16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 8: Convective Heat Transfer: Other Effects
Overall Heat Loss and Performance Effects of Heat Loss
(1)
Overall Heat Loss
i
The local heat loss per unit area is qw = ρucp ( Taw − Tw ) St , and using m = ρuπR2 , the
integrated heat loss is
L
Qw
∫
L
Qw
2
⎛ dR ⎞
qw 2πR ds ; ds = 1 + ⎜
⎟ dx
⎝ dx ⎠
x =0
i
m
∫ πR
2
i
dx (small angles)
L
cp ( Taw − Tw ) St 2πR dx = m cp ( Taw − Tw ) St 2
0
∫
0
(1)
dx
R
(2)
For an approximate evaluation, assume the quantity cp ( Taw − Tw ) St is a weak
function of x, and treat it as a constant. We then obtain
L
Qw
i
m cp Tc
Taw − Tw
dx
2St
Tc
R (x)
⎛L ⎞
For many rockets, ⎜ ⎟ ≡
⎝ R ⎠eff
ratio
Qw
i
∫
0
L
dx
∫ R (x)
0
⎛
Tw ⎞
⎜⎜1 −
⎟ 2St
Tc ⎟⎠
⎝
L
dx
∫ R (x)
(3)
0
is of the order of 6-10, and
Tw
1 1
∼ − , so the
Tc
4 3
(heat loss divided by total enthalpy flux) is of the order of 8-16 times
m cp Tc
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 8
Page 1 of 6
the Stanton number. As we found before, St is itself ~ 0.001, leading to fraction
heat losses of the order of 1-2%. While this is a small fraction, its absolute value
may be large, because the total thermal power is enormous. As an example, for the
SSME engine
i
m cp Tc =
2 × 106 N
J
× 2770
× 3600 K ,
4500 m s
KgK
F
cp Tc
C
i
or m cp Tc = 4.4 × 109 W (the output power of four large power stations).
A 1.5% fraction of this means 66 MW lost to the walls (some 80,000 HP).
(2)
Effect on Performance
As a starting guess, we could imagine that all of the losses ( Qw ) are reflected in an
equal amount of kinetic energy loss in the exhaust. If ue0 is the exist velocity with
no losses,
2
i ⎛u
u2 ⎞
e
m⎜ 0 − e ⎟
⎜ 2
2 ⎟
⎝
⎠
(4)
Qw
But a little reflection shows that the kinetic energy loss must be less than Qw .
Indeed, heat losses that occur near the nozzle exit plane are almost irrelevant for
performance, because the thermodynamic efficiency of the remaining expansion
from the point of loss to the exhaust is very small, so very little of that loss is
reflected in a kinetic energy decrease.
So, for the time being, we simply acknowledge this by writing
2
i ⎛u
u2 ⎞
e
m ⎜ 0 − e ⎟ < Qw
⎜ 2
2 ⎟
⎝
⎠
or ue > u2e0 −
Remembering that
2Qw
u2e0
i
m
(5)
or
ue
2Q
> 1− i w
ue0
mu2e0
2 = cp ( Tc − Te )
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
⎡
⎛P
⎢
= cp Tc ⎢1 − ⎜⎜ e
⎝ Pc
⎢
⎣
(6)
⎞
⎟⎟
⎠
γ −1 ⎤
2 ⎥
⎥,
⎥
⎦
Lecture 8
Page 2 of 6
ue
ue0
i
⎛
⎞
⎜ Qw m cp Tc ⎟
⎠
> 1− ⎝
γ −1
⎛P ⎞
1−⎜ e ⎟
⎝ Pc ⎠
If the fractional loss
(7)
γ
Qw
i
is of order 1.5% and the expansion efficiency
m cp Tc
γ −1
⎛P ⎞ γ
u
1 0.015
η = 1 − ⎜⎜ e ⎟⎟
is of order 75%, then e > 1 −
= 1 − 0.01 (i.e., a loss of less
∼
P
u
2 0.75
e0
⎝ c⎠
than 1% in specific impulse, ignoring the exit pressure contribution).
