Transcript Slides

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
𝑆𝐿 Laminar Flame Speed
Burned
Products
𝛿
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 𝑆𝐿 and thickness 𝛿?
β€’ Depends on the pressure, fuel, equivalence ratio,…
β€’ Flame reference frame:
𝑣𝑏 , πœŒπ‘
𝑣𝑒 = 𝑆𝐿 , πœŒπ‘’ ,
𝛿~1 π‘šπ‘š
β€’ 1 π‘˜π‘” 𝐹𝑒 + 𝜈 π‘˜π‘” 𝑂π‘₯ β†’ 1 + 𝜈 π‘˜π‘” π‘ƒπ‘Ÿ
β€’ Conservation of mass: π‘š = πœŒπ‘’ 𝑣𝑒 =
𝑣𝑏
πœŒπ‘ 𝑣𝑏 ;
𝑣𝑒
=
πœŒπ‘’
πœŒπ‘
=
𝑃𝑒 𝑅𝑇𝑏
𝑅𝑇𝑒 𝑃𝑏
=
β€’ For hydrocarbon fuels at P = 1 atm, 𝑇𝑒 β‰ˆ 300𝐾, 𝑇𝑏 β‰ˆ 2100𝐾, 𝑣𝑏 β‰ˆ 7𝑣𝑒
β€’ What happens within a premixed flame?
𝑇𝑏
𝑇𝑒
Simplified Analysis
Assumptions
Heat and Radical
β€’ One dimensional flow
β€’ Kinetic energy = viscosity = radiation = 0
β€’ Constant pressure
β€’ π‘žπ‘₯"
=
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
and
π‘‘π‘Œπ‘–
= βˆ’πœŒπ’Ÿ
𝑑π‘₯
𝛼
π‘˜
= =
β‰ˆ
π’Ÿ
π’ŸπœŒπ‘π‘ƒ
π‘šπ‘–"
β€’ Lewis Number, 𝐿𝑒
1;
π‘˜
𝑐𝑃
β‰ˆ π’ŸπœŒ
β€’ 𝑐𝑃,𝑖 = 𝑐𝑃 β‰  𝑓𝑛 π‘‡π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘Ÿ 𝑆𝑝𝑒𝑐𝑖𝑒
β€’ Single Step Kinetics
β€’ Ξ¦ < 1, Fuel Lean, so fuel is completely consumed
Conservations Laws
β€’ Mass Conservation
"
β€’ π‘š = πœŒπ‘£π‘₯ = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘;
𝑑 πœŒπ‘£π‘₯
𝑑π‘₯
𝑑π‘₯
=0
β€’ Species Conservation
β€’
π΄π‘šπ‘–"
β€’
π‘šπ‘–β€²β€²β€²
+ 𝐴 𝑑π‘₯
=
𝑑 π‘šπ‘–"
