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 presence affect affectss the outcome outcome of the process process because because nitrogen nitrogen absorbs a large large proportion proportion of the heat heat released released during the chemical process. 15-3C Moisture, Moisture, in general, general, does not react react chemica chemically lly with any of the species species present present in the combustio combustion n 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-poin dew-pointt temperature temperature of the product product gases is the temperat temperature ure at which the water water vapor 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 are preserved preserved during a chemical chemical reaction, but the total mole numbers 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 ratio is the inverse inverse of the air-fue air-fuell ratio. ratio.
Because the the molar molar mass mass of the the fuel and and the molar molar mass mass of the the air, in general, general, are different different.. 15-7C No. Because
Theoretical and Actual Combustion Combustion Processes Processes 15-8C The causes of incomplete combustion are insufficient insufficient time, insufficient oxygen, insufficient insufficient mixing, and dissociation. 15-9C CO. Because oxygen is more strongly attracted to hydrogen than it is to carbon, carbon, and hydrogen hydrogen is usually burned to completion even when there is a deficiency 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, complete, but the products of theoretical combustion does not contain any uncombined uncombined oxygen. 15-12C Case (b).
15-2
15-13 Methane Methane is burned burned with with the stoichiom stoichiometric etric amount amount of air during during a combustio combustion n process. process. The The AF and and FA ratios ratios are to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This is a theoretica theoreticall combustio combustion n process process since since methane is burned completely with stoichiometric amount of air. The stoichiometric stoichiometric combustion combustion equation of CH4 is
CH 4 a th O 2 3.76N 2 CO 2 2H 2 O 3.76a th N 2 a th 1 1
O2 balance: Substituting,
a th 2
CH4 Products Air stoichiometric
CH 4 2O 2 3.76N2 CO 2 2H 2O 7.52N 2
The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair
mfuel
2 4.76 kmol29 kg/kmol 17.3 1 kmol12 kg/kmol 2 kmol2 kg/kmol
kg air/kg fuel
The fuel-air ratio is the inverse of the air-fuel ratio, ratio, FA
1 AF
1 17.3 kg air/kg fuel
0.0578 kg fuel/kg air
15-14 Propane is burned with 75 percent excess air during a combustion combustion process. The AF ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, O2, and N2 only.
The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively Properties The (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 stoi stoichi chiom ometr etric ic coeffi coefficie cient nt for for air. air. We have have automatically accounted for the 75% excess air by using the C3H8 factor 1.75a 1.75ath instead of a of ath for air. The stoichiometric amount of oxygen (a (athO2) will will be used to oxidize oxidize the the fuel, and the the remaining remaining Air excess amount (0.75a (0.75athO2) will appear appear in in the products products as free free oxygen. The coefficient ath is determined determined from the the O2 balance, 75% excess O2 balance:
1.75a th 3 2 0.75ath
Substituting,
C3 H8 8.75 O2 3.76 N2
a th 5
3CO2 4 H2 O 3.75O2 32.9 N2
The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
8.75 4.76 kmol29 kg/kmol 27.5 kg air/kg fuel 3 kmol12 kg/kmol 4 kmol 2 kg/kmol
Products
15-3
15-15 Acetylene Acetylene is burne burned d with the the stoichiom stoichiometric etric amount amount of air during during a combustion combustion process. process. The AF ratio is to be deter determin mined ed on a mass mass and and on a mole mole basi basis. s. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This This is a theo theore reti tica call comb combus usti tion on proc proces esss sinc sincee C2H2 is burned completely with stoichiometric amount amount of air. The stoichiome stoichiometric tric combustion combustion equation equation of C2H2 is
2CO 2 H 2 O 3.76a th N 2 C 2 H 2 a th O 2 3.76N 2 O2 balance:
a th 2 0.5
C2H2 Products
ath 2.5 100% theoretical air
Substituting,
2CO 2 H 2 O 9.4N 2 C 2 H 2 2.5O 2 3.76N 2 The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
2.5 4.76 kmol29 kg/kmol 13.3 2 kmol12 kg/kmol 1 kmol2 kg/kmol
kg air/kg fuel
On a mole basis, the air-fuel air-fuel ratio is expressed as the ratio of the mole mole numbers of the air to the mole numbers numbers of the fuel, fuel, AFmole basis
N air N fuel
(2.5 4.76) kmol 1 kmol fuel
11.9 kmol air/kmol fuel
15-16 Ethane is burned with an unknown amount amount of air during 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 combustion products contain contain CO2, H2O, O2, and N2 only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combu combustion stion equation equation in this this case case can be written written as
C2H6 Products
C 2 H 6 aO 2 3.76N 2 2CO 2 3H 2 O 3O 2 3.76a N 2
a 6.5 a 2 1.5 3
O2 balance: Substituting,
air
C 2 H 6 6.5O 2 3.76N 2 2CO 2 3H 2 O 3O 2 24.44N 2
The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
6.5 4.76 kmol29 kg/kmol 29.9 kg air/kg fuel 2 kmol12 kg/kmol 3 kmol2 kg/kmol
(b) To find find the percent percent theore theoretical tical air air used, used, we need need to know the theoreti theoretical cal amount amount of air, air, which which is determine determined d from the theoretic theoretical al combustio combustion n equation equation of C2H6, C2 H 6 a th O 2 3.76N 2 O2 balance: Then,
a th 2 1.5
2CO2 3H2 O 3.76ath N2
Percent theoretical air
mair,act mair,th
a th 3.5
N air,act N air,th
a a th
6.5 3.5
186%
15-4
15-17E Ethylene is burned burned with 200 percent theoretical air during during a combustion process. The AF ratio and the dew-poi dew-point nt temperat temperature ure of the the products products are are to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H 2O, O2, and N2 only. 3 Combustion Combustion gases gases are ideal gases. gases. The molar olar masse assess of of C, C, H2, and air are 12 12 lbm/lbmol, lbm/lbmol, 2 lbm/lbmol, lbm/lbmol, and and 29 lbm/lbmol, lbm/lbmol, Properties The respective respectively ly (Table (Table A-1E). A-1E). Analysis (a) The combus combustion tion equation equation in this this case case can be written written as C2H4 C2 H 4 2a th O 2 3.76N 2 2CO2 2H2 O ath O2 (2 3.76)ath N2 Products where ath is the stoichiometric stoichiometric coefficient coefficient for air. 200% It is determin determined ed from theoretical air O2 balance: 2a th 2 1 a th ath 3
C2 H 4 6 O2 3.76 N2
Substituting,
2 CO2 2 H2 O 3O2 22.56N2
The air-fue air-fuell ratio is determ determined ined by taking the the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
6 4.76 lbmol29 lbm/lbmol 29.6 lbm air/lbm fuel 2 lbmol12 lbm/lbmol 2 lbmol2 lbm/lbmol
(b) The dew-point dew-point tempe temperatur raturee of a gas-va gas-vapor por mixtur mixturee is the saturat saturation ion temperat temperature ure of the water water vapor vapor in the product gases corresponding corresponding to its partial pressure. pressure. That That is,
N v 2 lbmol P prod 14.5 psia 0.981 psia 29.56 lbmol N prod T dp T sat @ 0.981 psia 101 F
P v
Thus,
15-18 Propylene is burned with 50 percent excess air during a combustion combustion process. The AF ratio 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, H 2O, O2, and N2 only. 3 Combustion Combustion gases gases are ideal gases. gases. Properties The molar molar masses masses of C, H2, and air are are 12 kg/kmol, kg/kmol, 2 kg/kmol, kg/kmol, and 29 kg/kmol, kg/kmol, respectively respectively (Table (Table A-1). Analysis (a) The combus combustion tion equation equation in this case case can can be written as
C3H6
Products
50% excess air
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 stoichiometric coefficient coefficient for air. It is determined determined from from O2 balance:
1.5a th 3 1.5 0.5a th
Substituting, C3 H 6 6.75 O2 3.76 N 2
ath 4.5
3CO2 3H2 O 2.25O2 25.38N2
The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
6.75 4.76 kmol29 kg/kmol 22.2 3 kmol12 kg/kmol 3 kmol2 kg/kmol
kg air/kg fuel
(b) The dew-point dew-point tempe temperatur raturee of a gas-va gas-vapor por mixtur mixturee is the saturat saturation ion temperat temperature ure of the water water vapor vapor in the product gases corresponding corresponding to its partial pressure. pressure. That That is,
N v 3 kmol P 105 kPa 9.367 kPa prod N prod 33.63 kmol T dp T sat @9.367 kPa 44.5C
P v Thus,
15-5
15-19 Propal Propal alcohol alcohol C3H7OH C3H7OH is burned burned with 50 percent percent excess excess air. air. The balanced balanced reaction reaction equatio equation n for complete complete combustion combustion is to to be written written and the air-to-f air-to-fuel uel ratio is to be determined determined.. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, O2, and N2 only. Properties The The mola molarr mass masses es of C, H2, O2 and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respective respectively ly (Table (Table A-1). A-1). Analysis The combustion equation in this case can be written as
B CO 2 D H 2 O E O 2 F N 2 C 3 H 7 OH 1.5a th O 2 3.76N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 50% excess air by usin using g the factor factor 1.5 1.5ath instead of ath for air. The coefficient ath and other other coeff coefficien icients ts are are to be determined from the mass mass balances Carbon balance:
B=3
Hydrogen balance:
D 4 2 D 8
C3H7OH Products Air
1 2 1.5a th 2 B D 2 E
Oxygen balance:
0.5a th E
50% eccess
1.5a th 3.76 F
Nitrogen balance:
Solving the above equations, we find the coefficients E ( E = 2.25, F = 25.38, and ath = 4.5) and write the balanced balanced reaction reaction equation equation as C 3 H 7 OH 6.75O 2 3.76N 2 3 CO 2 4 H 2 O 2.25 O 2 25.38 N 2 The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
(6.75 4.75 kmol)(29 kg/kmol) (3 12 8 1 1 16) kg
15.51 kg air/kg fuel
15-6
15-20 Buta Butane ne C4H10 is burned with 200 percent theoretical air. The kmol of water that needs needs to be sprayed into the combustio combustion n chamber chamber per kmol kmol of fuel is to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, O2, and N2 only. Properties The The mola molarr mass masses es of C, H2, O2 and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respective respectively ly (Table (Table A-1). A-1). Analysis The reaction reaction equation equation for 200% theoret theoretical ical air air without without the additio additional nal water is is
B CO 2 D H 2 O E O 2 F N 2 C 4 H 10 2a th O 2 3.76N 2 where ath is the stoichiometric stoichiometric coefficient for air. We have automatically accounted for the 100% excess air by using the factor 2a 2ath instead of a of ath for air. The coefficient ath and other other coefficient coefficientss are to be determ determined ined from the the mass balances balances Carbon balance:
B=4
Hydrogen balance:
D 5 2 D 10
Oxygen balance:
2 2a th 2 B D 2 E
C4H10 Products Air 200% theoretical
a th E 2a th 3.76 F
Nitrogen balance:
Solving the above equations, we find the coefficients E (E = 6.5, F = 48.88, and ath = 6.5) and write the balanced balanced reaction reaction equation equation as C 4 H 10 13O 2 3.76N 2 4 CO 2 5 H 2 O 6.5 O 2 48.88 N 2 With the additional water sprayed sprayed into the combustion combustion chamber, the balanced reaction equation is
4 CO 2 (5 N v ) H 2 O 6.5 O 2 48.88 N 2 C 4 H 10 13O 2 3.76N 2 N v H 2 O The partial partial pressure pressure of water water in the saturated saturated product product mixture mixture at the dew point is P v , prod P sat@60C 19.95 kPa The vapor mole mole fraction fraction is y v
P v , prod P prod
19.95 kPa 100 kPa
0.1995
The amount of water that needs to be sprayed into the combustion combustion chamber can be determined from from y v
N water N total, product
0.1995
5 N v 4 5 N v 6.5 48.88
N v 9.796 kmol
15-7
15-21 A fuel mixtur mixturee of 20% by by mass mass methane, methane, CH4, and 80% by mass ethanol, C2H6O, is burned completely with theoretical air. The The required flow rate rate of air is to be determined. determined. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The The mola molarr mass masses es of C, H2, O2 and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respective respectively ly (Table (Table A-1). A-1). Analysis The combustion equation in this case can be written as
B CO 2 D H 2 O F N 2 x CH 4 y C 2 H 6 O a th O 2 3.76N 2 where ath is the stoichiometric coefficient for air. The coefficient ath and other other coeff coefficient icientss are to be determined from the mass mass balances Carbon balance:
x 2 y B
Hydrogen balance:
4 x 6 y 2 D
Oxygen balance:
2ath y 2 B D
Nitrogen Nitrogen balance: balance:
3.76a th F
20% CH4 80% C2H6O Products
Air 100% theoretical
Solving Solving the above equations, equations, we find the coefficie coefficients nts as x 0.4182
B 1.582
y 0.5818
D 2.582
ath 2.582
F 9.708
Then, we write write the balanced balanced reaction reaction equation 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-fue air-fuell ratio is determ determined ined by taking the the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel (2.582 4.76 kmol)(29 kg/kmol) (0.4182 kmol)(12 4 1) kg/kmol (0.5818 kmol)( 2 12 6 1 16)kg/kmol
Then, the the required required flow rate of air becomes becomes air AFm fuel (10.64)(31 kg/s) 330 kg/s m
10.64 kg air/kg fuel
15-8
15-22 Octane is is burned burned with 250 percent percent theoretical theoretical air air during during a combustio combustion n process. process. The The AF ratio ratio and the dew-pint dew-pint temperat temperature ure of the products products are to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H 2O, O2, and N2 only. 3 Combustion Combustion gases gases are ideal gases. gases. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combu combustion stion equation equation in this this case case can be written written as
C8 H18 2.5a th O 2 3.76N 2
8CO2 9H2 O 1.5ath O2 (2.5 3.76)ath N2
where ath is the stoichiometric stoichiometric coefficient coefficient for air. It is determin determined ed from O2 balance:
C8H18
2.5a th 8 4.5 1.5a th a th 12.5
Air
Substituting, C 8 H 18 31.25O 2 3.76N 2 8CO 2 9H 2 O 18.75O 2 117.5N 2 Thus,
AF
mair mfuel
25C
31.25 4.76 kmol29 kg/kmol 37.8 8 kmol12 kg/kmol 9 kmol2 kg/kmol
Combustion chamber P = 1 atm
Products
kg air/kg fuel
(b) The dew-point dew-point tempe temperatur raturee of a gas-va gas-vapor por mixtur mixturee is the saturat saturation ion temperat temperature ure of the water water vapor vapor in the product gases corresponding corresponding to its partial pressure. pressure. That That is,
N v 9 kmol P P v prod 153.25 kmol 101.325 kPa 5.951 kPa N prod Thus,
T dp T sat @5.951 kPa 36.0C
15-23 Gasoline Gasoline is burned burned steadil steadily y with air in a jet jet engine. engine. The AF ratio ratio is given. given. The percentage percentage of excess excess air used is to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The The mola molarr mass masses es of 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
Gasoline
8CO2 9H2 O 3.76ath N2
where ath is the stoichiometric stoichiometric coefficient coefficient for air. It is determined determined from from O2 balance:
a th 8 4.5
a th 12.5
(C8H18)
Jet engine
Products
Air
The air-fuel ratio for the theoretical reaction is determined 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 kmol29 kg/kmol 15.14 kg air /kg fuel 8 kmol12 kg/kmol 9 kmol2 kg/kmol
Then the percent theoretical theoretical air air used can be determined from Percent theoretical air
AFact AFth
18 kg air/kg fuel 15.14 kg air/kg fuel
119%
15-9
15-24 Ethane is burned with air steadily. The mass flow rates rates of ethane and air are given. The The percentage of excess excess air used is to be determined determined.. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The theoretical theoretical combustion combustion equation in this case can be written as
C2 H 6 a th O 2 3.76N 2
2CO2 3H2 O 3.76ath N2
C2H6
where ath is the stoichiometric stoichiometric coefficient coefficient for air. It is determined determined from from O2 balance:
a th 2 1.5
a th 3.5
The air-fue air-fuell ratio for for the theoretic theoretical al reaction reaction is determine determined d by taking the the ratio of of the mass mass of the air to to the mass mass of the the fuel for, AFth
m air,th m fuel
air m fuel m
Air
3.5 4.76 kmol29 kg/kmol 16.1 kg air /kg fuel 2 kmol12 kg/kmol 3 kmol2 kg/kmol
The actual actual air-fuel air-fuel ratio ratio used used is is AFact
Combustion chamber
176 kg/h 8 kg/h
22 kgair /kgfuel
Then the percent theoretical theoretical air air used can be determined from Percent theoretical air
AFact AFth
22 kg air/kg fuel 16.1 kg air/kg fuel
Thus the exces excesss air used during this this process process is 37%.
137%
Products
15-10
15-25 Butane is burned with air. The masses masses of butane and air are given. The percentage percentage of theoretical theoretical air used and the dew-point dew-point temperat temperature ure of the products products are to be determine determined. d.
combust ustion ion produc products ts contai contain n CO2, H2O, an and N2 only. 3 Assumptions 1 Combustion is complete. 2 The comb Combustion Combustion gases gases are ideal gases. gases. Properties The The mola molarr mass masses es of 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
C4 H10 a th O 2 3.76N 2
4CO2 5H2 O 3.76ath N2
where ath is the stoichiometric coefficient for air. It is determined from a th 4 2.5
O2 balance:
C4H10 Combustion chamber
ath 6.5
The air-fuel ratio for the theoretical reaction is determined by taking the the ratio of of the mass mass of the air to to the mass mass of the the fuel for, AFth
m air,th m fuel
Products
Air
6.5 4.76 kmol29 kg/kmol 15.5 kg air /kg fuel 4 kmol12 kg/kmol 5 kmol 2 kg/kmol
The actual actual air-fuel air-fuel ratio ratio used used is is AFact
mair mfuel
25kg 1 kg
25 kg air / kg fuel
Then the percent theoretical theoretical air air used can be determined from Percent theoretical air
AFact AFth
25 kg air/kg fuel 15.5 kg air/kg fuel
161%
(b) The The combusti combustion on is complete, complete, and thus thus products products will contain contain only CO CO2, H2O, O2 and N2. The air-fuel ratio for for this combustion combustion process process on a mole basis basis is AF
N air N fuel
m air / M air m fuel / M fuel
25 kg / 29 kg/kmol 50 kmol air /kmol fuel 1 kg / 58 kg/kmol
Thus the combusti combustion on equation equation in this case can be written written as C 4 H10 50/4.76O 2 3.76N 2 4CO 2 5H 2 O 4.0O 2 39.5N 2 The dew-point dew-point temperat temperature ure of a gas-vapor gas-vapor mixture mixture is the saturation saturation temperat temperature ure of the water vapor vapor in the product product gases correspon corresponding ding to its partial partial pressur pressure. e. That is, is,
N v 5 kmol P prod 90 kPa 8.571 kPa N prod 52.5 kmol
P v Thus,
T dp T sat @8.571
kPa
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 dew-point temperat temperature ure of the products products are to be determine determined. d.
combust ustion ion produc products ts contai contain n CO2, H2O, an and N2 only. 3 Assumptions 1 Combustion is complete. 2 The comb Combustion Combustion gases gases are ideal gases. gases. Properties The The molar olar mass masses es of C, H2, and air are 12 12 lbm/lbmol, lbm/lbmol, 2 lbm/lbmol, lbm/lbmol, and and 29 lbm/lbmol, lbm/lbmol, respective respectively ly (Table (Table A-1). 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
4CO2 5H 2 O 3.76ath N2
where ath is the stoichiometric coefficient for air. It is determined from a th 4 2.5
O2 balance:
C4H10 Combustion chamber
ath 6.5
The air-fue air-fuell ratio for for the theoretic theoretical al reaction reaction is determine determined d by taking the the ratio of of the mass mass of the air to to the mass mass of the the fuel for, AFth
m air,th mfuel
Products
Air
6.5 4.76 lbmol29 lbm/lbmol 15.5 lbm air /lbm fuel 4 lbmol12 lbm/lbmol 5 lbmol2 lbm/lbmol
The actual actual air-fuel air-fuel ratio ratio used used is is AFact
m air mfuel
25 lbm 1 lbm
25 lbm air /lbm fuel
Then the percent theoretical theoretical air air used can be determined from Percent theoretical air
AFact AFth
25 lbm air/lbm fuel 15.5 lbm air/lbm fuel
161%
(b) The combusti combustion on is complete, complete, and and thus products products will will contain contain only CO2, H2O, O2 and N2. The air-fuel ratio for for this combustion combustion process process on a mole basis basis is AF
N air N fuel
m air / M air m fuel / M fuel
25 lbm/ 29 lbm/lbmol 50 lbmol air /lbmol fuel 1 lbm/ 58 lbm/lbmol
Thus the combusti combustion on equation equation in this case can be written written as
4CO 2 5H 2 O 4O 2 39.5N 2 C 4 H 10 50/4.76O 2 3.76N 2 The dew-point dew-point temperat temperature ure of a gas-vapor gas-vapor mixture mixture is the saturation saturation temperat temperature ure of the water vapor vapor in the product product gases correspon corresponding ding to its partial partial pressur pressure. e. That is, is,
N v 5 lbmol P prod 14.7 psia 1.4 psia N prod 52.5 lbmol
P v Thus,
T dp T sat @1.4 psia 113.2F
15-12
15-27 The volumetric fractions fractions of the constituents of a certain natural gas are given. The AF ratio ratio is to be determine determined d if this gas is burned burned with the stoichiom stoichiometric etric amount amount of dry air. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The molar molar masses 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, kg/kmol, respective respectively ly (Table (Table A-1). Analysis Considering 1 kmol of fuel, the combustion combustion equation can 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 z N 2 The unknown coefficients in the above equation equation are determined from mass balances, balances, C : 0.65 0.06 x
x 0.71
H : 0.65 4 0.08 2 2 y
Natural gas
y 1.38
Combustion chamber
a th 1.31 O 2 : 0.03 0.06 a th x y / 2 N 2 : 0.18 3.76a th z
z 5.106
Products
Dry air
Thus, (0.65CH 4 0.08H 2 0.18N 2 0.03O2 0.06CO2 ) 1. 31( O2 3. 76N 2 )
0.71CO 2
1 .38H 2 O
5.106N 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, mair 1.31 4.76 kmol 29 kg/kmol 180.8 kg mfuel 0.65 16 0.08 2 0.18 28 0.03 32 0.06 44kg 19.2 kg and AFth
m air,th mfuel
180.8 kg 19.2 kg
9.42 kg air/kg fuel
15-13
15-28 The compos composition ition of a certain certain natural natural gas gas is given. given. The gas is burned burned with stoichio stoichiometr metric ic amount amount of moist moist air. air. The AF ratio is to to be determ determined. ined. Assumptions 1 Combustion is complete. 2 The combustion combustion products contain contain CO2, H2O, and N2 only. Properties The molar molar masses 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, kg/kmol, respective respectively ly (Table (Table A-1). Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will contai contain n only only H2O, CO2 and N2, but no free O2. The moisture in the air does not react react with anything; it simply shows up as additional H2O in the products. products. Therefore, we can simply simply balance the combustion combustion equation equation using dry air, air, and then then add the moistur moisturee to both sides sides of the the equation. equation. Consideri Considering ng 1 kmol kmol of fuel, the combustion equation can be written as as
(0.65CH 4 0.08H2 0.18N2 0.03O2 0.06CO2 ) a th (O2 3.76N2 ) x CO2 y H2 O z N2 The unknown coefficients in the above equation equation are determined from mass balances, balances, C : 0.65 0.06 x
x 0.71
H : 0.65 4 0.08 2 2 y
Natural gas
y 1.38
Combustion chamber
O 2 : 0.03 0.06 a th x y / 2 a th 1.31 N 2 : 0.18 3.76a th z
z 5.106
Products
Moist air
Thus, (0.65CH 4 0.08 H 2 0.18 N 2 0.03O 2 0.06CO 2 ) 1.31( O 2 3.76 N 2 )
0.71CO 2
1.38 H 2 O 5.106 N 2
Next we determine determine the the amount amount of moistur moisturee that accomp accompanies anies 4.76 4.76a ath = (4.76)(1.31) = 6.24 kmol of dry air. The partial pressure pressure of the moisture in the air is
P v,in air P sat@25C (0.85)(3.1698 kPa) 2.694 kPa Assuming Assuming ideal gas behavio behavior, r, the number number of moles moles of the moisture moisture in the air (Nv, in) is determined to be
P v,in 2.694 kPa N v,air 0.17 kmol P N total 101.325 kPa 6.24 N v,in total
N v ,in
The balanced balanced combusti combustion on equation equation is obtained obtained by substituting substituting the coefficient coefficientss determined determined earlier earlier and adding adding 0.17 kmol of of H2O to both sides of the the equation equation,, (0.65CH 4 0.08 H 2 0.18 N 2 0.03O 2 0.06CO 2 ) 1.31( O 2 3.76 N 2 ) 0.17 H 2 O
0.71CO 2
1.55H 2 O 5 .106N 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 1.31 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 19.2kg
9.58 kg air/kg fuel
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 fraction of water vapor vapor that would would condense condense if the product product gases were were cooled are to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion products contain contain CO2, H2O, O2, and N2 only.
