Lecture –1: Basics of Combustion EngineeringCombustion Stoichiometry

1.1 Introduction

1.2 Definitions

1.3 Combustion stoichiometry for gaseous fuels

1.4 Combustion stoichiometry for liquid and solid fuels

1.5 Combustibles burnout for solid fuels

Examples of industrial combustion systems

Boiler fired with pulverised coal

Pulverised coal flame

Examples of industrial combustion systems

Modern reheating furnace

Examples of industrial combustion systems

Flameless oxidation

1.2 Definitions

Chemical Reactions, Atoms and Molecules in Combustion

H12

O H O2 2 2+ →

The atoms are conserved (neither created nor destroyed)

Molecules are not conserved

Water is the productHydrogen and oxygen arereactants

Atoms relevant in combustion are: C,H,O,N,S,Cl

1.1 Definitions

Compounds of carbon and hydrogen are called hydrocarbons.

HYDROCARBONS

Aliphatic

Alkanes CnH2n+2Alkenes CnH2nAlkynes CnH2n-2

Alicyclic

CH2

CH2 CH2

Aromatic

BenzeneBenzene derivativesNaphtalene

The first ten members of the unbranched-chain alkanes series are:

CH4 methane C6H14 hexaneC2H6 ethane C7H16 heptaneC3H8 propane C8H18 octaneC4H10 butane C9H20 nonaneC5H12 pentane C10H22 decane

Table 1.1 Names of aliphatic hydrocarbons

CnH2n+1-CnH2n-2CnH2nCnH2n+2n

C5H11- PentylC5H8 PentyneC5H10 PenteneC5H12 Pentane5

C4H9-ButylC4H6 ButyneC4H8 ButeneC4H10 Butane4

C3H7-PropylC3H6 ButyneC3H10PropeneC3H8 Propane3

C2H5- EthylC2H2 EthyneC2H4 EtheneC2H6 Ethane2

CH3- MethylCH4 Methane1

Alkylgroup

AlkyneAlkeneAlkaneNo.of Catoms

Other molecules relevant in combustion are:

Haloalkanes R-X CH3Cl (chloromethane)

Alcohols R-OH C2H5OH (ethanol)

Amines R-NH2 CH3NH2 (methylamine)

Aldehyde R-COH CH3COH (ethanal)

Ketons R-CO-R CH3COCH3 (propanone)

Carboxylic Acid R-COOH CH3COOH (ethanoic acid)

Amount of substances, mole and mass fractions

1 mole contains 6.023 x 1023 particles (atoms, molecules)

For a mixture of species:

n (total number of moles) =∑ni

xnnii= M x Mmean i i=∑

Mole (volume) fractions and mass fractions:

wi = =num ber of kg of species " i"

total num ber of kg in the system

= =∑ ∑

n Mn M

x Mx M

i i

k kk

i i

k kk

xi = =number of moles of species "i" in 1kg of mixture

total number of moles in 1kg of mixture

= =∑

w MM

w Mw M

i i

m e a n

i i

k kk

//

//1

Equation of state for gases and gas mixtures

F p T c( , , ) = 0or

F p T( , , )ρ = 0

The perfect gas equation:

pV nR T= or cp

R T=

ρ = =∑

pMRT

p

RTwM

mean

i

i

The perfect gas law:

pV nR T=Under constant pressure and temperature one mole (kmol) of any ideal gas occupies the same volume.

At normal conditions:p = 760 Tr (1 Tr=133,322 N/m2)T = 273.15 K (0 C)

1 kmol of gas = 22.418 mn3

1 mol of gas = 22.418 dmn3

R = 8,314 J/kmol/K

1.2 Combustion Stoichiometry for Gaseous Fuels

Stoichiometric Combustion

Combustion is said to be stoichiometric if fuel and oxidizerconsume each other completely forming only CO2 and H2O

CH 2O 2H O CO4 2 2 2+ → + stoichiometric

CH 3O 2H O CO O4 2 2 2 2+ → + + lean

CH O H O 0.5CO 0.5CH4 2 2 2 4+ → + + rich

1.3 Combustion Stoichiometry for Gaseous Fuels

Mole fraction of fuel in a stoichiometric mixture

1 kmol fuel O products (CO H O)2 2+ → +ν 2

xfuel stoich in oxygen, _ _ =+

=number of moles of fuel

total number of moles (fuel oxygen)

=+1

1 ν

x fuel stoich in air, _ _ / . .=

+=

+1

1 0 211

1 4 762ν ν

For example:

CO 0.5O CO2 2+ →xfuel,stoich_in_oxygen = 1/1.5=2/3xfuel,stoich_in_air = 1/(1+0.5/0.21)=0.2958

C H 5O 3CO 4H O3 8 2 2 2+ → +

Xfuel,stoich_in_oxygen = 1/(1+5)=1/6

xfuel,stoich_in_air = 1/(1+5/0.21)=0.0403

Excess air ratio (air equivalence ratio)

λ = =( / )

( / )( / )

( / )x xx x

w ww w

air fuel

air fuel stoich

air fuel

air fuel stoich

Fuel equivalence ratio Φ =1λ

Rich combustion Φ > 1λ < 1Stoichiometric combustion λ = 1Lean combustion λ > 1 Φ < 1

