Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics
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616 Chapter 12 Ideal Gas Mixture and Psychrometric Applications
12.40 A closed, rigid tank having a volume of 1 m
3
contains a
mixture of carbon dioxide (CO
2
) and water vapor at 75C. The
respective masses are 12.3 kg of carbon dioxide and 0.5 kg of
water vapor. If the tank contents are cooled to 20C, determine
the heat transfer, in kJ, assuming ideal gas behavior.
12.41 Moist air at 20C, 1.05 bar, 85% relative humidity and a
volumetric flow rate of 0.3 m
3
/s enters a well-insulated com-
pressor operating at steady state. If moist air exits at 100C,
2.0 bar, determine
(a) the relative humidity at the exit.
(b) the power input, in kW.
(c) the rate of entropy production, in kW/K.
12.42 Moist air at 20C, 1 atm, 43% relative humidity and a
volumetric flow rate of 900 m
3
/h enters a control volume at
steady state and flows along a surface maintained at 65C,
through which heat transfer occurs. Liquid water at 20C is in-
jected at a rate of 5 kg/h and evaporates into the flowing stream.
For the control volume, , and kinetic and potential
energy effects are negligible. Moist air exits at 32C, 1 atm.
Determine
(a) the rate of heat transfer, in kW.
(b) the rate of entropy production, in kW/K.
12.43 Using the psychrometric chart, Fig. A-9, determine
(a) the relative humidity, the humidity ratio, and the specific
enthalpy of the mixture, in kJ per kg of dry air, corre-
sponding to dry-bulb and wet-bulb temperatures of 30 and
25C, respectively.
(b) the humidity ratio, mixture specific enthalpy, and wet-bulb
temperature corresponding to a dry-bulb temperature of
30C and 60% relative humidity.
(c) the dew point temperature corresponding to dry-bulb and
wet-bulb temperatures of 30 and 20C, respectively.
(d) Repeat parts (a)–(c) using Interactive Thermodynamics:
IT.
12.44 A fixed amount of air initially at 52C, 1 atm, and 10%
relative humidity is cooled at constant pressure to 15C. Us-
ing the psychrometric chart, determine whether condensation
occurs. If so, evaluate the amount of water condensed, in kg
per kg of dry air. If there is no condensation, determine the
relative humidity at the final state.
12.45 A fan within an insulated duct delivers moist air at the
duct exit at 22C, 60% relative humidity, and a volumetric flow
rate of 0.5 m
3
/s. At steady state, the power input to the fan is
1.3 kW. Using the psychrometric chart, determine the temper-
ature and relative humidity at the duct inlet.
12.46 The mixture enthalpy per unit mass of dry air, in kJ/kg(a),
represented on Fig. A-9 can be approximated closely from the
expression
Noting all significant assumptions, develop the expression.
H
m
a
1.005 T 1°C2 v 32501.7 1.82 T 1°C24
W
#
cv
0
Considering Air-Conditioning Applications
12.47 Each case listed gives the dry-bulb temperature and
relative humidity of the moist-air stream entering an air-
conditioning system: (a) 30C, 40%, (b) 17C, 60%, (c) 25C,
70%, (d) 15C, 40%, (e) 27C, 30%. The condition of the
moist-air stream exiting the system must satisfy these cons-
traints: 22 T
db
27C, 40 60%. In each case, de-
velop a schematic of equipment and processes from Sec. 12.9
that would achieve the desired result. Sketch the processes on
a psychrometric chart.
12.48 Moist air enters an air-conditioning system as shown in
Fig. 12.11 at 26C, 80% and a volumetric flow rate of
0.47 m
3
/s. At the exit of the heating section the moist air is at
26C, 50%. For operation at steady state, and neglecting
kinetic and potential energy effects, determine
(a) the rate energy is removed by heat transfer in the dehu-
midifier section, in tons.
(b) the rate energy is added by heat transfer in the heating sec-
tion, in kW.
12.49 Air at 35C, 1 atm, and 50% relative humidity enters a
dehumidifier operating at steady state. Saturated moist air and
condensate exit in separate streams, each at 15C. Neglecting
kinetic and potential energy effects, determine
(a) the heat transfer from the moist air, in kJ per kg of dry air.