molar masse massess of C, C, H2, N2, and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, kg/kmol, 28 kg/kmol, kg/kmol, and 29 kg/kmol, kg/kmol, Properties The molar respectivel respectively y (Table (Table A-1). Analysis (a) The fuel is burned burned completely with excess air, air, and thus the products will contain H2O, CO2, N2, and some free O 2. Considering Considering 1 kmol of fuel, fuel, the combustion combustion equation equation can can be written written as
x CO2 y H2 O 0.3ath O2 z N2 (0.60CH 4 0.30H2 0.10N2 ) 1.3a th (O2 3.76N2 ) The unknown coefficients coefficients in the above equation are determined determined from mass balances, C : 0.60 x
x 0.60
H : 0.60 4 0.30 2 2 y
y 1.50
O 2 : 1.3a th x y / 2 0.3a th N 2 : 0.10 3.76 1.3a th z
a th 1.35 z 6.70
Gaseous fuel Air
Combustion chamber
Products
30% excess
Thus,
0.6CO2 1.5H2 O 0.405O2 6.7N2 (0.60CH 4 0.30H2 0.10N2 ) 1.755(O2 3.76N2 ) 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 28kg 13.0 kg
and AF
mair mfuel
242.3 kg 13.0 kg
18.6 kg air/kg fuel
(b) For each kmol kmol of fuel burned, burned, 0.6 + 1.5 + 0.405 + 6.7 = 9.205 kmol kmol of products are formed, formed, including 1.5 1.5 kmo kmoll of H2O. Assuming Assuming that that the the dew-poin dew-pointt tempera temperature ture of the the products products is is above above 20C, some of the water vapor vapor will will condense condense as the products products are are cooled cooled to 20C. If N If N w kmol of H2O condenses, there will be 1.5 - N w kmol of water vapor left in the products. products. The mole number number of the products products in in the gas phase will also decrease to 9.205 - N - N w as a result. result. Treating the product product gases (including the remaining water vapor) as ideal gases, N w is determined by equating the mole mole fraction of the water vapor to to its pressure fraction, fraction, N v N prod,gas
P v P prod
1.5 N w 9.205 N w
2.3392 kPa 101.325 kPa
N w 1.32 kmol
since P v = P sat sat @ 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 Problem 15-29 15-29 is reconsi reconsidered dered.. The effects effects of varying varying the percent percentages ages of CH4, H2 and N2 making making up the fuel and and the product product gas temperat temperature ure are are to be studied. studied. Analysis The problem is solved solved using EES, and the solution is given 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 ((a*y/2)+b) )+b) H2O + (c+3.76 (a*y/4 + a*x+b/2 a*x+b/2)) (Theo_ (Theo_air/100 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$) P_v = Moles_H2O/(M_other+Moles_H2O)*P_prod T_DewPoint = temperature(steam,P=P_v,x=0) IF T_DewPoint T_DewPoint <= T_prod T_prod then Moles_H2O_vap = Moles_H2O Moles_H2O_liq=0 Result$='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: "React ion: aCxHy+ aCxHy+bH2+cN2 bH2+cN2 + A_th Theo_ai Theo_air/100 r/100 (O2 + 3.76 N2) <--> a*xCO2 + (a*y/2+ (a*y/2+b) b) H2O + (c+3.7 (c+3.76 6 A_th Theo_ai Theo_air/100) r/100) N2 + A_th (Theo_a (Theo_air/100 ir/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
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
O 2 H
1.2
Vapor
1
Liquid
0.8 0.6 0.4
Dew Point = 55.96 C
0.2 0 0
10
20
30
40
50
Tprod [C]
60
70
80
90
15-17
15-31 The composition of a certain coal is given. The coal is burned burned with 50 percent excess air. The AF ratio is to be determined.
Combustion n is compl complete. ete. 2 The combustio combustion n products products contain contain CO2, H2O, O2, N2, and ash Assumptions 1 Combustio only. Properties The molar molar masse massess of C, C, H2, O2, and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, kg/kmol, 32 kg/kmol, kg/kmol, and 29 kg/kmol, kg/kmol, respectivel respectively y (Table (Table A-1).
compos positi ition on of the coal coal is is given given on a mass mass basis, basis, but we we need need to know know the comp composi ositio tion n on a Analysis The com mole basis basis to balance balance the combustio combustion n equation. equation. Consideri Considering ng 1 kg of coal, the numbers numbers of mole 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 chamber
Products
50% excess
Considering 1 kg of coal, the combustion equation can be written as as (0.0683C 0.0028H2 O 0.01H2 0.00031O2 ash) 1.5ath (O2 3.76N2 )
x O2 H y 2 O 0.5 tah O2 1.5 3.76 a C th N2 ash
The unknown coefficients coefficients in the above equation are determined determined from mass balances,
x 0.0683 C : 0.0683 x H : 0.0028 2 0.01 2 2 y
y 0.0128
a th 0.073 O 2 : 0.0028 / 2 0.00031 1.5a th x y / 2 0.5a th 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, coal, which which is taken taken to be 1 kg, mair 0.1095 4.76 kmol29 kg/kmol 15.1 kg mfuel 1 kg and AF
mair mfuel
15.1 kg 1 kg
15.1 kg air/kg fuel
15-18
15-32 Octane Octane is burned with dry dry air. The volumetric volumetric fractio fractions ns of the products products are given. given. The AF ratio and the percentage of theoretical theoretical air used are to be determined.
Combustion is comple complete. te. 2 The combustio combustion n products products contain contain CO2, CO, H2O, O2, and N2 Assumptions 1 Combustion only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis Consideri Considering ng 100 kmol kmol of dry dry products products,, the combustion combustion equation equation can can be written written as as
C8 H1x8
O 2 a3.76N 2
9.21CO 2 0.61CO 7.06O 2 83.12N 2 H 2O b
The unknown coefficients x, a, and b are determined from from mass 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 chamber
b 11.07
(Check O 2 : a 9.21 0.305 7.06 b / 2 22.11 22.10)
Products
Dry air
Thus, 1.23C 8 H 18 22.11O 2 3.76N 2 9.21CO 2 0.61CO 7.06O 2 83.12N 2 11.05H 2 O
The combustion combustion equation equation for 1 kmol of fuel fuel is obtained obtained by dividing dividing the the above above equation equation by 1.23, 1.23, C 8 H 18 18.0O 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 determined by taking the ratio of the mass mass of the air to the mass mass of the fuel, AF
mair mfuel
18.0 4.76 kmol29 kg/kmol 21.8 kg air/kg fuel 8 kmol12 kg/kmol 9 kmol 2 kg/kmol
(b) To find find the percent percent theoreti theoretical cal air air used, used, we need need to know the theoreti theoretical cal amount amount of air, air, which which is determine determined d from the theoretica theoreticall combustion combustion equation equation of the fuel, C8 H18 a th O 2 3.76N 2 O 2:
8CO 2 9H 2O 3.76 a th N 2
a th 8 4.5 a th 12.5
Then, Percent theoretical air
mair,act m air,th
N air,act N air,th
18.04.76 kmol 144% 12.54.76 kmol
15-19
15-33 Carbon Carbon is burned burned with dry air. air. The volumetric volumetric analys analysis is of the products products is given. given. The AF ratio ratio and the percentage of theoretical air used are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain contain CO2, CO, CO, O2, and N2 only.
The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively Properties The (Table A-1). Considering ng 100 kmol kmol of dry dry products products,, the combustion combustion equation equation can can be written written as as Analysis Consideri xC a O 2 3.76N 2
10.06CO 2 0.42 CO 10.69 O 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
Carbon
x 10.48
Combustion chamber
(Check O 2 : a 10.06 0.21 10.69 20.96 20.96)
Products
Dry air
Thus, 10.48C 20.96 O 2 3.76 N 2
10.06CO 2 0.42 CO 10.69 O 2 78.83 N 2
The combustion combustion equation equation for 1 kmol of fuel fuel is obtained obtained by dividing dividing the the above above equation equation by 10.48, 10.48, C 2.0 O 2 3.76 N 2
0.96CO 2 0.04 CO 1.02 O 2 7.52 N 2
(a) The air-fuel ratio is determined determined by taking the ratio of the mass mass of the air to the mass mass of the fuel, AF
m air m fuel
2.0 4.76 kmol29 kg/kmol 23.0 1 kmol12 kg/kmol
kg air/kg fuel
(b) To find find the the percen percentt theore theoretic tical al air air used, used, we need need to to know know the theo theoret retica icall amoun amountt of air, air, which which is determine determined d from the theoretica theoreticall combustion combustion equation 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
2.04.76 kmol 200% 1.04.76 kmol
15-20
15-34 Methane is burned burned with dry air. The volumetric analysis analysis of the products products is given. The AF ratio and the percentage of theoretical theoretical air used are to be determined.
Combustion is comple complete. te. 2 The combustio combustion n products products contain contain CO2, CO, H2O, O2, and N2 Assumptions 1 Combustion only. Properties The The mola molarr mass masses es of of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis Consideri Considering ng 100 kmol kmol of dry dry products products,, the combustion combustion equation equation can can be written written as as
CH 4 x
O 2 a 3.76N 2
5.20CO 2 0.33CO 11.24 O 2 83.23N 2 H 2O b
The unknown coefficients x, a, and b are determined from from mass 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 chamber
b 11.06
(Check O 2 : a 5.20 0.165 11.24 b / 2 22.14 22.14)
Products
Dry air
Thus, 5.53CH 4 22.14 O 2 3.76 N 2
5.20CO 2 0.33CO 11.24 O 2 83.23N 2 11.06 H 2O
The combustion combustion equation equation for 1 kmol of fuel fuel is obtained obtained by dividing dividing the the above above equation equation by 5.53, 5.53, CH 4 4.0 O2 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 m fuel
4.0 4.76 kmol29 kg/kmol 34.5 1 kmol 12 kg/kmol 2 kmol 2 kg/kmol
kg air/kg fuel
(b) To find find the percent percent theoreti theoretical cal air air used, used, we need need to know the theoreti theoretical cal amount amount of air, air, which which is determine determined d from the theoretica theoreticall combustion combustion equation equation of the fuel, CH 4 a th O 2 3.76N 2 O 2:
CO 2 2H 2O 3.76 a 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
4.04.76 kmol 200% 2.04.76 kmol
15-21
Enthalpy of Formation Formation and Enthalpy Enthalpy of Combustion 15-35C For combustion processes the enthalpy of reaction is referred referred to as the enthalpy of combustion, combustion, which represents represents the amount of heat heat released released during a steady-flow combustion process. 15-36C Enthalpy Enthalpy of formation formation is the the enthalp enthalpy y of a substan substance ce due to its its chemical chemical compositio composition. n. The enthalpy of formation is related to elements elements or compounds compounds whereas the enthalpy enthalpy of combustion is related to a particula particularr fuel. 15-37C The heating heating value is called called the the higher higher heating heating value when the 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 vapor form. form. The heating value of a fuel is equal to the absolute absolute value of the the enthalpy of combustion of that fuel. 15-38C If the combus combustion tion of a fuel results results in a single single compound, compound, the the enthalpy enthalpy of formati formation on of that compound compound is identical identical to the enthalpy of combustion combustion of that fuel. 15-39C Yes. 15-40C No. The enthal enthalpy py of formation formation of N2 is simply assigned a value of zero at the standard reference state for convenience. 15-41C 1 kmol kmol of of H2. This is evident evident from the observation that when chemical chemical bonds of H2 are destroyed to form H2O a large amount of energy is released.
enthalpy of combusti combustion on of methane methane at a 25 25C and 1 atm is to be determined using the data from from 15-42 The enthalpy Table A-26 and to be compared compared to the value value listed listed in Table A-27. Assumptions The water in the products products is in the liquid liquid phase. phase. Analysis The stoichiometric equation for this reaction is
CH 4 2O 2 3.76N 2 CO 2 2H 2 O 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 combustion of CH4 becomes hC H P H R
N h
P f , P
N h
R f , R
N h f
CO 2
N h f
H2O
N h f
CH 4
Using h f values values from Table Table A-26, A-26, hC 1 kmol 393,520 kJ/kmol 2 kmol285,830 kJ/kmol
1 kmol 74,850 kJ/kmol 890,330 kJ per kmol CH 4 The listed value in Table Table A-27 is -890,868 -890,868 kJ/kmol, which is almost identical to the calculated value. Since the water in the products is assumed assumed to be in the liquid liquid phase, this hc value corresponds to the higher heating heating value of CH4.
15-22
15-43 EES Problem 15-42 is reconsidered. reconsidered. The effect of temperature temperature on the enthalpy of combustion is to be studied. Analysis The problem is solved solved using EES, and the solution is given 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 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 h_bar_fg_H2O=enthalpy(Steam_iapws,T=298 _iapws,T=298,x=1)-enthalp ,x=1)-enthalpy(Steam_iapws y(Steam_iapws,T=298,x=0) ,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_ HP=1*enthalpy(CO2,T=T_prod)+2*(enthal prod)+2*(enthalpy(H2O,T=T_ py(H2O,T=T_prod)-h_bar_fg_ prod)-h_bar_fg_H2O)+A_th*3 H2O)+A_th*3.76* .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)/ABS(h_bar_comb_ (h_bar_comb_TableA27)*Con TableA27)*Convert(, vert(, %) "[%]"
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
-865000 -870000 -875000
S E E , b m o C
-880000 -885000 -890000 -895000 0
100
200
300
400
Tcomb [C]
500
600
15-23
enthalpy of combustio combustion n of gaseous gaseous ethane at at a 25C and and 1 atm is to be determ determined ined using the data 15-44 The enthalpy from Table Table A-26 A-26 and to be compared compared to the the value listed listed in Table Table A-27. Assumptions The water in the products products is in the liquid liquid phase. phase. Analysis The stoichiometric equation for this reaction is
C 2 H 6 3.5O 2 3.76N 2 2CO 2 3H 2 O 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 stable elements, elements, and and thus their their enthalpy enthalpy of formation formation is zero. Then Then the enthalpy enthalpy of combustio combustion n of C2H6 becomes hC H P H R
N h
P f , P
N h
R f , R
N h f
CO 2
N h f
H 2O
N h f
C2 H 6
Using h f values values from Table Table A-26, A-26, hC 2 kmol393,520 kJ/kmol 3 kmol285,830 kJ/kmol
1 kmol 84,680 kJ/kmol 1,559,850 kJ per kmolC 2 H 6 The listed value in Table Table A-27 is -1,560,633 kJ/kmol, which is almost almost identical to the calculated calculated value. Since the water in the products is assumed assumed to be in the liquid phase, thish this hc value corresponds to the higher heating heating value value of C2H6.
enthalpy of combusti combustion on of liquid liquid octane octane at a 25C and 1 atm is to be determined using the data 15-45 The enthalpy from Table Table A-26 A-26 and to be compared compared to the the value listed listed in Table A-27. Assumptions The water in the products products is in the liquid liquid phase. phase. Analysis The stoichiometric equation for this reaction is
8CO 2 9H 2 O 47N 2 C 8 H 18 12.5O 2 3.76N 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 h
P f , P
N h
R f , R
N h f CO N h f H 2
2O
N h f C H 8
18
Using h f values values from Table Table A-26, A-26, hC 8 kmol393,520 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 almost identical to the calculated value. Since the the water in the product productss is assumed assumed to be in the liquid phase, phase, thish this hc value corresp corresponds onds to the higher heating heating value value of C8H18.
15-24
First Law Analysis of Reacting Systems 15-46C In this case U + W b = H , and the conser conservation vation of energy energy relation relation reduces reduces to the form form of the steady-flow energy relation. 15-47C The heat transfer transfer will be the same same for all cases. The excess oxygen and nitrogen enters and leaves the combustion combustion chambe chamberr at the same state, state, and thus has no effect effect on the energy energy balance. balance.
(b), which contains contains the the maximum maximum amount amount of nonreactin nonreacting g gases. gases. This is because because part of 15-48C For case (b the chemical energy released released in the combustion combustion chamber chamber is absorbed and and transported transported out by the nonreacting nonreacting gases.
burned completely during a steady-flow steady-flow combustion combustion process. The heat transfer from the 15-49 Methane is burned combustion chamber is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. 4 Combustion is complete. 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 no fr free O2. Consideri Considering ng 1 kmol of of fuel, fuel, the theoretica theoreticall combustio combustion n equation can be written as Q
CH 4 a th O 2 3.76N 2 CO2 2H 2O 3.76a th N 2
CH4
where ath is determined determined from from the O2 balance,
25C
a th 1 1 2
Air
Substituting,
CO2 2H 2O 5.64N 2 CH 4 2O 2 3.76N 2
Combustion chamber P = P = 1 atm
Products 25C
100% theoretical
The heat transfer for this combustion combustion process is determined from from the energy balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N h
P f
h h P
N h R
f
h h R
N h
P f , P
N h
R f , R
since both the the reactant reactantss and the products products are at at 25C and both the air and the combustion gases can be treated treated as ideal ideal gases. gases. From the tables, tables,
Substance
CH4 O2 N2 H2O () CO2
h f kJ/kmol -74,850 0 0 -285,830 -393,520
Thus,
Qout 1 393 ,520 2 285 ,830 0 1 74 ,850 0 0 890,330 kJ / kmol CH 4 or Qout 890,330 kJ / kmol CH 4 If combustion is achieved with 100% excess air, the answer would still be the same since since it would enter and leave at 25C, and absorb absorb no energy. energy.
15-25
15-50 Hydrogen is burned completely completely during a steady-flow steady-flow combustion combustion process. The heat transfer from the combustion chamber is to be determined for two cases.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. Analysis The H2 is burned completely with the stoichiometric amount of air, and thus the products will contain only H2O and N2, but no free O2. Considering 1 kmol of H2, the theoretical combustion equation can be written as
H 2O 3.76a th N 2 H 2 a th O 2 3.76N 2
Q H2
where ath is determined determined from from the O2 balanc balancee to be Substituting, g, ath = 0.5. Substitutin
25C
H 2O 1.88N 2 H 2 0.5O 2 3.76N 2 The heat transfer for this combustion process is determined from the energy balance E in E out E system applied on
Air
Combustion chamber P = P = 1 atm
Products 25C
100% theoretical
the combustion combustion chamber chamber with W = It reduces reduces to W = 0. It
Qout
N h
P f
h h P
N h R
f
h h R
N h
P f , P
N h
R f , R
since both the the reactant reactantss and the products products are at at 25C and both the air and the combustion gases can be treated treated as ideal ideal gases. gases. From the tables, tables,
Substance
H2 O2 N2 H2O ()
h f kJ/kmol 0 0 0 -285,830
Substituting,
Qout 1 285 ,830 0 0 0 0 285,830 kJ / kmol H 2 or Qout 285,830 kJ / kmol H 2 If combustion is achieved with 50% excess air, the answer would still be the same since it would enter and leave at 25C, and absorb absorb no energy. energy.
