Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 37
Laminar fuel jets, Similarity of axial velocity and fuel concentration,
Diffusion flame length
Announcements
• HW16 Ch. 9 (8, 10, 12)
• Due Monday, 11/30/2015
• Term Project (3% of grade)
• Work in lab with a partner and Hasib to construct a homemade backpacking stove using
ethanol fuel and determine its water-boiling performance
• Parts are provided, please bring an aluminum can
•
•
•
•
•
Measure fuel mass before
Measure how long it takes to boil 𝑉 = 300 ml of room temperature water.
What color is the flame and is soot deposited on the beaker?
Measure fuel mass after test, and calculate the amount of fuel consumed, 𝑚𝑓𝑢𝑒𝑙 .
Calculate
• Energy to start 300 ml of room temperature water boiling: 𝑄𝐵 = 𝑉𝜌𝑐𝑝
• Energy release by combustion: 𝑄𝐵 = 𝑚𝑓𝑢𝑒𝑙 𝐿𝐻𝑉
• Fraction of combustion energy absorbed by water
• Stove/beaker effectiveness: E = 𝑄𝐵 /𝑄𝐵 (<1?)
• Finals schedule
• December 11, 2015, 2:45-4:45 PM (conflicts?)
𝑤𝑎𝑡𝑒𝑟
𝑇𝑏𝑜𝑖𝑙 − 𝑇𝑟𝑜𝑜𝑚
Ch. 9 Laminar Fuel Jet (not premixed or burning)
• A non-reacting, constant-density
laminar fuel jet in quiescent air
• No buoyancy
• Assume
Fuel
𝜌𝑒 , 𝑣𝑒 , 𝜇
𝑄𝐹 = 𝑣𝑒 𝜋𝑅2
𝑚𝐹 = 𝜌𝑒 𝑣𝑒 𝜋𝑅2
Centerline:
Dimensionless
𝑣
Speed, 𝑥,0
𝑣𝑒
Fuel Mass
Fraction 𝑌𝐹
Constant in Core
Then decrease
due to spreading
Axial Speed
Profiles
𝑣𝑥,0
versus r
𝑣𝑒
Spreads out as
x increases
Max magnitude
Decreases
• Temperature and Pressure are
constant
• 𝑀𝑊𝐹 = 𝑀𝑊𝑂𝑥 = 𝑀𝑊
• 𝜌𝐹 = 𝜌𝑂𝑥 = 𝜌
𝜈
𝜇
• Schmidt number, 𝑆𝑐 = =
=1
• before 𝐿𝑒 =
• 𝑥 < 𝑥𝑐
𝛼
𝒟
𝒟
𝜌𝒟
=1
• Potential core is not affected by
viscosity
Variables and Boundary Conditions
𝑥
Find
𝑟=0
0=
𝑣𝑟
𝑣𝑒
=
𝜕
𝑣𝑥
𝑣𝑒
𝜕𝑟
=
𝜕𝑌𝐹
𝜕𝑟
𝑣𝑥 𝑣𝑟
, ,𝑌
𝑣𝑒 𝑣𝑒 𝐹
= 𝑓𝑛(𝑥, 𝑟)
𝑣𝑥
𝑣𝑟
𝑥=0
𝑟<𝑅
𝑟=𝑅
𝑣𝑥
= 𝑌𝐹 = 1
𝑣𝑒
“Top Hat” Profile
𝑟 → ∞,
𝑣𝑥 = 𝑌𝐹 = 0
𝑟
𝑥=0
𝑟>𝑅
𝑣𝑥 = 𝑌𝐹 = 0
• Find axial and radial velocity 𝑣𝑥 , 𝑣𝑟 and fuel concentration 𝑌𝐹 versus 𝑥 and 𝑟
• Assume steady and axis-symmetric (use radial coordinates)
• Note that
𝑣𝑥
𝑣𝑒
and 𝑌𝐹 have the same boundary conditions
• Away from flame expect 𝑣𝑟 < 0 (because axial mass flow rate increases with x)
Conservation Equations (radial coordinates, p. 314)
• Mass:
• Axial Momentum:
• Species
𝜕𝑣𝑥
1 𝜕𝑣𝑟
−
𝑣𝑟 𝑟 = 0
𝜕𝑥
𝑟 𝜕𝑟
𝜕𝑣𝑥
𝜕𝑣𝑥
1 𝜕
𝑣𝑥
+ 𝑣𝑟
= 𝜈
𝜕𝑥
𝜕𝑟
𝑟 𝜕𝑟
𝜕𝑌𝐹
𝜕𝑌𝐹
1 𝜕
𝑣𝑥
+ 𝑣𝑟
=𝒟
𝜕𝑥
𝜕𝑟
𝑟 𝜕𝑟
1st order partial diff eqn.
