Combustion Engineering ME Dept NCHU 頁 1

Chapter 2 Basic Combustion Chemistry

Update: 2018/2/9

(2.1). Combustion Reaction

(2.1.1). Complete reaction of hydrocarbon fuel

The most common fuel that is used for power generation currently is a

chemical compound that contains carbon, hydrogen, and oxygen only,

and is called the hydrocarbon fuel.

The general expression of hydrocarbon fuel is CnHmOy.

Fuel is burned with sufficient amount of air such that all the carbon in the

fuel oxides to carbon dioxide, and all the hydrogen in the fuel oxides to

water. Besides, neither fuel nor oxygen remains.

Composition of air:

N2: 78.08%

O2: 20.95%

Argon: 0.93%

CO2: 0.03%

Others: 0.01%

Water vapor: variable

Average molecular weight: 28.97 kg/kmole

For simplicity, air is considered as a mixture that contains 79% N2 and

21% O2 only. Other components are neglected.

0.21O2 + 0.79N2 = 0.21(O2 + 3.76N2)

It is noted that the average molecular weight of the simplified air is 28.84

kg/kmole. However, the actual value of 28.97 kg/kmole is still used in

the calculation of combustion for accuracy.

The combustion formulas of several fuels with air for industry usage are

as following.

Methane(甲烷):

CH4 + 2(O2 + 3.76N2) → CO2 + 2H2O + 7.52N2

Combustion Engineering ME Dept NCHU 頁 2

Ethane(乙烷):

C2H6 + 3.5(O2 + 3.76N2) → 2CO2 + 3H2O + 13.16N2

Ethene(乙烯):

C2H4 + 3(O2 + 3.76N2) → 2CO2 + 2H2O + 11.28N2

Acetylene(乙炔 Ethyne):

C2H2 + 2.5(O2 + 3.76N2) → 2CO2 + H2O + 9.4N2

Propane(丙烷):

C3H8 + 5(O2 + 3.76N2) → 3CO2 + 4H2O + 18.8N2

Propene(丙烯):

C3H6 + 4.5(O2 + 3.76N2) → 3CO2 + 3H2O + 16.92N2

Butane(丁烷):

C4H10 + 6.5(O2 + 3.76N2) → 4CO2 + 5H2O + 24.44N2

Butene(丁烯):

C4H8 + 6(O2 + 3.76N2) → 4CO2 + 5H2O + 22.56N2

Benzene(苯):

C6H6 + 7.5(O2 + 3.76N2) → 6CO2 + 3H2O + 28.2N2

Coal:

C + O2 + 3.76N2 → CO2 + 3.76N2

Hydrogen:

H2 + 0.5(O2 + 3.76N2) →H2O + 1.88N2

In general, we have

CnHm + (n+m/4)(O2 + 3.76N2) → nCO2 + m/2H2O + 3.76(n+m/4)N2

If oxygen is contained in the fuel, the amount of air required to oxide the

fuel is reduced due to the presence of oxygen in the air fuel mixture.

Methanol(甲醇):

CH3OH + 1.5(O2 + 3.76N2) → CO2 + 2H2O + 5.64N2

Ethanol(乙醇):

C2H5OH + 3(O2 + 3.76N2) → 2CO2 + 3H2O + 11.28N2

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Propanol(丙醇):

C3H7OH + 4.5(O2 + 3.76N2) → 3CO2 + 4H2O + 16.92N2

Butanol(丁醇):

C4H9OH + 6(O2 + 3.76N2) → 4CO2 + 5H2O + 22.56N2

Dimethyl Ether( DME二甲醚):

CH3OCH3+ 3(O2 + 3.76N2) → 2CO2 + 3H2O + 11.28N2

Ethyl Ether(二乙醚):

C2H5OC2H5+ 6(O2 + 3.76N2) → 4CO2 + 5H2O + 22.56N2

In general, we have

2 2 2 2 2( )( 3.76 ) 3.76( )4 2 2 4 2

n m y

m y m m yC H O n O N CO H O n N

(2.1.2). Air fuel ratio

Molar ratio A/F：the molar ratio of air to fuel

Mass ratio A/F：the mass ratio of air to fuel

CH4 + 2(O2 + 3.76N2) → CO2 + 2H2O + 7.52N2

Molar A/F = 2*4.76/1 = 9.52

Mass A/F = 2*4.76*28.97/16 = 17.24

Gasoline and diesel fuels are mixtures of hydrocarbons rather than pure

substance. They don’t have chemical formula and fixed values of

molecular weight. In general, gasoline and diesel fuels are represented as

CH1.85

CH1.85 + 1.4625(O2 + 3.76N2) → CO2 + 0.925H2O + 5.499N2

A/F = 1.4625*4.76*28.97/13.85 = 14.56

Stoichiometric combustion:

The air fuel ratio for complete combustion is called stoichiometric air fuel

ratio.

e.g.：automotive gasoline engine at cruise condition.

Combustion Engineering ME Dept NCHU 頁 4

The reaction temperature is high at stoichiometric ratio which would

result in a high concentration of NOx. In most combustion appliance,

stoichiometric combustion is avoided to prevent overheat.

Lean combustion:

If the amount of air is more than that required for complete combustion,

there will be excess air after combustion. This is call lean combustion.

CH4 + 3(O2 + 3.76N2) → CO2 + 2H2O + O2 + 11.28N2

In a lean combustion, the air fuel ratio is greater than the stoichiometric

air fuel ratio.

A/F > (A/F)st

e.g.：Diesel engine, burner, steam boiler, and gas turbine combustor.

Lean combustion is the normal condition for most combustion appliance.

Rich combustion：

If the amount of air is less than that required for complete combustion,

part of the fuel would not be fully oxidized such that either there will be

excess fuel left after combustion or there will be incomplete combustion

products, such as carbon monoxide or unburned hydrocarbon.

CH4 + 1.5(O2 + 3.76N2) → CO2，H2O，O2，CO，HC，H2，N2

In a rich combustion, the air fuel ratio is less than the stoichiometric air

fuel ratio.

A/F < (A/F)st

e.g.：Moped gasoline engine at full load.

In most combustion appliance, rich combustion is avoided because of low

efficiency and high pollution.

Equivalence ratio：the ratio of stoichiometric air fuel ratio to actual air

fuel ratio, usually used in academic research.

/

/

stA F

A F

Combustion Engineering ME Dept NCHU 頁 5

1 ：lean combustion

1 ：stoichiometric combustion

1 ：rich combustion

Fuel air ratio：the reciprocal of air fuel ratio, usually used by Diesel

engine and gas turbine engine people.

Percentage of theoretical air：the ratio of actual air to stoichiometric air,

usually used by boiler and burner people.

% of theoretical air = 100%

Percentage of excess air：Percentage of theoretical air – 100%, usually

used by boiler and burner people.

% of excess air = 1

100%

Lambda(λ)：the reciprocal of equivalence ratio, usually used in

automotive engine control.

λ>1：lean combustion

λ=1：stoichiometric combustion

λ<1：rich combustion

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Example：The molar analysis of producer gas is N2 50.9%, CO 27.0%,

H2 14.0%, CO2 4.5%, CH4 3.0%, and O2 0.6%. Determine the

stoichiometric air fuel ratio of producer gas. ((A/F)st = 1.443)

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Example：Methane is burned with dry air. The molar analysis of the

products on a dry basis is CO2 9.7%，CO 0.5%，O2 2.95%，and N2

86.85%。Determine the air fuel ratio, the equivalence ratio, the

percentage of theoretical air, and the percentage of excess air.

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Example：The molar analysis of the products of coal on a dry basis is

CO2 5%。Determine the percentage of excess air.

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Example：E50 is a bio fuel which is blended 50% (in volume) of ethanol

with 50% of gasoline. Suppose that the density of gasoline is 750 kg/m3,

and that of ethanol is 800 kg/m3, and gasoline can be expressed as CH1.85,

Combustion Engineering ME Dept NCHU 頁 6

calculate the stoichiometric air fuel ratio o E50.

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Assignment 2.1

M85 is a bio fuel that was promoted about twenty years ago. It contains

85% (in volume) of methanol and 15% of gasoline. The density of

methanol is 790 g/L, and that of gasoline is 750 g/L. Suppose gasoline

can be expressed as CH1.85, calculate the stoichiometric air fuel ratio of

M85.

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Assignment 2.2

Coal is burned with dry air with the feeding rate of 10 tons/hr. The

molar analysis of the products of combustion on a dry basis is CO2 10%.

Determine the flow rate of air.

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Assignment 2.3

Hydrogen is burned with dry air with the feeding rate of 10 m3/min.

The molar analysis of the products of combustion on a dry basis is O2

10%. Determine the flow rate of air.

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Combustion Engineering ME Dept NCHU 頁 7

Lambda sensors produces a voltage signal that recognises the amount of

unburned oxygen in the exhaust. An oxygen sensor is essentially a

battery that generates its own voltage. When hot (at least 250℃.), the

zirconium dioxide element in the sensor's tip produces a voltage that

varies according to the amount of oxygen in the exhaust compared to

the ambient oxygen level in the outside air. The greater the difference,

the higher the sensor's output voltage.

Sensor output ranges from 0.2 Volts (lean) to 0.8 Volts (rich). A

stoichiometric fuel mixture gives an average reading of around 0.45

Volts.

Lambda sensor output and the conversion efficiency of catalytic

converter：

Combustion Engineering ME Dept NCHU 頁 8

(2.2). Energy balance in a reacting system

(2.2.1). Energy equation

In a closed system, the conservation of energy and mass give the

following equations.

f a pm m m m

p p f f a aQ U W m u m u m u W

e.g.: internal combustion engine.

Q

W

In an open system,

f a pm m m

f f a a p pQ m h m h m h W

e.g.: gas turbine engine, steam boiler.

Q

f am m pm

Combustion Engineering ME Dept NCHU 頁 9

(2.2.2). Enthalpy of formation

For the following exothermic reaction, an amount of heat of 393520 kJ

may be released out if all the reactants and products are kept at 25℃ and

100 kPa.

C + O2→ CO2

Since the process does not produce any output work, the energy equation

of an open system gives the relationship between the enthalpy of products

and that of reactants.

r r p pQ m h m h

2 2C O COQ h h h

0

T

o o

f f p

T

h h h h c dT

2 2, , ,f C f O f COQ h h h

2 2, , ,393520 f C f O f COh h h

The enthalpy of reactants must be lower than that of reactants

because the amount of heat transfer is a negative value. In other words,

the enthalpy of carbon dioxide must be lower than those of carbon and

oxygen although all the reactants and products are kept at 25℃ and 100

kPa. This is the enthalpy associated with the chemical composition of

species.

We define that all elements at their naturally existing states have a

standard enthalpy of formation of zero.

, 0o

f Ch

2, 0o

f Oh

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Elements: O, H, S, H, Cu, C

Compound: CO2, H2O, CuO, SO2, CH4

Naturally existing states:

C: Carbon, graphite, diamond

O: O, O2, O3

Standard condition: 25℃ and 100 kPa.

The enthalpy of formation of compounds can be obtained by measuring

the heat release in the formation reaction at standard condition.

2, 393520 /f COh kJ kmole

The enthalpy of formation of CO2 is thus -393520 kJ/kmole. The

negative sign means that the formation reaction is exothermic.

If the formation reaction is endothermic, the enthalpy of formation is of

positive value.

Endothermic reaction: Reactions that absorb heat.

Exothermic reaction: Reactions that release heat.

Combustion Engineering ME Dept NCHU 頁 11

Thermochemical Properties of selected substances at 298 K and 1 atm

Species Phase Chemical Formula ΔHfo (kJ/mol)

Benzene Liquid C6H6 48.95

Carbon Solid C 0

Carbon(Diamond) Solid C 1.8

Carbon Gas C 716.67

Carbon Dioxide Gas CO2 -393.509

Carbon Monoxide Gas CO -110.525

Ethane Gas C2H6 -83.85

Ethanol Liquid C2H5OH -277.0

Ethanol Gas C2H5OH -235.3

Ethene Gas C2H4 52.3

Ethyne Gas C2H2 226.73

Methane Gas CH4 -74.87

Methanol Liquid CH3OH -238.4

Methanol Gas CH3OH -201.0

Biodiesel Gas C19H34O2 -356.3

Methyl Trichloride Liquid CHCl3 -134.47

Methyl Trichloride Gas CHCl3 -103.18

Propane Liquid C3H8 -104.7

Hydrogen Gas H2 0

Water Liquid H2O -285.830

Water Gas H2O -241.818

Ammonia Aqueous NH3 -80.8

Ammonia Gas NH3 -45.90

Nitrogen Dioxide Gas NO2 33.1

Nitrogen Monoxide Gas NO 90.29

Nitrous oxide Gas N2O 82.05

Monoatomic oxygen Gas O 249

Oxygen Gas O2 0

Ozone Gas O3 143

Combustion Engineering ME Dept NCHU 頁 12

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Example：Stoichiometric mixture of carbon monoxide and oxygen are at

298 K and 100 kPa. Find the heat release of the reaction if the products

are (1). at 298 K, (2). at 1000K.