15-26
15-51 Liquid propane is burned with 150 percent excess excess air during during a steady-flow combustion process. process. The mass flow rate of air and the rate of heat transfer transfer from the combustion combustion chamber are are to be determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. are 44 kg/kmol and 29 kg/kmol, respectively respectively (Table (Table A-1). Properties The molar masses of propane and air are Analysis The fuel is burned completely completely with excess air, and thus the products will contain only CO2, H 2O, N2, and some free O 2. Consid Consideri ering ng 1 kmol kmol of C3H8, the combustio combustion n equation equation can be written written as
3CO 2 4H 2 O 1.5a th O 2 2.53.76a th N 2 C 3 H 8 2.5a th O 2 3.76N 2 where ath is the stoichiometric coefficient and is determine determined d from the O2 balance, 2.5a th 3 2 1.5a th
Q
ath 5
C3H8 25C
Thus, C 3 H 8 12.5O 2 3.76N 2 3CO 2 4H 2 O 7.5O 2 47N 2 (a) The air-f air-fuel uel ratio ratio for for this combustio combustion n process process is AF Thus,
mair mfuel
Air 12C
Combustion chamber
Products 1200 K
P = P = 1 atm
12.5 4.76 kmol29 kg/kmol 39.22 kg air/kg fuel 3 kmol12 kg/kmol 4 kmol 2 kg/kmol
air AFm fuel 39.22 kg air/kg fuel 1.2 kg fuel/min 47.1 kg air/min m
(b) The heat heat transf transfer er for this this comb combust ustion ion proces processs is determ determine ined d from from the ener energy gy balanc balancee applied on the combustio combustion n chamber chamber with W = reduces to E in E out E system applied W = 0. It reduces
Qout
N h P
f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
C3H8 () O2 N2 H2O ( g g ) CO2
h f
h 285 K
h 298 K
h1200 K
kJ/kmol -118,910 0 0 -241,820 -393,520
kJ/kmol --8296.5 8286.5 -----
kJ/kmol --8682 8669 9904 9364
kJ/kmol --38,447 36,777 44,380 53,848
The h f of liquid liquid propan propanee is is obtain obtained ed by adding adding h fg of propane at at 25C to h f of gas propane. Substituting,
Qout 3393,520 53,848 9364 4 241,820 44,380 9904 7.5 0 38,447 8682 47 0 36,777 8669 1 118,910 h298 h298 12.5 0 8296.5 8682 47 0 8286.5 8669 190,464 kJ/kmol C 3 H 8 or
Qout 190,464 kJ/kmol C 3 H 8
Then the the rate of heat transf transfer er for a mass flow rate of 1.2 kg/min kg/min for the the propane propane becomes becomes
1.2 kg/min m 190,464 kJ/kmol 5194 kJ/min Qout N 44 kg/kmol
Q Q out N out
15-27
15-52E Liquid propane is burned with 150 percent excess air during a steady-flow combustion process. The mass mass flow rate rate of air and and the rate of heat transf transfer er from the combustion combustion chamber chamber are to to be determine determined. d.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. Properties The molar masses of propane and air are are 44 lbm/lbmol lbm/lbmol and 29 lbm/lbmol, respectively (Table (Table A-1E). Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2, H2O, N2, and some free O 2. Considering 1 kmol of C3H8, the combustion combustion equation can be written as
C3H 8 2.5a th O 2 3.76N 2 3CO2 4H 2O 1.5a th O2 2.53.76ath N 2 where ath is the stoichiometric coefficient and is determine determined d from the O2 balance,
Q
C3H8
ath 5 2.5a th 3 2 1.5a th
77F
Thus, C3H 8 12.5O 2 3.76N 2 3CO2 4H 2O 7.5O2 47N 2
Air 40F
Combustion chamber
Products 1800 R
P = P = 1 atm
(a) The air-f air-fuel uel ratio ratio for for this combustio combustion n process process is AF Thus,
mair mfuel
12.5 4.76 lbmol29 lbm/lbmol 39.2 lbmair/lbmfuel 3 lbmol 12 lbm/lbmol 4 lbmol2 lbm/lbmol
air AFm fuel 39.2 lbm air/lbm fuel 0.75 lbm fuel/min 29.4 lbm air / min m
(b) The heat heat transf transfer er for this this comb combust ustion ion proces processs is determ determine ined d from from the ener energy gy balanc balancee chamber withW with W = = 0. It reduces reduces to E in E out E system applied on the combustion chamber
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
C3H8 () O2 N2 CO2 H2O ( g g )
h f
h 500R
h 537 R
h1800 R
Btu/lbmol -51,160 0 0 -169,300 -104,040
Btu/lbmol --3466.2 3472.2 -----
Btu/lbmol --3725.1 3729.5 4027.5 4258.0
Btu/lbmol --13,485.8 12,956.3 18,391.5 15,433.0
The h f of liquid propane propane is obtained obtained by adding adding the h fg of propane at 77F to the h f of gas propane. Substituting,
Qout 3169,300 18,391.5 4027.5 4 104,040 15,433 4258 7.5 0 13,485.8 3725.1 47 0 12,959.3 3729.5 1 51,160 h537 h537 12.5 0 3466.2 3725.1 47 0 3472.2 3729.5 262,773 Btu / lbmol C 3 H 8 or
Qout 262 ,773 Btu / lbmol C3 H 8
Then the rate rate of heat transfer transfer for a mass flow flow rate of 0.75 kg/min for the propane becomes becomes
0.75 lbm/min m 262,773 Btu/lbmol 4479 Btu/min Qout 44 lbm/lbmol N
Q Q out N out
15-28
15-53 Acetylene gas is burned with 20 percent excess air during a steady-flow combustion process. The AF ratio ratio and the the heat heat transfe transferr are to be determ determine ined. d.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. molar masses masses of C2H2 and air are 26 kg/kmol and 29 kg/kmol, respectively (Table (Table A-1). A-1). Properties The molar Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O 2. Considering 1 kmol of C2H2, the combustion combustion equation can be written as
2CO2 H 2O 0.2ath O2 1.23.76ath N 2 C2 H 2 1.2ath O 2 3.76N 2 Q
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance, 1.2a th 2 0.5 0.2a th
C2H2 25C
ath 2.5
Air
Thus, C 2 H 2 3O 2 3.76N 2 2CO 2 H 2 O 0.5O 2 11.28N 2 (a)
AF
mair mfuel
20% excess air
Combustion chamber
Products 1500 K
P = P = 1 atm
3 4.76 kmol29 kg/kmol 15.9 kg air/kg fuel 2 kmol12 kg/kmol 1 kmol 2 kg/kmol
(b) The heat heat transf transfer er for this this comb combust ustion ion proces processs is determ determine ined d from from the ener energy gy balanc balancee W = 0. It reduces chamber with W = reduces to Ein Eout Esyst em applied on the combustion chamber
Qout
N h P
f
h h
P
N h R
f
h h
R
N h P
f
h h
P
N h
R f , R
since all of the reactants are at 25C. Assuming the air and the combustion products to be ideal gases, gases, we have h = h(T ). ). From the tables, tables,
Substance
C2H2 O2 N2 H2O ( g g ) CO2
h f
h 298 K
h1500 K
kJ/kmol 226,730 0 0 -241,820
kJ/kmol --8682 8669 9904
kJ/kmol --49,292 47,073 57,999
-393,520
9364
71,078
Thus,
Qout 2393,520 71,078 9364 1241,820 57,999 9904 0.5 0 49,292 8682 11.280 47,073 8669 1226,730 0 0 630,565 kJ/kmol C 2 H 2 or 630,565 kJ / kmol C2 H 2 Qout 630,565
15-29
15-54E Liquid octane is burned with 180 percent theoretical theoretical air during during a steady-flow steady-flow combustion combustion process. The AF ratio and the heat transfer transfer from the combustion combustion chamber are to be determined. determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. molar masses masses of C3H18 and air air are are 54 kg/kmol kg/kmol and 29 kg/kmol, kg/kmol, respect respectively ively (Table (Table A-1). A-1). Properties The molar Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C8H18, the combustion combustion equation can be written as
8CO2 9H 2O 0.8ath O 2 1.83.76ath N 2 C8 H18 1.8ath O 2 3.76N 2 Q
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance, 1.8a th 8 4. 4.5 0.8ath
C8H18 77F
ath 12.5
Air
Combustion chamber
Products 2500 R
P = P = 1 atm
80% excess air 77F
Thus,
8CO2 9H 2O 10O2 84.6N2 C8 H18 22.5O 2 3.76N 2 (a)
AF
mair mfuel
22.5 4.76 lbmol29 lbm/lbmol 27.2 8 lbmol 12 lbm/lbmol 9 lbmol2 lbm/lbmol
lbmair/lbmfuel
(b) The heat heat transf transfer er for this this comb combust ustion ion proces processs is determ determine ined d from from the ener energy gy balanc balancee applied on the combus combustion tion chambe chamberr with W = It reduces reduces to Ein Eout Esyst em applied W = 0. It
Qout
N h P
f
h h
P
N h R
f
h h
R
N h P
f
h h
P
N h
R f , R
since all of the reactants are at 77F. Assuming the air and the combustion combustion products to be ideal gases, gases, we have h = h(T ). ). From the tables, tables,
Substance
C8H18 () O2 N2 CO2 H2O ( g g )
h f
h 537 R
h 2500 R
Btu/lbmol -107,530 0 0 -169,300 -104,040
Btu/lbmol --3725.1 3729.5 4027.5 4258.0
Btu/lbmol --19,443 18,590 27,801 22,735
Thus,
Qout 8169,300 27,801 4027.5 9 104,040 22,735 4258 10 0 19,443 3725.1 84.6 0 18,590 3729.5 1 107,530 0 0 412,372 Btu/lbmol C 8 H 18 or Qout 412,372 Btu/lbmol C 8 H 18
15-30
15-55 Benzene Benzene gas is burned with 95 percent percent theoretical theoretical air during during a steady-f steady-flow low combustion combustion process. process. The mole fraction fraction of the the CO in the produc products ts and and the heat transf transfer er from from the combus combustion tion chamber chamber are are to be determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. Analysis (a) The fuel is burned with insufficient amount amount of air, and and thus the products will contain some CO as well as CO2, H2O, and N2. The theoreti theoretical cal combusti combustion on equation equation of C6H6 is Q
6CO2 3H 2O 3.76a th N 2 C6 H 6 a th O 2 3.76N 2
C6H6 25C
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance,
Air
a th 6 1.5 7.5 Then the actual actual combustion combustion equation equation can can be written written as
95% theoretical 25C
Combustion chamber
Products 1000 K
P = P = 1 atm
C6 H 6 0.95 7.5O 2 3.76N 2 xCO2 6 x CO 3H 2O 26.79N 2 0.95 7.5 x 6 x / 2 1.5 x 5.25
O2 balance: Thus,
5.25CO2 0.75CO 3H 2O 26.79N 2 C6 H 6 7.125O2 3.76N 2
The mole fraction fraction of CO in the products is yCO
N CO N total
0.75 5.25 0.75 3 26.79
0021 . or 2.1%
(b) The heat heat transf transfer er for this this comb combust ustion ion proces processs is determ determine ined d from from the ener energy gy balanc balancee W = 0. It reduces chamber with W = reduces to Ein Eout Esyst em applied on the combustion chamber
Qout
N h
P f
h h P
N h R
f
h h R
N h
P f
h h P
N h
R f , R
since all of the reactants are at 25C. Assuming the air and the combustion products to be ideal gases, gases, we have h = h(T ). ). From the tables, tables,
Substance
C6H6 ( g g ) O2 N2 H2O ( g g ) CO CO2
h f
h 298 K
h1000 K
kJ/kmol 82,930 0 0 -241,820 -110,530 -393,520
kJ/kmol --8682 8669 9904 8669 9364
kJ/kmol --31,389 30,129 35,882 30,355 42,769
Thus,
Qout 5.25393,520 42,769 9364 0.75 110,530 30,355 8669 3 241,820 35,882 9904 26.79 0 30,129 8669 182,930 0 0 2,112,779 kJ / kmol C 6 H 6 or
Q out 2,112,800 kJ/kmol C 6 H 6
15-31
15-56 Diesel Diesel fuel fuel is burned burned with with 20 percent percent exces excesss air during during a steady steady-f -flow low combus combustion tion proce process. ss. The required mass mass flow rate of the diesel fuel fuel to supply heat at a specified rate is to be determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C12H26, the combustion combustion equation can be written as
C12 H 26 1.2a th O 2 3.76N 2 12CO 2 13H 2 O 0.2a th O 2 1.2 3.76 a th N 2
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance, 1.2a th 12 6. 6.5 0. 0.2 a th
2
kJ/
C12H26 25C
ath 18.5
Combustion chamber
500 K
Air 20% excess air 25
Substituting,
Products
P = P = 1 atm
C12 H 26 22.2O 2 3.76N 2 12CO 2 13H 2O 3.7O 2 83.47N 2
The heat transfer for for this combustion combustion process is determined determined from the energy balance E in E out E system applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N
P
h f h h
P
N
R
h f h h
R
N
P
h f h h
P
N h
R f , R
since all of the reactants are at 25C. Assuming the air and the combustion products to be ideal gases, gases, we h h T have = ( ). ). From the tables, tables,
Substance
C12H26 O2 N2 H2O ( g g ) CO2
hf
h 298 K
h 500 K
kJ/kmol -291,010 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
kJ/kmol --14,770 14,581 16,828 17,678
Thus,
Q out 12 393,520 17,678 9364 13241,820 16,828 9904 3.7 0 14,770 8682 83.47 0 14,581 8669 1 291,010 0 0 6,869,110 kJ/kmol C 12 H 26 or
Q out 6,869,110 kJ/kmol C12 H 26
Then the required required mass flow rate of fuel for a heat heat transf transfer er rate of 2000 kJ/s becomes becomes
Q out 2000 kJ/s M 6,869,110 kJ/kmol 170 kg/kmol 0.0495 kg/s 49.5 g/s Q out
M N m
15-32
15-57E Diesel Diesel fuel is burned with with 20 percent excess excess air during during a steady-flow steady-flow combustion combustion process process.. The required required mass mass flow rate of the diesel diesel fuel for a specified specified heat transfer transfer rate rate is to be determined. determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C12H26, the combustion combustion equation can be written as
C12 H 26 1.2a th O 2 3.76N 2 12CO 2 13H 2 O 0.2a th O 2 1.2 3.76a th N 2 1800 Btu/s Q
where ath is the stoichiometric coefficient and is determined from the O2 balance, 1.2a th 12 6. 6.5 0. 0.2a th
C12H26 77F
ath 18.5
Air
Substituting,
20% excess air 77F
Combustion chamber
Products 800 R
P = P = 1 atm
C12 H 26 22.2O 2 3.76N 2 12CO 2 13H 2 O 3.7O 2 83.47N 2
The heat transfer for this combustion combustion process is determined from from the energy balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N
P
h f h h
P
N
R
h f h h
R
N
P
h f h h
P
N h
R f , R
since all of the reactants are at 77F. Assuming the air and the combustion combustion products to be ideal gases, gases, we have h = h(T ). ). From the tables, tables,
Substance
C12H26 O2 N2 g ) H2O ( g CO2
hf
h537 R
h800 R
Btu/lbmol -125,190 0 0 -104,040 -169,300
Btu/lbmol --3725.1 3729.5 4258.0 4027.5
Btu/lbmol --5602.0 5564.4 6396.9 6552.9
Thus,
Q out 12 169,300 6552 .9 4027.5 13104,040 6396.9 4258 3.7 0 5602.0 3725.1 83.47 0 5564.4 3729 .5 1 125,190 0 0 3,040,716 Btu/lbmol C12 H 26 or
Qout 3,040,716 Btu/lbmol C12 H 26
Then the required required mass flow rate of fuel for a heat heat transf transfer er rate of 1800 Btu/s becomes becomes
Q 1800 Btu/s M 3,040,716 Btu/lbmol 170 lbm/lbmol 0.1006 lbm/s Q
M N m
15-33
15-58 [ Also solved by EES on enclosed CD] Octane gas is burned with 30 percent excess air during a steady-fl steady-flow ow combustio combustion n process. process. The heat trans transfer fer per unit unit mass of octane octane is to be determined determined.. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. 4 Combustion is complete. Properties The molar molar mass of of C8H18 is 114 kg/kmol kg/kmol (Table (Table A-1). A-1). Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O2. The moistur moisturee in the air does not react react with with anything; anything; it simply simply shows shows up as additional H2O in the products. products. Therefor Therefore, e, for simpli simplicity, city, we will balanc balancee the combustio combustion n equation equation using using dry air, and then add the moisture moisture to both both sides of the equation. equation. Consideri Considering ng 1 kmol kmol of C8H18, the combustion equation can be written as
C8 H18 g 1.3ath O 2 3.76N 2 8CO 2 9H 2O 0.3ath O 2 1.33.76ath N 2
where ath is the stoichiometric stoichiometric coefficient coefficient for air. It is determined determined from from
ath 12.5 O 2 balance: 1.3a th 8 4.5 0.3a th
q
C8H18
Thus,
8CO 2 9H 2O 3.75O 2 61.1N 2 C8 H18 g 16.25 O 2 3.76N 2 Therefore, 16.25 4.76 = 77.35 77.35 kmol of dry air will will be used used per per kmol kmol of the fuel. fuel. The partia partiall pres pressur suree of the water water vapor vapor present in the incoming air is P v ,in air P sat @ 25C 0.60 3.1698 kPa 1.902 kPa
25C Air
30% excess air
Combustion chamber
Products 600 K
P = P = 1 atm
Assuming ideal gas behavior, the number number of moles of the moisture that accompanies accompanies 77.35 kmol of incoming dry air is determined to be
P v,in 1.902 kPa N total N v,in 1.48 kmol 101.325 kPa 77.35 N v,in P total
N v ,in
The balanced balanced combustion combustion equation equation is obtained obtained by adding 1.48 kmol of H2O to both sides of the equation, C 8 H 18 g 16.25O 2 3.76N 2 1.48H 2 O 8CO 2 10.48H 2 O 3.75O 2 61.1N 2
The heat transfer for this combustion combustion process is determined from from the energy balance Ein Eout Esyst em W = 0. It reduces applied applied on the combustio combustion n chamber chamber with W = reduces to
Qout
N
P
h f h h
P
N
R
h f h h
R
N
P
h f h h
P
N h
R f , R
since all all of the reactants reactants are at 25C. Assuming the air and the combustion products to be ideal gases, we have h = h(T ). ). From the tables, tables, h 298 K h 600 K hf Substance kJ/kmol kJ/kmol kJ/kmol C8H18 ( g ) -208,450 ----O2 0 8682 17,929 N2 0 8669 17,563 g ) H2O ( g -241,820 9904 20,402 CO2 -393,520 9364 22,280 Substituting, Qout 8393,520 22,280 9364 10.48 241,820 20,402 9904 3.750 17,929 8682 61.10 17,563 8669 1 208,450 1.48 241,820 0 0 4,324,643 kJ/kmol C8H18 Thus Thus 4,324, 4,324,643 643 kJ of heat heat is transf transferr erred ed from from the combus combustion tion chamb chamber er for each each kmol kmol (114 (114 kg) kg) of C8H18. Then the the heat heat transfer transfer per kg of C8H18 becomes q
Qout M
4,324,643 kJ 114kg
37,935 kJ/kg C8H18
15-34
15-59 EES Problem Problem 15-58 15-58 is reconsidere reconsidered. d. The effect effect of the amount amount of excess excess air on the heat transfer transfer for the combustion process is to be investigated. Analysis The problem problem is solved solved using EES, EES, and the solution solution is given given below.
Fuel$ = 'Octane (C8H18)' T_fuel = (25+273) "[K]" {PercentEX = 30 "[%]"} Ex = PercentEX/100 "[%Excess air/100]" P_air1 = 101.3 [kPa] T_air1 = 25+273 "[K]" RH_1 = 60/100 "[%]" T_prod = 600 [K] M_air = 28.97 [kg/kmol] M_water = 18 [kg/kmol] M_C8H18=(8*12+18*1) "[kg/kmol]" "For theoretical dry air, the complete combustion equation is" "C8H18 + A_th(O2+3. A_th(O2+3.76 76 N2)=8 N2)=8 CO2+9 CO2+9 H2O + A_th (3.76) N2 " A_th*2=8*2+9*1 "theoretical O balance" "now to find the amount of water vapor associated with the dry air" w_1=HUMRAT(AirH2O,T=T_air1,P=P_air1,R=RH_1) "Humidity ratio, kgv/kga" N_w=w_1*(A_th*4.76*M_air)/M_water "Moles of water in the atmoshperic air, kmol/kmol_fuel" "The balanced combustion equation with Ex% excess moist air is" "C8H18 + (1+EX) (1+EX)[A_th( [A_th(O2+3.76 O2+3.76 N2)+N_ N2)+N_w w H2O]=8 CO2+(9 CO2+(9+(1+Ex +(1+Ex)*N_w) )*N_w) H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " "Apply First Law SSSF" H_fuel = -208450 [kJ/kmol] "from Table A-26" HR=H_fuel+ (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air1)+(1+Ex)*N_w*enthalpy(H2O,T=T_air1) HP=8*enthalpy(CO2,T=T_prod)+(9+(1+Ex)*N_w)*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) Q_net=(HP-HR)"kJ/kmol" Q_net=(HP-HR)"kJ/kmol"/(M_C8H18 /(M_C8H18 "kg/kmol" "kg/kmol")) "[kJ/kg_C8H18]" Q_out = -Q_net "[kJ/kg_C8H18]" "This solution used the humidity ratio form psychrometric data to determine the moles of water vapor in in atomspheric atomspheric air. One should should calculate calculate the moles of water water contained contained in in the atmosphe atmospheric ric air by the method shown in Chapter 14 which uses the relative humidity to find the partial pressure of the water vapor and, thus, the moles of water vapor. Explore what happens to the results as you vary the percent excess air, relative humidity, and product temperature." Qout [kJ/kgC8H18]
PercentEX [%]
39374 38417 37460 36503 35546 34588 33631 32674 31717 30760 29803
0 20 40 60 80 100 120 140 160 180 200
40000
] 8 1 H 8 C
37800 35600 33400
t u o
31200 29000 0
40
80
120
PercentEX [%]
160
200
15-35
15-60 Ethane gas is burned with stoichiometric stoichiometric amount of air during a steady-flow combustion combustion process. process. The rate of heat transf transfer er from the combustio combustion n chamber chamber is to be determine determined. d.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 Combustion is complete. molar mass of of C2H6 is 30 kg/kmol kg/kmol (Table A-1). A-1). Properties The molar Analysis The theoretical theoretical combustion equation of C2H6 is
C 2 H 6 a th O 2 3.76N 2 2CO 2 3H 2 O 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, ath 2 1.5 3.5
Q
C2H6 25C Air
Combustion chamber
Products 800 K
P = P = 1 atm
500 K
Then the actual actual combustion combustion equation equation can be written written as
1.9CO 2 0.1CO 3H 2 O 0.05O 2 13.16N 2 C 2 H 6 3.5O 2 3.76N 2 The heat transfer for this combustion combustion process is determined from from the energy balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
C2H6 ( g g ) O2 N2 H2O ( g g ) CO CO2
hf
h 500 K
h 298 K
h 800 K
kJ/kmol -84,680 0 0 -241,820 -110,530 -393,520
kJ/kmol --14,770 14,581 -------
kJ/kmol --8682 8669 9904 8669 9364
kJ/kmol --24,523 23,714 27,896 23,844 32,179
Thus,
Qout 1.9393,520 32,179 9364 0.1110,530 23,844 8669 3 241,820 27,896 9904 0.050 24,523 8682 13.160 23,714 8669 1 84,680 h298 h298 3.50 14,770 8682 13.160 14,581 8669 1,201,005 kJ / kmol C 2 H 6 or
Qout 1,201,005 kJ / kmol C 2H 6
Then the the rate rate of heat transfe transferr for a mass mass flow flow rate of of 3 kg/h for the the ethane ethane becomes becomes
5 kg/h m 1,201,005 kJ/kmol 200,170 kJ/h Qout M 30 kg/kmol
Q Q out N out
15-36
mixture of methane methane and oxygen oxygen contained contained in a tank is 15-61 [ Also solved by EES on enclosed CD] A mixture burned at constant volume. The final pressure in the tank and the heat transfer during this process are to be determined. Assumptions 1 Air and combustion combustion gases are ideal gases. 2 Combustion is complete.
molar masses masses of CH4 and O2 are 16 kg/kmol kg/kmol and 32 kg/kmol, respectively (Table A-1). Properties The molar Analysis (a) The combustion is assumed to be complete, complete, and thus all the carbon in in the methane burns to CO2 and all of the hydrogen hydrogen to H2O. The number number of moles moles of CH4 and O2 in the tank are
mCH 4
N CH 4 N O 2
M CH 4
m O2 M O 2
0.12 kg 16 kg/kmol 0.6 kg
32 kg/kmol
7.5 10 3 kmol 7.5 mol
Q O2 + CH4 25C, 200 kPa
18.75 10 3 kmol 18.75 mol
Then the combustion combustion equation can be written as
1200 1200 K
7.5CH 4 18.75O 2 7.5CO 2 15H 2 O 3.75O 2
At 1200 K, water exists in the gas phase. Assuming both the reactants and the products to be ideal gases, the final pressure pressure in the tank is determined to be P N P T P RV N R Ru T R P P P R P P V N P Ru T P N R T R Substituting,
26.25 mol 1200 K 805 kPa 26.25 mol 298 K
P P 200 kPa
which is relatively low. Therefore, the ideal gas assumption assumption utilized earlier is appropriate. appropriate. (b) The heat transfer for this this constant volume combustion process is determined determined from the energy energy balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N
P
h f h h P v
P
N
R
h f h h P v
R
Since both the reactants and products are assumed assumed to be ideal ideal gases, all the internal energy and enthalpies depend on temperature only, and the P v terms terms in this equation equation can be replaced replaced by RuT . It yie yield ldss
Q out
N
P
h f h1200 K h298 K R u T
P
N
R
h f Ru T
R
since the reactants are at the standard reference reference temperature of 25C. From From the the tabl tables es,, Substance
CH4 O2 g ) H2O ( g CO2
hf
h 298 K
h1200 K
kJ/kmol -74,850 0 -241,820 -393,520
kJ/kmol --8682 9904 9364
kJ/kmol --38,447 44,380 53,848
Thus,
Qout 7.5393,520 53,848 9364 8.314 1200 15 241,820 44,380 9904 8.314 1200 3.750 38,447 8682 8.314 1200 7.5 74,850 8.314 298 18.75 8.314 298 5,251,791 J 5252 kJ Thus Qout 5252 kJ of heat is transferred from the combustion chamber as 120 g of CH4 burned in this combustion chamber.
