16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18: Solid Propellants: Other Topics
Combustion of Solid Propellants
For a general discussion, read Sutton, Chapter 13. A detailed model of combustion of
composite propellants is presented next.
Combustion of Composite Propellant
(Ref: Guy Lengellé, Jean-Robert Duterque, Jean-Claude Godon, Jean-Francois
Trubert, ONERA, “Solid Propellant Steady Combustion – Physical Aspects”. In
AGARD-LS-180-Combustion of Solid Proplellants,1991 (TL507.N867, no. 180))
Composite propellants are heterogenous mixtures of oxidizer grains and
powdered aluminum fuel, both embedded in a rubber-like binder, which is also a fuel.
The most common oxidizer by far is Ammonium Perchlorate (AP), (CℓO4NH4), a
crystalline substance with ρ = 1.95 g/cm3, cp = 0.31 cal/g/K, thermal diffusivity =
dp = 2.5x10−3 − 4.55x10−6 T 0C cm2/sec, and an estimated m.p of 835K. AP is ground
( )
to sizes from a few to around 100 µm . The finer grades are dangerous, so grinding is
done just prior to fabrication. AP has M = 116.5 g/mole and 55% by mass is oxygen.
The aluminum is also ground to similar sizes at the last minute. Al is a very
exothermic fuel, producing Aℓ2O3 which is liquid at the flame temperature ( ≈ 3500K),
and condenses later to a solid.
ρp
p
The binder is often polybutadiene (synthetic rubber), either Carboxyl
Terminated (CTPB) or Hydroxyl Terminated (HTPB). The composition of CTPB is
cal / sec
C7H11.24O0.2 , with thermal conductivity λ = 3.6x10−4
,
= 0.97 g / cm3 ,
(cm)(K )
cp = 0.39 cal / g / K
Best performance is obtained with very high percentage of AP, although
mechanical properties require a minimum of binder, and AP concentration ranges
from ≈ 70% by mass when there is Aℓ ( ≈ 16%), the balance being binder (14%), to
about 80%-85%, with no Aℓ (as in “smoke-less” compositions), the balance then
being all binder.
Overview of Combustion Mechanism – The burning of AP-binder propellants (no Aℓ) is
a complex series of phenomena, and the detailed geometry of the grains does matter
(size, particularly). At P ≥ 20 atm, AP itself can deflagrate exothermically, and it
decomposes partly in a thin liquid layer on the surface of a grain, partly in an “AP
flame” a few µm above it. The heat of decomposition raises T to ≈ 1205K by itself;
heat from the outer flame (more below) can raise the AP flame temperature well
above this, however.
Around the AP grains, the heat from the main flame decomposes the binder,
which generates a mixture of short-chain hydrocarbons, while absorbing about 360
cal/g, plus the energy to heat it to the surface temperature Ts ≅ 1000K-1100K.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 1 of 15
The O2-rich gas generated in the AP flame co-flows outwards with the binder
decomposition products, with interdiffusion along the way. This is a “diffusion flame”,
and the final combustion takes place in it, raising T to about 3540K.
When there is Aluminum, the Aℓ Particles
are ejected when the binder holding them
recedes; they then burn at several
hundred µm from it; during this burning
they agglomerate to several tens of
µm (they are liquid, m.p=930K, but
remain “encased” in Aℓ2O3 until this shell
breaks at 2300K, then the Aℓ spews ≈
1 µm microparticles which burn quickly)
Diffusion flame
(Main flame ~
3450 k)
AP flame
(1200-1600 k)
AP
liquid
binder
solid
t=0
t = 2x10-4s
t = 4x10-4s
t = 6x10-4s
t = 8x10-4s
10-3sec
From the surface regression point of view, Aℓ “burns” instantly, as its particles are
ejected.
