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AA283
Aircraft and Rocket Propulsion
Chapter 8 - Multistage Rockets
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7.1 Notation
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7.2 Analysis
(8.2)
(8.3)
(8.4)
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(8.5)
(8.6)
(8.7)
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8.3 The variational problem
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8.4 Example - Exhaust velocity andstructural coefficient the same for all stages.
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There is very littleadvantage to using
more than about
three stages.
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C 1 = 2500 C
2 = 4250 C
3 = 4250
V 3 =C
1 Ln
1+ ! 1
" 1+ !
1
# $ %
& ' ( +C
2 Ln
1+ ! 2
" 2 + !
2
# $ %
& ' ( +C
3 Ln
1+ ! 3
" 3 + !
3
# $ %
& ' (
V 3 = 2500 Ln
1+ 0.321
0.05 + 0.321
! " #
$ % & + 4250 Ln
1+ 0.466
0.071+ 0.466
! " #
$ % & + 4250 Ln
1+ 0.603
0.191+ 0.603
! " #
$ % & =10429 M / sec
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The final velocity of a multistage system can be expressed as
V n = C
i ln
M 0i / M
Pi
M 0i / M
Pi !1
" # $
% & ' i=1
n
(Consider a two stage design
V 2 = C 1 Ln
M 01 / M
P1
M 01 / M
P1 !1
" # $
% & ' +C 2 Ln
M 02 / M
P2
M 02 / M
P2 !1
" # $
% & '
M s1 =! 1
1" ! 1
# $ % & ' ( M P1 M s2 =
! 2
1" ! 2
# $ % & ' ( M P2
M 01 = M
S 1 + M
P1 + M
S 2 + M
P2 + M
L
M 02 = M
S 2 + M
P2 + M
L
Alternative approach.
Express payload mass in terms of propellant masses and payload fraction
M L =
!1" !
# $ %
& ' (
1
1" ) 1
# $ %
& ' ( M
P1+
!1" !
# $ %
& ' (
1
1" ) 2
# $ %
& ' ( M
P2
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M 01
M P1
=1
1! "# $ %
& ' (
1
1! ) 1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' ( M
P2
M P1
#
$ %&
' (
M 02
M P2
=1
1! "# $ %
& ' (
"1! )
1
# $ %
& ' (
1
M P2 / M
P1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' (
#
$ %&
' (
Express stage mass ratios in terms of propellant mass ratios
V 2 = C
1 Ln
M 01 / M
P1
M 01 / M
P1 !1
" # $
% & ' +C
2 Ln
M 02 / M
P2
M 02 / M
P2 !1
" # $
% & '
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V 2 = C
1 Ln
1
1! "# $ %
& ' (
1
1! ) 1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' ( M
P2
M P1
#
$ %&
' (
1
1! "# $ %
& ' (
1
1! ) 1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' ( M
P2
M P1
#
$ %&
' ( !1
#
$
%%%%
&
'
( ( ( (
+C 2 Ln
1
1! "# $ %
& ' (
"1! )
1
# $ %
& ' (
1
M P2
/ M P1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' (
#
$ %&
' (
1
1! "# $ %
& ' (
"1! )
1
# $ %
& ' (
1
M P2
/ M P1
# $ %
& ' ( +
1
1! ) 2
# $ %
& ' (
#
$ %&
' ( !1
#
$
%%%%
&
'
( ( ( (
For given values of
! ,C 1,C
2,"
1,"
2
The final velocity is a function of the propellant ratio.
V 2 = F
M P2
M P1
! " #
$ % &
It is now just a matter of differentiating with respect to the propellantratio to identify a maximum.
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Application - Mars Sample Return Campaign
2020-2026(ish)
Lander to Surface
MAV
Mars Orbit Rendezvous
Return to Earth
•
Next major step in Mars Science• Requires international collaboration
• Multiple new developments
– Mars Ascent Vehicle (MAV)
– Sample acquisition and handling
– Precision entry descent and landing
Ashley Chandler ’s
PhD project
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Critical Challenge: Mars Environmental Conditions
0 5 10 15 20 25!120
!100
!80
!60
!40
!20
0
20
40
Local Time of Day, hours
S u r f a c e T e m p e r a t u r e ,
C
Min: !111C
Max: 24C
Average:!
60C
•
Diurnal/seasonal minima and maxima (-111C to 24C) – Nitrous oxide freezing point is -90.8 C.
– Paraffin-based fuel is primarily crystalline with a low glass transitiontemperature at -108 C.
Data from the NASA AmesResearch Center Mars Global
Climate Model for HoldenCrater.
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Use a two stage design for the Mars ascent vehicle
! 1 = 0.13
! 2 = 0.155
C 1 = 2883
C 2 = 3026
! = 0.147
M P2
/ M P1
V 2
0.354
Note that the finalvelocity is actually
quite insensitive to
the propellant mass
ratio giving the
designer quite a bit
of flexibility.
Notional values.