2 2

1

2CO O CO

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Example：Methane at 25℃is mixed with air at 600K in the combustor of

a gas turbine engine. The gas temperature at the inlet of turbine should

be confined to 1200K. Determine the air fuel ratio. (69.53)

25℃Methane

1200K products

600K air

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Example：A mixture of liquid methanol and air is fed to the inlet of an

adiabatic fuel reformer to produce hydrogen. The inlet mixture is at 25℃,

and the outlet gas is at 600K. Determine the air fuel ratio.

(partial oxidation reforming process)

CH3OH + x(O2 + 3.76N2) → CO2 + 2H2 + (x-0.5)O2 + 3.76xN2

(Ans: x=3.017, A/F = 13.0)

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Example：In an electrolysis reaction, water is decomposed into hydrogen

and oxygen. Calculate the required power if 100 L/min of hydrogen is

produced. (19.23 kW)

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Combustion Engineering ME Dept NCHU 頁 13

Assignment 2.4

Hydrogen is produced in a steam reformer in which methanol is reacted

with steam as the following equation.

CH3OH + H2O→ CO2 + 3H2

Determine the heat transfer required to produce one mole of hydrogen if

both the reactants and the products are kept at 25℃.

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Assignment 2.5

Hydrogen is produced in a partial oxidation reformer in which methanol

is reacted with air as the following equation.

CH3OH +0.5 (O2+3.76N2)→ CO2 +2H2 +1.88N2

Determine the heat transfer required to produce one mole of hydrogen if

both the reactants and the products are kept at 25℃.

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Assignment 2.6

Hydrogen is produced in an autothermal reformer in which methanol is

reacted with air and steam to form carbon dioxide and hydrogen as the

following equation.

CH3OH +x(O2+3.76N2) +yH2O → Products (CO2 , H2 , N2)

This reaction may occur in an adiabatic way in which neither heat input

nor heat output is required. Determine the yield rate of hydrogen if 1

L/min of methanol is fed into the reformer.

(1). Assume that both the reactants and the products are kept at 25℃.

(2). The products are at 300K while the reactants are at 25℃.

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Assignment 2.7

Hydrogen is produced with the electrolysis of liquid water. Determine the

electric current required to produce hydrogen with the rate of 1 L/min at

100 kPa and 25℃ if the applied voltage is 110 volts. The efficiency of

electrolysis is assumed to be 75%.

H2O→ 0.5O2 + H2

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Combustion Engineering ME Dept NCHU 頁 14

(2.2.3). Heating value

Heating value is the heat release of a certain amount of fuel in a

complete combustion at standard condition. It is an important

characteristic of fuel.

Fuel + Oxidants (air) → Products

Measurement of heating value: Combustion Calorimeters for solid and

liquid fuels

Combustion Calorimeters for gaseous fuels

Continuous flows of air and fuel are feeding into an adiabatic burner

and the hot exhaust gas is cooled by water coiled around the exhaust pipe.

The flow rate and the temperature difference of water are measured to

calculate the thermal energy released by the combustion products.

fuel

air exhaust

R PQ H H

Combustion Engineering ME Dept NCHU 頁 15

2 2 2 2 2( )( 3.76 ) 3.76( )4 2 2 4 2

n m y

m y m m yC H O n O N CO H O n N

2 2( ) ( )3.76

4 2 4 2R F O N

m y m yH h n h n h

2 2 2( )3.76

2 4 2P CO H O N

m m yH h h n h

,F f Fh h

2 20O Nh h

2 2,CO f COh h

2 2,H O f H Oh h

2 2, , ,2

P R f CO f H O f F

mQ H H h h h

Lower heating value (LHV): the heating value of a fuel in which that the

water vapor in the product remains in the form of vapor at 25℃.

2,f H Oh = -241818 kJ/kmole

Higher heating value (HHV): the heating value of a fuel in which that the

water vapor in the product is condensed to liquid water at 25℃.

2,f H Oh = -285830 kJ/kmole

HHV is greater than LHV. The difference between LHV and HHV is the

latent heat of water at 25℃.

The enthalpy of formation of any fuel can be obtained if the heating value

is already known.

2 2, , ,2

f F f CO f H O

mh h h Q

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Example：Determine the lower heating value and the higher heating

value of ethylene.

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Example：Determine the lower heating value and the higher heating

value of hydrogen.

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Example：If the higher heating value of gasoline is 42000 kJ/kg, and the

H/C ratio is 1.85, determine the equivalent enthalpy of formation of

Combustion Engineering ME Dept NCHU 頁 16

gasoline.

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Example：Dmethyl ether (DME, 3 3CH OCH ) is considered to be an

alternative fuel to replace fossil diesel. The lower heating value of DME

is 28900 kJ/kg. Determine the enthalpy of formation of DME.

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Lower heating value for some fuels (at 101.3kPa, 15.4°C)

Fuel MJ/kg kJ/mol

Methane 50.009 802.34

Ethane 47.794 1,437.2

Propane 46.357 2,044.2

Butane 45.752 2,659.3

Pentane(C5H10) 45.357 3,272.6

Hexane(C6H14) 44.752 3,856.7

Heptane(C7H16) 44.566 4,465.8

Octane(C8H18) 44.427 5,074.9

Nonane(C9H20) 44.311 5,683.3

Decane(C10H22) 44.240 6,294.5

Undecane(C11H24) 44.194 6,908.0

Dodecane(C12H26) 44.147 7,519.6

Methanol 19.930 638.55

Ethanol 28.865 1,329.8

n-Propanol 30.680 1,843.9

Isopropanol 30.447 1,829.9

n-Butanol 33.075 2,501.6

Isobutanol 32.959 2,442.9

Carbon (graphite) 32.808 —

Hydrogen 120.971 241.942

Carbon monoxide 10.112 283.24

Ammonia 18.646 317.56

Sulfur (solid) 9.163 293.82

gasoline 44.400

diesel 44.800

It is noted that heating values of hydrocarbon fuels are about the same in

terms of weight.

Combustion Engineering ME Dept NCHU 頁 17

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Assignment 2.8

DME is produced by catalytic dehydration of methanol with the

following reaction.

3 3 3 22CH OH CH OCH H O

Is this an endothermic reaction or exothermic reaction? Find the heat that

should be added or removed from the reaction if both the reactants and

products are at 25℃ and 100 kPa.

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Assignment 2.9

Another way to produce is to synthesize DME directly from hydrogen

and carbon monoxide with the following reaction.

2 3 3 23 3CO H CH OCH CO

Is this an endothermic reaction or exothermic reaction? Find the heat that

should be added or removed from the reaction if both the reactants and

products are at 25℃ and 100 kPa.

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Assignment 2.10

The averaged molecular weight of a jet fuel is 126 kg/kmole. The

hydrogen to carbon ratio is 1.85:1. If the lower heating value of the fuel

is 10280 kcal/kg, calculate the equivalent enthalpy of formation of the jet

fuel.

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Assignment 2.11

A synthesized gas is composed of 30% CO, 50% H2, 10% C3H8, and 10%

CH4 in molar basis.

(1). Determine the stoichiometric air fuel ratio of the synthesized gas.

(2). Determine the lower heating value of the synthesized gas in kJ/m3 at

25℃ and 100 kPa.

(3). The synthesized gas is mixed with excess air to burn in a heater, and

the flame temperature is 1500K. Determine the molar fraction of CO2 in

the products on a dry basis.

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Assignment 2.12

A turbojet engine runs on the fuel mentioned above. The temperature

and the pressure of the inlet air are 0℃ and 90 kPa. Jet fuel is injected

into the combustion chamber at 25℃. The pressure ratio of the

compressor is 10. The efficiency of compressor is 0.85. Find the air fuel

Combustion Engineering ME Dept NCHU 頁 18

ratio of the mixture if the temperature at the outlet of combustor is

confined to 1200K. Assume that air is an ideal gas.

1.85 2 2 2 2 2 2( 3.76 ) 0.925 ( 1.4625) 3.76CH X O N CO H O X O XN

1.85 2 2 2

2 2 2 2 2 23

,

, ,

( 3.76 )

0.925( ) ( 1.4625) 3.76

f CH O N T

f CO CO f H O H O O NT

h X h h

h h h h X h X h

Discussions on the energy density of fuels

The amount of heat that can be released in real engines is not

directly related to the heating value of fuel. Fuels with high heating

value are not necessarily to generate more energy as the mixture of fuel

and air is burned inside engine.

The reaction of fuel with stoichiometric amount of air is as

following.

2 2 2 2 2( )( 3.76 ) 3.76( )4 2 2 4 2

n m l

m l m m lC H O n O N nCO H O n N

( ) 4.76 28.974 2( / )12 16

st

m ln

A Fn m l

The amount of heat that can be released expressed in terms of much fuel

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contained inside cylinder and the heating value of fuel fQ .

/

aH f f f

mQ m Q Q

A F

If the fuel is of liquid form like gasoline of methanol, the volume is filled

with air only because liquid fuel would occupy negligible fraction of

volume only.

m Da

a m

P Vm

R T

/

fm DH

a m

QP VQ

R T A F

However, if the fuel is of gaseous form like propane or methane, the

volume is filled with the gaseous mixture.

( )4.764 2

1 ( )4.764 2

m D m Da a

a m a m

m ln

P V P Vm x

m lR T R Tn

( )4.764 2

/1 ( )4.76

4 2

fm DH

a m

m lnQP V

Qm lR T A F

n

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Example：Fuel is burned with stoichiometric air in an internal

combustion engine. The engine displacement volume is 1 L. The

temperature of mixture is 300K and the pressure is 100 kPa. Find the

energy that can be released in the combustion process.

(1). Hydrogen. (2). Methane. (3). Liquid gasoline. (4). Liquid methanol.

uPV nR T

u

PVn

R T = 4.009×10-5 kmole

2 2 2 2 20.5( 3.76 ) 1.88H O N H O N

Heating value of hydrogen is 241942 kJ/kmole

2 2 1 0.5 4.76H H

nn nx

= 1.186×10-5 kmole

2 2H HE n Q =1.186×10-5 kmole×241942 kJ/kmole= 2.87 kJ

Combustion Engineering ME Dept NCHU 頁 20

4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N

Heating value of methane is 802340 kJ/kmole

4 4 1 2 4.76CH CH

nn nx

= 0.381×10-5 kmole

4 4CH CHE n Q =0.348×10-5 kmole×802340 kJ/kmole= 3.06 kJ

1.85 2 2 2 2 21.4625( 3.76 ) 0.925 5.499CH O N CO H O N

Heating value of gasoline is 44400 kJ/kg

a a am n M = 4.009×10-5 kmole×28.97 kg/ kmole= 1.161×10-3 kg

/( / )f a stm m A F = 1.161×10-3 kg/14.56=7.97×10-5 kg

f fE m Q =7.97×10-5 kg×44400 kJ/kg = 3.54 kJ

3 2 2 2 2 21.5( 3.76 ) 2 5.64CH OH O N CO H O N

Heating value of methanol is 19930 kJ/kg

/( / )f a stm m A F = 1.161×10-3 kg/6.46=17.961×10-5 kg

f fE m Q =17.961×10-5 kg×19930 kJ/kg = 3.58 kJ

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Combustion efficiency:

The energy released in a combustion process is only a fraction of the

heating value of that fuel because of incomplete combustion.

Rc

HV

Q

Q

RQ : actual heat release

HVQ : theoretical heat release

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Example：Methane is burned with dry air. The molar analysis of the

products on a dry basis is CO2 9.7%，CO 0.5%，O2 2.95%，and N2

86.85%。Determine the combustion efficiency.

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Combustion Engineering ME Dept NCHU 頁 21

Wobbe Index (WI)

The Wobbe Index (WI) is an indicator of the interchangeability of fuel

gases such as natural gas, liquefied petroleum gas (LPG), and town

gas and is frequently defined in the specifications of gas supply and

transport utilities. It is defined as the following.

W

s

QI

G

where Q (kJ/Nm3) is the volumetric heating value at standard condition

(T=273.15 K, P=101.325k Pa), sG is the specific gravity compared with

air at the same temperature and pressure.

