Velocity Change Optimization

Homework
ROCKET PROPULSION
Velocity Change Optimization in Multistage Rockets
Consider the optimization of a multistage rocket for minimum initial mass and given velocity
change and payload mass. Show that the optimized velocity change for each stage are
expressed by:
! v j = "c j ln
#j
1 " ! v kc j
where k is a solution of:
! v = " $ c j ln
j
#j
1 " ! v kc j
Consider a two-stage rocket with:
!v =
8500 m/s
(ascent to LEO) and:
Isp1 = c1 g =
Isp2 = c2 g =
!1 = ! 2 = ! =
340 s
450 s
0.13
for the 1st and 2nd stage, respectively. Determine:
1) the optimized values of the stage velocity changes;
2) the optimized value of the payload ratio.
Solve the same problem with the same Δv but (reverse order of the two propellants):
Isp1 = c1 g =
450 s
Isp2 = c2 g =
340 s
Compare with previous results. Which solution is better and why?
Luca d’Agostino, Dipartimento di Ingegneria Aerospaziale, Università di Pisa – Page 1
Homework
Solution – Velocity Change Optimization in Multistage Rockets
From the notes:
! vi = "ci ln %&# i + $i (1 " # i ) '( ) $i =
e " ! vi ci " # i
1 " #i
Therefore we have to maximize:
e # $ vi ci # % i
! = " !i = "
1 # %i
i
i
for given total velocity change:
! v = " ! vi
# 1$ "
i
i
! vi
=0
!v
Since:
ln ! = % ln
i
e " # vi ci " $ i
1 " $i
is maximum at the same location where λ is maximum, consider the augmented function:
%
e " ! vi ci " # i
!v (
F ( ! vi ) = $ ln
" k '1 " $ i *
&
1 " #i
i
i !v )
where k is a nondimensional Lagrangian multiplier. Differentiating w.r.t. Δvj :
# "v c
e i j cj
!F
k
= # # "v j cj
+
=0
e
# $ j "v
! "v j
( )
obtain:
e
! "v j cj
=
#j
1 ! " v kc j
$ " v j = !c j ln
#j
1 ! " v kc j
where k is solution of:
! v = " $ c j ln
i
#j
1 " ! v kc j
For the given two-stage rocket:
g=
c1 = gIsp1 =
c2 = gIsp2 =
and:
! v = "c1 ln
#1
#2
" c2 ln
1 " ! v kc1
1 " ! v kc2
or, in nondimensional form:
f (k ) = 1 +
c1
"1
c
"2
ln
+ 2 ln
=0
! v 1 # ! v kc1 ! v 1 # ! v kc2
where f(k) is not defined for:
Luca d’Agostino, Dipartimento di Ingegneria Aerospaziale, Università di Pisa – Page 2
9.81 m/s2
3335.4 m/s
4414.5 m/s
Homework
# "v "v &
0 < k < k! = max $ , ' =
% c1 c2 (
2.55
and:
lim f ( k ) = 1 +
k! ±"
c1
c
ln$ 1 + 2 ln$ 2 < 0
#v
#v
lim f ( k ) = "#
k!0 "
lim f ( k ) = +#
k!k"+
as shown in the figure:
Therefore let:
k = 3.6483045
and solve for:
f (k ) = 1 +
c1
"1
c
"2
ln
+ 2 ln
= -1.824E-07
! v 1 # ! v kc1 ! v 1 # ! v kc2
Compute the stage velocity changes and payload ratios:
#1
=
1 " ! v kc1
2806 m/s
#2
=
1 " ! v kc2
5694 m/s
! v1 = "c1 ln
! v2 = "c2 ln
!1 =
e " # v1 c1 " $ 1
=
1 " $1
0.346
!2 =
e " # v2 c2 " $ 2
=
1" $2
0.167
! = !1!2 = 0.0578185
Compute the propellant mass ratios:
mP1
= 1 ! e ! " v1
m0
c1
=
Luca d’Agostino, Dipartimento di Ingegneria Aerospaziale, Università di Pisa – Page 3
0.569
Homework
mP2
= 1 ! e ! " v2
m1
=
0.725
mP2
m
= !1 P2 =
m0
m1
0.251
c2
where:
m1
= !1
m0
and therefore:
Notice that this configuration corresponds to a 1st stage realtively small (high payload
ratio, small velocity change) and a bigger 2nd stage (small payload ratio, high velocity
change.
Reversing the order in the use of the two propellants would not change the overall
payload ratio, but would swap the velocity changes and payload ratios of each stage,
thereby leading to a bigger 1st stage and a lighter 2nd stage.
Hence:
c1 =
4414.5 m/s
c2 =
3335.4 m/s
and:
#1
=
1 " ! v kc1
5694 m/s
#2
=
1 " ! v kc2
2806 m/s
! v1 = "c1 ln
! v2 = "c2 ln
!1 =
e " # v1 c1 " $ 1
=
1 " $1
0.167
!2 =
e " # v2 c2 " $ 2
=
1" $2
0.346
! = !1!2 =
0.0578
Compute the propellant mass ratios:
mP1
= 1 ! e ! " v1
m0
mP2
= 1 ! e ! " v2
m1
c1
=
0.725
c2
=
0.569
where:
m1
= !1
m0
and therefore:
Luca d’Agostino, Dipartimento di Ingegneria Aerospaziale, Università di Pisa – Page 4
Homework
mP2
m
= !1 P2 =
m0
m1
0.095
The second configuration is clearly less cost-effective, since it uses larger quantities of
high specific impulse propellant.
Luca d’Agostino, Dipartimento di Ingegneria Aerospaziale, Università di Pisa – Page 5