16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7: Convective Heat Transfer: Reynolds Analogy
Heat Transfer in Rocket Nozzles
General
Heat transfer to walls can affect a rocket in at least two ways:
(a) Reducing the performance. This tends to be a 1-3% effect on Isp only, and is
therefore secondary.
(b) Creating great difficulties in the design of hot-side structures that have to
survive heat fluxes in the 107 − 108 w / m2 range.
The principal modes of heat transfer to nozzle and combustor walls are
convection and radiation. Of these, convection dominates, and radiation tends to be
important only for particle-laden flows from solid propellant rockets.
Convective Heat Transfer
We will review here the compressible 2D boundary layer equations in order to extract
information on wall heat transfer.
The governing equations are (in the B.L. approximation)
Continuity
∂ ( ρu )
∂x
+
∂ ( ρv)
∂y
=0
∂u
∂u ∂p ∂τxy
∂ ⎛ ∂u ⎞
+ ρv
+
=
=
⎜µ
⎟
∂x
∂y ∂x
∂y
∂y ⎝ ∂y ⎠
X-Momentum
ρu
Y-Momentum
∂P
=0
∂y
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(1)
(2)
(3)
Lecture 7
Page 1 of 16
ρu
Total enthalpy
∂ht
∂h
∂
∂ ⎛ ∂T ⎞
uτxy +
+ ρv t =
⎜k
⎟
∂x
∂y
∂y
∂y ⎝ ∂y ⎠
(
)
(4)
u2
is the specific total enthalpy, and µ is the viscosity. For a
2
laminar flow, µ = µ ( T ) is a fluid property. Rocket boundary layers are almost always
where ht = h +
turbulent, and µ is then the “turbulent viscosity”, where momentum transport is
effected by the random motion of turbulent “eddies”. If these eddies have a velocity
scale u' and a length scale l ' , we have, in order-of-magnitude.
µ turb. ∼ ρ u'l'
(5)
where u' is some fraction of the local u, and l ' tends to be of the order of the wall
distance y. The important points about (5) are
(a) µ turb.
µ , mostly because l '
mean free path and
(b) µ turb. is proportional to density (whereas µ is not, because the m.f.p. is inversely
proportional to ρ ).
Similarly, the last term on the right in the energy balance, representing the
convergence of heat flux, contains the “turbulent thermal conductivity” K ∼ ρ cpu'l' .
Once again, we notice that K is here proportional to density. We also note that the
“turbulent Prandtl number”
Pr =
µ t cp
kt
∼ 1 (from the orders of magnitude)
It is of some interest to note the origin and composition of the viscous term in
equation (4). If we collect the dot products u i τ t around a fluid element as shown (in
B.L. approximation),
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 2 of 16
we obtain the term
∂
uτxy
∂y
(
)
as written in (4). This can be expanded as
2
∂τxy
⎛ ∂u ⎞
∂
∂u
∂ ⎛ ∂u ⎞
+ τxy
=u
uτxy = u
⎜µ
⎟ + µ⎜
⎟
∂y
∂y
∂y
∂y ⎝ ∂y ⎠
⎝ ∂y ⎠
(
)
(6)
The 1st term in (6) is just the velocity times the viscous net force per unit volume, so
it is the part of the total viscous work that goes to accelerate the local flow. The
second term in (6) is positive definite, and it is the rate of dissipation of energy into
heat due to viscous effects. We will return later to this heating effect.
