Transcript Slides
ME 475/675 Introduction to Combustion Lecture 29 Shvab-Zeldovich form of Energy Conservation Announcements • Midterm 2 • November 13, 2015 • HW 12 (problem 8.2) • Due Wednesday, November 4, 2015 Broader Impact Assignment • Two important ABET Student Learning Objectives: • Students will demonstrate : • A recognition of a need for, and an ability to engage in, life long learning (graduate school, continuing education, short courses, technical training, self instruction by reading articles or textbook) • A knowledge of contemporary issues • Two choices, Both due Friday, November 6, 2015, two paragraph summaries • Seminar: Used Nuclear Fuel: Storage, Transportation, and Disposal – Technical, Political and Other Issues • John Wagner, Director, Reactor & Nuclear Systems Division, Oak Ridge National Laboratory • Today, Noon, November 2, 2015, DMS 102 • Hosted by UNR American Nuclear Society’s Student Chapter • President: Kodi Summers [email protected] • Or • Article: Dependence of Fire Time of Concern on Location of a One-assembly Transport Packages Tube filled with stationary premixed Oxidizer/Fuel Products shoot out as flame burns in 𝑆𝐿 Laminar Flame Speed Burned Products 𝛿 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 𝑆𝐿 and thickness 𝛿? • Depends on the pressure, fuel, equivalence ratio,… • Flame reference frame: 𝑣𝑏 , 𝜌𝑏 𝑣𝑢 = 𝑆𝐿 , 𝜌𝑢 , 𝛿~1 𝑚𝑚 • 1 𝑘𝑔 𝐹𝑢 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑘𝑔 𝑃𝑟 • Conservation of mass: 𝑚 = 𝜌𝑢 𝑣𝑢 = 𝑣𝑏 𝜌𝑏 𝑣𝑏 ; 𝑣𝑢 = 𝜌𝑢 𝜌𝑏 = 𝑃𝑢 𝑅𝑇𝑏 𝑅𝑇𝑢 𝑃𝑏 = • For hydrocarbon fuels at P = 1 atm, 𝑇𝑢 ≈ 300𝐾, 𝑇𝑏 ≈ 2100𝐾, 𝑣𝑏 ≈ 7𝑣𝑢 • What happens within a premixed flame? 