15-1
Chapter 15
CHEMICAL REACTIONS
Fuels and Combustion
15-1C Gasoline is C8H18, diesel fuel is C12H26, and natural gas is CH4.
15-2C Nitrogen, in general, does not react with other chemical species during a combustion process but its
presence affects the outcome of the process because nitrogen absorbs a large proportion of the heat released
during the chemical process.
15-3C Moisture, in general, does not react chemically with any of the species present in the combustion
chamber, but it absorbs some of the energy released during combustion, and it raises the dew point
temperature of the combustion gases.
15-4C The dew-point temperature of the product gases is the temperature at which the water vapor in the
product gases starts to condense as the gases are cooled at constant pressure. It is the saturation temperature
corresponding to the vapor pressure of the product gases.
15-5C The number of atoms are preserved during a chemical reaction, but the total mole numbers are not.
15-6C Air-fuel ratio is the ratio of the mass of air to the mass of fuel during a combustion process. Fuelair ratio is the inverse of the air-fuel ratio.
15-7C No. Because the molar mass of the fuel and the molar mass of the air, in general, are different.
Theoretical and Actual Combustion Processes
15-8C The causes of incomplete combustion are insufficient time, insufficient oxygen, insufficient mixing,
and dissociation.
15-9C CO. Because oxygen is more strongly attracted to hydrogen than it is to carbon, and hydrogen is
usually burned to completion even when there is a deficiency of oxygen.
15-10C It represent the amount of air that contains the exact amount of oxygen needed for complete
combustion.
15-11C No. The theoretical combustion is also complete, but the products of theoretical combustion does
not contain any uncombined oxygen.
15-12C Case (b).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-2
15-13 Methane is burned with the stoichiometric amount of air during a combustion process. The AF and
FA ratios are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis This is a theoretical combustion process since
methane is burned completely with stoichiometric amount of
air. The stoichiometric combustion equation of CH4 is
CH 4 + a th [O 2 + 3.76N 2 ] ⎯
⎯→ CO 2 + 2H 2 O + 3.76a th N 2
O2 balance:
Substituting,
a th = 1 + 1
⎯
⎯→
a th = 2
CH4
Products
Air
stoichiometric
CH 4 + 2[O 2 + 3.76N 2 ] ⎯
⎯→ CO 2 + 2H 2O + 7.52N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(2 × 4.76 kmol)(29 kg/kmol)
=
= 17.3 kg air/kg fuel
m fuel (1 kmol)(12 kg/kmol) + (2 kmol)(2 kg/kmol)
The fuel-air ratio is the inverse of the air-fuel ratio,
FA =
1
1
=
= 0.0578 kg fuel/kg air
AF 17.3 kg air/kg fuel
15-14 Propane is burned with 75 percent excess air during a combustion process. The AF ratio is to be
determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The combustion equation in this case can be written as
C 3 H 8 + 1.75a th [O 2 + 3.76N 2 ] ⎯
⎯→ 3CO 2 + 4H 2 O + 0.75a th O 2 + (1.75 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. We have
automatically accounted for the 75% excess air by using the
C3H8
factor 1.75ath instead of ath for air. The stoichiometric amount of
oxygen (athO2) will be used to oxidize the fuel, and the remaining
Air
excess amount (0.75athO2) will appear in the products as free
oxygen. The coefficient ath is determined from the O2 balance,
75% excess
O2 balance:
17
. 5a th = 3 + 2 + 0.75a th
Substituting,
C3H 8 + 8.75 O 2 + 3.76 N 2
⎯
⎯→
Products
a th = 5
⎯
⎯→ 3CO 2 + 4H 2O + 3.75O 2 + 32.9 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(8.75 × 4.76 kmol)(29 kg/kmol)
=
= 27.5 kg air/kg fuel
m fuel (3 kmol)(12 kg/kmol) + (4 kmol)(2 kg/kmol)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-3
15-15 Acetylene is burned with the stoichiometric amount of air during a combustion process. The AF ratio
is to be determined on a mass and on a mole basis.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis This is a theoretical combustion process since C2H2 is burned completely with stoichiometric
amount of air. The stoichiometric combustion equation of C2H2 is
C 2 H 2 + a th [O 2 + 3.76N 2 ] ⎯
⎯→ 2CO 2 + H 2 O + 3.76a th N 2
O2 balance:
a th = 2 + 0.5
⎯
⎯→
C2H2
Products
a th = 2.5
Substituting,
C 2 H 2 + 2.5[O 2 + 3.76N 2 ] ⎯
⎯→ 2CO 2 + H 2 O + 9.4N 2
100%
theoretical air
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(2.5 × 4.76 kmol)(29 kg/kmol)
=
= 13.3 kg air/kg fuel
m fuel (2 kmol)(12 kg/kmol) + (1 kmol)(2 kg/kmol)
On a mole basis, the air-fuel ratio is expressed as the ratio of the mole numbers of the air to the mole
numbers of the fuel,
AFmole basis =
N air
(2.5 × 4.76) kmol
=
= 11.9 kmol air/kmol fuel
N fuel
1 kmol fuel
15-16 Ethane is burned with an unknown amount of air during a combustion process. The AF ratio and the
percentage of theoretical air used are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
C2H6
Analysis (a) The combustion equation in this case can be written as
C 2 H 6 + a[O 2 + 3.76N 2 ] ⎯
⎯→ 2CO 2 + 3H 2 O + 3O 2 + 3.76aN 2
O2 balance:
Substituting,
a = 2 + 1.5 + 3 ⎯
⎯→ a = 6.5
Products
air
C 2 H 6 + 6.5[O 2 + 3.76N 2 ] ⎯
⎯→ 2CO 2 + 3H 2 O + 3O 2 + 24.44N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(6.5 × 4.76 kmol)(29 kg/kmol)
=
= 29.9 kg air/kg fuel
m fuel (2 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is
determined from the theoretical combustion equation of C2H6,
C 2 H 6 + a th O 2 + 3.76N 2
O2 balance:
Then,
a th = 2 + 15
.