/ / aF F Fs

a F a u u a

M PM P MP PG

R T R T R T R T M

aM 28.97kg/kmole

As a result, Wobbe Index (WI) may be defined as

aW

F

MI Q

M

Pressure gauge

Fuel supply

Control valve

It is noted that the amount of fuel supply is adjusted by a control

valve. The fuel flow rate is determined with a venture type flow meter

that the pressure difference between the upstream and downstream of the

valve are measured. The flow rate is as the following.

2d

PV c A

where V (Nm3/sec) is the fuel flow rate, A (m2) is the valve open area,

P (kPa) is the pressure difference, (kg/m3) is the fuel density, and dc

Combustion Engineering ME Dept NCHU 頁 22

is the flow coefficient.

The total energy supply is the product of fuel flow rate and fuel

heating value.

2 2 2 2d d d d W

a a as

a

P P Q P Q PE c A Q c A c A c A I

G

For the same pressure difference and valve open area, the total energy

supply is proportional to the Wobbe index of the fuel. That is, if two fuels

have the same value of Wobbe index, they can be interchanged without

any adjustment of control valve to maintain the same level of total energy

supply.

The Wobbe Index of common fuel gases are as the following table.

Fuel gas Upper index

MJ/Nm3

Lower index

MJ/Nm3

Hydrogen 48.23 40.65

Methane 53.28 47.91

Ethane 68.19 62.47

Ethylene 63.82 60.01

Natural gas 53.71 48.52

Propane 81.07 74.54

Propylene 77.04 71.88

n-butane 92.32 85.08

Iso-butane 91.96 84.71

Butylene-1 88.46 82.54

LPG 86.84 79.94

Acetylene 61.32 59.16

Carbon monoxide 12.80 12.80

Combustion Engineering ME Dept NCHU 頁 23

(2.3). Fuels

(2.3.1). Wood

Wood is composed primarily of cellulose(纖維素), lignine(木質素) and

water. Moisture content varies as wood is dried.

The chemical formula of cellulose is C6H10O5. The chemical formula of

lignine is C9H10O2. The molar analysis of cellulose theoretically is carbon

44.4%, hydrogen 6.2%, and oxygen 49.4%. However, the actual

composition would vary.

The heating value of wood is about 20000 kJ/kg. It is low compared

with other fossil fuel.

Heating values of wood：

Type C H O HHV(kJ/kg)

Cedar(杉) 48.8 6.37 44.46 19540

Cypress(柏) 54.98 6.54 38.08 22960

Fir(樅) 52.3 6.3 40.5 21050

Hemlock(松) 50.4 5.8 41.4 20050

Pine(松) 59.0 7.19 32.68 26330

Redwood(紅木) 53.5 5.9 40.3 21030

Ash(梣) 49.73 6.93 43.04 20750

Beech(山毛櫸) 51.64 6.26 41.45 20380

Birch(樺) 49.77 6.49 43.45 20120

Elm(榆) 50.35 6.57 42.34 20490

Hickory(山胡桃) 49.67 6.49 43.11 20170

Maple(楓) 50.64 6.02 41.74 19960

Oak(橡) 48.78 6.09 44.98 19030

Poplar(白楊) 51.64 6.26 41.45 20750

Source：E.L. Keating, "Applied Combustion", Marcel Dekker, 1992.

Reference:

K. W. Ragland, D. J. Aerts, & A. J. Baker, “Properties of Wood for

Combustion Analysis”, Bioresource Technology 37 (1991) 161-168

-----------------------------------------------------------------------------------------

Example：Determine the equivalent enthalpy of formation of cedar.

Combustion Engineering ME Dept NCHU 頁 24

C 48.8%, 48.8/12= 4.067

H 6.37%, 6.37/1 = 6.37

O 44.46%, 44.46/16=2.779

4.067+ 6.37/4-2.779/2 =4.27

48.8+6.37+44.46=99.63

C4.067H6.37O2.779 +4.27 (O2+3.76N2)→4.067 CO2 +3.185H2O+16.055N2

4.067×(-390509) +3.185×(-285830)-hf=(-19540)×99.63

hf= -3941.5 kJ/kg

-----------------------------------------------------------------------------------------

Example：Dry cedar log is burned in a fireplace. Determine the amount

of excess air if the burned temperature is 700 K.

-----------------------------------------------------------------------------------------

Effect of moisture content

As the wood is burning, part of the heat is used to heat up and

vaporize water content. It is known that the vaporization latent heat of

water is 2400 kJ/kg, if the water content is x (by weight), the heat of

burning would be

Q=20000(1-x)-2400x=(20000-22400x) kJ/kg

-----------------------------------------------------------------------------------------

Example：Cedar log is burned in a fireplace, determine the amount of

excess air if the burned temperature is 700 K and the moisture content is

30%.

-----------------------------------------------------------------------------------------

(2.3.2). Coal

Coal can be categorized as lignite(褐煤), sub-bituminous(次煙煤),

bituminous(煙煤), and anthracite(無煙煤).

Proximate analysis: Coal can be considered to be composed of four

different constituents, the moisture, volatile matter, fixed carbon, and ash.

Combustion Engineering ME Dept NCHU 頁 25

Total moisture is analyzed by loss of mass between an untreated sample

and the sample dried in a minimum free-space oven at 150 °C within a

nitrogen atmosphere.

Volatile matter is the components which are liberated at high temperature

in the absence of air. This is usually a mixture of short and long chain

hydrocarbons, aromatic hydrocarbons and some sulfur. The volatile

matter of coal is determined by heating the coal sample to 900°C for 7

minutes in a cylindrical silica crucible in a muffle furnace.

Ash content of coal is the non-combustible residue left after coal is burnt.

It represents the bulk mineral matter after carbon, oxygen, sulfur and

water (including from clays) has been driven off during combustion.

The fixed carbon content of the coal is the carbon found in the material

which is left after volatile materials are driven off. This differs from the

ultimate carbon content of the coal because some carbon is lost in

hydrocarbons with the volatiles. Fixed carbon is used as an estimate of

the amount of coke that will be yielded from a sample of coal.

Components and Heating value of coal

water volatile carbon ash HV(kJ/kg)

Anthracite 4.4 4.8 81.8 9.0 30540

Bituminous 3.1 23.4 63.6 9.9 31470

Sub-bituminous 13.9 34.2 41.0 10.9 24030

Lignite 36.8 27.8 30.2 5.2 16190

Source：E.L. Keating, "Applied Combustion", Marcel Dekker, 1992.

Ultimate Analysis: Elemental or ultimate analysis encompasses the

quantitative determination of carbon, hydrogen, nitrogen, oxygen, and

sulfur within the coal.

-----------------------------------------------------------------------------------------

Example：An Illinois coal that has the following ultimate analysis:

81.3% C; 5.3% H; 9.8% O; 1.7% N; 1.9% S. Find the stoiciometric air

fuel ratio of the Illinois coal.

-----------------------------------------------------------------------------------------

Heating value

Combustion Engineering ME Dept NCHU 頁 26

The heating value Q of coal is the heat liberated by its complete

combustion with oxygen. Q is a complex function of the elemental

composition of the coal. Q can be determined experimentally using

calorimeters. Dulong suggests the following approximate formula for Q

when the oxygen content is less than 10%:

Q = 337C + 1442(H - O/8) + 93S

where C is the mass percent of carbon, H is the mass percent of hydrogen,

O is the mass percent of oxygen, and S is the mass percent of sulfur in the

coal. With these constants, Q is given in kilojoules per kilogram.

-----------------------------------------------------------------------------------------

Example：An Illinois coal that has the following ultimate analysis:

81.3% C; 5.3% H; 9.8% O; 1.7% N; 1.9% S. Find the heating value of

the Illinois coal.

-----------------------------------------------------------------------------------------

(2.3.3)、Hydrocarbon fuels

Gasoline is produced from crude oil via distillation. Typical gasoline

consists of hydrocarbons with between four and 12 carbon atoms per

molecule (commonly referred to as C5-C12).

The typical composition of gasoline hydrocarbons (% volume) is as

follows: 4-8% alkanes; 2-5% alkenes; 25-40% isoalkanes; 3-7%

cycloalkanes; l-4% cycloalkenes; and 20-50% total aromatics (0.5-2.5%

benzene)

Alkane(烷類)

An alkane is a saturated hydrocarbon. Alkanes consist only of hydrogen

and carbon atoms, all bonds are single bonds, and the carbon atoms are

not joined in cyclic structures but instead form an open chain. They have

the general chemical formula CnH2n+2.

Alkene(烯類)

An alkene is an unsaturated hydrocarbon containing at least one carbon–

carbon double bond. They have the general chemical formula CnH2n.

A correlation of heating value for a variety of fuels was proposed by

Channiwala and Parikh* and is given as the following.

HHV= 0.3491C+1.1783H+0.1005S–0.1034O–0.0151N–0.0211A

Combustion Engineering ME Dept NCHU 頁 27

HHV: higher heating value (MJ/kg), 4.745<HHV<55.345

C: carbon mass percentage, 0%<C<92.25%

H: hydrogen mass percentage,0.43%<H<25.15%

S: sulfur mass percentage, 0%<S<94.08%

O: oxygen mass percentage, 0%<O<50%

N: nitrogen mass percentage, 0%<N<5.6%

A: ash mass percentage, 0%<A<71.4%

Reference:

Channiwala S.A. and P.P. Parikh, “A unified correlation for estimating

HHV of solid, liquid, and gaseous fuels”, Fuel 81 (2002) 1051-1063.

The members of the series (in terms of number of carbon atoms) are

named as follows:

Alkane Formula Boiling

point [°C]

Melting

point [°C]

Density [g·cm−3]

(at 20 °C)

Methane CH4 -162 -182 gas

Ethane C2H6 -89 -183 gas

Propane C3H8 -42 -188 gas

Butane C4H10 0 -138 gas

Pentane C5H12 36 -130 0.626 (liquid)

Hexane C6H14 69 -95 0.659 (liquid)

Heptane C7H16 98 -91 0.684 (liquid)

Octane C8H18 126 -57 0.703 (liquid)

Nonane C9H20 151 -54 0.718 (liquid)

Decane C10H22 174 -30 0.730 (liquid)

Undecane C11H24 196 -26 0.740 (liquid)

Dodecane C12H26 216 -10 0.749 (liquid)

Hexadecane C16H34 287 18 0.773

Icosane C20H42 343 37 solid

Triacontane C30H62 450 66 solid

Tetracontane C40H82 525 82 solid

Pentacontane C50H102 575 91 solid

Hexacontane C60H122 625 100 solid

Combustion Engineering ME Dept NCHU 頁 28

Fuel Property Comparison for Ethanol, Gasoline and No. 2 Diesel

Property Ethanol Gasoline No. 2 Diesel

Chemical Formula C2H5OH C4 to C12 C8 to C25

Molecular Weight 46.07 100–105 ≈200

Carbon 52.2 85–88 84–87

Hydrogen 13.1 12–15 33–16

Oxygen 34.7 0 0

Specific gravity, 60° F/60° F 0.796 0.72–0.78 0.81–0.89

Boiling temperature, °F 172 80–437 370–650

Reid vapor pressure, psi 2.3 8–15 0.2

Research octane no. 108 90–100 --

Motor octane no. 92 81–90 --

(R + M)/2 100 86–94 N/A

Cetane no.(1) -- 5–20 40–55

Freezing point, °F -173.2 -40 -40–30a

Centipoise @ 60° F 1.19 0.37–0.44b 2.6–4.1

Flash point, closed cup, °F 55 -45 165

Autoignition temperature, °F 793 495 ≈600

Btu/gal @ 60° F 2,378 ≈900 ≈700

Btu/lb @ 60° F 396 ≈150 ≈100

Btu/lb air for stoichiometric mixture @ 60° F

44 ≈10 ≈8

Higher (liquid fuel-liquid water) Btu/lb

12,800 18,800–20,400 19,200–20000

Lower (liquid fuel-water vapor) Btu/lb

11,500 18,000–19,000 18,000–19,000

Higher (liquid fuel-liquid water) Btu/gal

84,100 124,800 138,700

Lower (liquid fuel-water vapor) Btu/gal @ 60° F

76,000b 115,000 128,400

Mixture in vapor state, Btu/cubic foot @ 68° F

92.9 95.2 96.9c

Fuel in liquid state, Btu/lb or air

1,280 1,290 –

Specific heat, Btu/lb °F 0.57 0.48 0.43

Stoichiometric air/fuel, weight 9 14.7b 14.7

Volume % fuel in vaporized stoichiometric mixture

6.5 2 –

Combustion Engineering ME Dept NCHU 頁 29

Evaporation characteristics of gasoline

IBP：Initial boiling point.

T10：The temperature that 10% of gasoline evaporates.

