ME 475/675 Introduction to Combustion
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ME 475/675 Introduction to Combustion Lecture 1 Formalities, Introduction and Applications, General Combustion Reaction, Thermodynamic Properties, Equation of State Formalities • Enhanced development of this course funded by US Nuclear Regulator Commission (Why is the NRC interested in combustion?) • Professor Miles Greiner, greiner@unr.edu, 784-4873 • Course Website: • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm • Textbook • An Introduction to Combustion Concepts and Applications, 3rd ed., Turns • http://catalogs.mhhe.com/mhhe/viewProductDetails.do?isbn=0073380199 • www.mhhe.com/turns3e • Syllabus • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/COMBUSTION475.675.syllabus.pdf • Prerequisites • ENGR 360 Fluids; ME 311 Thermo I; ME 314 Heat Transfer or CHE 374 Transport Theory • Grading • Assignments involve Excel and MathCad • Extra Credit: Complete Lecture examples • Projects (describe later in term) • homemade backpacking stove • ME 451/2: Design an experiment to measure quantities that are calculated in this class • i.e. Diffusion Flame Length; Premixed Flame Speed • Create education experiment to acquire and compare data with expectations Combustion is a source of: • Useful Energy for • • • • • • Light Space Heating Material Processing Boilers, refineries, smelters, dryers Incinerators (waste “disposal”) Heat engines • Electric Power Generation • Transportation • Safety hazards • Building and Forest Fires • Heat and toxic fumes • Nuclear Waste Transport • Pollution • Un- and partially-burned hydrocarbon fuel (CXHY) • Oxides of nitrogen (NO, NO2) • Particulates (carbon soot) • Smog • Acid rain • Intellectual Challenges • Integrates thermal science, chemistry, computational methods • Professional Opportunities Combustion Types • Flaming Combustion • Rapid Oxidation of fuel, generating heat and light • Converts chemical bond to sensible energy • Non-flame combustion modes • Detonation • Supersonic, volumetric, • engine knock, explosives • Smoldering • Coal mine Types of Flaming Combustion • Diffusion Flames • Pre-mixed Flames • Oxidizer and fuel mixed at molecular level before reaction takes place Products Fuel + Oxidizer • Auto engine, kitchen stove • Blue flame Fuel Vapor Oxidizer Products • Combustion takes place where oxidizer/fuel “ratio” is favorable • Candle, campfire, pool fire • Orange/yellow flame (soot) Products Fuel + Oxidizer • Bunsen Burner Demo (propane C3H8) • • • • Premixed Flame: Blue with internal cone, flash-back when turned off (flame speed exceeds gas speed) Diffusion: Orange Yellow, flame length Hot flame ring where combustion takes place How hot are the diffusion and premixed flames? Used Nuclear Fuel Transportation Safety • Nuclear fuel assemblies become highly radioactive after they have been used in a reactor • Thick-walled packages are used to transport used assemblies for storage, processing or disposal • Federal Regulations require transport packages to maintain their containment, shielding and criticality-control function even after the following series of hypothetical accident conditions • • • • 30 ft drop onto an unyielding surface 40 inch drop onto a steel puncture bar Full engulfment in an 800°C fire for 30 minutes Water immersion UNR Research 800 Tests 1 and 2 Light Winds DT [°C] 600 Simulations Measurements Test 3 Strong Winds 400 200 0 • Develop and experimentally benchmark computational methods to predict heat transfer to massive objects from large-scale fires. • Use these methods to predict package response assuming they are in proximity to large, longduration fires. • Are these diffusion or premixed flames? 0 10 20 Time, t [min] 30 40 General Combustion Reaction • Fuel + Oxidizer + small activation energy ο Products + Energy • In this class we examine combustion of Hydrocarbon Fuels with Air or pure O2 • General hydrocarbon fuel, CxHy (or CxHyOz if oxygenated) • Air (most common oxidizer) • Mostly Nitrogen N2 and Oxygen O2 (plus traces of other gases, H2O, CO2,…) • Mole Fractions: ππ2 = ππ2 ππππ‘ππ = 0.21, ππ2 = 0.79 • Molecules of N2 per O2 molecule: ππ2 ππ2 = 0.79 0.21 = 3.76 • For Ideal Combustion (with no dissociation to CO, H, HO, NO, NO2, O) • • • • All products are CO2, H2O, N2, O2 All C atoms ο CO2 All H atoms ο H2O All N atoms ο N2 (do not participate) Chemical Equation for Hydrocarbon Combustion • CxHy + a(O2+3.76N2) ο _b_CO2 + _c_ H2O + _d_ N2 + _e_ CxHy + _f_ O2 • a is the number of moles of O2 per mole of CxHy • Number of moles of air is 4.76a. • Affected by air/fuel ratio, which is “user defined” • What does a need to be so there is no unreacted CxHy or O2 in the products? • e=f=0 • Stoichiometric fuel and air mixture (aST) • Atomic Balance (atoms are conserved during chemical reactions) • C: x = b so b = x (e=0) • H: y = 2c, so c = y/2, • O: 2aST = 2b + c = 2x + y/2, so aST = x + y/4 (f = 0) • Depends on fuel; This is the amount of air needed for complete Stoichiometric combustion • N: 2(3.76) aST = 2d, so d = 3.76aST = 3.76(x + y/4) • For a Stoichiometric Combustion • CxHy + (x + y/4)(O2+3.76N2) ο (x)CO2 + (y/2) H2O + 3.76(x + y/4) N2 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) • 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 • π΄ πΉ = ππ΄ππ ππΉπ’ππ = ππ΄ππ πππ΄ππ ππΉπ’ππ πππΉπ’ππ = • For Stoichiometric mixture: ππ2 ππ΄ππ ππ 2 πππ΄ππ (1+3.76)πππ΄ππ ππΉπ’ππ πππΉπ’ππ 1∗πππΉπ’ππ π΄ (1+3.76)πππ΄ππ = π ππ‘ πΉ ππ‘ 1∗πππΉπ’ππ • Need to find molecular weights =π 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) + (15.9994) • πππππππππ = πππΆ3 π»8 = 3 12.011 + 8 1.00794 • See page 701 for fuels ππ = 18.02 ππππ ππ = 44.097 ππππ x xx x x x End 2015 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 • Mass Fraction of species i • ππ = π ππ πππ‘ππ = • πππππ₯ = π π=1 ππ • ππ = ππ πππ = mass of species π • Total Mass • π πππ‘ππ = • Mixture Molar Weight: πππππ₯ = • πππππ₯ = ππ π π=1 ππ • Useful facts: • ππ=1 ππ = ππ=1 ππ = 1 • but ππ ≠ ππ 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 1 ππ /πππ • Example • πππ΄ππ = ππ πππ = 0.21πππ2 + 0.79πππ2 • = 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 • Relationship between ππ and ππ • ππ = π ππ πππ‘ππ • ππ = ππ =π πππππ₯ πππ ππ πππ πππ‘ππ πππππ₯ πππ = ππ ππ πππ₯ Stoichiometric Air/Fuel Mass Ratio • π΄ πΉ ππ‘ = πππ‘ (1+3.76)πππ΄ππ 1∗πππΉπ’ππ • πππ΄ππ = ππ πππ = ππ 28.85 πππππ • πππΉπ’ππ = πππΆπ₯π»π¦ = π₯ 12.011 + π¦(1.00794) • aSt = x + y/4 • π΄ πΉ ππ‘ π¦ = ππ π₯+ 4 (4.76)28.85πππππ ππ ππ π₯ 12.011 +π¦(1.00794 ) • For πΆπ₯ π»π¦ , 10 < πππππ π΄ πΉ ππ‘ πππππ < 35 • Constraints on y/x later π¦ π₯ = 136.24 1+ 4 11.92+ π¦ π₯ Equivalence Ratio Φ •Φ= 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= πππ‘ Φ 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: Fuel Rich, Fuel Lean, Stoichiometric? • Work on the board Ideal Gas Equation of State • Number of molecules • ππ = ππ π π • Universal Gas Constant • π π = 8.315 ππ½ πππππ πΎ = 8315 • Inside book front cover • N*NAV π½ πππππ πΎ • kJ = kN*m= 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 • Avogadro's Number • 26 ππππππ’πππ ππ΄π = 6.022 ∗ 10 πππππ ππππππ’πππ ππ΄ = 6.022 ∗ 1023 ππππ • • Number of molecules in 12 kg of C12 Thermodynamics can be used to predict combustion energy release and actual product composition • 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 Ideal Gas Equation of State • Avogadro's Number • ππ = ππ π π • Universal Gas Constant • π π = ππ½ 8.315 πππππ πΎ = π½ 8315 πππππ πΎ • Inside book front cover • kJ = kN*m= kPa*m3 • ππ = ππ π = π ∗ ππ (π π /ππ)π • Specific Gas Constant • R =π π /ππ • ππ£ = π π; π£ = • π = ππ π π π = 1 π • ππππππ’πππ 26 ππ΄π = 6.022 ∗ 10 πππππ ππππππ’πππ ππ΄ = 6.022 ∗ 1023 ππππ • • Number of molecules in 12 kg of C12 • Calorific Equations of State • 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 ππ’ ππ π£ • For ideal gas • 0; • ππ’ = ππ£ π ππ πβ • πβ = ππ π ππ + • For ideal gas • πβ = ππ π 0; πβ ππ π • πβ = ππ π ππ ππ’ ππ£ π ππ£ • β π = βπππ + = ππ£ π πβ ππ π = ππ π • π’ π = π’πππ + ππ π π ππππ π£ π π ππππ π π ππ π ππ 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 Appendix A (pp. 687-699, bookmark) • ππ π : π‘πππππ πππ ππ’ππ£π πππ‘π • Note ππ£ = ππ − π π’ • ππ = ππ ∗ ππ Ideal Gas Mixtures o x x xx x x x o o o x x • ππ = number of moles of species π in system • Mixture Molar Weight ππππ‘ππ ππ • ππππ‘ππ = ππ=1 ππ • πππππ₯ = = = ππππ‘ππ ππ • Mole Fraction of species i ππ ππππ‘ππ • ππ = ππ ππππ‘ππ = • πππππ₯ = ππ π π=1 ππ • ππ = π π=1 ππ • • Mass Fraction of species i • ππ = ππ ππππ‘ππ = ππ π π π=1 π • Useful fact: • ππ ≠ ππ ; but π π=1 ππ = • Conversion between ππ and ππ • Total Mass • π πππ‘ππ = ππ ππ πππ = ππππ‘ππ 1 = ππ /πππ ππ /πππ = π π=1 ππ =1 ππ = ππππ‘ππ πππππ₯ ππ = ππ πππ ππ πππ ππππ‘ππ πππππ₯ = πππ ππ πππππ₯ ππ πππ Partial Pressure ππ of a specie in a mixture of pressure π • 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
