16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6: Heat Conduction: Thermal Stresses
Effect of Solid or Liquid Particles in Nozzle Flow
An issue in highly aluminized solid rocket motors.
2Al +
3
O2 → Al2O3
2
m.p. 2072 C , b.p. 2980 C
In modern formulations, with ∼ 20% Al by mass, the Al2O3 mass fraction of the
exhaust can be 35-40%. This material does not expand, so there must be a loss in
exit velocity, hence in Isp.
i
i
Assume mass flows mg (gas) ms (solids), non-converting.
The momentum equation is
i
i
mg dug + ms dus + Adp = 0
Call ρs the (mass of solids)/(volume) (not the density of the solid, theory)
ρg ug dug + ρs us dus + dp = 0
Define a mass flux function
i
x=
ms
i
=
i
mg + ms
ρsus
ρgug + ρsus
x
⎛
⎞
⇒ ρgug ⎜ dug +
dus ⎟ + dp = 0
−
1
x
⎝
⎠
ugdug = −
dp
ρg
−
x
ugdus
1−x
The energy equation is similarly,
(1 − x ) ( cpgdTg + ugdug ) + x ( csdTs + usdus ) = 0
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 1 of 10
Substitute here ugdug from above:
cpgdTg −
dp
ρg
dp
ρg
−
x
x
ugdus +
( csdTs + usdus ) = 0
1−x
1−x
= cpgdTg +
x ⎡
cs dTs + us − ug dus ⎤
⎦
1−x ⎣
(
)
⎛ T ⎞
dp
P
= cp dT →
= ⎜⎜
with no particles (x=0), this gives R g T
⎟
P
P0 ⎝ T0 ⎟⎠
γ
γ −1
With particles, we need to know the history of the velocity slip us − ug and of the
temperature slip Ts − Tg . This is a difficult problem, requiring detailed modeling of
the motion and heating/cooling of the particle. But we can look at the extreme cases
easily.
(a) Very Small Particles → good contact. For sub-micro particles (not a bad
representation of reality), we can say that
us
ug = u ,
Ts
Tg = T . Then
dp
ρg
x
⎛
⎞
= ⎜ cpg +
cs ⎟ dT
1
−
x
⎝
⎠
Note that the mean specific heat ( cpg and cs are per unit mass) is
cp = (1 − x ) cpg + xcs ⎫⎪
⎬ R g = c p − c v = (1 − x ) cpg − cvg = (1 − x ) R g
c v = (1 − x ) cvg + xcs ⎪⎭
(
and also
So that
RgT
dp
ρg
=
cp
1−x
)
dT
cp
dp
dp cp dT
=
dT →
=
P
1−x
P
Rg T
and defining an effective γ by the usual γ =
cp
cv
,
γ
⎛ T ⎞ γ −1
P
= ⎜⎜
⎟
P0 ⎝ T0 ⎟⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 2 of 10
The equation of motion is now
( ρg + ρs ) udu + dp = 0
Or
ρg
1−x
ρg
1−x
P
RgT
udu + dp = 0
=
P
P
=
R g T (1 − x ) R g T
udu + dp = 0
From the two boxed equations we see that everything from here can proceed
as if the gas were simple, but with molecular mass
M=
Mg
1−x
(or R g = (1 − x ) R g ),
and with
c p = (1 − x ) cpg + x cs .
For example,
⎡
⎢
⎛P
ue = 2
R g Tc ⎢1 − ⎜⎜ e
γ −1
⎝ Pc
⎢
⎣
γ
⎞
⎟⎟
⎠
γ −1 ⎤
γ
⎥
⎥
⎥
⎦
Tc , Pc in chamber etc.
