Thermochemistry of fuel air mixtures
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Transcript of Thermochemistry of fuel air mixtures
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Combustion process
• Thermodynamic aspects of particular type process involving chemical reactions, is called combustion
• Usually occurs between fuel and an oxygen carrier (air)
• Energy stored in the bonds between constituent atoms of fuel and air (form of internal energy) and in the combustion process it will transformed to new molecules of lower energy level combustion products plus release heat (exothermic reaction)
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Combustion process
• Controls the engine power• Efficiency• Controls the emissions• Different for SI and CI engines
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Flames
• A flame is a combustion reaction propagate subsonically through space; motion of gas relative to unburn gas is important.
• The existence of flame motion implies that the reaction is confined to a zone which is small in thickness compared to the engine combustion chamber.
• The reaction zone is usually called the flame front• Flames can be categorized as premixed and diffusion flame (mixed
together at same place where the reaction takes place)• Flames also categorized as laminar (mixing and transport done by
molecular process) and turbulent (enhanced by eddies and lumps)• Flames also categorized by whether the flow is steady or unsteady
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Fuel - fossil fuels mainly consists of H and C
Motor Petrol
Vaporizing oil
Kerosene Diesel oil (gas oil)
Light Fuel oil
Heavy Fuel oil
C H S
85.5 14.4 0.1
86.8 12.9 0.3
86.3 13.6 0.1
86.3 12.8 0.9
86.2 12.4 1.4
86.1 11.8 2.1
H2 CO CH4 CnHm O2 N2 CO2 Coal gas
Producer gas Blast furnace North Sea gas
49.4 12.0 2.0 -
18.0 29.0 27.0
-
20.0 2.6 -
93.0
C4H8 2.0 C2H4 0.4
- C2H6 4.8
0.4 - - -
6.2 52.0 60.0 2.0
4.0 4.0 11.0 0.2
By volume
By mass
By massDry Coal C H O N+S Ash
Anthracite Bituminous
Lignite
90.27 74.00 56.52
3.00 5.98 5.72
2.32 13.01 31.89
1.44 2.26 1.62
2.97 4.75 4.25
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Chemical equation and conservation of mass
Consider a simple equation22 COOC
22 111 COkmolOkmolCkmol
22 443212 COkgOkgCkg
22 110 COVolOVolCVol
Note: all gasses occupy equal volume for kmol when they are at same pressure and temperature (exactly true for perfect gases, but for other gasses substantially true). Volume occupied by liquid and solids are negligibly small compared to gasses.
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Chemical equation and conservation of mass
22 COOC
22 111 COkmolOkmolCkmol
22 443212 COkgOkgCkg
22 110 COVolOVolCVol COOC 221
2221 COOCO
If insufficient O2 presents
OHOH 222 21
OkmolHOkmolHkmol 222 1211
OHkgOkgHkg 222 18162
)(0
)(1211
2
222
liquidOHVol
VapourOHVolOVolHVol
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Necessary Oxygen is mainly obtained by mixing fuel with are
Note: Molar mass of N2 is 28 kg/kmol (28.16), and that for air 29 kg/kmol (28.962)
O2 N2 Volumetric analysis % 21
(20.95) 79
(78.09) Gravimetric analysis % 23.3 76.7
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ExampleDetermine the stoichiometric air/fuel ratio for a petrol approximating to hexane C6H14. Hence deduce the chemical equation if the petrol is burnt in 20 percent excess air, and the wet volumetric analysis of the products• If all the water vapor is present•If products are cooled to an atmospheric pressure and temperature of 1 bar and 15 °C. •Determine also the dry volumetric analysis. •Estimate the chemical equation if only 80% of the air required for stoichiometric combustion is provided
The partial pressure of saturated water vapor at 15 °C is 0.01704 bar
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SolutionOHCOOHC 222146 Products
OHCOOHC 222146 762
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OHCOOHC 222146 765.9
OHkgCOkgOkgHCkg 222146 12626430486
airof100incontain of3.23 2 kgOkg
airof3043.23
100incontain of320 2 kgOkg
17.1586
3043.23
100
ratio air/fuel ricstochiomet
kg
kg
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SolutionWet volumetric analysis of the Products including N2
OHCONOHC 2222146 765.921795.9
OHCONOHC 2222146 765.9
21795.92.1
222222146 5.921792.15.92.0765.9
21795.92.1 NOOHCONOHC
Amount-of-substance in the product
2222 5.921792.15.92.076 NOOHCO
Volkmol 79.57or79.575.921792.15.92.076
222222146 5.921792.15.92.0765.9
21792.15.92.11 NVolOVolOHVolCOVolNVolOVolHCVol
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Solution
Amount-of-substance in the product
2222 5.921792.15.92.076 NOOHCO
Volkmol 79.57or79.575.921792.15.92.076
222222146 5.921792.15.92.0765.9
21792.15.92.11 NVolOVolOHVolCOVolNVolOVolHCVol
The wet volumetric analysis
%38.101005.57
62 CO %11.12100
79.577
2 OH
%29.310079.57
5.92.02
O%22.74100
79.57
5.921792.1
2
N
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Solution
Volume fraction=Mole fraction=partial pressure/total pressure
If products are cooled to an atmospheric pressure and temperature of 1 bar and 15 °C.
Part of water will be condensed, if new amount of water is y, total substance is = 50.79 + y
yy
79.501
01704.0fraction mole
kmoly 88.0
67.5188.079.5079.50 volumetotal y
Above volumetric analysis repeats base on the total volume of 51.67
%01.83%;68.3%;7.1%;61.11 2222 NOOHCO
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SolutionDetermine also the dry volumetric analysis
Analysis is done by assuming no water present, then the amount of substance becomes = 50.79
Above volumetric analysis repeat base on total volume of 50.79
%45.84%;74.3%;81.11 222 NOCO
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Solution
When insufficient O2 is given there will be unburned C and H2. H2, However, has a greater affinity for O2. If mixture is not too rich in fuel, it is reasonable to assume that all the H2 will be burnt. Some of C will be burnt to CO and other to CO2.
Chemical reaction with insufficient air (80%)
222222146 5.921792.15.92.0765.9
21795.92.1 NOOHCONOHC
22222146 5.921798.075.9
21795.98.0 NOHbCOaCONOHC
ba 67225.98.0 ba 8.32.2 banda
22222146 5.921798.078.32.25.9
21795.98.0 NOHCOCONOHC
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General combustion stoichiometry
22222146 5.92179765.9
21795.9 NOHCONOHC
22222146 5.9217976
21795.9 NOHCONOHC
22222 ............95.2005.79.....................
95.2005.79.......... NOHCONOHC ba
22222 ............773.3.....................773.3.......... NOHCONOHC ba
22222 4773.3
2773.3
4NbaOHbaCONObaHC ba
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General combustion stoichiometry
22222 4773.3
2773.3
4NbaOHbaCONObaHC ba
bas HC
NOba
FA 22 773.3
4
ba
ba
FA
s
008.1011.12
16.28773.3324
Take, y=b/a = ratio of H2 to C
yy
FA
s 008.1011.12)4(56.34
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Energy balance
Systems changes from reactants to products (since mass constant, can apply first law for a close system)
Initial state reactantsTR, PR, VR, UR
Combustion process; heat and work transfer interaction Final state products
TP, PP, VP, UP
QR-P
WR-P
RPPRPR UUWQ )( R
P
RPPR VVPPdVW
pressureconstant at reaction ofheat ,TP
RPRRPPRPRPPR
HHHUPVUPVUUPVPVQ