minimum-time and -fuel optimization6. minimum-time and fuel problems mae 546 2018.pptx created date:...
TRANSCRIPT
Minimum-Time and -Fuel Optimization !
Robert Stengel! Optimal Control and Estimation MAE 546 !
Princeton University, 2018
Copyright 2018 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE546.html
http://www.princeton.edu/~stengel/OptConEst.html
!! Climbing flight of an airplane!! Energy and power!! Choice of control variable!! Numerical optimization!! Minimum time-to-climb problem!! Minimum fuel-to-climb problem!! Stability and control (supplement)
1
Flight Envelope Determined by Available Thrust
Excess thrust provides the ability
to accelerate or climb
•! Flight Envelope: Encompasses all altitudes and airspeeds at which an aircraft can fly
–! in steady, level flight –! at fixed weight
2
Energy Height and Specific Excess Power!
3
•! Specific Energy •! = (Potential + Kinetic Energy) per Unit Weight•! = Energy Height
Energy Height
Could trade altitude for airspeed with no change in energy height if thrust and drag were zero
Total EnergyUnit Weight
! Specific Energy =mgh + mV 2 2
mg= h +
V 2
2g! Energy Height, Eh , ft or m
4
Specific Excess Power
dEh
dt= ddt
h + V2
2g!"#
$%&= dhdt
+ Vg
!"#
$%&dVdt
dEh
dt= Specific Excess Power (SEP) = Excess Power
Unit Weight!
Pthrust " Pdrag( )W
dEh
dt=V sin! + V
g"#$
%&'
T ( D( )m
( gsin!)*+
,-.
dEh
dt=V T ! D( )
W=V
CT !CD( ) 12"(h)V 2S
W
•! Specific Power
!V = T ! D( ) m ! gsin"!h =V sin"
5
Contours of Constant Specific Excess Power
•! Specific Excess Power is a function of altitude and airspeed•! SEP is maximized at each altitude, h, when d SEP(h)[ ]
dV= 0
6
Subsonic Energy ClimbObjective: Minimize time or fuel to climb to
desired altitude and airspeed
Approximate Optimal Trajectory produced by a Switching Curve 7
Angle of Attack Control“Point-Mass” Model
!V = Tmax ! CDo+ " CL#
#$% &'2( ) 12 (V 2S$
%)&'*m ! gsin+
!+ = 1V
CL## 12(V 2S,
-./01 m ! gcos+$
%)&'*
!h =V sin!!r =V cos!
!mfuel = " SFC( ) Tmax( )
8
Minimum Time-to-Climb Problem
Vo = 100 m/s; ! o = 0 rad; ho = 0 m (sea level); ro = 0 mInitial and final conditions
!! End time, tf, is open, thrust takes maximum value, and control variable is angle of attack,
!! Fuel expended, but simulated vehicle mass held constant
Vf = 200 m/s; ! f = open; hf = 10,000 m; rf = open
! t( )
9
1,000-sec Trajectory,!Simple Angle of Attack Control
Angle of Attack History
!! No optimization, open-loop control of point-mass model!! Interchange of kinetic and potential energy!! Lightly damped long-period oscillation 10
Altitude vs. Velocity, !Simple Angle of Attack Control,
tf = 1,000 sec
!! Lightly damped, long-period oscillation!! Kinetic/potential energy interchange
11
Alternative: Pitch Angle Control
!! = 1
VCL"
# $!( ) 12%V 2S&
'()*+ m $ gcos!,
-./01
! = " #$
!! = 1
VCL"
12#V 2S$ %CL"
12#V 2S!&
'()*+ m % gcos!,
-./01
Controlling pitch angle introduces flight path angle damping
(see Supplemental Material for details)
! = Angle of Attack, rad" = Pitch Angle, rad# = Flight Path Angle, rad
12
Angle of Attack or Pitch Angle Control?
!!
! =" + #
13
!h =V sin!
1,000-sec Trajectory,!Simple Pitch Angle Control
Note difference in pitch-angle and angle-of-attack profiles
! =" + #
14
Pitch Angle History
!! No optimization, open-loop control of point-mass model!! Inherent damping!! Long-period oscillation does not occur
Altitude vs. Velocity!Simple Pitch Angle Control, !
tf = 1,000 sec!! Increased damping eliminates oscillation!! Pitch angle chosen to produce kinetic energy increase
followed by potential energy increase
15
Minimum Time-to-Climb Optimization Problem!
