notes lecture 4 - princeton university

Spring 2013

Moshe Matalon University of Illinois at Urbana-‐Champaign 1

Lecture 4 Deflagrations and Detonations

Moshe Matalon

plane, steady, one-dimensional flow involving exothermic chemical reactions in

which the properties become uniform as |x| � ⇥.

One-dimensional flow

�v0

u0 =0

p1 , ⇢1

T1 , Yi1

p0 , ⇢0

T0 , Yi0

in the laboratory frame

heat conduction

species di↵usion

viscous dissipation

chemical reactions

u1

The objective is to determine (i) the state of the burned gas and (ii) the wave

speed, or propagation speed Vwave = �v0.

Moshe Matalon

Spring 2013


p0�0T0

Yi0

p1�1T1Yi1

v0 v1

coordinate attached to a wave propagating to the left at speed v0

fluid particle velocity u = v + Vwave

here Vwave = �v0

�v0

u0 =0 u1=v1�v0

p1 , ⇢1

T1 , Yi1

p0 , ⇢0

T0 , Yi0

in the laboratory frame

Moshe Matalon

Steady, one dimensional flow

h =NX

i=1

Yi

ho

i

+

ZT

T

o

cp

dT

p = �RT/W

d

dx

(⇢v) = 0

d

dx

(⇢v2) = �dp

dx

+d

dx

✓4

3µ

dv

dx

◆

d

dx

(⇢vh) = v

dp

dx

+4

3µ

✓dv

dx

◆2� dq

dx

d

dx

(⇢vYi) +d

dx

(⇢YiVi) = !i i = 1, 2, . . . , N

Moshe Matalon

Spring 2013


p = ⇢RT/W

d

dx

(⇢vYi + ⇢YiVi) = !i

d

dx

✓⇢v(h+

1

2v

2)� 4

3µv

dv

dx

+ q

◆= 0

d

dx

✓p+ ⇢v

2 � 4

3µ

dv

dx

◆= 0

d

dx

(⇢v) = 0

p0/⇢0T0 = p1/⇢1T1 = ��1� cp

Note that cp � cv = R/W implies

R/W = ��1� cp

where � = cp/cv

⇢0v0 = ⇢1v1

p0 + ⇢0v20 = p1 + ⇢1v21

h0 +12v

20 = h1 + 1

2v21

!i0

!= 0 , !i1 = 0 i = 1, 2, . . . , N

Moshe Matalon

h1 � h0 =NX

i=1

(Yi1�Y

i0)ho

i

+ cp

(T1 � T0)

= �Q+ cp

(T1 � T0)

assume cpi⌘ cp and Wi ⌘ W for all i (or take some average value)

Q here is the heat of combustion per unit mass of combustible mixture

h =NX

i=1

Yi

ho

i

+

ZT

T

o

cp

dT

⇢0v0 = ⇢1v1

p0 + ⇢0v20 = p1 + ⇢1v21

cpT0 +12v

20 = cpT1 + 1

2v21 �Q

p0/⇢0T0 = p1/⇢1T1 = ��1� cp

eliminate the temperature, using the equation of state,

yields two equations for p1 and v1

Moshe Matalon

Spring 2013


after some manipulations (and use of the equation of state to eliminate T )

let m = �0v0 = �1v1

they determine the relation between the end and initial states

Rayleigh Line

Hugoniot curverelation between thermodynamic variables only

p1 � p01/⇢1 � 1/⇢0

= �m2

�

� � 1

✓p1

⇢1

� p0⇢0

◆� 1

2

✓1

⇢1

+1

⇢0

◆(p1 � p0) = Q

Moshe Matalon

upstream state corresponds to p = v = 1

(also the specific volume)

possible downstream states are intersections of the Hugoniot and Rayleigh line

Dimensionless variables

p = p1/p0 , ⇢ = ⇢1/⇢0 , v = v1/v0

µ = m2/p0⇢0 , q = Q/(p0/⇢0) = Q/(�a20)

