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ME 475/675 Introduction to Combustion Lecture 29 Announcements β’ Midterm 2 β’ November 12, 2014 (New Date, not Nov. 5) β’ HW 12 (problem 8.2) β’ 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 β’ 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) β’ How to estimate the laminar flame speed ππΏ and thickness πΏ? β’ Depends on the pressure, fuel, equivalence ratio,β¦ β’ Flame reference frame: π£π , ππ π£π’ = ππΏ , ππ’ , πΏ~1 ππ β’ 1 ππ πΉπ’ + π ππ ππ₯ β 1 + π ππ ππ β’ Conservation of mass: π = ππ’ π£π’ = π£π ππ π£π ; π£π’ = ππ’ ππ = ππ’ π ππ π ππ’ ππ = β’ For hydrocarbon fuels at P = 1 atm, ππ’ β 300πΎ, ππ β 2100πΎ, π£π β 7π£π’ β’ What happens within a premixed flame? ππ ππ’ Simplified Analysis Assumptions Heat and Radical β’ One dimensional flow β’ Kinetic energy = viscosity = radiation = 0 β’ Constant pressure β’ ππ₯" = ππ βπ ππ₯ and πππ = βππ ππ₯ πΌ π = = β π ππππ ππ" β’ Lewis Number, πΏπ 1; π ππ β ππ β’ ππ,π = ππ β ππ ππππππππ‘π’ππ ππ ππππππ β’ Single Step Kinetics β’ Ξ¦ < 1, Fuel Lean, so fuel is completely consumed Conservations Laws β’ Mass Conservation " β’ π = ππ£π₯ = ππππ π‘πππ‘; π ππ£π₯ ππ₯ ππ₯ =0 β’ Species Conservation β’ π΄ππ" β’ ππβ²β²β² + π΄ ππ₯ = π ππ" ππ₯ = ππβ²β²β² π ππ₯ =π΄ π΄ππ" " " π ππ ππ + ππ₯ ππ₯ π" ππ β ππ πππ ππ₯ π΄ (using Flickβs Law) β’ Apply to: 1 ππ πΉ + π ππ ππ₯ β 1 + π ππ; Air/Fuel ratio π = 1 π β²β²β² β’ ππΉβ²β²β² = πππ₯ =β 1 β²β²β² πππ π+1 β’ π = 1, 2, β¦ , π = 3; Fuel, Oxidizer, Products β’ β’ β’ π ππ₯ π ππ₯ π ππ₯ πππΉ π ππΉ β ππ ππ₯ ππππ₯ " π πππ₯ β ππ ππ₯ ππππ " π πππ β ππ ππ₯ " ππβ²β²β² = ππΉβ²β²β² β²β²β² = πππ₯ = π ππΉβ²β²β² β²β²β² = πππ = β 1 + π ππΉβ²β²β² ππ ππ₯ ππ πΉ " π π π π΄ ππ" + ππ₯ ππ₯ Energy Conservation (Ch. 7, pp. 239-244) ππ₯ π΄πβ²β² β π΄ππ₯β²β² πππ β’ πππ β πππ = π βππ’π‘ β βππ β’π΄ β’ ππ₯β²β² π ππ₯β²β² β ππ₯ β ππ₯β²β² = πβ β²β² π ππ₯ + π ππ₯β²β² ππ₯ ππ₯ = π΄π β²β² β+ πβ ππ₯ ππ₯ ββ β’ Decreasing heat flux in x-direction increases enthalpy in the +x-direction