Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 28
Problem 6.11 numerical solution, Premixed flames, Flame speed and
cone vertex angle, Flame zones
Announcements
• Today is the final day to drop classes and receive a “W”
• Friday: Holiday
• Midterm 2
• November 13, 2015
• HW 11 (6.11a numerical solution)
• Due Monday, Nov. 2, 2015
General Plug-Flow Reactor
• Characteristics:
𝑑𝐴
𝑑𝑥
• Area 𝐴 𝑥 ,
≠0
• Flow kinetic energy is not necessarily small compared to enthalpy
𝑣𝑥2
2
• ≪ℎ
• Pressure 𝑃 is not necessarily constant
• Species can have different, temperature-dependent properties
• Assume you are given
• Operating Conditions 𝑄 " 𝑥 , 𝐴 𝑥 and 𝑚
• Inlet conditions: 𝑇𝑖𝑛 , 𝑌𝑖,𝑖𝑛 , and 𝜌𝑖𝑛 (or 𝑃𝑖𝑛 )
Based on conservation of Species, Momentum, and energy
•
•
•
𝑑𝑌𝑖
𝑑𝑥
𝑑𝜌
𝑑𝑥
𝑑𝑇
𝑑𝑥
=
=
𝜔𝑖 𝑀𝑊𝑖 𝐴
,𝑖
𝜌𝑣𝑥
= 1,2, … , 𝑀 (eqn. 6.53)
𝑅𝑢
1−
𝑐𝑃 𝑀𝑊𝑚𝑖𝑥
1 𝑑𝐴
𝜌2 𝑣𝑥2
𝐴 𝑑𝑥
𝜌𝑅𝑢
+
𝑣𝑥 𝑐𝑃 𝑀𝑊𝑚𝑖𝑥
𝑃
=
𝑣𝑥2 𝑑𝜌
𝜌𝑐𝑃 𝑑𝑥
+
𝑣𝑥2 𝑑𝐴
𝑐𝑃 𝐴 𝑑𝑥
−
𝜔𝑖 𝑀𝑊𝑖 ℎ𝑖
𝜌𝑣𝑥 𝑐𝑃
• Integrate to find 𝑇 𝑥 , 𝑌𝑖 𝑥 , 𝜌 𝑥
• Use
• 𝜔𝑖 = 𝑓𝑛 𝑌𝑖 , 𝑇, 𝑃 ; 𝑃 = 𝜌𝑅𝑇; 𝑅 =
𝑣2
1+𝑐 𝑥𝑇
𝑃
−
• 𝑣𝑥 𝑥
• 𝑃 𝑥
𝜌𝑅𝑢 𝑄" 𝒫
+
𝑣𝑥 𝐴𝑐𝑃 𝑀𝑊𝑚𝑖𝑥
−𝜌𝑣𝑥2
𝑄" 𝒫
𝑚𝑐𝑃
𝑅𝑢
1
;
𝑀𝑊𝑚𝑖𝑥 𝑀𝑊𝑚𝑖𝑥
• At each location also need to calculate
𝑚
=
𝜌 𝑥 𝐴(𝑥)
𝜌 𝑥 𝑅𝑢 𝑇(𝑥)
=
𝑀𝑊𝑚𝑖𝑥
𝑀𝑊𝑚𝑖𝑥
𝑀𝑊𝑖 𝜔𝑖 ℎ𝑖 −
𝑐 𝑇
𝑀𝑊𝑖 𝑃
(Eqn. 6.51 is for 𝑄 " = 0)
"
(Eqn. 6.52 is for 𝑄" = 𝑄𝑜𝑢𝑡
= 0)
=
𝑌𝑖
;𝑣
𝑀𝑊𝑖 𝑥
=
𝑚
;𝑐
𝜌𝐴 𝑃
=
𝑌𝑖 𝑐𝑃𝑖
What do we expect?