Overall Burn Rate from Burn Rates of Constituents
Let υ p be the mean surface regression speed (cm/sec), and υ AP , υb the
corresponding rates for the AP and the binder individually and in isolation. Although
the geometry is more complex, we can idealize the propellant as a layered medium,
with alternative thicknesses δ AP , δ b . The time to burn through both δ AP and δ b is
δ AP + δ b
δ
δ
= t AP + tb = AP + b
υp
υ AP υb
t =
Calling ξ AP =
δ AP
δ AP + δ b
1
υp
=
, ξ b = 1 − ξ AP =
ξ AP
υ AP
+
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
ξb
υb
δb
δ AP + δ b
the volume fractions of the constituents
(1)
Lecture 17-18
Page 2 of 15
The mass flux (g/cm2/sec) burnt is
m
•
p
= ρp υ p
(2)
where ρp is the mean density
ρp =
mAP + mb
mAP mb
+
ρAP
1
ρp
ρb
=
α AP α b
+
ρAP
ρb
(3)
where α AP is the mass fraction of AP, and α b = 1 − α AP is that of the binder.
Note: ρAP = 1.95 g/cm3, ρb = 0.91 g/cm2
The mass fluxes of the individual constituents are, similarly,
AP
= ρAP υ AP
•
;
m
m
•
b
= ρb υb
(4)
From (1) and (2),
ρ
ρAP
ξb b
ρ
ρ
1
1
p
p
=
=
+
ρpυ p •
ρAP υ AP
ρb υb
ξ AP
m
p
and since ξ i
ρi ⎛ Vi ⎞ ⎛ Mi
=
⎜
ρp ⎜⎝ V ⎟⎠ ⎝ Vi
p
AP
AP
AP
b
AP
Al
b
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
αb
+•
(5)
m
α AP
=•
m
m
1
•
⎞ M Mi
=
= αi
⎟/
M
⎠ V
b
For aluminum loaded propellants, we noted that
the Aℓ particles are ejected when the binder
holding them burns through. Looking at the
simplified geometry below, it can be seen that the
mean burning speed υ p would be the same if the
binder were really “filling in” for the Aℓ.
Lecture 17-18
Page 3 of 15
We then get the approximate expression
1
υp
=
ξ AP ξ b + ξ AA
+
,
υ AP
υb
place of α b . The only difference is that the mean density ρp in
•
m
identical to the Eq. (1), even with Aℓ particles. Eq. (5) also follows, with 1 − α AP in
p
= ρp υ p is modified
by the Aℓ
⎛1
α
α
α ⎞
= AP + b + AA ⎟
⎜⎜
ρAP
ρb ρAI ⎟⎠
⎝ ρp
Separate Burning of AP
Heat penetrates into the receding AP particle to a small depth only. This can be seen
from the heat balance written in the receding frame:
x
Surface
fixed
Ts
AP moving
υAP
T = To
ρAP υ AP c AP
dT
d ⎛
dT ⎞
−
=0
λAP
⎜
dx dx ⎝
dx ⎟⎠
υ AP T − d AP
d (T − T0 )
dT
= υ AP T0 ; d AP
= υ AP (T − T0 )
dx
dx
υ AP
T − T0 = c e dAP
x
Define d =
λ
(heat diff., cm2/sec)
ρc
⎛ dAP ⎞
→
x / ⎜⎜
⎟
T − T0
υ ⎟
= e ⎝ AP ⎠
Ts − T0
Note: dp ≅ 1.2 x 10−3 cm2 / s, c ≅ 0.31 cal / g / K .
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 4 of 15
This shows an exponential temperature decay into the solid, with a
⎛d ⎞
characteristic thermal thickness x* = ⎜ ⎟ . Assuming υ = 1 cm / s,
⎝ υ ⎠ AP
−3
2
*
dAP ≅ 1.2 × 10 cm / sec , this gives x = 1.2 × 10−3 cm = 12 µ m .
The top few microns of the AP are molten when its temperature exceeds
TAP ≅ 835 K (which may not happen if the pressure is so low, under ≈ 20atm, that the
flame-surface distance is too large to provide sufficient heating to it. Note this rate is
⎛ dT ⎞
qs = −λAP ⎜
= − λAP / x * (Ts − T0 ) = − ρAP υ AP c AP (Ts − T0 ) , so
⎟
⎝ dx ⎠o
(
qs
MIN
)
= ( ρ υ c ) AP ( 835 − 298 ) typically.
In this molten layer, about 70% of the AP undergoes complete decomposition
to final (onydizing) gaseous products, according to
NH4CℓO4 → 0.285N2 + 0.12N2O+ 0.23NO + 1.62H2O + 0.76HCℓ + 0.12Cℓ2 + 1.015O2
(7)
m
whereas the remaining 30% sublimates as a mixture of ammonia, NH3, and
perchloric acid, HCℓO4; this mixture then completes the decomposition to the final
products of (7) in a premixed (AP) flame about 1µ
from the surface.