T50：The temperature that 50% of gasoline evaporates.

T90：The temperature that 90% of gasoline evaporates.

V70：The fraction of gasoline that evaporates at 70℃.

V100：The fraction of gasoline that evaporates at 100℃.

V150：The fraction of gasoline that evaporates at 150℃.

EBP：End of boiling point.

An example of the evaporation characteristics of China Petroleum.

(%) T (℃)

0 36.5 IBP

10 51.5 T10

33.7 70 V70

50 92.6 T50

54.1 100 V100

85.2 150 V150

90 161.2 T90

100 199.4 FBP

Temperature os saturation

0

50

100

150

200

0 20 40 60 80 100

Ratio (%)

T (

C)

T

Combustion Engineering ME Dept NCHU 頁 30

(2.3.4). Producer Gas(煤氣)

Wood gasification is the process of heating wood in an oxygen-

starved environment until volatile pyrolysis gases (carbon monoxide and

hydrogen) are released from the wood. Depending on the final use of the

typically low-energy wood (producer) gas (~5.6 MJ/m³), the gases can be

mixed with air or pure oxygen for complete combustion and the heat

produced transferred to a boiler for energy distribution.

Typical components of producer gas：N2 50.9%，CO 27.0%，H2

14.0%，CO2 4.5%，CH4 3.0%，O2 0.6%. Producer gas is toxic because

it contains a high percentage of CO.

Oxidation zone: C+O2→CO2

Primary reduction zone: C+CO2→2CO

Secondary reduction zone: CO+H2O→CO2+H2

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 31

Example：The producer gas is burned with 20% excess air. Find the air

fuel ratio.

-----------------------------------------------------------------------------------------

Assignment 2.13

Find the heating value of producer gas in kJ/m3, and make a comparison

with that of natural gas. Natural gas is composed of 85% CH4, 10% C2H6,

and 5% CO2.

-----------------------------------------------------------------------------------------

天然氣和煤氣的優劣

近年來國內天然氣產業大力發展，城鎮氣化率快速提高，不過大中城市

天然氣已經普及，經濟較為發達的縣城、鄉鎮也已經鋪設了大量的天然氣管

道。

我們江西省新餘市也開始了煤氣改造成天然氣的步伐。可是很多用戶不

知道煤氣和天然氣的差別，不知道為什麼要將煤氣改造成天然氣，再加上用

天然氣比用煤氣的收費要高，使家裡的每月支出多了一些，很多用戶都有意

見。在此，我向大家介紹一下煤氣和天然氣的優缺點，幫大家更好地認識用

天然氣的必要性。

一、什麼是煤氣

煤氣是以煤為原料加工制得的含有可燃成分的氣體。我們一般用的煤氣

是水煤氣，它是由水蒸氣通過熾熱的焦炭而生成的氣體，主要成分是一氧化

碳、氫氣，燃燒後排放水和二氧化碳，有微量 CO、HC 和 NOX。

將水蒸氣通過熾熱的煤層可制得較潔淨的水煤氣（主要成分是 CO 和

H2），現象為火焰騰起更高，而且變為淡藍色（氫氣和 CO 燃燒的顏色）。

煤氣中有一氧化碳，家庭中煤氣中毒主要指一氧化碳中毒。由於人們用煤氣

不當或煤氣洩漏，也常造成人員傷亡，所以人們要更加小心煤氣的使用情

況。煤氣廠常在家用水煤氣中特意摻入少量難聞氣味的氣體，因為 CO 和 H2

為無色無味氣體，目的是為了當煤氣洩漏時能聞到並及時發現。

二、什麼是天然氣

從廣義的定義來說，天然氣是指自然界中天然存在的一切氣體，包括大

氣圈、水圈、生物圈和岩石圈中各種自然過程形成的氣體。它主要存在於油

田和天然氣田，也有少量出於煤層。天然氣燃燒後無廢渣、廢水產生，相較

煤炭、石油等能源有使用安全、熱值高、潔淨等優勢。

三、煤氣和天然氣的優缺點對比

1.煤氣和天然氣在燃燒後的成分差別對比

煤氣主要是一氧化碳和氫氣的成分，天然氣主要是甲烷等烴類和非烴類

氣體。煤氣和天然氣如果都完全燃燒，生成的都是二氧化碳和水，它們對空

氣污染極少，差別不大。但是水煤氣要是燃燒不完全，就會有大量的一氧化

碳、碳、氫氣進入到大氣當中。

從環境污染上講天然氣優於水煤氣，現在全球都在節能減排，所以用天

然氣是大勢所趨。

2.煤氣和天然氣的每月用量和收費

Combustion Engineering ME Dept NCHU 頁 32

每個家庭都很關注家裡的用氣量和收費多少，是比原來用水煤氣多了多

少錢。我做了一個調查和對比。

三口之家一般每月的用水煤氣量大約是 30～45 立方米，每立方米的價錢

是 1.15 元，合計每月繳納 35～50 元。要是家裡有 5～6 人，用氣量大約是 60

～70 立方米，合計每月要繳納 70～80 元。

而使用天然氣以後經過煤氣公司對燃氣灶和熱水器的改造，用氣量要減

少一些，主要是對進氣口進行調整的原因。還有就是天然氣的燃燒放熱比水

煤氣要高，使用天然氣燒水和炒菜的時間要更短。三口之家一般每月用的天

然氣量大約是 20～30 立方米，每立方米的價錢是 3.00 元，合計每月繳納 60

～80 元。要是家裡有 5～6 人，用氣量大約是 40～50 立方米，合計每月要繳

納 120～150 元。整體來說三口之家用天然氣要比用水煤氣每月多繳納 25 元

左右，五六口的家庭每月要多繳納 50 元左右。每人每年要多支出大約 120

元。再加上煤氣改造時，要繳納的改造費用大約 120 元。

3.水煤氣和天然氣的燃燒熱值

水煤氣的主要成分是一氧化碳和氫氣，它們的燃燒熱化學方程式如下：

氫氣的燃燒熱化學方程式：

可見，水煤氣的燃燒熱值約為 10 500 千焦/標準立方米，天然氣的燃燒熱

值約為 35 000 千焦/標準立方米。

4.水煤氣和天然氣的安全性

家庭中煤氣中毒主要指一氧化碳中毒，多見於冬天用煤爐取暖，門窗緊

閉，排煙不良時，也常見於液化灶具漏泄或煤氣管道漏泄等。煤氣中毒時病

人最初感覺為頭痛、頭昏、噁心、嘔吐、軟弱無力，當他意識到中毒時，常

掙扎下床開門、開窗，但一般僅有少數人能打開門，大部分病人迅速發生抽

筋、昏迷，兩頰、前胸皮膚及口唇呈櫻桃紅色，如救治不及時，可很快呼吸

抑制而死亡。

可見，水煤氣比天然氣更容易使人中毒，毒性較強。水煤氣比天然氣更

容易爆炸，給人們帶來危害，所以使用天然氣要比水煤氣安全得多。

(2.3.5). Coke oven gas (焦爐氣)

The coke oven gas is the by-product of the process to convert coal into

coke in the coke oven. The volatile matter in the coal is vaporized and

leaves the coke oven chambers as hot, raw coke oven gas. After leaving

the coke oven chambers, the raw coke oven gas is cooled which results in

a liquid condensate stream and a gas stream. The gas stream can be used

as a fuel in the integrated iron and steel works.

-----------------------------------------------------------------------------------------

Example：The composition of coke oven gas is as the following.

H2 54.0%，CH4 30.6%，CO 7.4%，N2 5.6%，CO2 2.0%，O2 0.4%.

Find the heating value of coke oven gas in kJ/m3.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 33

Combustion Engineering ME Dept NCHU 頁 34

(2.4). Adiabatic flame temperature

Adiabatic flame temperature is the temperature of the combustion

products if the combustion process is adiabatic and without doing work.

0Q ， 0W

(2.4.1). Constant pressure adiabatic flame temperature

The combustion process is at a constant pressure. The combustor

in the gas turbine engine, furnace, and boiler are examples of constant

pressure combustion process.

F F a a p pm h m h m h ， R pH H

0

0 0

, ,

T

F f F F f F pF

T

h h h h c dT

2 2 2 2

0

0

,

T

O f O O pO

T

h h h c dT

2 2 2 2

0

0

,

T

N f N N pN

T

h h h c dT

The enthalpy of products equals to the enthalpy of reactants.

-----------------------------------------------------------------------------------------

Example：Hydrogen is mixed with air at an equivalence ratio of 1.0 at

25℃and 100 kPa. Determine the adiabatic flame temperature.

2 2 2 2 20.5( 3.76 ) 1.88H O N H O N

-----------------------------------------------------------------------------------------

Example：Methane is mixed with air at an equivalence ratio of 0.8 at 25

℃and 100 kPa. Determine the adiabatic flame temperature.

4 2 2 2 2 2 22.5( 3.76 ) 2 0.5 9.4CH O N CO H O O N

-----------------------------------------------------------------------------------------

Assignment 2.14

Determine the constant pressure adiabatic flame temperature of the

following flames. Assume that the mixture is at 25℃ before

combustion, and equivalence ratio is 1.0.

(1). Propane flame

3 8 2 2 2 2 25( 3.76 ) 3 4 18.8C H O N CO H O N

(2). Ethylene flame

Combustion Engineering ME Dept NCHU 頁 35

2 2 2 2 2 2 22.5( 3.76 ) 2 9.4C H O N CO H O N

(3). Carbon monoxide flame

2 2 2 20.5( 3.76 ) 1.88CO O N CO N

(4). Coke oven gas: CO 7.0%，H2 58.0%，CO2 2.5%，CH4 25.0%，O2

0.5%, and N2 7.0%.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 36

(2.4.2). Constant volume adiabatic flame temperature

The combustion process is at a constant volume. The internal

engine is an example of constant volume combustion process.

F F a a p pm u m u m u ， R pU U

0

0

T

f p

T

u h RT h c dT RT

b bb

n RTP

V

0 0 0

b b bP n T

P n T

-----------------------------------------------------------------------------------------

Example：Hydrogen is mixed with air at an equivalence ratio of 1.0 in a

constant volume chamber at 25℃and 100 kPa. Determine the adiabatic

flame temperature and the final pressure.

2 2 2 2 20.5( 3.76 ) 1.88H O N H O N

2 2 2 2 20.5 1.88 1.88H O N H O Nu u u u u

2 2 2 2 2 21 ,0.5 3.76 3.38 1.88 2.88H O N f H O H O N bh h h RT h h h RT

2.88

100 3.38 298

b bP T

-----------------------------------------------------------------------------------------

Example：Hythane is a mixture of hydrogen and methane at the ratio of

1:1 in volume. Hythane is mixed with stoichiometric amount of air at

25℃ and 100 kPa in a constant volume chamber. Determine the flame

temperature and the final pressure if the combustion process is adiabatic

and the volume remains the same.

-----------------------------------------------------------------------------------------

Assignment 2.15

An internal combustion engine runs on the fuel of propane with an

equivalence ratio of 0.9. The temperature and the pressure at the

end of compression stroke are 500℃ and 2.6 MPa. If a constant

volume combustion process occurs in the cylinder, find the

temperature and the pressure after combustion.

-----------------------------------------------------------------------------------------

Assignment 2.16

A synthesized gas is composed of 30% CO, 50% H2, 10% C3H8, and 10%

Combustion Engineering ME Dept NCHU 頁 37

CH4 in molar basis.

(1). Determine the stoichiometric air fuel ratio of the synthesized gas.

(2). Determine the lower heating value of the synthesized gas in kJ/m3 at

25℃ and 100 kPa.

(3). The synthesized gas is mixed with excess air to burn in a heater, and

the flame temperature is 1500K. Determine the molar fraction of CO2 in

the products on a dry basis.

(4). The synthesized gas is mixed with 20% excess air to burn in a

constant pressure adiabatic combustor. Both the synthesized gas and air

are at 100 kPa and 25℃. Determine the adiabatic lame temperaure.

(5). The synthesized gas is mixed with 20% of excess air in a constant

volume adiabatic combustor. Both the synthesized gas and air are at 100

kPa and 25℃. Determine the final temperature and pressure of the

products.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 38

(2.5). STANJAN

This is a computer program to solve the problems of combustion

chemistry developed by Prof. Reynolds of the Stanford University.

-----------------------------------------------------------------------------------------

Example：Methane is burned with stoichiometric air at 25℃and 100 kPa

in a constant pressure process.

4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N

-----------------------------------------------------------------------------------------

Example：Methane is burned with 25% excess air at 25℃and 100 kPa in

a constant volume process.

