Lecture 11- Diffusion Flames - Princeton University Copyright©2011%by%Moshe%Matalon.%This%...

6/26/11

Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 1

Lecture 11

Diffusion Flames

The candle flame is a classical example of a diffusion flame.

The flame reaches a steady state almost immediately after a match is broughtup close to the wick, and if the air in the room is sufficiently still, it does littleflickering. The flame basically remains constant in shape and size.

The flame has a distinct structure: a yellow luminous zone, which is responsiblefor most of the light provided by the candle, blue regions on the sides of theflame and near the wick, and a dark cone that starts below the curl of the wick.

The wax first melts due to the heat radiated from the flame to the candle. Asa result a pool of melted wax forms at the top of the candle, and liquid wax isdrawn up the wick by capillary action.

The candle flame

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At a critical temperature the wax molecules break into fragments, which freescarbon atoms that agglomerate into soot particles as they enter the yellow tongueof the flame. There the solid carbon particles become incandescent, which pro-duces the yellowish light.

Chemical reaction occurs only when fuel and oxygen molecules mix and if thetemperature is high enough.

The chemical reaction is extremely fast and occurs on a much shorter timescale than diffusion. Diffusion is rate controlling, which is the reason thatnon-premixed combustion is also referred to as “diffusion flames” .

Most of the burning proceeds in the blue reaction zone and upward on thesurface of the flame, which is where the highest temperatures are reached.

The liquid wax on the wick vaporizes by the heat transported to the wick fromthe surrounding reaction zone (primarily by radiation) and the wax vapor dif-fuses outwards and make contact with the oxygen diffusing in from the sur-rounding air.

1g µg

As a result of natural convection, the heat and combustion products flow up-ward and fresh oxygen from the ambient air is carried toward the flame. Thisgives the flame its elongated tear drop shape.

A candle can burn in µ-gravity conditions with the oxygen supplied solely bydiffusion (NASA TM 106858), and the flame shape is nearly a hemisphere.

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Oxidizer

Fuel

Counterflow diffusion flame

L

d Fuel

Jet diffusion flame

Oxidizer

Liquid fuel drop and Burner-generated

spherical diffusion flame

Fuel

Oxidizer

Fuel vapor

Oxidizer

drop

Diffusion flame configurations commonly studied

The steady planar unstrained flame

P. Monkewitz, EPFL

Diffusion flame

Fuel and Oxidizer are supplied from opposite ends where conditions are main-

tained uniform.

Fuel is supplied in the stream flowing from the bottom of the chamber at a con-stant speed, and oxidizer diffuses inwards from the top boundary. The reverseproblem, in which oxidizer is supplied in the stream and fuel diffuses against it,is readily available by exchanging the roles played by the two reactants.

x

xf

Fuel

Oxidizer

0

m̃

Products

Products

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The chemical reaction is described by a one-step irreversible step

νF F + νO O → Products

The planar steady unstrained flame

for simplicity the temperatures are takento be the same on both supply ends

x = −∞ : T = T̃0, YF = YF 0, YO = 0

x = 0 : T = T̃0, YF = 0, YO = YO1

A steady one-dimensional planar flame in the unlimited space (−∞ < x < ∞) isnot possible. The only solution of the reaction-diffusion equation for YO behindthe flame is a constant, so that the oxidizer cannot be consumed at the reactionzone, as required.

The mass flux m̃ is constant and prescribed.

x

xf

Fuel

Oxidizer

m̃

mdT

dx− d2T

dx2= ω

mdYF

dx− Le−1

F

d2YF

dx2= −ω

mdYO

dx− Le−1

O

d2YO

dx2= −νω

ν = νOWO/νFWF β =E/Rq/cp

Nondimensionalization

The incoming velocity U is used as a unit speed and the Dth/U as a unitlength

The temperature is scaled with respect to Q/cpνFWF , the heat release perunit mass of fuel consumed.

ω = D ρ2YFYOeβ/Ta−β/T

D =(Dth/U2)

[νF (ρ0/WO)B]−1 eE/RT̃a

Here T̃a is the adiabatic temperature (i.e., associated with complete combus-

tion), to be determined, and ˜ denotes dimensional quantities.

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Infinitely-fast chemistry

The Burke-Schumann limit of complete combustion

D → ∞ ⇒ YFYO ∼ 0

D =flow time

reaction time→ ∞

the chemistry is infinitely fast

YFYO ∼ 0 can be satisfied in a diffusion flame if YF vanishes on one side of thereaction sheet, where the oxidizer is being supplied (the oxidizer region) and YO

vanishes on the other side, where the fuel is being supplied (the fuel region).