(b) the amount of water condensed, in kg per kg of dry air.
(c) Check your answers using data from the psychrometric chart.
(d) Check your answers using Interactive Thermodynamics: IT.
12.50 An air conditioner operating at steady state takes in moist
air at 28C, 1 bar, and 70% relative humidity. The moist air
first passes over a cooling coil in the dehumidifier unit and
some water vapor is condensed. The rate of heat transfer be-
tween the moist air and the cooling coil is 11 tons. Saturated
moist air and condensate streams exit the dehumidifier unit at
the same temperature. The moist air then passes through a heat-
ing unit, exiting at 24C, 1 bar, and 40% relative humidity. Ne-
glecting kinetic and potential energy effects, determine
(a) the temperature of the moist air exiting the dehumidifer
unit, in C.
(b) the volumetric flow rate of the air entering the air condi-
tioner, in m
3
/min.
(c) the rate water is condensed, in kg/min.
(d) the rate of heat transfer to the air passing through the heat-
ing unit, in kW.
12.51 An air-conditioning system consists of a spray section
followed by a reheater. Moist air at 32C and 77% enters
the system and passes through a water spray, leaving the spray
section cooled and saturated with water. The moist air is then
heated to 25C and 45% with no change in the amount
of water vapor present. For operation at steady state, determine
(a) the temperature of the moist air leaving the spray section,
in C.
(b) the change the amount of water vapor contained in the moist
air passing through the system, in kg per kg of dry air.
Locate the principal states on a psychrometric chart.
Problems: Developing Engineering Skills 617
12.52 At steady state, moist air is to be supplied to a classroom
at a specified volumetric flow rate and temperature T. Air is
removed from the classroom in a separate stream at a temper-
ature of 27C and 50% relative humidity. Moisture is added to
the air in the room from the occupants at a rate of 4.5 kg/h.
The moisture can be regarded as saturated vapor at 33C. Heat
transfer into the occupied space from all sources is estimated
to occur at a rate of 34,000 kJ/h. The pressure remains uni-
form at 1 atm.
(a) For a supply air volumetric flow rate of 40 m
3
/min, deter-
mine the supply air temperature T, in C, and the relative
humidity.
(b) Plot the supply air temperature, in C, and relative hu-
midity, each versus the supply air volumetric flow rate
ranging from 35 to 90 m
3
/min.
12.53 Air at 35C, 1 bar, and 10% relative humidity enters
an evaporative cooler operating at steady state. The volu-
metric flow rate of the incoming air is 50 m
3
/min. Liquid
water at 20C enters the cooler and fully evaporates. Moist
air exits the cooler at 25C, 1 bar. If there is no significant
heat transfer between the device and its surroundings,
determine
(a) the rate at which liquid enters, in kg/min.
(b) the relative humidity at the exit.
(c) the rate of exergy destruction, in kJ/min, for T
0
20C.
Neglect kinetic and potential energy effects.
12.54 Using Eqs. 12.56b and 12.56c, show that
Employ this relation to demonstrate on a psychrometric chart
that state 3 of the mixture lies on a straight line connecting the
initial states of the two streams before mixing.
12.55 For the adiabatic mixing process in Example 12.14, plot
the exit temperature, in C, versus the volumetric flow rate of
stream 2 ranging from 0 to 1400 m
3
/min. Discuss the plot as
(AV)
2
goes to zero and as (AV)
2
becomes large.
12.56 A stream consisting of 35 m
3
/min of moist air at 14C,
1 atm, 80% relative humidity mixes adiabatically with a
stream consisting of 80 m
3
/min of moist air at 40C, 1 atm,
40% relative humidity, giving a single mixed stream at 1 atm.
Using the psychrometric chart together with the procedure of
Prob. 12.58, determine the relative humidity and temperature,
in C, of the exiting stream.
m
#
a1
m
#
a2
v
3
v
2
v
1
v
3
1h
a3
v
3
h
g3
2 1h
a2
v
2
h
g2
2
1h
a1
v
1
h
g1
2 1h
a3
v
3
h
g3
2
12.57 Atmospheric air having dry-bulb and wet-bulb tempera-
tures of 33 and 29C, respectively, enters a well-insulated
chamber operating at steady state and mixes with air entering
with dry-bulb and wet-bulb temperatures of 16 and 12C,
respectively. The volumetric flow rate of the lower tempera-
ture stream is three times that of the other stream. A single
mixed stream exits. The pressure is constant throughout at
1 atm. Neglecting kinetic and potential energy effects, deter-
mine for the exiting stream
(a) the relative humidity.