𝑇𝑏 𝑇𝑢 Simplified Analysis Assumptions • Assumptions • • • • • • • One dimensional flow Neglect kinetic energy, viscosity, radiation Constant pressure 𝑐𝑃,𝑖 = 𝑐𝑃 ≠ 𝑓𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑟 𝑆𝑝𝑒𝑐𝑖𝑒 Single Step Kinetics Φ < 1, Fuel Lean, so fuel is completely consumed 𝛼 𝑘 𝑘 Lewis Number, 𝐿𝑒 = = ≈ 1, so ≈ 𝒟𝜌 𝒟 𝒟𝜌𝑐𝑃 𝑐𝑃 • Calculate heat and species transport per unit area normal to flow • 𝑞𝑥" = 𝑚′′ ℎ − 𝑘 𝑑𝑇 𝑑𝑥 and 𝑚𝑖" = 𝑚′′ 𝑌𝑖 − 𝜌𝒟 𝑑𝑌𝑖 𝑑𝑥 • Transport of heat and species ahead of reaction zone allows the flame to move forward. • Use this to calculate laminar flame speed and flame thickness • Thinner flames have bigger gradients and so diffuse and move faster Conservations Laws Heat and Radical • Mass Conservation • 𝑚" = 𝜌𝑣𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡; 𝑑 𝜌𝑣𝑥 𝑑𝑥 =0 • Species Conservation (in + generation = out) • 𝐴𝑚𝑖" 𝑚𝑖′′′ + 𝐴 𝑑𝑥 𝑚𝑖′′′ " =𝐴 𝑑𝑚 𝑚𝑖" + 𝑖 𝑑𝑥 𝑑𝑥 𝑑 𝑚𝑖" • = 𝑑𝑥 • Generation of 𝑖 𝑚𝑖′′′ causes mass flux of 𝑖 𝑚𝑖" to increase with 𝑥. • Apply to stoichiometric combustion 1 𝑘𝑔 𝐹 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑃𝑟 𝑘𝑔 𝑂𝑥 = Stoichiometric Air/Fuel 𝑘𝑔 𝐹 1 ′′′ 1 ′′′ • 𝑚𝐹′′′ = 𝑚𝑂𝑥 =− 𝑚𝑃𝑟 𝜈 𝜈+1 " 𝑑 𝑚𝐹𝑢 ′′′ • 𝜈= • 𝑑𝑥 ratio = 𝑚𝐹𝑢 (< 0) • " 𝑑 𝑚𝑂𝑥 𝑑𝑥 ′′′ ′′′ = 𝑚𝑂𝑥 = 𝜈 𝑚𝐹𝑢 (< 0) • " 