𝑑π‘₯
=
π‘šπ‘–β€²β€²β€²
𝑑
𝑑π‘₯
=𝐴
π΄π‘šπ‘–"
"
" 𝑑 π‘šπ‘–
π‘šπ‘– +
𝑑π‘₯
𝑑π‘₯
π‘š" π‘Œπ‘– βˆ’ πœŒπ’Ÿ
π‘‘π‘Œπ‘–
𝑑π‘₯
𝐴
(using Flick’s Law)
β€’ Apply to: 1 π‘˜π‘” 𝐹 + 𝜈 π‘˜π‘” 𝑂π‘₯ β†’ 1 + 𝜈 π‘ƒπ‘Ÿ; Air/Fuel ratio 𝜈 =
1
𝜈
β€²β€²β€²
β€’ π‘šπΉβ€²β€²β€² = π‘šπ‘‚π‘₯
=βˆ’
1
β€²β€²β€²
π‘šπ‘ƒπ‘Ÿ
𝜈+1
β€’ 𝑖 = 1, 2, … , 𝑀 = 3; Fuel, Oxidizer, Products
β€’
β€’
β€’
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
π‘‘π‘ŒπΉ
π‘š π‘ŒπΉ βˆ’ πœŒπ’Ÿ
𝑑π‘₯
π‘‘π‘Œπ‘‚π‘₯
"
π‘š π‘Œπ‘‚π‘₯ βˆ’ πœŒπ’Ÿ
𝑑π‘₯
π‘‘π‘Œπ‘ƒπ‘Ÿ
"
π‘š π‘Œπ‘ƒπ‘Ÿ βˆ’ πœŒπ’Ÿ
𝑑π‘₯
"
π‘šπ‘–β€²β€²β€²
= π‘šπΉβ€²β€²β€²
β€²β€²β€²
= π‘šπ‘‚π‘₯
= 𝜈 π‘šπΉβ€²β€²β€²
β€²β€²β€²
= π‘šπ‘ƒπ‘Ÿ
= βˆ’ 1 + 𝜈 π‘šπΉβ€²β€²β€²
π‘˜π‘” 𝑂π‘₯
π‘˜π‘” 𝐹
"
𝑑
π‘š
𝑖
𝐴 π‘šπ‘–" +
𝑑π‘₯
𝑑π‘₯
Energy Conservation (Ch. 7, pp. 239-244)
𝑑π‘₯
π΄π‘šβ€²β€² β„Ž
𝐴𝑄π‘₯β€²β€²
π‘Šπ‘„π‘‰
β€’ 𝑄𝑄𝑉 βˆ’ π‘Šπ‘„π‘‰ = π‘š β„Žπ‘œπ‘’π‘‘ βˆ’ β„Žπ‘–π‘›
‒𝐴
β€’
𝑄π‘₯β€²β€²
𝑑 𝑄π‘₯β€²β€²
βˆ’
𝑑π‘₯
βˆ’
𝑄π‘₯β€²β€²
=
π‘‘β„Ž
β€²β€²
π‘š
𝑑π‘₯
+
𝑑 𝑄π‘₯β€²β€²
𝑑π‘₯
𝑑π‘₯
= π΄π‘š
β€²β€²
β„Ž+
π‘‘β„Ž
𝑑π‘₯
𝑑π‘₯
βˆ’β„Ž
β€’ Decreasing heat flux in x-direction increases enthalpy in the +x-direction
π΄π‘šβ€²β€²
π‘‘β„Ž
β„Ž+
𝑑π‘₯
𝑑π‘₯
β€²β€²
𝑑
𝑄
π‘₯
𝐴 𝑄π‘₯β€²β€² +
𝑑π‘₯
𝑑π‘₯
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€’
𝑄π‘₯β€²β€²
=
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
β€’ Note:
β€’
𝑄π‘₯β€²β€²