15-37
reconsidered. The effect of the final temperature temperature on the final pressure pressure and the 15-62 EES Problem 15-61 is reconsidered. heat transfe transferr for the combus combustion tion process process is to be investigat investigated. ed. Analysis The problem is solved solved using EES, and the solution is given given below.
"Input Data" T_reac = (25+273) "[K]" "reactant mixture temperature" P_reac = 200 [kPa] "reactant mixture pressure" {T_prod = 1200 [K]} "product mixture temperature" m_O2=0.600 [kg] "initial mass of O2" Mw_O2 = 32 [kg/kmol] m_CH4 = 0.120 [kg] "initial mass of CH4" Mw_CH4=(1*12+4*1) "[kg/kmol]" R_u = 8.314 [kJ/kmol-K] "universal gas constant" "For theoretical oxygen, the complete complete combustion equation is" "CH4 + A_th O2=1 CO2+2 H2O H2O " 2*A_th=1*2+2*1"theoretical 2*A_th=1*2+2*1"theoretical O balance" "now to find the actual actual moles of O2 supplied supplied per mole of fuel" N_O2 = m_O2/Mw_O2/N_CH4 m_O2/Mw_O2/N_CH4 N_CH4= m_CH4/Mw_CH4 "The balanced complete combustion combustion equation with Ex% excess excess O2 is" "CH4 + (1+EX) A_th O2=1 CO2+ 2 H2O + Ex( A_th) O2 " N_O2 = (1+Ex)*A_th "Apply First First Law to the closed closed system system combustion chamber chamber and assume ideal gas behavior. (At 1200 K, water exists in the gas phase.)" E_in - E_out = DELTAE_sys E_in = 0 E_out = Q_out "kJ/ "kJ/kmol_ kmol_CH4" CH4" "No work work is done beca because use volume volume is constant" constant" DELTAE_sys = U_prod - U_reac "neglect KE and PE and note: U = H - PV = N(h - R_u T)" U_reac = 1*(enthalpy(CH4, T=T_reac) - R_u*T_reac) +(1+EX)*A_th*(enthalpy +(1+EX)*A_th*(enthalpy(O2,T=T_reac) (O2,T=T_reac) - R_u*T_reac) U_prod = 1*(enthalpy(CO2, T=T_prod) - R_u*T_prod) +2*(enthalpy(H2O, T=T_prod) R_u*T_prod)+EX*A_th*(enthalpy(O2,T=T_pr R_u*T_prod)+EX*A_th*(enthalpy(O2,T=T_prod) od) - R_u*T_prod) "The total heat transfer out, in kJ, is:" Q_out_tot=Q_out"kJ/kmol_CH4" Q_out_tot=Q_out"kJ/kmol_CH4"/(Mw_CH4 /(Mw_CH4 "kg/kmol_CH4" "kg/kmol_CH4")) *m_CH4"k *m_CH4"kg" g" "k "kJ" J" "The final pressure in the tank is the pressure of the product product gases. Assuming ideal gas behavior for the gases in the constant volume tank, the ideal gas law gives:" P_reac*V =N_reac * R_u *T_reac P_prod*V = N_prod * R_u * T_prod N_reac = N_CH4*(1 + N_O2) N_prod = N_CH4*(1 + 2 + Ex*A_th) 5900
Tprod [K] 500 700 900 1100 1300 1500
Qout,tot [kJ] 5872 5712 5537 5349 5151 4943
Pprod [kPa] 335.6 469.8 604 738.3 872.5 1007
1100 1000
5700
t o t; t u o
900
] a P 700 k [
5500
800
5300
600 d 500
5100
400 4900 500
700
900
1100
Tprod [K]
1300
300 1500
o r p
P
15-38
stoichiometric etric mixture mixture of octane octane gas and and air contained contained in in a closed closed combu combustion stion chamber chamber is ignited. ignited. 15-63 A stoichiom The heat transfer from the combustion chamber chamber is to be determined. Assumptions 1 Both the reactants and products are ideal gases. 2 Combustion is complete. Analysis The theoretical theoretical combustion equation of C8H18 with stoichiometric amount of air is
C8 H18 g ath O 2 3.76N 2 8CO2 9H 2O 3.76 ath N 2
where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, a th 8 4.5 12.5
P = P = const. C8H18+ Air 25C, 300 kPa
Thus, C 8 H 18 g 12.5 O 2 3.76N 2 8CO 2 9H 2 O 47N 2
The heat transfer transfer for this constant volume combustion combustion process is determined from the energy balance Ein Eout Esyst em applied on the combustion chamber with W other other = 0,
Qout
N
P h f
h h P v P
N
R
Q
1000 K
h f h h P v
R
For a constant pressure quasi-equilibrium process process U + first law relation relation in this this case case is U + W b = H . Then the first
Qout
N
P h f
h1000 K h298K P
N h
R f , R
since the reactants reactants are at the standard reference reference temperature of 25C. Since Since both both the reacta reactants nts and the products behave as ideal gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
C8H18 ( g ) O2 N2 H2O ( g g ) CO2
hf
h 298 K
h1000 K
kJ/kmol -208,450 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
kJ/kmol --31,389 30,129 35,882 42,769
Thus,
Qout 8393,520 42,769 9364 9 241,820 35,882 9904 47 0 30,129 8669 1 208,450 0 0 3,606,428 kJ per kmol of C 8 H 18 or
Qout 3,606,428 kJ per kmol of C8 H18 .
Total mole numbers numbers initially present present in the combustion chamber chamber is determined determined from the ideal ideal gas relation,
300 kPa 0.5 m 3 0.06054 kmol N 1 RuT 1 8.314 kPa m 3/kmol K 298 K P 1V 1
Of these, these, 0.06054 0.06054 / (1 (1 + 12.5 12.54.76) = 1.00110-3 kmol of them is C8H18. Thus Thus the amoun amountt of heat heat -3 transferred from from the combustion chamber chamber as 1.00110 kmol of C8H18 is burned is
Qout 1.001 10 3 kmol C 8 H 18 3,606,428 kJ/kmol C 8 H 18 3610 kJ
15-39
15-64 A mixture of benzene gas and 30 percent excess air contained in a constant-volume tank is ignited. The heat transfer from the combustion chamber chamber is to be determined. Assumptions 1 Both the reactants and products are ideal gases. 2 Combustion is complete.
theoretical combustion equation of C6H6 Analysis The theoretical with stoichiometric amount of air is
Q C6H6+ Air 25C, 1 atm
6CO 2 3H 2O 3.76 ath N 2 C6 H 6 g ath O 2 3.76N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, a th 6 1.5 7.5
1000 1000 K
Then the actual combustion combustion equation with 30% excess air becomes becomes C6 H 6 g 9.75 O 2 3.76N 2 5.52CO 2 0.48CO 3H 2O 2.49O 2 36.66N 2
The heat transfer transfer for this constant constant volume volume combus combustion tion process process is is determ determined ined from the energy energy balance balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces to W = 0. It reduces
Qout
N
P
h f h h P v
P
N
R
h f h h P v
R
Since both the reactants and the products behave behave as ideal gases, gases, all the internal energy and enthalpies enthalpies depend on temperature only, and the P v terms terms in in this this equation equation can be replaced replaced by RuT . It yields
Q out
N
P
h f h1000 K h298 K R u T
P
N
R
h f R u T
R
since the reactants are at the standard reference reference temperature of 25C. From From the the tabl tables es,,
Substance
C6H6 ( g g ) O2 N2 H2O ( g g ) CO CO2
h f
h 298 K
h1000 K
kJ/kmol 82,930 0 0 -241,820 -110,530 -393,520
kJ/kmol --8682 8669 9904 8669 9364
kJ/kmol --31,389 30,129 35,882 30,355 42,769
Thus,
Qout 5.52393,520 42,769 9364 8.314 1000 0.48 110,530 30,355 8669 8.314 1000 3 241,820 35,882 9904 8.314 1000 2.490 31,389 8682 8.314 1000 36.660 30,129 8669 8.314 1000 182,930 8.314 298 9.754.76 8.314 298 2,200,433 kJ or
Qout 2,200,433 kJ
15-40
mixture of benzene benzene gas and 30 percent percent excess excess air contained contained in a constant-vo constant-volume lume tank tank is ignited. 15-65E A mixture The heat transfer from the combustion chamber chamber is to be determined. Assumptions 1 Both the reactants and products are ideal gases. 2 Combustion is complete.
theoretical combustion equation of C6H6 Analysis The theoretical with stoichiometric amount of air is
Q C6H6+ Air 77F, 1 atm
6CO 2 3H 2O 3.76 ath N 2 C6 H 6 g ath O 2 3.76N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, a th 6 1.5 7.5
180 1800 R
Then the actual combustion combustion equation with 30% excess air becomes becomes C6 H 6 g 9.75 O 2 3.76N 2 5.52CO 2 0.48CO 3H 2O 2.49O 2 36.66N 2
The heat transfer transfer for this constant constant volume volume combus combustion tion process process is determ determined ined from the energy energy balance balance Ein Eout Esyst em applie applied d on the combus combustio tion n chamb chamber er with with W = 0. It redu reduces ces to
Qout
N
P
h f h h P v
P
N
R
h f h h P v
R
Since both the reactants and the products behave behave as ideal gases, gases, all the internal energy and enthalpies enthalpies depend on temperature only, and the Pv terms in this equation can be replaced replaced by RuT . It yields
Qout
N
P h f
h1800 R h537 R RuT P
N
R
h f RuT
R
since the reactants are at the standard reference reference temperature of 77F. From the tables, tables,
Substance
C6H6 ( g g ) O2 N2 H2O ( g g ) CO CO2
h f
h537 R
h1800 R
Btu/lbmol 35,6860 0 0 -104,040 -47,540 -169,300
Btu/lbmol --3725.1 3729.5 4258.0 3725.1 4027.5
Btu/lbmol --13,485.8 12,956.3 15,433.0 13,053.2 18,391.5
Thus,
Qout 5.52169,300 18,391.5 4027.5 1.986 1800 0.48 47,540 13,053.2 3725.1 1.986 1800 3 104,040 15,433.0 4258.0 1.986 1800 2.490 13,485.8 3725.1 1.986 1800 36.660 12,956.3 3729.5 1.986 1800 135,680 1.986 537 9.754.76 1.986 537 946,870 Btu or
Qout 946,870 Btu
15-41
15-66 A high efficiency gas furnace burns gaseous propane C3H8 with 140 percent theoretical air. The volume flow rate of water condensed from the product product gases is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain contain CO2, H2O, O2, and N2 only. Properties The molar molar masses masses of C, C, H2, O2 and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectivel respectively y (Table (Table A-1). Analysis The reaction reaction equation equation for 40% excess excess air (140% (140% theoretica theoreticall air) is
B CO 2 D H 2O E O 2 F N 2 C3H 8 1.4ath O 2 3.76N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 40% excess air by usin using g the factor factor 1.4 1.4ath instead of ath for air. The coefficient ath and other other coeff coefficien icients ts are are to be determined from the mass mass balances Q Carbon balance: B=3 C3H8 D 4 Hydrogen balance: 2 D 8 25C Combustion Products Oxygen balance: 2 1.4a th 2 B D 2 E chamber 0.4a th E Air Nitrogen balance: 1.4a th 3.76 F 40% excess Solving Solving the above equations equations,, we find the coeffici coefficients ents E = (E = 2, F = F = 26.32, and ath = 5) and write the balanced reaction equation as
3 CO 2 4 H 2 O 2 O 2 26.32 N 2 C 3 H 8 7 O 2 3.76N 2 The partial pressure of water in the saturated product mixture at the dew point is P v , prod P sat@40C 7.3851 kPa The vapor mole mole fraction fraction is P v , prod 7.3851 kPa 0.07385 y v 100 kPa P prod The kmoles kmoles of water condensed condensed is determ determined ined from N water 4 N w 0.07385 N w 1.503 kmol yv 3 4 N w 2 26.32 N total, product The steady-flow steady-flow energy balance is expressed expressed as N fuel H R Qfuel N fuel H P
where
Q fuel
Q out furnace
H R h f o
31,650 kJ/h
fuel@25 C
0.96
32,969 kJ/h
7hO2@25C 26.32h N2@25C
(103,847 kJ/kmol) 7(0) 26.32(0) 103,847 kJ/kmol H P 3hCO2@25C 4hH2O@25C 2hO2@25C 26.32h N2@25C N w ( h f o
H2O(liq)
)
3(393,520 kJ/kmol) 4(-241,820 kJ/kmol) 2(0) 26.32(0) 1.503(285,830 kJ/kmol) 2.577 10 6 kJ/kmol Substituting into the energy balance equation, equation, N fuel H R Qfuel N fuel H P 6 N N fuel (103,847 kJ/kmol) 32,969 kJ/h N fuel (2.577 10 kJ/kmol) fuel 0.01333 kmol/h
The molar and mass mass flow rates of the liquid water are N w N w N fuel (1.503 kmol/kmol fuel)(0.01333 kmol fuel/h) 0.02003 kmol/h M (0.02003 kmol/h)(18 kg/kmol) 0.3608 kg/h w N m w w
The volume volume flow rate rate of liquid liquid water is is V w
(v f @25C )m w (0.001003 m 3 /kg)(0.3608 kg/h) 0.0003619 m 3 /h 8.7 L/day
15-42
steady-flow combustion chamber chamber with 40 percent 15-67 Liquid ethyl alcohol, C2H5OH (liq), is burned in a steady-flow excess excess air. The required required volume volume flow rate of the liquid ethyl ethyl alcohol is to be determine determined. d. Assumptions 1 Combustion is complete. 2 The combustion products contain contain CO2, H2O, O2, and N2 only.
molar masses masses of C, C, H2, O2 and air are 12 kg/kmol, kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, Properties The molar respectivel respectively y (Table (Table A-1). reaction equation equation for 40% excess excess air is Analysis The reaction
B CO 2 D H 2O E O 2 F N 2 C2 H 5OH 1.4ath O 2 3.76N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 40% excess air by usin using g the factor factor 1.4 1.4ath instead of ath for air. The coefficient ath and other other coeff coefficien icients ts are are to be determined from the mass mass balances Carbon balance:
B=2
Hydrogen balance:
D 3 2 D 6
Oxygen balance:
1 2 1.4a th 2 B D 2 E
Q
25C Air
0.4a th E Nitrogen balance:
C2H5OH (liq)
1.4a th 3.76 F
Combustion chamber
Products 600 K
40% excess
Solving the above equations, we find the coefficients E (E = 1.2, F = 15.79, and ath = 3) and write the balanced balanced reaction reaction equation equation as C 2 H 5 OH 4.2O 2 3.76N 2 2 CO 2 3 H 2 O 1.2 O 2 15.79 N 2 The steady-flow steady-flow energy balance is expressed expressed as N fuel H R Qout N fuel H P
where H R (h f o h fg ) fuel
[email protected] K ( 4.2 3.76)h
[email protected] K
235,310 kJ/kmol - 42,340 kJ/kmol 4.2(-4.425 kJ/kmol) (4.2 3.76)(4.376 kJ/kmol) 277,650 kJ/kmol H P 2hCO2@600 K 3hH2O@600 K 1.2hO2@600 K 15.79h N2@600 K
2(380,623 kJ/kmol) 3(-231,333 kJ/kmol) 1.2(9251 kJ/kmol) 15.79(8889 kJ/kmol) 1.304 10 6 kJ/kmol The enthalpies are obtained from EES EES except for the enthalpy of formation formation of the fuel, which is obtained in Table A-27 of the book. Substituting into the energy energy balance equation, N fuel H R Qout N fuel H P 6 N N fuel (277,650 kJ/kmol) 2000 kJ/s N fuel ( 1.304 10 kJ/kmol) fuel 0.001949 kmol/s
The fuel mass mass flow flow rate is fuel N m fuel M fuel (0.001949 kmol/s)(2 12 6 1 16) kg/kmol 0.08966 kg/s
Then, the the volume volume flow rate of the fuel fuel is determined determined to be V fuel
fuel m
fuel
0.08966 kg/s 6000 L/min
790 kg/m 3
1 m 3 /s
6.81 L/min
15-43
Adiabatic Flame Temperature Temperature 15-68C For the case case of stoichi stoichiomet ometric ric amount amount of pure oxygen since we have the same same amount amount of chemi chemical cal energy released but a smaller amount of mass to absorb it. 15-69C Under the conditions of complete combustion with stoichiometric amount of air. 15-70 [ Also Also solved by EES on enclosed CD ] Hydrogen Hydrogen is burned burned with 20 percent percent excess excess air during a steadysteadyflow combustion combustion process. process. The exit temperature of product product gases is to be determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic. Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber chamber under under adiabati adiabaticc conditions conditions (Q = 0) with with no work work inte intera ract ctio ions ns (W (W = 0). Under steady-flow conditions the energy balance E in E out E system applied on the combustion chamber reduces to
N h P
f
h h P
N h R
f
h h R
H2 7C
The combustion combustion equation equation of H2 with 20% excess excess air air is H 2 0.6O 2 3.76N 2 H 2 O 0.1O 2 2.256N 2
Air
Combustion chamber
Products T P
20% excess air 7C
From the tables,
Substance
H2 O2 N2 H2O ( g g )
h f
h 280 K
h 298 K
kJ/kmol 0 0 0 -241,820
kJ/kmol 7945 8150 8141 9296
kJ/kmol 8468 8682 8669 9904
Thus,
1 241,820 hH O 9904 0.10 hO 8682 2.2560 h N 8669 10 7945 8468 0.6 0 8150 8682 2.256 0 8141 8669 2
It yields
2
2
hH 2O 0.1hO 2 2.256h N 2 270,116 kJ
The adiabatic flame temperature temperature is obtained from a trial and error solution. A first guess guess is obtained obtained by dividing the right-hand side of the equation by the total number of moles, which yields 270,116/(1 + 0.1 + 2.256) = 80,488 kJ/kmol. This enthalpy value corresponds to about 2400 K for N2. Noting that the majority majority of the moles are N2, T P will be close to 2400 K, but somewh somewhat at under it because because of the higher higher specific specific heat of H2O. At 2300 K:
hH 2O 0.1hO 2 2.256h N 2 198,199 0.179,316 2.256 75,676
276,856 kJ Higher than 270,116 kJ At 2250 K:
hH 2O 0.1hO 2 2.256h N 2 195,562 0.177,397 2.256 73,856
269,921 kJ Lower than 270,116 kJ By interpolation,
T P = 2251.4 K
15-44
15-71 EES Problem Problem 15-70 15-70 is reconsi reconsidere dered. d. This problem problem is to be modifie modified d to include include the fuels butane butane,, ethane, methane, methane, and propane propane as well as as H2; to include the effects of inlet air and fuel temperatures; temperatures; and the percent theoretical air supplied. Analysis The problem problem is solved solved using EES, EES, and the solution solution is given given below.
Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHy + (y/4 + x) (Theo_air/100) (O2 + 3.76 N2) <--> xCO xCO2 2 + (y/2) H2O + 3.76 (y/4 + x) x) (Theo (Theo_air/ _air/100) 100) N2 + (y/4 + x) x) (Theo_ (Theo_air/1 air/100 00 - 1) 1) O2 O2 T_prod is the adiabatic combustion combustion temperature, assuming assuming no dissociation. Theo_air is the % theoretical theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$:x,y,Name$) Fuel(Fuel$:x,y,Name$) "This procedure takes the fuel name and returns the moles of C and moles moles of H" If fuel$='C2H fuel$='C2H6' 6' then x=2;y=6 Name$='ethane' else Product temperature vs % theoretical air for hydrogen If fuel$='C3H fuel$='C3H8' 8' then 3000 x=3; y=8 Name$='propane' else Calculated point If fuel$='C4H10' then x=4; x=4; y=10 y=10 Name$='butane' 2000 else if fuel$='CH4' then x=1; y=4 d Name$='methane' o r else p 1000 if fuel$='H2' then x=0; y=2 Name$='hydrogen' endif; endif; endif; endif; endif end {"Input data from the diagram window" 0 T_fuel = 280 [K] 100 200 300 400 500 T_air = 280 [K] Theoair [%] Theo_air = 200 "%" Fuel$='H2'} Call Fuel(fuel$:x,y,Name$) Fuel(fuel$:x,y,Name$) HR=enthalpy(Fuel$,T=T_fuel)+ HR=enthalpy(Fuel$,T=T _fuel)+ (y/4 + x) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 *enthalpy(O2,T=T_air)+3.76*(y/4 + x) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=x*enthalpy(CO2,T=T_prod)+(y/2)*e HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod) nthalpy(H2O,T=T_prod)+3.76*(y/4 +3.76*(y/4 + x)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 (Theo_air/100)*enthalpy(N 2,T=T_prod)+(y/4 + x) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x)* (Theo_air/100) Moles_CO2=x; Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air "arr "array ay variable variable are plotted plotted in Plot Window Window 1" Theoair [%] 100 144.4 188.9 233.3 277.8 322.2 366.7 411.1 455.6 500
Tprod [K] 2512 2008 1693 1476 1318 1197 1102 1025 960.9 907.3
15-45
15-72E Hydrogen is burned with 20 percent excess air during a steady-flow steady-flow combustion process. process. The exit temperatu temperature re of product product gases is to be determ determined. ined. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic. Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber chamber under under adiabati adiabaticc conditions conditions (Q = 0) with with no work work inte intera ract ctio ions ns (W (W = 0). Under steady-flow conditions the energy balance Ein Eout Esyst em applied on the combustion chamber reduces to
N h
P f
h h P
N h R
f
h h R
H2 40F
The combustion combustion equation equation of H2 with 20% excess excess air air is
H 2O 0.1O2 2.256N2 H 2 0.6O2 3.76N 2
Air
Combustion chamber
Products T P
20% excess air 40F
From the tables,
Substance
H2 O2 N2 g ) H2O ( g
h f
h 500 R
h 537 R
Btu/lbmol 0 0 0 -104,040
Btu/lbmol 3386.1 3466.2 3472.2 3962.0
Btu/lbmol 3640.3 3640.3 3725.1 3729.5 4258.0
Thus,
1 104,040 hH O 4258 0.10 hO 3725.1 2.2560 h N 3729.5 10 3386.1 3640.3 0.6 0 3466.2 3725.1 2.256 0 3472.2 3729.5 2
It yields
2
2
hH 2O 0.1hO 2 2.256h N 2 116,094 Btu
The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing the right-hand side of the equation by the total number number of moles, moles, which yields 116,094/(1 + 0.1+ 2.256) = 34,593 Btu/lbmol. Btu/lbmol. This enthalpy enthalpy value corresponds to about 4400 R for N2. Noting Noting that that the majority of the moles are N2, T P will be close to 4400 R, but somewhat under it because of the higher specific heat of H2O. At 4020 R:
hH 2O 0.1hO 2 2.256h N 2 140,740 0.132,989 2.256 31,503
115,110 Btu Lower than 116,094 Btu At 4100 R:
hH 2O 0.1hO 2 2.256h N 2 141,745 0.133,722 2.256 32,198
117,756 Btu Higher than 116,094 Btu By interpolation,
T P = 4054 R
15-46
15-73 Acetylene gas is burned with 30 percent excess air during a steady-flow combustion process. The exit temperatu temperature re of product product gases is to be determined. determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. Analysis The fuel is burned burned comple completely tely with with the excess excess air, air, and thus thus the the products products will contain contain only CO CO2, H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C2H2, the combustion combustion equation can can be written as
2CO2 H 2O 0.3ath O2 1.33.76ath N 2 C 2 H 2 1.3ath O2 3.76N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance, 1.3a th 2 0.5 0.3a th
75,000 kJ/kmol
C2H2
ath 2.5
25C Air
Thus, C2 H 2 3.25O 2 3.76N 2 2CO2 H 2O 0.75O2 12.22N2
Combustion chamber
Products T P
30% excess air 27C
Under steady-flow steady-flow conditions the energy balance Ein Eout Esyst em applied applied on the combustio combustion n chamber chamber with W = reduces es to W = 0 reduc
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,, Substance
C2H2 O2 N2 H2O ( g g ) CO2
h f
h 298 K
h 300 K
kJ/kmol 226,730 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
kJ/kmol --8736 8723 -----
Thus,
75,000 2 393,520 hCO 2 9364 1 241,820 hH 2O 9904
0.750 hO2 8682 12.220 h N 2 8669 1226,730 3.250 8736 8682 12.22 0 8723 8669
2hCO 2 hH2O 0.75hO 2 12.22hN 2 1,321,184 kJ
It yields
The temperature of the product gases gases is obtained from a trial and error solution. A first guess is obtained by dividing the the right-hand side of the equation by the total number of moles, moles, which yields 1,321,184/(2 + 1 + 0.75 + 12.22) 12.22) = 82,729 82,729 kJ/kmol. kJ/kmol. This enthalpy enthalpy value value corresp corresponds onds to about 2500 2500 K for N2. Noting that the majority of the moles are N 2, T P will be close to to 2500 K, but somewhat somewhat under under it it because because of the higher higher specific heats of CO2 and H2O. At 2350 K: 2hCO2 hH 2 O 0.75hO 2 12.22h N 2 2 122,091 1100,846 0.75 81,243 12.22 77,496
1,352,961 kJ Higher than 1,321,184 kJ At 2300 K: 2hCO2 hH 2 O 0.75hO 2 12.22h N 2 2 119,035 198,199 0.7579,316 12.22 75,676
1,320,517 kJ Lower than 1,321,184 kJ By interpolation,
T P = 2301 K
15-47
15-74 A mixture of hydrogen and the stoichiometric amount of air contained in a constant-volume tank is ignited. The final temperature in the tank is to be determined. Assumptions 1 The tank is adiabatic. 2 Both the reactants and products are ideal gases. 3 There are no work interactions. 4 Combustion is complete.
combustion n equation of H2 with stoichiometric amount of air is Analysis The combustio H 2 0.5O2 3.76N 2 H 2O 1.88N2
H2, AIR
The final temperature temperature in the tank is determined determined from the energy balance relation Ein Eout Esyst em for reacting closed systems under
25C, 1 atm
adiabatic adiabatic conditions conditions (Q = 0) with no work interac interactions tions (W (W = = 0),
N h P
f
h h P v P
N h R
f
T P
h h P v R
Since both the reactants and the products behave behave as ideal gases, gases, all the internal energy and enthalpies enthalpies depend on temperature only, and the P v terms terms in this equation equation can be replaced replaced by RuT . It yields
N h P
f
hT P h298 K Ru T P
N h R T
R
f
u
R
since the reacta reactants nts are at the standard standard referen reference ce temperatu temperature re of 25°C. From the tables, tables,
Substance
H2 O2 N2 H2O ( g g )
h f
h 298 K
kJ/kmol 0 0 0 -241,820
kJ/kmol 8468 8682 8669 9904
Thus,
1 241,820 hH O 9904 8.314 T P 1.880 h N 8669 8.314 T P 10 8.314 298 0.50 8.314 298 1.88 0 8.314 298 2
It yields
2
hH 2 O 1.88hN 2 23.94 T P 259,648 kJ
The temperature of the product gases is obtained from a trial and error solution, At 3050 K:
hH 2O 1.88h N 2 23.94 T P 1139,051 1.88103,260 23.94 3050
260,163 kJ Higher than 259,648 kJ At 3000 K:
hH 2O 1.88h N 2 23.94 T P 1136,264 1.88101,407 23.94 3000
255,089 kJ Lower than 259,648 kJ By interpolation,
T P = 3045 K
15-48
15-75 Octane gas is burned with 30 percent excess air during a steady-flow steady-flow combustion process. process. The exit temperatu temperature re of product product gases is to be determ determined. ined. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic. Analysis Under steady-flow conditions the energy balance Ein Eout Esyst em applied on the combustion
chambe chamberr with with Q = W = reducess to W = 0 reduce
N h P
f
h h P
N h R
f
h h R
N h P
f
h h P
N h
R f , R
since all the reactants reactants are at the standard standard reference reference temperat temperature ure of 25°C. Then, C8 H18 g 1.3ath O 2 3.76N2 8CO2 9H 2O 0.3ath O2 1.33.76 ath N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, 1.3a th 8 4. 4.5 0. 0.3a th
ath 12.5
8CO 2 9H 2 O 3.75O 2 61.1N 2 Thus, C 8 H 18 g 16.25O 2 3.76N 2 Therefore, 16.254.76 4.76 = 77.35 77.35 kmol kmol of dry air will will be be used used per kmol of the fuel. The partial partial pressure of the water water vapor present in the incoming air is P v ,in air P sat @ 25C 0.60 3.1698 kPa 1.902 kPa
C8H18 25C Air
Assuming ideal gas behavior, the number of moles of the moisture that accompanies 77.35 kmol of incoming dry air is determi determined ned to be
Combustion chamber
Products T P
30% excess air 25C
P v,in 1.902 kPa N v,in 1.48 kmol P N total 101.325 kPa 77.35 N v,in total
N v ,in
The balanced balanced combustion combustion equation equation is obtained obtained by adding 1.48 kmol of H2O to both sides of the equation,
8CO2 10.48H2O 3.75O2 61.1N2 C8 H18 g 16.25O2 3.76N 2 1.48H2O From the tables, Substance
C8H18 ( g ) O2 N2 g ) H2O ( g CO2 Thus,
h f
h 298 K
kJ/kmol -208,450 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
8 393,520 hCO 9364 10.48 241,820 hH O 9904 3.750 hO 8682 61.10 h N 8669 1 208,450 1.48 241,820 0 0 2
2
2
2
It yields
8hCO 2 10.48hH 2O 3.75hO 2 61.1h N 2 5,857,029 kJ
The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing dividing the right-hand right-hand side side of the equation equation by the total number number of moles, moles, which yields 5,857,02 5,857,029/(8 9/(8 + 10.48 + 3.75 + 61.1) = 70,287 kJ/kmol. kJ/kmol. This enthalpy value corresponds to about 2150 K for N2. Noting that that the major majority ity of of the moles moles are are N2, T P will be close to 2150 K, but somewhat under it because of the higher specific heat of H2O. At 2000 K: 8hCO 2 10.48hH 2O 3.75hO 2 61.1h N 2 8100,804 10.4882,593 3.75 67,881 61.164,810 5,886,451 kJ Higher than 5,857,029 kJ At 1980 K: 8hCO 2 10.48hH 2O 3.75hO 2 61.1h N 2 899,606 10.4881,573 3.75 67,127 61.164,090 5,819,358 kJ Lower than 5,857,029 kJ By interpolation,
T P = 1991 K
15-49
15-76 EES Problem 15-75 is reconsidered. reconsidered. The effect of the relative humidity humidity on the exit temperature of the product product gases is is to be investiga investigated. ted. Analysis The problem problem is solved solved using EES, EES, and the solution solution is given given below.
"The percent excess excess air and relative humidity are input by the diagram diagram window." {PercentEX = 30"[%]"} {RelHum=60"[%]"} "Other input data:" Fuel$ = 'Octane (C8H18)' T_fuel = (25+273) "[K]" Ex = PercentEX/100 "[%Excess air/100]" P_air1 = 101.3 [kPa] T_air1 = 25+273 "[K]" RH_1 = RelHum/100 M_air = 28.97 [kg/kmol] M_water = 18 [kg/kmol] M_C8H18=(8*12+18*1) "[kg/kmol]" "For theoretical dry air, the complete complete combustion equation is" is" "C8H18 + A_th(O2+3.76 N2)=8 N2)=8 CO2+9 H2O + A_th (3.76) N2 " A_th*2=8*2+9*1 "theoretical O balance" "now to find the amount of water vapor associated with the dry air" w_1=HUMRAT(AirH2O,T=T_air1,P=P_ w_1=HUMRAT(AirH2O,T=T_air1,P=P_air1,R=RH_1) air1,R=RH_1) "Humidity ratio, kgv/kga" N_w=w_1*((1+Ex)*A_th*4.76*M_air)/M_water"Moles N_w=w_1*((1+Ex)*A_th*4.76*M_air)/M_water "Moles of water in the atmoshperic air, kmol/kmol_fuel" "The balanced combustion equation with with Ex% excess moist moist air is" "C8H18 + (1+EX)[A_th(O2+3.76 N2)+N_w N2)+N_w H2O]=8 CO2+(9+N_w) H2O + (1+Ex) (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " "Apply First Law SSSF" H_fuel = -208450 [kJ/kmol] "from Table A-26" HR=H_fuel+ (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex)*A_th*3.76 )*A_th*3.76 *enthalpy(N2,T=T_air1)+N_w*enthalpy(H2O,T=T_air1) HP=8*enthalpy(CO2,T=T_prod)+(9+N_w)* HP=8*enthalpy(CO2,T=T_prod)+(9+N_w)*enthalpy(H2O,T=T_prod) enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* +(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) "For Adiabatic Combustion:" HP = HR "This solution used used the humidity ratio form psychrometric psychrometric data to determine the moles of water vapor in atomspheric air. One should calculate the moles moles of water contained contained in the atmospheric air by the the method shown in Chapter 14, which uses the relative humidity to find the partial pressure of the water vapor and, thus, the moles of water vapor. Explore what happens to the results as you vary the percent excess air, relative humidity, and product temperature. " RelHum [%] 0 10 20 30 40 50 60 70 80 90 100
Tprod [K] 2024 2019 2014 2008 2003 1997 1992 1987 1981 1976 1971
2030
Adiabatic Flame Temperature for 30% Excess air
2020 2010 2000 1990
d o r p
1980 1970 1960 1950 0
20
40
60
RelHum [%]
80
100
15-50
Entropy Change and Second Second Law Analysis of Reacting Systems Systems 15-77C Assuming the system exchanges heat with the surroundings at T 0, the increase-in-entropy principle can be express expressed ed as
Sgen
N
Ps P
N s
R R
Qout T 0
15-78C By subtracting Rln( P P / P P 0) from the tabulated value at 1 atm. Here P Here P is is the actual actual pressur pressuree of the substance and P 0 is the atmospheric pressure. 15-79C It represent representss the reversi reversible ble work associ associated ated with the forma formation tion of that compound. compound.
15-80 Hydrogen is burned steadily with oxygen. The reversible reversible work and exergy destruction destruction (or irreversibility) are to be determined. determined. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible.
H 2 0.5O 2 H 2 O.
Analysis The combustion equation is
The H2, the O2, and the H2O are are at at 25C and 1 atm, which which is the standard standard referenc referencee state and also the the state of the surroundings. surroundings. Therefore, the reversible work in this case is simply the difference difference between the Gibbs function of formation of the reactants reactants and that of the products, products, W rev
N R g f , R
N P g f , P N H 2 g f , H 2
0
N O2 g f ,O 2
0
N H 2 O g f , H 2O N H 2O g f , H 2O
1 kmol 237,180 kJ/kmol 237,180 kJ (per kmol of H 2 ) since the g f of stable elements at 25C and 1 atm atm is zero. zero. There Therefor fore, e, 237,18 237,180 0 kJ kJ of of work work could could be done done as as
1 kmol kmol of H2 is burn burned ed with with 0.5 0.5 kmol kmol of O2 at 25C and 1 atm in an environment at the same state. The reversible work in this case case represents the exergy of the reactants since the product (the H2O) is at the state of the surroundings. This process process involve involvess no actual actual work. work. Therefor Therefore, e, the the revers reversible ible work and exergy exergy destructi destruction on are are identical, X destruction destruction = 237,180 kJ
(per (per kmol kmol of H2)
We could also determine determine the reversible reversible work without without involving involving the Gibbs function, function,
N h N h
W rev
R
f
R
f
h h T 0 s R T 0 s R
N h
N h
P f
P f
T 0 s P
h h T 0 s P
N H 2 h f T 0 s H N O 2 h f T 0 s O N H 2O h f T 0 s H 2
2
2O
Substituting, W rev 10 298 130.58 0.50 298 205.03 1285,830 298 69.92 237,204 kJ which is almost almost identical to the result result obtained before.
15-51
15-81 Ethylene gas is burned steadily with 20 percent excess air. The temperature temperature of products, the entropy generation, generation, and the exergy destruction destruction (or irreversi irreversibility bility)) are to be determine determined. d. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible. is burned burned complete completely ly with the the excess excess air, air, and thus thus the products products will contain contain only CO2, Analysis (a) The fuel is H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C2H4, the combustion combustion equation can can be written as
2CO2 2H 2O 0.2ath O 2 1.2 3.76 ath N 2 C2 H 4 g 1.2ath O2 3.76N2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, 1.2a th 2 1 0.2ath Thus,
ath 3
2CO2 2H 2O 0.6O2 13.54N 2 C2 H 4 g 3.6O 2 3.76N 2
Under steady-flow conditions, the exit temperature of the product gases can be determined from the steadysteadyflow energy energy equatio equation, n, which reduces reduces to
N h
P f
h h P
N h
R f , R
N h f
C2H 4
since all all the reactants reactants are are at the standard standard refere reference nce state, state, and for O2 and N2. From the tables, h 298 K h f Substance C2H4 kJ/kmol kJ/kmol g ) C2H4 ( g 52,280 --25C Combustion O2 0 8682 chamber Air N2 0 8669 H2O ( g -241,820 9904 g ) 20% excess air CO2 -393,520 9364 25C Substituting,
Products T P
2 393,520 hCO 9364 2 241,820 hH O 9904 0.60 hO 8682 13.540 h N 8669 152,280 2
2
2
2
2hCO 2 2hH 2O 0.6hO 2 13.54h N 2 1,484,083 kJ
or,
By trial and error, T P = 2269.6 K (b) The entropy entropy genera generation tion during during this adiabati adiabaticc process process is determ determined ined from Sgen S P S R
N
Ps P
N
Rs R
The C2H4 is at 25C and 1 atm, atm, and and thus thus its absolu absolute te entrop entropy y is is 219.83 219.83 kJ/km kJ/kmol· ol·K K (Table (Table A-26). A-26). The entropy entropy values values listed in the ideal ideal gas gas tables tables are are for 1 atm pressure pressure.. Both the the air and the product product gases are at a total pressure pressure of 1 atm, atm, but the entropies entropies are to be calculated calculated at the partial partial pressure pressure of the componen components ts y i which is equal equal to P i = yi P total , where is the mole fraction frac tion of compo c omponent nent . Also, total i
S i N i s i T , P i N i s i T , P 0 Ru ln y i P m The entropy entropy calculations calculations can be presented presented in tabular form form as Ni yi s i T,1atm C2H4 O2 N2
1 3.6 13.54
1.00 0.21 0.79
219.83 205.14 191.61
CO2 H2O O2 N2
2 2 0.6 13.54
0.1103 0.1103 0.0331 0.7464
316.881 271.134 273.467 256.541
R u lny i Pm
N i si
--219.83 -12.98 784.87 -1.96 2620.94 3625.6 .64 4 kJ/K kJ/K S R = 3625 -18.329 670.42 -18.329 578.93 -28.336 181.08 -2.432 3506.49 S P = 4936 4936.9 .92 2 kJ/K kJ/K
Thus, S gen S P S R 4936.92 3625.64 1311.28 kJ/kmol K and (c (c) X destroyed T 0 S gen 298 K 1311.28 kJ/kmol K C 2 H 4 390,760 kJ per kmol C 2 H 4
15-52
15-82 Liquid octane is burned steadily with 50 percent excess air. The heat transfer rate from the combustion combustion chambe chamber, r, the entropy entropy generation generation rate, rate, and the reversib reversible le work and exergy exergy destruction destruction rate are are to be determined. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible. Analysis (a) The fuel is burned burned completely completely with the excess excess air, and thus thus the the products products will contain contain only CO2, H2O, N2, and some some free free O2. Considering 1 kmol C8H18, the combustio combustion n equation equation can be written written as
8CO2 9H 2O 0.5ath O 2 1.53.76ath N 2 C8 H18 1.5ath O 2 3.76N 2 where ath is the stoichiom stoichiometric etric coeffi coefficient cient and is determ determined ined from the the O2 balance, 1.5a th 8 4. 4.5 0. 0.5ath
ath 12.5
Thus,
8CO2 9H 2O 6.25O2 70.5N 2 C8 H18 18.75O 2 3.76N 2 Under steady-flow conditions the energy balance Ein Eout Esyst em applied on the combustion chamber W = 0 reduces to with W =
Qout
N h P
f
h h
P
N h R
f
h h
R
N h
P f , P
N h
R f , R
since all of the reactants are at 25°C. Assuming the air and the combustion combustion products to be ideal gases, we have h = h(T ). ). From the tables, tables, T 0 = 298 K
h f
Substance
kJ/kmol -249,950 0 0 -285,830 -393,520
C8H18 () O2 N2 H2O (l (l ) CO2
Q
C8H18 () 25C Air
Combustion chamber
Products 25C
50% excess air 25C
Substituting, or
Qout 8393,520 9285,830 0 0 1249,950 0 0 5,470,680 kJ/kmol of C8H18 Qout 5,470,680 kJ/kmol of C 8 H 18
The C8H18 is burned burned at a rate of 0.25 0.25 kg/min kg/min or N
Thus,
m
M
0.25 kg/min
812 181 kg/kmol
2.193 10 3 kmol/min
3 Q Q out N kmol/min 5,470,680 kJ/kmol 11,997 kJ/min out 2.193 10
The heat transf transfer er for this this process process is also also equivale equivalent nt to the enthalpy enthalpy of combu combustion stion of liquid liquid C8H18, which could easily be de determined from from Table A-27 A-27 to be hC = 5,470,740 5,470,740 kJ/kmol kJ/kmol C8H18.
15-53
(b) The entrop entropy y generati generation on during during this this proces processs is determi determined ned from from S gen S P S R
Qout T surr
S gen
N s N s P P
R R
Qout T surr
The C8H18 is at 25°C and and 1 atm, and thus thus its absolute entropy is s C8H18 = 360.79 kJ/kmol. kJ/kmol.K K (Table A-26). A-26). The entropy entropy values values listed listed in the ideal ideal gas tables tables are for 1 atm pressure pressure.. Both the air air and and the the product product gases are at a total total press pressure ure of 1 atm, atm, but but the entrop entropies ies are to be calc calcula ulated ted at the the par partia tiall press pressure ure of the the component componentss which is equal equal to P i = yi P total fraction of compon component ent i. Also, total, where yi is the mole fraction
S i N i s i T , P i N i s i T , P 0 Ru ln y i P m The entropy entropy calculations calculations can be presented presented in tabular form form as Ni
yi
s i T,1atm
C8H18 O2 N2
1 18.75 70.50
1.00 0.21 0.79
360.79 205.14 191.61
CO2 H2O () O2 N2
8 9 6.25 70.50
0.0944 --0.0737 0.8319
213.80 69.92 205.04 191.61
R u lny i Pm
N i si
--360.79 -12.98 4089.75 -1.96 13646.69 18,097 97.2 .23 3 kJ/K kJ/K S R = 18,0 -19.62 1867.3 --629.3 -21.68 1417.6 -1.53 13,616.3 S P = 17,531 kJ/K
Thus, S gen S P S R and
Qsurr T surr
17,531 18,097
5,470,523 kJ 298 K
17,798 kJ/kmol K
3 kmol/min 17,798 kJ/kmol K 39.03 kJ/min K S gen N S gen 2.193 10
(c) The exergy exergy destruct destruction ion rate associ associated ated with with this process process is determin determined ed from from X destroyed T 0 S gen 298 K 39.03 kJ/min K 11,632 kJ/min 193.9 kW
15-54
15-83 Acetylene gas is burned steadily with 20 percent excess air. The temperature of the products, the total entrop entropy y change, change, and and the exergy exergy destruction destruction are to to be determined determined.. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible. Analysis (a) The fuel is is burned burned complete completely ly with the the excess excess air, air, and thus thus the products products will contain contain only CO2, H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol C2H2, the combustion equation can be written as
C 2 H 2 g 1.2ath O 2 3.76N 2 2CO 2 H 2O 0.2ath O 2 1.2 3.76 ath N 2
where ath is the stoichiom stoichiometric etric coeffi coefficient cient and is determ determined ined from the the O2 balance, 1.2ath 2 0.5 0.2a th
300,000 kJ/kmol
ath 2.5
Substituting,
C2H2
C 2 H 2 g 3O 2 3.76N 2 2CO 2 H 2 O 0.5O 2 11.28N 2
25C Air
Under steady-flow conditions the exit temperature of the product gases can be determined determined from the energy energy balance Ein Eout Esyst emapplied on the
Combustion chamber
Products T P
20% excess air 25C
combustion combustion chamber, chamber, which which reduces reduces to
Qout
N h P
f
h h P
N h
R f , R
N h P
f
h h
P
N h f
C2H 2
since all the reactants are at the standard reference state, and h f = 0 for O2 and N2. From the tables, tables,
Substance
C2H2 ( g g ) O2 N2 H2O ( g g ) CO2
h f
h 298 K
kJ/kmol 226,730 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
Substituting,
300,000 2 393,520 hCO 2 9364 1 241,820 hH 2O 9904
0.50 hO 2 8682 11.280 h N 2 8669 1226,730
or,
2hCO 2 hH 2O 0.5hO 2 11.28h N 2 1,086,349 kJ
By trial and error,
T P = 2062.1 K
(b) The entropy entropy generation generation during during this process process is determine determined d from S gen S P S R
Qout T surr
N s N s P P
R R
Qout T surr
The C2H2 is at 25°C and 1 atm, and thus its absolute entropy is s C2 H 2 200.85 kJ/kmol K (Table A-26). The entropy entropy values values listed listed in the ideal ideal gas tables tables are for for 1 atm pressure. pressure. Both the the air and and the product product gases are at a total total press pressure ure of 1 atm, atm, but but the entrop entropies ies are to be calc calcula ulated ted at the the par partia tiall press pressure ure of the the component componentss which is equal equal to P i = yi P total fraction of compon component ent i. Also, total, where yi is the mole fraction S i N i si T , P i N i si T , P 0 Ru ln yi P m
15-55
The entropy entropy calculations calculations can be presented presented in tabular form form as Ni
yi
s i T,1atm
C2H2 O2 N2
1 3 11.28
1.00 0.21 0.79
200.85 205.03 191.502
CO2 H2O O2 N2
2 1 0.5 11.28
0.1353 0.0677 0.0338 0.7632
311.054 266.139 269.810 253.068
R u lny i Pm
N i si
--200.85 -12.98 654.03 -1.96 2182.25 3037.1 .13 3 kJ/K kJ/K S R = 3037 -16.630 655.37 -22.387 288.53 -28.162 148.99 -2.247 2879.95 3972.8 .84 4 kJ/K kJ/K S P = 3972
Thus, S gen S P S R
Qsurr T surr
3972.84 3037.13
300,000 kJ 298 K
1942.4 kJ/kmol K
(c) The exergy exergy destruct destruction ion rate associ associated ated with with this process process is determin determined ed from from X destruction T 0 S gen 298 K 1942.4 kJ/kmol K 578,835 kJ per kmol C 2 H 2
15-56
15-84 CO gas is burned steadily with air. The heat transfer transfer rate from the combustion combustion chamber chamber and the rate of exergy exergy destruction destruction are are to be determine determined. d. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible.
