Expert answer:Combustion questions laminar non premixed flames
Expert answer:My professor added one section that will be covered on the conceptual part of the exam. Could you make a quick study guide for this section? Specifically explaining the graphs/flame diagram (slides 18, 21, 22, 23, 24) and answering the questions on slide 25. Anything else you can think of would be appreciated too.
laminar_non_premixed_flames_14.pdf
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Non-Premixed
Laminar
Flames
Introduction
• Design of burners, in general, relies heavily on “art
and craft” of burner design
• After stability, the next main concern is emission
• NOx and CO are toxic gases
• Flame geometry, particularly compactness, is a
primary concern
• Fuel type must be considered
• Methane vs propane grill, for example
Non – reacting laminar fuel jet flowing into an infinite oxidizer reservoir
Uniform jet velocity profile
Free shear layer: fuel and air mix by molecular diffusion
Oxidizer is “entrained” as jet momentum is transferred to it
Pure fuel inside, pure oxidizer outside
Potential core: region free from viscous effects, no mixing
Initial jet momentum and mass conserved throughout the flow field
∞
2π ∫ ρ ( r , x ) v x2 ( r , x ) rdr = ρev e2 π R 2
0
∞
2π ∫ ρ ( r , x ) v x ( r , x )YF ( r , x ) rdr = ρev eπ R 2YF ,e
0
Assumptions
Jet and oxidizer MW’s are equal,
Ideal gas, P, T, and ρ are constant
Fick’s law diffusion
Equal momentum and species diffusivities,
=
= ν= 1
ν D, Sc
D
Neglect axial diffusion, radial only
Boundary conditions
along the jet centerline, r = 0
v r ( 0, x ) = 0
Mass
∂v x 1 ∂ (v r r )
+
=
0
∂x r ∂r
Momentum
∂v x
(0, x ) = 0
∂r
∂YF
(0, x ) = 0
∂r
vx
far from jet, r → ∞
∂YF
∂YF
1 ∂ ∂YF
+ vr
=
D
r
∂x
∂r
r ∂r ∂r
Yox = 1 − YF
v x ( ∞, x ) =
0
YF ( ∞, x ) =
0
at jet exit, x = 0
v x ( r ≤ R, 0 ) =
ve
v x ( r > R, 0 ) =
0
YF ( r ≤ R, 0 ) =YF ,e =1
YF ( r > R, 0 ) =
0
∂v x
∂v
1 ∂ ∂v x
+ vr x =
ν
r
∂x
∂r
r ∂r ∂r
Species
vx
Note: for ν = D species and momentum
have the same functional solution form
Solution
Similarity solution: profiles are “similar”,
i.e., intrinsic shape of velocity profiles are
the same everywhere
Profiles depend only on the similarity
variable r
x
vx
3 Je ξ 2
1 +
8π µ x
4
2
QF = v eπ R 2
3ρ J
ξ = e e
16π
1
( )
( )
along the centerline, r = 0 → ξ = 0
2
3 QF ξ
YF =
1 +
8π Dx
4
3
ξ
1
ξ−
2
4
3Je 1
vr =
16
x ξ 2 2
πρ
e
1 + 4
Je = ρev e2 π R 2
2
ρ v R 1
1
= 0.375 e e
2
ve
µ x
ξ2
R 1 +
r
ρ v R 1
1
YF = 0.375 e e
2
µ x
ξ2
R 1 +
r
vx
vx
ve
= 0.375
YF ,0 = 0.375
Re j
(x R)
Re j
(x R)
∴ centerline velocity decays with 1
and increases with Re j
Solutions valid far from jet, i.e.,
2
r
µx
( x R ) > 0.375 Re
j
x
Jet half – width, spreading rate
and angle
Jet half-width: jet velocity is half the
centerline value
vx
v x ,0
=1
r1 2
x
2
and solve for r
= 2.97
Re j
r1 2
α = tan−1
x
Jet flame physical description
Flame surface is the locus of points
where φ = 1
Overventilated: excess O2 in surroundings
Underventilated: deficient O2 in surroundings
Laminar non – premixed flame length
Flame length depends on QF but not on
v e or R independently
Lf ≈
QF
3
8π DYF ,stoic
Buoyancy tends to accelerate and narrow flame
increasing concentration gradient and diffusion.