4 2 2 2 2 2 22.5( 3.76 ) 2 0.5 9.4CH O N CO H O O N

-----------------------------------------------------------------------------------------

Example：Methane is burned in a gas turbine engine with an

equivalence ratio of 0.4. The pressure ratio of the engine is 10.0. Find

the specific work of the engine and the thermal efficiency.

4 2 2 2 2 2 25( 3.76 ) 2 3 18.8CH O N CO H O O N

T (K) P (kPa) H (kJ/kg) S (kJ/kg-K)

1 300 100

2 1000

3 1000

4 100

2 1cw h h = kJ/kg, 3 4tw h h = kJ/kg,

net t cw w w = kJ/kg, work output per kg of mixture.

/A F = 43.09

1

1 /

fm

m A F

=0.0227, mass of fuel per kg of mixture.

fQ = 50009 kJ/kg, heating value of methane.

H f fq m Q = 1135.2 kJ/kg, heat of combustion per kg of mixture.

net

H

w

q = , thermal efficiency.

2

1

P

P = 10, pressure ratio.

1

11Brayton k

= 0.6019, theoretical efficiency of Brayton cycle.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 39

Example：Methane is burned with stoichiometric air at 25℃and 100 kPa

in an Otto engine. The compression ratio of the engine is 10.0.

4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N

T (K) P (kPa) u (kJ/kg) S (kJ/kg-K) V (m3/kg)

1 300 100

2

3

4

2 1cw u u , 3 4ew u u ,

net e cw w w

2net qW w V T , 2

netq

w VT

, engine torque

-----------------------------------------------------------------------------------------

Example：A jet engine runs on the fuel of propane with an

equivalence ratio of 0.3. The pressure ratio is 9.0. The compressor

efficiency is 85%, and the turbine efficiency is 90%. Find the

thrust of the engine.

T (K) P (kPa) H (kJ/kg) S (kJ/kg-K)

1 313 90

2s

2

3

4s

4

5 90

1 2s , isentropic compression, P is specified and s the same as last run

2 1cw h h , 2 1

2 1

cs sc

c

w h h

w h h

, 2 1

2 1s

c

h hh h

1 2 , compression, P and h are specified

2 3 , constant pressure combustion, P and h are the same as last run

3 4tw h h , t cw w

2 1 3 4h h h h , 1 4h h

3 4

3 4

tt

ts s

w h h

w h h

Combustion Engineering ME Dept NCHU 頁 40

3 44 3s

t

h hh h

3 4s , isentropic expansion, h and s are specified

Since STANJAN does not have the function of specified h and s, iteration

process is required to find the final pressure.

4 5 , isentropic expansion, P is specified and s the same as last run 2

4 52

Vh h

engine thrust: 4 52( )F mV m h h

-----------------------------------------------------------------------------------------

Assignment 2.17

An internal combustion engine runs on the fuel of propane with an

equivalence ratio of 0.9. The compression process is assumed to be

isentropic and the compression ratio is 9.0. The temperature and

the pressure before the compression stroke are 40℃ and 90 kPa. If

a constant volume combustion process occurs in the cylinder, find

the IMEP of the engine.

T (K) P (kPa) u (kJ/kg) S (kJ/kg-K) V (m3/kg)

1 313 90

2

3

4

Combustion Engineering ME Dept NCHU 頁 41

(2.6). Applications of combustion calculation

(2.6.1) Determination of A/F

Reaction of gasoline and air inside the internal combustion engine

can be expressed as the following.

2 2 2 2 2 2 2

14 ( 3.76 )n n

n

CH O N aCO bCO cCH xO yH zH O wN

If the concentrations of the components in the engine exhaust are

measured with analyzer, the air fuel ratio can be determined.

There are eight variables to be determined. , , , , , , ,a b c x y z w

From conservations of atoms, we have four equations.

C: 1 a b c

O: 2

(1 ) 2 24

na b x z

H: 2 2n cn y z

N: 3.76

(1 )4

nw

We need four more equations to find the values of those unknowns.

These four equations may be in the form of concentration which can be

obtained by direct measurement.

Usually, four components are measured, including carbon monoxide,

carbon dioxide, unburned hydrocarbon, and oxygen.

[ ]a

CO

, 2[ ]

bCO

,

1[ ]

cHC

,

2[ ]x

O

a b c x y w

Water has been distilled out with a filter ahead of the inlet of

analyzer to avoid the problem of clogging and corrosion.

[ ]a CO , 2[ ]b CO , [ ]c HC ,

2[ ]x O

2[ ] [ ] [ ] 1a b c CO CO HC

2

1

[ ] [ ] [ ]CO CO HC

2

[ ]

[ ] [ ] [ ]

COa

CO CO HC

Combustion Engineering ME Dept NCHU 頁 42

2

2

[ ]

[ ] [ ] [ ]

COb

CO CO HC

2

[ ]

[ ] [ ] [ ]

HCc

CO CO HC

2

2

[ ]

[ ] [ ] [ ]

Ox

CO CO HC

2(1 ) ( 2 2 )

4

nz a b x

1 1 2( ) ( ) (1 ) ( 2 2 )

2 2 4

ny n cn z n cn a b x

3.76(1 )

4

nw

1 2 3.76( ) (1 ) ( 2 2 ) (1 )

2 4 4

n na b c x n cn a b x

1.762 3 (1 ) 3 (1 )

2 2 4

n n na b c x

2 2

2

1 2[ ] 3[ ] (1 )[ ] 3[ ]1.762 (1 )

[ ] [ ] [ ] 2 4

nCO CO HC O

n n

CO CO HC

2 2

2

1 2(1 )[ ] (3 )[ ] [ ] 3[ ]1.76 2 2(1 )

4 [ ] [ ] [ ]

n nCO CO HC O

n

CO CO HC

2 2

2

1 1 11 3 (2 ) (3 )

2 21.76(1 )

4

n nO CO CO HC

n CO CO HC

( / )

/

stA F

A F

( / )/ stA F

A F

-----------------------------------------------------------------------------------------

Example：Find the air fuel ratio of the following measured data.

CO(%) HC(ppm) CO2(%) O2

2.88 1632 13.18 1.15

[ ]CO 0.0288, 2[ ]CO 0.1318, [ ]HC 0.0176,

2[ ]O 0.0115

2

1

[ ] [ ] [ ]CO CO HC

=5.612

Combustion Engineering ME Dept NCHU 頁 43

2

[ ]

[ ] [ ] [ ]

COa

CO CO HC

=0.1616

2

2

[ ]

[ ] [ ] [ ]

COb

CO CO HC

=0.7396

2

[ ]

[ ] [ ] [ ]

HCc

CO CO HC

=0.0988

2

2

[ ]

[ ] [ ] [ ]

Ox

CO CO HC

=0.0645

1.85n

2 2

2

1 1 11 3 (2 ) (3 )

2 21.76(1 )

4

n nO CO CO HC

n CO CO HC

=0.755 2

(1 ) ( 2 2 )4

nz a b x

=0.4388

1( )

2y n cn z 0.3948

3.76(1 )

4

nw

=4.1522

1.85 2 2 2

1.85 2 2 2 2

1.1043( 3.76 ) 0.1616 0.7396

0.0988 0.0645 0.3948 0.4388 4.1522

CH O N CO CO

CH O H H O N

-----------------------------------------------------------------------------------------

Assignment 2.18

Find the air fuel ratio of the following measured data.

(1). Four stroke motorcycle

CO(%) HC(ppm) CO2(%) O2(%)

1.02 1408 13.02 2.27

0.61 1232 12.44 2.99

1.40 1028 12.4 2.42

1.06 978 11.5 2.00

(2). Two stroke motorcycle

CO(%) HC(ppm) CO2(%) O2(%)

1.82 3330 8.34 4.91

4.40 5010 5.82 5.52

4.51 6900 3.9 6.84

3.90 5660 5.1 6.56

3.17 4140 3.44 9.18

Combustion Engineering ME Dept NCHU 頁 44

-----------------------------------------------------------------------------------------

(2.6.2) Combustion efficiency：

1.85 2 20.925CH CO H O Complete combustion

1.85 1.85 1.850.9012 0.0988CH CH CH

2 2 2 1.850.1616 0.7396 0.3948 0.4388 0.0988CO CO H H O CH

Partial burned energy

0.1616*(393509-110525)+0.3948*241818=141200 kJ

Unburned energy

0.0988*13.85*42000=57472 kJ

Total energy

42000kJ*13.85=581700 kJ

(141200+57472)/581700=0.0.3415

34.15% of energy has not been released during the combustion process.

The combustion efficiency is 65.85%.

The unreleased energy is composed of unburned fuel and partial burned

fuel. The unburned fuel takes 9.88% of energy. The partial burned energy

take another 24.27% of energy.

-----------------------------------------------------------------------------------------

Example：Find the combustion efficiency of a gasoline engine with the

following measured data.

CO(%) HC(ppm) CO2(%) O2(%)

2.88 1632 13.18 1.15

-----------------------------------------------------------------------------------------

Example：Find the combustion efficiency of a gasoline engine with the

following measured data.

CO(%) HC(ppm) CO2(%) O2(%)

4.40 5010 5.82 5.52

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 45

10.96kW 1.85kW

4.54kW

3.32kW

1.26kW

9.7kW

Work

Pumping

Friction

2.41kW

0.22kW

0.69kW

-----------------------------------------------------------------------------------------

Energy distribution in a motorcycle engine at full load

Mechanical

Burned Heat loss

Fuel Exhaust loss

Unburned

-----------------------------------------------------------------------------------------

(2.6.3) Oxygen rich combustion

The oxygen concentration in normal air is about 21%. The oxygen rich

combustion is to increase the concentration of oxygen artificially to

enhance combustion. The flame temperature could be raised due to the

less content of dilute gas.

2 2 2 2 2 2

1 1(1 )( ) ( 1)(1 ) (1 )

4 2 4 4n

n n n X nCH O XN CO H O O N

1

1y

X

: the molar fraction of oxygen

11X

y

The amount of oxygen required for oxygen rich combustion:

2 2 2 2 2 2 23.76 (1 ) 3.76O XN O N zO z O N

11

1

1 3.76 3.76

X y

z

Combustion Engineering ME Dept NCHU 頁 46

3.76 11

yz

y

For example, for oxygen rich combustion of methane:

4 2 2 2 2 22( ) 2 2CH O XN CO H O XN

For one mole of methane, if oxygen rich air is used as the oxidizer, the

amount of pure oxygen that has to be mixed with normal air is

2 7.52 21

yn z

y

For example, 40% oxygen rich air is used, 3.013 kmole of pure oxygen is

required for 1 kmole of methane.

-----------------------------------------------------------------------------------------

Example：Find the products of methane mixed with oxygen rich air and

the associated adiabatic flame temperature.

(1). y 30%

(2). y 40%

(3). y 80%

-----------------------------------------------------------------------------------------

Heat of combustion is determined by the amount of fuel only. It has

nothing to do with the amount of air.

2 2 2 2 2 2

1 1(1 )( ) ( 1)(1 ) (1 )

4 2 4 4n

n n n X nCH O XN CO H O O N

2 2, , ,2

f F HV f CO f H O

nh Q h h

2 2, , ,2

f F f CO f H O HV

nh h h Q

2 2 2 2 2 2, , ,

1( 1)(1 ) (1 )

2 2 4 4f F f CO CO f H O H O O N

n n n X nh h h h h h h

2 2 2 2

1( 1)(1 ) (1 )

2 4 4HV CO H O O N

n n X nQ h h h h

If the products are cooled to the room temperature, then the heat of

combustion would be the same no matter how much excess air is.

Combustion Engineering ME Dept NCHU 頁 47

2 22HV CO H O

nQ h h

However, if the load of combustor is at temperature higher than that of

room temperature, energy could be saved for oxygen rich combustion.

Heat of load: ( )L p b LQ mc T T

Heat of waste: ( )e p L aQ mc T T

Required flow rate: ( )

L

p b L

Qm

c T T

L ae L

b L

T TQ Q

T T

Since both TL and QL are fixed values, the higher the value of Tb is, the

lower the waste heat would be. The advantage of rich oxygen

combustion is that the burned gas temperature could be higher than that

of normal air.

-----------------------------------------------------------------------------------------

Example：In a steam boiler, coal is fired with 150% excess air. The flue

gas temperature is 200℃. Find the saving of fuel if 50% oxygen rich air

is used instead of normal air.

-----------------------------------------------------------------------------------------

Assignment 2.19

In a foundry, molten iron is kept warm with stoichiometric propane flame.

Find the saving of fuel if 40% oxygen rich air is used instead of normal

air. The melting point of iron is 1536℃.