(b) the temperature, in C.
12.58 Air at 30C, 1 bar, 50% relative humidity enters an in-
sulated chamber operating at steady state with a mass flow rate
of 3 kg/min and mixes with a saturated moist air stream en-
tering at 5C, 1 bar with a mass flow rate of 5 kg/min. A sin-
gle mixed stream exits at 1 bar. Determine
(a) the relative humidity and temperature, in C, of the exit-
ing stream.
(b) the rate of exergy destruction, in kW, for T
0
20C.
Neglect kinetic and potential energy effects.
Analyzing Cooling Towers
12.59 In the condenser of a power plant, energy is discharged
by heat transfer at a rate of 836 MW to cooling water that ex-
its the condenser at 40C into a cooling tower. Cooled water
at 20C is returned to the condenser. Atmospheric air enters
the tower at 25C, 1 atm, 35% relative humidity. Moist air ex-
its at 35C, 1 atm, 90% relative humidity. Makeup water is
supplied at 20C. For operation at steady state, determine the
mass flow rate, in kg/s, of
(a) the entering atmospheric air.
(b) the makeup water.
Ignore kinetic and potential energy effects.
12.60 Liquid water at 50C enters a forced draft cooling tower
operating at steady state. Cooled water exits the tower with a
mass flow rate of 80 kg/min. No makeup water is provided. A
fan located within the tower draws in atmospheric air at 17C,
0.098 MPa, 60% relative humidity with a volumetric flow rate
of 110 m
3
/min. Saturated air exits the tower at 30C, 0.098
MPa. The power input to the fan is 8 kW. Ignoring kinetic and
potential energy effects, determine
(a) the mass flow rate of the liquid stream entering, in kg/min.
(b) the temperature of the cooled liquid stream exiting, in C.
Design & Open Ended Problems: Exploring Engineering Practice
12.1D For comfort in indoor natatoriums, air and water tem-
peratures should be between 24 and 30C, and the relative hu-
midity in the range 50 to 60%. A major contributor to dis-
comfort and inefficient energy use is water evaporating from
the pool surface. At 30C, an olympic-size pool can evaporate
up to 90 kg/h. The airborne water vapor has energy that can
be extracted during dehumidification and used to heat the pool
water. Devise a practical means for accomplishing this energy-
618 Chapter 12 Ideal Gas Mixture and Psychrometric Applications
12.6D Savings can be realized on domestic water heating bills
by using a heat pump that exhausts to the hot water storage
tank of a conventional water heater. When the heat pump evap-
orator is located within the dwelling living space, cooling and
dehumidification of the room air also result. For a hot water
(60C) requirement of 0.25 m
3
per day, compare the annual
cost of electricity for a heat-pump water heater with the an-
nual cost for a conventional electrical water heater. Provide a
schematic showing how the heat-pump water heater might be
configured. Where might the heat pump evaporator be located
during periods of the year when cooling and dehumidifying
are not required?
12.7D A laboratory test facility requires 1.4 m
3
/s of condi-
tioned air at 38C and relative humidities ranging from 20
Exit moist air,
T
2
< T
1
Inlet moist air,
T
1
1
3
2
Water
circulation
piping
Water
chiller
Spray
head
Figure P12.2D
12.3D Control of microbial contaminants is crucial in hospital
operating rooms. Figure P12.3D illustrates the proposed design
of a hospital air-handling system aimed at effective infection
control. Write a critique of this design, taking into considera-
tion issues including, but not necessarily limited to the
following: Why is 100% outside air specified rather than a
recirculated air-filtration arrangement requiring minimum out-
side air? Why is steam humidification used rather than the more
common water-spray method? Why are three filters with
increasing efficiency specified, and why are they located as
shown? What additional features might be incorporated to pro-
mote air quality?