𝑑 𝑚𝑃𝑟 𝑑𝑥 ′′′ = 𝑚𝑃𝑟 = − 𝜈 + 1 𝑚𝐹′′′ (> 0) 𝑑𝑥 𝑚𝑖′′′ 𝐴𝑚𝑖" 𝐴 " 𝑑 𝑚 𝑖 𝐴 𝑚𝑖" + 𝑑𝑥 𝑑𝑥 Energy Conservation (Ch. 7, pp. 239-244) 𝐴𝑚′′ ℎ 𝐴𝑄𝑥′′ • 𝑄𝑄𝑉 − 𝑊𝑄𝑉 = 𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 •𝐴 • 𝑄𝑥′′ 𝑑 𝑄𝑥′′ − 𝑑𝑥 − 𝑄𝑥′′ = 𝑑ℎ ′′ 𝑚 𝑑𝑥 + 𝑑 𝑄𝑥′′ 𝑑𝑥 𝑑𝑥 = 𝑑𝑥 𝐴𝑚′′ ℎ+ 𝑑ℎ 𝑑𝑥 𝑑𝑥 −ℎ 𝑊𝑄𝑉 • Decreasing heat flux in x-direction increases enthalpy in the +x-direction 𝑑𝑥 𝑄𝑥′′ ℎ 𝐴𝑚′′ 𝑑ℎ ℎ+ 𝑑𝑥 𝑑𝑥 ′′ 𝑑 𝑄 𝑥 𝐴 𝑄𝑥′′ + 𝑑𝑥 𝑑𝑥 What is the Heat Flux? • Heat: Energy transfer at a boundary due to temperature difference • When there are species and temperature gradients, diffusion contributes to heat flux • 𝑄𝑥′′ = • But • 𝑑𝑇 −𝑘 𝑑𝑥 𝑑ℎ 𝑑𝑥 𝑑ℎ −𝜌𝒟 𝑑𝑥 = ′′ 𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 ℎ𝑖 + 𝑑 𝑑𝑥 𝑌𝑖 ℎ𝑖 = = −𝜌𝒟 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 − + = 𝑑𝑇 −𝑘 𝑑𝑥 − 𝑌𝑖 𝑑ℎ𝑖 𝑑𝑥 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 𝑑𝑇 𝜌𝒟𝑐𝑝 𝑑𝑥 = = −𝜌𝒟 𝑑𝑌𝑖 𝜌𝒟𝑖 ℎ𝑖 𝑑𝑥 + 𝑌𝑖 𝑐𝑝,𝑖 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 − = 𝑑𝑇 𝑑𝑥 𝑑𝑇 −𝑘 𝑑𝑥 = 𝑑𝑇 𝑘 𝑑𝑥 − 𝜌𝒟 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 = 𝑄𝑥′′ • Shvab-Zeldovich assumption: 𝐿𝑒 ≈ 𝑂 1 , valid for most combustion gases • 𝐿𝑒 = • 𝑄𝑥′′ = 𝛼 𝒟 = 𝑘 𝒟𝜌𝑐𝑃 𝑑ℎ −𝜌𝒟 𝑑𝑥 ≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃 • heat flux in x-direction due to both conduction and diffusion + 𝑐𝑝 𝑑𝑌𝑖 ℎ𝑖 𝑑𝑥 𝑑𝑇 𝑑𝑥 Shvab-Zeldovich form of Energy Conservation • 𝑑 𝑄𝑥′′ − 𝑑𝑥 •ℎ= • • 𝑄𝑥′′ • • • • 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑ℎ 𝑑𝑥 = = 𝑑ℎ ′′ 𝑚 𝑑𝑥 