel with 250% theoretical air. Both gases initially are at
101.3 kPa, 25 °C. Determine the temperature of the products of combustion if the reaction occurs at
constant volume (in °C).
P 15-80 Methane (CH4) is burned in an insulated vessel with 250% theoretical air. Both gases initially are at
101.3 kPa, 25 °C. Determine the temperature of the products of combustion if the reaction occurs at
constant pressure in a piston-cylinder assembly (in °C).
P 15-81 Consider a simple Rankine cycle power plant. Steam enters the turbine at 600 lbf/in.2, 1000 °F and
expands to 2 lbf/in.2, 96% quality. Saturated liquid leaves the condenser, and the temperature rise of
the cooling water is 15 °F. In the boiler, methane (CH4) is burned with 200% theoretical air; both
gases enter at 14.7 lbf/in.2, 77 °F, and the products of combustion leave at 460 °F. Determine:
a) the balanced reaction equation
b) the vapor mass flow rate (in lbm/lbm fuel)
c) the cooling water mass flow rate (in lbm/lbm fuel).
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