Let us look at the energetics of these effects:
(a) Enthalpy per gram to bring AP to its surface (molten) temperature (including
some intermediate phase transitions):
P
A
,
hH
∆
= 266 + 0.328 (Ts, AP − 835)
(cal/g)
(8)
(b) Heat of the sublimation into NH3 + HCℓO4
P
A
,
hs
∆
= 476 to 510 cal/(g/sublimed)
(493 average)
(9)
(heat released)
(10)
(c) Heat of “combustion” of NH3 with HCℓO4
P
A
,
hC
∆
= −850 to − 885 cal/(g.reacted)
(d) Heat released per gram of AP directly degraded in the liquid phase:
P
A
,
hD
∆
= −375 cal / g
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(11)
Lecture 17-18
Page 5 of 15
hs
Note: We must have ∆
+ ∆hc = ∆hD )
We can combine these values to obtain the Adiabatic Flame Temperature Tfad
, AP of the
“AP flame”:
∆hD,AP
hC
) + c (T
P
A
,
+∆
P
A
+ 0.3 ( ∆
P
A
,
P
A
,
+ 0.7∆
,
hs
hD
hH
∆
g
ad
f , AP
)
Ts, AP = 0
(12)
which, using for the gas a specific heat
cg = 0.3cal / g / K
gives Tfad
, AP = 1205 K
(13)
Rates and AP flame structure. The rate of pyrolization is found experimentally to
depend on the surface temperature according to an Arrhenius-type expression
m
•
= AS , AP e
AP
− Es , AP / RTs
(14)
AS , AP = 96000 g/cm2/sec
with ES , AP = 20 Kcal/mol
The reaction rate for the premixed AP flame obeys a similar law, except that, being
primarily a bimolecular reaction, its rate is proportional to p2:
ω
•
= p2 Ag, AP e
P
,
Eg
A
•
(15)
ω
with p in atm,
− Eg , AP / RTf , AP
in g/cm3/sec, and with
= 15Kcal / mol; Ag, AP = 650g / cm3 / sec / atm2
of the gas normal to the surface were known, Eq. (15) would allow
. The time to “burn” the gas is
ω
= ρg /
, and then
h
c
τ
•
P
A
,
calculation of the flame standoff distance,
xf
g
If the velocity υ
,
xf
AP
•
= υ gτ ch = υ g ρg / ω =
υg ρg
2
p Ag, AP
e
+
Eg , AP
RTg
(16)
•
(and notice ρg υ g = m AP )
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 6 of 15
A separate expression for x f can be obtained from the fact that the AP flame has to
supply to the vaporizing surface the required heat of reaction, so if the flame moves
too far, the reaction is too slow, and vice versa. The net (convection + conduction)
heat flux in the gas between surface and flame is constant (no heat evolves there)
•
mAP cg
dT
d
−
dx dx
⎛ dt ⎞
−4
⎜ λg dx ⎟ = 0 ( cg ≅ 0.3cal / g / K , λg = 1.9 x 10 cal / sec / cm / K ) (17)
⎝
⎠
with the boundary conditions T=TS,AP at x=0 and
•
•
⎛ dT ⎞
= mAP Qc = m AP ( ∆hH , AP + QS )
⎜ λg dx ⎟
⎝
⎠ x =0
(18)
Here Qs is the net heat required to take the AP from liquid at Ts to gaseous products
(before the AP “flame”)
Qs = 0.3 x 493 + 0.7(-375) = -115 cal/g
(19)
Integrating (17) with (18),
•
m AP cgT − λg
•
•
dT
= m AP cgTS,AP − mAP Qc
dx
•
•
dT m AP
−
(T − TS, AP ) = mλAP Qc
dx
λg
g
and imposing T= TS, AP at x = 0 again, c =
Qc
, so
cg
⎡ mAP cg
⎤
Q ⎢ λ x
⎥
= c ⎢e g
− 1⎥
cg
⎢
⎥
⎣
⎦
•
T - TS, AP
(20)
The “AP flame” is at x = x f, AP, where T reaches TfAP (adiabatic Tf for AP alone, but
maybe higher if there is heat supply from the main flame). Solving for xf,AP then,
xf, AP =
λg
•
m AP
⎡
cg (Tf , AP − Ts, AP ) ⎤
⎥
An ⎢1 +
Qc
⎢
⎥⎦
cg
⎣
(21)
This must be the same as (16). Equating them,
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 7 of 15
•
Eg, AP
λg
m AP
RT
e f, AP = •
An ⎡⎣⋅ ⋅ ⋅⎤⎦
2
p Ag, AP
m AP cg
λg A
•
or
m AP = p
g , AP
cg
e
−
Eg, AP
RTf, AD
⎡
cg (Tf , AP − Ts, AP ) ⎤
⎥
An ⎢1 +
Qc
⎢⎣
⎥⎦
(22)
So, for AP alone, where Tf , AP ≅ 1205 K is known, and Ts,AP can be estimated (not too
much higher than 835 K), the regression rate is proportional to pressure ( nAP ≅ 1 ).
