16.512, Rocket Propulsion,
Prof. Manuel Martinez-Sanchez
Lecture 2: Rocket Nozzles and Thrust
Rocket Thrust (Thermal rockets)
i
m=
∫∫ ρ u dA
n
e
Ae
∫∫
P dSx −
Solid int .
surfaces
∫∫ P dA
e
ex
=
Ae
∫∫ u
x
i
dm
(Tanks included)
Note:
∫∫
∫∫
s.,int
PadSx −
( P − Pa )
( ρ un )dAe
Ae
∫∫ P dA
a
ex
= 0, so subtract,
Ae
dSx =
Solid int .
∫∫ ( P
e
Ae
− Pa ) dAex +
∫∫ ρu u dA
x n
e
Ae
Thrust ≡ F
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 2
Page 1 of 7
∫∫ ρu u dA
x n
In general then, define ue =
e
Ae
∫∫ ρu
n
dAe
Ae
∫∫ P dA
e
and P e =
⇒
ex
Ae
Aex
i
(
)
F = m u e + P e − Pa Aex
If things are nearly constant on spherical caps, modify control volume to
spherical wedge:
i
m=
∫∫ ρu dA
r
Ae
∫∫
( P − Pa ) dSx
−
int .
solids.
dAex = dA cos θ
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
∫∫ ( P
e
− Pa ) dAex =
Ae
∫∫ ( ρu ) u dA
r
x
Ae
ux = ur cos θ
Lecture 2
Page 2 of 7
Define
ue =
∫∫
Ae
ρ ur ux dA
i
m
∫∫
Ae
; Pe =
Pe dAex
Aex
and use
dA = 2π r sin θ rdθ
For ideal conical flow, ρ , ur , P are constant over Ae . Then
ue =
ρ ur2
∫∫
ρ ur
Ae
cos θ dA
∫∫
Ae
dA
= ur
∫
α
0
2π r sin θ cos θ dθ
∫
α
0
2π r sin θ dθ
1
sin2 α
= ur 2
1 − cos α
or
ue = ur
1 + cos α
2
Also, since Pe = const on the exit surface,
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
P e = Pe
Lecture 2
Page 3 of 7
∫ ( P − P )dA
a
x
+
s.s.
∫ (P
e
− Pa )dAx =
Ae
∫ ( ρu dA)u
n
x
Ae
F
(
i
)
F = m u e + P e − Pa Aex
i
m=
ue =
∫
ρundA
Ae
∫
∫
Ae
ρ unux dA
Ae
Pe
∫
=
∫
Ae
Pe dAx
Ae
Ax =
∫
Ae
ρ un dA
dAx
dAx
At design, Pe = Pa (and parallel flow beyond). Also ue x
i
Then uniform → F = m uex
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 2
Page 4 of 7
Energy Considerations
So, momentum balance gives the Thrust Equation. What does an Energy Balance
give?
Start with a near-stagnant flow in the upstream plenum (“combustion chamber”, or
“nuclear heater” or “arc heated plenum”). The total specific enthalpy
1
htc = hc + υc2 ≅ hc may be different for different streamlines, due to combustion
2
“streaks:, arc constriction, etc., But along the flow expansion in the nozzle, ht is
conserved for each streamline. At the exit,
he +
1 2
υe = hto
2
(each streamtube)
(
)
υe = 2 htc − he ≅ 2 ( hc − he )
or
For a well-expanded nozzle, with large area ratio, he → o by adiabatic expansion, and
υe tend to a max. υe MAX = 2 h tc . In any real, finite expansion, he ≠ o, so some of
htc is wasted as thermal energy in the exhaust. Define a nozzle efficiency.
ηN =
htc − he
htc
⎛P
h
T
For ideal gas, e = e = ⎜⎜ e
hc
Tc ⎝ Pc
=1−
⎞
⎟⎟
⎠
γ −1
γ
he
h
≅1− e
htc
hc
. But, in any case,
υe = υe MAX ηN = ηN 2 htc
Since Pe ≅ uniform, so is ηN , even when
(i.e., ηN =
υe2 2
htc
)
htc is not. Also, υe is non-uniform if htc is
htc ).
(in proportion to
The Jet Power is the kinetic energy flow out of the nozzle
PJet =
(
)
1
m htc − he = ηN htc m
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 2
Page 5 of 7
Effect of Stagnation Enthalpy Non-uniformities
Consider a case where htc varies from streamtube ( dm ) to streamtube (but
Pe=const., so ηN = const.). Then
F =
∫∫ υ
e
dm + ( Pe − Pa ) Ae
For Pa = o (vacuum operation) and PaAe << F (large expansion), (or if Pe = Pa)
F ≅
∫∫ υ
e
dm = 2 ηN
and the input power is P =
∫∫ h
∫∫
htc dm
(1)
dm
tc
(2)
⎧⎪P is minimum(For a given F, m)⎫⎪
It can be shown that ⎨
⎬ if the flow is uniform
⎩⎪or F is maximum ( given P, m) ⎭⎪
( htc =const.). If it were, we would have
FUNIF . = 2 ηN m htc
;
PUNIF = m htc
2
Eliminating
htc , PUNIF
⎛ F
⎞
F2
F2
= m ⎜ UNIF ⎟ = UNIF =
⎜ 2η m ⎟
2 ηN m 2 ηN m
N
⎝
⎠
Define an “efficiency” ηUNIF =
PUNIF
(for a given thrust)
PACTUAL
Now, express in general F by (1) and P by (2)
ηUNIF =
(
2 ηN
2 ηN
∫∫
htc dm
( ∫∫ dm)( ∫∫ h
tc
Define “generalized vectors” u = 1
Then ηUNIF =
(u
2
u
i υ)
2
i υ
2
≤1
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
( = cos
2
)
2
dm
)
υ = htc in the space of the dm values.
)
θu iυ .
Lecture 2
Page 6 of 7
Equality applies only when υ is a constant, i.e.,
htc =const. This proves the “ansatz”.
50% of flow has
htc = 0.5 htc
50% of flow has
htc = 1.5 htc
Example:
2
ηUNIF
1
⎛1
⎞
⎜ 2 0.5 + 2 1.5 ⎟
⎠ = 0.933 (6.7% energy loss due to nonunif.
= ⎝
1
⎛1
(1) ⎜ 2 0.5 + 2 1.5 ⎞⎟
⎝
⎠
Important in arcjets, less in film-cooled chemical rockets.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 2
Page 7 of 7