16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5: Nozzle Design: Method of Characteristics
The Method of Characteristics (Ideal Gas)
(Ref. Phillip Thompson Compressive Fluid Dynamics, McGraw Hill, 1972, Ch. 9)
2-D or axisymmetric
Homentropic as well as isentropic
ur
→ ∇×u = o
plus ∇
ur
(ρ u ) = 0
0
ur ur 1
⎛ u2 ⎞ ur ur 1
and u ∇u + ∇p = o ⇒ ∇ ⎜ ⎟ + ω × u + ∇p = 0
ρ
ρ
⎝ 2 ⎠
Use intrinsic co-ordinates
r
Eq. of motion along s :
u
r
Eq. of motion along n :
u2
1 ∂p
∂u2 2
=
=−
R
∂n
ρ ∂n
Now
1 ∂ϑ
≡
R ∂s
u2
∂ϑ
∂u
= −u
∂s
∂n
∂u 1 ∂p
+
=o
∂s ρ ∂s
∂u
∂ϑ
+u
=o
∂n
∂s
(good no p in it)
(1)
(2)
ur
(can also get this from ∇ × u = o )
Continuity:
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
∂
( ρ u 2πr δn ) = o
∂s
Lecture 8
Page 1 of 20
1 ∂ρ 1 ∂u 1 ∂r
1 ∂δn
+
+
+
=o
ρ ∂s u ∂s r ∂s ∂n ∂s
( δs ) sin ϑ =
∂r
1
∂ϑ
( ∂s ) sin ϑ
∂s
r
∂n
∂δn
⎛ ∂ϑ ⎞
δn ⎟ δs
ds = ⎜ −
∂s
⎝ ∂n ⎠
1 ∂ρ 1 ∂u ∂ϑ
sin ϑ
+
−
=−
r
ρ ∂s u ∂s ∂n
Homentropic: dp = c 2 d ρ
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
⎛ 2 ⎛ ∂p ⎞ ⎞
⎜⎜ c = ⎜
⎟ ⎟⎟
⎝ ∂ρ ⎠s ⎠
⎝
Lecture 8
Page 2 of 20
so
and, from s eq. of motion,
−
1 ∂p 1 ∂u ∂ϑ
sin ϑ
+
−
=−
2
r
ρ c ∂s u ∂s ∂n
∂p
1 ∂p
∂u
:
so you can eliminate
= −u
∂s
ρ ∂s
∂s
u ∂u 1 ∂u ∂ϑ
sin ϑ
+
−
=−
r
c 2 ∂s u ∂s ∂n
1⎛
u2 ⎞ ∂u ∂ϑ sin ϑ
1
−
+
+
=
⎜
⎟
r
u⎝
c 2 ⎠ ∂s ∂n
2
M -1
M 2 − 1 ∂u ∂ϑ sin ϑ
+
=
u
∂s ∂n
r
Introduce the Mach angle μ = sin−1
tan μ
Then (2)
1
= tan−1
M
1
2
M −1
(4)
tan μ =
1
M2 − 1
M 2 − 1 ∂u ∂ϑ
+
=o
u
∂n ∂s
M 2 − 1 ∂u
∂ϑ tan μ sin ϑ
+ tan μ
=
u
∂s
∂n
r
And (4)
Introduce the Prandtl-Meyer function ω (M) by
dω =
M2 − 1
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
du
u
(to be integrated later)
Lecture 8
Page 3 of 20
tan μ
∂ω ∂ϑ
+
=o
∂n ∂s
then
∂ω
∂ϑ tan μ sin ϑ
+ tan μ
=
∂s
∂n
r
add and subtract to obtain the “Characteristics form” (single differential operator per
equation)
∂ ⎞
tan μ sin ϑ
⎛ ∂
⎜ ∂s + tan μ ∂n ⎟ ( ω + ϑ ) =
r
⎝
⎠
⇒
∂ ⎞
tan μ sin ϑ
⎛ ∂
⎜ ∂s − tan μ ∂n ⎟ ( ϑ − ω)
r
⎝
⎠
∂
∂ ⎞
sin μ sin ϑ
⎛
⎜ cos μ ∂s + sin μ ∂n ⎟ ( ω + ϑ ) =
r
⎝
⎠
∂
∂ ⎞
sin μ sin ϑ
⎛
⎜ cos μ ∂s − sin μ ∂n ⎟ ( ϑ − ω) = −
r
⎝
⎠
r + ⎧cos μ ⎫
In (s, n) coordinates, 1 = ⎨
⎬
⎩ sin μ ⎭
cos μ
r+
∂
∂
∂
+ sin μ
=1 ∇ =
∂s
∂n
∂m+
cos μ
r−
∂
∂
∂
− sin μ
=1 ∇ =
∂s
∂n
∂m−
r − ⎧ cos μ ⎫
1 =⎨
⎬
⎩− sin μ ⎭
so
(m+, m- are lengths along characteristics)