Pyrolysis of Inert Binder. A similar formulation can be used for the calculation of the
heating rate due to the main flame, which serves to pyrolyze the surface of the
binder (and also to elevate the temperature of the AP flame).
Wf
Tf
Xf
Wf, AP
TF,AP
XF,
AP
AP
Ts, AP
Binder
Ts,b
The paper by Lengellè et al.
Simplifies the model by
treating for this purpose the
flow of gas as 1-D (even
though it really is 2-D or 3D due to the heterogeneity
of the surface. It also
assigns a uniform mass flux
mp above both, AP and
binder. This is questionable,
but we’ll press on.
Similar to Eq. (17), we have now
•
mp cg
and, defining q = λg
dT
d ⎛ dT ⎞
−
=0
λg
dx dx ⎜⎝ dx ⎟⎠
dT
the magnitude of the (surface-directed) conduction heat flux,
dx
•
dq mp cg
q=0
−
dx
λg
•
We integrate the condition mp Qf = q ( xf
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(23)
)
(24)
Lecture 17-18
Page 8 of 15
where xf is the location of the main flame, at which the energy of combustion
•
Qf ≅ 700 cal / g is released (q=0 above it, q = mp Qf below it):
•
•
q = mp Qf e
−
m p c g ( xf − x )
λg
(25)
At x=0, on the binder surface, the heat flux q(0) is used to pyrolyze the binder at the
•
•
rate mb (p.u. area), and if Qc,b is the required heat (cal/g), we obtain q(0) = mb Qc ,b ,
or
•
•
Qf
e
Qc ,b
•
mb = m p
−
m p cg xf
λg
(26)
The pyrolisation heat Qc,b is composed of that required to heat the binder form T0 to
Ts,b, plus the heat of decomposition Qs=360 cal/g:
Qc ,b = cb (Ts,b − T0 ) + 360
( cb ≅ 0.39 cal / g / k )
(27)
For Ts,b=1100K, Qc,b=675 cal/g.
Eq. (25) is also useful to estimate the rate of arrival of heat from the main flame at
the location of the AP flame (x=xf,AP):
m
qf , AP =
p
−
m
•
•
p
(
cg xf − xf , AP
Qf e
)
λg
(28)
and, from this, the AP flame temperature, using a modification of Eq. (12):
•
m AP ⎡⎣ ∆hH , AP + ∆hD, AP + cg (Tf , AP − Ts, AP ) ⎤⎦ = qf , AP
(29)
or
•
(
)
ad
m AP cg Tf ,AP − Tf,AP
= qf ,AP
(29b)
ad
ad
is the value of Tf,AP with no qf,AP present ( Tf,AP
where Tf,AP
≅ 1205K , as we found).
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 9 of 15
Estimating the Main Flame Distance. Let Dox be the diameter of the oxidizer particle.
The surface of the particle evolves oxidizing gas (after AP flame), while the binder
around it generates fuel gas.
fuel
gas
These inter-diffuse to form a
diffusion-flame, similar to that
from a Bunsen burner (but insideout). The radial distance covered
by a substance diffusing with a
diffusivity D, in a time t, is of the
Xf
oxidizer
gas
order of 2Dt . We say the
flame’s end is at the point when
this equals Dox/2 (times some
factor of order 1, to account for
real geometry):
AP
Dox
≅ Ad1 / 2 2Dt
2
Dox
Equating the time t to xf / υg = xf
ρg
•
( Ad
≈ 1)
•
(notice we again use the overall mass flux m p
mp
here),
•
Dox
≅ Ad1 / 2 2D ρ g xf / mp
2
or
•
xf =
2
mp DOX
(31)
8 Ad ( ρ g D )
The diffusivity D is inversely proportional to the gas density, so that ρg D is
independent of P at a given T (and weakly dependent on T). Thus, (31) gives a main
flame distance which is independent of pressure, and scales with the square of AP
particle diameter.