⎯
⎯→ 2CO 2 + 3H 2O + 3.76a th N 2
⎯
⎯→
Percent theoretical air =
m air,act
m air, th
a th = 35
.
=
N air,act
N air, th
=
a
6.5
=
= 186%
a th 3.5
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-4
15-17E Ethylene is burned with 200 percent theoretical air during a combustion process. The AF ratio and
the dew-point temperature of the products are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 lbm/lbmol, 2 lbm/lbmol, and 29 lbm/lbmol,
respectively (Table A-1E).
Analysis (a) The combustion equation in this case can be written as
C2H4
C 2 H 4 + 2a th O 2 + 3.76N 2 ⎯
⎯→ 2CO 2 + 2H 2 O + a th O 2 + (2 × 3.76)a th N 2
Products
where ath is the stoichiometric coefficient for air.
200%
It is determined from
theoretical air
O2 balance:
2a th = 2 + 1 + a th
⎯
⎯→ a th = 3
Substituting,
C 2 H 4 + 6 O 2 + 3.76N 2
⎯⎯→ 2CO 2 + 2 H 2 O + 3O 2 + 22.56N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
m
(6 × 4.76 lbmol)(29 lbm/lbmol)
AF = air =
= 29.6 lbm air/lbm fuel
m fuel (2 lbmol)(12 lbm/lbmol) + (2 lbmol)(2 lbm/lbmol)
(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in
the product gases corresponding to its partial pressure. That is,
Thus,
⎛ Nv ⎞
⎛ 2 lbmol ⎞
⎟P
⎟(14.5 psia ) = 0.981 psia
=⎜
Pv = ⎜
⎜ N prod ⎟ prod ⎜⎝ 29.56 lbmol ⎟⎠
⎠
⎝
Tdp = Tsat @ 0.981 psia = 101°F
15-18 Propylene is burned with 50 percent excess air during a combustion process. The AF ratio and the
temperature at which the water vapor in the products will start condensing are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 kg/kmol,
Products
C3H6
2 kg/kmol, and 29 kg/kmol, respectively (Table A-1).
Analysis (a) The combustion equation in this case can be
50% excess air
written as
C 3 H 6 + 1.5a th [O 2 + 3.76N 2 ] ⎯
⎯→ 3CO 2 + 3H 2 O + 0.5a th O 2 + (1.5 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
15
. a th = 3 + 15
. + 0.5a th
Substituting, C 3 H 6 + 6.75 O 2 + 3.76N 2
⎯
⎯→
a th = 4.5
⎯⎯→ 3CO 2 + 3H 2 O + 2.25O 2 + 25.38N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
m
(6.75 × 4.76 kmol)(29 kg/kmol)
AF = air =
= 22.2 kg air/kg fuel
m fuel (3 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in
the product gases corresponding to its partial pressure. That is,
Thus,
⎛ Nv ⎞
⎛ 3 kmol ⎞
⎟P
⎟(105 kPa ) = 9.367 kPa
Pv = ⎜
=⎜
⎜ N prod ⎟ prod ⎜⎝ 33.63 kmol ⎟⎠
⎝
⎠
Tdp = Tsat @9.367 kPa = 44.5°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-5
15-19 Propal alcohol C3H7OH is burned with 50 percent excess air. The balanced reaction equation for
complete combustion is to be written and the air-to-fuel ratio is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol,
respectively (Table A-1).
Analysis The combustion equation in this case can be written as
C 3 H 7 OH + 1.5a th [O 2 + 3.76N 2 ] ⎯
⎯→ B CO 2 + D H 2 O + E O 2 + F N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 50% excess air
by using the factor 1.5ath instead of ath for air. The coefficient ath and other coefficients are to be
determined from the mass balances
Carbon balance:
B=3
Hydrogen balance:
2D = 8 ⎯
⎯→ D = 4
Oxygen balance:
1 + 2 × 1.5a th = 2 B + D + 2 E
0.5a th = E
Nitrogen balance:
C3H7OH
Products
Air
50% eccess
1.5a th × 3.76 = F
Solving the above equations, we find the coefficients (E = 2.25, F = 25.38, and ath = 4.5) and write the
balanced reaction equation as
C 3 H 7 OH + 6.75[O 2 + 3.76N 2 ] ⎯
⎯→3 CO 2 + 4 H 2 O + 2.25 O 2 + 25.38 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(6.75 × 4.75 kmol)(29 kg/kmol)
=
= 15.51 kg air/kg fuel
(3 × 12 + 8 × 1 + 1× 16) kg
m fuel
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-6
15-20 Butane C4H10 is burned with 200 percent theoretical air. The kmol of water that needs to be sprayed
into the combustion chamber per kmol of fuel is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol,
respectively (Table A-1).
