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ME 475/675 Introduction to Combustion

ME 475/675 Introduction to Combustion. Lecture 2. Stoichiometric Hydrocarbon Combustion. air. C x H y + a(O 2 +3.76N 2 )  (x)CO 2 + (y/2) H 2 O + 3.76a N 2 a = number of oxygen molecules per fuel molecule Number of air molecules per fuel molecule is a(1+3.76)

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ME 475/675 Introduction to Combustion

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  1. ME 475/675 Introduction to Combustion Lecture 2

  2. Stoichiometric Hydrocarbon Combustion air • CxHy + a(O2+3.76N2)  (x)CO2 + (y/2) H2O + 3.76a N2 • a = number of oxygen molecules per fuel molecule • Number of air molecules per fuel molecule is a(1+3.76) • If a= aST = x + y/4, then the reaction is Stoichiometric • No O2 or Fuel in products • This mixture produces nearly the hottest flame temperature • If a < x + y/4, then reaction is fuel-rich (oxygen-lean) • If a> x + y/4, then reaction is fuel-lean (oxygen-rich) • Air to fuel mass ratio [kg air/kg fuel] of reactants • Need to find molecular weights

  3. x x Molecular Weight of a Pure Substance x x x x x x x • Only one type of molecule: • AxByCz… • Molecular Weight • MW = x(AWA) + y(AWB) + z(AWC) + … • AWi = atomic weights • Inside front cover of book • Examples • See page 701 for fuels

  4. x o x Mixtures containing n components x x x x x x x o o o • number of moles of species • Total number of moles in system • Mole Fraction of species i • mass of species • Total Mass • MassFraction of species i • Useful facts: • but • Mixture Molar Weight: • Example • write inside front cover of your book • Relationship between and

  5. Stoichiometric Air/Fuel Mass Ratio • aSt= x + y/4 • Generally

  6. Equivalence Ratio • Stiochiometric • Fuel Lean • % Stoichiometric Air (%SA) • % Excess Oxygen (%EO) = (%SA)-100%

  7. Example • For extra credit, this problem may be clearly reworked and turned in at the beginning of the next class period. • Problem 2.11, Page 91: In a propane-fueled truck, 3 percent (by volume) oxygen is measure in the exhaust stream of the running engine. Assuming “complete” combustion without dissociation, determine the air-fuel ratio (mass) supplied to the engine. • Also find • Equivalence Ratio: • % Stoichiometric air: %SA • % Excess Oxygen: %EA • ID: Fuel Rich, Fuel Lean, Stoichiometric? • Work on the board

  8. Ideal Gas Equation of State • Universal Gas Constant • Inside book front cover • kJ = kN*m= kPa*m3 • Specific Gas Constant • R = • MW = Molecular Weight of that gas • Number of molecules • N*NAV • Avogadro's Number, • Number of molecules in 12 kg of C12

  9. Partial Pressure of a specie in a mixture of pressure • Each specie acts as if it was the only component at the given V and T • Specie : • Mixture: • Ratio:

  10. Thermodynamics can be used to predict combustion energy release and actual product composition • Extensive thermodynamic properties depend on System Size (extent) • Examples • Volume V [m3] • Internal Energy E [kJ] • Enthalpy H = E + PV [kJ] • Test: cut system in half • Denoted with CAPITAL letters • Intensive Properties • Independent of system size • Examples • Per unit mass (lower case) • v = V/m [m3/kg] • u = U/m [kJ/kg] • h = H/m [kJ/kg] • Denoted using lower-case letters • Exceptions • Temperature T [°C, K] • Pressure P [Pa] • Molar Basis (use bar ) • V = vm = • U = um = • H = hm= • N number of moles in the system • Useful because chemical equations deal with the number of moles, not mass

  11. Thermodynamic Systems (reactors) • Closed systems • Rigid tanks, piston/cylinders • 1 = Initial state; 2 = Final state • Mass: • Chemical composition changes • But atoms are conserved • 1st Law • How to find internal energy for mixtures, and change when composition changes due to reactions (not covered in Thermodynamics I) m, E

  12. Open Systems (control volume) Inlet i Outlet o • Steady State, Steady Flow (SSSF) • Fixed volume, no moving boundaries • Properties are constant and uniform • Inside CV (Dm=DE=0), and • At ports () • One inlet and one outlet: • Atoms are conserved • Energy: • Composition and temperature of inlet and outlet are not the same due to reaction • Need to find (not covered in Thermodynamics I) Dm=DE=0

  13. Calorific Equations of State for a pure substance For ideal gases • Differentials (small changes) • For ideal gas • = 0; • For ideal gas • = 0; • Specific Heat [kJ/kg C] • Energy inpuy to increase temperature of one kg of a substance by 1°C at constant volume or pressure • How are and measured? • Calculate • = ; = w m, T m, T Q Q V = constant P = wg/A = constant

  14. Molar Specific Heat Dependence on Temperature • Monatomic molecules: Nearly independent of temperature • Only translational kinetic energy • Multi-Atomic molecules: Increase with temperature and number of molecules • Also possess rotational and vibrational kinetic energy

  15. Internal Energy and Enthalpy • Once cp(T) and cv(T) are known, internal energy change can be calculated by integration • Appendix A (pp. 687-699, bookmark) • Note = • =

  16. Mixture Properties • Enthalpy • Internal Energy • Use these relations to calculate mixture enthalpy and internal energies (per mass or mole) as functions of the properties of the individual components and their mass or molar fractions. • u and h depend on temperature, but not pressure • Individual gas properties are on pp. 687-699 as functions of gas and T

  17. Standardized Enthalpy and Enthalpy of Formation • Needed for chemically-reacting systems because energy is required to form and break chemical bonds • Not considered in Thermodynamics I • Needed to find and • Standard Enthalpy at Temperature T = • Enthalpy of formation at standard reference state: Tref and P°+ • Sensible enthalpy change in going from Trefto T = • Appendices A and B pp 687-702

  18. Normally-Occurring Elemental Compounds • For example: • O2, N2, C, He, H2 • = 0 • Use these as bases to tablulate the energy for form of more complex compounds • Example: • At 298K (1 mole) O2 + 498,390 kJ  (2 mole) O • To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond • for other compounds are in Appendices A and B

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