16
Minimum-Time Cost Function
Terminal cost provides trajectory
objective
J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }
+ 12
1+ xT t( )Qx t( ) + uT t( )Ru t( )"# $%{ }t0
t f
& dt
17
•! Integrand•! 1 is the integrand for minimizing time•! Small quadratic term
•! provides non-singular trajectory control with ad hoc damping and regulation [good], but
•! penalizes non-zero values of state and control [not so good]
V0 = 100 m/s! 0 = 0 radh0 = 0 m (sea level)r0 = 0 m
xdes t f( ) =Vf = 200 m/s
! f = open
hf = 10,000 m
rf = open
"
#
$$$
%
$$$
Minimum-Time Cost Function with Augmented Trajectory Damping
J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }
+ 12
1+ xT t( )Qx t( ) + !xT t( )Q !x !x t( ) + uT t( )Ru t( )"# $%{ }t0
t f
& dt
•! State rate weighting •! is unbiased by non-zero state or control•! provides damping
!x t( ) = f x t( ),u t( )!" #$
J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }
+ 12
1+ xT t( )Qx t( ) + f T x t( ),u t( )"# $%Q !xf x t( ),u t( )"# $% + uT t( )Ru t( )"# $%{ }
t0
t f
& dt18
Weights Used in Optimization
P =
100 0 0 00 20 0 00 0 0.4 00 0 0 0
!
"
####
$
%
&&&&
Q =
10!2 0 0 00 1 0 00 0 0 00 0 0 0
"
#
$$$$
%
&
''''
Q !x =
1 0 0 00 1 0 00 0 0 00 0 0 0
!
"
####
$
%
&&&&
R = 1
Terminal Penalty State Penalty
State-Rate Penalty
Control Penalty
x1 =V : Velocity, m/sx2 = ! : Flight path angle, radx3 = h : Height, mx4 = r : Range, mu = " : Pitch angle, rad
19
•! Optimization algorithm is still minimizing at iteration cutoff
Fixed End-Time Cost History, tf = 575 sec, 36 Iterations
•! Adaptive steepest-descent optimization algorithm
20
•! Ad hoc variation of final time (TBD)
Why is tf = 575 sec Approximately the Minimum-Time Solution?
tf = 600 secJ* = ~750
tf = 550 secJ* = ~3480tf = 575 sec
J* = ~935
Desired final state not achievable with shorter time
21
~Minimum-Time Altitude vs. Velocity, tf = 575 sec, 36 Iterations
Recall Energy-Height Approximation
Kinetic energy increaseTrade for potential energy increase
Velocity Increase and Shallow Dive to satisfy terminal condition
22
Power loss with altitude greater in the simulation than in E-H approximation
~Minimum-Time Trajectory, tf = 575 sec, 36 Iterations
Velocity Flight Path Angle
Height Range
23
Pitch Angle
Angle of Attack
Minimum Fuel-to-Climb Optimization Problem!
24
Minimum-Fuel Cost Function with Augmented Trajectory Damping
J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }
+ 12
!m + xT t( )Qx t( ) + !xT t( )Q !x !x t( ) + uT t( )Ru t( )"# $%{ }t0
t f
& dt
•! Same cost-function weights•! Different Integrand
–! Fuel-flow rate in the integrand for minimizing total fuel use
25
Minimum-Fuel Cost History, tf = 575 sec, 36 Iterations
Adaptive optimization algorithm takes two bad iterations
… but regains convergence, with small oscillation in terminal cost
26
Minimum-Fuel Altitude vs. Velocity, tf = 575 sec, 36 Iterations
27
~Minimum-Fuel Trajectory, tf = 575 sec, 36 Iterations
Virtually identical to minimum-time solution
… but less fuel is used. Minimum time: 849 kg. Minimum fuel: 831 kg
Velocity Flight Path Angle
Height Range
Pitch AngleAngle of Attack
28
Minimum-Fuel Adjoint Vector History, "(t), tf = 575 sec, 36 Iterations
!! t f( ) = "#"x
t f( )$%&
'()
T
= *xT t f( )PfCost sensitivity to state perturbations over time
Velocity Adjoint, "1
Flight Path Angle Adjoint, "2
Height Adjoint, "3
Range Adjoint, "4
29
Next Time:!Neighboring-Optimal Control via Linear-Quadratic Feedback!
!
Reading!OCE: Section 3.7 !
30
SSuupppplleemmeennttaall MMaatteerriiaall!!
31
Modal Properties of the System !
32
Linearized Equations for Velocity and Flight Path Angle Perturbations, Using
Angle of Attack as the Control
! !V! !"
#
$%%
&
'((=
TV ) DV( ) )g
LVVo
0
#
$
%%%
&
'
(((
!V!"
