the master equation dr. dimitri mavris boeing associate professor for advanced aerospace systems...
TRANSCRIPT
The Master Equation
Dr. Dimitri MavrisBoeing Associate Professor for Advanced Aerospace Systems
Analysis
Compiled by the Loyal FYGS of the Class of 2003.
The Master Equation
• Derived from Energy Considerations.
2
2V
dt
d
g
W
dt
dhWVRDT
o
(2.1)
Rate of Mechanical EnergyInput
Storage RateOf Potential Energy
Storage RateOf Kinetic Energy
The Master Equation
• The master equation is most useful in its dimensionless form.
dt
dz
Vg
Vh
dt
d
VW
RDT e
o
1
2
1 2
(2.2a)
The Master Equation
• The left hand side, when multiplied by Velocity (V), becomes the equation for “weight specific excess power.”
dt
dz
g
Vh
dt
dP e
os
2
2
(2.2b)
The Master Equation
• Correct thrust using lapse rate, – T=TSL
• Correct weight using fuel/payload correction, .– W=WSL
oTOTO
SL
g
Vh
dt
d
VW
RD
W
T
2
1 2
(2.5)
(2.4)
(2.3)
Use Lift Relationships to find CL
SqCnWL L
Where n is the “load factor” (number of g’s). n=1 for straight and level flight.
(2.6)
S
W
q
n
qS
nWC TO
L
(2.8)
Solving for CL and substituting equation 2.4 to get the takeoff wing loading.
Use the Equations for Drag
SqCD D
Equation 2.7 yields the relationship between drag coefficient and drag.
(2.7)
oDLLD CCKCKC 22
1 (2.8)
The parabolic lift-drag polar equation is given above in equation 2.9.
Combine the CL, CD, and D Equations
DqSCD (2.7)
oDLL CCKCKqSD 2
21
Substitute Equation 2.9 into Equation 2.7 for CD.
oDTOTO CS
W
q
nK
S
W
q
nKqSD
2
2
1
Then substitute Equation 2.8 for CL in the above Equation to yield Equation 2.10
(2.9b)
(2.10)
Back to the Master Equation
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
q
nK
S
W
q
nK
W
qS
W
To 2
1 2
2
2
1
(2.11)
oTOTO
SL
g
Vh
dt
d
VW
RD
W
T
2
1 2
(2.5)
Now, substitute Equation 2.10 back into Equation 2.5 (the Master Equation) for D.
The above form of the equation can be manipulated to set the constraints on thedesign plot.
Using the Master Equation• Case 1: Constant Altitude/Speed Cruise• Case 2: Constant Speed Climb• Case 3: Constant Altitude/Speed Turn• Case 4: Horizontal Acceleration• Case 5: Takeoff Ground Roll (lots of Thrust)
• Case 6: Takeoff Ground Roll (not so much thrust)– A: Do NOT Clear the Obstacle During Transition– B: Clear the Obstacle During Transition
• Case 7: Braking Roll• Case 8: Service Ceiling
Case 1: Constant Altitude/Speed Cruise (Ps=0)
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
qK
S
W
qK
W
qS
W
To 2
111 2
2
2
1
Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are Given
Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are Given
oDTOTO
TOTO
SL CS
W
qK
S
W
qK
W
qS
W
T
2
2
1(2.11b)
Case 1: Constant Altitude/Speed Cruise (Ps=0)
oDTOTO
TOTO
SL CS
W
qK
S
W
qK
W
qS
W
T
2
2
1 (2.11b)
Equation 2.11b can be simplified by canceling WTO and qS
SW
q
CK
S
W
qK
W
T
TO
DTO
TO
SL o
21 (2.12)
Equation 2.12, a function of wing loading and 1/wing loading, is used for case 1.
Case 1: Constant Altitude/Speed Cruise (Ps=0)
For very large and very small values of wing loading, the T/W ratio goes to infinity.To find the point at which the T/W ratio is minimum, take the partial derivative of Equation 2.12 with respect to WTO/S and set it equal to zero.
SW
q
CK
S
W
qK
SW
d
d
TO
DTO
TO
o
210
1K
Cq
S
WoDTO
Wing loading for
minimum T/W
Case 1: Constant Altitude/Speed Cruise (Ps=0)
Also, for very large q (when CD=CDo), a hyperbolic relationship between T/W and W/S is observed. If q is extremely large, the first term goes to zero.
SW
q
CK
S
W
qK
W
T
TO
DTO
TO
SL o
21(2.12)
Term 1
SW
q
C
W
T
TO
D
TO
SL o
K2 very small
oDTO
TO
SLqC
S
W
W
T
(2.13)
Case 2: Constant Speed Climb (Ps=dh/dt)
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
qK
S
W
qK
W
qS
W
To 2
111 2
2
2
1
Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given
Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given
(2.14)
dt
dh
VS
Wq
CK
S
W
qK
W
T
TO
DTO
TO
SL o1
21
Case 2: Constant Speed Climb (Ps=dh/dt)
(2.14)
dt
dh
VS
Wq
CK
S
W
qK
W
T
TO
DTO
TO
SL o1
21
SW
q
CK
S
W
qK
W
T
TO
DTO
TO
SL o
21 (2.12)
Note that the only difference between Case 1 and Case 2 is the final term, for climb rate.
Case 1:
Case 2:
Case 2: Constant Speed Climb (Ps=dh/dt)
(2.14)
dt
dh
VS
Wq
CK
S
W
qK
W
T
TO
DTO
TO
SL o1
21
Since the final term is not a function of W/S, taking the derivative as was done in case 1 will yield the same W/S value at which the minimum T/W occurs; however, the value of T/W at this point will be different based on the desired rate of climb.
1K
Cq
S
WoDTO
Wing loading for
minimum T/W
Case 3: Constant Altitude/Speed Turn (Ps=0)
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
q
nK
S
W
q
nK
W
qS
W
To 2
1 2
2
2
1
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are Given
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are Given
(2.15)
SW
q
CnK
S
W
qnK
W
T
TO
DTO
TO
SL o
22
1
Case 3: Constant Altitude/Speed Turn (Ps=0)
1K
C
n
q
S
WoDTO
Wing loading for
minimum T/W
The behavior of Equation 2.15 is the same as that of Case 1 and Case 2. The point of minimum thrust/weight is now a function of the load factor, n.
Case 3: Constant Altitude/Speed Turn (Ps=0)
The relationships for n in the turn can be calculated from the Pythagorean theorem and centripetal force equations.
2
12
1
og
Vn
2
12
1
coRg
Vn(2.16) (2.17)
Where is the rotation rate and Rc is the radius of curvature for a load factor, n.
RcW
L=nW
Bank angle(W/go)V=(W/go)V2/Rc
Case 4: Horizontal Acceleration
dt
dV
gS
Wq
CK
S
W
qK
W
T
TO
DTO
TO
SL
021
10
dtgVdVPs 0/Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given
Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given
Under the current above conditions, equation 2.11 becomes:
(2.18)
Rearranged it will yield:
SW
q
CK
S
W
qK
W
T
dt
dV
g TO
DTO
TO
SL
0
210
1
(2.19)
Case 4: Continued
Equation 2.9 must be then integrated to =
Allowable
InitialFinal
t
VV
gdt
dV
g 00
11
Case 5: Takeoff Ground Roll
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
q
nK
S
W
q
nK
W
qS
W
To 2
1 2
2
2
1
Assumptions:dh/dt=0 Constant AltitudeR <> 0 On the ground!sG, , CLmax, kTO GivenTSL >> (D+R) Loads of Thrust
Assumptions:dh/dt=0 Constant AltitudeR <> 0 On the ground!sG, , CLmax, kTO GivenTSL >> (D+R) Loads of Thrust
Note: sG is the distance of the takeoff ground roll in ft.
