16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10: Ablative Cooling, Film Cooling
Transient Heating of a Slab
Typical problem: Uncooled throat of a solid propellant rocket
Inner layer retards heat flux to the heat sink. Heat sink’s T gradually rises during
firing (60-200 sec). Peak T of heat sink to remain below matl. limit. Back T of heat
sink to remain below weakening point for structure.
Prototype 1-D problem:
Can be solved exactly, or can do transient 1-D numerical computation. But it is
useful to look at basic issues first.
Thermal conductance of B.L.=hg
Thermal conductance of front layer =
Thermal conductance of layer i =
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
ki
δi
k1
δ1
( δi = thickness, ki = thermal conductivity)
Lecture 10
Page 1 of 12
Want layer 1 to have
k1
δ1
hg to protect the rest.
⎛ k1 1 W / m / K ⎞
k1
W
= 330 2 compared to
(Say, porous, Oriented graphyte, ⎜⎜
⎟⎟ →
δ1
mK
⎝ δ1 = 3 mm
⎠
W
hg ∼ 50, 000 2 , so OK here).
mK
Also, from governing equation
ρc
∂T
∂2 T
∂T
∂2 T
=k 2 →
=α 2
∂t
∂t
∂x
∂x
(α =
k
, thermal diffusivity, m 2 / s )
ρc
we see that
x2 ∼ αt , or x ∼
x2
.
α
αt , or t ∼
So the layer 1 will “adapt” to its boundary conditions in a time t ∼
Say, c
so α =
710
J
and ρ
KgK
1100
Kg
(
3
m
δ12
.
α1
1
solid graphyte),
2
1
= 1.3 × 10−6 m2 / s .
710 × 1100
(3 × 10 )
The layer “adapts” in t ∼
−3
1.3 × 10
2
−6
= 7.0 sec
(more like
δ2
= 1.8 sec ).
4α
⇒ Treat front layer quasi-statically, i.e., responding instantly to changes in heat flux:
k1
T
(t)
wh1
−T
(t)
wc1
δ1
q (t)
This also means we can lump the thermal resistances of BL and 1st layer in series:
1
(hg ) eff
δ
1
+ 1
hg k1
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 2 of 12
and since
k1
δ1
(hg )eff
hg ,
∼
k1
δ1
hg
( )eff
For layer 2 (the heat sink), k2 is high (metal) and hg
is now small (thanks to 1st
layer) so, very likely,
k2
δ2
(hg )eff
(For instance, say Copper, k2
(hg )eff
360
W
, with δ2 = 2 cm. We now have
mK
k2
k
k
W
W
= 350 2 , but 2 = 36, 000 2 , so indeed, 2
δ2
δ2
δ2
mK
mK
(hg )eff ).
Under these conditions, the heat sink is being “trickle charged” through the high
thermal resistance of layer 1. Most likely, heat has time to redistribute internally, so
that T2 is nearly uniform across the layer. We can then write a lumped equation.
ρ2c2 δ2
Define τ =
dT2
= q = hg
dt
( )eff ( Taw − T2 )
ρ2c2 δ1δ2
k1
τ
T2 = Taw − ( Taw − T0 ) e
k1
( Taw − T2 )
δ1
dT2
+ T2 = Taw
dt
−
( T2 (0 ) = T0 )
t
τ
For our example, say ρ2 = 8900 Kg / m3 (Copper), c2 = 430
τ=
J
, δ2 = 2 cm
KgK
8900 × 430 × 3 × 10−3 × 2 × 10−2
= 230 sec
1
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 3 of 12
This is comfortable. Suppose Taw = 3300 K , T0 = 300 K , and we fire for 120 sec:
(60)
(60)
T2 (120 ) = 3300 − 3000 e
−
120
230
(989)
= 1520 K
May need 4 cm
which is still (OK) for Copper (melts at 1360K, but no stress bearing, so can go to
~900. Also OK for steel on Carbon str member).
NOTE:
(0.02)
δ22
=
= 1.1 sec , so, indeed, layer 2 “adapts” quickly to B.C.’s
4α2
4 × 9.4 × 10−5
2
→ uniform
k2
360
=
= 9.4 × 10−5 m2 / s .
