Combustion and burners

Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

HEAT PROCESSESHP10

Combustion and burners (pulverized coal, biofuels, oil and gas burners, NOx reduction, CFD analysis of gas burner). Properties of fuels, reaction enthalpy, combustion heat. Enthalpy balances, adiabatic flame temperature. Heat transfer by radiation, emissivity and absorptivity of flue gases. Hottel’s diagram.

Combustors, burners, boilers,

can be classified according to size of fuel particles

Large lumps (Stoker fired furnaces, bio-fuels, wastes)

Medium particles (fluidised beds)

Fine particles (conveying burners)

Liquid fuels (atomizers)

Gas burners

Combustion and burnersHP10

Tomasso

HP10 Fluidised bed boilerExample: Babcock&Wilcox bubbled fluidised bed boiler

HP10 Pulverised fuel boilerExample: Babcock&Wilcox spiral wound universal pressure (SWUP™) boiler

HP10 Burner - Pulverised fuel

Control of secondary air swirling

Control of secondary air

Primary air

HP10 Liquid fuels burners

Vortex chamber nozzle Steam atomizer

air Oil Oil

Oil Oil

Steam

Ultrasound atomizer Rotating cup

A nice video: Boilers and Their Operation 1956 US Navy Instructional Film

HP10 Gaseous fuels burners

air gas

• Fuel composition and Heating value

• Statics of combustion

• Mass and enthalpy balancing

• Heat transfer - radiation

HP10 COMBUSTION - fundamentals

Benson

HP10 Fuels calorific value

1. qv high heating value HHV MJkg-1, heat released by by combustion of 1 kg

fuel, when all products are cooled down to initial temperature and water in flue gas condenses (latent heat of evaporation is utilised).

2. qn low heating value LHV MJkg-1, less by the enthalpy of evaporation

Element composition (C-carbon, atomic mass AC=12,01), (O-oxygen, AO=16), (H-

hydrogen, AH=1,008), (N-nitrogen, AN=14,01), (S-sulphur, AS=32,06) and free water

explicitly (W-water, MW=18,015 kgkmol-1, moisture is determined by drying of sample

at 1050C) and ash (A-ash, minerals).

Composition is expressed in mass fractions C (kg carbon in kg of fuel), O, … and

these values enable to estimate LHV assuming prevailing chemical reactions

OSNHCvq 84,909,1928,632,12403,34

Enthalpy of evaporation OkgHkgH

OHOH

2

222

364

22

Jigisha Parikh, S.A. Channiwala, G.K. Ghosal:

A correlation for calculating HHV from proximate analysis of solid fuels. Fuel, Volume 84, Issue 5, March 2005, Pages 487-494.

WHvn qq 951,2

HP10 Fuels air consumption-flue gas production

222,3

12 4 32 32C S OH

OV

2

"

"1 ,

0,21O

Air

VpV

p p

222,3

12 32 28 2 18C S N WH

fg air OV V V

Consumption of oxygen necessary for combustion of 1 kg of fuel with known elemental composition (expressed as volume Nm3/kg)

Volume of 1 kmol of gas at normal conditions (0,1013

MPa and 00C) in m3

12kg C requires 1 kmol of O2 (C+O2CO2)

4kg of H requires 1kmol of O2 (2H2+O22H2O)

Consumption of pure oxygen can be easily recalculated to consumption of humid air ( is relative mumidity, p” pressure of saturated steam)

<1 lean fuel combustion =1 stoichiometric combustion >1 rich fuel combustion

In the same way production of flue gases can be expressed

Example: Combustion chamber f-fuel, o-oxidiser, fg-flue gas streams

{inlet flowrate} = {outlet flowrate}

f Air fgm m m

2 2 2

2 2 2{inlet flowrate of O } = {outlet flowrate of O } + {rate of production of O inside the black box}

O ,Air O , fg Om = m m

,Air ,fg2 2 2O Air O fg Om m m

Mass balance of mixture

Mass balances of individual components (chemical compounds)

Mass flowrate [kg/s]. Streams are composed of O2,N2,CO2,CO,CH4,H2O

4 2

4 4 2

, ,( )C C Cf fg CH e CO e

CH CH CO

M M Mm m

M M M

Mass balances of elements (C,H,O,N - four equations)

HP10 Mass balancing

fuelfm

airmCombustion chamber

flue gasfgm

Example: Simplified combustion chamber f-fuel, o-oxidiser, fg-flue gas streams

f o fgm m m Mass balance of mixture

4 2

4 4 2

2 2 2

2 2

4 2

4 4 2

4 2 2 2

, ,

, , ,

, ,

, , , ,

sum of mass fractions must b

: ( )

