HEAT PROCESSES

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HEAT PROCESSES
HP10
Combustion and burners
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. Hottels diagram.
Rudolf Žitný, Ústav procesní a zpracovatelské
techniky CVUT FS 2010
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Combustion and burners
HP10

• 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

Tomasso
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Fluidised bed boiler
HP10
Example BabcockWilcox bubbled fluidised bed
boiler
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Pulverised fuel boiler
HP10
Example BabcockWilcox spiral wound universal
pressure (SWUP) boiler
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Burner - Pulverised fuel
HP10
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Liquid fuels burners
HP10
A nice video Boilers and Their Operation 1956 US
Navy Instructional Film
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Gaseous fuels burners
HP10
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COMBUSTION - fundamentals
HP10

• Fuel composition and Heating value
• Statics of combustion
• Mass and enthalpy balancing
• Heat transfer - radiation

Benson
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Fuels calorific value
HP10

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
AC12,01), (O-oxygen, AO16), (H-hydrogen,
AH1,008), (N-nitrogen, AN14,01), (S-sulphur,
AS32,06) and free water explicitly (W-water,
MW18,015 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
Enthalpy of evaporation
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.
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Fuels air consumption-flue gas production
HP10
Consumption of oxygen necessary for combustion of
1 kg of fuel with known elemental composition
(expressed as volume Nm3/kg)
12kg C requires 1 kmol of O2 (CO2?CO2) 4kg of
H requires 1kmol of O2 (2H2O2?2H2O)
Volume of 1 kmol of gas at normal conditions
(0,1013 MPa and 00C) in m3
Consumption of pure oxygen can be easily
recalculated to consumption of humid air (? is
relative mumidity, p pressure of saturated steam)
?lt1 lean fuel combustion ?1
stoichiometric combustion ?gt1 rich fuel
combustion
In the same way production of flue gases can be
expressed
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Mass balancing
HP10
Example Combustion chamber f-fuel, o-oxidiser,
fg-flue gas streams
Mass flowrate kg/s. Streams are composed of
O2,N2,CO2,CO,CH4,H2O
Combustion chamber
Mass balance of mixture
Mass balances of individual components (chemical
compounds)
Mass balances of elements (C,H,O,N - four
equations)
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Mass balancing Example 1/2
HP10
Example Simplified combustion chamber f-fuel,
o-oxidiser, fg-flue gas streams
Combustion chamber
Mass balance of mixture
Mass balances of elements (C,O,H-3 equations)
Remark Notice the fact that the mass balances
can be written without knowledge of actual
chemical reactions, e.g.
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Mass balancing Example 2/2
HP10
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).
a33 12 0 00 72 99 880 0 1 01 1 1
1 b33.599.511 for i110
b(3)i0.1 omginv(a)b v(,i)omg end
Results for fixed ?O,fg0.1, 0.2, 0.3, , 1
(notice, that ?O,fggt0.5 results to negative mass
fractions of CO2 and H2O)
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
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Enthalpy balancing, temperatures
HP10
Enthalpy balance of a combustion chamber
Boiler RUN video
mass flowrate of flue gas kg/s
relative flowrate of flue gas dimensionless
It would be heating value of fuel if the
temperatures Tf,Tair,Tfg will be the same
Relative consumption of air
Relative production of flue gases
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Enthalpy balancing, temperatures
HP10
So that it could be possible to express
enthalpies by temperatures it is necessary to
modify the previous equation formally as
This term is heating value qn for TfTairTfgT0
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
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Enthalpy balancing, temperatures
HP10
Adiabatic flame temperature is the temperature of
flue gases for the case that the combustor
chamber is thermally insulated (Q0). This
maximal temperature follows directly from the
previous enthalpy balance
mairVair?air
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
c01300 J.m-3.K-1 c10,175 J.m-3.K-1
Substituting this linear relationship results to
a quadratic equation for adiabatic flame
temperature with the following solution
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Enthalpy balancing, temperatures
HP10
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 .
Stefan Boltzman
Heat transfer by radiation dominates at high
temperatures. In this case the heat flux is
proportional to 4th power of thermodynamic
temperature and
?-emisivity A-absorptivity
Irradiated heat transfer surface
Heat flow emitted by hot gas and absorbed by wall
Heat flow emitted by wall and absorbed by
molecules of gas
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Enthalpy balancing, temperatures
HP10
Photon absorbed by opposite wall no
contribution to Q
Photon absorbed by molecule of water
Photon is not absorbed by oxygen
Tw
Tfg
TS
Most photons emitted by gas are absorbed by wall
Photon absorbed by CO-no net contribution to Q
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.
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Enthalpy balancing, temperatures
HP10
Let us return back to the expression for
resulting power exchanged between the hot gas and
the wall of combustion chamber
Absorptivity of gas corresponding to wall
temperature Tw
Emissivity of gas corresponding to temperature of
gas Ts
According to Kirchhoffs law EmissivityAbsorptivi
ty (?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.
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Enthalpy balancing, temperatures
HP10
Hottels 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 L3.5V/S empirical
approximation)
Instead diagrams this approximation can be used
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Enthalpy balancing, temperatures
HP10
Subtractinq equations (enthalpy balance for real
and insulated combustors)
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
Quadratic equation for flue gas temperature
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Enthalpy balancing, temperatures
HP10
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.
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Example steam reforming (1/2)
HP10
Furnace for steam reforming (reaction proceeds
inside a set of vertical tubes) makes use a row
of gas burners, consuming natural gas as fuel.
For given mass flowrate of fuel It is possible
to evaluate consumption of air and production of
flue gases
For temperature of methane and preheated air
TCH4291 K, Tair573 K, and for heating value of
methane qn49,9 MJkg-1 it is possible to
evaluate temperature of adiabatic flame
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Example steam reforming (2/2)
HP10
The relative emisivity ?g(TS) is calculated for
estimated flame temperature TS?2000 K The
relative absorptivity Ag?g(Tw) is calculated
for estimated temperature of wall Tw?1200 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 L3,5
V/S3,5(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 VO22,9 (m3 oxygen/kg methane) and
stoichiometry of reaction VH2O VO22,9
VCO20,5 VO21,45 VN2 Vvz-VO211 Corresponding
ratio of partial pressures is 2,91,4511 and
because sum of pressures is atmospheric pressure
ppH2OpCO2pN2 the partial pressures of
heteropolar gases are pH2O0,0192 MPa,
pCO20,0096 MPa. Using these values in Hottels
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 VCO2VN2
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HP10
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EXAM
HP10
Combustion
Equations describing static of combustion follow
from the assumed chemical reactions CO2?CO2
2H2O2?2H2O SO2?SO2 You also need to know
atomic masses of participating elements
MC12g/mol, MO16, MS32, MH1, MN14,
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What is important (at least for exam)
HP10
Mass balancing
(For example ?N2,fg ?O2,fg, ?CH4,fg, ?CO2,fg,
?H2O,fg, ?O2,fg)
Overall balance
Balance of Species, e.g. O2
Balance of elements, e.g. C
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What is important (at least for exam)
HP10
Enthalpy balancing and temperature of flue gas
Maximum temperature of flue gas (for Q0,
adiabatic flame temperature)
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What is important (at least for exam)
HP10
Relationship between power Q and temperature of
flame and flue gas
Stefan Boltzmann law
Kirchhoff law (emissivityabsorptivity but only
at monochromatic radiation)