provide p, v (and consequently ⇢, T ) for a given µ

p� 1

v � 1= �µ

p =2q + �+1

��1 � v�+1��1v � 1

Moshe Matalon

Spring 2013


µ =m2

⇢0p0=

⇢20v20

⇢0p0=

v20p0/⇢0

=�v20a20

= �M20

µ =m2

⇢0p0=

⇢21v21⇢0p0

=p1p0

⇢1⇢0

v21p1/⇢1

=p1p0

⇢1⇢0

�v21a21

=�p

vM2

1

µ =�p

vM2

1

M21 =

v

pM2

0

a relation between upstream and downstream Mach numbers

where M0 = v0/a0 is the upstream Mach number

µ = �M20

where M1 = v1/a1 is the downstream Mach number

Moshe Matalon

1

1

q

q = 0

Hugoniot curves

Hugoniot is a rectangular hyperbola that asymptotes tov = (� � 1)/(� + 1) as p ! 1p = �(� � 1)/(� + 1) as v ! 1

solutions restricted to

��1�+1 v 2q + �+1

��1

0 p 1

p

v = 1/⇢

Moshe Matalon

Spring 2013


1

1

solutions further restricted to

��1�+1 v < 1 and 1+(� � 1)q p 1

1+

��1� q < v 2q + �+1

��1 and 0 p < 1

Detonations

Deflagrations

Ralyleigh line always goes through (1,1) and has a negative slope

v = 1/⇢

p

Moshe Matalon

1

1

the other solution, for a subsonic incident wave, would correspond to a

rarefaction wave, but is not acceptable because it violates the 2

ndlaw.

there is only one solution for each supersonic incident velocity

it is a shock wave, with p, � and T increasing behind the shock,

the gas slows down (v < 1) behind the shock and is subsonic.

q = 0

v = 1/⇢

p

shock

wave

Moshe Matalon

Spring 2013


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CJ

CJ

q > 0

v = 1/⇢

p

Moshe Matalon

Deflagrations

there is a maximum value of µ and, as a consequence, a maximum wave speed

corresponding to Chapman-Jouguet (CJ) deflagration;

for µ < µCJ there are two solutions, weak and strong deflagrations

Detonations

there is a minimum value of µ and, as a consequence, a minimum wave speed

corresponding to Chapman-Jouguet (CJ) detonation

for µ > µCJ there are two solutions, strong and weak detonations

passing through a detonation wave

v < 1 the gas slows down

⇢ > 1 the gas is compressed

p > 1 the pressure increases

passing through a deflagration wave

v > 1 the gas speeds up

⇢ < 1 the gas expands

p < 1 the pressure decreases

Moshe Matalon

Spring 2013


� (� + 1)p+ (� � 1)

(� + 1)v � (� � 1)=

p� 1

v � 1

[(� + 1)p+ (� � 1)][v � 1] + [p� 1][(� + 1)v � (� � 1)] = 0

[2(� + 1)v � 2�)]p� 2v = 0

Chapman-Jouget Waves

p� 1

v � 1= �µ , p =

2q + �+1��1 � v

�+1��1v � 1

Rayleigh line Hugoniot

equate the slopes

dp

dv

✓� + 1

� � 1v � 1

◆+ p

✓� + 1

� � 1

◆= �1

dp

dv

��R

= �µ =p� 1

v � 1

dp

dv

��H

= � (� + 1)p+ (� � 1)

(� + 1)v � (� � 1)

Moshe Matalon

p =v

(� + 1)v � � )v

�p

1� p

v � 1= 1 µ =

�p

v

Since µ =�p

vM2

1 ) M21 = 1

For Chapman-Jouguet (CJ) waves the downstream gas velocity relative to the

wave (for both, detonations and deflagrations) is sonic; i.e. (M1)CJ = 1

)

p =v

(� + 1)v � �

p =2q + �+1

��1 � v�+1��1v � 1

) � + 1

� � 1v2 � � + 1

�

2q +

2�

� � 1

�v + 2q +

� + 1

� � 1= 0

and a similar equations for p.

Then µ is calculated from µ�p/v = 1.