Case 5: Takeoff Ground Roll
• Rearranging yields:
dVVT
W
gds
SL
TO
o
• The master equation (2.5) can be reduced to:
Vds
dV
gdt
dV
gW
T
ooTO
SL
/
Case 5: Takeoff Ground Roll
• This is then integrated over s=0, V=0 to takeoff (s=sG, V=VTO):
o
TO
SL
TOG g
V
T
Ws
2
2
• Given the takeoff values of and are used.
Case 5: Takeoff Ground Roll
• Defining:
STALLTOTO VkV (2.20)
• Where kTO >1 and VSTALL is the stall speed of the aircraft, then:
TOLSTALLL WSCVSqC max2
21
max
Case 5: Takeoff Ground Roll
• Or….
S
W
C
kVk
V TO
L
TOSTALLTO
TO
max
222
2
22
(2.21)
• And finally:
S
W
Cgs
k
W
T TO
LoG
TO
TO
SL
max
22
(2.22)
• But….
Case 5: Takeoff Ground Roll
…in this limiting case:
TO
SL
W
TIs proportional to:
S
WTO
And inversely proportional to: Gs
Case 6: Takeoff Ground Roll (sG)
TO
TOTOLTODRD
TO W
WqSCCC
W
RD
Then:
Note: sG is the distance of the takeoff ground roll in ft.
Assumptions:dh/dt=0 Constant AltitudedV/dt=0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundsG, , CLmax, kTO GivenD=qCDS Equation 2.7
Assumptions:dh/dt=0 Constant AltitudedV/dt=0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundsG, , CLmax, kTO GivenD=qCDS Equation 2.7
SCqWSCqR LTOTODR
Case 6: Takeoff Ground Roll
dt
dV
gW
Sq
W
T
oTO
TOTO
TO
SL 1
(2.24)
Eq. 2.5 becomes:
Where:
)( LTODRDTO CCC
(2.23)
kTO
LTO
TO
SL
TO
TOo
TOG
CWTg
SWs
2max
1ln)/(
(2.25)
After rearranging and integrating the equation, the outcome is:
provided that the representative takeoff values for α and β are used.
It is assumed that TSL/WTO increases continuously with WTO/S. The limit, as all terms in ζTO go to zero and ln(1-ε) goes to – ε, is then taken in order to obtain the final result
kTO
L
TO
SL
TO
TOo
TOG
CWTg
SWs
2max
)/(
Case 6: Takeoff Ground Roll
S
W
CgsW
T TO
LoG
TO
TO
SL kmax
22
(2.22)
From the previous page then we get
On the next slide are the assumptions for Total Takeoff Distance
Case 6: Takeoff Ground Roll
SG
STO
SR Sobs
hTR
V = 0Ground Roll
Rotation
Transition
Climb
RC
VTO
VTO
VCL
Fig. 2.6b Takeoff Terminology ( )obsTR hh
Takeoff Field Length (A)
hOBS
CL
In Case A, the obstacle is NOT clearedDuring the transition.
Distance to pitch plane to liftoff CL
SWCktVts TOLTORTORR /)/(2 max
Multiply time to pitch and velocity to obtain distance
tR is roughly 3 seconds for modern aircraft – varies with rotational inertia and magnitude of control moments. This is the total aircraft rotation time.
max
2
2
2
1
L
TOTO
TOLTOTO
CS
WV
WSCVL
Take-off velocity determined by speed needed at TO CL to make lift equal to the weight
Multiply take-off velocity by a constant to account for the fact that CL lift-off is not CLmax
(2.26)
Distance to bring plane to reach climb angle
1sin
sin0
2
ng
VRs CLTO
CLCTR
S
W
Ckg
k
kg
Vs TO
LTO
CLTO
TO
CLTOTR
max2
0
2
20
2 2
18.0
sin
18.0
sin
The distance is the horizontal of the climb arc
Climb arc radius is defined by centripedal force needed to move the vehicle through the arc – excess lift tightens the arc
Breaking out velocity again links back to design variables
(2.27)
Distance to clear obstacleOutside the turn to climb, the flight path is a straight line at an angle to the ground defined by the vehicle climb angle
Climb angle defined by climb rate and velocity -> excess power
Height to climb through is the starting point at the arc from previous slide to the obstacle – geometry gives the horizontal distance
S
W
Ckg
k
kg
Vh TO
LTO
CLTO
TO
CLTOTR
max2
0
2
20
2 2
18.0
cos1
18.0
cos1
W
DT
dt
dh
V CL
sin
1CL
TRobsCL
hhs
tan
(2.28)
(2.29)
If hobs>htr:
If hobs>htr:
Case 6: Takeoff Ground Roll (B)
SG
STO
SR Sobs
hobs
V = 0Ground Roll
Rotation
Transition
Climb
RC
VTO
VTO
VCL
Fig. 2.6b Takeoff Terminology ( )obsTR hh
CL
In Case B, the obstacle is clearedDuring the transition.
Takeoff Ground Roll (B)B. (Fig. 2.6b) The distance sG and sR are the same as in case A, but the obstacle is cleared during transition so:
18.0
sinsin 2
0
2
TO
obsTOobsCobs
kg
VRs
obsRGTO ssss
(2.30)
where sobs is the distance from the end of rotation to the point where the height hobs is attained. There follows:
where
C
obsobs R
h1cos 1
2.2.7 Case 7: Braking Roll (sB)
TO
TOLDRD
TO W
WqSCCC
W
RD
(2.31)
Given: (reverse thrust), dh/dt = 0and values of , VTD = kTDVSTALL, D = qCDS,and
Given: (reverse thrust), dh/dt = 0and values of , VTD = kTDVSTALL, D = qCDS,and
0a
CLTOBDR qSWSqCR
Under these conditions:
So that Eq. (2.5) becomes
dt
dV
gW
Sq
W
T
oTOL
TO
SL 1
is defined in Equation 2.32L
Case 7: Braking Roll
• where…
• and after rearranging and integrating…
• as long as representative landing values of α and β are used.
)( LBDRDL CCC
2max0 )(
1ln)/(
TD
L
TO
SLB
L
L
TOB
kC
WTag
SWs
(2.32)
(2.33)
Case 7: Braking Roll• Reverse thrust can be used to an advantage in this case.
If “a” is made very large…
• or…
2max0
)()/(
TD
L
TO
SL
L
L
TOB
kC
WTag
SWs
S
W
Cgs
k
W
T TO
LB
TD
TO
SL
max0
22
(2.34)
Note: Total Landing Distance
hobs
Vobs
Approach
VTDVTD
Free Roll
Braking V=0
sA sFR sB
sL
Figure 2.7 Landing Terminology
• The total landing distance, sL, is the sum of the distance to clear an obstacle, sA, the free roll before the brakes are applied, sFR, and the braking roll, sB.
