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
ππΏ Laminar Flame Speed
Burned
Products
πΏ
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 ππΏ and thickness πΏ?
β’ Depends on the pressure, fuel, equivalence ratio,β¦
β’ Flame reference frame:
π£π , ππ
π£π’ = ππΏ , ππ’ ,
πΏ~1 ππ
β’ 1 ππ πΉπ’ + π ππ ππ₯ β 1 + π ππ ππ
β’ Conservation of mass: π = ππ’ π£π’ =
π£π
ππ π£π ;
π£π’
=
ππ’
ππ
=
ππ’ π
ππ
π
ππ’ ππ
=
β’ For hydrocarbon fuels at P = 1 atm, ππ’ β 300πΎ, ππ β 2100πΎ, π£π β 7π£π’
β’ What happens within a premixed flame?
ππ
ππ’
Simplified Analysis
Assumptions
Heat and Radical
β’ One dimensional flow
β’ Kinetic energy = viscosity = radiation = 0
β’ Constant pressure
β’ ππ₯"
=
ππ
βπ
ππ₯
and
πππ
= βππ
ππ₯
πΌ
π
= =
β
π
ππππ
ππ"
β’ Lewis Number, πΏπ
1;
π
ππ
β ππ
β’ ππ,π = ππ β ππ ππππππππ‘π’ππ ππ ππππππ
β’ Single Step Kinetics
β’ Ξ¦ < 1, Fuel Lean, so fuel is completely consumed
Conservations Laws
β’ Mass Conservation
"
β’ π = ππ£π₯ = ππππ π‘πππ‘;
π ππ£π₯
ππ₯
ππ₯
=0
β’ Species Conservation
β’
π΄ππ"
β’
ππβ²β²β²
+ π΄ ππ₯
=
π ππ"
ππ₯
=
ππβ²β²β²
π
ππ₯
=π΄
π΄ππ"
"
" π ππ
ππ +
ππ₯
ππ₯
π" ππ β ππ
πππ
ππ₯
π΄
(using Flickβs Law)
β’ Apply to: 1 ππ πΉ + π ππ ππ₯ β 1 + π ππ; Air/Fuel ratio π =
1
π
β²β²β²
β’ ππΉβ²β²β² = πππ₯
=β
1
β²β²β²
πππ
π+1
β’ π = 1, 2, β¦ , π = 3; Fuel, Oxidizer, Products
β’
β’
β’
π
ππ₯
π
ππ₯
π
ππ₯
πππΉ
π ππΉ β ππ
ππ₯
ππππ₯
"
π πππ₯ β ππ
ππ₯
ππππ
"
π πππ β ππ
ππ₯
"
ππβ²β²β²
= ππΉβ²β²β²
β²β²β²
= πππ₯
= π ππΉβ²β²β²
β²β²β²
= πππ
= β 1 + π ππΉβ²β²β²
ππ ππ₯
ππ πΉ
"
π
π
π
π΄ ππ" +
ππ₯
ππ₯
Energy Conservation (Ch. 7, pp. 239-244)
ππ₯
π΄πβ²β² β
π΄ππ₯β²β²
πππ
β’ πππ β πππ = π βππ’π‘ β βππ
β’π΄
β’
ππ₯β²β²
π ππ₯β²β²
β
ππ₯
β
ππ₯β²β²
=
πβ
β²β²
π
ππ₯
+
π ππ₯β²β²
ππ₯
ππ₯
= π΄π
β²β²
β+
πβ
ππ₯
ππ₯
ββ
β’ Decreasing heat flux in x-direction increases enthalpy in the +x-direction
π΄πβ²β²
πβ
β+
ππ₯
ππ₯
β²β²
π
π
π₯
π΄ ππ₯β²β² +
ππ₯
ππ₯
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’