Approximate Analysis Let us manipulate the right hand side of equation (4):
∂ ⎛
∂u ⎞ ∂ ⎛ ∂T ⎞
∂ ⎡ ⎛ ∂u K ∂T ⎞ ⎤
+
⎢µ ⎜ u
⎜ uµ
⎟+
⎜K
⎟=
⎟⎥
∂y ⎝
∂y ⎠ ∂y ⎝ ∂y ⎠ ∂y ⎣ ⎝ ∂y µ ∂y ⎠ ⎦
and, since
∂h
∂T
= cp
, this yields
∂y
∂y
⎡ ⎛ u2
⎞⎤
∂ ⎢ ⎜∂ 2
1 ∂h ⎟ ⎥
µ
+
∂y ⎢ ⎜⎜ ∂y
Pr ∂y ⎟⎟ ⎥
⎢⎣ ⎝
⎠ ⎥⎦
µcp ⎞
⎛
⎜⎜ Pr ≡
⎟
k ⎟⎠
⎝
We note here that, both for laminar and turbulent flows, Pr is a constant,
independent of P and T to a good approximation. In fact, as we noted before, it is
also of order unity ( ∼ 0.9 for turbulent flows). So, the RHS of the energy equation
becomes
∂
∂y
⎡ ∂ ⎛ h u2 ⎞ ⎤
⎢µ
⎜ +
⎟⎥
2 ⎠⎟ ⎥⎦
⎢⎣ ∂y ⎜⎝ Pr
(7)
If we made the approximation Pr = 1 , then this would reduce
∂ ⎛ ∂ht ⎞
u2
. If in addition, we made
⎜µ
⎟ with ht = h +
∂y ⎝ ∂y ⎠
2
∂P
the “flat plate” approximation
0 , then the pair of equations
∂x
(1), (4) would become
further to
∂u
∂u
∂ ⎛ ∂u ⎞
+ ρv
=
⎜µ
⎟
∂x
∂y ∂y ⎝ ∂y ⎠
∂h
∂h
∂ ⎛ ∂ht
ρu t + ρ v t =
⎜µ
∂x
∂y
∂y ⎝ ∂y
ρu
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
⎫
⎪
⎪
⎬
⎞⎪
⎟⎪
⎠⎭
approximations
(8)
Lecture 7
Page 3 of 16
These are identical equations for u and ht . The same equation would also govern the
linearly transformed variables
u=
u
;
ue
where the
( )e
h=
ht − htw
(9)
hte − htw
subscript denotes the value of a variable in the local “external”
flow (just outside the boundary layer). Both u and h satisfy identical boundary
conditions:
uw = hw = 0;
ue = he = 1
(10)
and, as noted, identical governing equations. We conclude that, under the
assumption
∂P
⎛
⎞
⎜ Pr = 1, ∂x = 0 ⎟ ,
⎝
⎠
ht − hw
u
=
hte − hw
ue
(11)
u2w
= hw . This similarity relation between velocity
2
and total enthalpy profiles is known as Crocco’s analogy.
where we also noticed htw = hw +
Approximate heat flux at the wall
We are interested in the magnitude of the wall heat flux
⎛ ∂T ⎞
qw = ⎜ K
⎟
⎝ ∂y ⎠w
⎛ K ∂h ⎞
⎛ K
∂h
qw = ⎜
µ t
⎟ =⎜
⎜ cp ∂y ⎟
⎜
⎝
⎠w ⎝ µcp ∂y
(12)
⎞
⎟
⎟
⎠w
0, since uw = 0
⎛ ∂h ⎞
⎛ ∂h ⎞
⎛ ∂h ⎞
u ⎞
∂ ⎛
where we used ⎜ t ⎟ =
⎜⎜ h +
⎟⎟ = ⎜ ⎟ + ⎜ u
⎟
2 ⎠
⎝ ∂y ⎠w ∂y ⎝
⎝ ∂y ⎠w ⎝ ∂y ⎠w
w
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 4 of 16
The group
K
1
should be set equal to unity, for consistency with the stated
=
µcp
Pr
approximations. Thus
⎛ ∂h ⎞
qw = ⎜ µ t ⎟
⎝ ∂y ⎠w
Use now equation (11):
ht − hw ⎛ ∂u ⎞
⎛ ∂ ⎡
u ⎤⎞
qw = ⎜ µ
hw + hte − hw
⎟ = e
⎢
⎥
⎜µ
⎟
⎜ ∂y
ue ⎦ ⎟⎠
ue
⎝ ∂y ⎠w
⎣
⎝
w
(
)
⎛ ∂u ⎞
and notice that ⎜ µ
⎟ is the wall shear stress, τw . So
⎝ ∂y ⎠w
qw =
hte − hw
ue
(13)
τw
which is also called Reynolds analogy. A more compact form of this can be written in
terms of the Friction Coefficient
cf ≡
τw
(14)
1
ρ u2
2 e e
and the Stanton number
St =
(
qw
ρeue hte − hw
)
(15)
with the result (from (13))
St =
cf
2
(16)
One important point can be made about the result (13):
The heat flux to the wall is driven by the enthalpy (or temperature) difference
between Total external and Wall values, not between static values. This can be nonintuitive. Consider the situation near the exit of a highly expanded space nozzle,
where the bulk temperature Te may have dropped to, say, 300K due to the strong
expansion from a chamber temperature of, say, 3000K. The wall could be made of
Tungsten so as to be able to sustain relatively high temperature and cool itself by
radiation to space, so Tw could be, say, 1500K. Is the nozzle wall being heated or
cooled by the 300K gas? The answer is that it is being heated, because
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 5 of 16
Tte = Tc = 3000K, while Tw = 1500 < Tt . Are we violating the 2nd Principle of
e
Thermodynamics. Read on.