To a good approximation, the mass diffusivity D is equal to the heat diffusivity:
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 10 of 15
D ≅ dg =
λg
λg
⇒ ρg D ≅
ρ g cg
cg
(32)
which can be substituted into (31):
•
xf ≅
mp cg
8 Ad λg
2
DOX
(32)
The final gas temperature (after the main flame) can be estimated now from
α AP Tf ,AP + α bTs ,b +
Qf
= Tf
cg
(34)
Ts ,b = 1100 K, cg = 0.3cal/g/K,
Using α AP = 0.8, α b = 0.2, Tfad
,AP = 1205 K,
Qf = 700 cal/g. We calculate Tf = 3250 K.
Actually, Eq.(33) is appropriate only when the diffusion time (distance) is much more
than the reaction time (distance), as at high P (short reaction times), and/or large
AP particle diameters (long diffusion times). In the opposite limit, xf is really dictated
by the chemical reaction time (distance); similar to Eq. (16), this distance can be
written as
•
xf ,r
Eg ,f
+
mp
= 2
e RTf
p Ag ,f
(35)
The values of Ag,f and Eg,f are not given by Langellé et al. This is especially
regrettable for Eg,f which is very sensitive. We here take tentatively
Ag,f = 650g/cm3/sec/atm2, as for AP, and determining Eg,f by matching approximately
one of the small-diameter data points quoted in the paper. This leads to
Eg ,f ≅ 29.4Kcal / mol
(36)
For the general case, then, we take xf to be the sum of the reaction and diffusion
distances:
Eg ,f
⎡ c
+
1
g
2
x f = mp ⎢
Dox + 2
e RTf
p Ag ,f
⎢⎣ 8 Ad λg
•
⎤
⎥
⎥⎦
(37)
•
Solution Procedure. Given P ,T0 , Dox , α AP , we want to calculate mp , as well as several
of the intermediate variables. The equations are fairly complex, so some iteration
must be devised. First, from the averaging law (Eq. (5)) with α b = 1 − α AP ,
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 11 of 15
•
mb
•
=
mp
1 − α AP
(38)
•
mp
1 − α AP
•
m AP
•
But
mb
is also calculable from Eq. (26); this involves xf, which is given by Eq. (37):
•
mp
•
mb
•
mp
•
Q
= f e
Qc ,b
−
m p cg
λg
(
• ⎡ c
exp + Eg ,f / QTs
g
mp ⎢
D2 +
⎢ 8 Ag λg ox
p2 Ag ,f
⎣
) ⎤⎥
⎥
⎦
(39)
•
Equate (38) and (39) and solve for mp :
•
•
⎡
⎤
Q
1
α
m
−
p / m AP
c
AP
⎥
λg A n ⎢
⎢ Qc ,b
⎥
1 − α AP
⎣
⎦
⎡ cg
exp ( − E g ,f / RTf
2
cg ⎢
Dox
+
p2 Ag ,f
⎢⎣ 8 Ad λg
•
mp =
(40)
) ⎤⎥
⎥⎦
•
This is similar to the expression Eq. (22) for m AP . In fact, the ratio of both equations
•
provides a relationship between
mp
•
, and Tf,AP with Dox and p as parameters:
m AP
•
mp
•
m AP
=
e
E g , AP / RTf , AP
Ag ,AP
•
•
⎡
⎤
Qf 1 − α AP m p / m AP ⎥
⎢
An
⎢ Qc ,b
⎥
1 − α AP
⎣
⎦•
⎡
cg (Tf ,AP − Ts ,AP ) ⎤
An ⎢1 +
⎥
Qc
⎢⎣
⎥⎦
1
cg
2
p2 Dox
+
8 Ad λg
exp ( + E g ,f / RTf
)
(41)
Ag ,f
•
mp
The AP flame temperature Tf,AP depends, in turn, on
, as can be seen by
•
m AP
combining Eqs. (28) and (29):
•
•
•
m AP ⎡⎣ ∆hH , AP + ∆hD, AP + cg (Tf , AP − Ts, AP ) ⎤⎦ = mp Qf e
−
(
mp cg xf − xf , AP
λg
)
(42)
•