molar masses masses of CO and and air are 28 kg/kmol kg/kmol and 29 kg/kmol, kg/kmol, respectivel respectively y (Table (Table A-1). Properties The molar Analysis (a) We first first need need to calculate calculate the the amount amount of air air used used per kmol kmol of CO before we can write the combustion equation, equation, v CO
CO m
RT P
0.2968 kPa m
3
/kg K 310 K
110 kPa
V CO
0.4 m 3 /min
v CO
0.836 m 3 /kg
800 K
0.836 m /kg
N air N fuel
air / M air m fuel / M fuel m
CO 37C
0.478 kg/min
110 kPa
Air
Products 900 K
25C
Then the molar molar air-fue air-fuell ratio becomes becomes AF
Q
3
1.5 kg/min / 29 kg/kmol 3.03 kmol air/kmol fuel 0.478 kg/min / 28 kg/kmol
Thus Thus the numb number er of moles moles of O2 used per mole mole of CO is 3.03/4.76 3.03/4.76 = 0.637. Then the combustio combustion n equation equation in this case case can be written written as
CO 2 0.137O 2 2.40N 2 CO 0.637O 2 3.76N 2 Under steady-flow steady-flow conditions the energy balance Ein E out E system applied applied on the combustion combustion chamber chamber with W = W = 0 reduces to
Qout
N
P h f
h h P
N
R
h f h h
R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
CO O2 N2 CO2
h f
h 298 K
h 310 K
h 900 K
kJ/kmol -110,530 0 0 -393,520
kJ/kmol 8669 8682 8669 9364
kJ/kmol 9014 -------
kJ/kmol --27,928 26,890 37,405
Substituting,
Qout 1393,520 37,405 9364 0.137 0 27,928 8682 2.4 0 26,890 8669 1 110,530 9014 8669 0 0 208,929 kJ/kmol of CO Thus 208,929 208,929 kJ of heat heat is transferr transferred ed from the the combustion combustion chamber chamber for for each kmol kmol (28 kg) of CO. This corres correspon ponds ds to 208,92 208,929/2 9/28 8 = 7462 kJ of heat heat transf transfer er per kg of CO. CO. Then Then the rate rate of heat heat trans transfe ferr for a mass mass flow rate of 0.478 kg/min kg/min for CO becomes becomes q out 0.478 kg/min 7462 kJ/kg 3567 kJ/min Q out m
15-57
(b) This process process involves involves heat transfer transfer with with a reservoir reservoir other than than the surround surroundings. ings. An exergy exergy balance balance on the combusti combustion on chamber chamber in this this case case reduces reduces to the following following relation relation for reversib reversible le work, work, W rev
N
R
h f h h T 0 s
R
N
h h T 0 s P Qout 1 T 0 / T R
P h f
The entropy values listed listed in the ideal gas tables are are for 1 atm = 101.325 kPa pressure. The entropy of each each reactant and the product is to be calculated at the partial pressure of the components components which is equal toP to P i = 110/101.325 5 = 1.0856 1.0856 atm. atm. Also, yi P total total, where yi is the mole fraction of component i, and P m = 110/101.32
S i N i s i T , P i N i s i T , P 0 Ru ln y i P m
The entropy entropy calculations calculations can be presented presented in tabular tabular form as Ni
yi
s i T,1atm
R u lny i Pm
CO O2 N2
1 0.637 2.400
1.00 0.21 0.79
198.678 205.04 191.61
0.68 -12.29 -1.28
CO2 O2 N2
1 0.137 2.400
0.2827 0.0387 0.6785
263.559 239.823 224.467
-9.821 -26.353 -2.543
N i si
198.00 138.44 462.94 S R = 799.38 kJ/K 273.38 36.47 544.82 S P = 854.67 kJ/K
The rate of exergy exergy destruction destruction can be determine determined d from
X destroyed T 0 S gen T 0 m S gen / M
where S gen S P S R Thus,
Qout T res
854.67 799.38
208,929 kJ 800 K
316.5 kJ/kmol K
X destroyed 298 K 0.478 kg/min 316.5/28 kJ/kmol K 1610 kJ/min
15-58
15-85E Benzene Benzene gas is burne burned d steadily steadily with with 95 percent percent theoretica theoreticall air. The The heat heat transfer transfer rate rate from from the combustio combustion n chamber chamber and the exergy exergy destruction destruction are are to be determined determined..
Steady operating operating conditio conditions ns exist. exist. 2 Air and the combustion gases are ideal gases. 3 Assumptions 1 Steady Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible. Analysis (a) The fuel is burned with insufficient amount amount of air, and and thus the products will contain some CO as well as CO2, H2O, and N2. The theoreti theoretical cal combusti combustion on equation equation of C6H6 is Q
C6 H 6 ath O 2 3.76N 2 6CO 2 3H 2O 3.76 ath N 2
C6H6
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance,
77F Air
a th 6 1.5 7.5
Products
Combustion chamber
1500 R
95% theoretical
Then the actual actual combustion combustion equation equation can be written written as
C6 H 6 0.95 7.5O 2 3.76N 2 xCO 2 6 x CO 3H 2O 26.79N 2
The value value of x of x is determine determined d from an an O2 balance,
0.957.5 x 6 x/2 1.5 x 5.25 Thus, C 6 H 6 7.125O 2 3.76N 2 5.25CO 2 0.75CO 3H 2 O 26.79N 2 Under steady-flow conditions the energy balance Ein E out E system applied on the combustion chamber with W = W = 0 reduces to
Q out
N
P
h f h h
P
N
R
h f h h
R
N
P
h f h h
P
N h
R f , R
since all of the reactants are at 77°F. Assuming the air and the combustion products to be ideal gases, we have h = h(T ). ). From the tables,
Substance
C6H6 ( g g ) O2 N2 H2O ( g g ) CO CO2
hf
h537 R
h1500 R
Btu/lbmol 35,680 0 0 -104,040 -47,540 -169,300
Btu/lbmol --3725.1 3729.5 4258.0 3725.1 4027.5
Btu/lbmol --11,017.1 10,648.0 12,551.4 10,711.1 14,576.0
Thus,
Q out 5.25169,300 14,576 4027.5 0.75 47,540 10,711.1 3725.1 3 104,040 12,551.4 4258 26.79 0 10,648 3729.5 135,680 0 0 1,001,434 Btu/lbmol of C 6 H 6
15-59
(b) The entropy entropy generation generation during during this process process is determine determined d from S gen S P S R
Qout T surr
N s N s P P
R R
Qout T surr
The C6H6 is at 77°F and 1 atm, and thus its absolute entropy is sC6H6 = 64.34 Btu/lbmol·R (Table A-26E). The entropy entropy values values listed listed in the ideal ideal gas tables tables are are for 1 atm pressure. pressure. Both the air air and the the product product gases are at a total total pres pressur suree of 1 atm, atm, but but the entrop entropies ies are to be calc calcula ulated ted at the the par partia tiall press pressure ure of the the component componentss which is equal equal to P i = yi P total fraction of component component i. Also, total, where yi is the mole fraction S i N i s i T , P i N i s i T , P 0 Ru ln y i P m The entropy entropy calculations calculations can be presented presented in tabular form form as Ni
yi
s i T,1atm
R u lny i Pm
C6H6 O2 N2
1 7.125 26.79
1.00 0.21 0.79
64.34 49.00 45.77
---3.10 -0.47
CO2 CO H2O ( g g ) N2
5.25 0.75 3 26.79
0.1467 0.0210 0.0838 0.7485
61.974 54.665 53.808 53.071
-3.812 -7.672 -4.924 -0.575
N i si
64.34 371.21 1238.77 1674..32 Btu/ Btu/R R S R = 1674 345.38 46.75 176.20 1437.18 2005..51 Btu/ Btu/R R S P = 2005
Thus, S gen S P S R
Qout T surr
2005.51 1674.32
1,001,434 537
2196.1 Btu/R
Then the exergy destroyed is determined from X destroyed T 0 S gen 537 R 2196.1 Btu/lbmol R 1,179,306 Btu/R per lbmol C 6 H 6
15-60
15-86 [ Also solved by EES on enclosed CD ] Liquid propane is burned steadily with 150 percent excess air. The mass flow rate rate of air, the heat transfer rate from the combustion combustion chamber, and the rate of entropy generation are are to be determined. Assumptions 1 Combustion is complete. 2 Steady operating conditions exist. 3 Air and the combustion gases are are ideal gases. gases. 4 Changes in kinetic and potential energies are negligible.
molar masses masses of C3H8 and air are 44 kg/kmol and 29 kg/kmol, respectively (Table (Table A-1). A-1). Properties The molar burned comple completely tely with the excess excess air, air, and and thus the produc products ts will contain contain only only CO2, Analysis (a) The fuel is burned H2O, N2, and some free O 2. Cons Consid ider erin ing g 1 kmol kmol of of C3H8, the combustion equation can be written as C3H 8 2.5ath O 2 3.76N 2 3CO 2 4H 2O 1.5ath O 2 2.53.76 ath N 2
where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, 2.5a th 3 2 1.5a th
ath 5
Substituting, C3H 8 12.5 O 2 3.76N 2 3CO 2 4H 2O 7.5O 2 47N 2
The air-fuel ratio for this combustion process process is AF Thus,
mair mfuel
12.5 4.76 kmol29 kg/kmol 39.2 kg air/kg fuel 3 kmol12 kg/kmol 4 kmol 2 kg/kmol
air AFm fuel 39.2 kg air/kg fuel0.4 kg fuel/min 15.7 kg air/min m
(b) Under steadysteady-flow flow conditions conditions the energy energy balance balance Ein E out E system applied on the combustion W = 0 reduces to chamber with W =
Qout
N
P h f
h h P
N
R
h f h h
R
Assuming Assuming the the air air and and the combustio combustion n products products to be ideal ideal gases, gases, we haveh have h = h(T ). ). From the tables, (The gaseous propane). propane). h f of liquid propane is obtained by adding the h fg at 25°C to h f of gaseous
Substance
C3H8 () O2 N2 H2O ( g g ) CO2
h f
h 285 K
h 298 K
h1200 K
kJ/kmol -118,910 0 0 -241,820 -393,520
kJ/kmol --8296.5 8286.5 -----
kJ/kmol --8682 8669 9904 9364
kJ/kmol --38,447 36,777 44,380 53,848
Thus,
Qout 3393,520 53,848 9364 4241,820 44,380 9904 7.50 38,447 8682 470 36,777 8669 1 118,910 h298 h298 12.50 8296.5 8682 470 8286.5 8669 190,464 kJ/kmol of C 3 H 8 Thus 190,464 kJ of heat is transferred from the combustion combustion chamber for each kmol (44 kg) of propane. This corre correspond spondss to 190,464/ 190,464/44 44 = 4328.7 4328.7 kJ of of heat transfer transfer per per kg of propane. propane. Then Then the rate rate of heat heat transfer transfer for a mass flow flow rate of 0.4 kg/min kg/min for the propane propane becom becomes es q out 0.4 kg/min 4328.7 kJ/kg 1732 kJ/min Q out m
15-61
(c) The entrop entropy y generatio generation n during during this process process is is determ determined ined from Sgen S P S R
Qout T surr
N
Ps P
N
Rs R
Qout T surr
The C3H8 is at 25°C and 1 atm, and and thus its absolute absolute entropy for for the gas phase is s C3H8 269.91 kJ/kmol·K (Table (Table A-26). A-26). Then Then the entr entropy opy of of C3H8() is obtained from s C3H 8 s C3H8 g s fg s C 3H8 g
h fg T
269.91
15,060 298.15
219.4 kJ/kmol K
The entropy entropy values values listed listed in the ideal ideal gas tables tables are are for 1 atm pressure. pressure. Both the air air and the the product product gases are at a total total pres pressur suree of 1 atm, atm, but but the the entro entropie piess are are to be calc calcula ulated ted at the the part partial ial pressu pressure re of of the the y i component componentss which is equal equal to P i = yi P total , where is the mole fraction of component . T h e n , total i S i N i s i T , P i N i s i T , P 0 Ru ln y i P m The entropy entropy calculations calculations can be presented presented in tabular form form as Ni
yi
s i T,1atm
R u lny i Pm
C3H8 O2 N2
1 12.5 47
--0.21 0.79
219.40 203.70 190.18
---12.98 -1.96
CO2 H2O ( g g ) O2 N2
3 4 7.5 47
0.0488 0.0650 0.1220 0.7642
279.307 240.333 249.906 234.115
-25.112 -22.720 -17.494 -2.236
N i si
219.40 2708.50 9030.58 S R = 11,9 11,958 58.4 .48 8 kJ/K kJ/K 913.26 1052.21 2005.50 11108.50 15,079 79.4 .47 7 kJ/K kJ/K S P = 15,0
Thus, S gen S P S R
Qout T surr
15,079.47 11,958.48
190,464 298
3760.1 kJ/K per kmol C 3 H 8
Then the rate of entropy generation generation becomes
0.4 kmol/min 3760.1 kJ/kmol K 34.2 kJ/min K 44
S gen N S gen
15-62
15-87 EES Problem 15-86 is reconsidered. The effect of the surroundings temperature on the rate of exergy destruction destruction is to be studied. studied. Analysis The problem is solved solved using EES, and the solution is given given below.
Fuel$ = 'Propane (C3H8)_liq' T_fuel = (25 + 273.15) "[K]" P_fuel = 101.3 [kPa] m_dot_fuel = 0.4 [kg/min]*Convert(kg/min, kg/s) Ex = 1.5 "Excess air" P_air = 101.3 [kPa] T_air = (12+273.15) "[K]" T_prod = 1200 [K] P_prod = 101.3 101.3 [kPa] [kPa] Mw_air = 28.97 "lbm/lbmol_air" Mw_C3H8=(3*12+8*1) "kg/kmol_C3H8" {TsurrC {Tsu rrC = 25 [C]} [C]} T_surr = TsurrC+273.15 "[K]" "For theoretical dry air, the complete combustion equation is" "C3H8 + A_th(O2+ A_th(O2+3.76 3.76 N2)=3 CO2+4 CO2+4 H2O + A_th (3.76) (3.76) N2 " 2*A_th=3*2+4*1"theoretical 2*A_th=3*2+4*1"theoretical O balance" "The balanced combustion equation with Ex%/100 excess moist air is" "C3H8 + (1+EX)A_th(O2+3.76 N2)=3 CO2+ 4 H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) A_th) O2 " "The air-fuel ratio on a mass basis is:" AF = (1+Ex)*A_th*4.76*Mw_air/(1*Mw_C3H8) "kg_air/kg_fuel" "The air mass flow rate is:" m_dot_air = m_dot_fuel * AF "Apply First Law SSSF SSSF to the combustion process per kilomole of fuel:" E_in - E_out = DELTAE_cv E_in =HR "Since EES gives the enthalpy of gasesous components, we adjust the EES calculated enthalpy to get the liquid enthalpy. Subtracting the enthalpy of vaporizatio vaporization n from the gaseous enthal enthalpy py gives the enthalpy enthalpy of the the liquid liquid fuel. h_fuel(liq) = h_fuel(gas) - h_fg_fuel" h_fg_fuel = 15060 "kJ/km "kJ/kmol ol from Table Table A-27" A-27" HR = 1*(enthalpy(C3H8, 1*(enthalpy(C3H8, T=T_fuel) T=T_fuel) - h_fg_fuel)+ (1+Ex)*A_th*enthalpy(O2,T=T_air)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air) E_out = HP + Q_out HP=3*enthalpy(CO2,T=T_prod)+4*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) DELTAE_cv = 0 "Steady-flow requirement" "The heat transfer rate from the combustion combustion chamber is:" Q_dot_out=Q_out"kJ/kmol_fuel" Q_dot_out=Q_out"kJ/kmol_fuel"/(Mw_C3H8 /(Mw_C3H8 "kg/kmol_fuel" "kg/kmol_fuel")*m_dot_fuel )*m_dot_fuel"k "kg/ g/s" s" "k "kW" W" "Entopy Generation due to the combustion process and heat rejection to the surroundings:" "Entopy of the reactants per kilomole of fuel:" P_O2_reac= 1/4.76*P_air "Dalton's law of partial pressures for O2 in air" s_O2_reac=entropy(O2,T=T_air,P=P_O2_reac)
15-63
P_N2_reac= 3.76/4.76*P_air "Dalto "Dalton's n's law of partial pressu pressures res for N2 in air" s_N2_reac=entropy(N2,T=T_air,P=P_N2_reac) s_C3H8_reac=entropy(C3H8, T=T_fuel,P=P_fuel) - s_fg_fuel "Adjust the EES gaseous value by s_fg" "For phase change, s_fg s_fg is given by:" s_fg_fuel = h_fg_fuel/T_fuel SR = 1*s_C3H8_reac + (1+Ex)*A_th*s_O2_reac + (1+Ex)*A_th*3.76*s_N2_reac "Entopy of the products per kilomle of fuel:" "By Dalton's law the partial pressures of the product gases is the product of the mole fraction and P_prod" N_prod = 3 + 4 + (1+Ex)*A_th* (1+Ex)*A_th*3.76 3.76 + Ex*A_th Ex*A_th "total kmol of products" P_O2_prod = Ex*A_th/N_prod*P_prod "Patrial pressure O2 in products" s_O2_prod=entropy(O2,T=T_prod,P=P_O2_prod) P_N2_prod = (1+Ex)*A_th*3.76/N_prod*P_prod "Patrial pressure N2 in products" s_N2_prod=entropy(N2,T=T_prod,P=P_N2_prod) P_CO2_prod = 3/N_prod*P_prod "Patrial pressure CO2 in products" s_CO2_prod=entropy(CO2, T=T_prod,P=P_CO2_prod) P_H2O_prod = 4/N_prod*P_prod "Patrial pressure H2O in products" s_H2O_prod=entropy(H2O, T=T_prod,P=P_H2O_prod) SP = 3*s_CO2_prod + 4*s_H2O_prod 4*s_H2O_prod + (1+Ex)*A_th*3.76*s_N2_prod + Ex*A_th*s_O2_prod "Since Q_out is the heat rejected to the surroundings per kilomole fuel, the entropy of the surroundings is:" S_surr = Q_out/T_surr "Rate of entropy generation:" S_dot_gen S_dot_gen = (SP - SR +S_surr) +S_surr)"kJ/kmol_fuel" "kJ/kmol_fuel"/(Mw_C3H8 /(Mw_C3H8 "kg/kmol_fuel" "kg/kmol_fuel")*m_dot_fuel )*m_dot_fuel"kg/s" "kg/s" "kW/K" X_dot_dest = T_surr*S_dot_gen"[kW]" T_surr*S_dot_gen"[kW]" TsurrC [C] 0 4 8 12 16 20 24 28 32 36 38
Xdest [kW] 157.8 159.7 161.6 163.5 165.4 167.3 169.2 171.1 173 174.9 175.8
177.5
173.5
169.5
[ t s e d
165.5
161.5
157.5 0
5
10
15
20
25
TsurrC [C]
30
35
40
15-64
Review Problems
sample le of a cert certai ain n fluid fluid is bur burne ned d in a bomb bomb calo calori rime mete ter. r. The The hea heatin ting g valu valuee of the the fuel fuel is to to be 15-88 A samp determined. Properties The specifi specificc heat of water water is 4.18 4.18 kJ/kg. kJ/kg.C (Table A-3). Analysis We take the the water as the system, system, which is a closed closed system system,, for which which the energy energy balance balance on the system Ein Eout Esyst em with W = W = 0 can be written as
Qin U or Qin mcT
2 kg 4.18 kJ/kg C 2.5C 20.90 kJ per gram of fuel Theref Therefore ore,, heat heat transf transfer er per kg of the the fuel fuel would would be be 20,900 kJ/kg fuel. Disregarding the slight energy stored in the gases of the combustion chamber, this value corresp corresponds onds to the heating heating value of the fuel.