Intermediate
filter
Supply
fan
Cooling
coil
Terminal
reheat
unit
Humidifier-
steam
injector
To other zones
Operating
room
Terminal
filter
Preheat
coil
Prefilter
100%
outside
air
Figure P12.3D
Moist air:
2100 m
3
/min at
22°C, = 60%
φ
Outside air
at 35°C,
= 55%
φ
Recirculated air 27°C, = 49%
φ
Damper
A
Damper
B
Heating coil
Cooling coil
Air-conditioned
space
Condensate
Figure P12.5D
saving measure. Also propose means for preventing conden-
sation on the natatorium windows for outdoor temperatures as
low as 23C. For each case, provide an equipment schematic
and appropriate supporting calculations.
12.2D Figure P12.2D shows a spray cooler included on the
schematic of a plant layout. Arguing that it can only be the
addition of makeup water at location 3 that is intended, an en-
gineer orders the direction of the arrow at this location on the
schematic to be reversed. Evaluate the engineer’s order and
write a memorandum explaining your evaluation.
12.4D Biosphere 2, located near Tucson, Arizona, is the world’s
largest enclosed ecosystem project. Drawing on what has been
learned from the project, develop a plan for the air-condition-
ing systems required to maintain a healthy environment for a
12-person biosphere on Mars. Estimate the total power required
by the systems of your plan.
12.5D Figure P12.5D shows a system for supplying a space
with 2100 m
3
/min of conditioned air at a dry-bulb tempera-
ture of 22C and a relative humidity of 60% when the outside
air is at a dry-bulb temperature of 35C and a relative humid-
ity of 55%. Dampers A and B can be set to give three alter-
native operating modes: (1) Both dampers closed (no use of
recirculated air). (2) Damper A open and damper B closed.
One-third of the conditioned air comes from outside air.
(3) Both dampers open. One-third of the conditioned air comes
from outside air. One-third of the recirculated air bypasses the
dehumidifier via open damper B, and the rest flows through
the damper A. Which of the three operating modes should be
used? Discuss.
Design & Open Ended Problems: Exploring Engineering Practice 619
to 50%. As shown in Fig. P12.7D, the supply system takes
in ambient air at temperatures ranging from 21 to 32C and
relative humidities ranging from 50 to 100%. As some de-
humidification is needed to achieve the temperature and hu-
midity control required by the test facility, two strategies are
under consideration: (1) Use a conventional refrigeration sys-
tem to cool the incoming air below its dew point tempera-
ture. (2) Use a rotary desiccant dehumidification unit.
Develop schematics and design specifications, based on each
of these strategies, that would allow the supply system to pro-
vide the required temperature and humidity control for this
application.
Figure P12.7D
Process
steam
Injection
steam
Return
water
Net
Power
Heat-recovery
steam generator
Combustor
Fuel
Air
8
5
7
1
2
6
4
Compressor
3
Turbine
Figure P12.9D
12.10D Most supermarkets use air-conditioning systems de-
signed primarily to control temperature, and these typically
produce a relative humidity of about 55%. At this relative hu-
midity a substantial amount of energy must be used to keep
frost and condensation from forming inside refrigerated dis-
play cases. Investigate technologies available for reducing the
overall humidity levels to 40–45% and thereby reducing prob-
lems associated with frost and condensation. Estimate the po-
tential cost savings associated with such a strategy for a
supermarket in your locale.
12.11D Is Stale Airplane Air Making Us Sick? (see box
Sec. 12.4). Investigate typical equipment schematics providing
a blend of fresh and recirculated filtered air to the passenger
cabin of a commercial airplane. Write a report including at
least three references.
12.12D How Cold Is Cold? (see box Sec. 12.6). Investigate
the development of the heat index used to alert us to possible
hot weather health dangers. Write a report including at least
three references.
Supply
system
Ambient
air
Laboratory
test
facility
Exhaust
21 ≤ T
amb
≤ 32C
50 ≤
amb
≤ 100%
φ
1.4 m
3
/s,
38C,
20 ≤ ≤ 50%
φ
12.8D As a consulting engineer, you have been retained to
recommend means for providing condenser cooling for a
proposed 25% expansion of an electrical-generating plant in
your locale. Consider the use of wet cooling towers, dry cool-
ing towers, and cooling lakes. Also, consider relevant envi-
ronmental issues, the effect on power plant thermodynamic
performance, and costs. Write a report detailing your recom-
mendations.