Simplified Profiles, Across the Boundary Layer
To better understand this situation, let us return to Crocco’s analogy (equation 11)
2
and write ht = h + u
, and solve for h:
2
(
h = hte − hw
) uu
−
e
u2
2
(17)
This is a quadratic relationship between h and u. For low subsonic flows h ht , so
the last term is not strong, and the relationship becomes linear in the limit.
The relationship between slopes at the wall flows from (17):
0
⎛ dh ⎞
⎛ ∂u ⎞
1 ⎛ du ⎞
⎜
⎟ = hte − hw
⎜
⎟ − ⎜u
⎟
dy
u
dy
⎝
⎠w
⎠w ⎝ ∂y ⎠w
e ⎝
(
or
)
⎛ hte − hw
⎛ dh ⎞
⎜ du ⎟ = ⎜⎜ u
⎝
⎠w ⎝
e
⎞
⎟
⎟
⎠
(18)
We can use (17) and (18) to sketch h vs. u across the boundary layer. For a case
with he > hw , this looks like
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 6 of 16
Note that whenever ue >
hte − hw there develops an intermediate temperature
maximum. But in any case, the wall slope is as if the line were coming from h t e , not
from he . The case when he < hw is more revealing even:
Now the wall slope is seen to be positive (heat into the wall), despite he < hw (as
long as ht > hw )
e
So, the quadratic portion of the Crocco relationship is responsible for the extra wall
heat; this can in turn be traced to viscous dissipation, which accumulates in the
boundary layer and elevates its temperature, so that the wall is heated even when
the outside temperature is low (as long as the flow has high speed).
Modification for Pr ≠ 1
We leave for now the issue of the non zero pressure gradient, except to note that it
introduces small modifications down to the throat. The deviations of Pr from unity
are small, and, for gases Pr < 1 (~ 0.9 for turbulent flow). This breaks the perfect
balance between dissipation and conduction responsible for Crocco’s analogy, in the
sense of favoring conduction of the dissipated heat. As a second consequence, the
temperature overshoot is reduced, and so is the wall slope of T and the heat flux to
the wall. The direct effect of higher conduction ( Pr < 1 ) is accounted for
approximately by modifying Reynolds analogy to
St =
cf
2Pr 0.6
(19)
The secondary effect (reduced overshoot) is accounted for by replacing the driving
enthalpy difference ht − hw by haw − hw , where haw is the “Adiabatic-wall enthalpy”,
e
defined as
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 7 of 16
haw = he + r
u2e
2
;r
(20)
0.9
(turbulent)
and r is the “Recovery factor”.
With these changes, the heat flux is now
qw = ρeue (haw − hw )
cf
2Pr 0.6
(21)
The Bartz heat flux formula
A very crude, but surprisingly effective representation for the friction factor cf is that
supplied by the well-studied case of fully developed turbulent flow in a pipe.