For sensitivity analysis it may be of interest to linearize this for x
algebra is tedious, but one gets,
1 . The
⎡
(1 − η0 ) ln (1 − η0 ) ⎤ ⎪⎫
ue
1 ⎪⎧
≅ 1 − × ⎨1 − c ⎢1 +
⎥⎬
ue0
2 ⎪
η0
⎢⎣
⎥⎦ ⎪⎭
⎩
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 3 of 10
with c =
cps
cpg
⎛P
, η0 = 1 − ⎜⎜ e
⎝ Pc
⎞
⎟⎟
⎠
γ −1
γ
and, of course,
ue0
⎡
⎢
⎛P
γ
R g Tc ⎢1 − ⎜⎜ e
= 2
γ −1
⎝ Pc
⎢
⎣
⎞
⎟⎟
⎠
γ −1 ⎤
γ ⎥
⎥
⎥
⎦
We see from this that if c < 1 ( cps < cpg , which is common), then ue < ue0 (and
∼
∼
vice-versa).
For a numerical example, look at Problem 2 (attached)
(b) Very Large Particles Hard to quantify, but probably for diameter > 100 µm or
∼
so, the particles have too much inertia (and thermal inertia) to follow the gas
acceleration and cooling. We then have
dus
dug ;
or dus
0;
Ts
Tc ( ≅ Tg at chamber)
dTs
0
dp
Returning to the
dp
ρg
ρg
equation, it now looks as if there were no particles:
= cpgdTg
(i.e. , particles just do not participate in the dynamics or in the thermal
γ γ −1
⎛ T ⎞
P
=⎜
. This does not mean zero
balances). So, we still have
⎟
P0 ⎜⎝ T0 ⎟⎠
performance effect, though. We do not get the full gas exit velocity
γ −1
⎡
⎛ Pe ⎞ γ
γ
⎢
ue = 2
R g Tc ⎢1 − ⎜⎜ ⎟⎟
γ −1
⎝ Pc ⎠
⎢⎣
⎤
⎥
⎥
⎥⎦
but the particulates do not contribute to thrust, because they exit at
us
ue :
i
g Isp =
i
mg ue + ms us
i
i
mg + ms
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
= (1 − x ) g Isp0
Lecture 6
Page 4 of 10
This is actually more loss than in the small particle case (about twice as much,
depending on c).
From the example, this is a serious loss in solid rockets.
Criterion for Slip
2
4
πRp3ρs
3
mp
dup
dup
dt
dug
dt
dt
=
−
(
= 6πµgRp ug − up
ug − up
call ug − up = s
τR
Say τR and
dt
τR =
2
2 ρsRp
9 µg
up = ug − s
dug
ds
s
+
=
dt τR
dt
ds
s
=
dt
τR
dug
2
2 Rp ρs dup
= ug − up
9 µg dt
)
= ag are constant → s
ag τR + C e
−
t
τR
s ( 0 ) = 0 → C = −ag τR
⎛
⎞
ε2
1 − ⎜1 − ε +
....⎟
⎜
⎟
2
⎝
⎠ = 1 − ε + ...
=
ε
2
t
⎛
−
τR
⎜
s = ag τR 1 − e
⎜
⎝
⎞
⎟
⎟
⎠
and ug = ag t
t
⎛
−
τR ⎜
s
τR
1−e
=
ug
t ⎜
⎝
1 t
...
2 τR
t
τR
1−
t
τR
1
t τR
⎞
⎟
⎟
⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 5 of 10
So, small slip for t
L
u
L
ug
τR
τR
2
2 ρs Rp
9 µg
9 µgL
2 ρs ug
Rp
Say
µg ∼ 3 × 10−5 Kg / m / s
L ∼ 0.3 m
ρs ∼ 3 × 103 Kg / m3
Rp
4.5
3 × 10−5 × 0.3
3 × 103 × 1.5 × 103
= 0.9 × 10−11 = 3 × 10−6 m = 3µm
ug ∼ 1.5 × 103 m / s
So,
Rp
3µm → no slip
Rp
3µm → full lag
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 6 of 10
Problems
Problem 2
As noted in class, the effect of carrying a mass fraction x of fine solid particles in the
expanding gas in a rocket nozzle can be accounted for by using an average specific
heat ratio
γ =
(1 − x ) cpg + xcs
(1 − x ) cvg + xcs
and an average molecular mass
M =
Mg
1− x
For Al2O3 the high temperature specific heat is cs = 1260 J/Kg/K.
Consider a solid rocket with γ = 1.17 (1.25) , Pe Pc = 0.01, M g = 18 g / mol.
For a 20% aluminum loading in the propellant, x is of the order of 37%.