12.9D Figure P12.9D shows a steam-injected gas-turbine
cogeneration system that produces power and process steam:
a simplified Cheng cycle. The steam is generated by a heat-
recovery steam generator (HRSG) in which the hot gas exit-
ing the turbine at state 4 is cooled and discharged at state 5. A
separate stream of return water enters the HRSG at state 6 and
steam exits at states 7 and 8. The superheated steam exiting at
7 is injected into the combustor. In what ways does the steam-
injected system shown in the figure differ from the humid air
turbine (HAT) cycle? Draw the schematic of a HAT cycle.
Discuss appropriate applications of each cycle.
620
13
C
H
A
P
T
E
R
Reacting
Mixtures and
Combustion
ENGINEERING CONTEXT The objective of this chapter is to
study systems involving chemical reactions. Since the combustion of hydrocarbon fuels
occurs in most power-producing devices (Chaps. 8 and 9), combustion is emphasized.
The thermodynamic analysis of reacting systems is primarily an extension of principles
introduced thus far. The concepts applied in the first part of the chapter dealing with com-
bustion fundamentals remain the same: conservation of mass, conservation of energy, and
the second law. It is necessary, though, to modify the methods used to evaluate specific
enthalpy, internal energy, and entropy, by accounting for changing chemical composition.
Only the manner in which these properties are evaluated represents a departure from
previous practice, for once appropriate values are determined they are used as in earlier
chapters in the energy and entropy balances for the system under consideration. In the
second part of the chapter, the exergy concept of Chap. 7 is extended by introducing
chemical exergy.
The principles developed in this chapter allow the equilibrium composition of a mixture
of chemical substances to be determined. This topic is studied in the next chapter. The
subject of dissociation is also deferred until then. Prediction of reaction rates is not within
the scope of classical thermodynamics, so the topic of chemical kinetics, which deals with
reaction rates, is not discussed in this text.
COMBUSTION FUNDAMENTALS
13.1 Introducing Combustion
When a chemical reaction occurs, the bonds within molecules of the reactants are broken,
and atoms and electrons rearrange to form products. In combustion reactions, rapid
oxidation of combustible elements of the fuel results in energy release as combustion prod-
ucts are formed. The three major combustible chemical elements in most common fuels
are carbon, hydrogen, and sulfur. Sulfur is usually a relatively unimportant contributor to
the energy released, but it can be a significant cause of pollution and corrosion problems.
Combustion is complete when all the carbon present in the fuel is burned to carbon dioxide,
all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other
chapter objective
reactants
products
complete combustion
13.1 Introducing Combustion 621
combustible elements are fully oxidized. When these conditions are not fulfilled, combus-
tion is incomplete.
In this chapter, we deal with combustion reactions expressed by chemical equations of
the form
or
When dealing with chemical reactions, it is necessary to remember that mass is conserved,
so the mass of the products equals the mass of the reactants. The total mass of each chemi-
cal element must be the same on both sides of the equation, even though the elements exist
in different chemical compounds in the reactants and products. However, the number of moles
of products may differ from the number of moles of reactants.
for example... consider the complete combustion of hydrogen with oxygen
(13.1)
In this case, the reactants are hydrogen and oxygen. Hydrogen is the fuel and oxygen is
the oxidizer. Water is the only product of the reaction. The numerical coefficients in the
equation, which precede the chemical symbols to give equal amounts of each chemical
element on both sides of the equation, are called stoichiometric coefficients. In words,
Eq. 13.1 states
Note that the total numbers of moles on the left and right sides of Eq. 13.1 are not equal.
However, because mass is conserved, the total mass of reactants must equal the total mass
of products. Since 1 kmol of H
2
equals 2 kg, kmol of O
2
equals 16 kg, and 1 kmol of H
2
O
equals 18 kg, Eq. 13.1 can be interpreted as stating
In the remainder of this section, consideration is given to the makeup of the fuel, oxidizer,
and combustion products typically involved in engineering combustion applications.
FUELS
A fuel is simply a combustible substance. In this chapter emphasis is on hydrocarbon fuels,
which contain hydrogen and carbon. Sulfur and other chemical substances also may be pres-
ent. Hydrocarbon fuels can exist as liquids, gases, and solids.