Here, we notice that the factor e
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
−
m p cg xf
λg
is
Lecture 17-18
Page 12 of 15
•
e
−
m p cg xf
λg
•
Q mb
Q
= c ,b • = c ,b
Qf
Qf
mp
1 − α AP
(43)
•
1 − α AP
mp
•
m AP
•
Also, the factor e
•
−
e
+
mp cg xf ,AP
λg
•
mp m AP cg xf, AP
λg
•
m AP
can be expressed, using (21) as
⎡
cg (Tf ,AP − Ts ,AP ) ⎤
⎥
= ⎢1 +
Qc
⎢⎣
⎥⎦
•
•
mp / m AP
(44)
Combining (42), (43) and (44)
•
mp
1 − α AP
∆hH ,AP + ∆hD ,AP + cg (Tf ,AP − Ts ,AP ) = Qc ,b
•
1 − α AP
mp
•
•
⎡
cg (Tf ,AP − Ts ,AP ) ⎤ mAP ⎛ mp
⎜
⎢1 +
⎥
× •
⎜
Qc
⎢⎣
⎥⎦
⎝ m AP
•
⎞
⎟
⎟
⎠
(45)
m AP
Here we recall that Qc depends on Ts,AP (Eqs. (18), (19) and (8))
Qc ≅ 266 + 0.328 (Ts ,AP − 835) − 115 = 0.328Ts ,AP − 123 ( cal / g )
(46)
For the present purposes, it is sufficient to assume a value for Ts,AP somewhat above
835K; this could be refined later by matching heat fluxes at the liquid surface. We
adopt for most of the following Ts,AP=925K.
•
•
Given Ts,AP, we can see that Eq. (45) uniquely relates to Tf,AP to mp / m AP . The
following calculational procedure can than be followed:
•
•
(a) Take a value of mp / m AP (typically in the range 0.2-0.6)
(b) Solve (45) for Tf,AP (non-linear equation, requires some internal iteration)
(c) Solve (41) for the group
ψ ≡
cg
8 Ad λg
( pDox )
2
+
e
+ E g ,f / RTf
Ag ,f
(47)
and, hence, for the product pDox. Given pDox, This gives p.
•
(d) Calculate m AP from Eq. (22).
⎛
•
⎞
•
•
p
⎟ m AP
(e) m p = ⎜ m
•
⎜
⎟
⎝ m AP ⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 13 of 15
•
•
This produces a curve of mp vs. p, with
mp
•
as the running parameter (for a fixed
m AP
Dox, particle diameter).
•
Notice how Tf,AP and m AP depend on the combination pDox rather than on the
•
separate variables. However, since m AP contains the factor p directly, the final
•
burning rate of mp is of the form
•
mp = p f(p Dox)
(48)
where f is some complicated function. This is an important scaling law (not pointed
out in Lengelle et al’s paper). To verify it, we can use the data reported in Fig. 31 of
the paper.
Burning Rate (mm/s)
100
80% 5µm AP-CTPB
80% 90µm AP-CTPB
Computed burning rate
10
1
100
10
Fig.3
1000
Pressure (atm)
For Dox=5 µm , we have
P(cm)
υ p (mm/sec)
10
7.8
30
21
100
61
300
92
PDox (atm µm )
50
150
500
1500
0.78
0.70
0.61
0.307
10
5
30
7.2
100
10.2
300
20.0
PDox (atm µm )
900
2700
9000
27,000
υ p / p ( mm / sec / atm )
0.5
0.24
0.102
0.0667
υ p / p ( mm / sec / atm )
For Dox=90 µm ,
P(cm)
υ p (mm/sec)
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 17-18
Page 14 of 15
Plotting now both curves as ( υ p /p) vs. (pDox), they do coincide:
1
υp / P
0.5
(mm/sec/atm)
Dox = 5µm
Dox = 90 µm
0.2
0.1
100
200
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
500 1000 2000 5000 10,000
pDox (atm x µm)
Lecture 17-18
Page 15 of 15