-----------------------------------------------------------------------------------------

Assignment 2.20

An internal combustion engine runs on the fuel of propane with an

equivalence ratio of 0.9. The compression process is assumed to be

isentropic and the compression ratio is 9.0. The temperature and

the pressure before the compression stroke are 40℃ and 90 kPa. If

40% oxygen rich air is used instead of normal air, find the IMEP of the

engine.

T (K) P (kPa) u (kJ/kg) S (kJ/kg-K) V (m3/kg)

1 313 90

Combustion Engineering ME Dept NCHU 頁 48

2

3

4

-----------------------------------------------------------------------------------------

(2.6.4). Performance boosting with nitrous oxide

In vehicle racing, nitrous oxide (often referred to as just "nitrous") allows

the engine to burn more fuel by providing more oxygen than air alone,

resulting in a more powerful combustion.

2 2 2

1

2N O N O

The gas itself is not flammable at a low pressure/temperature, but it

delivers more oxygen than atmospheric air by breaking down at elevated

temperatures. Therefore, it is often mixed with another fuel that is easier

to deflagrate.

Nitrous oxide is stored as a compressed liquid; the evaporation and

expansion of liquid nitrous oxide in the intake manifold causes a large

drop in intake charge temperature, resulting in a denser charge, further

allowing more air/fuel mixture to enter the cylinder. Nitrous oxide is

sometimes injected into (or prior to) the intake manifold, whereas other

Combustion Engineering ME Dept NCHU 頁 49

systems directly inject right before the cylinder (direct port injection) to

increase power.

One of the major problems of using nitrous oxide in a reciprocating

engine is that it can produce enough power to damage or destroy the

engine. Very large power increases are possible, and if the mechanical

structure of the engine is not properly reinforced, the engine may be

severely damaged or destroyed during this kind of operation.

-----------------------------------------------------------------------------------------

Example：In an insulated rigid chamber, propane is mixed with

Combustion Engineering ME Dept NCHU 頁 50

stoichiometric amount of nitrous oxide. The temperature and pressure

in the chamber are 250℃ and 100 kPa. Find the adiabatic flame

temperature of the system.

C3H8+14N2O→3CO2+4H2O+14N2

Instead, if propane is mixed with stoichiometric amount of air, the

adiabatic flame temperature would be lower because more nitrogen

is present in the products.

C3H8 + 5(O2 + 3.76N2) → 3CO2 + 4H2O + 18.8N2

-----------------------------------------------------------------------------------------

Assignment 2.21

In an internal combustion engine, liquid gasoline is mixed with

stoichiometric amount of nitrous oxide instead of air. For the same

displacement of cylinder, find the increase in energy content at 100 kPa

and 300K.

CH1.85+2.925N2O→CO2+0.925H2O+2.925N2

CH1.85+1.4625(O2 + 3.76N2)→CO2+0.925H2O+5.499N2

For 1 liter of air, about 0.07977 grams of gasoline may be contained,

which corresponds to 3.35 kJ of energy. However, 1 liter of nitrous oxide

may contain much more energy.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 51

(2.7). Criteria of Equilibrium

An equilibrium state is the state that a system undergoes no more changes.

In an isolated system, the entropy increases for any spontaneous process.

( ) 0isolateddS

An isolated system always proceeds from states of lower entropy to states

of higher entropy until the maximum is reached. At this state, no further

natural process is possible and the system is in a state of equilibrium.

In a closed system, the system exchanges energy with its environment via

heat transfer and work. If an amount of heat Q is transferred from

system to its environment, the entropy change associated with the transfer

of heat is

QdS

T

, or TdS Q

The inequality holds for irreversible process occurring in the system.

However, according to the first law of thermodynamics, we have

Q dU PdV

Combing the inequality of entropy and the first law results the following

equation.

TdS dU PdV , or 0dU PdV TdS

The inequality implies that ,( ) 0U VdS , which means that entropy would

increase for all processes in a system in which U and V are kept constant.

As the system reaches the equilibrium state, in which no more changes

would occur, the entropy would be at a maximum value.

Since H U PV , dU dH PdV VdP , the inequality may be

converted to

0dH VdP TdS

The inequality implies that ,( ) 0H PdS , which means that the entropy

would reach the maximum value at the equilibrium state for a system in

which H and P are kept constant.

Combustion Engineering ME Dept NCHU 頁 52

There is another way to interpret the inequality, i.e., ,( ) 0S PdH . It

means that the enthalpy reaches a minimum value as the system is at the

equilibrium state if S and P are kept constant in the system.

Besides, A U TS , dU dA TdS SdT , the inequality may be also

converted to

0dA PdV SdT

The inequality implies that ,( ) 0T VdA , which means that the Helmholtz

function would reach the minimum value at the equilibrium state for a

system in which T and V are kept constant.

And G H TS , dG dH TdS SdT , the inequality may be also

converted to

0dG VdP SdT

The inequality implies that ,( ) 0T PdG , which means that the Gibbs

function would reach the minimum value at the equilibrium state for a

system in which T and P are kept constant.

When phase equilibrium is reached for a pure substance at constant

pressure and temperature, the Gibbs functions of different phase must be

equal to each other.

-----------------------------------------------------------------------------------------

Example: Calculate the Gibbs function of saturated water and that of

saturated vapor at 100℃.

f f s fg h T s = 419.04-373.14×1.3069 = -68.617

g g s gg h T s = 2676.1-373.14×7.3549 = -68.307

The error is caused by the uncertainty of the measured data.

( ) ( ) 0g f g s g f s f fg s fgg g h T s h T s h T s

Note: When two phases are in equilibrium, their Gibbs functions are the

same.

-----------------------------------------------------------------------------------------

(2.7.1). Properties of Gibbs Function

Combustion Engineering ME Dept NCHU 頁 53

For a system in which T and P are kept constant, any spontaneous process

would cause a decrease of the Gibbs function, and the Gibbs function

would reach the minimum value at the equilibrium state. As a result, the

criterion for the equilibrium state of a system in which T and P are kept

constant is that ,( ) 0T PdG .

It is thus of necessity to know how to evaluate the variations of the Gibbs

function of a system in which reactions occur.

The Gibbs function of pure substance is defined as

G H TS

dG VdP SdT

Assume that Gibbs function is a function of temperature and

pressure, we have

( , )G G T P

P T

G GdG dT dP

T P

P

GS

T

，

T

GV

P

Since 0S , we may conclude that 0P

G

T

. That is, G

decreases when T is raised at constant pressure.

Since 0V , we may conclude that 0T

G

P

. That is, G

increases when P is raised at constant temperature.

G HS

T

，

P

G G H

T T

P

G G H

T T T

Combustion Engineering ME Dept NCHU 頁 54

1 1

PP

G G G H

T T T T T T T

Gibbs-Helmholtz Equation

2

P

G H

T T T

T

GV

P

,

T

gv

P

For ideal gas, RT

vP

T

g RT

P P

0 0

ln

P

P

RT Pg dP RT

P P

0

0

lnP

g g RTP

For solid and liquid, v const.

0

P

P

g vdP vP

0

0( )g g v P P

-----------------------------------------------------------------------------------------

Example: Steam is compressed from a state of 200℃ and 1.5 bars to

another state of 10 bars adiabatically, calculate the change of the Gibbs

function.

h1=2872.9 kJ/kg, s1=7.6433 kJ/kg-K

s2=s1=7.6433 kJ/kg-K

h2=3390.2, T2=459 ℃

Combustion Engineering ME Dept NCHU 頁 55

g2-g1 = h2-T2s2 – (h1-T1s1) = (3390.2 – 2872.9) – (732×7.6433 -

473×7.6433) = -1462.3 kJ/kg

-----------------------------------------------------------------------------------------

(2.7.2). Gibbs function of mixture

If a system is composed n different species with the mole numbers ni

respectively.

1 2( , , , , , )rG G T P n n n

, , , , j

i

P n T n i T P n

G G GdG dT dP dn

T P n

Define

, , j

i

i T P n

G

n

，chemical potential of component i

i idG SdT VdP dn

By use of the relationship of exact differential

, , , ,j j

i k

k iT P n T P nn n

，the effect of the mole number of component j

on the Gibbs function of the component i.

Why does the mole number of component j affect the Gibbs function of

the component i ?

For a system in which T and P are kept constant, the variations of Gibbs

function in terms of the variations in the mole fractions of each

component is

i idG dn , i idg dx

Euler’s Theorem：

1 2( , , , )rf f z z z

f is homogeneous of degree m if

1 2 1 2( , , , ) ( , , , )m

r rf z z z f z z z

Combustion Engineering ME Dept NCHU 頁 56

then 1 2( , , , )

j

r i

i z

fmf z z z z

z

1 2( , , , , , )rH H T P n n n

i iH TS n

i iG H TS n

When 1in , G g

i i i ig x g x , where ig is the molar Gibbs function of the

component of the mixture. However, the molar Gibbs function of each

component can be expressed as the following.

0

0

ln ii i

Pg g RT

P

As a result, the Gibbs function of a mixture is determined by the molar

fraction as well as the Gibbs function at the reference state of each

component.

0

0

ln ii i

Pg g RT x

P

(2.7.3). Gibbs function of formation

The Gibbs function of each component at the reference state will not

be the same.

g h Ts

Consider the reaction of carbon and oxygen to form carbon dioxide.

2 2C O CO

C C Cg h Ts

2 2 2O O Og h Ts

Combustion Engineering ME Dept NCHU 頁 57

2 2 2CO CO COg h Ts

At 25℃ and 1 atm pressure

2 2 2 2 2 2( ) ( ) ( )CO C O CO CO C C O Og g g h Ts h Ts h Ts

2 2 2 2( ) ( )CO C O CO C Oh h h T s s s

2 2 2 2

0 0 0

, , ,( ) ( )f CO f C f O CO C Oh h h T s s s

2,f COh －393520 kJ/kmole，2

0

COs 213.69 kJ/kmole-K

,f Ch 0， 0

Cs 5.74 kJ/kmple-K

2,f Oh 0，2

0

Os 205.03 kJ/kmole-K

2 2CO C Og g g -393520 - 298.14×(213.69-5.74-205.03)

= -394390 kJ/kmole

The Gibbs functions can be defined as

2,f COg -394390 kJ/kmole

,f Cg 0，2,f Og 0

Gibbs function of formation is the change of Gibbs function when a

compound is formed from its elements.

The standard Gibbs function of formation is Gibbs function of any

compound at 25℃ and 1 atm pressure.

The Gibbs function of formation for element at natural state is defined to

be zero.

It is noted that the Gibbs function of formation does not equal to the

enthalpy of formation minus the product of temperature and entropy.

2 2 2, , 0f CO f CO COg h T s

, , 0f C f C Cg h T s

The Gibbs function of formation is the Gibbs function of one compound

relative to the Gibbs function of another compound.

Combustion Engineering ME Dept NCHU 頁 58

Since a compound is formed from elements, it is important to know the

relative relationships among the compound and its forming elements.

However, an element can not be formed from another element, it is

meaningless to discuss the relative Gibbs function between two distinct

elements. As a result, the Gibbs functions of formation for all elements

at natural state can be set to zero.

-----------------------------------------------------------------------------------------

Example: Calculate the Gibbs function of formation H2O at 298K and 1

bar.

2 2 2 2 2 2 2 2 20.5 ( ) ( ) 0.5( )H O H O H O H O H H O Og g g h Ts h Ts h Ts

2

0

,f H Og -241820- 298×(188.72-0.5×205.03-130.57) = -228599 kJ/kmole

-----------------------------------------------------------------------------------------

The Gibbs function of formation of a compound at states other than 25℃

and 1 atm pressure can be obtained from its standard Gibbs function of

formation and the deviations of pressure and temperature.

0 0

,0 0 0 0( ) ( )f fg g h h Ts T s

The Gibbs function of formation at elevated temperature for the can be

found in Table A.1 ~ A.12 in the text book.

-----------------------------------------------------------------------------------------

Example: Calculate the Gibbs function of formation H2O at 1000K and

1 bar.

2,f H Og -228599 + (25978 – 11354 – 20686) - 1000×(232.597-

243.471×0.5-166.114) + 298×(188.72-0.5×205.03-130.57) = -192630

kJ/kmole

-----------------------------------------------------------------------------------------

Example: Methane is burned with stiochiometric amount of air.

Assume the reaction is complete, and the products are at 1000 K and 1

bar. Calculate the Gibbs function of the products.