WATER 2 kg
Reaction chamber Fuel 1g T = 2.5 C
15-65
15-89E Hydrogen is burned burned with 100 percent excess excess air. The AF ratio and the volume flow rate of air are to be determined. Assumptions 1 Combustion is complete. 2 Air and the combustion combustion gases are are ideal gases. Properties The molar molar masses masses of H2 and air air are are 2 kg/kmol kg/kmol and 29 kg/kmol, kg/kmol, respec respectivel tively y (Table (Table A-1). A-1).
combustio tion n is is com comple plete, te, and thus thus produc products ts will will cont contain ain only only H2O, O2 and N2. The Analysis (a) The combus moisture in the air does not react with anything; it simply shows up as additional H2O in the products. Therefore, for simplicity, we will balance the combustion equation using using dry air, and then add the moisture to both sides sides of the equation equation.. The combusti combustion on equation equation in this case case can be written written as H 2 2ath O 2 3.76N2 H 2O ath O 2 23.76ath N 2 where ath is the stoichiometric stoichiometric coefficient coefficient for air. It is determined determined from from O2 balance:
2a th 0.5 ath
ath 0.5
Substituting,
H 2 O2 3.76N2 H 2O 0.5O2 3.76N2
Therefore, 4.76 lbmol of dry air will be used per kmol of the fuel. fuel. The partial partial pressure pressure of the water vapor present present in the the incoming incoming air air is
Q
H2 Combustion chamber Air
Products
P = P = 14.5 psia
90F
P v ,in air P sat @90 F 0.60 0.69904 psi 0.419 psia The number of moles moles of the moisture that accompanies accompanies 4.76 lbmol of incoming dry air (Nv, in) is determined to be
P v,in 0.419 psia N total N v,in 0.142 lbmol 14.5 psia 4.76 N v,in P total
N v ,in
The balanced balanced combustion combustion equation equation is obtained obtained by substituti substituting ng the coefficie coefficients nts determ determined ined earlie earlierr and adding 0.142 lbmol of H2O to both both sides sides of the equatio equation, n,
1.142H 2 O 0.5O 2 3.76N 2 H 2 O 2 3.76N 2 0.142H 2 O The air-fue air-fuell ratio is determ determined ined by taking taking the ratio of the mass mass of the air air to the mass mass of the fuel, fuel, AF
mair mfuel
4.76 lbmol29 lbm/lbmol 0.142 lbmol18 lbm/lbmol 70.3 lbm air/lbmfuel 1 lbmol2 lbm/lbmol
(b) The mass mass flow rate of H2 is given given to be 10 10 lbm/h. lbm/h. Since we need need 70.3 lbm lbm air air per lbm of H2, the required mass flow rate of air is air AFm fuel 70.3 25 lbm/h 1758 lbm/h m
The mole fractions fractions of water vapor and the dry air in the incoming air are are y H 2O
N H 2O N total
0.142 4.76 0.142
0.029 and y dryair 1 0.029 0.971
Thus, M yM H 2O yM dryair 0.029 18 0.97129 28.7 lbm/lbmol v V
RT P
10.73/28.7 psia ft
3
/lbm R 550 R
14.5 psia
14.18 ft 3 /lbm
m v 1758 lbm/h 14.18 ft 3 /lbm 24,928 ft 3 /h
15-66
composition ion of a gaseous gaseous fuel is given. given. The fuel fuel is burned with 120 percent percent theoretical theoretical air. air. The 15-90 The composit AF ratio and the volume volume flow rate of air intake are to be determine determined. d. Assumptions 1 Combustion is complete. 2 Air and the combustion combustion gases are are ideal gases.
molar masses masses of C, H2, N2, O2, and air are are 12, 2, 28, 32, and and 29 kg/kmol kg/kmol (Table (Table A-1). A-1). Properties The molar is burned burned comple completely tely with with excess excess air, and and thus thus the products products will will contain contain H2O, CO2, Analysis (a) The fuel is N2, and some free O2. The moisture in the air does not react with anything; it simply shows up as additional H2O in the products. products. Therefor Therefore, e, we can can simply simply balance balance the the combustio combustion n equation equation using dry air, air, and then add the the moistur moisturee to both sides of the the equation equation.. Consideri Considering ng 1 kmol of fuel, fuel, the the combus combustion tion equati equation on can can be written as
0.80CH4 0.15N2 0.05O2 1.2ath O2 3.76N2 xCO 2 yH 2O 0.2ath O 2 z N 2 The unknown coefficients in the above equation are determined determined from from mass mass balances, balances,
80% CH4 15% N2 5% O2
C : 0.80 x
x 0.80 y 1.6 H : 0.804 2 y O 2 : 0.05 1.2a th x y / 2 0.2a th a th 1.55 z 7.14 3.76a th z N 2 : 0.15 1.2
air
Combustion chamber
Pr duct
120% theoretical
Next we determine determine the amount of moisture that accompanies accompanies 4.761.2a 1.2ath = 4.761.21.55 1.55 = 8.85 8.85 kmol kmol of dry air. air. The partial partial press pressure ure of the moistur moisturee in the air air is P v ,in air P sat @30C 0.604.247 kPa 2.548 kPa N v, in) is determined to be The number number of moles moles of the moisture moisture in the air air ( N
P v,in 2.548 kPa N total N v,in 0.23 kmol 100 kPa 8.85 N v,in P total
N v ,in
The balanced balanced combustion combustion equation equation is obtaine obtained d by substitut substituting ing the coefficie coefficients nts determ determined ined earlie earlierr and adding 0.23 kmol of H2O to both sides of the the equatio equation, n,
0.80CH 4 0.15N 2 0.05O 2 1.86O2 3.76N 2 0.23H 2O 0.8CO 2 1.83H 2O 0.31O 2 7.14N 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, mair 1.864.76kmol29 kg/kmol 0.23 kg 18 kg/kmol 260.9 kg mfuel 0.816 0.1528 0.05 32 kg 18.6 kg and
AF
mair mfuel
260.9 kg 18.6 kg
14.0 kg air/kg fuel
(b) The mass mass flow rate of the the gaseou gaseouss fuel fuel is given given to be 2 kg/min. kg/min. Since we need need 14.0 14.0 kg kg air per kg of fuel, the required mass flow rate of air is air AFm fuel 14.0 2 kg/min 28.0 kg/ min m
The mole fractions fractions of water vapor and the dry air in the incoming air are are y H 2O
N H 2 O N total
0.23 8.85 0.23
0.025 and y dryair 1 0.025 0.975
Thus, M yM H 2O yM dryair 0.02518 0.975 29 28.7 kg/kmol v V
RT P
8.314/28.7 kPa m
3
/kg K 303 K
100 kPa
0.878 m 3 /kg
m v 28.0 kg/min 0.878 m 3 /kg 24.6 m 3 /min
15-67
15-91 A gaseous gaseous fuel with with a known compositio composition n is burned burned with dry air, and and the volumetric volumetric analysis analysis of products products gases is determi determined. ned. The AF ratio, ratio, the percent percent theoreti theoretical cal air used, used, and and the volume volume flow rate rate of air air are to be determined. Assumptions 1 Combustion is complete. 2 Air and the combustion combustion gases are are ideal gases. Properties The molar molar masses masses of of C, H2, N2, O2, and air are 12, 2, 28, 28, 32, and 29 kg/km kg/kmol, ol, resp respect ective ively ly (Table A-1). Analysis Consideri Considering ng 100 kmol of dry dry products, products, the combustio combustion n equation equation can can be written written as x0.80CH 4 0.15N 2 0.05O 2 a O 2 3.76N 2
3.36CO 2 0.09CO 14.91O2 81.64N 2 bH 2O The unknown coefficients x, a, and b are determined from mass balances,
80% CH4 15% N2 5% O2
C : 0.80 x 3.36 0.09 x 4.31 H : 3.2 x 2b
b 6.90 a 21.54 N 2 : 0.15 x 3.76a 81.64
Check O
2:
Combustion chamber
Air
3.36 3.36% % CO2 0.09% CO 14.91% O2 81.64% N2
0.05 x a 3.36 0.045 14.91 b / 2 a 21.54
Thus, 4.310.80CH 4 0.15N 2 0.05O2 21.54O 2 3.76N 2
3.36CO 2 0.09CO 14.91O2 81.64N 2 6.9H 2O The combustion combustion equation equation for 1 kmol of fuel fuel is obtaine obtained d by dividing dividing the above equation equation by 4.31, 4.31,
0.80CH 4 0.15N 2 0.05O2 5.0O 2 3.76N 2 0.78CO 2 0.02CO 3.46O 2 18.95N 2 1.6H 2O (a) The air-fuel air-fuel ratio ratio is determined determined from from its definition, definition, AF
mair mfuel
5.0 4.76 kmol29 kg/kmol 0.8 16 0.15 28 0.05 32
37.1 kg air/kg fuel
(b) To find find the percent percent theore theoretical tical air air used, used, we need need to know the theoreti theoretical cal amount amount of air, air, which which is determine determined d from the theoretica theoreticall combustion combustion equation equation of the fuel,
0.80CH 4 0.15N 2 0.05O 2 a th O 2 3.76N 2 0.8CO 2 1.6H 2 O 0.15 3.76a th N 2 a th 1.55 O 2 : 0.05 a th 0.8 0.8 Then,
Percent theoretical air
mair,act mair,th
N air,act N air,th
5.0 4.76 kmol 323% 1.554.76 kmol
(c) The specific volume, mass flow rate, rate, and the volume flow rate of air at the inlet conditions are v
RT P
0.287 kPa m
3
/kg K 298 K
100 kPa
0.855 m 3 /kg
3 air (AF) m fuel 37.1 kg air/kg fuel 1.4 kg fuel/min 51.94 m /min m
V air
m v air 51.94 kg/min 0.855 m 3 /kg 44.4 m 3 /min
15-68
15-92 CO gas is burned burned with air air during during a steady-f steady-flow low combusti combustion on process. process. The rate of heat transf transfer er from the combustio combustion n chamber chamber is to be determine determined. d.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. 5 Combustion is complete. molar masses masses of CO and and air are 28 kg/kmol kg/kmol and 29 kg/kmol, kg/kmol, respectivel respectively y (Table (Table A-1). Properties The molar Analysis We first need to calculate the amount of air used per kmol of CO before we can write the combustion equation, v CO
CO m
RT P V CO v CO
0.2968 kPa m
3
/kg K 310 K
110 kPa 0.4 m 3 /min
0.836 m 3 /kg
Q
0.836 m 3 /kg
CO 37C
0.478 kg/min
Air
N air N fuel
air / M air m fuel / M fuel m
Products 900 K
25C
Then the molar molar air-fue air-fuell ratio becomes becomes AF
Combustion chamber
1.5 kg/min / 29 kg/kmol 3.03 kmol air/kmol fuel 0.478 kg/min / 28 kg/kmol
Thus Thus the numb number er of moles moles of O2 used per mole mole of CO is 3.03/4.76 3.03/4.76 = 0.637. Then the combustio combustion n equation equation in this case case can be written written as CO 0.637O 2 3.76N 2 CO 2 0.137O 2 2.40N 2 Under steady-flow conditions the energy balance E in E out E system
applied on the combustion
chamber with W = reduce cess to W = 0 redu
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,, h f
h 298 K
h 310 K
h 900 K
kJ/kmol
kJ/kmol
kJ/kmol
kJ/kmol
CO
-110,530
8669
9014
27,066
O2
0
8682
---
27,928
N2
0
8669
---
26,890
CO2
-393,520
9364
---
37,405
Substance
Thus,
Qout 1393,520 37,405 9364 0.137 0 27,928 8682 2.40 26,890 8669 1 110,530 9014 8669 0 0 208,927 kJ/kmol of CO Then the the rate rate of heat trans transfer fer for for a mass mass flow flow rate of 0.956 kg/min kg/min for CO CO becomes becomes
0.478 kg/min m 208,927 kJ/kmol 3567 kJ/min Qout 28 kg/kmol N
Q Q out N out
15-69
15-93 Methane gas is burned burned steadily with dry air. air. The volumetric analysis analysis of the products is given. The percentage of theoretical air used and the heat transfer transfer from the combustion chamber chamber are to be determined.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. equation can be written as Analysis (a) Considering 100 kmol of dry products, the combustion equation CH 4 x
O2 a 3.76N 2
5.20CO2 0.33CO 11.24O2 83.23N2 H2 Ob
The unknown coefficients x, a, and b are determined from from mass mass balances, N 2 : 3.76a 83.23
a 22.14
C : x 5.20 0.33 H : 4 x 2b
CH4
x 5.53
25C
b 11.06
Air
5.53CH4 22.14 O2 3.76N2
Combustion chamber
Products 700 K
17C
(Check O 2 : a 5.20 0.165 11.24 b / 2 22.14 22.14) Thus,
Q
5.20CO2 0.33CO 11.24O2 83.23N2 11.06H2 O
The combustion combustion equation equation for 1 kmol of fuel fuel is obtaine obtained d by dividing dividing the above equation equation by 5.53 5.53 CH 4 4O 2 3.76N 2 0.94CO 2 0.06CO 2.03O 2 15.05N 2 2H 2 O To find the percent percent theoretical air used, we need to know the theoretical theoretical amount of air, which is determined from the theoretic theoretical al combustion combustion equation equation of the fuel, fuel, CH 4 ath O 2 3.76N 2
ath 2.0 a th 1 1
O 2: Then,
CO2 2H2 O 3.76ath N2
Percent theoretical air
mair,act mair,th
N air,act N air,th
4.04.76 kmol 200% 2.04.76 kmol
(b) Under steady-flow conditions, energy balance balance applied on the combustion combustion chamber reduces to
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h(T ). ). From From the tables tables,,
Substance
CH4 ( g g ) O2 N2 g ) H2O ( g CO CO2
h f
h 290 K
h 298 K
h 700 K
kJ/kmol -74,850 0 0 -241,820 -110,530 -393,520
kJ/kmol --8443 8432 -------
kJ/kmol --8682 8669 9904 8669 9364
kJ/kmol --21,184 20,604 24,088 20,690 27,125
Thus,
Qout 0.94 393,520 27,125 9364 0.06 110,530 20,690 8669 2 241,820 24,088 9904 2.030 21,184 8682 15.04 0 20,604 8669 1 74,850 h298 h298 4 0 8443 8682 15.04 0 8432 8669 530,022 kJ/kmolCH 4 or
Qout 530,022 kJ/kmol CH 4
15-70
15-94 A mixture of hydrogen hydrogen and the stoichiometric amount amount of air contained in a rigid rigid tank is ignited. The fracti fraction on of H2O that condense condensess and the heat transfe transferr from the combustio combustion n chamber chamber are to be determine determined. d.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. 5 Combustion is complete. theoretical combustion equation of H2 with stoichiometric amount of air is Analysis The theoretical H 2 ath O2 3.76N 2 H 2O 3.76ath N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance,
Q
H2, Air
ath 0.5 Thus,
25C 25C
H 2O 1.88N 2 H 2 0.5O 2 3.76N 2
(a) At 25C part part of the water water (say, (say, N w moles) moles) will condense, condense, and the the number number of moles moles of products products that that remain remainss in the gas gas phase phase will will be 2.88 2.88 - N w. Neglec Neglecting ting the the volume volume occu occupie pied d by the liqui liquid d water water and and treating treating all the product gases as ideal gases, the final pressure in the t ank can be expressed as P f
N f ,gas Ru T f
V
2.88 N w kmol8.314 kPa m 3 /kg K 298 K 6 m3
412.92.88 N w kPa Then, N v N gas
P v P total
1 N w
2.88 N w
3.169 kPa 412.92.88 N w kPa
N w 0.992 kmol
Thus 99.2% of the the H2O will will condense condense when the produc products ts are are cooled cooled to to 25C. (b) The energy energy balance balance Ein Eout Esyst em applied applied for this constant constant volume volume combustion combustion process process with W = W = 0 reduc reduces es to
Qout
N h P
f
h h P v P
N h R
f
h h P v R
With the exception exception of liquid liquid water water for which the P v term is negligible, both the reactants and and the products are assum assumed ed to be ideal ideal gases, gases, all the the internal internal energy energy and enthalpie enthalpiess depend depend on temperatu temperature re only, only, and the P v terms in this equation can be replaced by RuT . It yie yields lds
N h N h
Qout
P
f
Ru T P
P f , P
N h
N h
R
R f , R
f
Ru T R
N
Ru T
P , gas
N
R R
since the reactants are at the standard reference reference temperature of 25C. From From the the tabl tables es,,
Substance
H2 O2 N2 g ) H2O ( g H2O ()
h f kJ/kmol 0 0 0 -241,820 -285,830
Thus,
Qout 0.008241,820 0.992 285,830 0 0 0 0 8.314 2981.89 3.38 281,786 kJ per kmol H 2 or
Qout 281,786 kJ (per kmol H 2 )
15-71
15-95 Propa Propane ne gas is burned burned with with air air during during a stead steady-f y-flow low combus combustio tion n proce process. ss. The adiaba adiabatic tic flame flame temperatu temperature re is to be determined determined for different different cases. cases.
combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and Assumptions 1 Steady operating conditions exist. 2 Air and combustion potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic. Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber chamber under under adiabati adiabaticc conditions conditions (Q = 0) with with no work work inte intera ract ctio ions ns (W (W = 0). Under steady-flow conditions the energy balance E in E out E system applied on the combustion chamber reduces to
N h P
f
h h P
N h R
f
h h R
N h P
f
hT h P N h f C H 3
8
since all the reactants are at the standard reference reference temperature temperature of 25C, and h f 0 for O2 and N2. (a) The theore theoretic tical al comb combust ustion ion equa equatio tion n of C3H8 with stoichiometric stoichiometric amount of air is
3CO2 4H 2O 18.8 N 2 C3H8 g 5O 2 3.76N2 From the tables, Substance
C3H8 ( g g ) O2 N2 H2O ( g g ) CO CO2
h f
h 298 K
C3H8
kJ/kmol -103,850 0 0 -241,820 -110,530 -393,520
kJ/kmol --8682 8669 9904 8669 9364
25C
Combustion Products chamber T P Air
25C
Thus,
3 393,520 hCO 9364 4 241,820 hH O 9904 18.80 h N 8669 1 103,850 2
2
2
It yields 3hCO 2 4hH 2O 18.8h N 2 2,274,675 kJ The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing the right-hand side of the equation by the total number of moles, which yields 2,274,675 / (3 + 4 + 18.8) = 88,165 kJ/kmol. This enthalpy value value corresponds corresponds to about 2650 K for N2. Noting that the majority majority of the moles are N2, T P will be close close to 2650 K, but somewhat somewhat under under it because because of the higher higher specific specific heats of CO2 and H2O. At 2400 K:
3hCO 2 4hH 2O 18.8h N 2 3125,152 4 103,508 18.8 79,320 2,280,704 kJ Higher than 2,274,675 kJ
At 2350 K:
3hCO 2 4hH 2O 18.8h N 2 3122,091 4 100,846 18.8 77,496 2,226,582 kJ Lower than 2,274,675 kJ
By interpolation,
T P = 2394 K
(b) The balanced balanced combu combustion stion equation equation for complete complete combustion combustion with 300% theoretica theoreticall air air is is C3H8 g 15O2 3.76N2 3CO2 4H 2O 10O2 56.4 N 2 Substituting known numerical values,
3 393,520 hCO 9364 4 241,820 hH O 9904 100 hO 8682 56.40 h N 8669 1 103,850 2
2
2
which yields
2
15-72
3hCO 2 4hH 2O 10hO 2 56.4h N 2 2,687,450 kJ The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing the right-hand side of the equation by the total number of moles, moles, which yields 2,687,449 / (3 + 4 + 10 + 56.4) 56.4) = 36,614 36,614 kJ/kmol. kJ/kmol. This enthalp enthalpy y value corre correspond spondss to about about 1200 K for N2. Noting that the majority of the moles are N2, T P will be close close to 1200 1200 K, but somewhat somewhat under under it becaus becausee of the higher higher specific heats of CO2 and H2O. At 1160 K: 3hCO 2 4hH 2O 10hO 2 56.4h N 2 351,602 4 42,642 10 37,023 56.4 35,430 2,693,856 kJ Higher than 2,687,450 kJ At 1140 K: 3hCO 2 4hH 2O 10hO 2 56.4h N 2 350,484 4 41,780 10 36,314 56.4 34,760 2,642,176 kJ Lower than 2,687,450 kJ By interpolation,
T P = 1158 K
(c) The balanced balanced combustion combustion equation equation for incomplete incomplete combustion combustion with 95% theoretica theoreticall air is C3H8 g 4.75O 2 3.76N2 2.5CO2 0.5CO 4H 2O 17.86 N 2 Substituting known numerical values,
2.5 393,520 hCO 9364 0.5 110,530 hCO 8669 4 241,820 hH O 9904 17.86 0 h N 8669 1 103,850 2
2
2
which yields 2.5hCO 2 0.5hCO 4hH 2O 17.86h N 2 2,124,684 kJ The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing the right-hand side of the equation by the total number of moles, which yields 2,124,684 / (2.5 + 4 + 0.5 + 17.86) 17.86) = 85,466 85,466 kJ/kmol. kJ/kmol. This enthalp enthalpy y value corre correspond spondss to about about 2550 K for N2. Noting that the majority of the moles are N2, T P will be close close to 2550 2550 K, but somewhat somewhat under under it becaus becausee of the higher higher specific heats of CO2 and H2O. At 2350 K: 2.5hCO 2 0.5hCO 4hH 2O 17.86h N 2 2.5122,091 0.578,178 4 100,846 17.86 77,496 2,131,779 kJ Higher than 2,124,684 kJ At 2300 K: 2.5hCO 2 0.5hCO 4hH 2 O 17.86h N 2 2.5119,035 0.576,345 4 98,199 17.86 75,676 2,080,129 kJ Lower than 2,124,684 kJ By interpolation,
T P = 2343 K
15-73
15-96 The highest possible temperatures temperatures that can be obtained when liquid gasoline is burned steadily with air and with pure oxygen oxygen are to be determine determined. d. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic. highest possible possible temperatu temperature re that that can can be be achieved achieved during during a combustio combustion n process process is the the Analysis The highest temperatur temperaturee which occurs occurs when a fuel is burned burned completel completely y with stoichiom stoichiometric etric amount amount of air in an adiabatic adiabatic combus combustion tion chamb chamber. er. It is deter determined mined from
N h P
f
h h P
N h R
f
h h R
N h P
f
hT h P N h f C H 8
18
since all the reactants are at at the standard reference reference temperature of 25C, and for O2 and N2. The theoretical combustion equation of C8H18 air is C8 H18 12.5O 2 3.76N2 8CO2 9H 2O 47 N 2 From the tables, Substance
C8H18 () O2 N2 g ) H2O ( g CO2 Thus,
h f
h 298 K
C8H18
kJ/kmol -249,950 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
25C
Combustion chamber
Air
Products T P, max
25
8 393,520 hCO 9364 9 241,820 hH O 9904 470 h N 8669 1 249,950 2
2
2
It yields 8hCO 2 9hH 2O 47 h N 2 5,646,081 kJ The adiabatic adiabatic flame flame temperatur temperaturee is obtained obtained from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing dividing the right-han right-hand d side of the equation equation by the total number number of moles, moles, which yields yields 5,646,081/(8 5,646,081/(8 + 9 + 47) = 88,220 kJ/kmol. This enthalpy value corresponds to about 2650 K for N2. Noting that the majority majority of the moles are N2, T P will will be close close to to 2650 2650 K, but somewh somewhat at unde underr it becaus becausee of the the higher higher specif specific ic heat heat of H2O. At 2400 K:
8hCO 2 9hH 2O 47h N 2 8125,152 9 103,508 47 79,320 5,660,828 kJ Higher than 5,646,081 kJ
At 2350 K:
8hCO 2 9hH 2O 47h N 2 8122,091 9 100,846 47 77,496 5,526,654 kJ Lower than 5,646,081 kJ
By interpolation, T P = 2395 K If the fuel is burned burned with stoichiom stoichiometric etric amount amount of pure O2, the combustio combustion n equation equation would be
8CO 2 9H 2 O C 8 H 18 12.5O 2 Thus,
8 393,520 hCO 9364 9 241,820 hH O 9904 1 249,950 2
2
It yields 8hCO 2 9hH 2O 5,238,638 kJ The adiabatic adiabatic flame flame temperature temperature is obtaine obtained d from a trial trial and error solution solution.. A first guess guess is obtained obtained by dividing dividing the right-han right-hand d side of the equation equation by the total number number of moles, moles, which yields yields 5,238,638/ 5,238,638/(8 (8 + 9) = 308,155 308,155 kJ/kmol. kJ/kmol. This enthalp enthalpy y value value is higher higher than than the highest highest enthalpy enthalpy value value listed listed for H2O and and CO2. Thus an estimate estimate of the adiabatic adiabatic flame temperatu temperature re can be obtained obtained by extrapolation extrapolation.. At 3200 K:
8hCO 2 9hH 2O 8174,695 9 147,457 2,724,673 kJ
At 3250 K:
8hCO 2 9hH 2O 8177,822 9 150,272 2,775,024 kJ
By extrapolati extrapolation, on, we get T P = 3597 K . However However,, the soluti solution on of this this proble problem m using using EES gives gives 5645 K . The large difference between these these two values is due to extrapolation.