cf =
0.046
R 0.2
e
;
Re =
ρeueD
(22)
µe
where R e is the Reynolds number based on diameter D, and µe is the laminar
viscosity. Putting also h = cp T + constant, equation (21) now gives
qw = ρeuecp ( Taw
0.2
0.023 ⎛ µe ⎞
− Tw )
⎜
⎟
Pr 0.6 ⎜⎝ ρeueD ⎟⎠
=
0.026
0.2
D
( ρeue )0.8 µ0.2
e cp ( Taw − Tw )
(23)
0.026
It is common practice to define a heat transfer “gas-side film coefficient”, hg (not an
enthalpy!) by
hg ≡
qw
Taw − Tw
(24)
And, so far, we have
hg =
0.026
D0.2
( ρeue )0.8 µ0.2
e cp
(25)
At this point we note that the formulation so far has ignored the strong variations of
ρ and µ across the boundary layer since these quantities depend on temperature as
ρ ∼
1
(at P=constant) ; µ ∼ T w ( w
T
0.6 )
(26)
A commonly used approach to including these variations is to replace ρe and µe in
equation (25) by their values at some intermediate temperature <T>:
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 8 of 16
⎛< T >⎞
µe → µe ⎜⎜
⎟⎟
⎝ Te ⎠
Te
;
<T>
ρe → ρe
w
(27)
and <T> can be evaluated by several empirical rules. For Mach numbers not much
higher than 1, we can simply use
<T>
Te + Tw
2
(28)
Making the replacements of (27) in equation (25), we obtain
hg =
0.026
D0.2
( ρeue )
0.8
0.8 − 0.2w
⎛ Te ⎞
⎜
⎟
⎝< T >⎠
µ0.2
e cp
(29)
which is one form of Bartz’ formula. A more useful form follows from the continuity
equation:
i
P At
m
ρeue =
= c
, with
A
c* A
R g Tc
Γ (γ)
,
and where A is the local cross-section, and A t the throat cross-section. Substituting
2
A
⎛D ⎞
in (29), and using t = ⎜ t ⎟ , the final form is
A
⎝D⎠
0.8
hg =
0.026 ⎛ Pc ⎞
⎜
⎟
D0.2
⎝ c *⎠
t
1.8
⎛ Dt ⎞
⎜
⎟
⎝D⎠
⎛ Te ⎞
cp µ0.2
⎟
e ⎜
⎝< T >⎠
0.8 − 0.2w
(30)
Several important trends and observations can be made now:
⎛
1
(a) Smaller throat diameter leads to larger heat flux ⎜ ∼ 0.2
⎜ D
t
⎝
straight from the Reynolds no. dependence of cf .
(
⎞
⎟ . This comes
⎟
⎠
)
(b) Heat flux is almost linear in chamber pressure ∼ Pc0.8 . This limits the
feasibility of high chamber pressures, which are otherwise very desirable.
⎛ ⎛ D ⎞1.8 ⎞
(c) Maximum heat flux occurs at the throat ⎜ ∼ ⎜ t ⎟ ⎟ . One critical design
⎜ ⎝D⎠ ⎟
⎝
⎠
consideration is therefore the thermal integrity of the throat structure.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 9 of 16
(d) Lighter gasses lead to higher heat fluxes, through the combined effects of cp
1 ⎞
⎛
and c * ⎜ hg ∼ 0.6 ⎟
M ⎠
⎝
0.8 − 0.2w
0.68
⎛ T ⎞
⎛ Te ⎞
(e) The factor ⎜ e ⎟
is greater than unity. This
⎜
⎟
⎝< T >⎠
⎝< T >⎠
enhancement of heat flux follows mainly from the fact that the gas in the
boundary layer is mostly cooler than in the core, hence denser. We showed
before that the turbulent heat conductivity is proportional to density.
Example
Consider the Space Shuttle Main Engine (SSME), which is a Hydrogen-Oxygen rocket
with (roughly) these characteristics:
2.2 × 107 Pa
Pc = 220 atm
Tc = 3600 K
M = 15g / mol
r
1.25
c* =
R g Tc
Γ (γ)
2600 m / s
cp =
γ R
γ −1 M
µe
3 × 10−5 Kg / m / s
Tthroat =
2800 J / Kg / K
2
Tc
γ +1
3200 K = Te
Tw = 1000 K
We calculate then
Te + Tw
3200 + 1000
=
= 2100 K
2
2
< T >=
0.8 − 0.2w
⎛ Te ⎞
⎜
⎟
⎝< T >⎠
0.61
⎛ 3000 ⎞
⎜ 2100 ⎟
⎝
⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(at the throat)
1.3
Lecture 7
Page 10 of 16
and so, using equation (30),
160, 000 w / m2 / K
hg
and qw
(
160, 000 Tawt − 1000
)
1
( Taw )t
qw
γ −1 2⎞
⎛
= Tt ⎜1 + r
µ ⎟ = 1.057 × 3200
2
⎝
⎠
0.9
3400 K
(slightly less than Tc )
160, 000 × 2400 = 3.8 × 108 W / m2
This is a very high level of heat flux. To visualize the implications, suppose this qw
had to be transmitted through a thin metal plate (thickness δ , thermal conductivity
k).