⎛ m− inclined ϑ + μ ⎞
⎜⎜ +
⎟⎟
⎝ m inclined ϑ − μ ⎠
sin μ sin ϑ ⎫
⎧ ∂
⎪ ∂m+ ( ϑ + ω) = +
⎪
r
⎪⎪
⎪⎪
⎨
⎬
⎪ ∂
sin μ sin ϑ ⎪
⎪
⎪
ϑ − ω) = −
− (
r
⎩⎪ ∂m
⎭⎪
For 2-D, r → ∞
⇒
ϑ + ω =const. along m+ ≡ I+ (inclined ϑ − μ )
ϑ − ω =const. along m- ≡ I- (inclined ϑ + μ )
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
Lecture 8
Page 4 of 20
2-D Simple Regions
Consider a uniform region; the flow from it enters
some disturbed region, like a wall turning. One of
the m families originate in the uniform region (m+
in example) and carries a constant invariant, e.g.
ϑ + ω = ϑo + ωo everywhere downstream. Along one
of the other characteristics (m- here), we carry a
constant (which varies from ch. To ch. of that
family); we can evaluate it at the wall, for instance
ϑ − ω = ϑw − ωw along each m- line and
ωw = ϑo + ωo − ϑw so ϑ − ω = 2ϑw − ( ϑo + ωo )
then
2ϑ = ϑo + ωo + ϑw − ωw = 2ϑw along each m− →
2ω = ϑo + ωo − ( ϑw − ωw ) along each m
−
→
ϑ = ϑw
ω = ϑo + ωo − ϑw = ωw
so ϑ and ω (and hence M, μ) are constant along each m-, and these m- lines are
straight (const. ϑ + μ ).
This is a Simple Region (one of the invariants is constant). The turning of the flow is
dictated by how ϑ changes on a m+ line, as different m- lines are crossed. Since on a
m+ we have ϑ + ω = ϑo + ωo , changes of ϑ are equal and opposite to those in ω (ω
increases as M does in the expansion, so ϑ decreases (increase negatively) at the
same rate.) So, we can interpret ω as the magnitude of the isentropic flow turning in
a simple region, i.e., when nothing varies along one characteristic family.
Calculation of ω (M)
dω =
M2 − 1
du
=
u
⎛ dM 1 dT ⎞
M2 − 1 ⎜
+
2 T ⎟⎠
⎝ M
but
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
Lecture 8
Page 5 of 20
dω =
γ −1 2
⎛
M
M2 − 1 ⎜
2
⎜1 −
γ −1 2
M
⎜⎜
M
1+
⎝
2
⎞
⎟
⎟ dM =
⎟⎟
⎠
M2 − 1
M
dM
γ −1 2
M
1+
2
Integrates (with ω =0 at M=1) to
ω = K tan−1
For M → ∞
M2 − 1
− tan−1 M 2 − 1
K
ω→K
γ
ω (M → ∞ )
⎛
⎜⎜ K =
⎝
γ +1⎞
⎟
γ − 1 ⎟⎠
⎞
π π π
π⎛ γ +1
− = ( K − 1) = ⎜
− 1⎟
⎜
⎟
2 2 2
2 ⎝ γ −1
⎠
1.2
208o
1.25
180o
1.4
130.5o
5/3
90o
(Max. turning from M=1)
So rocket exhaust ( γ
1.2 − 1.3 ) can turn backwards at a sonic nozzle exit to
vacuum)
(but very long density, long mfp, so continuum
approach fails at some point in the expansion,
molecules then continue in straight line).