Analysis The reaction equation for 200% theoretical air without the additional water is
C 4 H 10 + 2a th [O 2 + 3.76N 2 ] ⎯
⎯→ B CO 2 + D H 2 O + E O 2 + F N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 100% excess air
by using the factor 2ath instead of ath for air. The coefficient ath and other coefficients are to be determined
from the mass balances
Carbon balance:
B=4
Hydrogen balance:
2 D = 10 ⎯
⎯→ D = 5
Oxygen balance:
2 × 2a th = 2 B + D + 2 E
a th = E
2a th × 3.76 = F
Nitrogen balance:
C4H10
Products
Air
200%
theoretical
Solving the above equations, we find the coefficients (E = 6.5, F = 48.88, and ath = 6.5) and write the
balanced reaction equation as
C 4 H 10 + 13[O 2 + 3.76N 2 ] ⎯
⎯→ 4 CO 2 + 5 H 2 O + 6.5 O 2 + 48.88 N 2
With the additional water sprayed into the combustion chamber, the balanced reaction equation is
C 4 H 10 + 13[O 2 + 3.76N 2 ] + N v H 2 O ⎯
⎯→ 4 CO 2 + (5 + N v ) H 2 O + 6.5 O 2 + 48.88 N 2
The partial pressure of water in the saturated product mixture at the dew point is
Pv ,prod = Psat@60°C = 19.95 kPa
The vapor mole fraction is
yv =
Pv ,prod
Pprod
=
19.95 kPa
= 0.1995
100 kPa
The amount of water that needs to be sprayed into the combustion chamber can be determined from
yv =
N water
5 + Nv
⎯
⎯→ 0.1995 =
⎯
⎯→ N v = 9.796 kmol
N total,product
4 + 5 + N v + 6.5 + 48.88
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-7
15-21 A fuel mixture of 20% by mass methane, CH4, and 80% by mass ethanol, C2H6O, is burned
completely with theoretical air. The required flow rate of air is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol,
respectively (Table A-1).
Analysis The combustion equation in this case can be written as
x CH 4 + y C 2 H 6 O + a th [O 2 + 3.76N 2 ] ⎯
⎯→ B CO 2 + D H 2 O + F N 2
where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be
determined from the mass balances
Carbon balance:
x + 2y = B
Hydrogen balance:
4 x + 6 y = 2D
Oxygen balance:
2ath + y = 2 B + D
Nitrogen balance:
3.76a th = F
20% CH4
80% C2H6O
Air
Products
100% theoretical
Solving the above equations, we find the coefficients as
x = 0.4182
B = 1.582
y = 0.5818
D = 2.582
ath = 2.582
F = 9.708
Then, we write the balanced reaction equation as
0.4182 CH 4 + 0.5818 C 2 H 6O + 2.582 [O 2 + 3.76N 2 ] ⎯
⎯→ 1.582 CO 2 + 2.582 H 2O + 9.708 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
=
m air
m fuel
(2.582 × 4.76 kmol)(29 kg/kmol)
= 10.64 kg air/kg fuel
(0.4182 kmol)(12 + 4 × 1)kg/kmol + (0.5818 kmol)(2 × 12 + 6 × 1 + 16)kg/kmol
Then, the required flow rate of air becomes
m& air = AFm& fuel = (10.64)(31 kg/s) = 330 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-8
15-22 Octane is burned with 250 percent theoretical air during a combustion process. The AF ratio and the
dew-pint temperature of the products are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis (a) The combustion equation in this case can be written as
C8 H18 + 2.5a th O 2 + 3.76N 2
⎯
⎯→ 8CO 2 + 9H 2 O + 1.5a th O 2 + (2.5 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air.
It is determined from
O2 balance:
2.5a th = 8 + 4.5 + 1.5a th ⎯
⎯→ a th = 12.5
Substituting,
C 8 H 18 + 31.25[O 2 + 3.76N 2 ] → 8CO 2 + 9H 2 O + 18.75O 2 + 117.5N 2
Thus,
AF =
C8H18
Air
25°C
Combustion Products
chamber
P = 1 atm
m air
(31.25 × 4.76 kmol)(29 kg/kmol)
=
= 37.8 kg air/kg fuel
m fuel (8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)
(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in
the product gases corresponding to its partial pressure. That is,
⎛ N ⎞
⎛ 9 kmol ⎞
⎟⎟(101.325 kPa ) = 5.951 kPa
Pv = ⎜ v ⎟ Pprod = ⎜⎜
⎜ N prod ⎟
⎝ 153.25 kmol ⎠
⎝
⎠
Thus,
Tdp = Tsat @5.951 kPa = 36.0°C
15-23 Gasoline is burned steadily with air in a jet engine. The AF ratio is given. The percentage of excess
air used is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The theoretical combustion equation in this
case can be written as
C8 H18 + a th O 2 + 3.76N 2
⎯
⎯→ 8CO 2 + 9H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
a th = 8 + 4.5
⎯
⎯→
a th = 12.5
Gasoline
(C8H18)
Jet engine
Products
Air
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the
mass of the fuel for,
AFth =
m air,th
m fuel
=
(12.5 × 4.76 kmol)(29 kg/kmol)
= 15.14 kg air/kg fuel
(8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)
Then the percent theoretical air used can be determined from
Percent theoretical air =
AFact
AFth
=
18 kg air/kg fuel
= 119%
15.14 kg air/kg fuel
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-9
15-24 Ethane is burned with air steadily. The mass flow rates of ethane and air are given. The percentage of
excess air used is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The theoretical combustion equation in this case can be written as
C 2 H 6 + a th O 2 + 3.76N 2
⎯
⎯→ 2CO 2 + 3H 2O + 3.76a th N 2
C2H6
Combustion Products
chamber
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
a th = 2 + 15
.
⎯
⎯→
a th = 35
.