Case 7: Braking Roll
STALLobsobs VkV
22max
22
22
0
22
TDobs
obs
DRD
L
TDobs
TDobsTO
DRDA kk
h
CC
C
kk
kk
S
W
CCgs
(2.35)
Where hobs is the height of the obstacle and the velocity at the obstacle is
(2.36)
SWCktVts TOLTDFRTDFRFR /)/(2 max (2.37)
where tFR is a total system reaction time based on experience (normally 3 s) that allows for the deployment of a parachute or thrust reverser.
and
Case 8: Service Ceiling (Ps=dh/dt)
dt
dh
VS
Wq
CK
S
W
qK
W
T
TO
DTO
TO
SL 1021
TOL WSqC
(2.38)
Under these conditions, Eq. (2.11) becomes
where or
S
W
qC TO
L
(2.39)
Given:dV/dt = 0 No AccelerationN=1 Straight and Level (L=W)R=0 Not on the grounddh/dt > 0 Value is givenh (i.e., σ) and CL Given
Given:dV/dt = 0 No AccelerationN=1 Straight and Level (L=W)R=0 Not on the grounddh/dt > 0 Value is givenh (i.e., σ) and CL Given
Case 8: Service Ceiling
S
W
C
Vq TO
L
2
2
and
S
W
CV TO
LSL2or
(2.40)
Case 8: Service Ceiling
dt
dh
VC
CKCK
W
T
L
DL
TO
SL 10
21
Equation 2.41 can be used to find
(2.41)
TO
SL
W
Tas a function of
S
WTO
• V may be computed from given conditions and Eq. 2.40
• CL may be computed with V and Eq. 2.39
• α is a given function (Eq. 2.3)
NOTHING BEYOND THIS POINT
HAS BEEN CHECKED OR FORMATTED!!!
2.3 Preliminary Estimates For Constraint Analysis
• Preliminary Estimates for aerodynamic characteristics of the airframe and the installed engine thrust lapse are needed for constraint analysis
• 2.3.1 Aerodynamics– Maximum coefficient of lift (CLmax) is important in constraint analysis
during takeoff and landing
– For Fighter aircraft, a clean wing has CLmax between 1.0 and 1.2, and a wing with a leading edge slat has a CLmax between 1.2 and 1.6
– Table 2.1 has typical CLmax values for cargo and passenger aircraft
2.3 Preliminary Estimates For Constraint Analysis
High Lift Device Typical Flap Angle (deg) CLmax/cos(Λc/4)
Trailing Leading Edge Takeoff Landing Takeoff Landing
Plain -- 20 60 14-16 17-20
Single Slot -- 20 40 15-17 18-22
Fowler -- 15 40 20-22 25-29
Double sltd -- 20 50 17-20 23-27
Double sltd slat 20 50 23-26 28-32
Triple sltd slat 20 40 24-27 32-35
Λc/4 is the sweep angle at the quarter chord
Table 2.1 CLmax for High Lift Device
Lift-drag polar estimation• The lift-drag polar for most large cargo and passenger aircraft we can be
approximate use Fig.2.8 and Eq. (2.9) with
03.0001.0 K
eARnK
1
3.01.0 min LC
where wing planform efficiency factor (e) is between 0.75 and 0.85 and AR (wing aspect ratio) is between 7 and 10.
Lift-drag polar estimation
KKK 1
2minmin LDDo CKCC
2minmin LDDo CKCC
Lift-drag polar estimation
0
0.01
0.02
0.03
0.04
minDC
oM
0.2 0.4 0.6 0.8 1.0
Turboprop
Turbofan/jet
Fig 2.8 CDmin for Cargo and Passenger Aircraft
Lift-drag polar estimation
Fig 2.9 K1 for Fighter Aircraft
The lift-drag polar for high-performance fighter type aircraft can be
approximate use Eq.(2.9), K2=0, and Figs 2.9 and 2.10
0
0.2
0.4
0.6
1K
oM
0.5 1.0 1.5 2.0
Future
Current
Future
Current
Lift-drag polar estimation
Fig. 2.10 CDo for Fighter Aircraft
0
0.01
0.02
0.03
0.04
DoC
oM
0.5 1.0 1.5 2.0
Current
Future
PropulsionTrust of installed engine varies with Mach number, altitude, and afterburner operation can be described with simple algebraic equation that has been fit to either existing data of company-published performance curves or predicted data based on the output of off-design cycle analysis with estimates made for installation losses. The following algebraic equations for installed engine trust lapse are based on expected performance of advanced engines in the 1990 era and beyond
(2.43a)
6.032.125.0568.0 Ma (2.42)
For a High bypass ratio turbofan engine (M<0.9):
For a Low bypass ratio mixed turbofan engine with afterburner:
7.04.16.0245.088.072.0 Maa mil
7.02max 4.038.094.0 Maawet (2.43b)
(2.45a)
7.05.15.0262.0907.076.0 Maa mildry
(2.44a)For an advanced turbojet with afterburning:
For an advanced turboprop:
7.02max 4.03.0952.0 Maawet
(2.44b)
a For M ≤ 0.1
02.0
12.0
Ma For 0.1 < M < 0.8 (2.45b)
WEIGHT FRACTION
• Instantaneous weight fraction , , is needed in computing the constraints. is a fuel/payload correction.– Initial value for can be based on experience– Typical values for different mission phases are shown on the
mission profile figures on the following slides.• Fig 2.11: for Typical Fighter Aircraft• Fig 2.12: for Typical Cargo and Passenger Aircraft
Cruise
Descend
Descend & Land
Warm-up & Takeoff
Descend
Combat Air Patrol
Penetration
Subsonic Cruise Climb
= 1
= 0.85
= 0.9
= 0.78 = 0.68
= 0.60
Loiter
Accelerate & Climb
Instantaneous weight fraction, , for a Typical Fighter Aircraft
Combat
Deliver Expendables
Climb
Escape Dash
(Figure 2.11)
Cruise
Descend
Descend & Land
Warm-up & Takeoff
Descend & Land
Warm-up & Takeoff
Climb
Cruise
= 1
= 0.95 = 0.86
= 0.81 = 0.73
= 0.70
Loiter
Climb
Instantaneous weight fraction, , for a Typical Cargo and Passenger Aircraft
(Figure 2.12)
2.4 Example Constraint Analysis
• Creating constraint boundary on a diagram of takeoff thrust loading (TSL/WTO) vs wing loading (WTO/S)
• Based on RFP for Air-to-Air Fighter given in Ch. 1
• Done by creating equation:
f T W W SS L T O T O{ ( / ), ( / )} 0
Table 2.E1 Selected AAF SpecificationsMission Phases & Segments
Performance Requirements
1-2 Takeoff
Acceleration
Rotation
2000 ft PA, 100oF, STO = SG + SR ≤ 1500 ft
kTO = 1.2, TO=0.05, max power
VTO, tR = 3 s, max power
6-7 Supersonic
Penetration
And Escape Dash
1.5 M/ 30 k ft, no afterburning (if possible)
7-8 Combat
Turn 1
Turn 2
Acceleration
30,000 ft
1.6M, one 360 deg 5g sustained turn, with afterburning
0.9M, two 360 deg 5g sustained turns, with afterburning
0.8 1.6M, μt ≤ 50 s, max power
13-14 Landing
Free roll
Braking
2000 ft PA, 100˚F, sL = sFR + sB ≤ 1500 ft
kTD = 1.15, tFR = 3 s, B = 0.18
Drag chute diameter 15.6 ft, deployment ≤ 2.5 s
Max Mach Number
RFP para. C.1
2.0M/40k, max power
In order to proceed, you need:• CLmax • Lift drag polar• Engine data (fig. 2.E1)
These are obtained from the figures and the equation to come…
00.4 0.8 1.2 1.6 2.0 0.4 0.8 1.2 1.6 2.0
0
0.1
0.2
0.3
0.4K1
Mach Number
0.01
0.02
0.03
0.04CDo
Mach Number
CD = K1CL2 + CDo
CLmax = 2.0
(a) Aerodynamic Data
00.4 0.8 1.2 1.6 2.0 0.4 0.8 1.2 1.6 2.0
0
0.2
0.4
0.6
0.8adry
Mach Number
0.01
0.02
0.03
0.04awet
Mach Number(b) Installed Thrust Lapse
40k
30k
20k
10k
SL
40k
30k
20k
10kSL
T
Ta Md ry
S Ld ry 0 7 2 0 8 8 0 2 4 5 0 6 1 4 0 7. { . . ( | . | ) } *. .
