AME 513 Principles of Combustion Lecture 8 ... - Paul D. Ronney

Deflagrations - burning velocity"! Estimate of iConduction heat transfer rate = -kA(ΔT/δ)k = gas thermal conductivity, A = cross-sectional area of flameΔT = temperature rise across front = T products - T reactantsδ = thickness of front (unknown at this point) Estimate of iiEnthalpy flux through front = (mass flux) x C p x ΔTMass flux = ρuA (ρ = density of reactants = ρ ∞ , u = velocity = S L )Enthalpy flux = ρ ∞ C p S L AΔT Estimate of iiiHeat generated by reaction = Q R x (d[fuel]/dt) x M fuel x VolumeVolume = AδQ R = C P ΔT/ff ==Fuel mass (Mass fuel / volume)=Total mass (Mass total / volume)(Moles fuel / volume)(mass fuel / moles fuel)= [F] M " fuel(Mass total / volume)[F] ∞ = fuel concentration in the cold reactantsAME 513 - Fall 2012 - Lecture 8 - Premixed flames I13# "14Deflagrations - burning velocity" Combine (i) and (ii) δ = k/ρC p S L = α/S L (δ = flame thickness) (same as Lecture 7) Recall α = k/ρC p = thermal diffusivity (units length 2 /time) For air at 300K & 1 atm, α ≈ 0.2 cm 2 /s For gases α ≈ ν (ν = kinematic viscosity) For gases α ~ P -1 T 1.7 since k ~ P 0 T .7 , ρ ~ P 1 T -1 , C p ~ P 0 T 0 For typical stoichiometric hydrocarbon-air flame, S L ≈ 40 cm/s,thus δ ≈ α/S L ≈ 0.005 cm (!) (Actually when properties aretemperature-averaged, δ ≈ 4α/S L ≈ 0.02 cm - still small!) Combine (ii) and (iii)S L = (αω) 1/2ω = overall reaction rate = (d[fuel]/dt)/[fuel] ∞ (units 1/s) With S L ≈ 40 cm/s, α ≈ 0.2 cm 2 /s, ω ≈ 1600 s -1 1/ω = characteristic reaction time = 625 microseconds Heat release rate per unit volume = (enthalpy flux) / (volume)= (ρC p S L AΔT)/(Aδ) = ρC p S L /k)(kΔT)/δ = (kΔT)/δ 2= (0.07 W/mK)(1900K)/(0.0002 m) 2 = 3 x 10 9 W/m 3 !!! Moral: flames are thin, fast and generate a lot of heat!AME 513 - Fall 2012 - Lecture 8 - Premixed flames I• 7