π΄πβ²β² πβ β+ ππ₯ ππ₯ β²β² π π π₯ π΄ ππ₯β²β² + ππ₯ ππ₯ Heat Flux β’ Heat: Energy transfer at a boundary due to temperature difference β’ When there is a large species gradient, diffusion contributes to heat flux β’ ππ₯β²β² = ππ βπ ππ₯ β’ Note: β’ ππ₯β²β² πβ ππ₯ = + β²β² ππ,πππππ’π πππ βπ π ππ₯ ππ βπ = β’ πππ β ππ₯ π = ππ₯ β ππ ππ₯ πβ ππ = ππ βπ ππ₯ β ππ πβ ππ₯ β’ Heat Flux = πππ β ππ₯ π β ππ ππ ππ₯ ππ βπ ππ₯ = + ππ = πβπ ππ₯ = ππ βπ ππ₯ β πππ ππ β ππ₯ π πππ β ππ₯ π + πβ β ππ ππ₯ = ππ ππ,π + ππ βπ ππ₯ ππ ππ₯ = ππ ππππ ππ₯ β’ Flux due to conduction + β’ Flux of standardized enthalpy due to species diffusion + β’ Flux of sensible enthalpy due to species diffusion β’ For πΏπ = πΌ π = π ππππ β 1; π β ππππ β’ Shvab-Zeldovich assumption (πΏπ β π(1) for most combustion gases) πβ β²β² β’ ππ₯ = βππ (due to both conduction and diffusion) ππ₯ β ππ πππ β ππ₯ π + ππ πππ β ππ₯ π ππ ππ₯ Shvab-Zeldovich form of Energy Conservation β’ π ππ₯β²β² β ππ₯ β’β= β’ β’ ππ₯β²β² β’ β’ β’ β’ π ππ₯ π ππ₯ πβ ππ₯ = = πβ β²β² π ππ₯ π ππ βπ,π ππ βπ = = πππ π β ππ₯ π,π πβ βππ ππ₯ π ππ₯β²β² β ππ₯ = π ππ₯ (Energy Equation) + π πβπ,π π π ππ + π ππ ππ₯ ππ₯ ππππ π + = βππ πππ π β ππ₯ π,π ππ πππ π ππ βπ,π ππ₯ πππ π ππ β ππ₯ π,π π π β βπ,π πβ²β² ππ ππ₯ π ππ π ππ,π ππ πππ + β β 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 β²β² β²β² π ππ βπ,π = π ππ β ππππ ππ₯ ππ₯ ππ₯ πππ ππ π ππ β²β² ππ = π ππ β ππππ ππ₯ ππ₯ ππ₯ ππ₯ πβ²β² Shvab-Zeldovich form of Energy Conservation β’ π β ππ₯ π βπ,π β’ π β ππ₯ π βπ,π ππ" = πππ ππ π β²β² β ππ = π ππ β ππ₯ ππ₯ ππ₯ ππ" " π π π π π β βπ,π = β βπ,π ππβ²β²β² ππ₯ πβ²β² ππ β’ Species conservation: β’ π ππ" ππ₯ ππ ππππ ππ₯ = ππβ²β²β² π π π π β²β²β² β²β²β² βπ,π ππβ²β²β² = βπ,πΉ ππΉβ²β²β² + βπ,ππ₯ πππ₯ + βπ,ππ πππ β’ π π π = βπ,πΉ ππΉβ²β²β² + βπ,ππ₯ πππΉβ²β²β² β βπ,ππ 1 + π ππΉβ²β²β² β’ π π π = ππΉβ²β²β² βπ,πΉ + βπ,ππ₯ π β βπ,ππ 1+π = ππΉβ²β²β² ΞβπΆ π π π β’ ΞβπΆ = βπ,πΉ + βπ,ππ₯ π β βπ,ππ 1 + π : Heat of combustion β’ For πΏπ = πΌ π = π ππππ β 1; π β ππππ β’ βππΉβ²β²β² ΞβπΆ = πβ²β² ππ ππ ππ₯ β π ππ₯ π ππ ππ₯ nd 2 order differential equation for T(x) β’ β²β² ππ π ππ₯ β 1 π ππ ππ₯ ππ π ππ₯ β’ Where πβ²β² = ππ’ ππΏ = β²β²β² ππΉ ΞβπΆ β ππ β’ Only accepts 2 boundary conditions, β’ But we have 4 (Eigenvalue problem) β’ For π₯ β ββ: π β ππ’ πππ β’ For π₯ β +β: π β ππ΅ πππ ππ ππ₯ ππ ππ₯ β0 β0 β’ For an approximate solution, assume a simple profile β’ Find flame thickness πΏ and laminar flame speed ππΏ = conditions can be satisfied πβ²β² ππ’ so that all four boundary Approximate Solution β’ ππ β²β² π ππ₯ β 1 π ππ ππ₯ β’ Integrate β’ πβ²β² π ππ ππ’ β ππ π ππ₯ +β ββ 1 ππ = β²β²β² ππΉ ΞβπΆ β ππ ππ₯ ππ 0 π ππ₯ 0 = ΞβπΆ +β β²β²β² β ππΉ ππ₯ ππ ββ β’ ππΉβ²β²β² = ππ π = 0 πππ π₯ < 0 ππ π₯ > πΏ β’ Inside 0 < π₯ < πΏ, β’ πβ²β² β’ πβ²β² ππ ππ₯ = ππ βππ’ , πΏ so ππ₯ = ππ β ππ’ = ΞβπΆ +β β²β²β² β ππΉ ππ ββ ππ β ππ’ = ΞβπΆ πΏ β²β²β² β ππΉ ππ π πΏ ππ ππ βππ’ eqn. 1 β’ Two unknowns: πβ²β² = ππΏ ππ’ and πΏ β’ Need another equation πΏ ππ ππ βππ’ = ππ β²β²β² ΞβπΆ πΏ 1 β ππΉ ππ ππ βππ’ ππ’ π ππ Average over temperature: ππΉβ²β²β² Approximate Solution β’ ππ β²β² π ππ₯ β 1 π ππ ππ₯ β’ Integrate β’ β’ β’ β’ ππ π ππ₯ πΏ/2 ββ = β²β²β² ππΉ ΞβπΆ β ππ ππ₯ ππ βππ’ πΏ ππ +ππ’ 2 ππ’ π ππ ΞβπΆ πΏ/2 β²β²β² π π β =β ππΉ ππ₯ ππ ππ₯ 0 ππ ββ π +π π ππ βππ’ πβ²β² π π’ β ππ’ β =0 2 ππ πΏ β²β² ππ βππ’ π ππ βππ’ β²β² π = 2 ππ πΏ 2π πΏ = β²β² eqn. 2 π ππ β’ From eqn. 1 β’ πβ²β² = πβ²β² ππ β ππ’ = 2πΞβπΆ ππ2 ππ βππ’ ΞβπΆ πΏ β²β²β² β ππΉ ππ βππΉβ²β²β² = ππΏ ππ’ ; ππΏ = 1 ππ’ = β0 ΞβπΆ β²β²β² 2π β ππΉ β²β² ππ π ππ 2πΞβπΆ ππ2 ππ βππ’ βππΉβ²β²β² Approximate Solution β’ ππΏ = 1 ππ’ 2πΞβπΆ ππ2 ππ βππ’ βππΉβ²β²β² β’ But ΞβπΆ = 1 + π ππ ππ β ππ’ β’ Show this in HW (problem 8.2) β’ ππΏ = β’ β’ 2π 1+π ππ ππ βππ’ ππ’ ππ’ ππ2 ππ βππ’ βππΉβ²β²β² 2πΌ 1+π ππΏ = βππΉβ²β²β² ππ’ 2π 2π eqn. 2: πΏ = β²β² = π ππ ππΏ ππ’ ππ β’πΏ= 2πΌππ’ β²β²β² 1+π βπ πΉ = 2πΌ ππΏ = ; where πΌ = 2πΌ ππΏ = π ππ’ ππ 2πΌ 2πΌ 1+π ππ’ β²β²β² β ππΉ (Fast flames are thin) Example 8.2 β’ 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. β’ Find: ___?