T
Ta Mw et
S Lw e t { . . ( . ) } * .0 9 4 0 3 8 0 4 2 0 7
NOTE: TSL = Std.sea level max power static thrust
Fig. 2.E1 Preliminary AAF Data
Procedure with Takeoff• The airplane accelerated by thrust with no resisting forces in the ground roll
• Thrust is balanced by drag forces during the constant velocity rotation
Given these conditions and eqn 2.2 becomes:
sk
g C aT
w
w
St k
C
w
ST OT O
o L w e tS L
T O
T OR T O
L
T O { }* * *
{ *
m a xm a x( )
( ) ( )2 2
1
22
(2.E1)
Where:
a W S b W S cT O T O( / ) / 0
From table 2.E1, Fig 2.E1 and standard atmosphere of Appendix B
kTO = 1.2β = 1.0ρ = 0.002047 sl/ft3
go = 32.17 ft/s2
CLmax = 2.0σ = 0.8613
(awet)M=1.0 = 0.8775tR = 3.0 ssTO = 1500 ft
sk
g C aT
w
w
St k
C
w
ST OT O
o L w e tS L
T O
T OR T O
L
T O { }* * *
{ *
m a xm a x( )
( ) ( )2 2
1
22
(2.E1)
Evaluate coefficients a, b, and c from eqn 2.E1
Rewritten eqn 2.E1:
( ) { }w
S
b b a c
aT O
224
2(2.E2)
at wS L T O
1 2 4 7.
/
b = 79.57
c = 1500
To obtain the data:
TSL/WTO 0.4 0.8 1.2 1.6 2.0 2.4
WTO/S (lb/ft2) 33.4
57.5
77.1
93.7
108
121
Maximum Mach Number Cruise Constraint for Fighter Aircraft
Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightT = D Thrust equals DragR=0 Not on the groundh & v Values are GivenK2 Small
Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightT = D Thrust equals DragR=0 Not on the groundh & v Values are GivenK2 Small
Pertinent Data:= 0.78(wet)M=2 = 0.7189q = 1101 lb/ft2
K1 = 0.36CDo = 0.028
Pertinent Data:= 0.78(wet)M=2 = 0.7189q = 1101 lb/ft2
K1 = 0.36CDo = 0.028
SW
q
C
S
W
qK
W
T
TO
DTO1
TO
SL o
(2.12)
Begin with equation 2.12
Maximum Mach Number Cruise Constraint for Fighter Aircraft
SWS
W
W
T
TO
TO
TO
SL 42.88102.767 4
Substitution of pertinent data into equation 2.12 yields the following
(2.E3)
The maximum Mach number cruise constraint combined with the takeoff constraint enclose a solution space as illustrated in figure 2.E2.
Figure 2.E2 Constraints for Takeoff and Max Mach Number
Selection of the Air-to-Air Fighter Design Point
2TO ft/lb64S
W
Figure 2.E3 AAF Design Constraints
The requirements defined for a new air-to-air fighter result in the constraint diagram shown in figure 2.E3.
Based on the problem assumptions, the constraints that are dominate the design are the takeoff, landing, and supercruise constraints.
To keep cost and practicality of the AAF design under control, thrust loading is chosen near the region of previous experience.
Reasonable first choice:
2.1W
T
TO
SL
• Capability requirements for short takeoff and landing and nonafterburning supercruise drive design
• Relaxing these constraints would not significantly change the design choice because other constraints are important just outside of the takeoff, landing, and supercruise constraints.
• For many flight segments, the aircraft will be operating at full thrust; thus, no one requirement will cause the size to become excessively large. This is a result of careful consideration and balancing of requirements.
Selection of the Air-to-Air Fighter Design Point
Selection of the Air-to-Air Fighter Design Point
If thrust loading and wing loading are selected, it is possible to determine weight specific excess power (Ps) for a given flight condition by combining equations 2.2b and 2.11. Taking R = 0, this combination yields the following relationship:
S
W
q
CK
S
W
qK
W
TVP
TO
D2
TO1
TO
SLs
0
(2.E4)
Selection of the Air-to-Air Fighter Design Point
This relationship is used to generate the contours of constant Ps in figure 2.E5 for specified loadings, , and flight conditions. This figure can be used to determine the minimum time-to-climb flight path; this flight path corresponds to the maximum Ps attainable for a given line of constant energy height, ze. This time-to-climb relationship is a modified version of equation 2.2b:
Figure 2.E5 AAF Ps Contours 2e
1e
z
zs
e2
1 P
dzdtt
Mission Phase 1-2: Takeoff
22
2
4
a
acbb
S
WTO
2
1ln
TO
LTO
TO
SL
TO
TOo
k
C
WTg
aMAX
MAXLTOR C
ktb
2
Assumptions:2000 ft. PA100ºFkTO=1.2µTO=0.05tR=3 ssTO=sG+sR1500 ftMax Power
From (2.25) and (2.26), with STO=sG+sR:
where:
TOsc
Mission Phase 1-2: Takeoff
• =1.0 • K1=0.18
• =0.002047 slug/ft3 • =0.8613
• CLmax=2.0 • (wet)M=0.1=0.8775
• kTO=1.2 • µTO=0.05
• CDo=0.014 • tR=3.0s
• TO=0.36 • sTO=1500 ft
05.08775.0
2601.01ln03.42
TO
SL
WT
a
57.79b
We estimate and to be constant at appropriate mean values. A conservative estimate for TO for the AAF is obtained by assuming (CDR - µTOCL) =0 and evaluating CD at CL=CLmax/kTO
2.