Combustion Engineering ME Dept NCHU 頁 59

CH4 + 2 (O2 + 3.76 N2 ) → CO2 + 2 H2O + 7.52 N2

The molar fraction of each component is

CO2 9.5%, H2O 19%, N2 71.5%

2

0

COg -395939 kJ/kmole

2

2 2

0

0

lnCO

CO CO

Pg g RT

P = -395939 + 8.314×1000 ln(0.095) = -415509

2

0

H Og -192652 kJ/kmole

2

2 2

0

0

lnH O

H O H O

Pg g RT

P = -192652 + 8.314×1000 ln(0.19) = -206459

2

0

Ng 0 kJ/kmole

2

2 2

0

0

lnN

N N

Pg g RT

P = 0 + 8.314×1000 ln(0.715) = -2789

2 2 2

1( 2 7.52 )

10.52i i CO H O Ng x g g g -80741 kJ/kmole

-----------------------------------------------------------------------------------------

(2.8). Equilibrium constant

Assuming that a chemical reaction takes place in that A and B reacts to

become C and D with the associated mole numbers as following.

A B C DA B C D

As the equilibrium is reached, the Gibbs function should be at a minimum

value, and the criterion of equilibrium is

0A B C DdA dB dC dD

However, the reaction equation shows that consumption of A moles of

A would accompany consumption of B moles of B, as well as

appearance of C moles of C and D moles of D. As a result, the

rates of change of each component in the reaction can be represented as

AdA d ，BdB d ，

CdC d ，DdD d

The equilibrium criterion can thus be represented as

Combustion Engineering ME Dept NCHU 頁 60

0A A B B C C D Dd d d d

0A A B B C C D D

0A A B B C C D Dg g g g

Since the Gibbs function of each component is related to its Gibbs

function of formation and partial pressure,

0

0

( ) ln AA A u

Pg g T R T

P

0 0

0 0

0 0

0 0

ln ln

ln ln 0

A BA u A B u B

C DC u C D u D

P Pg R T g R T

P P

P Pg R T g R T

P P

0 0 0 0 0 0

0 0

ln

C D

A B

C D

C C D D A A B B u

A B

P P

P Pg g g g R T

P P

P P

Let 0 0 0 0

C C D D A A B BG g g g g , and note that the partial pressure

can be decomposed as product of molar fraction and absolute pressure

ratio.

0 0 0

A AA

P P P Px

P P P P

The equilibrium criterion becomes as

0

ln

C D A BC D

A B

C D

u A B

x xG P

R T Px x

We may have the final form of the equilibrium criterion as

Combustion Engineering ME Dept NCHU 頁 61

0

exp

C D A BC D

A B

C D

p

uA B

x x P GK

P R Tx x

Where pK is denoted as the equilibrium constant.

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2

1+

2CO CO O

2 2

0 0 0

1

1

2O CO COG g g g = -285948 – (-396410) = 110462 kJ/kmole

1 expp

u

GK

R T

1.303×10-3

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2 2

1

2H O H O

2 2 2

0 0 0

2

1

2H O O HG g g g -135643 – (0) = -135643 kJ/kmole

2 expp

u

GK

R T

3489.47

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2 2H CO H O CO

2 2 2 2 2 2 2 2

0 0 0 0 0 0 0 0 0 01 1( ) ( )2 2

H O CO CO H O CO CO H O O HG g g g g g g g g g g

1 1G G G

1 2 1 2

1 2exp e e eu u u

G G G G

R T R T R T

p p p

u

GK K K

R T

pK 4.547

-----------------------------------------------------------------------------------------

Assignment 2.21

Calculate the equilibrium constant of the reaction at temperature of 2000

K and pressure of 10 bars.

2 2 2N O NO

Combustion Engineering ME Dept NCHU 頁 62

-----------------------------------------------------------------------------------------

If an equilibrium equation can be composed with two other equilibrium

equations, then the equilibrium constant of this equation can be expressed

as the product of the equilibrium constants of the other equations.

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2 2H O CO H CO

-----------------------------------------------------------------------------------------

If a reaction is the reverse of other reaction, then the equilibrium constant

of this equation is the reciprocal of the equilibrium constant of the other

equation.

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2 2H CO H O CO

-----------------------------------------------------------------------------------------

If a reaction is expressed as N multiple of another reaction, the

equilibrium constant of this equation is the Nth order of that of the other

equation.

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 22 2 +CO CO O

-----------------------------------------------------------------------------------------

Equilibrium constants of come commonly used reactions

2log ln( )1000

p

T BK A C DT ET

T

1: 2

1

2H H

2: 2

1

2O O

3: 2 2

1 1

2 2H O OH

Combustion Engineering ME Dept NCHU 頁 63

4: 2 2

1 1

2 2N O NO

5: 2 2 2

1

2H O H O

6: 2 2

1

2CO O CO

A B C D E

1 0.432168e+00 -0.112464e+05 0.267269e+01 -0.745744e-04 0.242484e-08

2 0.310805e+00 -0.129540e+05 0.321779e+01 -0.738336e-04 0.344645e-08

3 -0.141784e+00 -0.213308e+04 0.853461e+00 0.355015e-04 -0.310227e-08

4 0.150879e-01 -0.470959e+04 0.646096e+00 0.272805e-05 -0.154444e-08

5 -0.752364e+00 0.124210e+05 -0.260286e+01 0.259556e-03 -0.162687e-07

6 -0.415302e-02 0.148627e+05 -0.475746e+01 0.124699e-03 -0.900227e-08

-----------------------------------------------------------------------------------------

Example: Calculate the equilibrium constant of the reaction at

temperature of 2000 K and pressure of 10 bars.

2 2 2H CO H O CO

-----------------------------------------------------------------------------------------

Assignment 2.22

Calculate the equilibrium constant of the following reaction at

temperature in the range of 1000~3000K.

2 2 2H CO H O CO

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 64

(2.9). Equilibrium calculations

Procedures of equilibrium calculation

For any reaction in consideration, the procedure to calculate the

equilibrium state should include the following steps.

(1). Conservation of each kind of atom that participates the reaction

before and after reaction. If there are n kinds of atoms present in the

reaction, there would be n conservation equations.

(2). Criteria of equilibrium. If m reactions are in equilibrium, there

would be m equilibrium equations.

The molar fraction and concentration.

uPV nR T

[ ] i i ii i

u u u

n P P P PC x

V R T P R T R T

-----------------------------------------------------------------------------------------

Example: Carbon monoxide reacts with excess oxygen at temperature of

2000 K and pressure of 10 bars. Calculate the CO remaining in the

exhaust.

2 2 2+CO O aCO bCO cO

Balance in C atoms: 1a b , 1b a

Balance in O atoms: 2 2 3a b c , 12

ac

1 1 22 2

a aa b c

2 2

1+

2CO CO O

2

2

42

2

CO

a a ax

aa b c a

1 2 2

42

2

CO

b a ax

aa b c a

Combustion Engineering ME Dept NCHU 頁 65

2

1224

22

O

ac a

xaa b c a

2 2

4CO

aP P

a

2

2

4CO

aP P

a

2

2

4O

aP P

a

0.50.5

0.5

0

(2 2 )(2 )

2 (4 )P

a a PK

a a P

=1.303×10-3

1 2

4

a a

a a

= 4.12×10-4

a= 0.9994

2 2 2+ 0.9994 0.0006 0.5003CO O CO CO O

-----------------------------------------------------------------------------------------

Example: Hydrogen is produced from steam and carbon monoxide in a

water shift reaction. It is known that steam and carbon monoxide are

mixed with the mole ratio of 1:1 at 800 K and 1 atm, calculate the mole

fraction of hydrogen at the outlet.

2 2 2H O CO H CO

1 1 0 0

1-y 1-y y y

2 2

2

1 1 1 1

0 (1 )(1 )

CO H

p

CO H O

x x P y yK

x x P y y

=0.22

y=0.319

2Hx 16%

-----------------------------------------------------------------------------------------

Assignment 2.23

Hydrogen is reacting with 140% of theoretical air in an insulated constant

volume chamber. The initial temperature and pressure are 298K and

100kPa.

1. Determine the adiabatic flame temperature.

Combustion Engineering ME Dept NCHU 頁 66

2. Determine the final pressure.

3. Determine the equilibrium molar fractions of O, N2, NO, N, O2, and

OH.

2 2O O

2 2N N

2 2 2O N NO

2 2

12

2H O O OH

-----------------------------------------------------------------------------------------

(4.2.2). Equilibrium of hydrocarbon fuel burning

2 2

1( )( 3.76 )

4n m

mC H n O N

2 2 2 2 2, , , , , , , , , , ,.........CO CO H O H O N OH O H N NO

For a lean combustion, CO2, H2O, and O2 are abundant in the products of

reaction, dissociation reactions do not affect the concentrations of these

components. As a result, it is reasonable to obtain the concentrations of

CO2, H2O, and O2 directly from atom balance.

-----------------------------------------------------------------------------------------

Example: A gas turbine engine runs on natural gas with an equivalence

ratio of 0.3. The inlet air is at 25℃ and 101.3 kPa. Fuel is injected into

the combustion chamber at 25℃. The pressure ratio of the compressor

is 10. The efficiency of compressor is 0.85. Assume that NO is in

equilibrium. Find the molar fraction of NO and CO at the exit of

combustor. Assume that air is an ideal gas.

2 2

1 1

2 2N O NO

2 2

1+

2CO CO O

-----------------------------------------------------------------------------------------

However, for rich and stoichiometric mixtures, O2 and H2 are rare in the

products of reaction, dissociation reactions play an important role to

determine the concentrations of these components.

Combustion Engineering ME Dept NCHU 頁 67

-----------------------------------------------------------------------------------------

Example: Methane is burned with stiochiometric amount of air.

Assume the products are at 2000 K and 10 bar. Calculate the mole

fraction of each component assuming that the products contain CO, CO2,

H2O, H2, O2, and N2 only.

4 2 2 2 2 2 2 22( 3.76 )CH O N aCO bCO cH O dH eO fN

Carbon balance: a + b = 1

Hydrogen balance: 2c + 2d = 4

Oxygen balance: 2a + b + c + 2e =4

Nitrogen balance: f = 7.52

There are six unknowns in the reaction equation. However, there are

only four balance equations. We need two more equations. These two

equations should be derived from the equilibrium between components.

We propose the following equations.

2 2

1

2CO CO O

2 2 2

1

2H O H O

2

2

0.5 0.50.5

1

0 0

CO O

p

CO

x x P b e PK

x P Pa

2 2

2

0.5 0.50.5

2

0 0

H O

p

H O

x x P d e PK

x P Pc

a + 1 + c + 2e =4, e = 1.5 – 0.5(a+c)

1 2 7.52 10.52 12.02 0.5( )a b c d e f e e a c 0.5

01

(1 ) 1.5 0.5( )

12.02 0.5( )p

a a c PK

Pa a c

0.5

02

(2 ) 1.5 0.5( )

12.02 0.5( )p

c a c PK

Pc a c

This is a set of nonlinear simultaneous equations that has to be solved

numerically.

Combustion Engineering ME Dept NCHU 頁 68

0.5

01(1 ) (1 ) (2 ) 24.04 ( )p

Pa a c K a a c

P

0.5

02(2 ) (1 ) (2 ) 24.04 ( )p

Pc a c K c a c

P

0.5

01 2(1 ) (2 ) (1 ) (2 ) 24.04 ( ) p p

Pa c a c a c K a K c

P

20.5 3

01 2(1 ) (2 ) 24.04 ( ) p p

Pa c a c K a K c

P

10.5 0.53

0 01 2 1(1 ) 24.04 ( ) 24.04 ( )p p p

P Pa a c K a K c K a a c

P P

0.5

01

10.5 3

01 2

24.04 ( )

(1 )

24.04 ( )

p

p p

PK a a c

Pa

Pa c K a K c

P

0.5

02

10.5 3

01 2

24.04 ( )(1 )

(2 )

24.04 ( )

p

p p

PK c a c a

Pc

Pa c K a K c

P

Initial guess of a and c would be 1a and 2c . Several iterations

would converge the values of a and c to the final solutions.

-----------------------------------------------------------------------------------------

Assignment 2.23

Propane is burned with insufficient air with an equivalence ratio of

1.1. If the temperature is 2000 K and the pressure is 10 bar, calculate

the equilibrium constants and find the equilibrium concentrations of

the products. Assume the combustion products contain CO, H2,

H2O, CO2, and N2 only.

-----------------------------------------------------------------------------------------

Example: Methane is burned with stiochiometric amount of air.

Combustion Engineering ME Dept NCHU 頁 69

Assume the products are at 1000 K and 10 bar. Calculate the mole

fraction of each component assuming that the products contain CO, CO2,

H2O, H2, O2, H, O, OH, NO, N, and N2.