ρ2c2 8900 × 430
A More Exact Solution
Consider Taw “turned on” at t=0. The B.L. has a film coefficient hg , and the first
( )eff
layer has δ1 , k1 , so that hg
=
hg
δ
1 + hg 1
k1
∼
k1
. Layer 2 has thickness δ2 , and has
δ1
k2 , ρ2 , σ2 , α2 . The back is insulated.
Then one can prove that layer 2 has a temperature distribution
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 4 of 12
Taw − T2 ( x, t )
Taw − T0
where an =
∞
=
∑a e
−λn2
n
α2t
δ22
n =1
⎛δ −x ⎞
cos ⎜ 2
λn ⎟
⎝ δ2
⎠
2 sin λn
λn + sin λn cos λn
and λn (n=1,2,…) are the roots of
λn tan λn =
(hg )eff δ2
k1 δ2
k2 δ1
k2
Graphically,
For small ∆ ≡
λ12
and also a1
k1 δ2
, small λ1 , so tan λ1
k 2 δ1
∆ λ1
∆ =
1
λ12
λ1 , so
k1 δ2
k2 δ1
α2
δ22
k1 δ2
k δ1
2
k /ρ c
k1
2 2 2 =
≡τ
2
ρ2c2 δ1δ2
δ2
from before
So, leading term is then
Taw − T2 ( x, t )
Taw − T0
e
−
t
τ
⎛δ −x ⎞
λ1 ⎟
cos ⎜ 2
⎝ δ2
⎠
1
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 5 of 12
which is what we found before. The other terms are much smaller, except at very
small time.
For thermal protection of solid rocket nozzles read sec. 14.2 (pp. 550-563) of
Sutton-Biblarz, 7th ed., especially, pp. 556-563.
A key concept is ablative materials. They contain a C-based homogeneous matl.
embedded in reinforcing fibres of strong (anisotropic) C. Best is C/C, strong
expensive fibre since nozzle can get to 3600 K, can be 2D or 3D. Also good is C or
Kelvin (Aramid) fibres +phenolic plastic resins (for large nozzles)
For the shuttle RSRM, the throat insert (C cloth phenolic) regresses ~ 1 inch/120 sec,
and the char depth is ~ 0.5° inch/120 s.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 6 of 12
Film Cooling of Rockets
For application of data on slot-injected films, we need to define the initial film
thickness s, velocity uF , density ρF , or at least mass flux uF ρF .
i
i
i
i
Assume we know the flow rates mc and mF , where mc is the “core” flow and mF the
“film” flow. We also know the fully-burnt temperatures and molecular weights
( Tc , TF ; Mc , MF ).
The areas occupied at the “fully burnt” section are not known; let them be Ac , AF .
From continuity,
i
uc A c =
mc
ρc
i
mc R
=
Tc
P Mc
(1)
P = Pc is common to both
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 7 of 12
i
i
mF
uF AF =
mF R
=
TF
P MF
ρF
(2)
and the total cross-section is known:
Ac + AF = A
(3)
We need some additional information to find uF . The two momentum equations are
(neglecting friction):
duc dP
⎫
+
= 0⎪
dx
dx
⎪⎪
duc
duF
= ρFuF
⎬ ρcuc
dx
dx
⎪
duF dP
+
= 0⎪
ρFuF
⎪⎭
dx
dx
ρcuc
ρ
uF duF
= c
ρF
uc duc
(4)
Both, ρF and ρc , have been evolving as drops evaporate and burn. We make now
the approximation of assuming their ratio to remain constant (equal to the fullyburnt value). Then (4) integrates to
uF2
u2c
=
ρc
ρF
uF
=
uc
ρc
ρF
(5)
Substitute into the ratio (2)/(1)
i
ρFuF AF mF
ρ
= i → F
ρcuc Ac
ρc
mc
i
or
i
ρc AF mF
= i
ρF Ac
mc
AF
mF
= i
Ac
mc
ρc
ρF
(6)
ρFuF
=
ρcuc
ρF
ρc
(7)
and also
⎛ρu
This last ratio ⎜⎜ F F
⎝ ρcuc
⎞