Liquid hydrocarbon fuels are commonly derived from crude oil through distillation and
cracking processes. Examples are gasoline, diesel fuel, kerosene, and other types of fuel oils.
Most liquid fuels are mixtures of hydrocarbons for which compositions are usually given in
terms of mass fractions. For simplicity in combustion calculations, gasoline is often mod-
eled as octane, C
8
H
18
, and diesel fuel as dodecane, C
12
H
26
.
Gaseous hydrocarbon fuels are obtained from natural gas wells or are produced in certain
chemical processes. Natural gas normally consists of several different hydrocarbons, with the
major constituent being methane, CH
4
. The compositions of gaseous fuels are usually given
in terms of mole fractions. Both gaseous and liquid hydrocarbon fuels can be synthesized
from coal, oil shale, and tar sands.
Coal is a familiar solid fuel. Its composition varies considerably with the location from
which it is mined. For combustion calculations, the composition of coal is usually expressed
as an ultimate analysis. The ultimate analysis gives the composition on a mass basis in terms
2 kg H
2
16 kg O
2
S 18 kg H
2
O
1
2
1 kmol H
2
1
2
kmol O
2
S 1 kmol H
2
O
1H
2
1
2
O
2
S 1H
2
O
fuel oxidizer S products
reactants S products
stoichiometric coefficients
ultimate analysis
622 Chapter 13 Reacting Mixtures and Combustion
of the relative amounts of chemical elements (carbon, sulfur, hydrogen, nitrogen, oxygen)
and ash.
MODELING COMBUSTION AIR
Oxygen is required in every combustion reaction. Pure oxygen is used only in special
applications such as cutting and welding. In most combustion applications, air provides
the needed oxygen. The composition of a typical sample of dry air is given in Table 12.1.
For the combustion calculations of this book, however, the following model is used for
simplicity:
All components of air other than oxygen are lumped together with nitrogen. Accord-
ingly, air is considered to be 21% oxygen and 79% nitrogen on a molar basis. With this
idealization the molar ratio of the nitrogen to the oxygen is 0.790.21 3.76. When
air supplies the oxygen in a combustion reaction, therefore, every mole of oxygen is
accompanied by 3.76 moles of nitrogen.
We also assume that the nitrogen present in the combustion air does not undergo chemi-
cal reaction. That is, nitrogen is regarded as inert. The nitrogen in the products is at the
same temperature as the other products, however, so the nitrogen undergoes a change of
state if the products are at a temperature other than the air temperature before combus-
tion. If high enough temperatures are attained, nitrogen can form compounds such as
nitric oxide and nitrogen dioxide. Even trace amounts of oxides of nitrogen appearing
in the exhaust of internal combustion engines can be a source of air pollution.
AIR–FUEL RATIO. Two parameters that are frequently used to quantify the amounts of fuel
and air in a particular combustion process are the air–fuel ratio and its reciprocal, the fuel–air
ratio. The air–fuel ratio is simply the ratio of the amount of air in a reaction to the amount
of fuel. The ratio can be written on a molar basis (moles of air divided by moles of fuel) or
on a mass basis (mass of air divided by mass of fuel). Conversion between these values is
accomplished using the molecular weights of the air, M
air
, and fuel, M
fuel
,
or
(13.2)
where is the air–fuel ratio on a molar basis and AF is the ratio on a mass basis. For
the combustion calculations of this book the molecular weight of air is taken as 28.97.
Table A-1 provide the molecular weights of several important hydrocarbons.
THEORETICAL AIR. The minimum amount of air that supplies sufficient oxygen for the
complete combustion of all the carbon, hydrogen, and sulfur present in the fuel is called
the theoretical amount of air. For complete combustion with the theoretical amount of air,
the products would consist of carbon dioxide, water, sulfur dioxide, the nitrogen accompa-
nying the oxygen in the air, and any nitrogen contained in the fuel. No free oxygen would
appear in the products.