𝑑𝑣𝑥
𝑟
𝑑𝑟
𝑑𝑌𝐹
𝑟
𝑑𝑟
2nd order partial diff eqn.
2nd order partial diff eqn.
𝑌𝑂𝑥 = 1 − 𝑌𝐹
𝜈
𝜇
• For Schmidt number 𝑆𝑐 = =
=1
𝒟
𝜌𝒟
• Axial momentum and species equations are the same (and have same BC’s)
• So expect 𝑌𝐹 =
𝑣𝑥
𝑣𝑒
= 𝑓𝑛(𝑥, 𝑟)
𝑣𝑥 𝑥,𝑟
𝑣 𝑥,𝑟
𝑟
= 𝑥
= 𝑓𝑛
𝑣𝑥 𝑥,𝑟=0
𝑣𝑥,0 𝑥
𝑥
• Similar but expanding shape for all x (this suggestions that a similarity solution may be used to solve)
• “Reasonable” to assume:
• Define
• Jet fuel volume flow rate: 𝑄𝐹 = 𝑣𝑒 𝜋𝑅 2
• Jet initial momentum: 𝐽𝑒 = 𝑄𝐹 𝜌𝑒 𝑣𝑒 = 𝑣𝑒 𝜋𝑅 2 𝜌𝑒 𝑣𝑒 = 𝜌𝑒 𝑣𝑒2 𝜋𝑅 2
Dimensionless Similarity Variable
•𝜉=
3𝜌𝑒 𝐽𝑒 1 2 1 𝑟
16𝜋
𝜇𝑥
(Greek letter Xi)
• Velocity Solutions (Take my word for it, p. 315)
• 𝑣𝑥 =
• 𝑣𝑟 =
• 𝑣𝑥 =
•
•
𝑣𝑥
𝑣𝑒
=
−2
3 𝐽𝑒
𝜉2
1+
where
8𝜋 𝜇𝑥
4
𝜉3
1/2
3𝐽𝑒
1 𝜉− 4
2
16𝜋𝜌𝑒
𝑥
𝜉2
1+
4
3 𝜌𝑒 𝑣𝑒2 𝜋𝑅 2
8𝜋
𝜇𝑥
3 𝜌𝑒 𝑣𝑒 𝑅 𝑅
8
𝜇
𝑥
𝑣𝑥,𝑟=0
𝑣𝑒
1+
1+
𝜉2
4
−2
𝜉2
4
−2
𝐽𝑒 = 𝜌𝑒 𝑣𝑒2 𝜋𝑅2
1
=
𝑣𝑥
𝑣𝑥,𝑟=0
• Jet Reynolds number 𝑅𝑒𝑗 =
𝑅
0.375𝑅𝑒𝑗
𝑥
1
𝜉2
+
4
0.9
−2
0.8
= 𝑌𝐹
0.7
0.6
0.5
0.5
𝑣𝑥
0.4
0.4
0.3
0.3
0.2
𝜌𝑒 𝑣𝑒 𝑅
𝜇
𝑣𝑥,𝑟=0
0.1
-0.1
-0.2
-0.3
0.2
𝑣𝑟
0
0.1
𝑣𝑥,𝑟=0
0
-0.4
-0.5
0
1
2
3
4
5
0
1
𝜉
2
3
4
5
Dimensionless Solutions
•
𝑣𝑥
𝑣𝑒
=
𝑅
0.375𝑅𝑒𝑗
𝑥
1
𝜉2
+
4
−2
, where 𝜉 =
3𝜌𝑒 𝐽𝑒 1 2 1 𝑟
16𝜋
𝜇𝑥
𝑅𝑒𝑗,2 > 𝑅𝑒𝑗,1
• At the centerline 𝑟 = 0, 𝜉 = 0, 𝑣𝑥 = 𝑣𝑥,0 ,
•
𝑣𝑥,𝑟=0
𝑣𝑒
=
𝑣𝑥,0
𝑣𝑒
=
𝑅
0.375𝑅𝑒𝑗
𝑥
• Centerline speed decreases with x
𝑥
𝑅
• Find for a given decrease
•
𝑥
𝑅
=
0.375𝑅𝑒𝑗
𝑣𝑥,0 𝑣𝑒
𝑣𝑥,0