Minimum oxygen requirement for a mixture of gaseous fuels

Fuel composition

Molar fractions of species are known

xH2, xCO, xCH4,

xO2, xCnHm

Chemical reactionsH 0.5O H O2 2 2+ →CO 0.5O CO2 2+ →

CH O CO H O4 2 2 2+ → +2 2Om/2HnCOm/4)O(nHC 222mn +→++

Minimum oxygen requirement is then uniquely determined

lO2,min = 0.5xH2+ 0 5. xCO+2xCH4 mnHCm/4)x(n++ − xO2

lO2,min is in kmol O2/kmol fuel

Minimum air requirement for a mixture of gaseous fuels

ll0.21

4.7619 lair,minO ,min

O ,min2

2= = ×

λ = =amount of dry air supplied per kmol of fuel

minimum dry air requirement per kmol of fuell

lair

air,min

The excess air ratio

Amount of combustion products

Fuel composition Chemical reactions

Composition and amount (per a unit of fuel) of combustion products are uniquely determined for a given excess air ratio

Fuel component

Combustion productsCO2 H2O N2 O2

H2 - 1 - -CO 1 - - -CH4 1 2 - -CnHm n m/2 - -

O2 - - - -N2 - - 1 -

Vwet = xH2+ xCO+ 3xCH4

+ +(n m / 2)xC Hn m+ xN2 +0.79lair,min

+ −( )λ 1 lair,min

Vdry = + xCH4 mnHCnx+ + xN2+0.79lair,minCOx min,)1( airl⋅−+ λ

Composition of combustion products

Fuel composition Chemical reactions

Composition and amount (per a unit of fuel) of combustion products are uniquely determined for a given excess air ratio

Fuel component

Combustion productsCO2 H2O N2 O2

H2 - 1 - -CO 1 - - -CH4 1 2 - -CnHm n m/2 - -

O2 - - - -N2 - - 1 -

Vwet = xH2+ xCO+ 3xCH4

+ +(n m / 2)xC Hn m+ xN2 +0.79lair,min

+ −( )λ 1 lair,min

Vdry = + xCH4+ nxC Hn m

+ xN2+0.79lair,min min,)1( airl⋅−+ λCOx

1.3 Combustion stoichiometry for liquid and solid fuels

For solid and liquid fuels the fuel composition is usually expressesin mass fraction (percentage)

The following applies:

c+h+s+o+n+moisture+ash =1

cc

1 moisturedryas_received=

−

cc

1 moisture ashdry_ash_freeas_received=

− −

1%3%1%

4%

82%

4%

4%

Coal Fettnuss mvb

onh

ash

H2O

Carbon

Coal Analysis – “as fired” (“as received”)

Minimum oxygen and air requirements for a solid fuel

Fuel composition

Mass fractions of species are known: c,h,s,o,n,

moisture,ash

Chemical reactionsC O CO2 2+ →H 0.5O H O2 2 2+ →S O SO2 2+ →

Minimum oxygen and air requirements are then uniquely determined

lO ,min2= c

12 + ×12

h2 +

s32

−o

32 ll0 .2 1air, m inO , m in2=

in kmol O2/kg of fuel in kmol air/kg of fuel

Amount of combustion products

Fuel composition

c,h,s,o,n,

moisture,ash

Chemical reactions

Composition and amount (per unit of fuel) of combustion products are uniquely determined for a given excess air ratio

C O CO2 2+ →H 0.5O H O2 2 2+ →S O SO2 2+ →

Vwet = +h2

+moisture

18 +s

3 2+ 0.79 lair,minλ + −( )λ 1 lO ,min2

Vdry = +s

3 2+ 0.79 lair,minλ + −( )λ 1 lO ,min2

12c

12c

28n+

28n+

1.0 1.1 1.2 1.3 1.40

2

4

6

coal Fettnuss

Dutch Natural Gas

C2H6

CH4

O2 %

vol

, dry

Excess air ratio λλλλ

Oxygen (%,vol,dry) content in combustion products as a functionof excess air ratio

1.0 1.1 1.2 1.3 1.40

2

4

6

8

10

12

14

16

18

20

Dutch Natural Gas

C2H6

CH4

coal Fettnuss

CO

2 % v

ol, d

ry

Excess air ratio λλλλ

Carbon dioxide concentration (%,vol,dry) as a function ofexcess air ratio

1.4 Combustibles burnout for solid fuels

Total combustibles = 1 - ash

FURNACESolid fuel

Combustion air

Co+Ash0=1

mo (kg/s)

Combustion products

Unburned solids

C1+Ash1=1

m1 (kg/s)

Question: What fraction of combustibles has been burned ?

FURNACESolid fuel

Combustion air

Combustion products

Unburned solids

Co+Ash0=1 mo (kg/s) C1+Ash1=1 m1 (kg/s)

Mass balance of combustibles:

moCo = m1C1 + bmoCo

Mass balance of ash (ash is assumed to be an inert):

m0Ash0 = m1Ash1

bC A s hC A s h

A s hA s hA s h

= −××

=−

−1

1

11 0

0 1

0

1

0

0 2 4 6 8 1097.00

97.25

97.50

97.75

98.00

98.25

98.50

98.75

99.00

99.25

99.50

99.75

100.00

Ash0=30%

Ash0=20%

Ash0=16%

Ash0=12%

Ash0=8%

Ash0=4%

Com

bust

ible

s bu

rnou

t in

%

Carbon in ash in %

Burnout of combustibles