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MAV System
Two StageSolid
40% MarginHybrid
20% MarginHybrid
S t a g e
1
Mass (kg) 215.6 145.89 128.86
Structural Coefficient 0.170 0.189 0.167
Max. Pchamber (MPa) 4.83 1.72 1.72
Delivered Isp (m/sec) 2813 2952 2951
!V (m/sec) 2530 1675 1675
S t a g e
2
Mass (kg) 86.2 91.37 83.57
Structural Coefficient 0.175 0.169 0.147
Max. Pchamber (MPa) 6.45 1.38 1.38
Delivered Isp (m/sec) 2813 2977 2976
!V (m/sec) 1845 2700 2700
Total Ideal !V (m/sec) 4375 4375 4375
Maximum Diameter (m) 0.442 0.541 0.524
Vehicle Length (m) 2.56 3.829 3.747
Payload Mass (kg) 36 36 36
Gross Lift Off Mass (kg) 301.8 273.3 248.4
Igloo Mass (kg) 50 0 0
System Total Mass (kg) 351.8 273.3 248.4
System Parameters - proposed design
! = 0.132
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Volumetric Constraint
•
Hybrid MAV can be repackaged to fit within the volumetric and
center of gravity constraints
Volume Constraint
In order to confirm the design it is necessary to fly it to orbit.
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Kepler ’s equations govern the motion of objects near gravitating
bodies. This is called the two body problem.
!! x(t )+ M Planet G
x(t )
r(t )3 = 0
!! y(t )+ M Planet G y(t )
r(t )3 = 0
!! z(t )+ M Planet G z(t )
r(t )3 = 0
Kepler ’s Equations
G = 6.67 !10"11
m3/kg-sec2
r t ( ) = x t ( )2
+ y t ( )2
+ z t ( )2
F = !G Mmr2
r
r" # $ % & '
Universal gravitational
constant
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Orbital Period
Kepler ’s theory gives
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Orbital Periods of the Planets about the Sun
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Mars Ascent Vehicle - launch to orbit
Equations of motion
!! x(t )+ m MARS
G x(t )
r(t )3 +
F x DRAG
(t )
m(t )!
F xTHRUST
(t )
m(t )= 0
!! y(t )+m MARS
G y(t )
r(t )3 +
F y DRAG (t )
m(t )!
F yTHRUST (t )
m(t )= 0
!! z(t )+m MARS
G z(t )
r(t )3 +F z DRAG (t )
m(t )!
F zTHRUST (t )
m(t )= 0
Mars radius=3.376 x 106 m
Mars mass=6.418 x 1023 kg
g MARS =
m MARS
G / r2= 3.756 m/sec2
! MARS
=
r3
mG= 948.03Mars time scale: sec
Mars velocity scale: U MARS
=
mG
r= 3561 m/sec
x
y z
G = 6.67 !10"11 m3/kg-sec2
Universal gravitational constant
Vehicle mass
m(t )
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Aerodynamic drag
C D
C D0
M
Drag
t /!
CD0=0.2
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Angle between velocity vector and planet radius
V
r
x
y z
Launch point
!
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Surface speed at the
equator=241.17 m/sec
Vertical launch12.5 sec
First stage
gravity turn
36.6 sec
Coast620 sec
Second stagegravity turn to orbit
39.4 sec
d !
dt ROCKET
= 0.038 degrees/sec
d !
dt
= 0.132 degrees/sec
d !
dt = 0.0007 degrees/sec
Gravity turn
Launch trajectory
F THRUST
!V = 0
V = ( ! x, ! y, ! z)
See the paper Universal
Gravity Turn Trajectories onmy website.
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Launch and orbit trajectory
Maximum altitude
557.9 km
Minimum altitude 527.5 km
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The mass ratios can be written in terms of the payload fraction as follows.
M 01 / M
P1 =
1
1! "# $ %
& ' (
1
1! ) 1
+1
1! ) 2
M P2 / M
P1( )+1
1! ) 3
M P3 / M
P1( )# $ %
& ' (
M 03 / M
P3 =
1
1! "# $ %
& ' (
1
1! ) 3
+"
1! ) 1
1
M P3 / M
P1
# $ %
& ' ( +
"1! )
2
M P2 / M
P1
M P3 / M
P1
# $ %
& ' (
#
$ %&
' (
M 02 / M
P2 =
1
1! "
#
$ %
&
' (
1
1! ) 2
+"
1! ) 1
1
M P2 / M
P1
#
$ %
&
' ( +
1
1! ) 3
M P3 / M
P1
M P2 / M
P1
#
$ %
&
' (
#
$ %
&
' (
V 3 = C
1ln
M 01 / M
P1
M 01 / M
P1 !1
" # $
% & ' +C
2 ln
M 02 / M
P2
M 02 / M
P2 !1
" # $
% & ' +C
3ln
M 03 / M
P3
M 03 / M
P3 !1
" # $
% & '
Three stage design
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For given values of
! ,C 1,C
2,C
3,"
1,"
2,"
3
The final velocity is a function of of the propellant ratios.
Differentiate with respect to the two variables toidentify a maximum.
V 3 = F
M P2
M P1
, M
P3
M P1
! " #
$ % &
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! 1 = 0.116
! 2 = 0.117
! 3 = 0.148
C 1 = 3040 m /sec
C 2 = 3210 m / sec
C 3 = 3350 m /sec
" = 0.01125
#V = 10,800 m /sec
Payload mass = 85 kg
M 01
= 7557 kg
Three stage launch vehicle for small satellites