Actually, one starts from some high Me>1, so
the actual turning is through ω∞ − ω (Me ) , not
ω∞ , even in a vacuum:
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
Lecture 8
Page 6 of 20
Example of Application: Ideal 2-D plug Nozzle at Design Condition
Along m− , ϑ − ω = ϑa − ωa
Simple region (Prandtl-Meyer fan
centered at lip L)
In particular, at inlet M=1, so
ϑthroat = −ωa
tan μ =
From geometry, μ − ϑ + ϕ =
ϑt = −ωa
ϕ=
π
+ K tan−1
2
ϕ=
M2 − 1
−
π
+ ωa
2
π
+ω−μ
2
M2 − 1
− tan−1 M 2 − 1 − tan−1
K
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
1
π
ϑt
2
μ − ω + ωa + ϕ =
Sub. ϑ = ω − ωa
r dϕ
=
dr
1
2
M −1
⎛
⎜⎜ K =
⎝
γ +1⎞
⎟
γ − 1 ⎠⎟
π
2
Lecture 8
Page 7 of 20
M2 − 1
K
ϕ = K tan−1
tan
⇒
ϕ
=
K
M2 − 1
1
=
K
K tan μ
r dϕ
=
dr
1
K tan
dr
=K
r
ϕ
K
ϕ⎞
⎛
ϕ
d ⎜ cos ⎟
K
⎠
K dϕ = −K 2 ⎝
ϕ
ϕ
cos
cos
K
K
sin
r =
For ϕ = 0,r = ht (throat height)
In particular, at end of expansion
ra =
ht
K2
ϕa ⎞
⎛
⎜ cos K ⎟
⎝
⎠
and
r =
const.
K2
ϕ⎞
⎛
⎜ cos K ⎟
⎝
⎠
ht
K2
ϕ⎞
⎛
⎜ cos K ⎟
⎝
⎠
M = Ma → ϕa = K tan−1
Ma2 − 1
K
ha = ra sin μa =
, then
ra
Ma
xa = ra cos μa = ra 1 −
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
1
Ma2
Lecture 8
Page 8 of 20
Numerical Application
γ
Po
γ − 1 2 ⎞ γ −1
⎛
= 100 = ⎜ 1 +
⇒ Ma =
Ma ⎟
Pa
2
⎝
⎠
Take γ = 1.3 ,
Then
Ma =
γ −1
⎡
⎤
2 ⎢⎛ Po ⎞ γ
⎥
⎜ ⎟ − 1⎥
γ − 1 ⎢⎝ Pa ⎠
⎥⎦
⎣⎢
0.3
⎤
2 ⎡
1.3
⎢100 − 1⎥ = 3.554
0.3 ⎣
⎦
⎛ M2 − 1 ⎞
−1
M2 − 1
ωa = K tan−1 ⎜
⎟ − tan
K
⎝
⎠
ωa = 67.35o
⇒
K=
γ +1
=
γ −1
2.3
= 2.769
0.3
ϑt = −67.35o
⎛ M2 − 1 ⎞
o
Also ρa = K tan−1 ⎜
⎟ = 141.01
⎝ K ⎠
and so
ra
1
1
=
=
K2
2.7692
ht
ρa ⎞
141.01 ⎞
⎛
⎛
⎜ cos 2.769 ⎟
⎜ cos K ⎟
⎝
⎠
⎝
⎠
From the geometry, ha=ra sin μa =
ra
Ma
ha
1
=
34.41
ht
3.554
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
ra
= 34.41
ht
ha
=9.684
ht
Lecture 8
Page 9 of 20
2
and
xa
⎛ 1 ⎞
= ra cos μa = 34.41 1 − ⎜
⎟
ht
⎝ 3.554 ⎠
xa
=33.02
ht
Very long and pointy, should be truncated.
Non-Simple Regions. When the characteristics of both families intersect some
upstream disturbance, they affect each other’s invariant, and characteristics are no
longer straight, and no longer carry constant flow properties (except for their own
invariant)
0-1-4 is a simple region
0-4-41 is uniform
1-4-10 is non-simple
4-41--11 is a simple region
Then we need to calculate in a step-by-step manner, carrying to each point the two
invariants I+, I- from neighboring upstream points, along the m+, m- lines from them
to us. After this is done, we know the new segments of m+, m- from our point (slopes
ϑ + μ , ϑ − μ ), so we can extend the grid as we go. Notice the flowfield properties can
be found first everywhere and only then we need to come back and place the points
geometrically.