𝑜 𝑌𝑖 ℎ𝑓,𝑖 𝑌𝑖 ℎ𝑖 = = 𝑑𝑌𝑖 𝑜 ℎ 𝑑𝑥 𝑓,𝑖 𝑑ℎ −𝜌𝒟 𝑑𝑥 𝑑 𝑄𝑥′′ − 𝑑𝑥 = 𝑑 𝑑𝑥 (Energy Equation) + 𝑜 𝑑ℎ𝑓,𝑖 𝑑 𝑇 𝑌𝑖 + 𝑐 𝑑𝑇 𝑑𝑥 𝑑𝑥 𝑇𝑟𝑒𝑓 𝑝 + = −𝜌𝒟 𝑑𝑌𝑖 𝑜 ℎ 𝑑𝑥 𝑓,𝑖 𝜌𝒟 𝑑𝑌𝑖 𝑜 𝜌𝒟 ℎ𝑓,𝑖 𝑑𝑥 𝑑𝑌𝑖 𝑜 𝜌𝒟 ℎ 𝑑𝑥 𝑓,𝑖 𝑑 𝑜 − ℎ𝑓,𝑖 𝑚′′ 𝑌𝑖 𝑑𝑥 𝑇 𝑌𝑖 𝑇 𝑐𝑝,𝑖 𝑑𝑇 𝑟𝑒𝑓 + − − 𝑚𝑖" 𝑑𝑇 𝑐𝑝 𝑑𝑥 𝑑𝑌𝑖 𝑜 ℎ 𝑑𝑥 𝑓,𝑖 + + = = 𝑜 𝑌𝑖 ℎ𝑓,𝑖 𝑑𝑌𝑖 𝑜 ℎ 𝑑𝑥 𝑓,𝑖 + + 𝑇 𝑐 𝑑𝑇; 𝑇𝑟𝑒𝑓 𝑝 𝑑𝑇 𝑐𝑝 𝑑𝑥 𝑑𝑇 𝑐𝑝 𝑑𝑥 𝑑𝑇 𝑐𝑝 𝑑𝑥 𝑑𝑌𝑖 𝑜 𝑑𝑇 = ℎ𝑓,𝑖 + 𝑐𝑝 𝑑𝑥 𝑑𝑥 𝑑𝑇 𝑑 𝑑𝑇 𝑜 ′′ ′′ 𝑚 𝑌𝑖 ℎ𝑓,𝑖 = 𝑚 𝑐𝑝 − 𝜌𝒟𝑐𝑝 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑌𝑖 𝑑𝑇 𝑑 𝑑𝑇 ′′ 𝜌𝒟 = 𝑚 𝑐𝑝 − 𝜌𝒟𝑐𝑝 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑚′′ 𝑐𝑝 = 𝑌𝑖 𝑐𝑝,𝑖 Shvab-Zeldovich form of Energy Conservation • 𝑑 − 𝑑𝑥 𝑜 ℎ𝑓,𝑖 𝑚′′ 𝑌𝑖 𝑑𝑌𝑖 𝜌𝒟 𝑑𝑥 − 𝑚𝑖" • Left hand side= 𝑑 − 𝑑𝑥 • Species conservation: = 𝑑𝑇 ′′ 𝑚 𝑐𝑝 𝑑𝑥 𝑜 ℎ𝑓,𝑖 𝑚𝑖" = − 𝑑𝑚𝑖" 𝑑𝑥 − 𝑑 𝑑𝑥 " 𝑑 𝑚 𝑜 𝑖 ℎ𝑓,𝑖 𝑑𝑥 𝑑𝑇 𝜌𝒟𝑐𝑝 𝑑𝑥 =− 𝑜 ℎ𝑓,𝑖 𝑚𝑖′′′ = 𝑚𝑖′′′ • • 𝑜 𝑜 𝑜 𝑜 ′′′ ′′′ ℎ𝑓,𝑖 𝑚𝑖′′′ = ℎ𝑓,𝐹 𝑚𝐹′′′ + ℎ𝑓,𝑂𝑥 𝑚𝑂𝑥 + ℎ𝑓,𝑃𝑟 𝑚𝑃𝑟 𝑜 𝑜 𝑜 = ℎ𝑓,𝐹 𝑚𝐹′′′ + ℎ𝑓,𝑂𝑥 𝜈𝑚𝐹′′′ − ℎ𝑓,𝑃𝑟 1 + 𝜈 𝑚𝐹′′′ • 𝑜 𝑜 𝑜 = 𝑚𝐹′′′ ℎ𝑓,𝐹 + ℎ𝑓,𝑂𝑥 𝜈 − ℎ𝑓,𝑃𝑟 1+𝜈 = 𝑚𝐹′′′ Δℎ𝐶 𝑜 𝑜 𝑜 • Δℎ𝐶 = ℎ𝑓,𝐹 + ℎ𝑓,𝑂𝑥 𝜈 − ℎ𝑓,𝑃𝑟 1+𝜈 • For • Heat of combustion for 1 𝑘𝑔 𝐹 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑃𝑟 𝛼 𝑘 𝐿𝑒 = = ≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃 𝒟 𝒟𝜌𝑐𝑃 𝑑𝑇 𝑑 𝑑𝑇 • −𝑚𝐹′′′ Δℎ𝐶 = 𝑚′′ 𝑐𝑝 − 𝑘 𝑑𝑥 𝑑𝑥 𝑑𝑥 • Generation = (net flow out) + (net conduction out) nd 2 order differential equation for T(x) • ′′ 𝑑𝑇 𝑚 𝑑𝑥 − 1 𝑑 𝑐𝑝 𝑑𝑥 𝑑𝑇 𝑘 𝑑𝑥 = ′′′ 𝑚𝐹 Δℎ𝐶 − 𝑐𝑝 • Where 𝑚′′ = 𝜌𝑢 𝑆𝐿 • Only accepts 2 boundary conditions, • But we have 4 (Eigenvalue