π‘‘β„Ž
𝑑π‘₯
=
+
β€²β€²
π‘šπ‘–,π‘‘π‘–π‘“π‘“π‘’π‘ π‘–π‘œπ‘›
β„Žπ‘–
𝑑
𝑑π‘₯
π‘Œπ‘– β„Žπ‘– =
β€’
π‘‘π‘Œπ‘–
β„Ž
𝑑π‘₯ 𝑖
= 𝑑π‘₯ βˆ’ 𝑐𝑝 𝑑π‘₯
π‘‘β„Ž
𝑑𝑇
=
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
βˆ’ πœŒπ’Ÿ
π‘‘β„Ž
𝑑π‘₯
β€’ Heat Flux =
π‘‘π‘Œπ‘–
β„Ž
𝑑π‘₯ 𝑖
βˆ’
𝑑𝑇
𝑐𝑝
𝑑π‘₯
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
=
+
π‘Œπ‘–
=
π‘‘β„Žπ‘–
𝑑π‘₯
=
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
βˆ’
π‘‘π‘Œπ‘–
πœŒπ’Ÿ
β„Ž
𝑑π‘₯ 𝑖
π‘‘π‘Œπ‘–
β„Ž
𝑑π‘₯ 𝑖
+
π‘‘β„Ž
βˆ’ πœŒπ’Ÿ
𝑑π‘₯
=
π‘Œπ‘– 𝑐𝑝,𝑖
+
𝑑𝑇
βˆ’π‘˜
𝑑π‘₯
𝑑𝑇
𝑑π‘₯
=
𝑑𝑇
πœŒπ’Ÿπ‘π‘
𝑑π‘₯
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 𝐿𝑒 =
𝛼
π’Ÿ
=
π‘˜
π’ŸπœŒπ‘π‘ƒ
β‰ˆ 1; π‘˜ β‰ˆ π’ŸπœŒπ‘π‘ƒ
β€’ Shvab-Zeldovich assumption (𝐿𝑒 β‰ˆ 𝑂(1) for most combustion gases)
π‘‘β„Ž
β€²β€²
β€’ 𝑄π‘₯ = βˆ’πœŒπ’Ÿ
(due to both conduction and diffusion)
𝑑π‘₯
βˆ’ πœŒπ’Ÿ
π‘‘π‘Œπ‘–
β„Ž
𝑑π‘₯ 𝑖
+ 𝑐𝑝
π‘‘π‘Œπ‘–
β„Ž
𝑑π‘₯ 𝑖
𝑑𝑇
𝑑π‘₯
Shvab-Zeldovich form of Energy Conservation
β€’
𝑑 𝑄π‘₯β€²β€²
βˆ’
𝑑π‘₯
β€’β„Ž=
β€’
β€’
𝑄π‘₯β€²β€²
β€’
β€’
β€’
β€’
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
π‘‘β„Ž
𝑑π‘₯
=
=
π‘‘β„Ž
β€²β€²
π‘š
𝑑π‘₯
π‘œ
π‘Œπ‘– β„Žπ‘“,𝑖
π‘Œπ‘– β„Žπ‘– =
=
π‘‘π‘Œπ‘– π‘œ
β„Ž
𝑑π‘₯ 𝑓,𝑖
π‘‘β„Ž
βˆ’πœŒπ’Ÿ
𝑑π‘₯
𝑑 𝑄π‘₯β€²β€²
βˆ’
𝑑π‘₯
=
𝑑
𝑑π‘₯
(Energy Equation)
+
π‘œ
π‘‘β„Žπ‘“,𝑖
𝑑 𝑇
π‘Œπ‘–
+
𝑐 𝑑𝑇
𝑑π‘₯
𝑑π‘₯ π‘‡π‘Ÿπ‘’π‘“ 𝑝
+
= βˆ’πœŒπ’Ÿ