With: Then:
1500c
TSL/WTO 0.4 0.8 1.2 1.6 2.0
WTO/S (lb/ft2) 14.3 45.1 67.2 85.3 101
whence:
Mission Phases 6-7 and 8-9: Supersonic Penetration and Escape Death
2
1
SW
q
C
S
W
qK
W
T
TO
DoTO
TO
SL
Assumptions:1.5M/30k ftNo afterburning
From (2.12):
Mission Phase 6-7: Supersonic Penetration and Escape Dash
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
q
nK
S
W
q
nK
W
qS
W
To 2
1 2
2
2
1
SW
q
C
S
W
qK
W
T
TO
DTO
TO
SL o
1
Assumptions (most from Case1):dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are GivenK2=0 Pure Parabolic Drag Polar
Assumptions (most from Case1):dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are GivenK2=0 Pure Parabolic Drag Polar
1
Supersonic Penetration and Escape Dash Example
WTO/S 20 40 60 80 100 120
TSL/WTO 2.35 1.77 1.2 0.913 0.746 0.638
SWS
W
W
T
TO
TO
TO
SL 25.7010*345.4 4
With:
=0.78 dry=0.3953 q=991.8 psf
=0.3747 K1=0.28 CDo=0.028
Mission Phase 7-8: Combat Turn 1
oD
TOTO
TOTO
SL
g
Vh
dt
d
VqS
RC
S
W
q
nK
S
W
q
nK
W
qS
W
To 2
1 2
2
2
1
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are GivenK2=0 Pure Parabolic Drag Polar
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are GivenK2=0 Pure Parabolic Drag Polar
SW
q
C
S
W
qnK
W
T
TO
DTO
TO
SL o
2
1
Combat Turn 1 Example
SWS
W
W
T
TO
TO
TO
SL 22.4210*407.5 3
With:
=0.78 wet=0.7481 q=1128 psf n=5
=0.3747 K1=0.30 CDo=0.028
WTO/S 20 40 60 80 100 120
TSL/WTO 2.22 1.27 1.03 0.96 0.963 1
2.4.3.4 Mission Phase 7-8: Combat Turn 2
--0.9 M/30K ft, two 360 degree 5g sustained turns, with afterburning.
From Eq. (2.15)
SWS
W
W
T
TO
TO
TO
SL 34.1201473.0
SW
q
C
S
W
qnK
W
T
TO
DoTO
TO
SL
2
1
with β = 0.78 αwet = 0.5206 n = 5 CDo = 0.018
σ = 0.3747 K1 = 0.18 q = 357.0 lb/ft2
then
whence
WTO/S (lb/ft2) 20 40 60 80 100 120
TSL/WTO 0.910 0.900 1.09 1.33 1.60 1.87
2.4.3.5 Mission Phase 7-8: Horizontal Acceleration
--0.8 → 1.6 M/30K ft, Δt ≤ 50s, max power.
From Eq. (2.18)
6484.079.28
10704.3 4
S
WS
W
W
T
TO
TO
TO
SL
tg
Ma
SW
q
C
S
W
qK
W
T
oTO
DoTO
TO
SL
1
We obtain approximate constant values of ρ, K1, CDo, and α at a mean Mach number of 1.2.
with β = 0.78 K1 = 0.23 a = 994.8 ft/s
σ = 0.3747 q = 634.7 lb/ft2 ΔM = 0.8 αwet M=1.2= 0.5952 CDo = 0.025 Δt = 50 s
then
Constituted from p. 60
whence,
20 40 60 80 100 120
2.10 1.38 1.15 1.04 0.973 0.933
)/(/ 2ftlbSWTO
TOSL WT /
Mission Phase 13-14: Landing
From Eqs. (2.33) and (2.37) with2
2
2
4
a
acbb
S
WTO
s
sss
st
k
BFRL
BFR
TD
2.5 deployment
ft,15.6diameter chute, drag GFE
ft1500
,18.0,3
,15.1 100F, PA,2000ft
BFRL sss
Mission Phase 13-14: Landing(cont.)
Where,
2
0 max)(
1ln
TD
L
TO
STB
L
L
k
C
WTag
a
max/2 LTDFR Cktb Lsc
Assumptions
Mission Phase 13-14: Landing(cont.)
5348.0
500ft area wingairplane Estimate
ft 191 area chute RFP
4.1 chute Drag
2
2
DRc
D
C
C
A conservative estimate of from at 0.8 of touchdown lift coefficient and by assuming
L DC0)( LBDRcDR CCC )/( 2
max TDL kC
0.18 0.18,K
,1500s ,8123.0 ,014.0C ,0.2
,3 ,5348.0C ,210.1C ,slug/ft002047.0
0a ,2775.0C ,15.1 0.56,
B1
LLD
DRcL3
D
0max
ftC
st
k
L
FR
TD
Mission Phase 13-14: Landing(cont.)
2/5.70/ ftlbSWTO
whence
1500c 57.06b 47.14 athen
Section 2.4.3.7
• This gives
S
WS
W
W
T
TO
TO
TO
SL 88.4210*767.2 4
W TO /S (lb/ft 2 ) 20 40 60 80 100 120
T SL /W TO 2.14 1.07 0.713 0.535 0.428 0.357
The table below shows a few calculations
Chapter 3 – Mission Analysis
• Section 3.1 - Concept– With the values of thrust loading (TSL/WTO) and wing
loading (WTO/S) determined, now the focus shifts to the estimation of the gross takeoff weight, which is a sum of component weight:
– Wp is specified in the RFP and is separated into payload weight (WPE) that consists of the cargo being delivered during a mission (ie ammunition) and the permanent payload weight (WPP), which consists of the crew, passengers, their equipment, and anything else that is carried the entire mission
FEPTO WWWW (Eqn 3.1)
Chapter 3 – Mission Analysis
– WE consists of the basic aircraft structure and any permanently attached equipment
– WF represents the required fuel weight to fly the mission
Chapter 3 – Mission Analysis
• WF depends on the rate of fuel consumption, which depends on the thrust specific fuel consumption (TSFC)
• Thrust can be found from (Eqn 2.1) while the TSFC depends on the engine cycle, flight conditions (altitude, mach #, etc…)
Chapter 3 – Mission Analysis
• Fuel consumption analysis results in multiple benefits:– Little information needed for calculation– Reveals optimum solution (minimum fuel) for
flying certain segments of the mission– Shows the fuel consumed during each mission
segment as a fraction of the aircraft weight at the beginning of the mission segment, which makes WF a calculable fraction of WTO
Chapter 3 – Mission AnalysisFigure 3.1 Weight Fractions of Cargo and Passenger Aircrafts
Figure 3.2 Weight Fractions of Fighter Aircrafts
Chapter 3 – Mission Analysis
• Section 3.2 – Design Tools
• This section deals with the thermodynamics of the flight and the way the thrust work of the engine is used by the aircraft
• Begin by considered the change in weight of aircraft from fuel burn
TTSFCdt
dW
dt
dW F *
(Eqn 3.2)
Chapter 3 – Mission Analysis
• (Eqn 3.2) can be rewritten as:
• Note that T is the installed thrust and TSFC is the installed thrust specific fuel consumption
• Also, since:
dtW
TTSFC
W
dW** (Eqn 3.3)
VW
Tdsds
ds
dt
W
Tdt
W
T
*
Chapter 3 – Mission Analysis
• The above equation shows the incremental weight-velocity specific engine thrust work done as an amount of fuel is consumed (dWF)
• This work is partly used in the mechanical energy (PE + KE) of the airplane mass and partly dissipated into the atmosphere (wing tip vortices, heating, etc…)
Weight Fractions
• The goal of subsequent section is to integrate Eq.(3.3) to get “weight fractions”
• Requires knowledge of TSFC behavior and “instaneous thrust loading” as a function of time along flight path.