The air-fuel ratio for the theoretical reaction is determined by
taking the ratio of the mass of the air to the mass of the fuel for,
AFth =
m air,th
m fuel
=
Air
(3.5 × 4.76 kmol)(29 kg/kmol)
= 16.1 kg air/kg fuel
(2 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
The actual air-fuel ratio used is
AFact =
m& air 176 kg/h
=
= 22 kgair/kgfuel
m& fuel
8 kg/h
Then the percent theoretical air used can be determined from
Percent theoretical air =
AFact
AFth
=
22 kg air/kg fuel
= 137%
16.1 kg air/kg fuel
Thus the excess air used during this process is 37%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-10
15-25 Butane is burned with air. The masses of butane and air are given. The percentage of theoretical air
used and the dew-point temperature of the products are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis (a) The theoretical combustion equation in this case can be written as
C 4 H10 + a th O 2 + 3.76N 2
⎯
⎯→ 4CO 2 + 5H 2O + 3.76a th N 2
where ath is the stoichiometric coefficient for air. It is
determined from
a th = 4 + 2.5
O2 balance:
⎯
⎯→
m air,th
m fuel
=
Combustion Products
chamber
a th = 6.5
The air-fuel ratio for the theoretical reaction is determined by
taking the ratio of the mass of the air to the mass of the fuel for,
AFth =
C4H10
Air
(6.5 × 4.76 kmol)(29 kg/kmol)
= 15.5 kg air/kg fuel
(4 kmol)(12 kg/kmol) + (5 kmol)(2 kg/kmol)
The actual air-fuel ratio used is
AFact =
mair
25 kg
=
= 25 kg air / kg fuel
mfuel
1 kg
Then the percent theoretical air used can be determined from
Percent theoretical air =
AFact
AFth
=
25 kg air/kg fuel
= 161%
15.5 kg air/kg fuel
(b) The combustion is complete, and thus products will contain only CO2, H2O, O2 and N2. The air-fuel
ratio for this combustion process on a mole basis is
AF =
N air
m / M air
(25 kg )/ (29 kg/kmol) = 50 kmol air/kmol fuel
= air
=
(1 kg )/ (58 kg/kmol)
N fuel m fuel / M fuel
Thus the combustion equation in this case can be written as
C 4 H10 + (50/4.76 )[O 2 + 3.76N 2 ] ⎯
⎯→ 4CO 2 + 5H 2 O + 4.0O 2 + 39.5N 2
The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the
product gases corresponding to its partial pressure. That is,
⎛ Nv
Pv = ⎜
⎜ N prod
⎝
⎞
⎟ Pprod = ⎛⎜ 5 kmol ⎞⎟(90 kPa ) = 8.571 kPa
⎜ 52.5 kmol ⎟
⎟
⎝
⎠
⎠
Thus,
Tdp = Tsat @8.571
kPa
= 42.8°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-11
15-26E Butane is burned with air. The masses of butane and air are given. The percentage of theoretical air
used and the dew-point temperature of the products are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 lbm/lbmol, 2 lbm/lbmol, and 29 lbm/lbmol,
respectively (Table A-1).
Analysis (a) The theoretical combustion equation in this case can be written as
C 4 H 10 + ath O 2 + 3.76N 2
⎯⎯→ 4CO 2 + 5H 2 O + 3.76ath N 2
where ath is the stoichiometric coefficient for air. It is
determined from
a th = 4 + 2.5
O2 balance:
⎯
⎯→
m air,th
m fuel
=
Combustion Products
chamber
a th = 6.5
The air-fuel ratio for the theoretical reaction is determined by
taking the ratio of the mass of the air to the mass of the fuel for,
AFth =
C4H10
Air
(6.5 × 4.76 lbmol)(29 lbm/lbmol)
= 15.5 lbm air/lbm fuel
(4 lbmol)(12 lbm/lbmol) + (5 lbmol)(2 lbm/lbmol)
The actual air-fuel ratio used is
AFact =
m air
25 lbm
=
= 25 lbm air/lbm fuel
m fuel
1 lbm
Then the percent theoretical air used can be determined from
Percent theoretical air =
AFact
AFth
=
25 lbm air/lbm fuel
= 161%
15.5 lbm air/lbm fuel
(b) The combustion is complete, and thus products will contain only CO2, H2O, O2 and N2. The air-fuel
ratio for this combustion process on a mole basis is
AF =
N air
m / M air
(25 lbm)/ (29 lbm/lbmol) = 50 lbmol air/lbmol fuel
= air
=
(1 lbm)/ (58 lbm/lbmol)
N fuel m fuel / M fuel
Thus the combustion equation in this case can be written as
C 4 H 10 + (50/4.76 )[O 2 + 3.76N 2 ] ⎯
⎯→ 4CO 2 + 5H 2 O + 4O 2 + 39.5N 2
The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the
product gases corresponding to its partial pressure. That is,
⎛ Nv
Pv = ⎜
⎜ N prod
⎝
⎞
⎟ Pprod = ⎛⎜ 5 lbmol ⎞⎟(14.7 psia ) = 1.4 psia
⎜ 52.5 lbmol ⎟
⎟
⎝
⎠
⎠
Thus,
Tdp = Tsat @1.4 psia = 113.2°F
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-12
15-27 The volumetric fractions of the constituents of a certain natural gas are given. The AF ratio is to be
determined if this gas is burned with the stoichiometric amount of dry air.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol,
and 29 kg/kmol, respectively (Table A-1).
Analysis Considering 1 kmol of fuel, the combustion equation can be written as
(0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯
⎯→ xCO 2 + yH 2 O + zN 2
The unknown coefficients in the above equation are determined from mass balances,
C : 0.65 + 0.06 = x
⎯
⎯→ x = 0.71
H : 0.65 × 4 + 0.08 × 2 = 2 y
O 2 : 0.03 + 0.06 + a th = x + y / 2
N 2 : 0.18 + 3.76a th = z
Natural gas
⎯
⎯→ y = 1.38
Combustion Products
chamber
⎯
⎯→ a th = 1.31
⎯
⎯→ z = 5.106
Dry air
Thus,
(0.65CH 4 + 0.08H 2 + 018
. N 2 + 0.03O 2 + 0.06CO 2 ) + 131
. (O 2 + 3.76N 2 )
⎯⎯→ 0.71CO 2 + 138
. H 2 O + 5106
. N2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of
the fuel,
m air = (1.31× 4.76 kmol)(29 kg/kmol) = 180.8 kg
m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44 )kg = 19.2 kg
and
AFth =
m air,th
m fuel
=
180.8 kg
= 9.42 kg air/kg fuel
19.2 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-13
15-28 The composition of a certain natural gas is given. The gas is burned with stoichiometric amount of
moist air. The AF ratio is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol,
and 29 kg/kmol, respectively (Table A-1).
Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will
contain only H2O, CO2 and N2, but no free O2. The moisture in the air does not react with anything; it
simply shows up as additional H2O in the products. Therefore, we can simply balance the combustion
equation using dry air, and then add the moisture to both sides of the equation. Considering 1 kmol of fuel,
the combustion equation can be written as
(0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯
⎯→ xCO 2 + yH 2O + zN 2
The unknown coefficients in the above equation are determined from mass balances,
C : 0.65 + 0.06 = x
⎯
⎯→ x = 0.71
H : 0.65 × 4 + 0.08 × 2 = 2 y
O 2 : 0.03 + 0.06 + a th = x + y / 2
N 2 : 0.18 + 3.76a th = z
Natural gas
⎯
⎯→ y = 1.38
Combustion Products
chamber
⎯
⎯→ a th = 1.31
⎯
⎯→ z = 5.106
Moist air
Thus,
(0.65CH 4 + 0.08H 2 + 018
. N 2 + 0.03O 2 + 0.06CO 2 ) + 1.31(O 2 + 3.76N 2 )
⎯⎯→ 0.71CO 2 + 138
. H 2 O + 5106
. N2
Next we determine the amount of moisture that accompanies 4.76ath = (4.76)(1.31) = 6.24 kmol of dry air.
The partial pressure of the moisture in the air is
Pv,in = φ air Psat@ 25°C = (0.85)(3.1698 kPa) = 2.694 kPa
Assuming ideal gas behavior, the number of moles of the moisture in the air (Nv, in) is determined to be
⎛ Pv ,in
N v ,in = ⎜⎜
⎝ Ptotal
⎞
⎛ 2.694 kPa ⎞
⎟ N total = ⎜
⎯→ N v,air = 0.17 kmol
⎜ 101.325 kPa ⎟⎟ 6.24 + N v ,in ⎯
⎟
⎝
⎠
⎠
(
)
The balanced combustion equation is obtained by substituting the coefficients determined earlier and
adding 0.17 kmol of H2O to both sides of the equation,
(0.65CH 4 + 0.08H 2 + 018
. N 2 + 0.03O 2 + 0.06CO 2 ) + 131
. (O 2 + 3.76N 2 ) + 017
. H 2O
⎯⎯→ 0.71CO 2 + 155
. H 2 O + 5106
. N2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of
the fuel,
m air = (1.31 × 4.76 kmol )(29 kg/kmol ) + (0.17 kmol × 18 kg/kmol ) = 183.9 kg
m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44 )kg = 19.2 kg
and
AFth =
m air, th
m fuel
=
183.9 kg
= 9.58 kg air/kg fuel
19.2 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-14
15-29 The composition of a gaseous fuel is given. It is burned with 130 percent theoretical air. The AF ratio
and the fraction of water vapor that would condense if the product gases were cooled are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, N2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, and 29 kg/kmol,
respectively (Table A-1).
Analysis (a) The fuel is burned completely with excess air, and thus the products will contain H2O, CO2,
N2, and some free O2. Considering 1 kmol of fuel, the combustion equation can be written as
(0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.3a th (O 2 + 3.76N 2 ) ⎯
⎯→ xCO 2 + yH 2O + 0.3a th O 2 + zN 2
The unknown coefficients in the above equation are determined from mass balances,
C : 0.60 = x
⎯
⎯→ x = 0.60
H : 0.60 × 4 + 0.30 × 2 = 2 y
O 2 : 1.3a th = x + y / 2 + 0.3a th
N 2 : 0.10 + 3.76 × 1.3a th = z
⎯
⎯→ y = 1.50
⎯
⎯→ a th = 1.35
⎯
⎯→ z = 6.70
Gaseous fuel
Air
Combustion Products
chamber
30% excess
Thus,
(0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.755(O 2 + 3.76N 2 ) ⎯
⎯→ 0.6CO 2 + 1.5H 2O + 0.405O 2 + 6.7N 2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of
the fuel,
m air = (1.755 × 4.76 kmol )(29 kg/kmol ) = 242.3 kg
m fuel = (0.6 × 16 + 0.3 × 2 + 0.1× 28)kg = 13.0 kg
and
AF =
mair
242.3 kg
=
= 18.6 kg air/kg fuel
mfuel
13.0 kg
(b) For each kmol of fuel burned, 0.6 + 1.5 + 0.405 + 6.7 = 9.205 kmol of products are formed, including
1.5 kmol of H2O. Assuming that the dew-point temperature of the products is above 20°C, some of the
water vapor will condense as the products are cooled to 20°C. If Nw kmol of H2O condenses, there will be
1.5 - Nw kmol of water vapor left in the products. The mole number of the products in the gas phase will
also decrease to 9.205 - Nw as a result. Treating the product gases (including the remaining water vapor) as
ideal gases, Nw is determined by equating the mole fraction of the water vapor to its pressure fraction,
Nv
P
1.5 − N w
2.3392 kPa
= v ⎯
⎯→
=
⎯
⎯→ N w = 1.32 kmol
N prod,gas Pprod
9.205 − N w 101.325 kPa
since Pv = Psat @ 20°C = 2.3392 kPa. Thus the fraction of water vapor that condenses is 1.32/1.5 = 0.88 or
88%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-15
15-30 EES Problem 15-29 is reconsidered. The effects of varying the percentages of CH4, H2 and N2
making up the fuel and the product gas temperature are to be studied.
Analysis The problem is solved using EES, and the solution is given below.