⎟⎟ is called the “film cooling parameter”, MF :
⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 8 of 12
MF =
ρF
=
ρc
MF Tc
Mc TF
(8)
The film thickness s (at complete burn up) follows from
2
AF = π ⎡D2 − (D − 2s ) ⎤ ⎫
⎢⎣
⎥⎦ ⎪
2
⎪ AF ⎛ D ⎞
=
⎬
⎜
⎟ −1
⎪ A c ⎝ D − 2s ⎠
2
A c = π (D − 2s )
⎪
⎭
i
s
4s
D
(if
s
D)
ρc
ρF
D AF
D mF
=
4 Ac
4 i
mc
(9)
From Rosenhow & Hartnett, Chapter 17-B, we characterize film cooling by the
change it induces to the driving temperature ( Taw ) for heat flow. In the absence of a
(
)
γ −1 2⎞
⎛
0
0
− Tw . The film
= Tc ⎜1 + r
film, Taw
Mc ⎟ , and we calculate ( qw )No Film = hg Taw
2
⎝
⎠
0
F
F
to Taw
(lower, presumably). The lowest we could Taw
to get is TF , so
changes Taw
we define a film cooling efficiency
η=
0
F
Taw
− Taw
Taw − TF
⎧⎪η = 0
Limits: ⎨
⎪⎩η = 1
(10)
F
0
if Taw
= Taw
if
F
Taw
= TF
(no effect)
(max imum effect)
If we can predict η , then
(
F
0
0
Taw
= Taw
− η Taw
− TF
)
(11)
and then
(
F
qw = hg Taw
− Tw
)
(12)
where hg is computed as if there were no film. To predict η , we first computes the
parameter
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 9 of 12
µF ⎞
x ⎛
ζ=
⎜⎜ ReF
⎟
µc ⎟⎠
MF s ⎝
−
1
4
(13)
where x is the distance downstream of the film injection (here we assume this is
from the burn-out section), and
ρFuF s
ReF =
(14)
µF
and ρFuF = MF ( ρcuc ) , from before
From ζ , there are several semi-empirical correlations for η . A recommendation from
R & H is
2
η=
1.9 Pr
3
⎛ cp
1 + 0.329 ⎜ c
⎜ cp
⎝ F
⎞ 0.8
⎟ζ
⎟
⎠
(15)
(or η = 1 if this gives >1)
which is supported by air data of Seban.
Example
Say
TF
ρ
1 M
0.8
= ; F = 0.8 → F =
= 1.6 → MF = 1.6 = 1.265
ρc 0.5
Tc
2 Mc
i
mF
i
mF
1
= 0.1 → i =
9
m (0.01) mc
(0.0101)
i
Say D=0.5m
x t − x com pl. com b = 0.5 m
P=70 atm=7.09 × 106 N / m2 ⎫
⎪
7.09 × 106 × 0.020
= 5.33Kg / m3 ; ρF = 8.53Kg / m3
Tc = 3200 K
⎬ ρc =
8.314
3200
×
⎪
Mc = 20 g / mol; γ c = 1.2
⎭
Mc = 0.2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 10 of 12
uc = 0.2 1.2 ×
8.314
× 3200 = 253 m / s
0.02
1
= 200 m / s
1.6
uF = 253
Say µF = 2 × 10−5 Kg / m / s → ReF =
8.53 × 200 × s
2 × 10
−5
= 8.53 × 107 s
ReF = 9.37 × 105
i
ρc
0.5
=
×
ρF
4
D mF
s=
4 i
mc
1
9
OR 0.0101
1
= 0.0110 m
1.6
0.000998
(8.51 × 10 )
4
0.6
µF ⎛ TF ⎞
=⎜
⎟
µc ⎜⎝ Tc ⎟⎠
ζ=
= 0.50.6 = 0.660
(
0.5
9.37 × 105 × 0.660
1.265 × 0.0110
8.51 × 104
0.000998
epc
µF
= 0.8
µc
c pF
)
(say, rF
2
η=
(
1.9 × 0.8
3
1 + 0.329 × 0.8 × 1.2820.8
(25.74 )0.8
)
−
1
4
= 1.282
(25.74)
rc ), Pr = 0.8
= 1.24 → η = 1
0.368 0.361
So, this offers perfect film cooling, meaning
Tc
= 1600 K
2
(3200-0.361(3200-700)=2296 K)
F
Taw
= TF =
If the wall is made of Cu, and is at Tw = 700K , the reduction in heat flow is
qFw
q0w
=
1600 − 700
= 0.360
3200 − 700
⎛ 2296 − 700
⎞
⎜ 3200 − 700 = 0.638 ⎟
⎝
⎠
which can be decisive.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 11 of 12
(This example shows one could get good film cooling with much less than 10% flow
in the film, maybe with only 2%).
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 10
Page 12 of 12