•
Problem 6.11
Extra credit-turn in first lines of Excel next meeting
Homework-turn in numerical solution
• Develop a plug-flow-reactor model using the same chemistry and
thermodynamics as in Example 6.1. Assume the reactor is adiabatic. Use the
model to:
A. Determine the mass flow rate such that the reaction is 99 percent complete in
a flow length of 10 cm for 𝑇𝑖𝑛 = 1000𝐾, 𝑃𝑖𝑛 = 0.2 𝑎𝑡𝑚, and Φ𝑖𝑛 = 0.2. The
circular duct has a diameter of 3 cm.
B. Explore the effects of 𝑇𝑖𝑛 , 𝑃𝑖𝑛 , and Φ𝑖𝑛 on the flow length required for 99
percent complete combustion using the flow rate determined in Part A.
• Does this reactor operate at constant volume, at constant pressure, as a wellstirred, or as a plug-flow?
Solution
• 𝜔𝐹𝑢𝑒𝑙 = −6.186 ∗ 109 𝑒𝑥𝑝
• 𝑃 = 𝜌𝑅𝑇; 𝑅 =
•
𝑑𝑌𝐹𝑢
𝑑𝑥
=
• 𝑣𝑥 =
•
•
𝑑𝜌
𝑑𝑥
𝑑𝑇
𝑑𝑥
𝜔𝐹𝑢 𝑀𝑊
𝜌𝑣𝑥
𝑌𝐹
0.1
𝑅𝑢
𝑀𝑊
;
𝑑𝑌𝑂𝑥
𝑑𝑥
=
𝑑𝑌𝐹𝑢
16
;
𝑑𝑥
𝑑𝑌𝑃𝑟
𝑑𝑥
1.65
𝑃
0.233𝑌𝑂𝑥 1.65
𝑅𝑢 𝑇
=
𝑑𝑌𝐹𝑢
−17
𝑑𝑥
𝑚
𝜌𝐴
𝑜
𝜌𝑅𝑢 𝜔𝐹𝑢 ℎ𝑓,𝐹𝑢
=
𝑣𝑥 𝑐𝑃 𝑃
=
−15,098𝐾
𝑇
𝑣𝑥2 𝑑𝜌
𝜌𝑐𝑃 𝑑𝑥
• 𝑐𝑃 =
𝑣2
1+𝑐 𝑥𝑇
𝑃
−
−𝜌𝑣𝑥2
𝑜
𝑀𝑊 ℎ𝑓,𝐹𝑢
𝜔𝐹𝑢
𝜌𝑣𝑥 𝑐𝑃
𝐽
𝑜
1200
; ℎ𝑓,𝐹𝑢
𝑘𝑔 𝐾
• Remember Units!
=
𝐽
7
4𝑥10
;
𝑘𝑔
𝑅𝑢 =
𝐽
8315
;
𝑘𝑚𝑜𝑙𝑒 𝐾
𝑀𝑊 =
𝑘𝑔
29
𝑘𝑚𝑜𝑙𝑒
Excel Solution Method
• Starting Point
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/Prob.6.11.start.xlsx
kg/s
0.00125
1
m
0.0001
m
kg/kg
kg/kg
0 0.058824 0.941176
kg/kg
kg/m3
0 0.07066
K
1000
kPa
m/s
kmole/m3s
20.26 25.02663 -0.002825094
kg/kgm
kg/kgm
kg/kgm
kg/m4
K/m
-0.046329 -0.74126467 0.787593712 -0.109302442 1543.494015
• Pay attention to
• Integration step size
• Avoiding raising negative numbers to a non-integer power
• From last lecture:
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/Prob.6.11.start.calc.xlsx
Results m for YF/YFi = 0.01 at x = 0.1 m
Ch. 7 Simplified Conservation Equations for Reacting Flows
(present as needed)
Ch. 8 Laminar Premixed Flames
• Flame: Subsonic propagation of a self-sustaining local (thin) combustion zone
• Recall Bunsen Burner Demo
• Propane (C3H8)/Air
• Air induction holes at bottom of burner
• Closed: Diffusion Flame
• Orange, tall, unsteady
• Open: Premixed inner cone
• Outer: diffusion flame of unburned components
• Turn off gas: Flash back
• Increase Flow rate: cones lengthen
• When mesh was put into flame
• Hot ring shows thin flame zone
• Cool interior
Bunsen Burner Inner Cone angle, 𝛼
𝛼
𝑣𝑢
𝑆𝐿
𝛼
• 𝑆𝐿 is the laminar flame speed, relative to the premixed reactants
• 𝑣𝑢 is the unburned reactant speed
• If 𝑣𝑢 > 𝑆𝐿 , then cone will adjust its angle 𝛼 so that 𝑆𝐿 = 𝑣𝑢 sin 𝛼
• The angel 𝛼 and its sine sin 𝛼 =
𝑆𝐿
𝑣𝑢
decrease as increases 𝑣𝑢 (inner cone length increases)
• If 𝑣𝑢 < 𝑆𝐿 , then flame will flash back to air holes (unless quenched in tube).
