1. ME 475/675 Introduction to Combustion
Lecture 37
Laminar fuel jets, Similarity of axial velocity and fuel concentration, Diffusion flame length

2. Announcements
HW16 Ch. 9 (8, 10, 12)
Due Monday

3. Ch. 9 Laminar Fuel Jet (not premixed or burning)
A non-reacting, constant-density laminar fuel jet in quiescent air
No buoyancy
Assume
Temperature and Pressure are constant
𝑀𝑊 𝐹 = 𝑀𝑊 𝑂𝑥 =𝑀𝑊
𝜌 𝐹 = 𝜌 𝑂𝑥 =𝜌
Schmidt number, 𝑆𝑐= 𝜈 𝒟 = 𝜇 𝜌𝒟 =1
before 𝐿𝑒= 𝛼 𝒟 =1
For 𝑥< 𝑥 𝑐
Potential core is not affected by viscosity
Centerline:
Dimensionless
Speed, 𝑣 𝑥,0 𝑣 𝑒
Fuel Mass
Fraction 𝑌 𝐹 = 𝑚 𝐹 𝑚 𝑡𝑜𝑡𝑎𝑙
Constant in Core = 1
Then decrease
due to spreading
Axial Speed and mass
fraction profiles
𝑣 𝑥,0 𝑣 𝑒 and 𝑌 𝐹 versus r
Spread out as
x increases
Max magnitude
Decreases
Fuel
𝜌 𝑒 , 𝑣 𝑒 ,𝜇
𝑄 𝐹 = 𝑣 𝑒 𝜋 𝑅 2
𝑚 𝐹 = 𝜌 𝑒 𝑣 𝑒 𝜋 𝑅 2

4. Variables and Boundary Conditions𝑥
Find
𝑣 𝑥 𝑣 𝑒 , 𝑣 𝑟 𝑣 𝑒 , 𝑌 𝐹 =𝑓𝑛(𝑥,𝑟)
𝑟=0
𝜕 𝑣 𝑥 𝑣 𝑒 𝜕𝑟 = 𝜕 𝑌 𝐹 𝜕𝑟 = 𝑣 𝑟 𝑣 𝑒 =0
𝑟→∞,
𝑣 𝑥 = 𝑌 𝐹 =0
𝑣 𝑥
𝑣 𝑟 𝑟
𝑟=𝑅
𝑥=0
𝑟<𝑅
𝑣 𝑥 𝑣 𝑒 = 𝑌 𝐹 =1
“Top Hat” Profile
𝑥=0
𝑟>𝑅
𝑣 𝑥 = 𝑌 𝐹 =0
Find axial and radial velocity 𝑣 𝑥 , 𝑣 𝑟 and fuel concentration 𝑌 𝐹 versus 𝑥 and 𝑟
Assume steady and axis-symmetric (use radial coordinates)
Note that 𝑣 𝑥 𝑣 𝑒 and 𝑌 𝐹 have the same boundary conditions
Away from flame expect 𝑣 𝑟 <0 (because axial mass flow rate increases with x)

5. Conservation Equations (radial coordinates, p. 314)
Mass: 𝜕 𝑣 𝑥 𝜕𝑥 − 1 𝑟 𝜕 𝑣 𝑟 𝜕𝑟 𝑣 𝑟 𝑟 =0 1st order partial diff eqn.
Axial Momentum: 𝑣 𝑥 𝜕 𝑣 𝑥 𝜕𝑥 + 𝑣 𝑟 𝜕 𝑣 𝑥 𝜕𝑟 = 𝜈 1 𝑟 𝜕 𝜕𝑟 𝑟 𝑑 𝑣 𝑥 𝑑𝑟 2nd order partial diff eqn.
Species 𝑣 𝑥 𝜕 𝑌 𝐹 𝜕𝑥 + 𝑣 𝑟 𝜕 𝑌 𝐹 𝜕𝑟 =𝒟 1 𝑟 𝜕 𝜕𝑟 𝑟 𝑑 𝑌 𝐹 𝑑𝑟 2nd order partial diff eqn.
𝑌 𝑂𝑥 =1− 𝑌 𝐹
For Schmidt number 𝑆𝑐= 𝜈 𝒟 = 𝜇 𝜌𝒟 =1
Axial momentum and species equations are the same (and have same BC’s)
So expect 𝑌 𝐹 = 𝑣 𝑥 𝑣 𝑒 =𝑓𝑛(𝑥,𝑟)
“Reasonable” to assume: 𝑣 𝑥 𝑥,𝑟 𝑣 𝑥 𝑥,𝑟=0 = 𝑣 𝑥 𝑥,𝑟 𝑣 𝑥,0 𝑥 =𝑓𝑛 𝑟 𝑥
Similar but expanding shape for all x (this suggestions that a similarity solution may be used to solve)
Define
Jet fuel volume flow rate: 𝑄 𝐹 = 𝑣 𝑒 𝜋 𝑅 2
Jet initial momentum: 𝐽 𝑒 = 𝑄 𝐹 𝜌 𝑒 𝑣 𝑒 = 𝑣 𝑒 𝜋 𝑅 2 𝜌 𝑒 𝑣 𝑒 = 𝜌 𝑒 𝑣 𝑒 2 𝜋 𝑅 2