AF
AF AF a
M
air
M
fuel
b
moles of air
moles of fuel
a
M
air
M
fuel
b
mass of air
mass of fuel
moles of air M
air
moles of fuel M
fuel
air–fuel ratio
METHODOLOGY
UPDATE
In this model, air is as-
sumed to contain no water
vapor. When moist air is
involved in combustion,
the water vapor present
must be considered in
writing the combustion
equation.
theoretical air
13.1 Introducing Combustion 623
for example. . . let us determine the theoretical amount of air for the complete
combustion of methane. For this reaction, the products contain only carbon dioxide, water,
and nitrogen. The reaction is
(13.3)
where a, b, c, and d represent the numbers of moles of oxygen, carbon dioxide, water, and
nitrogen. In writing the left side of Eq. 13.3, 3.76 moles of nitrogen are considered to ac-
company each mole of oxygen. Applying the conservation of mass principle to the carbon,
hydrogen, oxygen, and nitrogen, respectively, results in four equations among the four
unknowns
Solving these equations, the balanced chemical equation is
(13.4)
The coefficient 2 before the term (O
2
3.76N
2
) in Eq. 13.4 is the number of moles of
oxygen in the combustion air, per mole of fuel, and not the amount of air. The amount
of combustion air is 2 moles of oxygen plus 2 3.76 moles of nitrogen, giving a total of
9.52 moles of air per mole of fuel. Thus, for the reaction given by Eq. 13.4 the air–fuel
ratio on a molar basis is 9.52. To calculate the air–fuel ratio on a mass basis, use Eq. 13.2
to write
Normally the amount of air supplied is either greater or less than the theoretical amount.
The amount of air actually supplied is commonly expressed in terms of the percent of
theoretical air. For example, 150% of theoretical air means that the air actually supplied
is 1.5 times the theoretical amount of air. The amount of air supplied can be expressed al-
ternatively as a percent excess or a percent deficiency of air. Thus, 150% of theoretical air
is equivalent to 50% excess air, and 80% of theoretical air is the same as a 20% deficiency
of air.
for example. . . consider the complete combustion of methane with 150% theo-
retical air (50% excess air). The balanced chemical reaction equation is
(13.5)
In this equation, the amount of air per mole of fuel is 1.5 times the theoretical amount
determined by Eq. 13.4. Accordingly, the air–fuel ratio is 1.5 times the air–fuel ratio
determined for Eq. 13.4. Since complete combustion is assumed, the products contain only
carbon dioxide, water, nitrogen, and oxygen. The excess air supplied appears in the products
as uncombined oxygen and a greater amount of nitrogen than in Eq. 13.4, based on the
theoretical amount of air.
The equivalence ratio is the ratio of the actual fuel–air ratio to the fuel–air ratio for com-
plete combustion with the theoretical amount of air. The reactants are said to form a lean
mixture when the equivalence ratio is less than unity. When the ratio is greater than unity,
the reactants are said to form a rich mixture.
CH
4
11.521221O
2
3.76N
2
2S CO
2
2H
2
O O
2
11.28N
2
AF AF a
M
air
M
fuel
b 9.52 a
28.97
16.04
b 17.19
CH
4
21O
2
3.76N
2
2S CO
2
2H
2
O 7.52N
2
N:
d 3.76a
O:
2b c 2a
H:
2 c 4
C:
b 1
CH
4
a1O
2
3.76N
2
2S bCO
2
cH
2
O dN
2
percent of theoretical air
percent excess air
equivalence ratio
624 Chapter 13 Reacting Mixtures and Combustion
In Example 13.1, we use conservation of mass to obtain balanced chemical reactions. The
air–fuel ratio for each of the reactions is also calculated.
EXAMPLE 13.1 Determining the Air–Fuel Ratio
Determine the air–fuel ratio on both a molar and mass basis for the complete combustion of octane, C
8
H
18
, with (a) the the-
oretical amount of air, (b) 150% theoretical air (50% excess air).
SOLUTION
Known: Octane, C
8
H
18
, is burned completely with (a) the theoretical amount of air, (b) 150% theoretical air.
Find: Determine the air–fuel ratio on a molar and a mass basis.
Assumptions:
1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen.
