Section 5, Lecture 1: Review of Idealized Nozzle Theory: Summary of Fundamental Properties and Relationships
Section 5, Lecture 1: Review of Idealized Nozzle Theory: Summary of Fundamental Properties and Relationships
• Many of the historical data suggest that 50% of solid rocket failures
stemmed from nozzle problems.
Isentropic
Nozzle
• Effect known
as Choking in a
Duct or Nozzle
• Consider a “choked-flow”
Nozzle … (I.e. M=1 at Throat)
• “Newton’s Method”
CF ≡
⎧⎡ γ −1 1/2
⎤ ⎫
⎪⎢ ⎛ pexit ⎞ γ
⎥ + ⎪
⎪ ⎢1− ⎜⎝ P0 ⎟⎠ ⎥ ⎪
⎪⎣ ⎦ ⎪
γ +1 ⎪ ⎪
γ +1
2 ⎛ 2 ⎞ (γ −1) ⎪ ⎪
CF =γ ⋅⎨ ⎛ pexit ⎞ −γ ⎬
γ − 1 ⎜⎝ γ + 1 ⎟⎠ ⎪γ − 1 ⎜⎝ P ⎟⎠ ⎡ pexit p∞ ⎤ ⎪
⎢ P − P ⎥⎪
0
⎪ (γ −1)
⎪ 2γ ⎡ ⎤ ⎣ 0 0 ⎦⎪
⎪ ⎢⎛ pexit ⎞ −γ
−1⎥ ⎪
• CF is a function of Nozzle pressure ratio ⎪ ⎢⎜⎝ P0 ⎟⎠ ⎥ ⎪
⎩ ⎢⎣ ⎥⎦ ⎭
and5540
MAE back pressure
- Propulsion only
Systems
Thrust Coefficient Summary
Ideal Thrust Coefficient
γ+1 ⎛ γ−1 ⎞
Fthrust 2 ⎛⎜ 2 ⎞⎟ γ−1 ⎜⎜ ⎛ p ⎞ γ ⎟⎟ A ⎛ p − p ⎞
CF = = γ ⋅ ⋅⎜ ⎟ ⋅⎜⎜1−⎜⎜⎜ exit ⎟⎟⎟ ⎟⎟⎟ + exit* ⋅⎜⎜⎜ exit ∞⎟
⎟⎟
P0 ⋅ A* γ −1 ⎜⎝ γ +1⎟⎟⎠ ⎜⎜ ⎜⎝ P ⎟⎠ ⎟⎟ A ⎜
⎝ P ⎟⎠
⎜⎝ 0 ⎟⎠ 0
C * = ⎜ •0 ⎟ = ⎜ • e ⎟ = ⎜ • =
⎟ C = m exit
⎝ m exit ⎠ ⎜ m exit ⎟
⎜ m Thrust ⎟ F
γ +1
⎡ γ −1
⎤
⎜ ⋅C ⎟ exit
⋅ 2 ⎛ 2 ⎞ ( )
γ −1
⎢1− ⎛ p ⎞ γ
⎥ + Aexit ( pexit − p∞ )
⎝ P0 A
e
⎜
⎠ ⎝ P0 A m exit ⎠
•
⎟ γ ⋅ exit
γ − 1 ⎜⎝ γ + 1 ⎟⎠ ⎢ ⎜⎝ P0 ⎟⎠ ⎥ A*
* *
P0
⎣ ⎦
• Let nozzle expand infinitely in Vacuum….
• The characteristic velocity is a
2
Vexit ... pexit ,Texit , p∞ → 0 → Vexit → 2 ⋅ c p ⋅T0
figure of thermo-chemical merit
>> Texit for a particular propellant and
2 ⋅ cp may be considered to be
Indicative of the combustion
performance of propellants.
γ RgT0 γ Ru To
→C = *
γ +1
= γ +1 MW
⎛ 2 ⎞ (γ −1) ⎛ 2 ⎞ (γ −1) • Propellants that burn Hot
γ ⎜ ⎟⎠ γ ⎜ and have a low product Molecular
⎝ γ + 1
MAE 5540 - Propulsion Systems ⎝ γ + 1 ⎟⎠ weight … best C*
C* of Given Propellants
• from Earlier
ThrustThrust P0 A* CF ⋅C *
I sp = • = =
go m go m! P0 A *
go
• Assuming an infinitely expanded nozzle in a vacuum, Maximum Achievable Specific Impulse for
Selected propellants is
γ +1
CF ⋅C *
γ 2 ⎛ 2 ⎞ (γ −1) γ Ru To 1 2 ⋅ γ ⋅ Ru To
I(spI ideal
Max )
= = ⋅ =
go go γ − 1 ⎜⎝ γ + 1 ⎟⎠ γ +1 M W go γ −1 MW
⎛ 2 ⎞ (γ −1)
γ ⎜
⎝ γ + 1 ⎟⎠
m!