ππ₯β²β²
=
ππ
βπ
ππ₯
β’ Note:
β’
ππ₯β²β²
πβ
ππ₯
=
+
β²β²
ππ,πππππ’π πππ
βπ
π
ππ₯
ππ βπ =
β’
πππ
β
ππ₯ π
= ππ₯ β ππ ππ₯
πβ
ππ
=
ππ
βπ
ππ₯
β ππ
πβ
ππ₯
β’ Heat Flux =
πππ
β
ππ₯ π
β
ππ
ππ
ππ₯
ππ
βπ
ππ₯
=
+
ππ
=
πβπ
ππ₯
=
ππ
βπ
ππ₯
β
πππ
ππ
β
ππ₯ π
πππ
β
ππ₯ π
+
πβ
β ππ
ππ₯
=
ππ ππ,π
+
ππ
βπ
ππ₯
ππ
ππ₯
=
ππ
ππππ
ππ₯
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 πΏπ =
πΌ
π
=
π
ππππ
β 1; π β ππππ
β’ Shvab-Zeldovich assumption (πΏπ β π(1) for most combustion gases)
πβ
β²β²
β’ ππ₯ = βππ
(due to both conduction and diffusion)
ππ₯
β ππ
πππ
β
ππ₯ π
+ ππ
πππ
β
ππ₯ π
ππ
ππ₯
Shvab-Zeldovich form of Energy Conservation
β’
π ππ₯β²β²
β
ππ₯
β’β=
β’
β’
ππ₯β²β²
β’
β’
β’
β’
π
ππ₯
π
ππ₯
πβ
ππ₯
=
=
πβ
β²β²
π
ππ₯
π
ππ βπ,π
ππ βπ =
=
πππ π
β
ππ₯ π,π
πβ
βππ
ππ₯
π ππ₯β²β²
β
ππ₯
=
π
ππ₯
(Energy Equation)
+
π
πβπ,π
π π
ππ
+
π ππ
ππ₯
ππ₯ ππππ π
+
= βππ
πππ π
β
ππ₯ π,π
ππ
πππ π
ππ
βπ,π
ππ₯
πππ π
ππ
β
ππ₯ π,π
π
π
β
βπ,π
πβ²β² ππ
ππ₯
π
ππ π ππ,π ππ
πππ
+
β
β
ππ"
ππ
ππ
ππ₯
πππ π
β
ππ₯ π,π
+
+
ππ
ππ
ππ₯
=
=
π
ππ βπ,π
πππ π
β
ππ₯ π,π
+
+
π
π ππ
ππππ π
ππ
ππ
ππ₯
ππ
ππ
ππ₯
πππ π
ππ
=
βπ,π + ππ
ππ₯
ππ₯
ππ
π
ππ
π
β²β²
β²β²
π ππ βπ,π = π ππ β
ππππ
ππ₯
ππ₯
ππ₯
πππ
ππ
π
ππ
β²β²
ππ
= π ππ β
ππππ
ππ₯
ππ₯
ππ₯
ππ₯
πβ²β²
Shvab-Zeldovich form of Energy Conservation
β’
π
β
ππ₯
π
βπ,π
β’
π
β
ππ₯
π
βπ,π
ππ" =
πππ
ππ
π
β²β²
β ππ
= π ππ β
ππ₯
ππ₯
ππ₯
ππ"
"
π
π
π
π
π
β βπ,π
= β βπ,π
ππβ²β²β²
ππ₯
πβ²β² ππ
β’ Species conservation:
β’
π ππ"
ππ₯
ππ
ππππ
ππ₯
= ππβ²β²β²
π
π
π
π
β²β²β²
β²β²β²
βπ,π
ππβ²β²β² = βπ,πΉ
ππΉβ²β²β² + βπ,ππ₯
πππ₯
+ βπ,ππ
πππ
β’
π
π
π
= βπ,πΉ
ππΉβ²β²β² + βπ,ππ₯
πππΉβ²β²β² β βπ,ππ
1 + π ππΉβ²β²β²
β’
π
π
π
= ππΉβ²β²β² βπ,πΉ
+ βπ,ππ₯
π β βπ,ππ
1+π
= ππΉβ²β²β² ΞβπΆ
π
π
π
β’ ΞβπΆ = βπ,πΉ
+ βπ,ππ₯
π β βπ,ππ
1 + π : Heat of combustion
β’ For πΏπ =
πΌ
π
=
π
ππππ
β 1; π β ππππ
β’ βππΉβ²β²β² ΞβπΆ = πβ²β² ππ
ππ
ππ₯
β
π
ππ₯
π
ππ
ππ₯
nd
2
order differential
equation for T(x)
β’
β²β² ππ
