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2 Combustion and Thermochemistry

2 Combustion and Thermochemistry. Review of Property Relations. Extensive Properties V (m 3 ), U (J), H (J)(= U+PV ) Intensive Properties v (m 3 /kg), u (J/kg), h =(J/kg)(= u+Pv ) P,T V=mv;U=mu;H=mh. Equation of State. For Ideal-gas Behavior: PV=NR u T PV=mRT Pv=RT

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2 Combustion and Thermochemistry

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  1. 2 Combustion and Thermochemistry

  2. Review of Property Relations Extensive Properties V(m3), U(J), H(J)(=U+PV) Intensive Properties v(m3/kg), u(J/kg),h=(J/kg)(=u+Pv) P,T V=mv;U=mu;H=mh

  3. Equation of State For Ideal-gas Behavior: PV=NRuT PV=mRT Pv=RT P=RT where, R=Ru/MW Ru=8315J/kmol-K; MW is the gas molecular weight

  4. Calorific Equations of State u=u(T,v) h=h(T,P) du= dh=

  5. Constant-volume specific heats Constant-pressure specific heats

  6. For Ideal Gas

  7. Translation (a)Monatomic species Translation Rotation Vibration (b)Diatomic Species

  8. Fig.2.2

  9. Ideal-Gas Mixtures Constituent mole fraction and mass fraction Mole fraction of species i,xi

  10. Mass fraction of species i, Yi

  11. By definition

  12. Relation of xi and Yi

  13. Mixture molecular weight MWmix

  14. Partial pressure of ith species,Pi For ideal gas:

  15. Mass-(or Molar-) specific mixture properties The mixture entropy is calculated:

  16. The constituent entropies Standard-state(Pref P0=1 atm):

  17. Latent heat of Vaporization It is also called the enthalpy of vaporization Clausius-Clapeyron Equation

  18. First Law of Thermodynamics First Law-Fixed Mass Work done by system on surroundings in going from state 1 to state 2 Change in total system energy in going from state 1 to state 2 Heat added to system in going fgom state1 to state 2

  19. E1-2(E2-E1) Mass-specific system potential energy Mass-specific system internal energy Mass-specific kinetic energy

  20. Instantaneous rate of heat transferred into system Instantaneous rate of work done by system, or power Instantaneous time rate of change of system energy e E/m

  21. First Law-Control Volume-SSSF Rate of heat transferred across the control surface from the surroundings, to the control volume Rate of all work done by the control volume,including shaft work, but excluding flow work Rate of energy flowing out of the control volume Net rate of workj associated with pressure forces where fluid crosses the control surface, flow work Rate of energy flowing into the control volume

  22. The principal assumptions • The control volume is fixed relative to the coordinate system • The properties of the fluid at each point within the control volume,or the control surface, do not vary with time. • Fluid properties are uniform over the inlet and outlet flow areas • There are only inlet and one exit stream

  23. Total energy per unit mass Internal energy per unit mass Potential energy per unit mass Kinetic energy per unit mass

  24. Final form of energy conservation for a control volume

  25. Reactant and Product Mixtures • Stoichiometry • 3.76=79/21( by volume) • The stoichiometric air-fuel ratio

  26. Some combustion properties of methane, hydrogen, and solid carbon for reactants at 298K

  27. The equivalence ratio  >1, fuel rich mixtures <1, fuel lean mixtures =1, stoichiometric mixture

  28. Percent stoichiometric air Percent excess air

  29. Example 2.1 A small, low-emission, stationary gas-turbine engine operates at full load(3950kW) at an equivalence ratio of 0.286 with an air flowrate of 15.9kg/s. The equivalent composition of the fuel(natural gas) is C1.16H4.32. Determine the fuel mass flowrate and the operating air-fuel ratio for the engine.

  30. Solution Given: =0.286, MWair=28.85, mair=15.9 kg/s, MWfuel=1.16(12.01)+4.32(1.008)=18.286 Find: mfuel and (A/F) We will proceed by first finding (A/F) and then mfuel .The solution requires only the application of definitions expressed in Equs above

  31. where a=x+y/4=1.16+4.32/4=2.24. Thus, and, from Equ. above

  32. Since (A/F) is the ratio of the air flowrate to the fuel flowrate, Comment Note that even at full power, a large quantity of excess air is supplied to the engine.

  33. Example 2.2 A natural gas-fired industrial boiler operates with an oxygen concentration of 3 mole percent in the flue gases. Determine the operating air-fuel ratio and the equivalence ratio. Treat the natural gas as methane.

  34. Solution Given: xO2=0.03, MWfuel=16.04 MWair=28.85. Find : (A/F)and. We can use the given O2 mole fraction to find the air-fuel ratio by writing an overall combustion equation assuming “complete combustion,” i.e., no dissociation (all fuel C is found in CO2 and all fuel H is found in H2O):

  35. CH4+a(O2+3.76N2)CO2+2H2O+bO2+3.76aN2 where a and b are related from conservation of O atoms. 2a=2+2+2b From the definition of a mole fraction

  36. Substituting the known value of xO2(=0.03) and then solving for a yields or a=2.368 The mass air-fuel ratio, in general, is expressed as

  37. To find , we need to determine(A/F)stoic. For a=2,hence, • Apply the definition of ,

  38. Absolute(or Standardized) Enthalpy and Enthalpy of Formation Enthalpy of formation at standard reference state(Tref,P0) Sensible enthalpy change in going from Tref to T Absolute enthalpy at temperature T Where,

  39. A standard state A standard-state temperature: Tref=25C(298.15K) A standard-state pressure: Pref=P0=1 atm(101,325Pa)

  40. The enthalpy of formation are zero for the elements in their naturally occurring state at the reference state temperature and pressure. For example, at 25ºC and 1 atm, oxygen exists as diatomic molecules; hence,

  41. To form oxygen atoms at sandard state requires the breaking of a rather strong chemical bond. The bond dissociation energy for O2 at 298K is 498,390 kJ/kmol O2

  42. Enthalpy of formation have a clear physical interpretation as the net change in enthalpy associated with breaking the chemical bonds of the standard state elements and forming new bonds to create the compound of interest.

  43. Example 2.3 A gas stream at 1 atm contains a misture of CO, CO2, and N2 in which the CO mole fraction is 0.10 and the CO2 mole fraction is 0.20. The gas-stream temperature is 1200K. Determine the absolute enthalpy of the mixture on both a mole basis(kJ/kmol) and a mass basis(kJ/kg). Also determine the

  44. Solution Given Xco=0.10, T=1200K, Xco2=0.20,P=1 atm Find:hmix, hmix , Yco, Yco2 , and YN2 Finding hmix requires the straightforward application of the ideal-gas mixture law, thus, and

  45. Substitute values from Appendix A hmix=0.1[-110,540+28,440] +0.2[-393,546+44,488] +0.7[0+28,118] =-55,339.1kJ/kmolmix To find hmix, we need to determine the molecular weight of the mixture:

  46. MWmix=xiMWi=0.1(28.01)+0.20(44.01)+0.7(28.013)=31.212 Then,

  47. Enthalpy of Combustion and Heating Values For the steady-flow reactor, complete combustion: All C CO2 All H H2O Fig.2.7

  48. The definition of the enthalpy of reaction, or the enthalpy of Combustion, hR(per mass of mixture), is • or, in terms of extensive properties,

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