• Two distinct classes of integration– Ps > 0 (Type A)
– Ps > 0 (Type B)
i
f
initial
final
W
W
W
W
TO
SL
W
T
W
T
Instantaneous Thrust Loading Behavior: Type A
• Ps > 0; examples of Type A found in cases1. Constant speed climb2. Horizontal acceleration3. Climb and acceleration4. Takeoff acceleration
• Using Eq.(2.2a) to move towards solution:
• Combining Eqs.(3.3) and (3.4)
• u = how total engine thrust work is distributed between mechanical energy and dissipation
• Eq. 3.5 shows u is fraction of thrust work dissipated
edzuV
TSFC
W
dWor
g
Vhd
uV
TSFC
W
dW
121 0
2
T
RDuwhere
u
dz
u
gVhdds
W
TVdt
W
T e
11
2 02
(3.4), (3.5)
(3.6a), (3.6b)
Instantaneous Thrust Loading Behavior: Type A (continued)
• Integration of Eq.(3.6a) which largely depends on variation of {TSFC/V(1-u)} with altitude and velocity. When that is relatively constant:
• Δze = total change in energy height
• Eqs. (3.6) and (3.7) show how potential and kinetic energy can be “traded” within certain limits
• Example of when bad: No drag (u = 0) and constant ze
– Unpowered dive or zoom climb
– No fuel consumed and dW = 0 or Wf = Wi
ei
f
i
f
zuV
TSFC
W
Wor
g
Vh
uV
TSFC
W
W
1exp
21exp
0
2
(3.7a)
(3.7b)
(from Type A, cont.)Note: Minimum Fuel Path
• Fuel consumption analysis also shows the supposed “best” way to fly certain legs for minimum fuel used.
• For Type A legs, Eqs. (3.2) (2.2b) and (2.3) produce
es
SLF dzP
aTSFCTdtTSFCTdW
Pg 58
VuW
TV
W
RDTPs )1(
)(
Defining the fuel consumed specific work (fs) as:
then:
Eq. 3.9 shows that the minimum fuel-to-climb flight from ze1(h1,V1) to ze2(h2,V2)Corresponds to a flight path that produces the max. thrust work per unit weight of fuelAt each energy height level in the climb (i.e. max value of fs at each ze)
8.3TSFCa
Pf s
s
2
1
2
1
219.3
e
e
z
z s
eSLFF f
dzTdWW
Pg 58Graphical method for finding max. fs at any ze (thus min. fuel-to-climb path from ze1 to ze2)
• constant fs and ze contours in the altitude-velocity plane (fig. 3.E5)
• Analogous to using the Ps and ze contours (fig. 2.E5) to find min time-to-climb path (Fig. 3.E2) in conjunction with the time-to-climb equation from eq. 2.2b.
2
1
2
1
e
e
z
z s
e
P
dzdtt
Pg 58/9Instantaneous Thrust Loading Behavior: Type B
When Ps=0, it is generally true that the speed and altitude are essentially constant and specific information is given regarding the total amount of time (Δt or Δs/V) which elapses. Examples of type B cases are:
constant speed cruiseconstant speed turn
Best cruise Mach number and altitudeLoiterwarm-up
Take off rotationConstant energy height maneuver
For this type of flight, the thrust work is completely dissipated and the required thrust may not be known in advance. The thrust is modulated or throttled so that T=(D+R) or u=1 by Eq. 3.5 and dze=0 by eq. 3.4
Using Eq. 2.2 it follows that
)10.3(dtW
RDTSFC
W
dW
Pg. 59Often TSFC (D+R)/W remains relatively constant over the flight leg
So eq. 3.10 can be integrated
If desired, the required thrust may be computed after the fact.
timeflighttotalt
tW
RDTSFC
W
W
i
f
11.3exp
TSFC Behavior (Thrust Specific Fuel Consumption)
•Complex function:
•Altitude
•Speed
•Throttle setting (especially if has afterburner)
Pg. 59Good starting point at this stage is the assumption:
C = constant
θ = represents the usual thermodynamic cycle improvement due to a lower ambient temperature at higher altitude.
Basis of this assumption is the observation that:
TSFC is not a strong function of speed & throttle setting
C could be selected differently for different legs of the mission.
NOTE:
Assumptions for TSFC other than eq. 3.12 can be made and the weigh fraction analysis continued to completion but this model is a good for turbojet and turbofan. Different assumptions would be required for turboprop engines which TSFC is primarily proportional to flight Mach # (eq. 1.25). The same process could be applied.
12.3CTSFC
Summary of Weight Fraction Equations
• The equations previously mentioned for dW/W and Wf/Wi (Eqs. 3.6, 3.7, 3.10, 3.11) are combined with the equation for TSFC (Eq. 3.12) in order to get the equations for the Type A and Type B situations.
Summary of Weight Fraction Equations (cont.)
• TYPE A (Ps >0, T=αTSL given)
oi
f
o
g
Vh
uV
C
W
W
g
Vhd
uV
C
W
dW
2)1(exp
2)1(
2
2
Instantaneously exact
(3.13)
Exact when
sqrt(θ)/ V(1-u) constant
(3.14)
Summary of Weight Fraction Equations (cont.)
• TYPE B (Ps =0, T=D+R given)
tW
RDC
W
W
dtW
RDC
W
dW
i
f
exp
Instantaneously exact
(3.15)
Exact when
sqrt(θ)(D+R)/ W constant
(3.16)
Special Cases
• The following slides will discuss a various special cases to give both specific design tools and more insight into weight fraction analysis.
• Assumptions– α, β, K1, K2, CDo are known
– n=1 and L=W • except for Cases 4,8 (n>1), and 10
– R=0• except where the aircraft is on the ground (Cases 4 and 10)
CASE 1: Constant Speed Climb (Ps=dh/dt)
• Given:– dV/dt = 0– n=1(L=W)– R=0
– T = αTSL (Full Thrust)
– hinitial
– hfinal
– V
• Conditions
SL
TO
L
D
T
W
C
Cu
Case 1.
• Eq. (3.14) becomes Eq. (3.17)
where the level flight relationships (3.18)
have been used, and the change in altitude is
SLTOLDi
f
TWCC
h
V
C
W
W
/1exp
L
D
L
D
C
C
SqC
SqC
L
D
W
D
initialfinal hhh
Case 1.
• Notes on using Eq. 3.17• Typically, variables values are averaged for the altitude interval used in the
analysis although the variance in the quantity
should be checked to verify the accuracy of this averaging approach. If the accuracy is poor, the climb should be broken into intervals and Eq. (3.17) should be applied to each interval. The overall Wf / Wi the then the product of the interval results.
• Note that the fraction of engine thrust dissipated is
while is the fraction of the engine thrust invested
in potential energy.
u1
SLTOLD TWCC /
SLTOLD TWCC /1
Case 2.
• Horizontal Acceleration
• Under these conditions
so that Eq. (3.14) becomes Eq. (3.19)
where Eq. (3.18) has been used and
Similar to the previous analysis, average values for the speed intervals should be used unless the quantity V(1-u) varies too much.
dtg
VdVP
os
Given: dh/dt = 0, n =1 (L = W), R = 0, full thrust (T = aTSL) and the values of h, Vinitial, and Vfinal.