• A given
•
𝑣𝑥
𝑣𝑥,0
= 1+
𝑣𝑒
𝜉2
4
𝑅𝑒𝑗,1
𝑣𝑥,0
𝑣𝑒
(increases as 𝑅𝑒𝑗 increases)
moves downstream (Decays less) as 𝑅𝑒𝑗 increases
−2
= 𝑓𝑛 𝜉 = 𝑓𝑛
𝑟
𝑥
• Consistent with our “reasonable” assumption
Fuel Mass Fraction (dimensionless)
•
•
𝜈
For Schmidt number 𝑆𝑐 =
𝒟
−2
𝑅
𝜉2
𝑌𝐹 = 0.375𝑅𝑒𝑗 1 +
𝑥
4
• Recall
𝑣𝑥
𝑣𝑒
=
𝑅
0.375𝑅𝑒𝑗
𝑥
1
=
𝜉2
+
4
𝜇
𝜌𝒟
=1
−2
• Where
•
•
3𝜌𝑒 𝐽𝑒 1 2 1 𝑟
Dimensionless Similarity Variable: 𝜉 =
16𝜋
𝜇𝑥
𝜌 𝑣 𝑅
𝑣 𝑅
𝑣 𝑅
Jet Reynold number: 𝑅𝑒𝑗 = 𝑒 𝑒 = 𝑒 = 𝑒
𝜇
𝜈
𝒟
Jet “half” radius
• 𝑟1
•
𝑣𝑥
𝑣𝑥,0
2
= radius where
= 1+
• 𝜉12
• 𝜉1
•
2
2
𝑟1 2
𝑥
=4
=2
=
8
𝜉12 2
𝑣𝑥
𝑣𝑥,0
−2
=
4
=
1
2
𝑟1
2
2
2−1
2−1=
2−1
3𝑅𝑒𝑗
=
3𝜌𝑒 𝐽𝑒 1 2 1 𝑟1 2
16𝜋
𝜇 𝑥
2.97
−1
tan
,
𝑅𝑒𝑗
=
3𝜌𝑒 𝜌𝑒 𝑣𝑒2 𝜋𝑅2
16𝜋
1 2
2.97
𝑅𝑒𝑗
• Jet spreading half-angle 𝛼; tan 𝛼 =
• 𝛼=
𝑟1
1
2
𝑟1 2
𝑥
=
2.97
𝑅𝑒𝑗
angle decreases as 𝑣𝑒 and 𝑅𝑒𝑗 increase
1 𝑟1 2
𝜇 𝑥
=
3 𝜌𝑒 𝑣𝑒 𝑅 𝑟1 2
4
𝜇
𝑥
Fast
Slow
Example 9.1
• A jet of ethylene (C2H4) exits a 10-mm-diameter nozzle into still air at 300 K and 1
atm. Compare the spreading angles and axial locations where the jet centerline
mass fraction drops to the stoichiometric value for initial jet velocities of 10 cm/s
and 1.0 cm/s. The viscosity of ethylene at 300 k is 102.3x10-7 Ns/m2.
• This is a model for flame length (assuming buoyancy doesn’t effect it)
• Angle and flame length depend on flow rate, but not jet exit velocity or diameter
separately.