Moshe Matalon

Spring 2013


One findp± = 1 + q(� � 1)(1±�)

v± = 1 + q� � 1

�(1⌥�)

µ± = � + q(�2 � 1)(1±�)

where � = [1 + 2�/q(�2 � 1)]1/2, and +/� stand for detonation/deflagration,respectively.

detonations are supersonic waves (1 < M0 < 1) and

deflagrations are subsonic waves ( 0 < M0 < 1).

Using µ = �M20 we have

M0± =

s

1 +q(�2 � 1)

2�±

sq(�2 � 1)

2�

which implies that

Moshe Matalon

��dp

dv

��H

>

��dp

dv

��R

��dp

dv

��H

<

��dp

dv

��R

M21 < 1

M21 > 1

Note also that

for strong detonations and weak deflagrations:

for weak detonations and strong deflagrations:

Moshe Matalon

Spring 2013


�v0 �� u = v1 � v0u = 0v1�!v0 �!

as seen by an observer moving with the wave

as seen in the laboratory frame

u = v + Vwave is the fluid particle velocity

u < 0 for detonations and u > 0 for deflagrations

t

x

f

r

o

n

t

particle path

Unburned

t

x

f

r

o

n

t

particle path

Unburned

Detonations

Deflagrations

the flow follows the detonation wave, but

cannot catch up with it (since v1 < a1).

Moshe Matalon

Comment

The Rankine-Hugoniot analysis provides, for given conditions, the state of thegas downstream, but does not determine (except for the CJ waves) the wavespeed v0.

The possible existence of a final state based on this analysis does not guar-antee the existence of the solution. One needs to ensure that the initial andfinal states are connected smoothly by a solutions of the di�erential equationsand that this solution is stable.

Moshe Matalon

Spring 2013


Deflagrations

• Expansion waves that propagate subsonically (0 < M0 < 1)

• The burned gas behind the front expands and accelerates away from the

front

• In a frame attached to the wave, the downstream flow beyond CJ waves

is sonic (M1 = 1); for weak deflagrations M1 < 1 and for for strong

deflagrations M1 > 1

• Wave structure considerations forbid strong deflagrations (there is no

structure that connects the burned and unburned states)

• There is a unique solution for the (weak) deflagration, giving a definite

wave speed for each gas mixture

• Weak deflagrations (flames) travel typically at 10 - 100 cm/s and

p1/p0 ⇠ 1, T1/T0 ⇠ 6, ⇢1/⇢0 ⇠ 1/6

Moshe Matalon

Detonations

• Compression waves that propagate supersonically (1 < M0 < 1)

• The wave retards the burned gas that compresses behind the front

• In a frame attached to the wave the downstream flow beyond CJ waves is

sonic (M1 = 1); for weak detonations M1 > 1 and for strong detonations

M1 < 1.

• A strong detonation can be produced by driving a piston at an appropriate

speed into the mixture

• There is only one wave speed at which a weak detonation can propagate

(but weak detonations are seldom observed)

• Self-sustained detonations are CJ detonations; in a hydrogen/air mixture

v0 ⇠ 2000� 3000m/s, and

p1/p0 ⇠ 20, T1/T0 ⇠ 10, ⇢1/⇢0 ⇠ 2

Moshe Matalon

Spring 2013


Deflagrations (weak) are subsonic waves, so that disturbances behind the wave

can propagate ahead and a↵ect the state of the unburned gas before arrival of

the wave (this depends, of course, on the rear end boundary conditions, if open

or close, etc. . . ).

The expansion of the products can therefore cause a displacement of the re-

actants so that the wave moves faster than the flame speed (burning velocity

relative to the quiescent mixture).

In unconfined obstacle-free systems the mode of propagation is, generally speak-

ing, controlled by the ignition conditions; deflagrations are initiated by a mild

energy discharge (i.e, by a spark) while detonations are provoked by shock waves

via localized explosion.

Because of intrinsic instabilities, turbulence, obstacles, etc. . . the deflagra-

tion wave may accelerate and turn into a detonation, a process known as

Deflagration to Detonation Transition (DDT).

Moshe Matalon

notes lecture 4 - princeton university

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