2: ( )

4 4 2: ( )

1

C C Cf fg CH fg CO fg

CH CH CO

O Oo fg CO fg O fg H O fg

CO H O

H H Hf fg CH fg H O fg

CH CH H O

CH fg H O fg CO fg O fg

M M MC m m

M M M

M MO m m

M M

M M MH m m

M M M

e one

Mass balances of elements (C,O,H-3 equations)

HP10 Mass balancing Example 1/2

4 ,

pure methane

1f CH fm

2 ,

pure oxygen

1o O om

Combustion chamber

4 2 2 2, , , , , , ,fg CH fg CO fg O fg H O fgm

OHCOOCH 2224 22

Remark: Notice the fact that the mass balances can be written without knowledge of actual chemical reactions, e.g.

HP10 Mass balancing Example 2/2

116

4/

16

12

111116

200

16

418

161

44

320

0044

12

16

12

,

,

,

,

2

2

2

4

fg

f

fgo

fg

f

fgOH

fgO

fgCO

fgCH

m

mmm

m

m

Matrix form of element balances after substituting molecular masses

It is obvious that the matrix of system is singular (sum of the first 3 rows is the last row), therefore at least one more equation describing the mass balance of species is necessary (or any mass fraction can be fixed).

a=[33 12 0 0;0 72 99 88;0 0 1 0;1 1 1 1];

b=[33*.5;99*.5;1;1];

for i=1:10

b(3)=i*0.1;

omg=inv(a)*b;

v(:,i)=omg;

end

0.4000 0.4250 0.4500 0.4750 0.5000 0.5250 0.5500 0.5750 0.6000 0.6250

0.2750 0.2062 0.1375 0.0687 0 -0.0688 -0.1375 -0.2062 -0.2750 -0.3438

0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

0.2250 0.1688 0.1125 0.0563 0 -0.0563 -0.1125 -0.1688 -0.2250 -0.2813

4

2

2

2

,

,

,

,

CH fg

CO fg

O fg

H O fg

Results for fixed O,fg=0.1, 0.2, 0.3, …, 1 (notice, that O,fg>0.5 results to negative mass fractions of CO2 and H2O)

HP10 Enthalpy balancing, temperatures

Enthalpy balance of a combustion chamber

f f f air air air fg fg fg

f f air air air fg fg fgf

m h T m h T m h T Q

Qh T m h T m h T

m

Boiler RUN videoreleasedheat Q

Relative consumption of air Relative production

of flue gases

It would be heating value of fuel if the temperatures

Tf,Tair,Tfg will be the same

, fuelf fm T

,air airm T

Combustion chamber

, flue gasfg fgm T

ffgfg

fg

mmm

m

/mass flowrate of flue gas [kg/s]

relative flowrate of flue gas [dimensionless]

HP10 Enthalpy balancing, temperatures

So that it could be possible to express enthalpies by temperatures it is necessary to modify the previous equation formally as

0 0 0 .

fg

f f air air air fg fg fg nf

Q Tc T T m c T T m c T T q

m

0 0 0

0 0 0

f f f air air air air fg fg fg fg

fg

f air air fg fgf

h T h T m h T h T m h T h T

Q Th T m h T m h T

m

This term is heating value qn for Tf=Tair=Tfg=T0

qn is the low heating value as soon as the reference temperature T0 is above

the temperature of condensation of water in flue gases

Pierre-Alexandre Glaude, René Fournet, Roda Bounaceur, Michel Molière: Adiabatic flame temperature from biofuels and fossil fuels and derived effect on NOx emissions. Fuel Processing Technology, Volume 91, Issue 2, February 2010, Pages 229-235.

Kubota, N. (2007) Thermochemistry of Combustion, in Propellants and Explosives: Thermochemical Aspects of Combustion, Second Edition, Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim, Germany. doi: 10.1002/9783527610105.ch2

HP10 Enthalpy balancing, temperatures

0 0

,max 0 ,n f f air air air air

fgfg fg fg

q c T T V c T TT T

V c

Adiabatic flame temperature is the temperature of flue gases for the case that the combustor chamber is thermally insulated (Q=0). This maximal temperature follows directly from the previous enthalpy balance

0 1 ,fg fgc T c c T

2

0 0,max 0 0 0 0

1 1 1

1 4

2fg n f f air air air airfg

c cT T T q c T T V c T T

c c cV

mair=Vairair

Specific heat capacities of fuel and air (cf,cair) can calculated easily, but the specific heat capacity cfg depends upon temperature and upon unknown composition of flue gases. Fortunately the product of density and specific heat capacity depends upon composition only weakly and can be approximated by linear function of temperature c0=1300 [J.m-3.K-1] c1=0,175 [J.m-3.K-1]

Substituting this linear relationship results to a quadratic equation for adiabatic flame temperature with the following solution

HP10 Enthalpy balancing, temperatures

Actual flame temperature and actual temperature of flue gases cannot be calculated so easily. It is necessary to express the power Q in terms of mean temperature of flame TS and the temperature of wall Tw .

