HEAT PROCESSES
description
Transcript of HEAT PROCESSES
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
SteamUltrasound 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
36422
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, 21
OAir
VpVp 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 mM 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 mM M M
M MO m m
M M
M M MH m mM 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
1164
/1612
111116200
164
18161
44320
004412
1612
,
,
,
,
2
2
2
4
fg
f
fgo
fg
f
fgOH
fgO
fgCO
fgCH
mmmm
mm
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 Tm
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
fgf air air fg fg
f
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 42fg n f f air air air air
fg
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
fgf f air air air fg fg fg n
f
Q Tc T T m c T T m c T T q
m
0 0 ,max 00
f f air air air fg fg fg nf
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 12
,
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.
, / ,MCHQqn
kg s4
0 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
26
1 42
1 1300 1300 4273 273 49,9 10 2190 18 13,94 1, 276 1006 300 3752 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 Tc c cV
K
2
airNatural 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 12 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
4 18 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
fgf f air air air fg fg fg n
f
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 emittedcombustion by 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