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Thermochemistry of fuel air mixtures

Dr. Primal Fernandoprimal@eng.fsu.edu Ph: (081) 2393608

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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 22

1

222

1COOCO

If insufficient O2 presents

OHOH 222 2

1

OkmolHOkmolHkmol 222 12

11

OHkgOkgHkg 222 18162

)(0

)(12

11

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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Solution

OHCOOHC 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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Solution

Wet volumetric analysis of the Products including N2

OHCONOHC 2222146 765.921

795.9

OHCONOHC 2222146 765.9

21

795.92.1

222222146 5.921

792.15.92.0765.9

21

795.92.1 NOOHCONOHC

Amount-of-substance in the product

2222 5.921

792.15.92.076 NOOHCO

Volkmol 79.57or79.575.921

792.15.92.076

222222146 5.921

792.15.92.0765.9

21

792.15.92.11 NVolOVolOHVolCOVolNVolOVolHCVol

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Solution

Amount-of-substance in the product

2222 5.921

792.15.92.076 NOOHCO

Volkmol 79.57or79.575.921

792.15.92.076

222222146 5.921

792.15.92.0765.9

21

792.15.92.11 NVolOVolOHVolCOVolNVolOVolHCVol

The wet volumetric analysis

%38.101005.57

62 CO %11.12100

79.57

72 OH

%29.310079.57

5.92.02

O

%22.7410079.57

5.921

792.1

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

y

y

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 5.921

792.15.92.0765.9

21

795.92.1 NOOHCONOHC

22222146 5.921

798.075.9

21

795.98.0 NOHbCOaCONOHC

ba 6

7225.98.0 ba 8.32.2 banda

22222146 5.921

798.078.32.25.9

21

795.98.0 NOHCOCONOHC

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General combustion stoichiometry

22222146 5.921

79765.9

21

795.9 NOHCONOHC

22222146 5.921

7976

21

795.9 NOHCONOHC

22222 ............95.20

05.79.....................

95.20

05.79.......... NOHCONOHC ba

22222 ............773.3.....................773.3.......... NOHCONOHC ba

22222 4773.3

2773.3

4N

baOH

baCONO

baHC ba

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General combustion stoichiometry

22222 4773.3

2773.3

4N

baOH

baCONO

baHC ba

bas HC

NOb

a

F

A 22 773.34

ba

ba

F

A

s

008.1011.12

16.28773.3324

Take, y=b/a = ratio of H2 to C

y

y

F

A

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 productsTP, PP, VP, UP

QR-P

WR-P

RPPRPR UUWQ )( R

P

R

PPR VVPPdVW

pressureconstant at reaction ofheat ,TP

RPRRPPRPRPPR

H

HHUPVUPVUUPVPVQ