Let's modify this problem to include the fuels butane, ethane, methane, and propane in
pull down menu. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at
T_air:
Reaction: aCxHy+bH2+cN2 + (a*y/4 + a*x+b/2) (Theo_air/100) (O2 + 3.76 N2)
<--> a*xCO2 + ((a*y/2)+b) H2O + (c+3.76 (a*y/4 + a*x+b/2) (Theo_air/100)) N2 +
(a*y/4 + a*x+b/2) (Theo_air/100 - 1) O2
T_prod is the product gas temperature.
Theo_air is the % theoretical air. "
Procedure
H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Re
sult$)
P_v = Moles_H2O/(M_other+Moles_H2O)*P_prod
T_DewPoint = temperature(steam,P=P_v,x=0)
IF T_DewPoint <= T_prod then
Moles_H2O_vap = Moles_H2O
Moles_H2O_liq=0
Result$='No condensation occurred'
ELSE
Pv_new=pressure(steam,T=T_prod,x=0)
Moles_H2O_vap=Pv_new/P_prod*M_other/(1-Pv_new/P_prod)
Moles_H2O_liq = Moles_H2O - Moles_H2O_vap
Result$='There is condensation'
ENDIF
END
"Input data from the diagram window"
{P_prod = 101.325 [kPa]
Theo_air = 130 "[%]"
a=0.6
b=0.3
c=0.1
T_prod = 20 [C]}
Fuel$='CH4'
x=1
y=4
"Composition of Product gases:"
A_th = a*y/4 +a* x+b/2
AF_ratio = 4.76*A_th*Theo_air/100*molarmass(Air)/(a*16+b*2+c*28) "[kg_air/kg_fuel]"
Moles_O2=(a*y/4 +a* x+b/2) *(Theo_air/100 - 1)
Moles_N2=c+(3.76*(a*y/4 + a*x+b/2))* (Theo_air/100)
Moles_CO2=a*x
Moles_H2O=a*y/2+b
M_other=Moles_O2+Moles_N2+Moles_CO2
Call
H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Re
sult$)
Frac_cond = Moles_H2O_liq/Moles_H2O*Convert(, %) "[%]"
"Reaction: aCxHy+bH2+cN2 + A_th Theo_air/100 (O2 + 3.76 N2)
<--> a*xCO2 + (a*y/2+b) H2O + (c+3.76 A_th Theo_air/100) N2 + A_th (Theo_air/100 - 1) O2"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-16
AFratio
[kgair/ kgfuel]
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
18.61
Fraccond
[%]
95.54
91.21
83.42
69.8
59.65
46.31
28.75
17.94
5.463
0.5077
0.1679
0
0
MolesH2O,liq
MolesH2O,vap
1.433
1.368
1.251
1.047
0.8947
0.6947
0.4312
0.2691
0.08195
0.007615
0.002518
0
0
0.06692
0.1319
0.2487
0.453
0.6053
0.8053
1.069
1.231
1.418
1.492
1.497
1.5
1.5
Tprod
[C]
5
15
25
35
40
45
50
52.5
55
55.9
55.96
60
85
1.6
1.4
1.2
Vapor
Liquid
1
O
2
H
s
el
o
M
0.8
0.6
0.4
Dew Point = 55.96 C
0.2
0
0
10
20
30
40
50
60
70
80
90
Tprod [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-17
15-31 The composition of a certain coal is given. The coal is burned with 50 percent excess air. The AF
ratio is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, N2, and ash
only.
Properties The molar masses of C, H2, O2, and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol,
respectively (Table A-1).
Analysis The composition of the coal is given on a mass basis, but we need to know the composition on a
mole basis to balance the combustion equation. Considering 1 kg of coal, the numbers of mole of the each
component are determined to be
N C = (m / M )C = 0.82 / 12 = 0.0683 kmol
N H 2O = (m / M )H 2O = 0.05 / 18 = 0.0028 kmol
N H 2 = (m / M )H 2 = 0.02 / 2 = 0.01 kmol
N O 2 = (m / M )O 2 = 0.01 / 32 = 0.00031 kmol
Coal
Air
Combustion Products
chamber
50% excess
Considering 1 kg of coal, the combustion equation can be written as
(0.0683C + 0.0028H 2O + 0.01H 2 + 0.00031O 2 + ash) + 1.5a th (O 2 + 3.76N 2 )
⎯
⎯→ xCO 2 + yH 2O + 0.5a th O 2 + 1.5 × 3.76a th N 2 + ash
The unknown coefficients in the above equation are determined from mass balances,
C : 0.0683 = x
⎯
⎯→ x = 0.0683
H : 0.0028 × 2 + 0.01× 2 = 2 y
⎯
⎯→ y = 0.0128
O 2 : 0.0028 / 2 + 0.00031 + 1.5a th = x + y / 2 + 0.5a th ⎯
⎯→ a th = 0.073
Thus,
(0.0683C + 0.0028H 2O + 0.01H 2 + 0.00031O 2 + ash) + 0.1095(O 2 + 3.76N 2 )
⎯
⎯→ 0.0683CO 2 + 0.0128H 2O + 0.0365O 2 + 0.4117N 2 + ash
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of
the coal, which is taken to be 1 kg,
m air = (0.1095 × 4.76 kmol)(29 kg/kmol) = 15.1 kg
mfuel = 1 kg
and
AF =
mair 15.1 kg
=
= 15.1 kg air/kg fuel
mfuel
1 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-18
15-32 Octane is burned with dry air. The volumetric fractions of the products are given. The AF ratio and
the percentage of theoretical air used are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, CO, H2O, O2, and N2
only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis Considering 100 kmol of dry products, the combustion equation can be written as
xC8 H18 + a O 2 + 3.76N 2
⎯
⎯→ 9.21CO 2 + 0.61CO + 7.06O 2 + 83.12N 2 + bH 2 O
The unknown coefficients x, a, and b are determined from mass balances,
N 2 : 3.76a = 83.12
C:
⎯
⎯→ a = 22.11
8 x = 9.21 + 0.61
H : 18 x = 2b
C8H18
⎯
⎯→ x = 1.23
Combustion Products
chamber
⎯