Example: Design a 2-D ideal nozzle to expand from near sonic conditions (M0 = 1.1)
to Me = 3. Use only 4 characteristics. Use a corner expansion as a starter,
γ =1.25
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
Lecture 8
Page 10 of 20
K =
ωo = 3 tan−1
1.21 − 1
− tan−1 1.21 − 1 = 1.435o ; ϑo = 0
3
ωe = 3 tan−1
9 −1
− tan−1 9 − 1 = 59.413o ; ϑe = 0
3
2.25
=3
0.25
At inlet: I − = ϑ − ω = ϑo − ωo = −1.435o (also at 4)
At exit: I + = ϑ + ω = ϑe + ωe = 59.413o
7, 9, 10)
2ϑ4 = −1.435 + 59.413 = 57.978
At 4, then,
2ω4 = 59.413 + 1.435 = 60.848
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
(also at 4,
ϑ4 = 28.989o
ω4 = 30.424o
Lecture 8
Page 11 of 20
I1+ = 1.44o
Select:
I2+ = 20.76o
I3+ = 40.09o
()
()
Point
M
I+ = ϑ + ω
1
1.1
+1.44
2
3
4 (and 4w)
5
6
7 (and 7w)
8
9 (and 9w)
10 (and 11)
1.439
1.726
2.013
1.726
2.013
2.315
2.315
2.641
.3
20.76
40.09
59.41
20.76
40.09
59.41
40.09
59.41
59.41
o
I4+ = 59.41o
()
ϑ
-1.44
1.44
-1.44
-1.44
-1.44
-20.76
-20.76
-20.76
-40.09
-40.09
-59.41
11.10
20.77
30.43
20.76
30.43
40.09
40.09
49.75
59.41
I- = ϑ − ω
o
ω
o
()
()
o
ϑ+μ
0
65.38
65.38
9.66
19.33
28.99
0
9.67
19.33
0
9.66
0
44.02
35.41
29.79
35.41
29.79
25.59
25.59
22.25
19.47
53.68
54.74
58.78
35.41
39.46
44.92
25.59
31.91
19.47
o
μ
()
o
ϑ−μ
Note the very shallow angles of the m+ lines (from 4, 7, 9, 10) which will put point
10 far to the right, and point 11 (at slope (m+) of 19.47o even farther.
Locating the points geometrically
− x a ) tan α + ( x c − x b ) tan β = y a − y b
xc =
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
o
-65.38o
-34.36
-16.08
-0.80
-35.41
-20.12
6.26
-25.59
-12.59
-19.47
Notes
Points 1-4: same ϑ − ω
Point 5: Here ϑ = 0 (a boundary condition) and ϑ + ω = 20.76
Point 6-7: same ϑ − ω
Point 8: Here ϑ = 0 and ϑ + ω = 40.09
Point 9: same ϑ − ω
Point 10: Here ϑ = 0 and ϑ − ω = 59.41
(xc
()
ya − yb + xa tan α + xb tan β
tan α + tan β
Lecture 8
Page 12 of 20
yc − yb = ( xc − xb ) tan β = tan β
yc =
( ya − yb ) + ( xa − xb ) tan α
tan α + tan β
yb tan α + ya tan β + ( xa − xb ) tan α tan β
tan α + tan β
put xa, ya into 1, 2
tan α into 3
Run 2 →
xc (st. in 7)
xb, yb into 4.5
Run 3 →
yc (st. in 7)
tan β into 6
Finding the (x,y)’s is very laborious. Accuracy can be increased by averaging
together the angles at the ends of each segment, which can be done because those
α + αc
β + βc
, β = b
.
angles come from the first pass. For instance, α = a
2
2
16.512, Rocket Propulsion
Prof. M. Martinez-Sanchez
Lecture 8
Page 13 of 20
Boundary with Prescribed Pressure
P=
Pc
r −1 2⎞
⎛
⎜1 + 2 M ⎟
⎝
⎠
r
,
r −1
and M = M ( w ) ,
so P = P ( w )
(given Po )
or w = w (P )
So, if P is fixed on a boundary (contact surface), we can assign w there (just as we
assign θ on a solid boundary)
From known point a,
θ − w = θa − wa
So θ = w (P ) + θa − wa
and this determines the slope of the boundary, and that of the “reflected” m+
θ + w = 2w (P ) + θa − wa
More General Contact Surface Condition
If the outside fluid is also supersonic, we must solve on both sides of the contact
surface, making sure P and θ are common at each boundary point
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 14 of 20
P1 ( w1 ) = P2 ( w2 )
θ1 = θ2
From m1+ know I1+ = θ + w1+ (P )
From m2− know I2− = θ − w2− (P )
So, I1+ − I2− = w1+ (P ) + w2− (P )
⇒ Solve for P ⇒ w1 , w2
and
I1+
+
2
I2−
=θ+
w1+ (P ) + w2− (P )
known now
2
Modifications for Axisymmetric Conditions
∂
sin μ sin θ ⎫
⎪
r
⎪
∂m
⎬
∂
sin μ sin θ ⎪
( θ − w) = −
⎪⎭
r
∂m−
+
( θ + w) =
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 15 of 20
(1) Calculate ( x, r )c from the angle ( μ − θ )a , ( μ + θ )b
( x − xa ) tan α + ( x − xb ) tan β = ra − rb
⇒x−
x a tan α + xb tan β + ra − rb
tan α + tan β
and r − rb = ( x − xb ) tan β
(2) Δm+ =
x − xa
cos α
Δm− =
x − xb
cos β
(3) Advance invariants θ + w , θ − w based on μ, θ at a, b:
sin μa sin θa
⎫
Δm+ ⎪
ra
⎪
⎬
sin μb sin θb
−
Δm− ⎪
⎪
rb
⎭
θ + w = ( θ + w )a +
θ − w = ( θ − w )b
(r at c, computed in (1))
(4) M = M ( w ) , μ = μ (M)
(5) Iterate from (1) with α, β averaged between (a, b) and c:
α→
( μ − θ )a + ( μ − θ )c
β→
( μ + θ )b + ( μ + θ )c
2
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 16 of 20
ra + rb
2
⎫
⎪
⎪
⎪
μa + μb ⎪⎪
μ→
⎬ better
2 ⎪
⎪
⎪
θ + θb ⎪
θ→ a
⎪⎭
2
r→
(6) Continue to new point.