π‘‘π‘Œπ‘– π‘œ
β„Ž
𝑑π‘₯ 𝑓,𝑖
πœŒπ’Ÿ
π‘‘π‘Œπ‘– π‘œ
πœŒπ’Ÿ
β„Žπ‘“,𝑖
𝑑π‘₯
π‘‘π‘Œπ‘– π‘œ
πœŒπ’Ÿ
β„Ž
𝑑π‘₯ 𝑓,𝑖
𝑑
π‘œ
βˆ’
β„Žπ‘“,𝑖
π‘šβ€²β€² π‘Œπ‘–
𝑑π‘₯
𝑇
π‘Œπ‘– 𝑇 𝑐𝑝,𝑖 𝑑𝑇
π‘Ÿπ‘’π‘“
+
βˆ’
βˆ’
π‘šπ‘–"
𝑑𝑇
𝑐𝑝
𝑑π‘₯
π‘‘π‘Œπ‘– π‘œ
β„Ž
𝑑π‘₯ 𝑓,𝑖
+
+
𝑑𝑇
𝑐𝑝
𝑑π‘₯
=
=
π‘œ
π‘Œπ‘– β„Žπ‘“,𝑖
π‘‘π‘Œπ‘– π‘œ
β„Ž
𝑑π‘₯ 𝑓,𝑖
+
+
𝑇
𝑐 𝑑𝑇
π‘‡π‘Ÿπ‘’π‘“ 𝑝
𝑑𝑇
𝑐𝑝
𝑑π‘₯
𝑑𝑇
𝑐𝑝
𝑑π‘₯
π‘‘π‘Œπ‘– π‘œ
𝑑𝑇
=
β„Žπ‘“,𝑖 + 𝑐𝑝
𝑑π‘₯
𝑑π‘₯
𝑑𝑇
𝑑
𝑑𝑇
π‘œ
β€²β€²
β€²β€²
π‘š π‘Œπ‘– β„Žπ‘“,𝑖 = π‘š 𝑐𝑝 βˆ’
πœŒπ’Ÿπ‘π‘
𝑑π‘₯
𝑑π‘₯
𝑑π‘₯
π‘‘π‘Œπ‘–
𝑑𝑇
𝑑
𝑑𝑇
β€²β€²
πœŒπ’Ÿ
= π‘š 𝑐𝑝 βˆ’
πœŒπ’Ÿπ‘π‘
𝑑π‘₯
𝑑π‘₯
𝑑π‘₯
𝑑π‘₯
π‘šβ€²β€²
Shvab-Zeldovich form of Energy Conservation
β€’
𝑑
βˆ’
𝑑π‘₯
π‘œ
β„Žπ‘“,𝑖
β€’
𝑑
βˆ’
𝑑π‘₯
π‘œ
β„Žπ‘“,𝑖
π‘šπ‘–" =
π‘‘π‘Œπ‘–
𝑑𝑇
𝑑
β€²β€²
βˆ’ πœŒπ’Ÿ
= π‘š 𝑐𝑝 βˆ’
𝑑π‘₯
𝑑π‘₯
𝑑π‘₯
π‘šπ‘–"
"
𝑑
π‘š
π‘œ
π‘œ
𝑖
βˆ’ β„Žπ‘“,𝑖
= βˆ’ β„Žπ‘“,𝑖
π‘šπ‘–β€²β€²β€²
𝑑π‘₯
π‘šβ€²β€² π‘Œπ‘–
β€’ Species conservation:
β€’
𝑑 π‘šπ‘–"
𝑑π‘₯
𝑑𝑇
πœŒπ’Ÿπ‘π‘
𝑑π‘₯
= π‘šπ‘–β€²β€²β€²
π‘œ
π‘œ
π‘œ
π‘œ
β€²β€²β€²
β€²β€²β€²
β„Žπ‘“,𝑖
π‘šπ‘–β€²β€²β€² = β„Žπ‘“,𝐹
π‘šπΉβ€²β€²β€² + β„Žπ‘“,𝑂π‘₯
π‘šπ‘‚π‘₯
+ β„Žπ‘“,π‘ƒπ‘Ÿ
π‘šπ‘ƒπ‘Ÿ
β€’
π‘œ
π‘œ
π‘œ
= β„Žπ‘“,𝐹
π‘šπΉβ€²β€²β€² + β„Žπ‘“,𝑂π‘₯
πœˆπ‘šπΉβ€²β€²β€² βˆ’ β„Žπ‘“,π‘ƒπ‘Ÿ
1 + 𝜈 π‘šπΉβ€²β€²β€²
β€’
π‘œ
π‘œ
π‘œ
= π‘šπΉβ€²β€²β€² β„Žπ‘“,𝐹
+ β„Žπ‘“,𝑂π‘₯
𝜈 βˆ’ β„Žπ‘“,π‘ƒπ‘Ÿ
1+𝜈
= π‘šπΉβ€²β€²β€² Ξ”β„ŽπΆ
π‘œ
π‘œ
π‘œ
β€’ Ξ”β„ŽπΆ = β„Žπ‘“,𝐹
+ β„Žπ‘“,𝑂π‘₯
