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ME 475/675 Introduction to Combustion Lecture 29 Announcements • Midterm 2 • November 12, 2014 (New Date, not Nov. 5) • HW 12 (problem 8.2) • Due Wednesday, November 5, 2014 Tube filled with stationary premixed Oxidizer/Fuel Products shoot out as flame burns in ππΏ Laminar Flame Speed Burned Products πΏ Unburned Fuel + Oxidizer • Video: http://www.youtube.com/watch?v=CjGuHbsi3a8&feature=related (1.49; 3:19; 3:45; 4:05; 4:31; 5:05; 5:35; 6:17; 6:55) • How to estimate the laminar flame speed ππΏ and thickness πΏ? • Depends on the pressure, fuel, equivalence ratio,… • Flame reference frame: π£π , ππ π£π’ = ππΏ , ππ’ , πΏ~1 ππ • 1 ππ πΉπ’ + π ππ ππ₯ → 1 + π ππ ππ • Conservation of mass: π = ππ’ π£π’ = π£π ππ π£π ; π£π’ = ππ’ ππ = ππ’ π ππ π ππ’ ππ = • For hydrocarbon fuels at P = 1 atm, ππ’ ≈ 300πΎ, ππ ≈ 2100πΎ, π£π ≈ 7π£π’ • What happens within a premixed flame? ππ ππ’ Simplified Analysis Assumptions Heat and Radical • One dimensional flow • Kinetic energy = viscosity = radiation = 0 • Constant pressure • ππ₯" = ππ −π ππ₯ and πππ = −ππ ππ₯ πΌ π = = ≈ π ππππ ππ" • Lewis Number, πΏπ 1; π ππ ≈ ππ • ππ,π = ππ ≠ ππ ππππππππ‘π’ππ ππ ππππππ • Single Step Kinetics • Φ < 1, Fuel Lean, so fuel is completely consumed Conservations Laws • Mass Conservation " • π = ππ£π₯ = ππππ π‘πππ‘; π ππ£π₯ ππ₯ ππ₯ =0 • Species Conservation • π΄ππ" • ππ′′′ + π΄ ππ₯ = π ππ" ππ₯ = ππ′′′ π ππ₯ =π΄ π΄ππ" " " π ππ ππ + ππ₯ ππ₯ π" ππ − ππ πππ ππ₯ π΄ (using Flick’s Law) • Apply to: 1 ππ πΉ + π ππ ππ₯ → 1 + π ππ; Air/Fuel ratio π = 1 π ′′′ • ππΉ′′′ = πππ₯ =− 1 ′′′ πππ π+1 • π = 1, 2, … , π = 3; Fuel, Oxidizer, Products • • • π ππ₯ π ππ₯ π ππ₯ πππΉ π ππΉ − ππ ππ₯ ππππ₯ " π πππ₯ − ππ ππ₯ ππππ " π πππ − ππ ππ₯ " ππ′′′ = ππΉ′′′ ′′′ = πππ₯ = π ππΉ′′′ ′′′ = πππ = − 1 + π ππΉ′′′ ππ ππ₯ ππ πΉ " π π π π΄ ππ" + ππ₯ ππ₯ Energy Conservation (Ch. 7, pp. 239-244) ππ₯ π΄π′′ β π΄ππ₯′′ πππ • πππ − πππ = π βππ’π‘ − βππ •π΄ • ππ₯′′ π ππ₯′′ − ππ₯ − ππ₯′′ = πβ ′′ π ππ₯ + π ππ₯′′ ππ₯ ππ₯ = π΄π ′′ β+ πβ ππ₯ ππ₯ −β • Decreasing heat flux