16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11: Radiation Heat Transfer and Cooling
Radiative Losses
At throat of a RP1-LOX rocket, evaluate radiation heat flux
Pc = 70 atm
Dc = 0.21m
Tc = 3500 K
O / F = 2.2
γ=1.25
M=25 g/mol
x co
0.3 8
xH2O = 0.3 1
xH2
0.1 4
x co2
0.1 1
γ
⎛ 2 ⎞ γ −1
Pthroat = ⎜
⎟
⎝ γ + 1⎠
Pc = 38.85 atm
Pco = 14.8 atm
PH2 O = 12.0 at m
PH2 = 5.4 atm
Pco 2 = 4.3 at m
Tthroat =
2
Tc = 3111K = 5600 R
γ +1
Assume slab if thickness L=0.9 Dt = 0.191m = 0.63 ft
(PL )co = 9.2 ft atm
(PL )H O = 7.5 ft atm
(PL )H = 3.4 ft atm
(PL )co = 2.7 ft atm
2
2
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
Page 1 of 14
CO
Fig 4-22 for CO gas only to 2400 R
2 ft atm
So, extrapolated to 9.2 ft atm, ε (2400 R )
0.1 at 2400 R. But ε falls rapidly with T.
If we conservatively extrapolate linearly in Log ε (T). εco would appear to go to ~
0.005 or so. Hence, even though the gas is CO-rich, radiation by CO is negligible.
H2O
At PT = 1atm , PL = 7.5 ft atm , Fig 4.15 gives εH2 o (5000 R ) = 0.18 , and extrapolating
a bit to T=5600R, εH2O
0.15 .
Fig 4.15 gives εw for Pw → 0 ,PT = 1atm . To correct for finite Pw and higher PT , use
4.16. Here, for PwL = 7.5 ft atm , there is some significant effect of
Pw + PT
. We have
2
Pw + PT
12 + 38.9
=
= 25.5 atm way beyond the graph.
2
2
β=
0.5
0.8
1
1.2
0
Pw + PT
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
cw
1
1.18
1.23
1.28
0.57
(1)
(1.14)
(1.21)
(1.28)
Lecture 11
Page 2 of 14
(
cw = c P + B
)
n
n
1 ⎫
⎫ ⎛ B + 0.5 ⎞
⎛ 1 ⎞
⎛ 1.28 ⎞
⎪
ln ⎜
ln ⎜
⎟ =
⎪⎜
⎟
⎟
B
0.57
⎪
n
⎪⎝
⎠
⎝ 0.57 ⎠ =
⎝ 0.57 ⎠
1 = c (B + 0.5)
n=
⎬
⎬
n
0.5 ⎞
1.2 ⎞
⎛
⎛
1.28 ⎪
ln ⎜ 1 +
ln ⎜ 1 +
n ⎪ ⎛ B + 1.2 ⎞
⎟
=
1.28 = c (B + 1.2 ) ⎪⎭ ⎜
⎪
⎟
B ⎠
B ⎟⎠
⎝
⎝
0.57 ⎭
⎝ B
⎠
0.57 = c Bn
n=
0.562
0.809
=
0.5 ⎞
1.2 ⎞
⎛
⎛
ln ⎜ 1 +
ln ⎜ 1 +
⎟
B ⎠
B ⎟⎠
⎝
⎝
(
cw = 1.175 P + 0.105
⎛
ln ⎜ 1 +
⎝
⎛
ln ⎜ 1 +
⎝
1.2 ⎞
B ⎟⎠
= 1.439
0.5 ⎞
B ⎟⎠
B = 0.105
c = 1.175
n = 0.321
)
0.321
Then, for P = 25.5 , cw = 3.33 → εw = 3.33 × 0.15 = 0.499
Suspect!
too much
CO2
For PL=2.7 ft atm, T=5600 R ε CO2
From fig 4.14, Correction cc
0.056
1.1 → εCO2
0.062
For interference, use Fig 4.17
Pw
12
=
= 0.736
Pw + Pc
12 + 4.3
(Pr
+ Pw ) L = 7.5 + 2.7 = 10.2 ft atm
⇒ ∆ε
So εgas
0.06
0.499 + 0.062 − 0.06
0.5
This is likely to be an over estimate, because εH2O must saturate as PT increases, not
grow as PT0.3 .
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
Page 3 of 14
With this ε g σ Tt4 = 0.5 × 5.67 × 10 −8 × 31114 = 2.66 × 106 W / m2
Compare to Convection:
Say Tw = 1000 K
hg =
c* =
cp µ0.2
( ρu)0.8
<>
<>
D0.2
Pr0.6
t
(0.026 )
3111 + 1000
= 2056
2
< T >=
( ρu)e
=
( ρu)<>
Pc
70 × 105
=
= 4269 Kg / m2 / s
c*
1640
= 4269
3111
= 6460 Kg / m2 / s
2856
0.6
⎛ 2056 ⎞
6 × 10−5 ⎜
⎟
⎝ 3000 ⎠
= 4.8 × 10−5 Kg / m / sec
Pr = 0.8
Sp = 1663 J / Kg / K
hg =
(
64600.8 × 1663 × 4.8 × 10−5
( qw )conv
So
δ
8.314
× 3500
0.025
= 1640 m s
0.658
c* =
µ<>
Rs T
0.2
0.21
0.6
0.8
)
0.2
× 0.026
= 345, 000 × 0.026 = 8960 W / m2 / K
8960 (3500 − 1000 ) = 2.24 × 10 7 W / m 2
qrad
= 0.12
qconv
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
Page 4 of 14
As PT increases, each individual emission line is broadened by collisions, and ε
(
)
increases. However, when PL is relatively large > 2.5 ft atm , Figure 4.14 shows the
∼
effect is small; this is because at that PL the bands are largely overlapped already
and only the broadening of their edges matters anymore. So, we ignore the PT effect.
Effect of Particulates
2R
For
2πR
λ
1 , geometrical optics
For
2πR
λ
⎛ 2πR ⎞
1 , Rayleigh regime, particle appear to be smaller by ∼ ⎜
⎟ .
⎝ λ ⎠
4
For 3000 K, peak of spectrum at λ ∼ 1.2 µm
R cross over =
λ
∼ 0.4 µm
2π
Particles tend to be near (somewhat below) this value. For conservatism, assume
geometrical occultation.
α = Prob. of absorption = 1-Prob. of transmission= 1 − e
nP =
εp
ρgas
x
1 − x 4π 3
R ρ
3 p s
1−e
−
−
L
mfp
=1−e
−npQPL
Qp = πRp2
x ρg 3 L
1 − x ρs 4 Rp
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
Page 5 of 14
Say L = R t = 0.3 m
ρg =
R p = 1 µ m = 10 −6 m
38.9 × 105 × 0.025
= 3.75 Kg / m3
8.314 × 3111
ρs = 3000 Kg / m3
x=0.3
εp
1−e
−
0.3 3.75 3 0.3
0.7 3000 4 10−6
= 1 − e−120 = 1
⎛ 2πRp
Now, suppose R p = 0.1 µm instead. Exponent has a factor ⎜
⎜
⎝ λ
and has a 1 , which is another
Rp
4
4
⎞
1
⎛ 0.1 ⎞
⎟⎟ = ⎜
⎟ = 256 ,
0.4
⎝
⎠
⎠
1
= 1 0 → 2 5 .6 sm aller → 1 − e − 4.69 = 0 .9 9 1 still ∼ 1
0.1
In this case
(a) In flame looks solid (black body radiator)
(b) Radiative losses double ( ε =1 instead of 0.5), to ~20% of loss
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
Page 6 of 14
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 11
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