HEAT PROCESSES

1. 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. Hottel’s diagram. Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

2. 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

3. Fluidised bed boiler HP10 Example: Babcock&Wilcox bubbled fluidised bed boiler

4. Pulverised fuel boiler HP10 Example: Babcock&Wilcoxspiral wound universal pressure (SWUP™) boiler

5. Burner - Pulverised fuel Control of secondary air Primary air Control of secondary air swirling HP10

6. Liquid fuelsburners air Oil Oil Oil Steam Oil Steam atomizer Vortex chamber nozzle Ultrasound atomizer Rotating cup HP10 A nice video: Boilers and Their Operation 1956 US Navy Instructional Film

7. Gaseous fuels burners air gas HP10

8. COMBUSTION - fundamentals HP10 • Fuel composition and Heating value • Statics of combustion • Mass and enthalpy balancing • Heat transfer - radiation Benson

9. Fuels calorific value ( ) = - w + w q q 2 , 51 9 n v H W HP10 • qvhigh heating value HHVMJkg-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). • qnlow heating value LHVMJkg-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) andash (A-ash, minerals). Composition is expressed in mass fractions C (kg carbon in kg of fuel), O, … and these values enable to estimate LHVassuming 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.

10. 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 (C+O2CO2) 4kg of H requires 1kmol of O2 (2H2+O2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) <1 lean fuel combustion =1 stoichiometric combustion >1 rich fuel combustion In the same way production of flue gases can be expressed

11. 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)

12. 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.

13. 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). 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 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) 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

14. Enthalpy balancing, temperatures Combustion chamber 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

15. 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 Tf=Tair=Tfg=T0 qn is the low heating value as soon as the reference temperature T0 is above the temperature of condensationof 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

16. Enthalpy balancing, temperatures HP10 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 mair=Vair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 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

17. Enthalpy balancing, temperatures Tw Tfg TS 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

18. Enthalpy balancing, temperatures Photons emmited at high temperature TS (short wavelength) Photons emmited at low temperature Tw (long wavelength) 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.

19. 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 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.

20. Enthalpy balancing, temperatures HP10 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) Instead diagrams this approximation can be used

21. 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

22. 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.

23. Example: steam reforming (1/2) air Reaction mixture Natural gas Flue gas 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 TCH4=291 K, Tair=573 K, and for heating value of methane qn=49,9 MJkg-1it is possible to evaluate temperature of adiabatic flame

24. 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 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

25. HP10

26. EXAM HP10 Combustion Equations describing static of combustion follow from the assumed chemical reactions C+O2CO2 2H2+O22H2O S+O2SO2 … You also need to know atomic masses of participating elements MC=12g/mol, MO=16, MS=32, MH=1, MN=14,…

27. 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 Combustion chamber 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

28. Reaction heat of fuel low heating value Enthalpy of evaporation of water in flue gas high heating value ( ) = - w + w q q 2 , 51 9 n v H W What is important (at least for exam) HP10 Enthalpy balancing and temperature of flue gas Maximum temperature of flue gas (for Q=0, adiabatic flame temperature)

29. Tw Tfg TS What is important (at least for exam) HP10 Relationship between power Q and temperature of flame and flue gas Stefan Boltzmann law Kirchhoff law (emissivity=absorptivity but only at monochromatic radiation)