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ME 475/675 Introduction to Combustion Lecture 2 Announcement • On Friday I will modify HW 1, which will be due on Monday Ideal Stoichiometric Hydrocarbon Combustion air • CxHy + a(O2+3.76N2) ο (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) ππ2 ππΉπ’ππ • 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, fuel left over) • If a > x + y/4, then reaction is fuel-lean (oxygen-rich, O2 left over) • Equivalence Ratio • Φ= πππ‘ ππ΄ππ‘π’ππ = ππ2 ππΉπ’ππ ππ‘ • Φ = 1 → Stiochiometric • Φ > 1 → Fuel Rich • Φ < 1 → Fuel Lean ππ2 ππΉπ’ππ π΄ππ‘π’ππ = ππΉπ’ππ,π΄ππ‘π’ππ ππ2 ,ππ‘ ππΉπ’ππ,ππ‘ ππ2 ,π΄ππ‘π’ππ Air to fuel mass ratio [kg air/kg fuel] of reactants • • π΄ πΉ = ππ΄ππ ππΉπ’ππ π΄ πΉ ππ‘ = π π2 ππ΄ππ πππ΄ππ = = ππΉπ’ππ πππΉπ’ππ ππΉπ’ππ (1+3.76)πππ΄ππ π ππ‘ πππΉπ’ππ • Need to find molecular weights ππ΄ππ ππ 2 πππ΄ππ πππΉπ’ππ = (1+3.76)πππ΄ππ π πππΉπ’ππ Molecular Weight of a Pure Substance 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 • πππ2 = 2(AWπ ) = 2(15.9994) = 32.00 ππ ππππ • πππ»2π = 2(AWπ» ) + (AWπ ) = 2(1.00794) + • πππππππππ = πππΆ3 π»8 = 3 12.011 + 8 • See page 701 for fuels ππ (15.9994) = 18.02 ππππ ππ 1.00794 = 44.097 ππππ x xx x x x Mixtures containing n components • Total number of moles in system • ππππ‘ππ = π π=1 ππ • Mole Fraction of species i • ππ = π ππ πππ‘ππ = ππ π π=1 ππ • Mixture Molar Weight: πππππ₯ = • πππππ₯ = • πππππ₯ = ππ ππ ππ ππ = = x x xx x x x o o o x x • ππ = number of moles of species π • π = 1, 2, . . π o ππ πππ = ππππ‘ππ ππππ‘ππ = ππ /πππ ππππ‘ππ ππππ‘ππ ππ πππ (weighted average) 1 ππ /πππ • Example • ππ = ππ πππ = mass of species π • Total Mass • π πππ‘ππ = • πππ΄ππ = • = 0.21 ∗ 2 ∗ 15.9994 + 0.79 ∗ 2 ∗ 14.0067 ππ • = 0.21 ∗ 32.00 + 0.79 ∗ 28.00 = 28.85 πππππ • Remember and/or write inside front cover of your book π π=1 ππ • Mass Fraction of species i • ππ = π ππ πππ‘ππ = ππ π π=1 ππ • Useful facts: • ππ=1 ππ = ππ=1 ππ = 1 • but ππ ≠ ππ ππ πππ = 0.21πππ2 + 0.79πππ2 • Relationship between ππ and ππ • ππ = • ππ = ππππ‘ππ ππ ππ = ππ πππππ₯ π ππ πππ ππππ‘ππ πππππ₯ = ππ πππ πππππ₯ Stoichiometric Air/Fuel Mass Ratio • For Hydrocarbon fuel CxHy • π΄ πΉ ππ‘ = πππ‘ (1+3.76)πππ΄ππ πππΉπ’ππ ππ 28.85 πππππ • πππ΄ππ = ππ πππ = • πππΉπ’ππ = πππΆπ₯π»π¦ = π₯ 12.011 + π¦(1.00794) • aSt = x + y/4 • π΄ πΉ ππ‘ = π¦ ππ (4.76)28.85 4 πππππ ππ ππ 12.011πππππ +π¦(1.00794πππππ) π₯+ π₯ • For πΆπ₯ π»π¦ , 10 < π΄ πΉ ππ‘ < 35 • Constraints on y/x later 1 ππ π₯ πππππ 1 ππ 1.00794 π₯ πππππ 1.00794 = 136.24 1+ π¦ π₯ 4 11.92+ π¦ π₯ Equivalence Ratio Φ •Φ= πππ‘ ππ΄ππ‘π’ππ = πππ‘ (1+3.76)πππ΄ππ πππΉπ’ππ ππ΄ππ‘π’ππ (1+3.76)πππ΄ππ πππΉπ’ππ π΄ = π΄ πΉ ππ‘ πΉ π΄ππ‘π’ππ = πΉπ΄ππ‘π’ππ π΄ππ‘ πΉππ‘ π΄π΄ππ‘π’ππ • Φ = 1 → Stiochiometric • Φ > 1 → Fuel Rich • Φ < 1 → Fuel Lean π¦ •π= π₯+ 4 