ω ∼ 0 for x ≶ xf

where xf is the location of the reaction sheet

ω = D ρ2YFYOeβ/Ta−β/T

All variables must be continuous at x = xf , i.e.,

[YF ] = [YO] = [T ] = 0

YF ≡ 0 for x > xf

YO ≡ 0 for x < xf

and in addition

where [ · ] denotes the jump across xf

The jump in the gradients is obtained by adding any two of the equations, inorder to eliminate ω, then integrating across the reaction sheet, from x−

f to x+f .

md

dx(YF − ν−1YO)−

d2

dx2(Le−1

FYF − ν−1Le−1

OYO) = 0

md

dx(T + ν−1YO)−

d2

dx2(T + ν−1Le−1

OYO) = 0

�dT

dx

�= − 1

LeF

�dYF

dx

�= − 1

νLeO

�dYO

dx

�

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�dT

dx

�= − 1

LeF

�dYF

dx

�= − 1

νLeO

�dYO

dx

�[YF ] = [YO] = [T ] = 0Across x = xf :

and

In addition to the BCs, there are 7 conditions at the reaction sheet. Being a6th order system, we have enough conditions for determining the solution andthe reaction sheet location xf .

mdT

dx− d2T

dx2= 0

mdYF

dx− Le−1

F

d2YF

dx2= 0

mdYO

dx− Le−1

O

d2YO

dx2= 0

YF ≡ 0 for x > xf , YO ≡ 0 for x < xf

x = −∞ : T = T0, YF = YF 0, YO = 0

x = 0 : T = T0, YF = 0, YO = YO1

initial mixture strength

It is “like” an equivalence ratio based onthe mass fractions at the BCs.

1

LeF

�dYF

dx

�=

1

νLeO

�dYO

dx

�

φ =YF 0

/νFWF

YO1/νOWO

YF =

YF 0

�1− emLeF (x−xf )

�−∞ < x < xf ,

0 xf < x < 0

YO =

0 −∞ < x < xf ,

YO1

�1− 1− emLeOx

1− emLeOx

�xf < x < 0

xf = − 1

mLeO

ln(1 + φ−1)

Note that the location of the reaction sheet depends on the mixture strengthφ and only on the diffusivity of the oxidizer, which has to diffuse against thestream.

where continuity at xf has been satisfied.

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�dT

dx

�= − 1

LeF

�dYF

dx

�

stoichiometric temperature

The flame temperature and flame sheet location in dimensional form for LeO = 1

xf = −λ/cpm̃

ln

�1 +

YO1/νOWO

YF 0/νFWF

�

Ta = T0 + YF 0− YF 0

�1 + φ−1

�−1/LeO

For LeO = 1, Ta = T0 +YF 0

1 + φ

T̃a = T̃0 +Q/cp

νFWF /YF 0+ νOWO/YO1

T =

T0 + (Ta − T0)em(x−xf ) −∞ < x < xf ,

T0 + (Ta − T0)1− emx

1− emxfxf < x < 0

where continuity at xf has been applied and the common value denoted Ta.

which peaks at stoichiometry

x

YFYO

T

xf

For the solution to be valid, the discontinuity in slopes must be smoothed outin a thin region near xf , i.e., in the reaction zone. We shall discuss this later.

Hence, for φ < 1, or lean systems, γ < 0 and more heat is conducted to the fuel

side, whereas for φ > 1, or rich systems γ > 0 and more heat is transported to

the oxidizer side.

Note that ∆ = −[dT/dx] = QYF 0/cpνFWF which is the total heat released is

not equally conducted to both sides. The excess conducted to one side is

for LeO = 1γ = − 1

∆

�(dT/dx)x−

f− |dT/dx|x+

f

�=

φ− 1

φ+ 1

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The conditions across the reaction zone can be easily generalized to an arbitrarymulti-dimensional surface

[YF ] = [YO] = [T ] = 0

1

Q

�λ∂T

∂n

�= −

�ρD

F

νFWF

∂YF

∂n

�= −

�ρD

O

νOWO

∂YO

∂n

�

in dimensional form

• The temperature and concentrations are continuous at the reaction sheet

• The fuel and oxidizer are completely consumed at the sheet

• The net fluxes to the reaction sheet are diffusive (diffusive flux overcomethe convective flux) and are in stoichiometric proportions

n

Reaction sheetF(x, t) = 0

Fuel

Oxidizer

with [ · ] denoting the jump across the reaction sheet F (x, t) = 0

YO|n=0 = 0YF |n=0 = 0

The Burke-Schumann Problem Burke & Schumann, Ind. Eng. Chem. 1928

Assumptions:

•  The flow is in the axial direction everywhere; the velocity field is thus given by v = u ez

•  The total mass flux in the axial direction is constant everywhere; this is practically a constant-density assumption intended to uncouple the hydrodynamic from the transport equations

•  Axial diffusion is negligible compared to radial diffusion (this can be justified a-posteriori if the flame height is large compared with b ). This assumption implies that

∇2 =∂2

∂z2+

1

r

∂

∂r

�r∂

∂r

�≈ 1

r

∂

∂r

�r∂

∂r

�

z

u

F O O

r a

b

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z

u

F O O

r a

b

•  In studying this problem Burke and Schumann introduced the ingenious limit of infinitely fast chemistry, we have discussed earlier.

•  The Lewis numbers were assumed equal to one. This assumption is not essential but leads to significant simplification in the mathematical manipulations.

•  Gravity effects have been neglected. Indeed, buoyancy forces may cause distortion of the flame. The controlling parameter is the Froude number

Fr = u2/a g representing the ratio of inertial and buoyant forces. Flame heights are controlled by buoyancy when Fr << 1. •  The boundary conditions are:

YF = YF 0, YO = 0 at z = 0, 0 < r < a

YF = 0, YO = YO0at z = 0, a < r < b

∂YO

∂r=

∂YF

∂r= 0 at r = b, for z > 0

We write dimensionless equations using b as a unit of the radial distance, and(u/D)b2 a unit of the axial distance, and normalize both YF and Yo with YF 0

.

∂YF

∂z− 1

r

∂

∂r

�r∂YF

∂r

�= −ω

∂YO

∂z− 1

r

∂

∂r

�r∂YO

∂r

�= −νω

where the reaction term has been properly nondimensionalized, and ν = νOWO/νFWF .

YF = 1, YO = 0 at z = 0, 0 < r < c

YF = 0, YO = YO0/YF 0

at z = 0, c < r < 1

∂YO

∂r=

∂YF

∂r= 0 at r = 1, for z > 0

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We will make use of the “coupling function” ψ = YF − ν−1YO before invokingthe fast chemistry limit.

Clearly∂ψ

∂z− 1

r

∂

∂r

�r∂ψ

∂r

�= 0

is valid everywhere, and being a linear equation can be solved by standardtechniques.

ψ = 1 at z = 0, 0 < r < c

ψ = −φ−1 at z = 0, c < r < 1

∂ψ/∂r = 0 at r = 1, for z > 0

The boundary conditions are

where φ =YF 0

/νFWF

YO0/νOWO

is the initial mixture strength.

The solution, using separation of variables, is expressed as a Bessel-Fourier series

ψ(r, z) = A0 +∞�

n=1

AnJ0(λnr) eλ2nz

A0 = (1 + φ−1)c2 − φ−1

An =2(1 + φ−1)c

λnJ20 (λn)

J1(cλn)

where λn are the roots of J1.

The mass fractions can readily available. On the fuel side, YF = ψ, YO = 0,

and on the oxidizer side YO = −νψ, YF = 0.

In the fast chemistry limit, YF = YO = 0 along the reaction sheet, so that ψ = 0,or

A0 +∞�

n=1

AnJ0(λnr) eλ2nz = 0

determines the reaction sheet surface.

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z

u

F O O

r a

b

overventilated underventilated

F (r, z) ≡ A0 +∞�

n=1

AnJ0(λnr) eλ2nz = 0

H = F (0, z) = 0 for the overventilated case

H = F (1, z) = 0 for the underventilated case

Flame height

For example, if we approximate the infinite sumby taking only one term (n = 1),

A0 +A1J0(λ1r) eλ21z ≈ 0

The flame can either extend to the walls of the duct - the underventilated case(that results for larger values of φ), or converges to the axis - the overventilatedcase (that results for lower values of φ).

H ≈ 1

λ21

ln

�2(1 + φ)c J1(cλ1)

[1− (1 + φ)c2]λ1J0(cλ1)

�

the height of an underventilated flame.

The counterflow diffusion flame

A planar diffusion flame can be established with a reaction-sheet located at adistance x = xf , on one or the other side of the stagnation plane. The flame,however, is strained.

x =xf

YF =YF-!

