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ME 475/675 Introduction to Combustion. Lecture 14. Announcements. HW 5 Due Now Midterm 1 September 29, 2014 In-class review Today Tutorial: Sunday morning, Sept 25, 9 to 10 am, in PE 113 (who will come?). Midterm. Open book with bookmarks, 1 page of notes 3-4 HW-like problems
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ME 475/675 Introduction to Combustion Lecture 14
Announcements • HW 5 Due Now • Midterm 1 • September 29, 2014 • In-class review Today • Tutorial: Sunday morning, Sept 25, 9 to 10 am, in PE 113 (who will come?)
Midterm • Open book with bookmarks, 1 page of notes • 3-4 HW-like problems • Coverage • Chapter 1-3, HW 1-5 • All examples in class and book • Handout • Applicable problems from an only Midterm 1 • Intended to show problem style (like HW) • Please do not ask Rachel or me to work them for you
Likely Problem Types • Mixtures and their properties • Heat of Combustion • for a specified temperature, TProd = TReact • Adiabatic Flame Temperature • no dissociation • constant Pressure or Volume • Chemical Equilibrium • for specified T and P • Simple Combustion Equilibrium Products • water/shift reaction • Diffusion • Stefan (planar) or radial • mass flow rates and mass fraction profiles • Boundary mass fraction • Dependence of diffusion coefficient on T and P
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 • (generally ~20) • Need to find molecular weights
Equivalence Ratio • Stiochiometric • Fuel Lean • CxHy + a(O2+3.76N2) • % Stoichiometric Air (%SA) • % Excess Oxygen (%EO) = (%SA)-100%
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 • Hint: put in front cover of book • Fuels • Bookmark page 701 for fuels
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: • Put inside front cover of book • Relationship between and
Ideal Gas Equation of State • Universal Gas Constant • Inside book front cover • kJ = kPa*m3 • Specific Gas Constant • R = • MW = Molecular Weight of that gas • Number of molecules • N*NAV • Avogadro's Number,
Thermodynamic Systems (reactors) • Closed systems • Open Steady State, Steady Flow (SSSF) Systems • How to find changes, and , for mixture when temperatures and composition change due to reactions (not covered in Thermodynamics I) m, E Inlet i Outlet o Dm=DE=0
For a pure substances • Mass and Molar () Bases • N number of moles in the system • and • Tabulated: Appendix A, pp. 687-699, for combustion gases • bookmark • = • =; = • For mixtures • , • ,
Standardized Enthalpy and Enthalpy of Formation • Needed to find and for chemically-reacting systems because energy is required to form and break chemical bonds • Standard Enthalpy at Temperature T = • Enthalpy of formation from “normally occurring elemental compounds,” at standard reference state: Tref = 298 K and P° = 1 atm • Sensible enthalpy change in going from Trefto T = • Normally-Occurring Elemental Compounds • Examples: O2, N2, C, He, H2 • Their enthalpy of formation at are defined to be = 0 • Use these compounds as bases to tabulate the energy to form other compounds
Enthalpy of Combustion (or reaction) Products Complete Combustion CCO2 HH2O 298.15 K, 1 atm Reactants 298.15 K, P = 1 atm Stoichiometric • How much energy is released from a reaction if the product and reactant temperatures and pressures are the same? • 1st Law, Steady Flow Reactor • and Enthalpy of Reaction (< 0 for combustion) • Dependent on T and P of reaction • Heat of Combustion
Example: Stoichiometric Methane Combustion • CH4 + 2 (O2 + 3.76 N2) 1 CO2 + 2 H2O + 7.52 N2 • @ 25C and 1 kmol CH4 p 688 p 692 p 701 (Heat into system for TR = TP) Water Vapor
Other Bases • Per kg fuel • Heat of Combustion • (Heat out for TR = TP) • See page 701, LHV = Lower Heating Value • Corresponds to water vapor in the products p 692 p 692 • p. 701: Higher Heating Value = HHV = 55,528 (slightly larger than book, not due to dissociation since temp is low)
Per kg of reactant mixture • LHV =
Adiabatic () Flame Temperature, Complete Combustion Products CCO2 HH2O PP = PR, T = TAd Stoichiometric Reactants TR PR • 1st Law, Steady Flow Reactor • All chemical energy goes into heating the products • To find adiabatic flame temperature use • PP = PR and • Lower if we include dissociation
Constant Volume Adiabatic Flame Temperature V, m • Use definition: (since standard internal energy U is not tabulated, and Ideal ga)
Find equilibrium composition (reactants and products) for a given Temperature, Pressure & Mass Q T,P • aA + bB + … eE + fF + … • Use Q and boundary work to achieve P and T • Equilibrium Constant • , • since • Standard State Gibbs Function Change • ) • In terms of Gibbs functions of formation (tabulated in App. A and B) • Products are “favored” • As increases, which happens when decreases • If NProd > NRact , as P decreases
Equilibrium Products of Combustion • Combine Chemical Equilibrium (2nd law) & Adiabatic Flame Temperature (1st law) • For Example: Propane and air combustion • Ideal Stoichiomectric • Four products for a range of air/fuel ratios: • Now consider seven more possible dissociation products: • What happens as air/fuel (equivalence) ratio changes minor
Simple Product Calculation method • No minor species • Assume and are known • What is a good assumption for lean or stoichiometric mixtures ? • c = e = 0 (no CO or H2), but now include • Mole Fractions • ; ; ;
For Rich combustion • ; no (or fuel) • 4 unknowns: b, c, d and e • 3 Atom balances: C, H, O • Need one more constraint • Consider “Water-Gas Shift Reaction” equilibrium • Not dependent on P since number of moles of products and reactants are the same • ; • See plot from data on page 51 • KP = 0.22 to 0.1635 for T = 2000 to 3500 K
Solution • Since , use “-” root
Stefan Problem (no reaction) x L- • One dimensional tube (Cartesian) • Gas B is stationary: • Gas A moves upward • Want to find this • but treat as constant YB YA B+A Y A
Mass Flux and fraction of evaporating liquid A For =0.99 =0.9 =0.5 =0.1 =0.05
Liquid-Vapor Interface Boundary Condition • At interface need • Saturation pressure at temperature T • For water, tables in thermodynamics textbook • Or use Clausius-Slapeyron Equation (page 18 eqn. 2.19) • If given , we can use this to find • Page 701, Table B: , at A+B Vapor Liquid A
Spherical Droplet Evaporation • A is evaporating, find and • B is stagnant • If and do not change with time • Then decreases as decreases
Droplet Diameter versus time • Mass Conservation • , • Evaporation Const. • Constant slope for versus • Confirmed by experiment • Droplet life
Dependence of on Temperature and Pressure • Fairly independent of T and P
Flame temperature and major mole-fractions vs Tad[K] • Equivalence Ratio • At , O2, CO, H2 all present due to dissociation. Not present in “ideal” combustion • at .15 • at .05 • and decrease for • For decreases faster • For decreases faster % “Old” “New” Fuel Rich Get CO, H2 Fuel Lean O2
Minor-Specie Mole Fractions • NO, OH, H, O • < 4000 ppm = 0.4% • Peak near ppm