ME 475/675 Introduction to Combustion
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ME 475/675 Introduction to Combustion Lecture 21 Coupled chemical/thermal analysis, Constant pressure, Constant Volume, Start Ex 6.1 Announcements • HW 8, Numerical Solution to Example 6.1 • Due Monday, Oct. 19, 2015 Chapter 6 Coupling Chemical and Thermal Analysis of Reacting systems • Four simple reactor systems, p 184 1. Constant pressure and fixed Mass • Time dependent, well mixed 2. Constant-volume fixed-mass • Time dependent, well mixed 3. Well-stirred reactor • Steady, different inlet and exit conditions 4. Plug-Flow • Steady, dependent on location • Coupled Energy, species production, and state constraints • For plug flow also need momentum since speed and pressure vary with location Constant pressure and fixed mass reactor π • Constituents • reactants and products, π = 1, 2, … π (book uses π) • P and m constant • Find, as a function of time, t • Temperature π • To find use conservation of energy • Molar concentration π = ππ π π (book calls this ππ ) • use species generation/consumption rates from chemical kinetics • π= π , ππππ π π • state, mixture • Highly coupled • Assume we know “production rates” per unit volume • 1 πππ π ππ‘ = ππ = π(π) π π=1 π ππ = ππ 1 , 2 , … π • Rate depends on current molar concentration (per volume) of each constituent, and temperature • From chemical Kinetics First Law (Energy Conservation) • π−π = =π ππ’ ππ‘ ππ • Only boundary work: π = π ππ‘ ππ ππ = ππ‘ ππ‘ ππ ππ +π ππ‘ ππ‘ • π−π • π= π ππ ππ‘ = π π+ππ ππ‘ ππ» ππ‘ = • Where enthalpy π» = π + ππ, and for a mixture H = • π= ππ» ππ‘ = πππ ππ‘ βπ + ππ ππ βπ π π βπ ππ‘ 1 πππ ; ππ‘ πβπ • Production rate ππ = π • πππ ππ‘ • π= • π π π = = πππ ; ππ = V π ; ππ‘ πππ βπ + V π ππ,π ππ ππ‘ ππ βπ + π ππ,π ππ ππ‘ = = ππ,π π ππ ππ‘ ; Divide by π ππ βπ + ππ ππ‘ π ππ,π ; Solve for ππ ππ‘ π • − ππ βπ π ππ π = ππ‘ π ππ,π π • First order differential equation, Initial conditions (IC): π = π0 ππ‘ π‘ = 0 • At each time step, to find the change in π • Need π, π, ππ (which depends on π and T), π ,π, βπ π , and ππ,π (π) Change in Molar Concentrations • ππ ππ‘ π = π ππ ππ‘ = 1 πππ π ππ‘ − ππ ππ π 2 ππ‘ = ππ − π ππ π ππ‘ * • Species production and volume change affect molar concentration • Find the volume V from ideal gas equation of state • ππ = ππ π’ π = ππ π π’ π π • Take time derivative • ππ π ππ‘ = π π’ π π ππ ππ‘ to see how volume changes with time = π π’ π ππ‘ • Divide both sides by ππ = • 1 ππ π ππ‘ = 1 ππ πππ ππ‘ + 1 ππ π ππ‘ = π ππ ππ‘ ππ + ππ‘ ππ = π π’ π • = ππ − π ππ π + ππ‘ + ππ π π’ π 1 ππ π 1 πππ π ππ‘ + 1 ππ π ππ‘ = ππ π • Plug into * ππ ππ‘ πππ 1 ππ π ππ‘ • Initial Conditions: at t = 0, π = π 0 , π = 1, 2, … π + 1 ππ π ππ‘ ππ ππ‘ ππ π + 1 coupled System of 1st order differential equations • Initial Conditions, at t = 0 • π = π 0 , π = 1, 2, … π, and • π = π0 • Assume we also know ππ = π΄ππ₯π(− πΈπ΄,π π π’ π ) π π=1 π ππ • Use the first order differentials to find π and π at time t + Δπ‘ • ππ ππ‘ • ππ ππ‘ = ππ − π = π − π ππ βπ π π ππ,π π ππ π ≈ + 1 ππ π ππ‘ ≈ Δπ Δπ‘ ; π Δπ ; Δπ‘ π‘+Δπ‘ = π π‘+ ππ ππ‘ ππ ππ‘ Δπ‘ ππ‘+Δπ‘ = ππ‘ + Δπ‘ • System Volume • π π‘ = t T π π π = [1] [2] π ; π πππ π= π π = … [M] w1 w2 0 T0 [1]0 [2]0 … [M]0 Dt 2Dt ππ π = … ππ πππ π wM V = π πππ (algebraic not differential eqn.) Q d[1]/dt d[2]/dt … d[M]/dt dT/dt Constant-Volume Fixed-Mass Reactor • Constant V and m • Find T, P, π (πππ π = 1,2, … , π) versus time π‘ • 1st Law • π−π = • π= • π= • • • • = ππ ππ‘ = = π ππ π’π π ππ π’π π ; π’π πππ ππ‘ + ππ ππ‘ π π’π ππ‘ πππ 1 πππ ππ = ; so = πππ π ππ‘ ππ‘ π π’π ππ ππ = V π , = ππ£,π π ππ‘ ππ‘ • π= π π ππ ππ‘ π’π π ππ + V π ππ£,π π’π ππ + π π − π π’ π ππ π ππ£,π ; ππ ππ£,π ππ‘ ππ ππ‘ = , divide by V π’π ππ + ππ ππ‘ π ππ£,π , solve in contrast for constant pressure ππ ππ‘ ππ for ππ‘ π − π π βπ π π = π ππ,π π Tabulated Data • Need to evaluate ππ ππ‘ = π − π π’π ππ π ππ£,π (true, but not useful) • However, tables only have • ππ,π (= ππ£,π + π π’ ), so use • ππ£,π = ππ,π − π π’ • βπ (= π’π + ππ£ = π’π + π π’ π) , so use • π’π = βπ − π π’ π • ππ ππ‘ = π − π βπ ππ +π π’ π π ππ,π −π π’ ππ (true and useful) • Initial Condition: π = π0 at π‘ = 0 • Species Production (no volume change) • ππ ππ‘ = π ππ π ππ‘ = 1 πππ π ππ‘ = ππ = π(π) π π=1 π ππ Reactor Pressure • Ideal Gas Law • ππ = ππ π π’ π • Divide by π (constant) • π= ππ π π π’ π = π π π’ π • Pressure Rate of change (affects detonation) • ππ ππ‘ = π π’ π π π ππ‘ + π ππ ππ‘ = π π’ π π ππ π ππ‘ + ππ ππ‘ π = π π’ π ππ + ππ ππ‘ π Example 6.1 (p. 189) This will be HW • In spark-ignition engines, knock occurs when the unburned fuel-air mixture ahead of the flame reacts homogeneously, i.e., it auto-ignites. The rate-ofpressure rise is a key parameter in determining knock intensity and propensity for mechanical damage to the piston-crank assembly. Pressure-versus-time traces for normal and knocking combustion in a spark-ignition engine are illustrated in Fig. 6.2. Note the rapid pressure rise in the case of heavy knock. Figure 6.3 shows schleiren (index-of-refraction gradient) photographs of flame propagation for normal and knocking combustion • Example 6.1 • Create a simple constant-volume model of the autoignition process and determine the temperature and the fuel and product concentration histories. Also determine the dP/dt as a function of time. Assume initial conditions corresponding to compression of a fuel-air mixture from 300 K and 1 atm to top-dead-center for a compression ratio of 10:1. The initial volume before compression is 3.68*10-4 m3, which corresponds to an engine with both a bore and a stroke of 75 mm. Use ethane as fuel. Assume: • One-step global kinetics using the rate parameters for ethane C2H6 (Table 5.1) • Fuel, air, and products all have equal molecular weights: MWF= MWOx= MWP= 29 • The specific heats of the fuel, air and products are constants and equal: • cp,F= cp,Ox= cp,Pr= 1200 J/kg K • The enthalpy of formation of the air and products are zero, and that of the fuel is • 4*107 J/kg • The stoichiometric air-fuel ratio is 16.0 and restrict combustion to stoichiometric or lean conditions. Global and Quasi-global mechanisms • Empirical • πΆπ₯ π»π¦ + π₯ + π¦ 4 π2 ππΊ π¦ 2 π₯πΆπ2 + π»2 π • stoichiometric mixture with π2 , not air • π πΆπ₯ π»π¦ ππ‘ = −π΄ππ₯π πΈπ π π’ π πΆπ₯ π»π¦ π π2 π = πππππ ππ3 π • Page 157, Table 5.1: π΄, πΈπ π π’ , π πππ π for different fuels • These values are based on flame speed data fit (Ch 8) • In Table 5.1 units for π΄ = • However, we often want π΄ • 1 πππππ 1−π−π π ππ3 • π΄ 1 πππππ 1−π−π π π3 πππππ 1−π−π − π+π πππππ πππππ ππ3 = ππ3 π ππ3 π 1−π−π 1 πππππ in units of π π3 πππππ 1000 πππππ =π΄ 1−π−π 100 ππ 3 π 1 πππππ 1−π−π π ππ3 = 1 πππππ 1−π−π π ππ3 Sometimes Want These Units 1000 1−π−π = 1 πππππ 1−π−π π π3 10001−π−π = π΄ πππ₯π‘ππππ 10001−π−π Given in Table 5.1, p. 157 •
