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ME 475/675 Introduction to Combustion Lecture 30 Ex. 8.2, Flame speed and thickness calculations dependence on temperature and pressure Announcements • HW 13 • Due Monday, November 9, 2015 • Midterm 2 • Friday, November 13, 2015 • Review Monday, Nov. 9, 2015 • PE 104, 8 to 9:30 (or 9:45?) AM • Project • In groups of two, construct a backpacking stove • Materials will be supplied in lab • Determine how long it takes and how much fuel is required to boil water • Form your group by Monday, November 9 • Inform TA: Hasibul Alam [email protected] Broader Impact Assignment • 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 • November 2, 2015 • 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 • Are you having problems downloading this article? Premixed Flame Laminar Speed, SL, and Thickness, d • Flame reference frame: 𝑣𝐵 , 𝜌𝐵 𝑣𝑢 = 𝑆𝐿 , 𝜌𝑢 , 𝛿 • 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 = 𝑇𝑏 𝑇𝑢 • For hydrocarbon fuels at P = 1 atm, 𝑇𝑢 ≈ 300𝐾, 𝑇𝑏 ≈ 2100𝐾, 𝑣𝑏 ≈ 7𝑣𝑢 • Approximate Solution for Lewis Number, 𝐿𝑒 = • 𝑆𝐿 = 2𝛼 1+𝜈 𝜌𝑢 • 𝛿= 2𝛼𝜌𝑢 ′′′ 1+𝜈 −𝑚𝐹 −𝑚𝐹′′′ = 2𝛼 𝑆𝐿 𝛼 𝒟 = 𝑘 𝒟𝜌𝑐𝑃 ≈ 𝑂(1) (Heat and species diffusion allows flame to propagate) (Fast flames are thin) Example 8.2 (turn in next time for extra credit) • 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: ___? Detailed Analysis Heat and Radical • Multi-step chemical data • Temperature dependent thermodynamic properties • Integrate energy, species and momentum differential equations • Energy: • 𝑑𝑇 ′′ 𝑚 𝑑𝑥 − 1 𝑑 𝑐𝑝 𝑑𝑥 𝑑𝑇 𝑘 𝑑𝑥 = ′′′ 𝑚𝐹 Δℎ𝐶 − , 𝑐𝑝 • where 𝑚′′ = 𝜌𝑢 𝑆𝐿 • Apply temperature and species boundary conditions (Eigenvalue problem) • Get more detailed structure of premixed flame than we got from the simplified analysis • pp. 276-9 Stoichiometric CH4/air Flame • Figure 8.10 a. 𝐶𝐻4 (𝑓𝑢𝑒𝑙) → 𝐶𝑂 → 𝐶𝑂2 • • • • Most everything happens x = 0.5 to 1.5 mm Fuel gone at 1mm, temperature rise is 73% to Teq 𝐶𝑂 produced and consumed, precedes 𝐶𝑂2 𝐶𝑂2 → equilibrium at 2 mm b. C-containing intermediates 𝐶𝐻3 , 𝐶𝐻2 𝑂, 𝐻𝐶𝑂 • Produced and consumed 0.4 to 1.1 mm c. H intermediates a. 𝐻2 𝑂 produced at 0.9 mm a. before 𝐶𝑂2 d. 𝑁𝑂 needs CH first Pressure and temperature dependence of SL and 𝛿 (from approximate solution) • 𝑆𝐿 = • 2𝛼 1 + 𝜈 𝜔𝐹 − 𝜌𝑢 𝜔𝐹 𝜌𝑢 • 𝛼= • 𝑆𝐿 ~ •𝛿= = 𝜌𝑢 𝐸 𝑅 𝐴𝑒𝑥𝑝 − 𝑎 𝑢 𝐹 𝑚 𝑇𝑏 𝑃 𝑅𝑢 𝑇𝑢 𝑛 𝑁𝑖 𝑁 𝑛 𝑛 = 𝑉 = 𝜒𝑖 𝑉 𝑂𝑥 𝐴𝑒𝑥𝑝 = = −𝑀𝑊𝐹 𝜔𝐹 𝜌𝑢 2𝛼 1 + 𝜈 𝑛 = • 𝑖 • − ′′′ −𝑚𝐹 𝑛 𝑃 𝜒𝑖 𝑅 𝑇 ; 𝑢 𝑏 𝐸𝑎 𝑅𝑢 𝑇 𝑃𝑁−1 𝑢𝑁 𝑇𝑏 𝑇𝑏 𝑅𝑢 𝑇𝑢 𝑘 ~𝑃−1 𝑇𝑢 𝑇 0.75 , 𝑀𝑊 𝑃 𝑐𝑝 𝐸 𝑅 − 𝑎 𝑢 𝑇𝑏 𝑃 𝑚 𝑛 𝜒𝐹 𝜒𝑂𝑥 𝑅𝑢 𝑇𝑏 𝑃 𝑅𝑢 𝑇𝑢 𝑚+𝑛 Use 𝑇𝑏 to find 𝜔𝐹 . 𝑁 = 𝑚 + 𝑛 = 𝑒𝑥𝑝 − 𝑘 𝜌𝑢 𝑐𝑝 = 𝑃−1 𝑇𝑢 𝑇 0.75 2𝛼 ~ 𝑆𝐿 𝑒𝑥𝑝 where 𝑇 = 𝐸𝑎 𝑅𝑢 − 𝑇𝑏 𝑃−1 𝑇𝑢 𝑇 0.75 𝑁 𝑁−2 −2 𝑃 2 𝑇𝑢 𝑇 0.375 𝑇𝑏 𝑒𝑥𝑝 𝑇𝑢 𝑁−1 𝑃 𝑇𝑏𝑁 ~𝑃 𝐸 𝑅𝑢 − 𝑎 2𝑇 𝑏 𝑇𝑢 +𝑇𝑏 2 𝑁 −2 ~𝑃 𝑁−2 2 𝑁 2 𝑇 0.375 𝑇𝑏 𝑁 −2 𝑇𝑢 𝑇 0.375 𝑇𝑏 𝑒𝑥𝑝 𝐸𝑎 𝑅𝑢 2𝑇𝑏 𝑒𝑥𝑝 𝐸𝑎 𝑅𝑢 − 2𝑇𝑏 End 2015 For hydrocarbon fuels N~2 • 𝑆𝐿 ~𝑃 𝑁−2 2 𝑇𝑢 𝑇 0.375 − 𝑁 2 𝑇𝑏 𝑒𝑥𝑝 𝐸𝑎 𝑅𝑢 − 2𝑇𝑏 ~𝑃 0 𝑇𝑢 𝑇 0.375 𝑇𝑏−1 𝑒𝑥𝑝 • Approximate solution predicts 𝑆𝐿 ≠ 𝑓𝑛 𝑃 • Measurements for stoichiometric CH4/air • 𝑆𝐿 cm s = 43 𝑃 [𝑎𝑡𝑚] • 𝑆𝐿 actually decreases as pressure increases • Theoretical pressure-dependence is not reliable • What does detailed solution predict? • But from page 157, N = 1.75, • 1.75−2 2 = −0.125, 𝑃−0.125 𝐸𝑎 𝑅𝑢 − 2𝑇𝑏 ≠ 𝑓𝑛 𝑃 Methane CH4 flame speed data • Dependence on 𝑇𝑢 • Predicted: 𝑆𝐿 ~𝑃0 𝑇𝑢 𝑇 0.375 𝑇𝑏−1 𝑒𝑥𝑝 − • 𝑇𝑏 and 𝑇 increase with 𝑇𝑢 • Measured: 𝑆𝐿 cm s 𝐸𝑎 𝑅𝑢 2𝑇𝑏 = 10 + 3.71 × 10−4 𝑇𝑢2 𝐾 • Figure compares correlation with measurements • As predicted, 𝑆𝐿 increases as 𝑇 increases • Dependence on Equivalence ratio Φ • 𝑇𝑏 = 𝑇𝑎𝑑 𝑎𝑡 Φ = 1.05 • 𝑇𝑏 and 𝑆𝐿 decrease at other Φ • Premixed Flame Speed and Thickness 0.2 0.168 0.15 SL ( Tu ) 0.1 0.05 0.015 0 400 273 0.375 d ( Tu) Ta ( Tu) Tb ( Tu) 600 800 Tu EaRu 2 Tb ( Tu) exp 3 110 1000 Methane Flame thickness (N ~ 2) 2 1.88 1.5 d ( Tu ) 1 0.5 0.219 0 200 400 1 600 800 Tu 3 110 1000 • Expect burned temperature decreases as Φ increases or decrease from 1.05 • For N = 2: 𝛿~𝑃 𝑁 −2 𝑁 2 𝑇 0.375 𝑇𝑏 𝑒𝑥𝑝 𝐸𝑎 𝑅𝑢 2𝑇𝑏 𝐸𝑎 𝑅𝑢 1 −1 0.375 ~𝑃 𝑇 𝑇𝑏 𝑒𝑥𝑝 2𝑇𝑏 (see plot) • Theory predicts thickness increases as temperature decreases (which happens as Φ moves away from 1.05) • Data and predictions indicate fast flames are thin (see plot)