15 15-1
Chapter 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).
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
CH4 Products Air stoichiometric
O2 balance: Substituting,
a th = 1 + 1
a th = 2
CH 4 + 2[O 2 + 3.76N 2 ] CO 2 + 2H 2O + 7.52N 2 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) 1 1 = = 0.0578 kg fuel/kg air AF 17.3 kg air/kg fuel
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
The fuel-air ratio is the inverse of the air-fuel ratio,
FA =
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: Substituting,
175a th = 3 + 2 + 0.75a th . a th = 5
Products
C3H 8 + 8.75 O 2 + 3.76 N 2
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)
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
C2H2 Products 100% theoretical air
O2 balance: Substituting,
a th = 2 + 0.5
a th = 2.5
C 2 H 2 + 2.5[O 2 + 3.76N 2 ] 2CO 2 + H 2 O + 9.4N 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 (2.5 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
Products air
O2 balance: Substituting,
a = 2 + 1.5 + 3 a = 6.5
C 2 H 6 + 6.5[O 2 + 3.76N 2 ] 2CO 2 + 3H 2 O + 3O 2 + 24.44N 2
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)
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
(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 2CO 2 + 3H 2O + 3.76a th N 2 a th = 35 . = N air,act N air, th = a 6.5 = = 186% a th 3.5
O2 balance: Then,
a th = 2 + 15 .
m air,act m air, th
Percent theoretical air =
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,
Nv 2 lbmol P (14.5 psia ) = 0.981 psia = Pv = N prod prod 29.56 lbmol Tdp = Tsat @ 0.981 psia = 101F
Thus,
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:
15a th = 3 + 15 + 0.5a th . . a th = 4.5
Substituting, C 3 H 6 + 6.75 O 2 + 3.76N 2
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,
Nv 3 kmol P (105 kPa ) = 9.367 kPa Pv = = N prod prod 33.63 kmol Tdp = Tsat @9.367 kPa = 44.5C
Thus,
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: Hydrogen balance: Oxygen balance: B=3
2D = 8 D = 4 1 + 2 1.5a th = 2 B + D + 2 E 0.5a th = E 1.5a th 3.76 = F
C3H7OH Products Air 50% eccess
Nitrogen balance:
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
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: Hydrogen balance: Oxygen balance: B=4
2 D = 10 D = 5 2 2a th = 2 B + D + 2 E a th = E 2a th 3.76 = F
C4H10 Products Air 200% theoretical
Nitrogen balance:
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@60C = 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
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: Hydrogen balance: Oxygen balance: Nitrogen balance:
x = 0.4182 y = 0.5818 ath = 2.582 x + 2y = B 4 x + 6 y = 2D 2ath + y = 2 B + D 3.76a th = F B = 1.582 D = 2.582 F = 9.708
20% CH4 80% C2H6O Air 100% theoretical Products
Solving the above equations, we find the coefficients as
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
& & mair = AFmfuel = (10.64)(31 kg/s) = 330 kg/s
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: Substituting,
2.5a th = 8 + 4.5 + 1.5a th a th = 12.5
C8H18 Air 25C Combustion Products chamber P = 1 atm
C 8 H 18 + 31.25[O 2 + 3.76N 2 ] 8CO 2 + 9H 2 O + 18.75O 2 + 117.5N 2 AF =
Thus,
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 Pv = v Pprod = 153.25 kmol (101.325 kPa ) = 5.951 kPa N prod
Thus,
Tdp = Tsat @5.951 kPa = 36.0C
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 a th = 8 + 4.5 8CO 2 + 9H 2 O + 3.76a th N 2 a th = 12.5
Gasoline (C8H18) Air Jet engine Products
where ath is the stoichiometric coefficient for air. It is determined from O2 balance:
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)
AFact AFth = 18 kg air/kg fuel = 119% 15.14 kg air/kg fuel
Then the percent theoretical air used can be determined from
Percent theoretical air =
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 Air
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
=
(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 =
& mair 176 kg/h = = 22 kgair/kgfuel & mfuel 8 kg/h
AFact AFth
Then the percent theoretical air used can be determined from
Percent theoretical air = = 22 kg air/kg fuel = 137% 16.1 kg air/kg fuel
Thus the excess air used during this process is 37%.