#
$%%
&
'((+
~ 0L*Vo
#
$
%%%
&
'
(((!*
TV !! Thrust m( )
!V= Sensitivity of acceleration-due-to-thrust to velocity variation
= 0 for this problem because thrust = maximum thrust" #
DV !! Drag m( )
!V= Sensitivity of acceleration-due-to-drag to velocity variation > 0
State : !x = !V!"
#
$%%
&
'((
Control : !u = !)
LV !! Lift m( )
!V= Sensitivity of acceleration-due-to-lift to velocity variation > 0
L" !! Lift m( )
!"= Sensitivity of acceleration-due-to-lift to angle-of-attack variation > 0
33
Characteristic Equation and Stability, Using Angle of Attack as
the Control Variable
sI! F = sI!!DV !gLVVo
0
"
#
$$$
%
&
'''=
s + DV( ) g
! LV Vos
= s s + DV( ) + g LV Vo= s2 + DVs + g
LVVo
= s2 + 2() ns +) n2 = 0
0 = s2 + 2!" ns +" n2
" n = g LV Vo! 2 g
V0!
13.9Vo m / s( ) ; Period ! 0.453Vo, sec
! = DV
2 g LV Vo
!22
CD
CL
#$%
&'(
Characteristic Equation
Natural Frequency and Damping Ratio
34
"Total Damping"= 2!" n = DV
Linearized Equations for Velocity and Flight Path Angle Perturbations, Using Pitch Angle as the Control
Replace angle of attack by pitch angle for control
State : !x = !V!"
#
$%%
&
'((
Control : !u = !) = !* + !"
! !V! !"
#
$%%
&
'((=
)DV )gLVVo
0
#
$
%%%
&
'
(((
!V!"
#
$%%
&
'((+
~ 0L*Vo
#
$
%%%
&
'
(((!*
! !V! !"
#
$%%
&
'((=
)DV )gLVVo
0
#
$
%%%
&
'
(((
!V!"
#
$%%
&
'((+
~ 0L*Vo
#
$
%%%
&
'
(((!+ ) !"( )
=)DV )gLVVo
) L*Vo
#
$
%%%
&
'
(((
!V!"
#
$%%
&
'((+
~ 0L*Vo
#
$
%%%
&
'
(((!+
35
Characteristic Equation and Stability, Using Pitch Angle as the
Control Variable
sI! F = sI!!DV !gLVVo
! L"Vo
#
$
%%%
&
'
(((=
s + DV( ) g
! LV Vos + L"
Vo( )= s + L"
Vo( ) s + DV( ) + g LV Vo= s2 + L"
Vo+ DV( )s + g LV Vo
+ DVL"Vo( )
= s2 + 2)* ns +* n2 = 0
0 = s2 + 2!" ns +" n2
" n = g LV Vo+ DV
L#Vo( )
! =
L#Vo
+ DV( )2 g LV Vo
+ DVL#Vo( )!! Natural frequency is increased
!! Damping is increased
36"Total Damping"= 2!" n =
L#Vo
+ DV( )
Definitions and Numerical Values for Variables and Constants
V =Velocity, m/s! = Flight path angle, radh = Altitude (or height), mr = Range, mm = Mass, kg" = Angle of attack, rad; "max = 10°# = Air density = #SLe
$%h = 1.225e$h/9,042kg/m3
g = Gravitational acceleration = 9.801 m/s2
Angle of attack has linear effect on lift and quadratic effect on drag
T = TSL!!SL
"#$
%&'(T = TSL e)*h( )(T = Thrust, N;
(T = Throttle setting, %, = 100% for minimum-time/fuel problemSFC = Specific Fuel Consumption = 10 g/kN-sCL = CL+
+ = Lift coefficient = 5.7+
CD = CDo
+ ,CL2( )
1) V Vsound( )2= Drag coefficient
= 0.025 + 0.072CL2( ) 1) V Vsound( )2
S = Reference area = 21.5 m2
mE = Vehicle mass = 4,550 kg ~ constant
37
Supersonic Energy ClimbObjective: Minimize time or fuel to climb to
desired altitude and airspeed
Approximate Optimal Trajectory produced by a Switching Curve 38
Energy State Profile, tf = 570 sec
Energy State : E =1mg
mV 2
2+ mgh
!"#
$%&=V 2
2g+ h, meters
Specific Excess Power :
SEP = Vmg
Thrust ! Drag( )
!! Optimized Energy State is monotonic and always increasing
!! Rate of change decreases with altitude
39