.44wgSg TATSQ

TS

Tw

Q Tfg

Heat transfer by radiation dominates at high temperatures. In this case the heat flux is proportional to 4th power of thermodynamic temperature and

Heat flow emitted by hot gas and absorbed

by wall

Heat flow emitted by wall and absorbed by

molecules of gas

Irradiated heat transfer surface

-emisivity A-absorptivity

4Tq Stefan Boltzman

HP10 Enthalpy balancing, temperatures

TS

Tw

Tfg

Photon absorbed by molecule of water Photon is not

absorbed by oxygen

Photon absorbed by opposite wall – no contribution to Q

Most photons emitted by gas are absorbed by wall

Photon absorbed by CO-no net

contribution to Q

Photons emmited at high temperature TS (short wavelength)

Photons emmited at low temperature Tw (long wavelength)

Wall of combustor chamber is almost “black body”, therefore all photons impacting to wall are absorbed and not bounced off. On the other hand the photon emitted by wall has only limited probability to be absorbed by a heteropolar molecule (H2O, CO2, homeopolar molecules like O2,N2 are almost transparent for photons). The probability of absorption is proportional to density of heteropolar molecules (to their partial pressure) and to the length of ray L. Probability of catching depends also upon the photon energy (wavelength), the greater is energy the lower is probability of absorption.

HP10 Enthalpy balancing, temperatures

.44wgSg TATSQ

Emissivity of gas corresponding to

temperature of gas Ts

Absorptivity of gas corresponding to wall

temperature Tw

According to Kirchhoff’s law Emissivity=Absorptivity (g = Ag ) but this equivalence holds only at the same wavelength (monochromatic radiation). Emissivity and absorptivity of photons depends upon their wavelength (frequency, energy). The first term (g) should be evaluated for high energy photons emitted by hot gas, while the second term (Ag) for photons emitted by colder wall.

Let us return back to the expression for resulting power exchanged between the hot gas and the wall of combustion chamber

HP10 Enthalpy balancing, temperatures

Hottel’s diagram for emissivity of CO2 and H2O as a function of temperature and pL (partial

pressure pCO2 is calculated from composition of flue gas, and length of ray L=3.5V/S – empirical approximation)

,14

222 108,311608

10Tp

Lpp

gOH

COOH

eT

Instead diagrams this approximation

can be used

HP10 Enthalpy balancing, temperatures

4 4,max

g gfg fg fg fg S w

f g

S Am c T T T T

m

Subtractinq equations (enthalpy balance for real and insulated combustors)

0 0 0

fg

f f air air air fg fg fg nf

Q Tc T T m c T T m c T T q

m

0 0 ,max 0

0f f air air air fg fg fg n

f

c T T m c T T m c T T qm

we arrive to the equation for two unknown temperatures Tfg and TS

The flame temperature TS must be somewhere between Tfg and Tfg,max and can be approximated by geometric average of these two temperatures, giving

2 2 4,max ,max

g gfg fg fg fg fg fg w

f g

S Am c T T T T T

m

Quadratic equation for flue gas temperature

HP10 Enthalpy balancing, temperatures

4,max

4,max

3,max

41 1 1

2

,

fg g wfg

g fg

fg fg

g fg

T Bo A TT

Bo T Bo

m cBo

S T

The solution of quadratic equation for flue gas temperature can be expressed in terms of Boltzmann criterion (ratio of overall transferred heat to the heat transferred only by radiation)

Remark: this formula is only a rough approximation. Its application will be demonstrated on the following example.

HP10 Example: steam reforming (1/2)Furnace for steam reforming (reaction proceeds inside a set of vertical tubes) makes use a row of gas burners, consuming natural gas as fuel.