problem) • For 𝑥 → −∞: 𝑇 → 𝑇𝑢 𝑎𝑛𝑑 • For 𝑥 → +∞: 𝑇 → 𝑇𝐵 𝑎𝑛𝑑 𝑑𝑇 𝑑𝑥 𝑑𝑇 𝑑𝑥 →0 →0 • For an approximate solution, assume a simple profile • Find flame thickness 𝛿 and laminar flame speed 𝑆𝐿 = conditions can be satisfied 𝑚′′ 𝜌𝑢 so that all four boundary Approximate Solution • 𝑑𝑇 ′′ 𝑚 𝑑𝑥 − 1 𝑑 𝑐𝑝 𝑑𝑥 • Integrate • 𝑚′′ 𝑇 𝑇𝑏 𝑇𝑢 − 𝑑𝑇 𝑘 𝑑𝑥 +∞ −∞ 1 𝑐𝑝 = ′′′ 𝑚𝐹 Δℎ𝐶 − 𝑐𝑝 𝑑𝑥 𝑑𝑇 0 𝑘 𝑑𝑥 0 = Δℎ𝐶 +∞ ′′′ − 𝑚𝐹 𝑑𝑥 𝑐𝑝 −∞ • 𝑚𝐹′′′ = 𝑓𝑛 𝑇 = 0 𝑓𝑜𝑟 𝑥 < 0 𝑜𝑟 𝑥 > 𝛿 • Inside 0 < 𝑥 < 𝛿, ′′ • 𝑚 𝑇𝑏 − 𝑇𝑢 = • 𝑚′′ 𝑇𝑏 − 𝑇𝑢 = 𝑑𝑇 𝑑𝑥 = 𝑇𝑏 −𝑇𝑢 , 𝛿 so 𝑑𝑥 = Δℎ𝐶 +∞ ′′′ − 𝑚𝐹 𝑐𝑝 −∞ Δℎ𝐶 𝛿 ′′′ − 𝑚𝐹 𝑐𝑝 𝑇 𝛿 𝑑𝑇 𝑇𝑏 −𝑇𝑢 eqn. 1 • Two unknowns: 𝑚′′ = 𝑆𝐿 𝜌𝑢 and 𝛿 • Need another equation 𝛿 𝑑𝑇 𝑇𝑏 −𝑇𝑢 = 𝑇𝑏 ′′′ Δℎ𝐶 𝛿 1 − 𝑚𝐹 𝑐𝑝 𝑇𝑏 −𝑇𝑢 𝑇𝑢 𝑇 𝑑𝑇 Average over temperature: 𝑚𝐹′′′ Approximate Solution • 𝑑𝑇 ′′ 𝑚 𝑑𝑥 − 1 𝑑 𝑐𝑝 𝑑𝑥 • Integrate • • • • 𝑑𝑇 𝑘 𝑑𝑥 𝛿/2 −∞ = ′′′ 𝑚𝐹 Δℎ𝐶 − 𝑐𝑝 𝑑𝑥 𝑇𝑏 −𝑇𝑢 𝛿 𝑇𝑏 +𝑇𝑢 2 𝑇𝑢 𝑘 𝑑𝑇 Δℎ𝐶 𝛿/2 ′′′ 𝑚 𝑇 − =− 𝑚𝐹 𝑑𝑥 𝑐𝑝 𝑑𝑥 0 𝑐𝑝 −∞ 𝑇 +𝑇 𝑘 𝑇𝑏 −𝑇𝑢 𝑚′′ 𝑏 𝑢 − 𝑇𝑢 − =0 2 𝑐𝑝 𝛿 ′′ ≈0 𝑇𝑏 −𝑇𝑢 𝑘 𝑇𝑏 −𝑇𝑢 ′′ 𝑚 = 2 𝑐𝑝 𝛿 2𝑘 𝛿 = ′′ eqn. 2 𝑚 𝑐𝑝 • From eqn. 1 • 𝑚′′ = 𝑚′′ 𝑇𝑏 − 𝑇𝑢 = 2𝑘Δℎ𝐶 𝑐𝑝2 𝑇𝑏 −𝑇𝑢 Δℎ𝐶 𝛿 ′′′ − 𝑚𝐹 𝑐𝑝 −𝑚𝐹′′′ = 𝑆𝐿 𝜌𝑢 ; 𝑆𝐿 = 1 𝜌𝑢 = Δℎ𝐶 ′′′ 2𝑘 − 𝑚𝐹 ′′ 𝑐𝑝 𝑚 𝑐𝑝 2𝑘Δℎ𝐶 𝑐𝑝2 𝑇𝑏 −𝑇𝑢 −𝑚𝐹′′′ Approximate Solution • 𝑆𝐿 = 1 𝜌𝑢 2𝑘Δℎ𝐶 𝑐𝑝2 𝑇𝑏 −𝑇𝑢 −𝑚𝐹′′′ • But Δℎ𝐶 = 1 + 𝜈 𝑐𝑃 𝑇𝑏 − 𝑇𝑢 • Show this in HW (problem 8.2) • 𝑆𝐿 = • • 2𝑘 1+𝜈 𝑐𝑃 𝑇𝑏 −𝑇𝑢 𝜌𝑢 𝜌𝑢 𝑐𝑝2 𝑇𝑏 −𝑇𝑢 −𝑚𝐹′′′ 2𝛼 1+𝜈 𝑆𝐿 = −𝑚𝐹′′′ 𝜌𝑢 2𝑘 2𝑘 eqn. 2: 𝛿 = ′′ = 𝑚 𝑐𝑝 𝑆𝐿 𝜌𝑢 𝑐𝑝 •𝛿= 2𝛼𝜌𝑢 ′′′ 1+𝜈 −𝑚 𝐹 = 2𝛼 𝑆𝐿 = ; where 𝛼 = 2𝛼 𝑆𝐿 = 𝑘 𝜌𝑢 𝑐𝑃 2𝛼 2𝛼 1+𝜈 𝜌𝑢 ′′′ −𝑚𝐹 (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: ___? Conservations Laws 𝑑𝑥 𝐴𝑚𝑖" 𝐴 • 𝑚𝑖′′′ = 𝑑 𝑚𝑖" 𝑑𝑥 = 𝑑 𝑑𝑥 𝑚" 𝑌𝑖 − 𝜌𝒟 𝑑𝑌𝑖 𝑑𝑥 (using Flick’s Law) • 𝑖 = 1, 2, … , 𝑀 = 3; Fuel, Oxidizer, Products • • • 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑𝑌𝐹 𝑑𝑥 𝑑𝑌 𝜌𝒟 𝑂𝑥 𝑑𝑥 𝑑𝑌 𝜌𝒟 𝑃𝑟 𝑑𝑥 𝑚" 𝑌𝐹 − 𝜌𝒟 = 𝑚𝐹′′′ 𝑚" 𝑌𝑂𝑥 − ′′′ = 𝑚𝑂𝑥 = 𝜈 𝑚𝐹′′′ 𝑚" 𝑌𝑃𝑟 − ′′′ = 𝑚𝑃𝑟 = − 1 + 𝜈 𝑚𝐹′′′ 𝑚𝑖′′′ " 𝑑 𝑚 𝑖 𝐴 𝑚𝑖" + 𝑑𝑥 𝑑𝑥 What is the Heat Flux? • Heat: Energy transfer at a boundary due to temperature difference • When there are species and temperature gradients, diffusion contributes to heat flux • 𝑄𝑥′′ = 𝑑𝑇 −𝑘 𝑑𝑥 • Note: • 𝑄𝑥′′ 𝑑ℎ 𝑑𝑥 = + ′′ 𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 ℎ𝑖 𝑑 𝑑𝑥 𝑌𝑖 ℎ𝑖 = • 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 = 𝑑𝑥 − 𝑐𝑝 𝑑𝑥 𝑑ℎ 𝑑𝑇 = 𝑑𝑇 −𝑘 𝑑𝑥 − 𝜌𝒟 𝑑ℎ 𝑑𝑥 • Heat Flux = 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 − 𝑑𝑇 𝑐𝑝 𝑑𝑥 𝑑𝑇 −𝑘 𝑑𝑥 = + 𝑌𝑖 = 𝑑ℎ𝑖 𝑑𝑥 = 𝑑𝑇 −𝑘 𝑑𝑥 − 𝑑𝑌𝑖 𝜌𝒟𝑖 ℎ 𝑑𝑥 𝑖 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 + 𝑑ℎ − 𝜌𝒟 𝑑𝑥 = 𝑌𝑖 𝑐𝑝,𝑖 + 𝑑𝑇 𝑑𝑥 𝑑𝑇 −𝑘 𝑑𝑥 = − 𝜌𝒟 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 𝑑𝑇 𝜌𝒟𝑐𝑝 𝑑𝑥 • Flux due to conduction + • Flux of standardized enthalpy due to species diffusion + • Flux of sensible enthalpy due to species diffusion • For 𝐿𝑒 = 𝛼 𝒟 = 𝑘 𝒟𝜌𝑐𝑃 ≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃 • Shvab-Zeldovich assumption (𝐿𝑒 ≈ 𝑂 1 , valid for most combustion gases) 𝑑ℎ ′′ • 𝑄𝑥 = −𝜌𝒟 (due to both conduction and diffusion) 𝑑𝑥 + 𝑐𝑝 𝑑𝑌𝑖 ℎ 𝑑𝑥 𝑖 𝑑𝑇 𝑑𝑥