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
Moshe Matalon University of Illinois at Urbana-‐Champaign 2
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 3
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 4
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 5
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 6
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 7
1
1
s
t
r
o
n
g
d
e
t
o
n
a
t
i
o
n
w
e
a
k
d
e
t
o
n
a
t
i
o
n
w
e
a
k
d
e
fl
a
g
r
a
t
i
o
n
s
t
r
o
n
g
d
e
fl
a
g
r
a
t
i
o
n
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 8
� (� + 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
Moshe Matalon University of Illinois at Urbana-‐Champaign 9
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 10
�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
Moshe Matalon University of Illinois at Urbana-‐Champaign 11
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
Moshe Matalon University of Illinois at Urbana-‐Champaign 12
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