Now: Burning Fuel Jet
(Diffusion Flame)
• Laminar Diffusion flame structure
• T and Y versus r at different x
• Flame shape
• Assume flame surface is located
where Φ ≈ 1, stoichiometric mixture
• No reaction inside or outside this
• Products form in the flame “sheet”
and then diffuse outward (and
inward)
• No oxidizer inside the flame envelop
• Over-ventilated: enough oxidizer to
burn all fuel
Fuel
𝜌𝑒 , 𝑣𝑒 , 𝜇
𝑄𝐹 = 𝑣𝑒 𝜋𝑅2
𝑚𝐹 = 𝜌𝑒 𝑣𝑒 𝜋𝑅2
Soot
• Soot particles form from incomplete reaction of the HC
fuels
• Forms on the fuel side (inside) of the flame surface and
radiate orange and yellow
• Most soot is consumed as it flows through the hot flame
• “Wings” form when unburned soot breaks through burning zone
• Smoke is soot that breaks through
• Roughly how long will the flame be?
Flame length (a measurable quantity)
• Flame length 𝐿𝑓 :
• Equivalence ratio Φ 𝑟 = 0, 𝑥 = 𝐿𝑓 = 1;
𝑌𝐹 = 𝑌𝐹,𝑠𝑡
• For un-reacting fuel jet (no buoyancy)
• For Schmidt number 𝑆𝑐 =
𝜈
𝒟
= 1,
𝑅
𝑌𝐹 = 0.375𝑅𝑒𝑗 𝑥
3𝜌 𝐽 1 2 1 𝑟
• Dimensionless Similarity Variable: 𝜉 =
• Jet Reynold number: 𝑅𝑒𝑗 =
𝜌𝑒 𝑣𝑒 𝑅
𝜇
=
𝑣𝑒 𝑅
𝜈
1+
𝜉2
4
−2
•
•
=
𝑒 𝑒
16𝜋
𝑣 𝑅
= 𝑒
𝒟
𝜇𝑥
Y( x y) 
3 𝜌𝑒 𝑄𝐹
8𝜋 𝜇𝑌𝐹,𝑠𝑡
=
3 𝑚𝐹
8𝜋 𝜇𝑌𝐹,𝑠𝑡
=
• Decreases with increasing 𝒟 and 𝑌𝐹,𝑠𝑡 =
1
𝑚
1+ 𝑂𝑥
𝑚𝐹𝑢
=
3 𝑄𝐹
8𝜋 𝜈𝑌𝐹,𝑠𝑡
=
3 𝑄𝐹
8𝜋 𝒟𝑌𝐹,𝑠𝑡
0.095
1
𝑁 𝑀𝑊
1+𝑁𝑂𝑥 𝑀𝑊𝑂𝑥
𝐹𝑢
𝐹𝑢
=
Stoichiometric A/F
Mass Ratio
1
𝑀𝑊
1+𝑆𝑀𝑊𝑂𝑥
𝐹𝑢
0.06
Y ( x Ya ( x) )
X( x Ya ( x) )
Stoichiometric A/F
Mole Ratio
0.04
𝑦
4
,
𝑀𝑊𝑂𝑥
𝑀𝑊𝐹𝑢
=
28.85
12.011𝑥+1.00794𝑦
0.02
• For y = 2x+2 (alkanes), decreases with increasing x
• What is the effect of buoyancy?
1
y
28.85

1  4.76  x   
4  12.011 x  1.00794y

0.08
• Depend on fuel
• For 𝐶𝑥 𝐻𝑦 fuel, 𝑆 = 4.76 𝑥 +
y

4
Ya ( x)  ( 2 x)  2
• Increases with 𝑄𝐹 = 𝑣𝑒 𝜋𝑅2 (not dependent on 𝑣𝑒 𝑜𝑟 𝑅 separately)
• What about 𝒟?


1  4.76  x 
• Flame length, x = 𝐿𝐹 where 𝑌𝐹 = 𝑌𝐹,𝑠𝑡 at 𝑟 = 𝜉 = 0
−2
𝑅
02
𝑌𝐹,𝑠𝑡 = 0.375𝑅𝑒𝑗
1+
𝐿𝐹
4
3
𝑅
3 𝜌𝑒 𝑣𝑒 𝑅
𝑅𝜋
𝐿𝐹 = 𝑅𝑒𝑗
=
8
𝑌𝐹,𝑠𝑡
8
𝜇
𝑌𝐹,𝑠𝑡 𝜋
1
X( x y) 
0
0
2
1
4
6
x
8
10
11