Transcript ME 475/675 Introduction to Combustion

ME 475/675 Introduction to
Combustion
Lecture 30
Announcements
• Midterm 2
• Wednesday, November 12, 2014
• Review Monday, Nov. 10, 2014
• Extra Tutorials:
• Monday 10-11 MS 227
• Monday 5-6 PE 113
• Tuesday 4-5 PE 105
• HW 13
• Due Monday, November 10, 2014
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
(Fast flames are thin)
𝛼
𝒟
=
𝑘
𝒟𝜌𝑐𝑃
≈ 𝑂(1)
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 = 𝑥𝑖 𝑉
𝑂𝑥
𝐴𝑒𝑥𝑝
=
=
−𝑀𝑊𝐹 𝜔𝐹
𝜌𝑢
2𝛼 1 + 𝜈
𝑛
=
• 𝑖
• −
′′′
−𝑚𝐹
𝑛
𝑃
𝑥𝑖 𝑅 𝑇 ;
𝑢 𝑏
𝐸𝑎 𝑅𝑢
𝑇
𝑃𝑁−1 𝑢𝑁
𝑇𝑏
𝑇𝑏
𝑅𝑢 𝑇𝑢 𝑘
~𝑃−1 𝑇𝑢 𝑇 0.75 ,
𝑃 𝑐𝑝
𝐸 𝑅
− 𝑎 𝑢
𝑇𝑏
𝑃
𝑥𝐹 𝑚 𝑥𝑂𝑥 𝑛
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
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?
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
•
http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/PremixedFlameSpeedTickness.mcd
0.2
0.168
0.15
SL ( Tu ) 0.1
0.05
0.015
0
400
273
d ( Tu) 
Ta ( Tu)
0.375
Tb ( Tu)
600
800
Tu
 EaRu 

 2 Tb ( Tu) 
exp
3
110
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
110
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)