SL
TO
L
D
T
W
C
Cu
SLTOLDi
f
TWCC
gV
V
C
W
W
/1
2exp 0
2
222initialfinal VVV
Page 62
Case 4:
Where u can be obtained from Eq. (2.21) as
and VTO is as given by Eq. (2.20)
SL
TOTO
TOTO T
W
W
Squ
(3.22)
Case 5: Constant Altitude/Speed Cruise (Ps=0)
Given:
Under these conditions, Eqs. (3.16) and (3.18) become
sVh
R
WLndt
dV
dt
dh
range cruise and, , of valuesand
,0
),(1
0 ,0
s
C
C
V
C
W
W
L
D
i
f exp (3.23)
Case 5: Constant Altitude/Speed Cruise (Ps=0)
Where CL is obtained by setting lift equal to weight and CD is computed from Eq. (2.9). Since CD/CL will vary as fuel is consumed and the weight of the aircraft decreases, this result is not, strictly speaking, exact. Hence, it might become advisable to break the cruise into several intervals and apply Eq. (3.23) to each.The Type B case 6-10 that follow are, however, exact integrals.
Case 6: Constant Altitude/Speed Turn (Ps=0)
Given:
This situation can be treated much the same as Case 5, except that L=nW. The duration of the turning can be shown, with the help of Eq. (2.17), to equal
Under these conditions, Eqs. (3,16) and (3.24) become
NnVh
Rdt
dV
dt
dh
turnsofnumber and ,1, , of valuesand
,0
0 ,0
1
222
0
ng
nNV
V
NnRt c
1
2exp
20 ng
nNV
C
nCC
W
W
L
D
i
f (3.25)
(3.24)
Case 6: Constant Altitude/Speed Turn (Ps=0)
Given: Under these conditions, Eqs. (3,16) and (3.24) become
1
2exp
20 ng
nNV
C
nCC
W
W
L
D
i
f
Page 64: Equations 3.26-3.28
S
W
qC TO
L
SLL
d
a
Cds
MC
C
W
dW
1
3.2.7 Case 7: Best Cruise Mach Number and Altitude (BCM/BCA) (Ps = 0)
Given: dh/dt = 0, dV/dt = 0, n=1, R=0, and value of cruise range is Δs
Replacing dt with ds/V, Eq. (3.15) becomes:
(3.26)
3.27
L
DoLL
L
D
C
CCKCK
C
C 2
21
From the above equation, a minimum CD/CL can be found by differentiating the equation with respect to CL and setting the result equal to zero. This leads to:
21
*
4 KKCC
CDo
L
D
(3.27)
From Eq. (3.26) weight reduction can be minimized by operating at lowest value of (1/M)(CD/CL). This is known as the best cruise condition and the parameters associated with it contain a * superscript. The below equation shows that drag and lift do not depend on M below critical drag rise Mach number:
3.28
• Insert equation 3.28 here because I can’t see it because it is cut off. It is very short though!
Mission Analysis
1
0202*
K
CKCC D
DD
LD CC /*)//()/1( LD CCM
(3.29)
Since the lowest achievable value of is constant below the critical M, it follows that decreases as M increases. Once MCRIT is reached, the situation changes. Further increases in M cause to increase and finally cause to increase. Rather than search for the exact at which the minimum occurs, it is more practical to take , then
*)/( LD CC
BCMMM CRIT *
CRIT
D
L
D
M
KKC
MC
C 210* 4)
1(
CRITMM
*)//()/1( LD CCM
*)//()/1( LD CCM
(3.30)
This may be substituted into Eq. (3.27) to yield
Mission Analysiswhich may be substituted into Eq. (3.26) to yield
dsa
C
M
KKC
W
dW
SLCRIT
D
2104
(3.31)
This may be directly integrated to yield the equivalent of Eq. (3.16),
sa
C
M
KKC
W
W
SLCRIT
D
i
f 24exp 10
(3.32)
WSCPM
SqC LL 2
2
S
W
KCMPP
P TO
DCRITSLSL 102 /
12
Finally, since the aircraft must sustain its weight under this condition,
or,
(3.33)
Mission Analysis
Which may be used to find the altitude of “best” cruise or BCA. Since must gradually diminish as the mission progresses, this last result shows that the altitude must gradually increase. Since the flight Mach number is fixed at , it also follows that the speed, which is proportional to , must gradually decrease. In most cases, these changes occur so slowly that the assumption of PS = 0 remains true.
CRITM
Case 8: Loiter (Ps=0)Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundn=1 (L=W)
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundn=1 (L=W)
t Flight Duration
dtC
CC
W
dW
L
D
(3.34)
Putting these assumptions into Equation 3.15, we get:
dtKKCCW
dWDo 214 (3.35)
From inspection in case 7, we know that the optimum CD/CL for minimum fuel consumption is equal to:
214 KKCCC DoLD
Thus,
Case 8: Loiter (Ps=0)
tKKCCW
WDo
i
f 214exp (3.36)
S
W
KCMPTO
DoSL 12 /
12
(3.37)
Integrating equation 3.35 for weight from Wi to Wf and for the time interval Δt, we get:
Variation in pressure altitude can be found as a function of Mach number for loiter by equating Lift and Weight.
)(
Case 9: Warm-Up (Ps=0)Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=T=αTSL
V=0D=0Given:h AltitudeΔt Warm-up time
Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=T=αTSL
V=0D=0Given:h AltitudeΔt Warm-up time
Equation (3.15) under these assumptions becomes:
dtTCdW SL
Mission Analysis
• Derived from ¿¿PAGE 66??
tW
TC
W
W
TO
SL
i
f
1 (3.39)
Weight Fraction Where β is evaluated atthe beginning of warm-up
CASE 10: TAKEOFF ROTATION (PS = 0)
• Given: dh/dt = 0, dV/dt = 0, (D+R) = T = αTSL and values of h, V, and rotation time tR
(3.40) dtTCdW SL
(3.41)
RTO
SL
i
f tW
TC
W
W
1
Where β is evaluated at the beginning of rotation
Under these conditions, Eq. 3.15 becomes
CASE 11: CONSTANT ENERGY HEIGHT MANEUVER (PS = 0)
• Given: Δze = 0, n = 1 (L=W), R = 0 and thevalues of hinitial, hfinal, Vinitial, and Vfinal
(3.42)
t
C
CC
W
W
L
D
i
f exp
Where CL is obtained by setting lift equal to weight and CD is computed from Eq. 2.9. The maneuver time interval t is best [continued in next page]
In this case potential energy is exchanged for kinetic energy and any loss to aerodynamic drag is balanced by the engine thrust.Under these conditions, Eqs. 3.16 and 3.18 become
Calculation of Takeoff Weight
• Weight fraction for each mission segment is the ratio of final weight to the initial weight for that segment
1 fi
i
f
W
W(3.43)
• For the weight ratio of mission legs where only fuel is consumed
Calculation of Takeoff Weight (cont).
• For example, for several consecutive mission segments:
525443324
5
3
4
2
3
2
5
W
W
W
W
W
W
W
W
Expendable payload case
• For the special case where the expendable payload is delivered at some point j in the mission, we write instead:
j
PE
j
PEj
W
W
W
WW
1 (3.44)
Takeoff Weight
• Recall that takeoff weight is defined as
PPPEEFTO WWWWW (3.45a)
Fuel Weight• For a mission with n junction and payload
delivery at junction j (j<n), is expressed in terms of by the sum of the aircraft weight decrements between takeoff and junction j and between junction j (after delivery of expendable payload) and landing in the following manner:
FW
TOW
njPE
jTOPE
jTO
jTOTOF WWWWWWW
111
which reduces to
njPE
nTOF WWW 11
1(3.45b)
Historical Correlations• Figures 3.1 and 3.2 relate empty weight to
takeoff weight via
)( TOWf
)( TOTO
E WfW
W
(3.45c)
where is given by Eq-ns (3.47)-(3.50).