in x-direction increases enthalpy in the +x-direction π΄π′′ πβ β+ ππ₯ ππ₯ ′′ π π π₯ π΄ ππ₯′′ + ππ₯ ππ₯ Heat Flux • Heat: Energy transfer at a boundary due to temperature difference • When there is a large species gradient, diffusion contributes to heat flux • ππ₯′′ = ππ −π ππ₯ • Note: • ππ₯′′ πβ ππ₯ = + ′′ ππ,πππππ’π πππ βπ π ππ₯ ππ βπ = • πππ β ππ₯ π = ππ₯ − ππ ππ₯ πβ ππ = ππ −π ππ₯ − ππ πβ ππ₯ • Heat Flux = πππ β ππ₯ π − ππ ππ ππ₯ ππ −π ππ₯ = + ππ = πβπ ππ₯ = ππ −π ππ₯ − πππ ππ β ππ₯ π πππ β ππ₯ π + πβ − ππ ππ₯ = ππ ππ,π + ππ −π ππ₯ ππ ππ₯ = ππ ππππ ππ₯ • Flux due to conduction + • Flux of standardized enthalpy due to species diffusion + • Flux of sensible enthalpy due to species diffusion • For πΏπ = πΌ π = π ππππ ≈ 1; π ≈ ππππ • Shvab-Zeldovich assumption (πΏπ ≈ π(1) for most combustion gases) πβ ′′ • ππ₯ = −ππ (due to both conduction and diffusion) ππ₯ − ππ πππ β ππ₯ π + ππ πππ β ππ₯ π ππ ππ₯ Shvab-Zeldovich form of Energy Conservation • π ππ₯′′ − ππ₯ •β= • • ππ₯′′ • • • • π ππ₯ π ππ₯ πβ ππ₯ = = πβ ′′ π ππ₯ π ππ βπ,π ππ βπ = = πππ π β ππ₯ π,π πβ −ππ ππ₯ π ππ₯′′ − ππ₯ = π ππ₯ (Energy Equation) + π πβπ,π π π ππ + π ππ ππ₯ ππ₯ ππππ π + = −ππ πππ π β ππ₯ π,π ππ πππ π ππ βπ,π ππ₯ πππ π ππ β ππ₯ π,π π π − βπ,π π′′ ππ ππ₯ π ππ π ππ,π ππ πππ + − − ππ" ππ ππ ππ₯ πππ π β ππ₯ π,π + + ππ ππ ππ₯ = = π ππ βπ,π πππ π β ππ₯ π,π + + π π ππ ππππ π ππ ππ ππ₯ ππ ππ ππ₯ πππ π ππ = βπ,π + ππ ππ₯ ππ₯ ππ π ππ π ′′ ′′ π ππ βπ,π = π ππ − ππππ ππ₯ ππ₯ ππ₯ πππ ππ π ππ ′′ ππ = π ππ − ππππ ππ₯ ππ₯ ππ₯ ππ₯ π′′ Shvab-Zeldovich form of Energy Conservation • π − ππ₯ π βπ,π • π − ππ₯ π βπ,π ππ" = πππ ππ π ′′ − ππ = π ππ − ππ₯ ππ₯ ππ₯ ππ" " π π π π π − βπ,π = − βπ,π ππ′′′ ππ₯ π′′ ππ • Species conservation: • π ππ" ππ₯ ππ ππππ ππ₯ = ππ′′′ π π π π ′′′ ′′′ βπ,π ππ′′′ = βπ,πΉ ππΉ′′′ + βπ,ππ₯ πππ₯ + βπ,ππ πππ • π π π = βπ,πΉ ππΉ′′′ + βπ,ππ₯ πππΉ′′′ − βπ,ππ 1 + π ππΉ′′′ • π π π = ππΉ′′′ βπ,πΉ + βπ,ππ₯ π − βπ,ππ 1+π = ππΉ′′′ ΔβπΆ π π π • ΔβπΆ = βπ,πΉ + βπ,ππ₯ π − βπ,ππ 1 + π : Heat of combustion • For πΏπ = πΌ π = π ππππ ≈ 1; π ≈ ππππ • −ππΉ′′′ ΔβπΆ = π′′ ππ ππ ππ₯ − π ππ₯ π ππ ππ₯ nd 2 order differential equation for T(x) • ′′ ππ π ππ₯ − 1 π ππ ππ₯ ππ π ππ₯ • Where π′′ = ππ’ ππΏ = ′′′ ππΉ ΔβπΆ − ππ • Only accepts 2 boundary conditions, • But we have 4 (Eigenvalue problem) • For π₯ → −∞: π → ππ’ πππ • For π₯ → +∞: π → ππ΅ πππ ππ ππ₯ ππ ππ₯ →0 →0 • For an approximate solution, assume a simple profile • Find flame thickness πΏ and laminar flame speed ππΏ = conditions can be satisfied π′′ ππ’ so that all four boundary Approximate Solution • ππ ′′ π ππ₯ − 1 π ππ ππ₯ • Integrate • π′′ π ππ ππ’ − ππ π ππ₯ +∞ −∞ 1 ππ = ′′′ ππΉ ΔβπΆ − ππ ππ₯ ππ 0 π ππ₯ 0 = ΔβπΆ +∞ ′′′ − ππΉ ππ₯ ππ −∞ • ππΉ′′′ = ππ π = 0 πππ π₯ < 0 ππ π₯ > πΏ • Inside 0 < π₯ < πΏ, • π′′ • π′′ ππ ππ₯ = ππ −ππ’ , πΏ so ππ₯ = ππ − ππ’ = ΔβπΆ +∞ ′′′ − ππΉ ππ −∞ ππ − ππ’ = ΔβπΆ πΏ ′′′ − ππΉ ππ π πΏ ππ ππ −ππ’ eqn. 1 • Two unknowns: π′′ = ππΏ ππ’ and πΏ • Need another equation πΏ ππ ππ −ππ’ = ππ ′′′ ΔβπΆ πΏ 1 − ππΉ ππ ππ −ππ’ ππ’ π ππ Average over temperature: ππΉ′′′ Approximate Solution • ππ ′′ π ππ₯ − 1 π ππ ππ₯ • Integrate • • • • ππ π ππ₯ πΏ/2 −∞ = ′′′ ππΉ ΔβπΆ − ππ ππ₯ ππ −ππ’ πΏ ππ +ππ’ 2 ππ’ π ππ ΔβπΆ πΏ/2 ′′′ π π − =− ππΉ ππ₯ ππ ππ₯ 0 ππ −∞ π +π π ππ −ππ’ π′′ π π’ − ππ’ − =0 2 ππ πΏ ′′ ππ −ππ’ π ππ −ππ’ ′′ π = 2 ππ πΏ 2π πΏ = ′′ eqn. 2 π ππ • From eqn. 1 • π′′ = π′′ ππ − ππ’ = 2πΔβπΆ ππ2 ππ −ππ’ ΔβπΆ πΏ ′′′ − ππΉ ππ −ππΉ′′′ = ππΏ ππ’ ; ππΏ = 1 ππ’ = ≈0 ΔβπΆ ′′′ 2π − ππΉ ′′ ππ π ππ 2πΔβπΆ ππ2 ππ −ππ’ −ππΉ′′′ Approximate Solution • ππΏ = 1 ππ’ 2πΔβπΆ ππ2 ππ −ππ’ −ππΉ′′′ • But ΔβπΆ = 1 + π ππ ππ − ππ’ • Show this in HW (problem 8.2) • ππΏ = • • 2π 1+π ππ ππ −ππ’ ππ’ ππ’ ππ2 ππ −ππ’ −ππΉ′′′ 2πΌ 1+π ππΏ = −ππΉ′′′ ππ’ 2π 2π eqn. 2: πΏ = ′′ = π ππ ππΏ ππ’ ππ •πΏ= 2πΌππ’ ′′′ 1+π −π πΉ = 2πΌ ππΏ = ; where πΌ = 2πΌ ππΏ = π ππ’ ππ 2πΌ 2πΌ 1+π ππ’ ′′′ − ππΉ (Fast flames are thin) Example 8.2 • Estimate the laminar flame speed of a stoichiometric propane-air mixture using the simplified theory results (Eqn. 8.20). Use the global one-step reaction mechanism (Eqn. 5.2, Table 5.1, pp. 156-7) to estimate the mean reaction rate. • Find: ___?