15-74
15-97E The work potent potential ial of diesel diesel fuel fuel at a given state is is to be determine determined. d. Assumptions 1 Steady operating conditions exist. 2 Air and combustion combustion gases gases are ideal ideal gases. gases. 3 Kineti Kineticc and potential energies are negligible. potentiall or availabili availability ty of a fuel at a specifi specified ed state is is the reversibl reversiblee work that would be Analysis The work potentia obtained if that fuel were burned completely with stoichiometric amount of air and the products are returned returned to the state of the surroundin surroundings. gs. It is determine determined d from
or,
N h N h
W rev
R
f
W rev
R
f
h h T 0 s R T 0 s R
N h P
N h P
f
f
h h T 0 s P
T 0 s P
since both the reactants and the products are at the state of the surroundings. surroundings. Considering 1 kmol kmol of C12H26, the theoretical combustion equation can can be written as C12H 26 ath O 2 3.76N 2 12CO2 13H 2O 3.76ath N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, a th 12 6.5 Substituting,
a th 18.5
C12H 26 18.5O 2 3.76N 2 12CO2 13H2O 69.56N 2
For each lbmol of fuel burned, 12 + 13 + 69.56 = 94.56 lbmol lbmol of products are formed, including 13 lbmol of H2O. Assumi Assuming ng that that the the dew-po dew-point int temp tempera eratur turee of the the produ products cts is above above 77F, some of the water will exist in the liquid form in the products. If N condenses, there there will be 13 - N - N w lbmol of water N w lbmol of H2O condenses, vapor left in the products. The mole number of the products products in the gas phase will also decrease to 94.56 (including the remaining water water vapor) as ideal gases, N gases, N w is N w as a result. Treating the product gases (including determine determined d by equating equating the mole fraction fraction of the water vapor vapor to pressure pressure fraction, fraction, N v N prod,gas
P v P prod
13 N w 94.56 N w
0.4648 psia 14.7 psia
N w 10.34 lbmol
since P v = P sat 0.4648 psia. Then the the combustio combustion n equation equation can be written written as sat @ 77F = 0.4648
12CO2 10.34H2O 2.66H2Og 69.56N2 C12H 26 18.5O2 3.76N2 The entropy entropy values values listed listed in the ideal ideal gas tables tables are for 1 atm pressure pressure.. Both the air air and and the the product product gases are at a total total press pressure ure of 1 atm, atm, but but the entrop entropies ies are to be calc calcula ulated ted at the the par partia tiall press pressure ure of the the component componentss which is equal equal to P i = y P , where is the mole fraction frac tion of component comp onent . Also, y i i total total i
S i N i si T , P i N i si T , P 0 Ru ln yi P m The entropy entropy calculations calculations can be presented presented in tabular form form as Ni
yi
s i 77 F,1atm
R u lny i Pm
si
h f ,Btu/lbmol
C12H26 O2 N2
1 18.5 69.56
--0.21 0.79
148.86 49.00 45.77
---3.10 -0.47
148.86 52.10 46.24
-125,190 0 0
CO2 H2O ( g g )
12 2.66 10.34
0.1425 0.0316 ---
51.07 45.11 16.71
-3.870 -6.861 ---
54.94 51.97 16.71
-169,300 -104,040 -122,970
69.56
0.8259
45.77
-0.380
46.15
0
H2O () N2 Substituting,
W rev 1125,190 537 148.86 18.50 537 52.10 69.56 0 537 46.24 12 169,300 537 54.94 2.66 104,040 537 51.97 10.34 122,970 537 16.71 69.56 0 537 46.15 3,375,000 Btu per lbmol C12 H 26
15-75
15-98 Liquid Liquid octa octane ne is burn burned ed with with 200 200 perce percent nt excess excess air during during a steady steady-fl -flow ow combus combustio tion n proce process. ss. The heat transfer rate from the combustion chamber, the power output of the turbine, and and the reversible reversible work and exergy destruction destruction are to be determined. Q
Assumptions 1 Combustion is complete. 2 Steady operating operating conditions conditions exist. exist. 3 Air and the combustion gases are ideal gases. 4 Changes Changes in kinetic kinetic and potential potential energies energies are negligible. negligible.
C8H18 Combustion chamber
25C, 8 atm
Properties The molar molar mass mass of of C8H18 is 114 114 kg/k kg/kmo moll (Table A-1).
Air =
200% excess air
Analysis (a) The fuel fuel is burned burned comple completel tely y with with the the excess air, and thus the products will contain only CO2, H2O, N2, and some free O2. Considering 1 kmol of C8H18, the combusti combustion on equation equation can can be written written as
1300 1300 K 8 atm
8CO2 9H 2O 2ath O 2 33.76ath N 2 C8 H18 3ath O 2 3.76N 2
Combustion gases
W
where ath is the stoichiometric coefficient and is determine determined d from the O2 balance, 950 K 2 atm
ath 12.5 3ath 8 4.5 2ath Substituting,
8CO2 9H 2O 25O 2 141N 2 C8 H18 37.5O 2 3.76N 2 The heat transfer for this combustion process is determined from the energy balance E in E out E system applied applied on the combustio combustion n chamber chamber with W = W = 0,
Qout
N h
P f
h h P
N h R
f
h h R
Assuming Assuming the air and and the combu combustion stion products products to be be ideal ideal gases, gases, we haveh have h = h( ). From From the the tabl tables, es, h(T ).
Substance
C8H18 () O2 N2 H2O ( g g ) CO2 Substituting,
h f
h 500 K
h 298 K
h1300 K
h 950 K
kJ/kmol -249,950 0 0 -241,820 -393,520
kJ/kmol --14,770 14,581 -----
kJ/kmol --8682 8669 9904 9364
kJ/kmol --42,033 40,170 48,807 59,552
kJ/kmol --26,652 28,501 33,841 40,070
Qout 8393,520 59,522 9364 9 241,820 48,807 9904 250 42,033 8682 1410 40,170 8669 1 249,950 h298 h298 37.50 14,770 8682 1410 14,581 8669 109,675 kJ/kmol C 8 H 18 The C8H18 is burned at a rate of 0.8 kg/min or N
Thus,
m
M
0.8 kg/min
812 181 kg/kmol
7.018 10 3 kmol/min
3 Q kmol/min 109,675 kJ/kmol 770 kJ/min Q out N out 7.018 10
15-76
(b) Noting that no chemical chemical reactions occur in the turbine, the turbine is adiabatic, and the product gases gases can be treated as ideal gases, the power output of the turbine can be determined from the steady-flow energy energy balance balance equation equation for for nonreactin nonreacting g gas mixtures, mixtures,
W out
N h P
e
hi W out
N h P
1300 K h950 K
Substituting, W out 859,522 40,070 9 48,807 33,841 25 42,033 29,652 14140,170 28,501 2,245,164 kJ/kmol C 8 H 18 Thus Thus the power power output output of the the turbin turbinee is
3 kmol/min 2,245,164 kJ/kmol 15,756 kJ/min 262.6 kW W out N W out 7.018 10
(c) The entrop entropy y generatio generation n during during this this process process is is determ determined ined from S gen S P S R
Qout T surr
N s N s P P
R R
Qout T surr
where the entropy entropy of the products are to be evaluated at the turbine exit state. The C8H18 is at 25C and and 8 atm, and thus its absolute entropy is s C8 H18 =360.79 =360.79 kJ/kmol·K kJ/kmol·K (Table (Table A-26). A-26). The entropy entropy values values listed listed in the ideal ideal gas tables tables are for for 1 atm pressur pressure. e. The entropie entropiess are to be calculated calculated at at the partial partial pressur pressuree of the component componentss which is equal equal to P i = y P fraction of component component i. Also, i total total, where yi is the mole fraction S i N i si T , P i N i si T , P 0 Ru ln yi P m The entropy entropy calculations calculations can be presented presented in tabular tabular form as Ni
yi
s i T,1 atm
C8H18 O2 N2
1 37.5 141
1.00 0.21 0.79
360.79 220.589 206.630
CO2 H2O O2 N2
8 9 25 141
0.0437 0.0490 0.1366 0.7705
266.444 230.499 241.689 226.389
R u lny i Pm
N i si
17.288 343.50 4.313 8,110.34 15.329 26,973.44 S R = 35,4 35,427 27.2 .28 8 kJ/K kJ/K -20.260 2,293.63 -19.281 2,248.02 -10.787 6,311.90 3.595 31,413.93 42,267 67.4 .48 8 kJ/K kJ/K S P = 42,2
Thus, S gen 42,267.48 35,427.28
109,675 kJ 298 K
7208.2 kJ/K per kmol
Then the rate rate of entropy entropy generation generation becomes becomes
3 S gen N S gen 7.018 10 kmol/min 7208.2 kJ/kmol K 50.59 kJ/min K
and X destruction T 0 S gen 298 K 50.59 kJ/min K 15,076 kJ/min 251.3 kW W rev W X destruction 262.6 251.3 513.9 kW
15-77
15-99 Methyl Methyl alcohol alcohol vapor is burned burned with the the stoichio stoichiometr metric ic amount amount of air in a combustio combustion n chamber. chamber. The maximum pressure pressure that can occur in the combustion chamber if the combustion takes place at constant volume and the maximum volume of the combustion chamber chamber if the combustion occurs at constant pressure are to be determined.
combustion n gases are are ideal gases. gases. 4 Changes in Assumptions 1 Combustion is complete. 2 Air and the combustio kinetic kinetic and potential potential energies energies are negligible. negligible. Analysis (a) The The combus combustion tion equation equation of CH CH3OH(g) with stoichiometric amount of air is
CO 2 2H 2O 3.76ath N 2 CH3OH ath O 2 3.76N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, 1 2a th 2 2
ath 1.5
CH3OH( g g ) AIR 25C,98 kPa
Thus, CH3OH 1.5O 2 3.76N 2 CO 2 2H 2O 5.64N 2 The final temperature in the tank is determined from the energy balance relation E in E out E system for W = 0), reacting closed systems under adiabatic conditions (Q (Q = 0) with no work work interac interactions tions (W = 0
N h P
f
h h P v P
N h R
f
h h P v R
Assuming both the reactants and the products to behave as ideal ideal gases, all the internal internal energy and and enthalpies enthalpies depend depend on tempe temperatur raturee only, only, and and the the P v terms in this equation can be replaced by RuT . It yields
N h P
f
hT P h298 K Ru T P
N h R
f
Ru T R
since the reactants are at the standard reference reference temperature of 25C. From From the the tabl tables es,,
Substance
CH3OH O2 N2 H2O ( g g ) CO2
h f
h 298 K
kJ/kmol -200,670 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
Thus,
1 393,520 hCO 9364 8.314 T P 2 241,820 hH O 9904 8.314 T P 5.640 h N 8669 8.314 T P 1 200,670 8.314 298 1.50 8.314 298 5.640 8.314 298 2
2
2
It yields hCO2 2hH 2 O 5.64 hN 2 71.833 T P 7 34, 388 kJ The temperature of the product gases is obtained from a trial and error solution, At 2850 K: hCO 2 2hH 2O 5.64h N 2 71.833 T P 1152,908 2 127,952 5.64 95,859 71.8332850 744,733 kJ Higher than 734,388 kJ
15-78
At 2800 K: hCO 2 2hH 2O 5.64h N 2 71.833 T P 1149,808 2 125,198 5.64 94,014 71.833 2800 729,311 kJ Lower than 734,388 kJ By interpolation
T P = 2816 K
Since both the reactants reactants and the products products behave behave as ideal gases, gases, the the final final (maximum (maximum)) pressure pressure that can occur in the combustion chamber is determined to be P 1V P 2V
N 1 Ru T 1 N 2 Ru T 2
P 2
N 2 T 2
P 1
N 1T 1
8.64 kmol2816 K 98 kPa 983 8.14 kmol298 K
kPa
(b) The combustion combustion equation of CH3OH( g remains the same in the case of constant pressure. Further, the g ) remains boundary work in this case can be combined with the u terms so that the first law relation can be expressed in terms of enthalpies enthalpies just like the steady-fl steady-flow ow process, process, Q
N h P
f
h h P
N h R
f
h h R
Since both the reactants and the products behave as ideal gases, we have have h = h(T ). ) . Also Also not notin ing g that that Q = 0 for an an adiabatic adiabatic combustio combustion n process, process, the the 1st law relation relation reduces reduces to
N h P
f
hT P h298 K P
N h R
f R
since the reactants are at the standard reference temperature of 25 C. Then using data from from the mini table above, we get
1 393,520 hCO 9364 2 241,820 hH O 9904 5.640 h N 8669 1 200,670 1.5 0 5.64 0 2
2
2
It yields hCO2 2hH 2O 5.64h N 2 754,555 kJ The temperature of the product gases is obtained from a trial and error solution, At 2350 K:
hCO 2 2hH 2O 5.64h N 2 1122,091 2 100,846 5.64 77,496 760,860 kJ Higher than 754,555 kJ
At 2300 K:
hCO 2 2hH 2O 5.64h N 2 1119,035 2 98,199 5.64 75,676 742,246 kJ Lower than 754,555 kJ T P = 2333 K
By interpolation,
Treating both the reactants and the products as ideal gases, gases, the final (maximum) (maximum) volume that the combustion combustion chambe chamberr can have is determine determined d to be P V 1 P V 2
N 1 Ru T 1 N 2 Ru T 2
V 2
N 2T 2 N 1T 1
V 1
8.64 kmol2333 K 0.8 L 6.65 L 8.14 kmol298 K
15-79
15-100 Problem 15–99 is reconsidered. The effect effect of the initial volume of the combustion chamber on the maximum pressure pressure of the chamber chamber for constant volume combustion combustion or the maximum volume of the chamber for constant pressure combustion is to be investigated. Analysis The problem problem is solved solved using EES, EES, and the solution solution is given given below.
"Input Data" T_reac = (25+273) "[K]" "reacta "reactant nt mixtur mixture e temperat temperature" ure" P_reac = 98 [kPa] "reactant mixture pressure" {V_chamber_1 = 0.8 [L]} h_CH3OH = -200670 [kJ/kmol] Mw_O2 = 32 [kg/kmol] Mw_N2 = 28 [kg/kmol] Mw_CH3OH=(3*12+1*32+4*1) "[kg/kmol]" R_u = 8.314 [kJ/kmol-K] "universal gas constant" "For theoretical oxygen, the complete combustion equation is" "CH3OH "CH 3OH + A_th A_th O2= O2=1 1 CO2+ CO2+2 2 H2O " 1+ 2*A_th=1*2+2*1"theoretical 2*A_th=1*2+2*1"theoretical O balance" "The balanced complete combustion equation with theroetical air is" "CH3OH + A_th A_th (O2+3.7 (O2+3.76 6 N2)=1 N2)=1 CO2+ CO2+ 2 H2O + A_th(3 A_th(3.76) .76) N2 " "now to find the actual moles of reactants and products per mole of fuel" N_Reac = 1 + A_th*4.76 N_Prod=1+2+A_th*3.76 "Apply First Law to the closed system combustion chamber and assume ideal gas behavior. Assume the water formed in the combustion process exists in the gas phase." "The following is the constant volume, adiabatic solution:" E_in - E_out = DELTAE_sys E_in = 0 "No heat transfer for the adiabatic combustion combustion process" process" E_out = 0"kJ/kmol_CH3OH" 0"kJ/kmol_CH3OH" "No work is done because volume is constant" DELTAE_sys DELTAE_sys = U_prod U_prod - U_reac "neglect KE and PE and note: U = H - PV = N(h - R_u T)" U_reac = 1*(h_CH3OH - R_u*T_reac) +A_th*(enthalpy(O2,T=T_reac) R_u*T_reac)+A_th*3.76*(enthalpy(N2,T=T_reac) - R_u*T_reac) U_prod = 1*(enthalpy(CO2, 1*(enthalpy(CO2, T=T_prod) - R_u*T_prod) R_u*T_prod) +2*(enthalpy(H2O, +2*(enthalpy(H2O, T=T_prod) T=T_prod) R_u*T_prod)+A_th*3.76*(enthalpy(N2,T=T_prod) - R_u*T_prod) V_chamber_2 = V_chamber_1 "The final final pressure pressure and volume of the tank are are those of the product gases. Assumi Assuming ng ideal gas behavior for the gases in the constant volume tank, the ideal gas law gives:" P_reac*V_chamber_1*convert(L,m^3) =N_reac*N_fuel* R_u *T_reac P_prod*V_cha P_prod*V_chamber_2* mber_2*convert convert(L,m^3) (L,m^3) = N_prod*N_fuel N_prod*N_fuel** R_u * T_prod { "The following is the constant pressure, adiabatic solution:" P_prod = P_Reac H_reac=H_prod H_reac = 1*h_CH3OH +A_th*enthalpy(O2,T=T_reac) + A_th*enthalpy(O2,T=T_reac) +A_th*3.76*enthalpy(N2,T=T_reac) H_prod = 1*enthalpy(CO2,T=T_prod)+2*enthalpy(H2O,T=T_prod) +A_th*3.76*enthalpy(N2,T=T_prod) }
15-80
Nfuel [kmol] 4.859E-07 0.000001512 0.000002538 0.000003564 0.000004589 0.000005615 0.000006641 0.000007667 0.000008693 0.000009719
Pprod [kPa] 983.5 983.5 983.5 983.5 983.5 983.5 983.5 983.5 983.5 983.5
Tprod [K] 2817 2817 2817 2817 2817 2817 2817 2817 2817 2817
Vchamber,1 [L] 0 .1 0.3111 0.5222 0.7333 0.9444 1.156 1.367 1.578 1.789 2
Constant pressure combustion of CH3OH 18
Preac = Pprod = 98 kPa
16 14
Adiabatic Tprod = 2334 K
12
2, r e b m a h c
10 8 6 4 2 0 0
0.4
0.8
1.2
1.6
2
Vchamber,1 [L]
Constant volume combustion of CH3OH 1500
Preac = 98 kPa 1300
Adiabatic Tprod = 2817 K 1100
d o r p
900 700 500 0
0.4
0.8
1.2
Vchamber,1 [L]
1.6
2
15-81
15-101 Methane is burned burned with the stoichiometric stoichiometric amount of air in a combustion chamber. chamber. The maximum maximum pressure that can occur in the combustion chamber if the combustion combustion takes place at constant volume and the maximum maximum volume volume of the combusti combustion on chamber chamber if the combustion combustion occurs occurs at constant constant pressure pressure are to be determined.
combustion n gases are are ideal gases. gases. 4 Changes in Assumptions 1 Combustion is complete. 2 Air and the combustio kinetic kinetic and potential potential energies energies are negligible. negligible. The combus combustion tion equation equation of CH CH4( g ) with stoichio stoichiometr metric ic amount amount of air is Analysis (a) The CH 4 ath O 2 3.76N2 CO2 2H 2O 3.76ath N 2 where ath is the stoichiometric coefficient coefficient and is determined from the O2 balance, a th 1 1
g ) CH4 ( g AIR 25C,98 kPa
a th 2
Thus,
CO2 2H 2O 7.52N2 CH 4 2O2 3.76N 2 The final temperature in the tank is determined from the energy balance relation Ein Eout Esyst em for reacting closed systems under adiabatic conditions (Q (Q = 0) with no work work interac interactions tions (W = W = 0), 0
N h
P f
h h P v P
N h R
h h P v R
f
Since both the reactants and the products behave behave as ideal gases, gases, all the internal energy and enthalpies enthalpies depend on temperature only, and the P v terms terms in this equation equation can be replaced replaced by RuT . It yie yield ldss
N h
P f
hT P h298 K RuT P
N h R
f
RuT R
since the reactants are at the standard reference reference temperature of 25C. From From the the tabl tables es,,
Substance
CH4 O2 N2 g ) H2O ( g CO2
h f
h 298 K
kJ/kmol -74,850 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
Thus,
1 393,520 hCO 9364 8.314 T P 2 241,820 hH O 9904 8.314 T P 7.520 h N 8669 8.314 T P 1 74,850 8.314 298 2 0 8.314 298 7.52 0 8.314 298 2
2
2
It yields hCO2 2hH 2O 7.52h N 2 87.463 T P 870,609 kJ The temperature of the product gases is obtained from a trial and error solution, At 2850 K: hCO 2 2hH 2 O 7.52h N 2 87.463 T P 1152,908 2 127,952 7.52 95,859 87.463 2850 880,402 kJ Higher than 870,609 kJ At 2800 K: hCO 2 2hH 2 O 7.52h N 2 87.463 T P 1149,808 2 125,198 7.52 94,014 87.463 2800 862,293 kJ Lower than 870,609 kJ By interpolation,
T P = 2823 K
15-82
Treating both the reactants and the products as ideal gases, the final final (maximum) (maximum) pressure that can occur in the combustion chamber is determined to be P 1V P 2V
N 1 Ru T 1 N 2 Ru T 2
P 2
N 2T 2
P 1
N 1T 1
10.52 kmol2823 K 98 kPa 928 10.52 kmol 298 K
kPa
(b) The combustion equation of CH4( g case of constant pressure. pressure. Further, the g ) remains the same in the case boundary work in this case can be combined with the u terms so that the first law relation can be expressed in terms of enthalpies enthalpies just like the steady-fl steady-flow ow process, process, Q
N h P
f
h h
P
N h R
f
h h
R
Again since both the reactants and the products behave as ideal ideal gases, we haveh have h = h(T ). ). Also noting noting that that Q = 0 for an adiabatic combustion combustion process, the energy balance relation reduces to
N h P
f
hT P h298 K P
N h R
f R
since the reactants are at the standard reference temperature of 25 C. Then using data from from the mini table above, we get
1 393,520 hCO 9364 2 241,820 hH O 9904 7.520 h N 8669 1 74,850 2 0 7.52 0 2
2
2
It yields hCO2 2hH 2O 7.52h N 2 896,673 kJ The temperature of the product gases is obtained from a trial and error solution, At 2350 K:
hCO 2 2hH 2O 7.52h N 2 1122,091 2 100,846 7.52 77,496 906,553 kJ Higher than 896,673 kJ
At 2300 K:
hCO 2 2hH 2O 7.52h N 2 1119,035 2 98,199 7.52 75,676 884,517 kJ Lower than 896,673 kJ
By interpolation,
T P = 2328 K
Treating both the reactants and the products as ideal gases, gases, the final (maximum) (maximum) volume that the combustion combustion chambe chamberr can have is determine determined d to be P V 1 P V 2
N 1 Ru T 1 N 2 Ru T 2
V 2
N 2T 2 N 1T 1
V 1
10.52 kmol2328 K 0.8 L 6.25 10.52 kmol298 K
L