𝜈 βˆ’ β„Žπ‘“,π‘ƒπ‘Ÿ
1 + 𝜈 : Heat of combustion
β€’ For 𝐿𝑒 =
𝛼
π’Ÿ
=
π‘˜
π’ŸπœŒπ‘π‘ƒ
β‰ˆ 1; π‘˜ β‰ˆ π’ŸπœŒπ‘π‘ƒ
β€’ βˆ’π‘šπΉβ€²β€²β€² Ξ”β„ŽπΆ = π‘šβ€²β€² 𝑐𝑝
𝑑𝑇
𝑑π‘₯
βˆ’
𝑑
𝑑π‘₯
π‘˜
𝑑𝑇
𝑑π‘₯
nd
2
order differential
equation for T(x)
β€’
β€²β€² 𝑑𝑇
π‘š
𝑑π‘₯
βˆ’
1 𝑑
𝑐𝑝 𝑑π‘₯
𝑑𝑇
π‘˜
𝑑π‘₯
β€’ Where π‘šβ€²β€² = πœŒπ‘’ 𝑆𝐿
=
β€²β€²β€²
π‘šπΉ
Ξ”β„ŽπΆ
βˆ’
𝑐𝑝
β€’ Only accepts 2 boundary conditions,
β€’ But we have 4 (Eigenvalue problem)
β€’ For π‘₯ β†’ βˆ’βˆž: 𝑇 β†’ 𝑇𝑒 π‘Žπ‘›π‘‘
β€’ For π‘₯ β†’ +∞: 𝑇 β†’ 𝑇𝐡 π‘Žπ‘›π‘‘
𝑑𝑇
𝑑π‘₯
𝑑𝑇
𝑑π‘₯
β†’0
β†’0
β€’ For an approximate solution, assume a simple profile
β€’ Find flame thickness 𝛿 and laminar flame speed 𝑆𝐿 =
conditions can be satisfied
π‘šβ€²β€²
πœŒπ‘’
so that all four boundary
Approximate Solution
β€’
𝑑𝑇
β€²β€²
π‘š
𝑑π‘₯
βˆ’
1 𝑑
𝑐𝑝 𝑑π‘₯
β€’ Integrate
β€’
π‘šβ€²β€²
𝑇
𝑇𝑏
𝑇𝑒
βˆ’
𝑑𝑇
π‘˜
𝑑π‘₯
+∞
βˆ’βˆž
1
𝑐𝑝
=
β€²β€²β€²
π‘šπΉ
Ξ”β„ŽπΆ
βˆ’
𝑐𝑝
𝑑π‘₯
𝑑𝑇 0
π‘˜
𝑑π‘₯ 0
=
Ξ”β„ŽπΆ +∞ β€²β€²β€²
βˆ’
π‘šπΉ 𝑑π‘₯
𝑐𝑝 βˆ’βˆž
β€’ π‘šπΉβ€²β€²β€² = 𝑓𝑛 𝑇 = 0 π‘“π‘œπ‘Ÿ π‘₯ < 0 π‘œπ‘Ÿ π‘₯ > 𝛿
β€’ Inside 0 < π‘₯ < 𝛿,
β€’
π‘šβ€²β€²
β€’
π‘šβ€²β€²
𝑑𝑇
𝑑π‘₯
=
𝑇𝑏 βˆ’π‘‡π‘’
,
𝛿
so 𝑑π‘₯ =
𝑇𝑏 βˆ’ 𝑇𝑒 =
Ξ”β„ŽπΆ +∞ β€²β€²β€²
βˆ’
π‘šπΉ
𝑐𝑝 βˆ’βˆž
𝑇𝑏 βˆ’ 𝑇𝑒 =
Ξ”β„ŽπΆ 𝛿 β€²β€²β€²
βˆ’
π‘šπΉ
𝑐𝑝
𝑇
𝛿
𝑑𝑇
𝑇𝑏 βˆ’π‘‡π‘’
eqn. 1
β€’ Two unknowns: π‘šβ€²β€² = 𝑆𝐿 πœŒπ‘’ and 𝛿
β€’ Need another equation