→ C* ( ) ideal
= γ +1
= γ +1
To
MW
⎛ 2 ⎞ (γ −1) ⎛ 2 ⎞ (γ −1)
γ ⎜ γ ⎜
⎝ γ + 1 ⎟⎠ ⎝ γ + 1 ⎟⎠
Thrust P0 A* CF ⋅C *
Thrust
→ I sp = • = =
go m go !
m P0 A *
go
⎡ γ +1 ⎤
2 ⎛ 2 ⎞ γ −1) ⎥ 1
( 2γ Ru
( ) C ⎢
*
To
→ IIspsp Max = γ ⎥= g
ideal go ⎢ γ − 1 ⎜⎝ γ + 1 ⎟⎠ γ −1 MW
⎢⎣ ⎥⎦ o
• Combustion
• Transport
dS
Actual
Momentum
Thrust
Momentum
Thrust of
idealized
Nozzle
Application of
Correction
Factor
• Define
• Look At
• Assume
Shuttle has
A very long
Burn time
=1.525
= 0.995
MAE 5540 - Propulsion Systems
Pchamber Correction,
Chimaera Rocket Example
• Expected Thrust ~ 2300 lbf
• Expected Pchamber ~ 450 psia
• A* ~ 1.25 in
=0.969
MAE 5540 - Propulsion Systems
Chimaera Rocket Example (cont’d)
• Thrust Time History (Ax) curve
Total Impulse:
23.26 g-seconds
“Steady State”
Impulse:
22.10 g-seconds
~95%
… so we are
Pretty consistent
• Thrust Coefficient
• Specific Impulse
• Unlike other propellants, the optimum mixture ratio for liquid oxygen and
liquid hydrogen is not necessarily that which will produce the maximum
specific impulse. Because of the extremely low density of liquid hydrogen,
the propellant volume decreases significantly at higher mixture ratios.
⎡ 7.396 1 ⎤
O / F = 7.936 → 721,000 kg ⎢ + ⎥ = 15315m 3
⎢⎣ 1140 kg/m 3 67.78 kg/m 3 ⎥⎦
“best compromise”
⎡ 6.000 1 ⎤
O / F = 6.000 → 721,000 kg ⎢ + ⎥ = 14432m 3
⎢⎣ 1140 kg/m 3 67.78 kg/m 3 ⎥⎦
⎡ 6.000 1 ⎤
O / F = 3.5 → 721,000 kg ⎢ + ⎥ = 12850m 3
⎢⎣ 1140 kg/m 3 67.78 kg/m 3 ⎥⎦
P0=186.92 atm
=18940 Kpa
T0 ~
3615°K
Mw ~
13.6 kg/kg-mol
g~
1.196
M W ~ 13.6 kg/kg-mole
• Mass flow =
=438.15 kg/sec
= 8252.59 kg/sec-m2
• Solution methods
i) Plot A/A* versus mach
ii) Numerical Solution
= 77.49998 ----> M exit = 4.677084
= 4288.61 m/sec
=1149.90 °K
CF = =
• Vacuum
CF = =
Thrust ⎛ 26 ⎞ 2 π
6
Thrust = CF ⋅ = 1.52546 ·18.94 ·10 ⎝ ⎠ = 1.53397 x 106 N
P0 A* 100 4
• Vacuum
Thrust = CF ⋅
Thrust
= ⎛ 26 ⎞ 2 π
6
P0 A*
1.94006 ·18.94 ·10 ⎝ ⎠ = 1.95089 x 106 N
100 4
• @ Optimal Altitude?
= 1.87907 x 106 N
MAE 5540 - Propulsion Systems
Example: SSME Rocket Engine (cont’d)
Compute Characteristic Velocity C* in two ways
⎛ 26 ⎞ 2 π
6
* 18.95 ·10 ⎝ ⎠
P0 A 100 4
C =
*
= =2296.27 m/sec
m! 438.15
Close enough!
γ Ru To
C =*
γ +1
= =2296.25 m/sec
MW
⎛ 2 ⎞ (γ −1)
γ ⎜
⎝ γ + 1 ⎟⎠
I sp = =
max
0.5
⎛ 1.196 + 1 ⎞
⎜ 2 ⎛ 2 ⎞ 1.196 − 1 ⎟
2296 ·1.196 ⎜ ⎝ ⎠ ⎟
⎝ 1.196 − 1 1.196 + 1 ⎠
9.8067
= 529.80 sec
I sp =
max
0.5
⎛ 1.196 + 1 ⎞
⎜ 2 ⎛ 2 ⎞ 1.196 − 1 ⎟
2296 ·1.196 ⎜ ⎝ ⎠ ⎟
⎝ 1.196 − 1 1.196 + 1 ⎠
438.15 ·9.8067
9.8067
= 2.276 mNt
= 4452.53 m/sec
= 3500.98 m/sec
= 357.024 sec
Summary:
Ideal Calc. Calc. Actual Actual
Vac. S.L. Vac. S.L
________________________________________________________________
Isp 529.69 454.06 357.03 452.5 363
(sec):