∆T
where ∆T is the temperature drop through the metal .
δ
As an initial guess, suppose the metal were stainless steel (K 20 W / m / K ) , and
One would have qw = K
δ =1mm. Then
∆T =
qw δ 3.8 × 108 × 10−3
=
= 19, 000 K !!
K
20
Obviously, this is unacceptable. Try using Copper instead, with K 400 W / m / K
(twenty times better). This gives ∆T =950K, still not acceptable (copper would be
very soft then). The plate would have to be thinner and made of copper. Not an easy
problem.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 11 of 16
More rationally
The St or hg should depend on x, distance from start of nozzle, since the B.L. is still
developing (not fully developed). In addition, there should be some accounting for
•
•
•
acceleration
property variation through B.L.
cylindrical geometry
The article by Rubsin and Inonye (ch. 8 in Rosenhow and Hartnett’s Handbook of
Heat Transfer, McGraw-Hill, 1973) gives a general formula for turbulent B.L. In an
cylinder, with acceleration:
St ( x ) =
s=
A
(and hg = ρeuecpSt )
n
⎛ρu x ⎞
s ⎜ e e eff ⎟ Fc1−n FRnθ
⎝ µe
⎠
cf 2
ch
1 found walls.
A⎫
⎬ = constants, depending on Reynolds no. based on mom. th.
n⎭
R eθ > 4000 , A = 0.0131, n =
1
7
R eθ < 4000 , A = 0.0293 , n =
1
5
Fc ⎫⎪
⎬ = Factors for property variability. Can take several nearly equivalent forms. A
FR θ ⎪
⎭
simple one from Eckert, is
Fc =
FRθ =
ρe
ρ (< Τ >)
µe
µ (< Τ >)
Taw = Te + r
=
<Τ >
Te
⎛ Te ⎞
⎜
⎟
⎝< Τ >⎠
w
(w
⎫
⎪
⎪
T
T
⎪< Τ >
= 0.28 + 0.50 w + 0.22 aw
⎬
Te
Te
⎪ Te
⎪
0.6 ) ⎪
⎭
u2e
γ −1 2⎞
⎛
= Te ⎜ 1 + r
Me ⎟
2
2
⎝
⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 12 of 16
r
0.9 (recovery factor)
The “effective distance” xeff is related to the actual distance x through an integral
(accounting for memory of past acceleration)
xeff ( x ) =
∫
where f =
z=
x
0
f ( x ')
f (x)
dx '
(
ρeue zRµne
haw − hw
hte − hw
)
1
1−n
n
Fc FR1 −n
θ
⎛
u2
⎜ haw = he + r e
⎜
2
⎝
⎞
⎟
⎟
⎠
R=R(x)= body radius at x.
For a quick estimate of R eθ , we can simplify further to the flat-plate case, in which
dθ c f
,
=
dx
2
⎧0.0128 R1 4
eθ
cf
⎪
with
=⎨
2
⎪0.0065 R1e 6
θ
⎩
dR eθ
dR ex
or
(R
(R
eθ
eθ
) , and with
> 4000 )
< 4000
⎧ 0.0128
⎧4 5 4
⎪
14
⎪⎪ 5 R eθ = 0.0128 R ex
⎪ R eθ
=⎨
→⎨
→ R eθ
⎪ 0.0065
⎪ 6 R 7 6 = 0.0065R
ex
⎪ R1 6
⎪⎩ 7 e
eθ
⎩
R eθ
R ex
dR eθ
dθ
=
dx dR ex
⎧0.0366 R 4 5
ex
⎪
⎪
⎨
⎪
⎪0.0152 R 6e 7
x
⎩
R eθ < 4000
(R
ex
< 1.99 × 101
)
R eθ > 4000
⎧ 0.0366
⎪
0.2
θ ⎪ R ex
=
=⎨
x ⎪ 0.0152
⎪ R1 7
ex
⎩
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 13 of 16
For large rockets, R ex
throat
tends to be ∼ 107 − 108 , so the high R e formulas should
be better, despite the common use of Bartz’s formulae, which are based on the low
R e formulation. Fortunately, differences tend to be small, and are marked often by
other uncertainties (surface films, fluid properties).