So, computation of ( θ,M) is now coupled to that of ( x, r ) , whereas in 2-D ( θ,M)
can be found first. But the actual amount of computation is not much more (only
the iteration stops).
On the axis, r → 0 , but θ = 0 , so
1. x = xa + ra tan α
2. Δm+ =
ra
sin α
3. θc = 0; wc = ( θ + w )a +
sin μa sin θa
ra
4. Mc = M ( wc ) , μc = μ (Mc )
5. α →
( μ − θ )a + μ c
2
ra , θa ,
, go to (1) (once)
6. Continue
Extension to Cases with Weak Shocks
Since the entropy jump in a shock increases only as the cube of
ΔPshock
ρu20
, the
isentropic assumption can be approximately extended when characteristics of our
family show some mild convergence (in principle, that is always indicative of shock
formation, because they carry conflicting information). When is the convergence too
strong? Since characteristics are discretized, for weak convergence the ones of our
family will converge, but not cross, and as long as they don’t, it should be OK. Of
course, with finer resolution they will cross, but the loss of accuracy in ignoring that
is of the same order as that increased by the coarse discretization in the first place.
This allows us to calculate off-design nozzle performance, like an overexpanded plug
nozzle.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 17 of 20
2-D Spike Nozzle with Pa < Padesign
Pressure forces from hot-gas bathed surfaces are the same as at design. Net thrust
is increased because the Pa contribution PaA e is reduced:
(
)
F = Fdes. + Pades. − Pa A e
and so Fvac = Fdes. + Pades. A e
(just as for a bell)
“Practical” Nozzle Designs
Ideal nozzle are too long, last portion has small wall angle, so small thrust
contribution. With small ?, maybe negative contribution. So, options:
(a) Constrain length, ask for contour that gives highest thrust/given L.
Methods of calculus of variations (Raw nozzle, Ref : “ Exhaust Nozzle
Contour from Optimum Thrust”, Jet Propulsion 28 (June 1958): 377-382.
Exit flow non-parallel, non-uniform, computationally high.
(b) Ad-hoc method (widely used) is to truncate contract an ideal nozzle.
(1) Design wall contour for desired Pe Pc
(2) If longer than desired L truncate it to some intermediate area ratio.
(3) Contract this truncated nozzle to desired length
x
L
x'a = xb' c = xb' desired
xd
L ideal
(4) Translate profile to right metal kink at P smoothes out.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 18 of 20
Gives “adequate” performance but less than ? nozzle. Not really
justified. (Hoffman, J. D. J. of Propulsion 3 (March-April 1987): 150156.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
Page 19 of 20
XRS-2200 Engine Data
Thrust, lbf
At Sea Level
In Vacuum
206,500
268,000
Specific Impulse, sec.
At Sea Level
In Vacuum
339
439
Propellants
O2, H2
Mixture Ratio (O/F)
5.5
Chamber Pressure, psia
857
Cycle
Gas Generator
Area Ratio
58
Throttling, Percent Thrust
40-119
Differential Throttling
+/- 15%
Dimensions, Inches
Forward End
133 high x
88 wide
Aft End
46 high x
88 wide
Forward to Aft
79
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 4-5
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