, / ,MCHQ

qnkg s

40 0222

For given mass flowrate of fuel

It is possible to evaluate consumption of air and production of flue gases

4

2

3

3

3

22,3 22,3 0,75 0,251,05 13,94 / ,

0,21 12 4 0,21 12 4

0,31 / ,

22,3 22,3 22,312 2 12 2 12 4

22,3 15,34 / .4

C Hair

air CH air

C C CH H Hfg air O air

Hvz

V m kg

V M V m s

V V V V

V m kg

For temperature of methane and preheated air TCH4=291 K, Tair=573 K, and for heating value of methane qn=49,9 MJkg-1 it is possible to evaluate temperature of adiabatic flame

2

0 0,max 0 0 0 0

1 1 1

2

6

1 4

2

1 1300 1300 4273 273 49,9 10 2190 18 13,94 1,276 1006 300 375

2 0,175 0,175 0,175 15,333

fg n f f air air air airfg

c cT T T q c T T V c T T

c c cV

K

2

air

Natural gasReaction mixture

Flue gas

HP10 Example: steam reforming (2/2)The relative emisivity g(TS) is calculated for estimated flame temperature TS2000 K

The relative absorptivity Ag=g(Tw) is calculated for estimated temperature of wall Tw1200 K.

Mean path of ray L is estimated from geometry of combustion chamber (rectangular channel of height 10.8 m a width 2.5 m) as L=3,5 V/S=3,5(10,8.2,5)/(2.10,8)=4,4 m

Partial pressures are determined by composition of flue gas composed of H2O, CO2,a N2. This calculation follows from previously evaluated relative volume of air VO2=2,9 (m3 oxygen/kg methane) and stoichiometry of reaction

VH2O= VO2=2,9

VCO2=0,5 VO2=1,45

VN2= Vvz-VO2=11

Corresponding ratio of partial pressures is

2,9:1,45:11

and because sum of pressures is atmospheric pressure p=pH2O+pCO2+pN2 the partial pressures of heteropolar gases are

pH2O=0,0192 MPa, pCO2=0,0096 MPa.

Using these values in Hottel’s diagrams (or using mentioned correlation for relative emissivity follows

g(TS)=0,258 and Ag=g(Tw)=0,49, and final result (flue gas temperature)

check Vfg= VH2O+ VCO2+VN2

3 3 8 3

,max ,max

4 4,max

4,max

15,34 0,0222 1300 0,174 20000,133

2 10,8 5,67 10 0,258 2375

4 2375 0,133 4 0,49 12001 1 1 1 1

2 2 0,133 0,25

fg fg fg fg fg

g fg g fg

fg g wfg

g fg

M c V cBo

S T S T

T Bo A TT

Bo T Bo

41

8 2375 0,133

1056 .K

HP10

EXAMHP10

Combustion

Equations describing static of combustion follow from the assumed chemical reactions

C+O2CO2 2H2+O22H2O S+O2SO2 …

You also need to know atomic masses of participating elements MC=12g/mol, MO=16, MS=32, MH=1, MN=14,…

What is important (at least for exam)HP10

222,3

12 4 32 32C S OH

OV

Volume of oxygen necessary for combustion of 1 kg of fuel

22.3 (l) is volume of 1 mol of gas at normal conditions. C, H,… are mass fractions of elements obtained in proximate analysis (usually mass spectroscopy of fuel) Volume of flue gas produced by combustion of 1 kg of fuel

222,3

12 32 28 2 18C S N WH

fg air OV V V

fuelfmairm

Combustion chamber

flue gasfgm

f Air fgm m m

Mass balancing

,Air ,fg2 2 2O Air O fg Om m m

Overall balance

Balance of Species, e.g. O2

4 2

4 4 2

, ,( )C C Cf fg CH fg CO fg

CH CH CO

M M Mm m

M M M Balance of elements, e.g. C

(For example N2,fg O2,fg, CH4,fg, CO2,fg, H2O,fg, O2,fg)

What is important (at least for exam)HP10

Reaction heat of fuel WHvn qq 951,2

high heating value

low heating value

Enthalpy of evaporation

of water in flue gas

Enthalpy balancing and temperature of flue gas

0 0 0

low heatingenthalpy of air streamenthalpy of fuel enthalpy of flue gas valuepower of heat releasedby combustion chamber (without conden

fg

f f air air air fg fg fg nf

Q Tc T T m c T T m c T T q

m

sation of water)

.

0 0

,max 0 ,n f f air air air air

fgfg fg fg

q c T T V c T TT T

V c

Maximum temperature of flue gas (for Q=0, adiabatic flame temperature)

What is important (at least for exam)HP10

Relationship between power Q and temperature of flame and flue gas

TS

Tw

Q Tfg

4 4

surface of photons emitted photons emittedcombustionby gas at flame by wall and absorbedchambertemparature by molecules of gas

( ).g S g wS

Q S T A T

4Tq Stefan Boltzmann law

Kirchhoff law (emissivity=absorptivity but only at monochromatic radiation)

( ) ( )T A T