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

Deflagrations - burning velocity" How does S L vary with pressure?d[A]= !k f [ A] ! A[ B] ! B~P ! AdtP! B ~P! A+! B~P n (for example A = fuel, B = oxidant)!"~ 1 d[A][A] "dtThus S L ~ {αω} 1/2 ~ {P -1 P n-1 } 1/2 ~ P (n-2)/2 For typical n = 2, S L independent of pressure For real hydrocarbons, working backwards fromexperimental results, typically (e.g. stoichiometric CH 4 -air) S L~ P -0.4 , thus n ≈ 1.2 This suggests more reactions are one-body than two-body,but actually observed n is due to competition between twobodyH + O 2 branching vs. 3-body H + O 2 + M whichdecelerates reaction~ P !1 P n ~ P n!1 18AME 513 - Fall 2012 - Lecture 8 - Premixed flames I17Deflagrations - temperature effect" Since Zeldovich number (β) >> 1! T ad"!= E!(T ad) "T T=Tad"T adFor typical hydrocarbon-air flames, E ≈ 40 kcal/moleR = 1.987 cal/mole, T f ≈ 2200K if adiabatic⇒ β ≈ 10, at T close to T f , ω ~ T 10 !!!⇒ Thin reaction zone concentrated near highest temp.∴ In Zeldovich (or any) estimate of S L , overall reaction rate Ωmust be evaluated at T ad , not T ∞⇒ How can we estimate E? If reaction rate depends moreon E than concentrations [ ], S L ~ {αω} 1/2 ~ {exp(-E/RT)} 1/2~ exp(-E/2RT) - Plot of ln(S L ) vs. 1/T ad has slope of -E/2R If β isnt large, then ω(T ∞ ) ≈ ω(T ad ) and reaction occurseven in the cold gases, so no control over flame ispossible! Since S L ~ ω 1/2 , S L ~ (T β ) 1/2 ~ T 5 typically!AME 513 - Fall 2012 - Lecture 8 - Premixed flames I• 9