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 O2 balance:
a th = 4 + 2.5
C4H10 Combustion Products chamber Air
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 = m air,th m fuel =
(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 AFact AFth
Then the percent theoretical air used can be determined from
Percent theoretical air = = 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
kPa
Thus,
Tdp = Tsat @8.571 = 42.8C
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 O2 balance:
a th = 4 + 2.5 a th = 6.5
C4H10 Combustion Products chamber 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 =
(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 AFact AFth
Then the percent theoretical air used can be determined from
Percent theoretical air = = 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 lbmol 52.5
Thus,
Tdp = Tsat @1.4 psia = 113.2F
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 y = 1.38 a th = 1.31 z = 5.106 H : 0.65 4 + 0.08 2 = 2 y N 2 : 0.18 + 3.76a th = z
Natural gas Combustion Products chamber Dry air
O 2 : 0.03 + 0.06 + a th = x + y / 2
Thus,
(0.65CH 4 + 0.08H 2 + 018N 2 + 0.03O 2 + 0.06CO 2 ) + 131(O 2 + 3.76N 2 ) . . 0.71CO 2 + 138H 2 O + 5106N 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 fuel = (0.65 16 + 0.08 2 + 0.18 28 + 0.03 32 + 0.06 44 )kg = 19.2 kg m air = (1.31 4.76 kmol)(29 kg/kmol) = 180.8 kg
and
AFth = m air,th m fuel = 180.8 kg = 9.42 kg air/kg fuel 19.2 kg
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 y = 1.38 a th = 1.31 Moist air z = 5.106 Natural gas Combustion Products chamber
H : 0.65 4 + 0.08 2 = 2 y N 2 : 0.18 + 3.76a th = z Thus,
O 2 : 0.03 + 0.06 + a th = x + y / 2
(0.65CH 4 + 0.08H 2 + 018N 2 + 0.03O 2 + 0.06CO 2 ) + 1.31(O 2 + 3.76N 2 ) . 0.71CO 2 + 138H 2 O + 5106N 2 . .
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 = air Psat@ 25C = (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 = P total 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 + 018N 2 + 0.03O 2 + 0.06CO 2 ) + 131(O 2 + 3.76N 2 ) + 017 H 2 O . . . 0.71CO 2 + 155H 2 O + 5106N 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 fuel = (0.65 16 + 0.08 2 + 0.18 28 + 0.03 32 + 0.06 44 )kg = 19.2 kg m air = (1.31 4.76 kmol )(29 kg/kmol ) + (0.17 kmol 18 kg/kmol ) = 183.9 kg
and
AFth = m air, th m fuel = 183.9 kg = 9.58 kg air/kg fuel 19.2 kg
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 y = 1.50 a th = 1.35 z = 6.70 Gaseous fuel Air 30% excess Combustion Products chamber 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 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 fuel = (0.6 16 + 0.3 2 + 0.1 28)kg = 13.0 kg m air = (1.755 4.76 kmol )(29 kg/kmol ) = 242.3 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 20C, some of the water vapor will condense as the products are cooled to 20C. 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 @ 20C = 2.3392 kPa. Thus the fraction of water vapor that condenses is 1.32/1.5 = 0.88 or 88%.
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"
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 1.433 1.368 1.251 1.047 0.8947 0.6947 0.4312 0.2691 0.08195 0.007615 0.002518 0 0
MolesH2O,vap 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 1
Vapor Liquid
s e l o M
O 2 H
0.8 0.6 0.4 0.2 0 0 Dew Point = 55.96 C
10
20
30
40
50
60
70
80
90
Tprod [C]
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 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 N C = (m / M )C = 0.82 / 12 = 0.0683 kmol
Coal Air 50% excess Combustion Products chamber
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 y = 0.0128 H : 0.0028 2 + 0.01 2 = 2 y
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
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: 8 x = 9.21 + 0.61 H : 18 x = 2b a = 22.11 x = 1.23 b = 11.07 Dry air C8H18 Combustion Products chamber
22.11 22.10) (CheckO 2 : a = 9.21 + 0.305 + 7.06 + b / 2 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
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 C : x = 10.06 + 0.42 a = 20.965 x = 10.48
Carbon Combustion Products chamber Dry air
(Check O 2 : a = 10.06 + 0.21 + 10.69 20.96 = 20.96)
Thus,
10.48C + 20.96 O 2 + 3.76N 2 C + 2.0 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,
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
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 x = 5.53 CH4 Combustion Products chamber Dry air
b = 11.06
22.14 = 22.14) (CheckO 2 : a = 5.20 + 0.165 + 11.24 + b / 2 Thus,
5.53CH 4 + 22.14 O 2 + 3.76N 2 CH 4 + 4.0 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,
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
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 25C 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 25C 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) = -890,330 kJ (per kmol CH 4 ) - (1 kmol)(- 74,850 kJ/kmol)
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.
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] -890335 -887336 -884186 -880908 -877508 -873985 -870339 -866568 -862675 -858661
TComb [C] 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
-865000 -870000 -875000 -880000 -885000 -890000 -895000 0 100 200 300 400 500 600
h
Tcomb [C]
15-23
15-44 The enthalpy of combustion of gaseous ethane at a 25C 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 25C 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 ) = -1,559,850 kJ (per kmolC 2 H 6 ) - (1 kmol)(- 84,680 kJ/kmol )
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 25C 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 25C 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.