Calculate the matched specific impulse of the rocket and compare to what it would
be for the same Tc = 3300 K , but with no particles.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 7 of 10
Problem 2
Specific heat of clean gas
1.25
r R 1.17 8.314
=
= 3180 J / Kg / K
cpg =
r − 1 M 0.17 0.018 2309
0.25
cvg =
cpg
r
2309
3180 1.25
=
= 2718 J / Kg / K
1.17
1845
The specific heat of the solid (or liquid) Al2O3 is cs = 1260 J / Kg / K.
The average specific heat ratio is then
r=
(1 − x ) cpg + x cs
(1 − x ) cvg + x cs
2309
=
(1 − 0.37) × 3180 + 0.37 × 1260
(1 − 0.37) × 2718 + 0.37 × 1260
1845
And the average molecular mass ( Ms
M=
Mg
1−x
=
The exit speed for
∞ ) is
18
= 28.57 g / mol
1 − 0.37
Pe
= 0.01 and Tc = 3300 K is then
Pc
⎡
⎢
⎛P
ue = 2
Tc ⎢1 − ⎜⎜ e
γ −1 M
⎝ Pc
⎢
⎣
γ
= 1.1336
1.1795
Q
⎞
⎟⎟
⎠
γ −1 ⎤
γ
⎥
⎥ = 2613 m / sec
⎥ 2521
⎦
NOTE: Alternatively, and easier to do, you can use
⎡
⎢
⎛P
ue = 2 Cp Tc ⎢1 − ⎜⎜ e
⎝ Pc
⎢
⎣
⎞
⎟⎟
⎠
γ −1 ⎤
γ
⎥
⎥
⎥
⎦
with C p = (1 − 0 .3 7 ) 3 1 8 0 + 0 .3 7 × 1 2 6 0 = 2 4 6 9 J/Kg/K
(so M is not really needed)
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 8 of 10
As a check,
γ
R
γ −1 M
=
1.1336 8.314
= 2469 J/Kg/K
0.1336 0.02857
as it should.
Under pressure-matched conditions, there is no exit pressure contribution to thrust
or Isp , and hence
Isp =
2613
= 266.3 sec
9.81
257.3 s
Without particulate but with the same Pc , Pe and Tc , we would obtain
ue0 =
γ −1 ⎤
⎡
⎛ Pe ⎞ γ ⎥
2γ Q
⎢
Tc ⎢1 − ⎜⎜ ⎟⎟
⎥ = 3199 m / sec
γ − 1 Mg
⎝ Pc ⎠
⎢
⎥ 3029
⎣
⎦
and Isp0 =
3199
= 326.1 sec
9.81
309.1
16.8%
266.3 ⎞
⎛
There is therefore a loss of ⎜1 −
×
100
=
18.3% in Igp
326.1 ⎟⎠
⎝
257.3
1−
309.1
It is interesting to test the accuracy of the linear approximation given in class for
small x:
ue
ue0
x⎡
c
1 − ⎢1 −
2⎢
cpg
⎣
⎛
(1 − η0 ) ln (1 − η0 ) ⎞ ⎥⎤ ; η = 1 − ⎜⎛ Pe
⎜1 +
⎟
0
⎜
⎟⎥
⎜P
η0
⎝
⎠⎦
⎝ c
We find η0 = 0.4878 , and then
⎞
⎟
⎟
⎠
γ −1
γ
ue
= 0.837 (16.3% loss)
ue0
(not too different, despite large x)
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 9 of 10
NOTE: Alternatively, and easier to do, you can use
⎡
⎢
⎛P
ue = 2 Cp Tc ⎢1 − ⎜⎜ e
⎝ Pc
⎢
⎣
⎞
⎟⎟
⎠
γ −1 ⎤
γ
⎥
⎥
⎥
⎦
with C p = (1 − 0 .3 7 ) 3 1 8 0 + 0 .3 7 × 1 2 6 0 = 2 4 6 9 J/Kg/K
(So M is not really needed)
As a check,
γ
R
γ −1 M
=
1.1336 8.314
= 2469 J/Kg/K as it should.
0.1336 0.02857
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6
Page 10 of 10