The two effects tend to cancel so that simple
theories neglecting buoyancy well-predict Lf
Simplified theory
Burke-Schumann (1928 ) , and Roper (1977 )
Assumptions
1. Laminar, steady flow, radial port of radius R
2. Quiescent, infinite reservour of oxidizer
3. Species: fuel, oxidizer, products
4. Fuel and oxidizer react at φ = 1
5. Fast chemical kinetics (flame sheet approximation)
6. Fick’s law diffusion
7. Le= 1= α
D
8. Neglect thermal radiation and axial diffusion
9. Vertically oriented flame
Mass conservation
1 ∂ ( r ρv r ) ∂ ( ρv x )
+
=
0
∂r
∂x
r
Axial momentum
1 ∂ ( r ρv x v x ) 1 ∂ ( r ρv x v r ) 1 ∂
+
−
r
r
r ∂r
∂x
∂r
∂v x
( ρ∞ − ρ ) g
r µ ∂r =
Species conservation
1 ∂ ( r ρv xYi ) 1 ∂ ( r ρv r Yi ) 1 ∂
+
−
r
r
r ∂r
∂x
∂r
∂Yi
r
D
0
=
ρ
∂r
YPr =1 − YF − YOx
Energy conservation
(
)
(
)
∂
∂
∂
r ρv x ∫ c p dT +
r ρv r ∫ c p dT −
∂r
∂r
∂x
∂ c dT
r ρD ∫ p
∂r
=
0
Unknowns: v r , v x , T , YF , and YOx
Conserved scalar approach requires BCs only along flame
sheet, far from jet, and jet exit
A conserved scalar is any scalar quantity that is
conserved throughout the flow field
mass of material having it’s
f ≡
origin in the fuel stream
mass of mixture
1 kg fuel + ν kg oxidizer → (ν + 1 ) kg products
f =
1)
(
YF
kg fuel stuff kg fuel
kg mixture
kg fuel
1
YPr + ( 0 )
YOx
+
ν
+ 1 kg
product kg fuel stuff kg oxidizer
kg fuel stuff kg mixture
kg product
1
Y =
YF + fstYPr
ν + 1 Pr
f =
YF +
kg oxidizer kg mixture
Species conservation, conserved scalar approach
Replace two species conservation equations with one involving
the mixture fraction f , which is the fraction of material having its
origin in the fuel system
1 ∂ ( r ρv x f ) 1 ∂ ( r ρv r f ) 1 ∂
+
−
r
r
r ∂r
∂x
∂r
∂f
(0, x ) = 0 by symmetry
∂r
f ( ∞, x ) =
0
f ( r ≤ R, 0 ) =
1
f ( r > R, 0 ) =
0
Flame situates itself at f = fstoic
∂f
r
D
ρ
∂r
0
=
Energy conservation, conserved scalar approach
1 ∂ ( r ρv x h ) 1 ∂ ( r ρv r h ) 1 ∂
+
−
r
r
r ∂r
∂x
∂r
∂h
r
D
0
ρ
=
∂r
∂h
(0, x ) = 0 by symmetry
∂r
h ( ∞, x ) =
hOx
h ( r ≤ R, 0 ) =
hF
h ( r > R, 0 ) =
hOx
Continuity and axial momentum
No change in equations or BCs
Non – dimensional forms
x*
=
h − hOx ,∞
vx
vr
x
r
ρ
r*
v x*
v r*
h*
ρ*
=
=
=
=
=
ρe
R
R
ve
ve
hF ,e − hOx ,∞
Continuity
(
∂ ρ *v x*
) + 1 ∂ (r
*
ρ *v x* )
r*
∂x *
∂r *
Axial momentum
(
∂ r * ρ *v x* v x*
∂x *
) + ∂ (r
*
0
=
ρ *v x* v r* )
∂r *
∂
− *
∂r
µ * ∂v x* gR ρ∞
* *
r
=
−
ρ
r
*
2
ρev e R ∂r v e ρe
Mixture fraction
(
∂ r * ρ *v x* f
∂x *
) + ∂ (r
*
ρ *v r* f )
∂r *
−
∂
∂r *
ρ D * ∂f
0
=
r
*
ρev e R ∂r
Energy
(
∂ r * ρ *v x* h *
∂x *
) + ∂ (r
*
ρ *v r* h * )
∂r *
∂
− *
∂r
ρ D * ∂h *
0
=
r
*
v
R
ρ
r
∂
e e
Boundary conditions
( )
∂f
,
x
∞
=
( ) ∂r ( ∞, x ) =h ( ∞, x ) =0
v r* 0, x * = 0
v x*
*
*
*
*
∂v x*
∂f
∂h *
*
*
0, x =
0, x =
( ∞, x )= 0
∂r *
∂r *
∂r *
(
(
(r
)
(
) (
> 1, 0 ) = f ( r
)
) (
> 1, 0 ) = h ( r
)
> 1, 0 ) = 0
v x* r * ≤ 1, 0 = f r * ≤ 1, 0 = h * r * ≤ 1, 0 = 1
v x*
*
*
*
*
• Mixture fraction and standardized enthalpy equations are of the same
form, as are their boundary conditions. ∴ their solutions are identical.