π
ππ₯
β
1 π
ππ ππ₯
ππ
π
ππ₯
β’ Where πβ²β² = ππ’ ππΏ
=
β²β²β²
ππΉ
ΞβπΆ
β
ππ
β’ Only accepts 2 boundary conditions,
β’ But we have 4 (Eigenvalue problem)
β’ For π₯ β ββ: π β ππ’ πππ
β’ For π₯ β +β: π β ππ΅ πππ
ππ
ππ₯
ππ
ππ₯
β0
β0
β’ For an approximate solution, assume a simple profile
β’ Find flame thickness πΏ and laminar flame speed ππΏ =
conditions can be satisfied
πβ²β²
ππ’
so that all four boundary
Approximate Solution
β’
ππ
β²β²
π
ππ₯
β
1 π
ππ ππ₯
β’ Integrate
β’
πβ²β²
π
ππ
ππ’
β
ππ
π
ππ₯
+β
ββ
1
ππ
=
β²β²β²
ππΉ
ΞβπΆ
β
ππ
ππ₯
ππ 0
π
ππ₯ 0
=
ΞβπΆ +β β²β²β²
β
ππΉ ππ₯
ππ ββ
β’ ππΉβ²β²β² = ππ π = 0 πππ π₯ < 0 ππ π₯ > πΏ
β’ Inside 0 < π₯ < πΏ,
β’
πβ²β²
β’
πβ²β²
ππ
ππ₯
=
ππ βππ’
,
πΏ
so ππ₯ =
ππ β ππ’ =
ΞβπΆ +β β²β²β²
β
ππΉ
ππ ββ
ππ β ππ’ =
ΞβπΆ πΏ β²β²β²
β
ππΉ
ππ
π
πΏ
ππ
ππ βππ’
eqn. 1
β’ Two unknowns: πβ²β² = ππΏ ππ’ and πΏ
β’ Need another equation
πΏ
ππ
ππ βππ’
=
ππ β²β²β²
ΞβπΆ πΏ 1
β
ππΉ
ππ ππ βππ’ ππ’
π ππ
Average over
temperature: ππΉβ²β²β²
Approximate Solution
β’
ππ
β²β²
π
ππ₯
β
1 π
ππ ππ₯
β’ Integrate
β’
β’
β’
β’
ππ
π
ππ₯
πΏ/2
ββ
=
β²β²β²
ππΉ
ΞβπΆ
β
ππ
ππ₯
ππ βππ’
πΏ
ππ +ππ’
2
ππ’
π ππ
ΞβπΆ πΏ/2 β²β²β²
π π
β
=β
ππΉ ππ₯
ππ ππ₯ 0
ππ ββ
π +π
π ππ βππ’
πβ²β² π π’ β ππ’ β
=0
2
ππ
πΏ
β²β²
ππ βππ’
π ππ βππ’
β²β²
π
=
2
ππ
πΏ
2π
πΏ = β²β² eqn. 2
π ππ
β’ From eqn. 1
β’ πβ²β² =
πβ²β²
ππ β ππ’ =
2πΞβπΆ
ππ2 ππ βππ’
ΞβπΆ πΏ β²β²β²
β
ππΉ
ππ
βππΉβ²β²β² = ππΏ ππ’ ; ππΏ =
1
ππ’
=
β0
ΞβπΆ β²β²β² 2π
β
ππΉ β²β²
ππ
π ππ
2πΞβπΆ
ππ2 ππ βππ’
βππΉβ²β²β²
Approximate Solution
β’ ππΏ =
1
ππ’
2πΞβπΆ
ππ2 ππ βππ’
βππΉβ²β²β²
β’ But ΞβπΆ = 1 + π ππ ππ β ππ’
β’ Show this in HW (problem 8.2)
β’ ππΏ =
β’
β’
2π 1+π ππ ππ βππ’
ππ’ ππ’ ππ2 ππ βππ’
βππΉβ²β²β²
2πΌ 1+π
ππΏ =
βππΉβ²β²β²
ππ’
2π
2π
eqn. 2: πΏ = β²β² =
π ππ
ππΏ ππ’ ππ
β’πΏ=
2πΌππ’
β²β²β² 1+π
βπ πΉ
=
2πΌ
ππΏ
=
; where πΌ =
2πΌ
ππΏ
=
π
ππ’ ππ
2πΌ
2πΌ 1+π
ππ’
β²β²β²
β ππΉ
(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: ___?