Φ • CxHy + a(O2+3.76N2) •% •% 100% Φ πΉππ‘ π΄π΄ππ‘π’ππ Stoichiometric Air (%SA)= = ∗ 100% πΉπ΄ππ‘π’ππ π΄ππ‘ 1 Excess Oxygen (%EO) = (%SA)-100% = − 1 100% Φ 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: Φ • % Stoichiometric air: %SA • % Excess Oxygen: %EA • ID: Are reactants Fuel Rich, Fuel Lean, or Stoichiometric? • Work on the board Thermodynamic Systems (reactors) • Closed systems 1π2 • Rigid tanks, piston/cylinders • 1 = Initial state; 2 = Final state • Mass: π1 = π2 = π • Chemical composition changes 1π2 m, E • But atoms are conserved • 1st Law • 1π2 − 1π2 = ΔπΈ = π(π2 − π1 ) • π=π’+ • 1π2 π£2 2 + ππ§ − 1π2 = π π’2 − π’1 + π£22 2 − π£12 2 + π π§2 − π§1 • How to find internal energy π’ for mixtures, and change π’2 − π’1 when composition changes due to reactions (not covered in Thermodynamics I) Open Systems (control volume) • Steady State, Steady Flow (SSSF) Inlet i • Fixed volume, no moving boundaries Outlet o • Properties are constant and uniform • Inside CV (Dm=DE=0), and ππ π + ππ£ π Dm=DE=0 π0 π + ππ£ • At ports (β = π’ + ππ£) • One inlet and one outlet: ππ = ππ = π • Atoms are conserved • Energy: ππΆπ − ππΆπ = π βπ − βπ + ππΆπ π£π2 2 − π£π2 2 ππΆπ + π π§π − π§π • Composition and temperature of inlet and outlet are not the same due to reaction • Need to find βπ − βπ (not covered in Thermodynamics I) π Combustion Thermochemistry • We use thermodynamics to evaluate the internal energy, enthalpy and entropy of a systems at different states • The difference in energy and enthalpy between the products and reactants is used with the first law to predict the heat of combustion (energy released during combustion, due to changes in chemical bonds) if the products are assumed to be at the same temperature as the reactants • The adiabatic flame temperature (product temperature assuming all reaction heat release stays in the system) can be determined by assuming the products have the same energy level as the reactants • The steady state product composition for a reaction may be determined from entropy considerations • Need to find • The internal energy, enthalpy and entropy of mixtures of gases, and • How to account for effects of chemical bonds • These are not covered in ME 311 Thermodynamics I, but we we’ll cover them now Ideal Gas Equation of State • ππ = ππ π π • Universal Gas Constant • π π = 8.315 ππ½ πππππ πΎ = 8315 • Inside book front cover • kJ = kN*m= π½ πππππ πΎ kPa*m3 • ππ = ππ π = π ∗ ππ (π π /ππ)π • Specific Gas Constant • R =π π /ππ • MW = Molecular Weight of that gas • ππ£ = π π; π£ = • π = ππ π π π = 1 π • Number of molecules • N*NAV • Avogadro's Number, ππ΄π • 6.022 ∗ 26 ππππππ’πππ 10 πππππ ππππππ’πππ 1023 ππππ • 6.022 ∗ • Number of molecules in 12 kg of C12 Partial Pressure ππ and volume fraction mixture of pressure π and volume V • Each specie acts as if it was the only component at the given V and T • Specie π: ππ π = ππ π π’ π • Mixture: π π = π π π’ π ππ Ratio: π ππ π • = = ππ • ππ = ππ π • ππ = ππ π = π ππ = π ππ π of a specie in a • Volume Fraction • Each specie acts as if it was the only component at the given P and T • Specie π:πππ = ππ π π’ π • Mixture: π π = π π π’ π • ππ Ratio: π = ππ π = ππ End 2015 Extensive and Intensive System Properties • Intensive Properties • 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 • 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 = Nπ£ • U = um = Nπ’ • H = hm = Nβ • N number of moles in the system • Useful because chemical equations deal with the number of moles, not mass Calorific Equations of State for a pure substance • π’ = π’ π, π£ = π’(π) ≠ ππ(π£) • β = β π, π = β(π) ≠ ππ(π) For ideal gases • Differentials (small changes) • ππ’ = ππ’ ππ π£ ππ + ππ’ = ππ£ π 0; • For ideal gas • ππ’ ππ π£ ππ’ ππ£ π • Specific Heat [kJ/kg C] • Energy input to increase temperature of one kg of a substance by 1°C at constant volume or pressure • How are ππ£ π and ππ π measured? ππ£ ππ ππ£ T [K] Q [joules] w Q = ππ£ π m, T Q m, T • ππ’ = ππ£ π ππ • πβ = πβ ππ π ππ + • For ideal gas • πβ = ππ π 0; πβ ππ π • πβ = ππ π ππ πβ ππ π = ππ π ππ V = constant P = wg/A = constant • Calculate ππ ππ π£ = π πΔπ π ππ π£ • ππ = ππ ∗ ππ; ππ£ = ππ£ ∗ ππ ππ½ πππΎ ππ½ ππππ πΎ 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 Internal Energy and Enthalpy • Once cp(T) and cv(T) are known, internal energy change can be calculated by integration • π’ π = π’πππ + • β π = βπππ + π π ππππ π£ π π ππππ π π ππ π ππ • Appendix A (pp. 687-699, bookmark) • ππ π : π‘πππππ πππ ππ’ππ£π πππ‘π • Note ππ£ = ππ − π π’ • ππ =ππ /ππ Mixture Properties • Enthalpy • π»πππ₯ = ππ βπ = π πππ‘ππ βπππ₯ • ππππ (π») = • π»πππ₯ = π π βπ ππππ‘ππ = ππ ππ (π») ππ βπ = ππππ‘ππ βπππ₯ • ππππ (π») = ππ βπ ππππ‘ππ = • Internal Energy • ππππ (π») = ππ ππ (π») • ππππ π» = ππ ππ π» ππ ππ (π») • 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 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 π’2 − π’1 and βπ − βπ π • βπ π = βπ,π ππππ + Δβπ ,π (π) • Standard Enthalpy at Temperature T = • Enthalpy of formation at standard reference state: Tref and P° • Sensible enthalpy change in going from Tref to T = • Appendices A and B pp 687-702 π π ππππ π + π ππ Normally-Occurring Elemental Compounds • For example: • O2, N2, C, He, H2 π • βπ,π ππππ = 0 • ππππ = 298K • Use these as bases to tabulate the energy for form of more complex compounds • Example: • At 298K (1 mole) O2 + 498,390 kJ ο (2 mole) O • To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond • • π βπ,π π βπ,π ππππ = 498,390 kJ 2 ππππ π = ππ½ 249,195 πππππ ππππ for other compounds are in Appendices A and B