YO = 0

T=T!

y

YF =0

YO =YO!

T=T!

YF = YF0YF = 0

YO = YO0YO = 0T = T0T = T0

A stream of gas containing fuel with mass fraction YF−∞ impinges against an-other stream of gas containing an oxidizer with mass fraction YO∞ , forming astagnation-point at the origin

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The incoming flow is given by

{u, v} = {−2εx , 2εy}

with ε the strain rate (with units 1/s).

This is the potential flow solution near a stagnation point flow,valid if the velocity of the incoming streams is sufficiently large.

x =xf

YF =YF-!

YO = 0

T=T!

y

YF =0

YO =YO!

T=T!

x

For simplicity, we will assume that the flow remains unperturbed by the presenceof the flame, which is formally obtained by assuming that the density of themixture remains constant.

Density variations can be taken into accountwithout much difficulty by introducing adensity-weighted axial coordinate,

x̂ =

� x

0ρdx

Viscous effect may also be taken into consideration

by replacing u = −2εx with u = −2εf(x),a function obtained numerically as the solution of

a nonlinear ODE.

To write dimensionless equations, we chose (Dth/ε)1/2 as a unit of length, and

normalize the temperature with respect to the heat release (per unit mass offuel consumed) Q/cpνFWF .

2xdYF

dx+ Le−1

F

d2YF

dx2= ω

2xdYO

dx+ Le−1

O

d2YO

dx2= νω

2xdT

dx+

d2T

dx2= −ω,

YF = YF0, YO = 0, T = T0 as x → −∞

YF = 0, YO = YO0, T = T0 as x → ∞.

ω = DYFYOeβ/Ta−β/T

ν = νOWO/νFWF β =E/Rq/cp

D =ε−1eE/RT̃a

[νF (ρ0/WO)B]−1

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Frozen limit, D → 0

2xdYF

dx+ Le−1

F

d2YF

dx2= 0

2xdYO

dx+ Le−1

O

d2YO

dx2= 0

2xdT

dx+

d2T

dx2= 0

YF = YF0, YO = 0, T = T0 as x → −∞

YF = 0, YO = YO0, T = T0 as x → ∞.

The solution illustrates pure mixing between the two streams

computed for equal Lewis numbers

x

T = T∞

YO = 12YO0

�1 + erf(

�Le

Ox)�

YF = 12YF 0

�1− erf(

�LeF x)

�

Infinitely fast chemistry, D → ∞

YF =

YF 0

�1−

1 + erf(�LeF x)

1 + erf(�

LeF xf )

�(x < xf )

0 (x > xf )

YO =

0 (x < xf )

YO0

�1−

erf(�Le

Ox)− 1

erf(�Le

Oxf )− 1

�(x > xf )

1 + erf(�Le

Fxf )

1− erf(�

LeOxf )

− φ

�Le

O

LeF

e(LeO−Le

F)x2

f = 0

where φ =YF 0

/νFWF

YO0/νOWO

is the initial mixture strength.

the location of the reaction sheet is determined from

solving the equations and using the jump conditions across xf yields

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If we assume that LeF= Le

O= Le; i.e., retaining differential diffusion between

heat and mass while ignoring preferential diffusion between species, an explicitexpression results for xf

erf(√Lexf ) =

φ− 1

φ+ 1

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3

-1

-0.5

0.5

1

xf

the variations caused bychanging Le are relatively small

If fuel and oxidizer are supplied in stoichiometric proportions, φ = 1 the reactionsheet will locate along the stagnation surface, i.e., xf = 0. It will be located onthe fuel side (xf < 0) for a “lean” system and on the oxidizer side (xf > 0) fora “rich” system.

Le = 1 ν = 4

0.3

0.6

0.8

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We can also solve for the temperature

T =

T0 + (Tf − T0)1 + erf(x)

1 + erf(xf )(x < xf )

T0 + (Tf − T0)1− erf(x)

1− erf(xf )(x > xf )

and with the remaining jump condition of the temperature gradients, determinethe flame temperature Tf . The expression is cumbersome, but for Le = 1simplifies to

Tf = To +YF 0

1 + φ

referred to, earlier, as the adiabatic temperature.

Flam

e Te

mpe

ratu

re

Damköhler number D

Tf

Dext

BS limit Ta

Dign

T0

Recall that D ∼ ε−1, so that increasing the strain rate results in decreasing D.

Frozen limit

Lecture 11- Diffusion Flames - Princeton University Copyright©2011%by%Moshe%Matalon.%This%...

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