µ
ν
1 and neglecting buoyancy yields
• Assuming Sc =
=
=
ρD D
for axial momentum, mixture fraction, and enthalpy
(
∂ r * ρ *v x* ζ
∂x *
) + ∂ (r
*
ρ *v r*ζ )
∂r *
−
∂ 1 * ∂ζ
0
r
=
∂r * Re ∂r *
ζ = f= v x* = h *
• ρ * and v x* are coupled through continuity
• ρ * and f (or h * ) are coupled through state relations
State relations
Need to relate density to mixture fraction or other conserved scalars
=
YF YF ( =
f)
YPr YPr=
( f ) YOx YOx=
( f ) T T=
(f ) ρ ρ (f )
Inside flame sheet ( fst < f ≤ 1)
=
YF
f − fst
1−f
=
YOx 0=
YPr
1 − fst
1 − fst
At flame
=
YF 0 =
YOx 0=
YPr 1
Outside flame
f
YF ==
0
YOx 1 − f
YPr =
fst
fst
Note: fst =
1
ν +1
Temperature state relation
Assume:
c=
c p=
c=
cp
,Ox
p,Pr
p,F
hf0,P = hf0,Ox = 0 and hf0,F = ∆hc
Enthalpy
h=
∑Y h
i
i
= YF ∆hc + c p (T − Tref )
YF ∆hc + c p (T − TOx ,∞ )
h − hOx ,∞
h
=
=
= f
hF ,e − hOx ,∞ ∆hc + c p (TF ,e − TOx ,∞ )
*
c p (TOx ,∞ − Tref ) and hF ,e =
∆hc + c p (TF ,e − Tref )
using hOx ,∞ =
Solve for T
T =−
( f YF )
∆hc
+ f (TF ,e − TOx ,∞ ) + TOx ,∞
cp
Substituting appropriate expressions for YF inside, outside, and on
the flame sheet
Inside ( fst < f ≤ 1)
fst ∆hc
T =T ( f ) =TF ,e f + TOx ,∞ +
(1 − f )
−
1
f
c
st
p
At the flame ( f = fst )
∆hc
+ TF ,e − TOx ,∞ + TOx ,∞
T = T ( f ) = fst
c
p
Outside the flame ( 0 ≤ f < fst )
∆h
T = T ( f ) = f c + TF ,e − TOx ,∞ + TOx ,∞
c
p
Roper
Constant density solution
Lf ≈
3 QF
8π DYF ,st
Variable - density approximate solution
µ = µref T
Tref
F
m
ρ∞
3
1
8π YF ,st µref ρref I ρ∞
ρ
f
ρ
I ∞ is a momentum integral tabulated in Table 9.2
ρf
The variable density solution predicts lengths about 2.4 times longer
than constant density solutions.
Lf ≈
All theories predict that flame lengths depend on
Volume flow rate and 1
YF ,st
independent of port diameter
Circular port
Lf ,th
QF
TF
=
4π D∞ ln 1 + 1
T∞
(
S
)
T∞
T
F
0.67
T
QF ∞
TF
Lf ,exp = 1330
ln 1 + 1
S
where S is the molar oxidizer-fuel ratio
Square port
(
)
T
QF ∞
TF
Lf ,th =
(
16D∞ Inverf 1 + 1
Lf ,exp = 1045
S
)
T
QF ∞
TF
(Inverf
Slot burners
see text for expressions
1+ 1
S
2
)
T∞
T
F
2
0.67
S=
1 − ψ pri
ψ pri + 1
Spure
ψ pri is the ratio of primary air to the
amount required at stoichiometric
S=
x+y
χO
2
4
Diluent effect
S=
χ dil
x+y
4
1
χO2
1
−
χ
dil
is the diluent mole fraction in the fuel stream
Study Turns Example 9.4
• How does flame height change for increase/decrease of primary aeration?
• How does flame height change for increased jet flow rate?
• How does replacing methane with propane affect the burner design
assuming the total burner power remains the same?
• What is the effect of aeration with oxygen rather than air?
Soot formation
...
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