6. Dimensionless Similarity Variable
𝜉= 3 𝜌 𝑒 𝐽 𝑒 16𝜋 𝜇 𝑟 𝑥 (Greek letter Xi)
Velocity Solutions (Take my word for it, p. 315)
𝑣 𝑥 = 3 8𝜋 𝐽 𝑒 𝜇𝑥 𝜉 − where 𝐽 𝑒 = 𝜌 𝑒 𝑣 𝑒 2 𝜋 𝑅 2
𝑣 𝑟 = 3 𝐽 𝑒 16𝜋 𝜌 𝑒 1/2 1 𝑥 𝜉− 𝜉 𝜉
𝑣 𝑥 = 3 8𝜋 𝜌 𝑒 𝑣 𝑒 2 𝜋 𝑅 2 𝜇𝑥 𝜉 −2
𝑣 𝑥 𝑣 𝑒 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇 𝑅 𝑥 𝜉 −2 =0.375 𝑅𝑒 𝑗 𝑅 𝑥 𝜉 −2 = 𝑌 𝐹
𝑣 𝑥,𝑟=0 𝑣 𝑒 𝑣 𝑥 𝑣 𝑥,𝑟=0
Jet Reynolds number 𝑅𝑒 𝑗 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇
𝑣 𝑥 𝑣 𝑥,𝑟=0
𝑣 𝑟 𝑣 𝑥,𝑟=0 𝜉

7. Dimensionless Solutions
𝑣 𝑥 𝑣 𝑒 =0.375 𝑅𝑒 𝑗 𝑅 𝑥 𝜉 −2 , where 𝜉= 3 𝜌 𝑒 𝐽 𝑒 16𝜋 𝜇 𝑟 𝑥
At the centerline 𝑟=0, 𝜉=0, 𝑣 𝑥 = 𝑣 𝑥,0 ,
𝑣 𝑥,𝑟=0 𝑣 𝑒 = 𝑣 𝑥,0 𝑣 𝑒 =0.375 𝑅𝑒 𝑗 𝑅 𝑥
Centerline speed decreases with x
Find 𝑥 𝑅 for a given decrease 𝑣 𝑥,0 𝑣 𝑒
𝑥 𝑅 = 𝑅𝑒 𝑗 𝑣 𝑥,0 𝑣 𝑒 (increases as 𝑅𝑒 𝑗 increases)
A given 𝑣 𝑥,0 𝑣 𝑒 moves downstream (Decays less) as 𝑅𝑒 𝑗 increases
𝑣 𝑥 𝑣 𝑥,0 = 1+ 𝜉 −2 =𝑓𝑛 𝜉 =𝑓𝑛 𝑟 𝑥
Consistent with our “reasonable” assumption
𝑅𝑒 𝑗,2 > 𝑅𝑒 𝑗,1
𝑅𝑒 𝑗,1

8. Fuel Mass Fraction (dimensionless)
For Schmidt number 𝑆𝑐= 𝜈 𝒟 = 𝜇 𝜌𝒟 =1
𝑌 𝐹 =0.375 𝑅𝑒 𝑗 𝑅 𝑥 𝜉 −2
Recall 𝑣 𝑥 𝑣 𝑒 =0.375 𝑅𝑒 𝑗 𝑅 𝑥 𝜉 −2
Where
Dimensionless Similarity Variable: 𝜉= 3 𝜌 𝑒 𝐽 𝑒 16𝜋 𝜇 𝑟 𝑥
Jet Reynold number: 𝑅𝑒 𝑗 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇 = 𝑣 𝑒 𝑅 𝜈 = 𝑣 𝑒 𝑅 𝒟