Example and Comparisons:
Consider nozzle R = x tan x +
x=
R ± R 2 − R 2t
2 tan α
R c − R 2c − R 2t
with origin at x = xc =
and
R 2t
1
4 tan α x
2 tan α
Rc
= 1.5 , α = 15o
Rt
and going through throat at x = xt =
Rt
2 tan α
Using γ=1.25 and the R eθ > 4000 option, we find
⎛ xeff
⎜⎜
⎝ Rt
⎞
=
⎟⎟
⎠throat
where
xt
Rt
∫
xc
Rt
5
M 12
⎛
1.125
⎜⎜
2
⎝ 1 + 0.125M
R
x
=
tan15o +
Rt
Rt
⎛ x
⎜
⎝ Rt
1 ⎛ 1 + 0.125M2
and 1 ⎜
⎜
1.125
M 2⎝
2.25
⎞
⎟⎟
⎠
=
1.875
⎞
⎟⎟
⎠
⎛
0.6979
⎜⎜
2
⎝ 0.6515 + 0.0464 M
0.9
⎞
⎟⎟
⎠
⎛ x
d ⎜⎜
⎝ Rt
⎞
⎟⎟
⎠
1
R
4
⇒
(x)
R
⎞
t
o
⎟ tan15
⎠
R
⇒ M (x)
Rt
⎛x ⎞
The integration gives ⎜⎜ eff ⎟⎟
= 1.0892
⎝ R t ⎠throat
Compared to
x t − xc
= 1.153
Rt
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
⎛
⎞
xc
= 0.713 ⎟⎟
⎜⎜ and
Rt
⎝
⎠
Lecture 7
Page 14 of 16
Since xeff appears to the 1
power, the memory/acceleration effect (up to the
7
throat) is insignificant.
The throat St is then
( Taw
(0.9)
= 3263 )
(using r=1, Tw = 1000 K , Tc = 3300 K,
(St )throat
Te
2
=
3300 = 2933K )
t
2.25
0.0131
=
1
⎛ ρ u ( x t − xc ) ⎞ 7
⎜⎜
⎟⎟
µ
⎝
⎠throat
0.771
⎛< T >⎞
⎜
⎟
⎝ Te ⎠throat
(3263)
(0.6952)
1000
3300
<T >
= 0.28 + 0.50
+ 0.22
= 0.6979
Te
2933
2933
Take Pc = 2 × 107 N / m2 ,
M= 25 g/mol
*
c =
R g Tc
=
Γ
⇒ ( ρu ) t =
8.314
3300
0.025
2.25
= 1592 m / s
⎛ 2 ⎞ 0.5
1.25 ⎜
⎟
⎝ 2.25 ⎠
Pc
*
c
= 12560 Kg / s / m2
xc 1.5 − 1.52 − 1
=
= 0.7128
Rt
2 tan15o
xt
1
=
= 1.866
Rt
2 tan15o
and, with R t = 0.3 m ,
0.6
µ
⎛ T ⎞
6.8 × 10−5 ⎜
⎟
⎝ 3000 ⎠
⇒ µthroat = 6.70 × 10−5 Kg m sec
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 15 of 16
One gets,
(St )throat
(0.00124 using xt instead of xt − xc )
= 0.00133
Using the R eθ < 4000 option (small rockets)
(St )throat
=
0.0293
0.68
0.2
⎛< T >⎞
⎛ ρ uxeff ⎞
⎜
⎟
⎜
⎟
⎝ µ ⎠throat ⎝ Te ⎠throat
and using again xeff = xt − xc , etc,
we get ( St )throat = 0.00102
(0.000933 using xt )
For comparison, the “fully developed pipe flows” formulation would give
St =
0.2
Rg
ρ ue cp
=
0.026 ⎛ c* ⎞
⎜⎜ ⎟⎟
D0.2
⎝ Pc ⎠
t
0.8 − 0.2w
⎛ Te ⎞
µ0.2
⎟
e ⎜
⎝< T >⎠
⎛ At ⎞
⎜
⎟
⎝ A ⎠
0.9
1 at throat
(St )throat
= 0.000958
This is close to the R θ < 4000 results above (and, indeed, the coefficients are for
R θ < 4000 ). But this appears coincidental, based on the fact that for most nozzles,
∆x ∼ R t .
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 7
Page 16 of 16