4 2 22( 3.76 )CH O N

2 2 2 2 2aCO bCO cH O dH eO fN gOH hO iH jN kNO

Carbon balance: a + b = 1

Hydrogen balance: 2c + 2d +g + I = 4

Oxygen balance: 2a + b + c + 2e + g + h + k =4

Nitrogen balance: 2f + j + k = 15.04

There are eleven unknowns in the reaction equation. However, there are

only four balance equations. We need seven more equations. These

seven equations should be derived from the equilibrium between

components.

We propose the following equations.

2 2

1

2CO CO O

2 2 2

1

2H O H O

2 2O O

2 2H H

2 2N N

2 2 2H O OH

2 2 2N O NO

2

2

0.5 0.50.5

1

0 0

CO O

p

CO

x x P b e PK

x P Pa

2 2

2

0.5 0.50.5

2

0 0

H O

p

H O

x x P d e PK

x P Pc

2

2 2

3

0 0

Op

O

x P h PK

x P e P

2

2 2

4

0 0

Hp

H

x P i PK

x P d P

Combustion Engineering ME Dept NCHU 頁 70

2

2 2

5

0 0

Np

N

x P j PK

x P f P

2 2

2 2

6OH

p

O H

x gK

x x de

2 2

2 2

7NO

p

O N

x kK

x x fe

a b c d e f g h i j k

This is a set of nonlinear simultaneous equations that has to be solved

numerically.

-----------------------------------------------------------------------------------------

Example: Methane is burned with stiochiometric amount of air

adiabatically in a constant pressure combustor. Assume the reactants

are at 298 K and 10 bar. Calculate the mole fraction of each component

assuming that the products contain CO, CO2, H2O, H2, O2, H, O, OH, NO,

N, and N2.

4 2 22( 3.76 )CH O N

2 2 2 2 2aCO bCO cH O dH eO fN gOH hO iH jN kNO

There are twelve unknowns in the reaction equation. Besides the mole

fractions that are not known, the temperature is also unknown. We need

eight more equations. These seven equations should be derived from the

equilibrium between components. The last equation is derived from the

conservation of energy before and after combustion.

HR=HP

4 2 2 2 2 2 22( 3.76 )CH O N CO CO H O H Oh h h a h b h c h d h e h

2N OH O H N NOf h g h h h i h j h k h

The mole fractions to be determined are functions of temperature.

However, temperature should be solvable only when all the mole

fractions are already known. As a result, temperature should be guessed

before calculations proceed. The energy equation is used to check if the

guessed temperature is correct or not.

-----------------------------------------------------------------------------------------

Combustion Engineering ME Dept NCHU 頁 71

Example: Methane is burned with stiochiometric amount of air in a

constant volume combustor. Assume the reactants are at 298 K and 1 bar.

Calculate the mole fraction of each component assuming that the products

contain CO, CO2, H2O, H2, O2, H, O, OH, NO, N, and N2.

4 2 22( 3.76 )CH O N

2 2 2 2 2aCO bCO cH O dH eO fN gOH hO iH jN kNO

There are thirteen unknowns in the reaction equation. Besides the mole

fractions that are not known, the temperature and the pressure are also

unknown. We need nine more equations. These seven equations

should be derived from the equilibrium between components. One

equation is derived from the conservation of energy before and after

combustion, and the other equation is derived form the state equation of

ideal gas.

UR=UP

4 2 2 2 2 2 22( 3.76 )CH O N CO CO H O H Ou u u a u b u c u d u e u

2N OH O H N NOf u g u h u i u j u k u

in RTP

V

The mole fractions to be determined are functions of temperature and

pressure. However, temperature should be solvable only when all the

mole fractions are already known. As a result, temperature and pressure

should be guessed before calculations proceed. The energy equation is

used to check if the guessed temperature is correct or not, and the ideal

gas equation is used to check if the pressure is correct or not.

-----------------------------------------------------------------------------------------

(2.10). STANJAN for equilibrium calculations

The equilibrium calculation can be carried out with the STANJAN

program.

-----------------------------------------------------------------------------------------

Example: Methane is burned with air at 2000K and 10 bar. If the

combustion products contain CO, CO2, O2, H2, H2O, HO, H, O, NO,

N, and N2, find the equilibrium concentrations of products at

equilibrium ratio of 1.0.

-----------------------------------------------------------------------------------------

Example: Mixtures of methane and air at 300K and 100 kPa are

Combustion Engineering ME Dept NCHU 頁 72

burned in a constant pressure process. If the combustion products

contain CO, CO2, O2, H2, H2O, HO, H, O, NO, N, and N2, find the

equilibrium concentrations of products at equilibrium ratio of 1.0.

-----------------------------------------------------------------------------------------

Example: Mixtures of methane and air at 300K and 100 kPa are

burned in a constant volume process. If the combustion products

contain CO, CO2, O2, H2, H2O, HO, H, O, NO, N, and N2, find the

equilibrium concentrations of products at equilibrium ratio of 1.0.

-----------------------------------------------------------------------------------------

Example: An Otto engine runs with stoichiometric methane mixture.

The volume of the cylinder is 1L, and the compression ratio is 10. The

initial temperature and pressure inside cylinder are 300 K and 1 atm

respectively. Calculate the work output, assuming that the products

contain CO, CO2, H2O, H2, O2, H, O, OH, NO, N, and N2.

-----------------------------------------------------------------------------------------

Comparison of adiabatic flame temperature with difference assumptions

in the final species of equilibrium:

A：No dissociation

4 2 2 2 2 2 2

2 2 2( 3.76 ) 2 ( 2) 3.76CH O N CO H O O N

, 1

4 2 2 2 2 2

2 2( 3.76 ) 2 3.76CH O N aCO bCO H O N

, 1

B：Weak dissociation

4 2 2 2 2 2 2 2

2( 3.76 )CH O N aCO bCO cH O dH eO fN

C：Strong dissociation

4 2 2

2( 3.76 )CH O N

2 2 2 2 2aCO bCO cH O dH eO fN gOH hO iH jN kNO

A B C

0.8

0.9

1.0

1.1

Combustion Engineering ME Dept NCHU 頁 73

1.2

-----------------------------------------------------------------------------------------

Assignment 2.24

Propane is burned with air at 2000K and 10 bar. If the combustion

products contain CO, CO2, O2, H2, H2O, HO, H, O, NO, N, and N2,

find the equilibrium concentrations of products at equilibrium ratios

from 0.7 to 1.3 with an increment of 0.1. The calculations may be

conducted with STANJAN. Plot the results of calculations.

-----------------------------------------------------------------------------------------

Assignment 2.25

Mixtures of propane and air at 300K and 100 kPa are burned in a

constant pressure process. If the combustion products contain CO,

CO2, O2, H2, H2O, HO, H, O, NO, N, and N2, find the adiabatic

flame temperature and the equilibrium concentrations of products at

equilibrium ratios from 0.7 to 1.3 with an increment of 0.1. The

calculations may be conducted with STANJAN. Plot the results of

calculations.

-----------------------------------------------------------------------------------------

Assignment 2.26

Mixtures of propane and air at 300K and 100 kPa are burned in a

constant volume process. If the combustion products contain CO,

CO2, O2, H2, H2O, HO, H, O, NO, N, and N2, find the adiabatic

flame temperature, the final pressure, and the equilibrium

concentrations of products at equilibrium ratios from 0.7 to 1.3 with

an increment of 0.1. The calculations may be conducted with

STANJAN. Plot the results of calculations.

-----------------------------------------------------------------------------------------

Assignment 2.27

A gas turbine engine runs with mixtures of propane and air. The

equivalence ratio is 0.4. The inlet air is at 300 K and 1 bar. The pressure

ratio is 10.0. Fuel is injected into the combustion chamber at 300 K and

10 bars. Calculate the work output. Assume that the products contain CO,

CO2, H2O, H2, O2, H, O, OH, NO, N, and N2.

-----------------------------------------------------------------------------------------

Assignment 2.28

An Otto engine runs with stoichiometric mixtures of propane and air.

The inlet condition is 300K and 100 kPa. The compression ratio is

11. If the combustion products contain CO, CO2, O2, H2, H2O, HO,

Combustion Engineering ME Dept NCHU 頁 74

H, O, NO, N, and N2, find the equilibrium concentrations of

products during the expansion stroke. The calculations may be

conducted with STANJAN. Plot the results of calculations.

-----------------------------------------------------------------------------------------

(2.11). Applications of equilibrium calculations

(2.11.1) Determination of A/F with [ ]CO , 2[ ]CO , and [ ]HC

Reaction of gasoline and air inside the internal combustion engine

can be expressed as the following.

2 2 2 2 2 2 2

14 ( 3.76 )n n

n

CH O N aCO bCO cCH xO yH zH O wN

If only the concentrations of CO, CO2, and unburned hydrocarbons in the

engine exhaust are measured, just like the normal operation of emission

check in the motorcycle smoke test stand, the air fuel ratio can not be

determined because of insufficient conditions.

There are eight variables to be determined. , , , , , , ,a b c x y z w

From conservations of atoms, we have four equations.

C: 1 a b c

O: 2

(1 ) 2 24

na b x z

H: 2 2n cn y z

N: 3.76

(1 )4

nw

We need four more equations to find the values of those unknowns.

Three of them may be in the form of concentration which can be obtained

by direct measurement.

[ ]a

CO

, 2[ ]

bCO

,

1[ ]

cHC

a b c x y w

We need one more condition to solve these variables. Usually, it is

assumed that CO, CO2, H2, and H2O are in equilibrium in the exhaust.

Combustion Engineering ME Dept NCHU 頁 75

2 2 2H CO H O CO

2 2

2

[ ][ ]

[ ][ ]

H COK

H O CO

[ ]a CO , 2[ ]b CO , [ ]c HC

2[ ] [ ] [ ] 1a b c CO CO HC

2

1

[ ] [ ] [ ]CO CO HC

2

[ ]

[ ] [ ] [ ]

COa

CO CO HC

2

2

[ ]

[ ] [ ] [ ]

COb

CO CO HC

2

[ ]

[ ] [ ] [ ]

HCc

CO CO HC

2 2

2

[ ][ ]

[ ][ ]

H CO ybK

H O CO za

ay Kz

b

1( )

2

ay n cn z Kz

b

1( 1) ( )

2

aK z n cn

b

1

21

n cnz

aK

b

21

aK

a n cn by Kzab

Kb

1 1 1 1 1(1 ) ( 2 ) (1 ) ( 2 )

4 2 4 2 21

n n n cnx a b z a b

aK

b

3.76(1 )

4

nw

1 1 1 3.76(1 ) ( 2 ) (1 )

4 2 2 2 41 1

aK

n n cn n cn nba b c a ba a

K Kb b

Combustion Engineering ME Dept NCHU 頁 76

11 4.762 (1 )2 2 4

1

aK

n cn nba ca

Kb

2

2 2

1 11 [ ] [ ] [ ] [ ]

4.762 2 (1 )[ ] [ ] [ ] 2 [ ] [ ] 4

CO HC CO K COn cn n

CO CO HC CO K CO

2

2

2

2

1 1 1

4.76(1 )4

1 31 ( )

4 2 4 4

n CO CO HC

kn CO CO COn nCO CO HC

CO k CO

( / )/ stA F

A F

2

2 2

2 2

2

[ ] [ ]

1 1 [ ] 2[ ] 1 [ ] [ ] [ ](1 ) ( )

[ ]4 2 [ ] [ ] [ ] [ ] [ ] [ ] 21

[ ]

CO COn

n CO CO CO CO HCx

COCO CO HC CO CO HCK

CO

22

2

2

1 1 1 [ ] [ ](1 ) ([ ] 2[ ] )

[ ]4 2 [ ] [ ] [ ] 21

[ ]

n n CO COx CO CO

COCO CO HCK

CO

2[ ]x

O

-----------------------------------------------------------------------------------------

Example：The measured emission data of a motorcycle at idling are

shown as the following.

CO(%) HC(ppm) CO2(%) O2

2.88 1632 13.18 1.15

Find the air fuel ratio with and without the oxygen concentration.

-----------------------------------------------------------------------------------------

Assignment 2.29

Find the air fuel ratio of the following measured data without the oxygen

concentration, and then compare the oxygen concentration obtained with

the measured one.

(1). Four stroke motorcycle CO(%) HC(ppm) CO2(%) O2

Combustion Engineering ME Dept NCHU 頁 77

1.02 1408 13.02 2.27

0.61 1232 12.44 2.99

1.40 1028 12.4 2.42

1.06 978 11.5 2.00

(2). Two stroke motorcycle CO(%) HC(ppm) CO2(%) O2

1.82 3330 8.34 4.91

4.40 5010 5.82 5.52

4.51 6900 3.9 6.84

3.90 5660 5.1 6.56

3.17 4140 3.44 9.18

-----------------------------------------------------------------------------------------