9. Jet “half” radius 𝑟 1 2 = radius where 𝑣 𝑥 𝑣 𝑥,0 = 1 2
𝑟 1 2
𝑟 = radius where 𝑣 𝑥 𝑣 𝑥,0 = 1 2
𝑣 𝑥 𝑣 𝑥,0 = 1+ 𝜉 −2 = 1 2
𝜉 =4 2 −1
−1 = 𝜉 = 3 𝜌 𝑒 𝐽 𝑒 16𝜋 𝜇 𝑟 𝑥 = 3 𝜌 𝑒 𝜌 𝑒 𝑣 𝑒 2 𝜋 𝑅 2 16𝜋 𝜇 𝑟 𝑥 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇 𝑟 𝑥
𝑟 𝑥 = − 𝑅𝑒 𝑗 = 𝑅𝑒 𝑗
Jet spreading half-angle 𝛼; tan 𝛼 = 𝑟 𝑥 = 𝑅𝑒 𝑗
𝛼= tan − 𝑅𝑒 𝑗 , angle decreases as 𝑣 𝑒 and 𝑅𝑒 𝑗 increase
Fast
Slow

10. 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.

11. End 2017 This lecture was somewhat confusing
Next year, could derive conservation equations and then find similarity solution (this would take time but help students understand the value of the partial differential equations)
Students in this class generally have not taken ME 467, so do not know the Naiver-Stokes equations.

12. 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
𝜌 𝑒 , 𝑣 𝑒 ,𝜇
𝑄 𝐹 = 𝑣 𝑒 𝜋 𝑅 2
𝑚 𝐹 = 𝜌 𝑒 𝑣 𝑒 𝜋 𝑅 2

13. Soot Incomplete pre-reaction of HC fuels form Soot particles
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?

14. Flame length (a measurable quantity)
Equivalence ratio Φ 𝑟=0,𝑥= 𝐿 𝑓 =1; 𝑌 𝐹 = 𝑌 𝐹,𝑠𝑡
For un-reacting fuel jet (no buoyancy)
For Schmidt number 𝑆𝑐= 𝜈 𝒟 =1, 𝑌 𝐹 =0.375 𝑅𝑒 𝑗 𝑅 𝑥 𝜉 −2
Dimensionless Similarity Variable: 𝜉= 3 𝜌 𝑒 𝐽 𝑒 16𝜋 𝜇 𝑟 𝑥
Jet Reynold number: 𝑅𝑒 𝑗 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇 = 𝑣 𝑒 𝑅 𝜈 = 𝑣 𝑒 𝑅 𝒟
Flame length, x= 𝐿 𝐹 where 𝑌 𝐹 = 𝑌 𝐹,𝑠𝑡 at 𝑟=𝜉=0
𝑌 𝐹,𝑠𝑡 =0.375 𝑅𝑒 𝑗 𝑅 𝐿 𝐹 −2
𝐿 𝐹 = 3 8 𝑅𝑒 𝑗 𝑅 𝑌 𝐹,𝑠𝑡 = 𝜌 𝑒 𝑣 𝑒 𝑅 𝜇 𝑅𝜋 𝑌 𝐹,𝑠𝑡 𝜋 = 3 8𝜋 𝜌 𝑒 𝑄 𝐹 𝜇 𝑌 𝐹,𝑠𝑡 = 3 8𝜋 𝑚 𝐹 𝜇 𝑌 𝐹,𝑠𝑡 = 3 8𝜋 𝑄 𝐹 𝜈 𝑌 𝐹,𝑠𝑡 = 3 8𝜋 𝑄 𝐹 𝒟 𝑌 𝐹,𝑠𝑡
Increases with 𝑄 𝐹 = 𝑣 𝑒 𝜋 𝑅 2 (not dependent on 𝑣 𝑒 𝑜𝑟 𝑅 separately)
Decreases with increasing 𝒟 and 𝑌 𝐹,𝑠𝑡 = 𝑚 𝑂𝑥 𝑚 𝐹𝑢 = 𝑁 𝑂𝑥 𝑁 𝐹𝑢 𝑀𝑊 𝑂𝑥 𝑀𝑊 𝐹𝑢 = 1 1+𝑆 𝑀𝑊 𝑂𝑥 𝑀𝑊 𝐹𝑢
Depend on fuel
For 𝐶 𝑥 𝐻 𝑦 fuel, 𝑆=4.76 𝑥+ 𝑦 4 , 𝑀𝑊 𝑂𝑥 𝑀𝑊 𝐹𝑢 = 𝑥 𝑦
For y = 2x+2 (alkanes), decreases with increasing x
What about 𝒟?
What is the effect of buoyancy?
Stoichiometric A/F
Mass Ratio
Stoichiometric A/F
Mole Ratio