The calculation can be made more precise by tracking the evolution of the gas
temperature. The total energy equation, accounting for the losses, is
Tc
i
m
dht
du ⎞
⎛ dT
= ρuA ⎜ cp
+u
⎟
dx
dx
dx
⎝
⎠
or cp
−qw 2πR = − ρucp ( Taw − Tw ) St 2πR
2St
dT
du
+u
=−
cp ( Tc − Tw )
dx
dx
R
(8)
But the momentum equation (ignoring, somewhat inconsistently, the effects of
du
1 dp
du dp
friction), gives ρ u
=−
. Substituting in (8),
+
= 0 , or u
dx dx
dx
ρ dx
cp
dT
1 dp 2St
=
−
cp ( Tc − Tw )
dx
R
ρ dx
Divide by cp T and note that
1
dp
ρ cp T dx
=
γ − 1 1 dp
γ p dx
1 dT γ − 1 1 dp 2St Tc − Tw
=
−
T dx
γ p dx
R
T
⎛P ⎞
T
= ⎜⎜ ⎟⎟
Without the heat loss term, this would integrate to
Tc ⎝ Pc ⎠
(isentropic) relation. More generally now,
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(9)
γ −1
γ
, the ideal flow
Lecture 8
Page 3 of 6
⎛P ⎞
T
= ⎜⎜ ⎟⎟
Tc ⎝ Pc ⎠
γ −1
γ
⎡ L 2S T − T
⎤
t
c
w
exp ⎢ −
dx ⎥
R
T
⎢
⎥
⎣ 0
⎦
∫
(10)
and since the exponent is a small number,
⎛P ⎞
T
= ⎜⎜ ⎟⎟
Tc ⎝ Pc ⎠
γ −1
γ
⎡
⎢1 −
⎢
⎣
L
∫
0
⎤
2St Tc − Tw
dx ⎥
R
T
⎥
⎦
(11)
To evaluate the correction term, we use for T the undisturbed T, as if no heat loss
had happened. This gives
Tc − Tw
T − Tw Tc
= c
T
Tc
T
⎛
Tw ⎞ ⎛ Pc ⎞
⎜⎜1 −
⎟⎜ ⎟
Tc ⎟⎠ ⎝ P ⎠
⎝
γ −1
γ
⎛
T ⎞
and we also assume that St ⎜⎜1 − w ⎟⎟ is nearly constant:
Tc ⎠
⎝
T
Tc
⎛P
⎜⎜
⎝ Pc
⎞
⎟⎟
⎠
γ −1
γ
⎡
⎢
⎢1 − 2St
⎢⎣
⎛
Tw ⎞
⎜⎜1 −
⎟
Tc ⎟⎠
⎝
L
⎛ Pc ⎞
⎜ ⎟
P ⎠
0⎝
∫
γ −1
γ
⎤
dx ⎥
R ⎥
⎥⎦
(12)
γ −1
γ
⎤
dx ⎥
R ⎥
⎥⎦
(13)
and, in particular, at the exit plane,
Te
Tc
⎛ Pe
⎜⎜
⎝ Pc
⎞
⎟⎟
⎠
γ −1
γ
⎡
⎢
⎢1 − 2St
⎢⎣
⎛
Tw ⎞
⎜⎜ 1 −
⎟
Tc ⎟⎠
⎝
L
⎛ Pc ⎞
⎜ ⎟
P ⎠
0⎝
∫
We now express the exit kinetic energy as
(
u2e = 2cp Tte − Te
)
(14)
where both Tte and Te are affected by the losses. For the total energy loss, we have
i
(
)
m cp Tc − Tte = Qw
and so
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 8
Page 4 of 6
Tt e
⎛
T ⎞
=1− i
= 1 − 2St ⎜⎜1 − w ⎟⎟
Tc ⎠
⎝
m cp Tc
Qw
Tc
L
dx
∫R
(15)
0
where we have used the result in equation (3). For the loss of static energy, we have
the result in (13). Using both in (14),
⎡
⎢
2
ue = 2cp Tc ⎢1 − 2St
⎢
⎣
or
u2e
⎛
Tw ⎞
⎜⎜1 − T ⎟⎟
c ⎠
⎝
⎧
⎛P
⎪
= 2cp Tc ⎨1 − ⎜⎜ e
⎝ Pc
⎪
⎩
⎞
⎟⎟
⎠
γ −1
γ
L
∫
0
dx ⎛ Pe ⎞
− ⎜⎜ ⎟⎟
R
⎝ Pc ⎠
γ −1
γ
⎛
T ⎞
− 2St ⎜⎜ 1 − w ⎟⎟
Tc ⎠
⎝
⎛
T ⎞
+ 2St ⎜⎜ 1 − w ⎟⎟
Tc ⎠
⎝
L
⎛ Pe ⎞
⎜ ⎟
P ⎠
0⎝
∫
γ −1 ⎤
⎫
⎡
⎛ Pe ⎞ γ ⎥ dx ⎪
⎢
⎢1 − ⎜ P ⎟
⎥ R ⎬
⎝ ⎠
⎪
0 ⎢
⎥⎦
⎣
⎭
γ −1
γ
⎤
dx ⎥
R ⎥⎥
⎦
L
∫
(16)
γ −1
⎛P ⎞ γ
occurring in the integral of (16) is just the
We see now that the factor 1 − ⎜ e ⎟
⎝P ⎠
“thermodynamic relief” we had mentioned earlier, which makes the loss of kinetic
energy be less than the heat loss. Indeed, this factor becomes zero as P → Pe , so, as
anticipated, heat losses near the exit are irrelevant.
To simplify the writing, use u2e0
⎡
⎛P
⎢
= 2cp Tc ⎢1 − ⎜⎜ e
⎝ Pc
⎢
⎣
⎞
⎟⎟
⎠
γ −1 ⎤
γ ⎥
⎥
⎥
⎦
and define
η0,e
u2e
u2e0
⎛P ⎞
= 1 − ⎜⎜ e ⎟⎟
⎝ Pc ⎠
γ −1
γ
and ηx,e
⎛
T ⎞
= 1 − 2St ⎜⎜1 − w ⎟⎟
Tc ⎠
⎝
L
⎛ P ⎞
=1−⎜ e ⎟
⎜ P (x) ⎟
⎝
⎠
⎛ ηx,e
⎜
⎜η
0 ⎝ 0,e
∫
⎞ dx
⎟
⎟ R
⎠
γ −1
γ
:
(17)
⎛ ηx,e ⎞
and, again, the right-hand-side minus the ⎜
⎟ factor would be the relative heat
⎜ η0,e ⎟
⎝
⎠
loss (equation 3). Numerical evaluation shows that the modified integral in (17) is
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 8
Page 5 of 6
about 2
L
3
of the original integral
dx
∫R
. Remembering our earlier estimate of the
0
relative Isp loss ( < 1%), we conclude that a better estimate is about 0.67%. This
∼
amounts to 3 sec. out of Isp
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
400s .
Lecture 8
Page 6 of 6