𝛿
𝑑𝑇
𝑇𝑏 βˆ’π‘‡π‘’
=
𝑇𝑏 β€²β€²β€²
Ξ”β„ŽπΆ 𝛿 1
βˆ’
π‘šπΉ
𝑐𝑝 𝑇𝑏 βˆ’π‘‡π‘’ 𝑇𝑒
𝑇 𝑑𝑇
Average over
temperature: π‘šπΉβ€²β€²β€²
Approximate Solution
β€’
𝑑𝑇
β€²β€²
π‘š
𝑑π‘₯
βˆ’
1 𝑑
𝑐𝑝 𝑑π‘₯
β€’ Integrate
β€’
β€’
β€’
β€’
𝑑𝑇
π‘˜
𝑑π‘₯
𝛿/2
βˆ’βˆž
=
β€²β€²β€²
π‘šπΉ
Ξ”β„ŽπΆ
βˆ’
𝑐𝑝
𝑑π‘₯
𝑇𝑏 βˆ’π‘‡π‘’
𝛿
𝑇𝑏 +𝑇𝑒
2
𝑇𝑒
π‘˜ 𝑑𝑇
Ξ”β„ŽπΆ 𝛿/2 β€²β€²β€²
π‘š 𝑇
βˆ’
=βˆ’
π‘šπΉ 𝑑π‘₯
𝑐𝑝 𝑑π‘₯ 0
𝑐𝑝 βˆ’βˆž
𝑇 +𝑇
π‘˜ 𝑇𝑏 βˆ’π‘‡π‘’
π‘šβ€²β€² 𝑏 𝑒 βˆ’ 𝑇𝑒 βˆ’
=0
2
𝑐𝑝
𝛿
β€²β€²
𝑇𝑏 βˆ’π‘‡π‘’
π‘˜ 𝑇𝑏 βˆ’π‘‡π‘’
β€²β€²
π‘š
=
2
𝑐𝑝
𝛿
2π‘˜
𝛿 = β€²β€² eqn. 2
π‘š 𝑐𝑝
β€’ From eqn. 1
β€’ π‘šβ€²β€² =
π‘šβ€²β€²
𝑇𝑏 βˆ’ 𝑇𝑒 =
2π‘˜Ξ”β„ŽπΆ
𝑐𝑝2 𝑇𝑏 βˆ’π‘‡π‘’
Ξ”β„ŽπΆ 𝛿 β€²β€²β€²
βˆ’
π‘šπΉ
𝑐𝑝
βˆ’π‘šπΉβ€²β€²β€² = 𝑆𝐿 πœŒπ‘’ ; 𝑆𝐿 =
1
πœŒπ‘’
=
β‰ˆ0
Ξ”β„ŽπΆ β€²β€²β€² 2π‘˜
βˆ’
π‘šπΉ β€²β€²
𝑐𝑝
π‘š 𝑐𝑝
2π‘˜Ξ”β„ŽπΆ
𝑐𝑝2 𝑇𝑏 βˆ’π‘‡π‘’
βˆ’π‘šπΉβ€²β€²β€²
Approximate Solution
β€’ 𝑆𝐿 =
1
πœŒπ‘’
2π‘˜Ξ”β„ŽπΆ
𝑐𝑝2 𝑇𝑏 βˆ’π‘‡π‘’
βˆ’π‘šπΉβ€²β€²β€²
β€’ But Ξ”β„ŽπΆ = 1 + 𝜈 𝑐𝑃 𝑇𝑏 βˆ’ 𝑇𝑒
β€’ Show this in HW (problem 8.2)
β€’ 𝑆𝐿 =
β€’
β€’
2π‘˜ 1+𝜈 𝑐𝑃 𝑇𝑏 βˆ’π‘‡π‘’
πœŒπ‘’ πœŒπ‘’ 𝑐𝑝2 𝑇𝑏 βˆ’π‘‡π‘’
βˆ’π‘šπΉβ€²β€²β€²
2𝛼 1+𝜈
𝑆𝐿 =
βˆ’π‘šπΉβ€²β€²β€²
πœŒπ‘’
2π‘˜
2π‘˜
eqn. 2: 𝛿 = β€²β€² =
π‘š 𝑐𝑝
𝑆𝐿 πœŒπ‘’ 𝑐𝑝
‒𝛿=
2π›ΌπœŒπ‘’
β€²β€²β€² 1+𝜈
βˆ’π‘š 𝐹
=
2𝛼
𝑆𝐿
=
; where 𝛼 =
2𝛼
𝑆𝐿
=
π‘˜
πœŒπ‘’ 𝑐𝑃
2𝛼
2𝛼 1+𝜈
πœŒπ‘’
β€²β€²β€²
βˆ’ π‘šπΉ
(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: ___?