⎯→ b = 11.07
⎯→ 22.11 ≅ 22.10)
(CheckO 2 : a = 9.21 + 0.305 + 7.06 + b / 2 ⎯
Dry air
Thus,
1.23C 8 H 18 + 22.11[O 2 + 3.76N 2 ] ⎯
⎯→ 9.21CO 2 + 0.61CO + 7.06O 2 + 83.12N 2 + 11.05H 2 O
The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1.23,
C 8 H 18 + 18.0[O 2 + 3.76N 2 ] ⎯
⎯→ 7.50CO 2 + 0.50CO + 5.74O 2 + 67.58N 2 + 9H 2 O
(a) The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
mair
(18.0 × 4.76 kmol)(29 kg/kmol)
=
= 21.8 kg air/kg fuel
mfuel (8 kmol)(12 kg/kmol ) + (9 kmol)(2 kg/kmol)
(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is
determined from the theoretical combustion equation of the fuel,
C8 H18 + a th O 2 + 3.76N 2
O 2:
⎯
⎯→ 8CO 2 + 9H 2 O + 3.76a th N 2
a th = 8 + 4.5 ⎯
⎯→ a th = 12.5
Then,
Percent theoretical air =
m air,act
m air, th
=
N air,act
N air, th
=
(18.0)(4.76) kmol = 144%
(12.5)(4.76) kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-19
15-33 Carbon is burned with dry air. The volumetric analysis of the products is given. The AF ratio and the
percentage of theoretical air used are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, CO, O2, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis Considering 100 kmol of dry products, the combustion equation can be written as
xC + a O 2 + 3.76N 2
⎯⎯→ 10.06CO 2 + 0.42CO + 10.69O 2 + 78.83N 2
The unknown coefficients x and a are determined from mass balances,
N 2 : 3.76a = 78.83
⎯
⎯→ a = 20.965
C : x = 10.06 + 0.42
⎯
⎯→ x = 10.48
Carbon
Combustion Products
chamber
(Check O 2 : a = 10.06 + 0.21 + 10.69 ⎯
⎯→ 20.96 = 20.96)
Dry air
Thus,
10.48C + 20.96 O 2 + 3.76N 2
⎯⎯→ 10.06CO 2 + 0.42CO + 10.69O 2 + 78.83N 2
The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 10.48,
C + 2.0 O 2 + 3.76N 2
⎯⎯→ 0.96CO 2 + 0.04CO + 1.02O 2 + 7.52 N 2
(a) The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(2.0 × 4.76 kmol)(29 kg/kmol) = 23.0 kg air/kg fuel
=
(1 kmol)(12 kg/kmol)
m fuel
(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is
determined from the theoretical combustion equation of the fuel,
C + 1 O 2 + 3.76N 2
⎯⎯→ CO 2 + 3.76N 2
Then,
Percent theoretical air =
m air,act
m air, th
=
N air,act
N air, th
=
(2.0)(4.76) kmol = 200%
(1.0)(4.76) kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-20
15-34 Methane is burned with dry air. The volumetric analysis of the products is given. The AF ratio and
the percentage of theoretical air used are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, CO, H2O, O2, and N2
only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis Considering 100 kmol of dry products, the combustion equation can be written as
xCH 4 + a O 2 + 3.76N 2
⎯⎯→ 5.20CO 2 + 0.33CO + 11.24O 2 + 83.23N 2 + bH 2 O
The unknown coefficients x, a, and b are determined from mass balances,
N 2 : 3.76a = 83.23
C : x = 5.20 + 0.33
H : 4 x = 2b
⎯
⎯→ a = 22.14
CH4
⎯
⎯→ x = 5.53
Combustion Products
chamber
⎯
⎯→ b = 11.06
⎯→ 22.14 = 22.14)
(CheckO 2 : a = 5.20 + 0.165 + 11.24 + b / 2 ⎯
Dry air
Thus,
5.53CH 4 + 22.14 O 2 + 3.76N 2
⎯⎯→ 5.20CO 2 + 0.33CO + 11.24O 2 + 83.23N 2 + 11.06H 2 O
The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 5.53,
CH 4 + 4.0 O 2 + 3.76N 2
⎯
⎯→ 0.94CO 2 + 0.06CO + 2.03O 2 + 15.05N 2 + 2H 2O
(a) The air-fuel ratio is determined from its definition,
AF =
m air
(4.0 × 4.76 kmol)(29 kg/kmol)
=
= 34.5 kg air/kg fuel
m fuel (1 kmol)(12 kg/kmol) + (2 kmol)(2 kg/kmol)
(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is
determined from the theoretical combustion equation of the fuel,
CH 4 + a th O 2 + 3.76N 2
O 2:
⎯
⎯→ CO 2 + 2H 2O + 3.76a th N 2
a th = 1 + 1 ⎯
⎯→ a th = 2.0
Then,
Percent theoretical air =
m air,act
m air, th
=
N air,act
N air, th
=
(4.0)(4.76) kmol = 200%
(2.0)(4.76) kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-21
Enthalpy of Formation and Enthalpy of Combustion
15-35C For combustion processes the enthalpy of reaction is referred to as the enthalpy of combustion,
which represents the amount of heat released during a steady-flow combustion process.
15-36C Enthalpy of formation is the enthalpy of a substance due to its chemical composition. The
enthalpy of formation is related to elements or compounds whereas the enthalpy of combustion is related to
a particular fuel.
15-37C The heating value is called the higher heating value when the H2O in the products is in the liquid
form, and it is called the lower heating value when the H2O in the products is in the vapor form. The
heating value of a fuel is equal to the absolute value of the enthalpy of combustion of that fuel.