Tube filled with stationary premixed Oxidizer/Fuel
Products shoot
out as flame
burns in
𝑆𝐿 Laminar Flame Speed
Burned
Products
𝛿
Unburned Fuel + Oxidizer
• 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)
• How to estimate the laminar flame speed 𝑆𝐿 and thickness 𝛿?
• Depends on the pressure, fuel, equivalence ratio,…
• Flame reference frame:
𝑣𝑏 , 𝜌𝑏
𝑣𝑢 = 𝑆𝐿 , 𝜌𝑢 ,
𝛿~1 𝑚𝑚
• 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
=
• For hydrocarbon fuels at P = 1 atm, 𝑇𝑢 ≈ 300𝐾, 𝑇𝑏 ≈ 2100𝐾, 𝑣𝑏 ≈ 7𝑣𝑢
• What happens within a premixed flame?
𝑇𝑏
𝑇𝑢
Flame Zones
Heat and radicals diffuse
to unburned region
• Fast Chemical Zone
• Fuel  Intermediate Species by bi-molecular reactions
• Very thin (1 mm)
• large gradients drive heat and species ahead of reaction zone and produce self-sustaining behaviors
• Slow Chemical Zone
• 3 body radial recombination and burnout 𝐶𝑂 + 𝑂𝐻 → 𝐶𝑂2 + 𝐻
• Thicker, several mm
• Premixed Color
• Φ < 1 (excess air), Blue excitation of CH radical
• Φ ≈ 1 (stoichiometric), Blue green, radiation of 𝐶2
• Φ ≫ 1 (fuel rich), Soot, black body radiation
End 2015
Simplified Analysis Assumptions
• One dimensional flow
• Kinetic energy = viscosity = radiation = 0
• Constant pressure
•
𝑞𝑥"
=
𝑑𝑇
−𝑘
𝑑𝑥
• Usually,
𝑑𝑌𝑖
and
= −𝜌𝒟
𝑑𝑥
𝛼
𝑘
Lewis Number, 𝐿𝑒 = =
𝒟
𝒟𝜌𝑐𝑃
𝑚𝑖"
≈ 1;
𝑘
𝑐𝑃
≈ 𝒟𝜌
• 𝑐𝑃,𝑖 = 𝑐𝑃 ≠ 𝑓𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑟 𝑆𝑝𝑒𝑐𝑖𝑒
• Single Step Kinetics
• Φ < 1, Fuel Lean, so fuel is completely consumed
Conservations Laws
• Mass Conservation
"
• 𝑚 = 𝜌𝑣𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡;
𝑑 𝜌𝑣𝑥
𝑑𝑥
𝑑𝑥
=0
• Species Conservation
•
𝐴𝑚𝑖"
•
𝑚𝑖′′′
+ 𝐴 𝑑𝑥
=
𝑑 𝑚𝑖"
𝑑𝑥
=
𝑚𝑖′′′
𝑑
𝑑𝑥
𝐴𝑚𝑖"
"
=𝐴
" 𝑑 𝑚𝑖
𝑚𝑖 +
𝑑𝑥
𝑑𝑥
𝑚" 𝑌𝑖 − 𝜌𝒟
𝑑𝑌𝑖
𝑑𝑥
𝐴
(using Flick’s Law)
• Apply to: 1 𝑘𝑔 𝐹 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑃𝑟; Air/Fuel ratio 𝜈 =
1
𝜈
′′′
• 𝑚𝐹′′′ = 𝑚𝑂𝑥
=−
1
′′′
𝑚𝑃𝑟
𝜈+1
• 𝑖 = 1, 2, … , 𝑀 = 3; Fuel, Oxidizer, Products
•
•
•
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑𝑌𝐹
− 𝜌𝒟
𝑑𝑥
𝑑𝑌𝑂𝑥
"
𝑚 𝑌𝑂𝑥 − 𝜌𝒟
𝑑𝑥
𝑑𝑌𝑃𝑟
"
𝑚 𝑌𝑃𝑟 − 𝜌𝒟
𝑑𝑥
𝑚" 𝑌𝐹
𝑚𝑖′′′
= 𝑚𝐹′′′
′′′
= 𝑚𝑂𝑥
= 𝜈 𝑚𝐹′′′
′′′
= 𝑚𝑃𝑟
= − 1 + 𝜈 𝑚𝐹′′′
𝑘𝑔 𝑂𝑥
𝑘𝑔 𝐹
"
𝑑
𝑚
𝑖
𝐴 𝑚𝑖" +
𝑑𝑥
𝑑𝑥
Energy Conservation (Ch. 7, pp. 239-244)
𝑑𝑥
𝐴𝑚
𝐴𝑚′′ ℎ
′′′
𝑚𝑖
𝐴𝑄𝑥′′
𝑄𝑥′′
•𝐴
•
•
−
+
𝑑 𝑄𝑥′′
𝑑ℎ
′′
−
=𝑚
𝑑𝑥
𝑑𝑥
𝑑𝑇
′′
𝑄𝑥 = −𝑘 +
𝑑𝑥
• Note:
•
•
𝑄𝑥′′
𝑄𝑥′′
=
𝑑ℎ
𝑑𝑥
=
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑇
−𝑘
𝑑𝑥
𝑑
𝑑𝑥
=
𝑑ℎ
𝑑𝑥
𝑑 𝑄𝑥′′
𝑑𝑥
𝑑𝑥
− 𝑊𝑄𝑉 =
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑌𝑖 ℎ𝑖 =
𝑑𝑇
𝑑𝑥
− 𝜌𝒟
𝑄𝑥′′
𝑑𝑄𝑥′′
+
𝑑𝑥
𝑑𝑥
𝑊𝑄𝑉
′′
𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
ℎ𝑖
−
𝐴
𝑑ℎ
ℎ+
𝑑𝑥
𝑑𝑥
′′
𝑌𝑖 𝑐𝑝,𝑖 =
𝑑ℎ
𝑑𝑥
−
𝑑ℎ
𝑑𝑥
− 𝑐𝑝
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑇
−𝑘
𝑑𝑥
=
+
𝐴𝑚′′
𝑌𝑖
𝑑𝑇
𝑑𝑥
𝑑ℎ𝑖
𝑑𝑥
=
−
ℎ+
𝑑ℎ
𝑑𝑥
𝑑𝑥
−ℎ
𝑑𝑌𝑖
𝜌𝒟
ℎ𝑖
𝑑𝑥
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
+
=
𝑌𝑖 𝑐𝑝,𝑖
𝑑𝑇
−𝑘
𝑑𝑥
𝑑𝑇
𝑑𝑥
=
− 𝜌𝒟
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
+
𝑑𝑌𝑖
ℎ𝑖
𝑑𝑥
𝑑𝑇
𝑑𝑥
𝑌𝑖 𝑐𝑝,𝑖
Shvab-Zeldovich form
•
𝑄𝑥′′
=
𝑑𝑇
−𝑘
𝑑𝑥
𝑑ℎ
− 𝜌𝒟
𝑑𝑥
+
𝑑𝑇
𝜌𝒟𝑐𝑝
𝑑𝑥
• Heat Flux =
• Flux of sensible enthalpy due to conduction +
• Flux of standardized enthalpy due to species diffusion +
• Flu of sensible enthalpy due to species diffusion
• For 𝐿𝑒 =
•
𝑄𝑥′′
=
𝛼
𝒟
=
𝑘
𝒟𝜌𝑐𝑃
𝑑ℎ
−𝜌𝒟
𝑑𝑥
≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