2. The nitrogen is inert.
3. Combustion is complete.
Analysis:
(a) For complete combustion of C
8
H
18
with the theoretical amount of air, the products contain carbon dioxide, water, and
nitrogen only. That is
Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, gives
Solving these equations, a 12.5, b 8, c 9, d 47. The balanced chemical equation is
The air–fuel ratio on a molar basis is
The air–fuel ratio expressed on a mass basis is
(b) For 150% theoretical air, the chemical equation for complete combustion takes the form
Applying conservation of mass
N:
d 11.52112.5213.762
O:
2 b c 2e 11.52112.52122
H:
2 c 18
C:
b 8
C
8
H
18
1.5112.521O
2
3.76N
2
2S bCO
2
cH
2
O dN
2
eO
2
AF c59.5
kmol 1air2
kmol 1fuel2
d≥
28.97
kg 1air2
kmol 1air2
114.22
kg 1fuel2
kmol 1fuel2
¥ 15.1
kg 1air2
kg 1fuel2
AF
12.5 12.513.762
1
12.514.762
1
59.5
kmol 1air2
kmol 1fuel2
C
8
H
18
12.51O
2
3.76N
2
2S 8CO
2
9H
2
O 47N
2
N:
d 3.76a
O:
2 b c 2a
H:
2 c 18
C:
b 8
C
8
H
18
a1O
2
3.76N
2
2S bCO
2
cH
2
O dN
2
❶
13.1 Introducing Combustion 625
DETERMINING PRODUCTS OF COMBUSTION
In each of the illustrations given above, complete combustion is assumed. For a hydro-
carbon fuel, this means that the only allowed products are CO
2
,H
2
O, and N
2
, with O
2
also present when excess air is supplied. If the fuel is specified and combustion is complete,
the respective amounts of the products can be determined by applying the conservation
of mass principle to the chemical equation. The procedure for obtaining the balanced re-
action equation of an actual reaction where combustion is incomplete is not always so
straightforward.
Combustion is the result of a series of very complicated and rapid chemical reactions, and
the products formed depend on many factors. When fuel is burned in the cylinder of an in-
ternal combustion engine, the products of the reaction vary with the temperature and pres-
sure in the cylinder. In combustion equipment of all kinds, the degree of mixing of the fuel
and air is a controlling factor in the reactions that occur once the fuel and air mixture is
ignited. Although the amount of air supplied in an actual combustion process may exceed
the theoretical amount, it is not uncommon for some carbon monoxide and unburned oxygen
to appear in the products. This can be due to incomplete mixing, insufficient time for com-
plete combustion, and other factors. When the amount of air supplied is less than the theo-
retical amount of air, the products may include both CO
2
and CO, and there also may be
unburned fuel in the products. Unlike the complete combustion cases considered above, the
products of combustion of an actual combustion process and their relative amounts can be
determined only by measurement.
Among several devices for measuring the composition of products of combustion are the
Orsat analyzer, gas chromatograph, infrared analyzer, and flame ionization detector. Data
from these devices can be used to determine the mole fractions of the gaseous products of
combustion. The analyses are often reported on a “dry” basis. In a dry product analysis, the
mole fractions are given for all gaseous products except the water vapor. In Examples 13.2
and 13.3, we show how analyses of the products of combustion on a dry basis can be used
to determine the balanced chemical reaction equations.
Since water is formed when hydrocarbon fuels are burned, the mole fraction of water va-
por in the gaseous products of combustion can be significant. If the gaseous products of com-
bustion are cooled at constant mixture pressure, the dew point temperature is reached when
water vapor begins to condense. Since water deposited on duct work, mufflers, and other
metal parts can cause corrosion, knowledge of the dew point temperature is important. De-
termination of the dew point temperature is illustrated in Example 13.2, which also features
a dry product analysis.
Solving this set of equations, b 8, c 9, d 70.5, e 6.25, giving a balanced chemical equation
The air–fuel ratio on a molar basis is
On a mass basis, the air–fuel ratio is 22.6 kg (air)kg (fuel), as can be verified.
When complete combustion occurs with excess air, oxygen appears in the products, in addition to carbon dioxide, water,
and nitrogen.
AF
18.7514.762
1
89.25
kmol 1air2
kmol 1fuel2
C
8
H
18
18.751O
2
3.76N
2
2S 8CO
2
9H
2
O 70.5N
2
6.25O
2
❶
dry product analysis