15-38C If the combustion of a fuel results in a single compound, the enthalpy of formation of that
compound is identical to the enthalpy of combustion of that fuel.
15-39C Yes.
15-40C No. The enthalpy of formation of N2 is simply assigned a value of zero at the standard reference
state for convenience.
15-41C 1 kmol of H2. This is evident from the observation that when chemical bonds of H2 are destroyed
to form H2O a large amount of energy is released.
15-42 The enthalpy of combustion of methane at a 25°C and 1 atm is to be determined using the data from
Table A-26 and to be compared to the value listed in Table A-27.
Assumptions The water in the products is in the liquid phase.
Analysis The stoichiometric equation for this reaction is
CH 4 + 2[O 2 + 3.76N 2 ] ⎯
⎯→ CO 2 + 2H 2 O(l ) + 7.52N 2
Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, N2 and O2
are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of CH4
becomes
hC = H P − H R =
∑N
o
P h f ,P
−
∑N
o
R h f ,R
( )
= Nh fo
CO 2
( )
+ Nh fo
H 2O
( )
− Nh fo
CH 4
Using h fo values from Table A-26,
hC = (1 kmol)(−393,520 kJ/kmol) + (2 kmol)(−285,830 kJ/kmol)
− (1 kmol)(− 74,850 kJ/kmol)
= −890,330 kJ (per kmol CH 4 )
The listed value in Table A-27 is -890,868 kJ/kmol, which is almost identical to the calculated value. Since
the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher
heating value of CH4.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-22
15-43 EES Problem 15-42 is reconsidered. The effect of temperature on the enthalpy of combustion is to
be studied.
Analysis The problem is solved using EES, and the solution is given below.
Fuel$ = 'Methane (CH4)'
T_comb =25 [C]
T_fuel = T_comb +273 "[K]"
T_air1 = T_comb +273 "[K]"
T_prod =T_comb +273 "[K]"
h_bar_comb_TableA27 = -890360 [kJ/kmol]
"For theoretical dry air, the complete combustion equation is"
"CH4 + A_th(O2+3.76 N2)=1 CO2+2 H2O + A_th (3.76) N2 "
A_th*2=1*2+2*1 "theoretical O balance"
"Apply First Law SSSF"
h_fuel_EES=enthalpy(CH4,T=298) "[kJ/kmol]"
h_fuel_TableA26=-74850 "[kJ/kmol]"
h_bar_fg_H2O=enthalpy(Steam_iapws,T=298,x=1)-enthalpy(Steam_iapws,T=298,x=0)
"[kJ/kmol]"
HR=h_fuel_EES+ A_th*enthalpy(O2,T=T_air1)+A_th*3.76 *enthalpy(N2,T=T_air1) "[kJ/kmol]"
HP=1*enthalpy(CO2,T=T_prod)+2*(enthalpy(H2O,T=T_prod)-h_bar_fg_H2O)+A_th*3.76*
enthalpy(N2,T=T_prod) "[kJ/kmol]"
h_bar_Comb_EES=(HP-HR) "[kJ/kmol]"
PercentError=ABS(h_bar_Comb_EESh_bar_comb_TableA27)/ABS(h_bar_comb_TableA27)*Convert(, %) "[%]"
hCombEES
[kJ/kmol]
TComb
[C]
-890335
-887336
-884186
-880908
-877508
-873985
-870339
-866568
-862675
-858661
25
88.89
152.8
216.7
280.6
344.4
408.3
472.2
536.1
600
-855000
-860000
]l
o
m
k/
J
k[
S
E
E,
b
m
o
C
h
-865000
-870000
-875000
-880000
-885000
-890000
-895000
0
100
200
300
400
500
Tcomb [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
600
15-23
15-44 The enthalpy of combustion of gaseous ethane at a 25°C and 1 atm is to be determined using the data
from Table A-26 and to be compared to the value listed in Table A-27.
Assumptions The water in the products is in the liquid phase.
Analysis The stoichiometric equation for this reaction is
C 2 H 6 + 3.5[O 2 + 3.76N 2 ] ⎯
⎯→ 2CO 2 + 3H 2 O(l ) + 13.16N 2
Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, N2 and O2
are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C2H6
becomes
hC = H P − H R =
∑N
o
P h f ,P
−
∑N
o
R h f ,R
(
= Nh fo
)
CO 2
(
+ Nh fo
)
H 2O
( )
− Nh fo
C2H6
Using h fo values from Table A-26,
hC = (2 kmol)(−393,520 kJ/kmol ) + (3 kmol)(−285,830 kJ/kmol )
− (1 kmol)(− 84,680 kJ/kmol )
= −1,559,850 kJ (per kmolC 2 H 6 )
The listed value in Table A-27 is -1,560,633 kJ/kmol, which is almost identical to the calculated value.
Since the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher
heating value of C2H6.
15-45 The enthalpy of combustion of liquid octane at a 25°C and 1 atm is to be determined using the data
from Table A-26 and to be compared to the value listed in Table A-27.
Assumptions The water in the products is in the liquid phase.
Analysis The stoichiometric equation for this reaction is
C 8 H 18 + 12.5[O 2 + 3.76N 2 ] ⎯
⎯→ 8CO 2 + 9H 2 O(l ) + 47N 2
Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, N2 and O2
are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C8H18
becomes
hC = H P − H R =
∑N
o
P h f ,P
−
∑N
o
R h f ,R
( )
= Nh fo
CO 2
( )
+ Nh fo
H 2O
( )
− Nh fo
C8 H18
Using h fo values from Table A-26,
hC = (8 kmol )(−393,520 kJ/kmol ) + (9 kmol )(−285,830 kJ/kmol )
− (1 kmol )(− 249,950 kJ/kmol )
= −5,470,680 kJ
The listed value in Table A-27 is -5,470,523 kJ/kmol, which is almost identical to the calculated value.
Since the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher
heating value of C8H18.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.