• Combining Eq-ns (3.45a) – (3.45c) and solving for yieldsTOW
n
njPEPP
TO
WWW
1
(3.46)
Summary• Takeoff weight is proportional to and
(where matters more because it is carried the whole way)
• Improved structural materials, which reduce (and therefore ), can reduce dramatically
• An initial estimate of must be made in order to place values upon the of the payload deliveries and of the structural weight.
• If the initial estimate of takeoff weight is far from the value given by the Eq-n (3.46), an iterative solution of Eq-n (3.46) is will be necessary.
PPW
PEW PEW
EW TOW
TOW
TO
PEW
W
Pages 70-73 go here
• Arvind
Pages 74-77 go here
• Joan
Table 3.E3 Summary of Results – Mission Analysis
Mission Phases & Segments β=W/WTO End of Leg
Wf/Wi=∏ if
1-2 Warm-up and Takeoff 0.9759 0.9759 2-3 Accelerate and Climb 0.9445 0.9678 3-4 Subsonic Cruise Climb 0.9141 0.9678 4-5 Descend 0.9141 1.0000 5-6 Combat Air Patrol 0.8780 0.9605 6-7 Supersonic Penetration 0.8035 0.9152 7-8 Combat 0.7441 0.9261
Deliver Expendables 0.6917 ------- 8-9 Escape Dash 0.6757 0.9769 9-10 Minimum Time Climb 0.6743 0.9979
10-11 Subsonic Cruise Climb 0.6487 0.9620 11-12 Descend 0.6487 1.0000 12-13 Loiter 0.6210 0.9573 13-14 Descend and Land 0.6210 1.0000
Figure 3.E3 Fraction of Takeoff Weight vs Mission Phase
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 2 3 4 5 6 7 D 8 9 10 11 12 13 14
Mission Phase Number
W/W
TO
3.4.2 Determination of WTO, TSL, and S
• Now compute the takeoff weight for the AAF (Air-to-Air Fighter) using Equation 3.46 shown below.
148
141
PP PETO
W WW
Where,
WPP= 200 lb Pilot plus equipment270 Cannon405 Ammunition feed system275 Returning ammunition198 Casings weight
1,348 lb
(3.46)
3.4.2 ContinuedWPE = 382 lb Sidewinder missiles
652 AMRAAMs275 Spent Ammunition
1,309 lb
0.6273E
TO
W
W
14 9 148 8 13
14 2 141 1 13
Source: AAF RFP Sec. D
Source: Eq. (3.49)Assumed WTO = 25,000 lb
Thus,1348 (1309)(0.8978)
62,0000.6680 0.6273TOW lb
Source: Table 3.E3
Example Mission Analysis (cont.)
Fig. 3.E4 Weight Fractions of Fighter Aircraft
0.3
0.4
0.5
0.6
0.7
0.8
10 20 30 40 50 60 70 80 90 100
WTO (1000 lbs)
WE/W
TO
X-29
F-102A
F-16C/ D
F-16A/ B
F-106AF-15
F-15C
F-105GF-14A
F-4E
F-111F
F-101B
F/ A-18
F-100D
F-104C
YF16YF17
Notice (Figure 3.E4) that this conventional metal version of the AAF would be considerably heavier than existing lightweight fighters (e.g., F-16).
Example Mission Analysis (cont.)
lbsWTO 400,24)90.0(6273.06680.0
)8978.0)(1309(1348
Reducing the payload provides little relief because the large WTO is the result of the very small denominator.
Another assumption can be the reducing the fuel usage, for example by shortening the range or duration of mission. Yet, another assumption would be reducing the empty weight, for example by using advanced lightweight construction materials as composites.
With the latter assumption, the aircraft’s Γ will be reduced by approximately 10%. Therefore, with this reduction, WTO is recalculated for the assumed WTO of 25,000 lbs,
Example Mission Analysis (cont.)The conclusion is that AAF can have a practical size. This is true only if reliable non metallics with competitive strength, durability, and repairability become available according to schedule. With the choices of TSL/WTO = 1.20 and WTO/S = 64.0 made, the description of the AAF at this stage of design, using WF/WTO = 0.3265, is:
WTO = 24,400 lb
TSL = 29,300 lb
S = 381 ft2
WP = 2,660 lb
WE= 13,800 lb
WF = 7,970 lb
This information will give perspective and insight regarding the nature and shape of the corresponding aircraft.
3.4.3 FIRST REPRISEHere, the key assumptions made so far will be evaluated. Hopefully, the assumptions made are accurate and the results acceptable. If not, the whole process will have to be reiterate with new assumptions.
Earlier, the wing area (S) was assumed to have a value of 500 ft2, whereas the recently derived wing area is 381 ft2. This in turn will increase the solution space in Fig. 2.E3, but it is important to note that it wont provide any design points superior to the one already selected.
Warm-up
tW
TC
TO
SLupwarm
1
Assumptions: 2000 ft Pressure Altitude, 100 °F Temp, 60 second duration, min power
Δt = 60 sec, σ = 0.8613, β = 1.0, TSL/WTO = 1.2,
αdry = 0.6484, θ = 1.08, C = 1.35 h-1
9818.0
9818.0
upwarm
Takeoff AccelerationAssumptions: 2000 ft Pressure Altitude, 100 °F Temp, kTO = 1.2,
μTO = 0.05, max power
SL
TOTO
TOTO
TOAccelTakeoff
T
W
W
Squ
g
V
u
C
01exp
Takeoff Acceleration
β = 0.9818, a = 1160 ft/s, μTO = 0.05, WTO/S = 64 lb/ft2, MTO = 0.1812,
q = 11.31 lb/ft2, σ = 0.8613, C = 2.0 h-1, ξTO = 0.36 (Estimated from 2.4.3.1),
CL max = 2.0, θ = 1.08, u = 0.1067, kTO = 1.2, TSL/WTO = 1.2, VTO = 210.2 ft/s, αwet = 0.8795
9777.0
9958.0
acceltakeoff
Pages 84-85 go here
• Matt daniell
∆ h
(ft)
∆(V2/2go)
(ft)
∆ ze
(ft)
T∆s/W
(ft)
∆t
(min)
∆s
(nmi)
c
(ft/s)-1
∏
a 14,000 2208 16,210 23,250 0.6452 5.683 0.001557 0.9906
b 14,000 3.105 14,000 20,420 0.7589 6.749 0.001485 0.9922
c 13,000 -661.6 12,340 18,520 0.9810 8.437 0.001494 0.9931
∑ 41,000 1550 42,550 62,190 2.385 20.87 n/a n/a
Segment E – Climb/Acceleration Results
9761.0 cbaE
The results do NOT differ insignificantly, therefore the phase should be broken down into finer intervals. In summary, Mission Phase 2-3:
9445.0
9678.0
min892.2
43.23
23
23
23
ED
ED
ED
ttt
nmisss
Example Mission Analysis
Mission Phase 3-4: Subsonic Cruise Climb
BCM/BCA, Δs23+Δs34 = 150 nmi
Assume: (Ps = 0 , T = D+R , K2 = 0)
SLCRIT
DO
a
sC
M
KC341
34
4exp
Where: Δs34 = 126.6 nmi K1 = 0.18
C = 1.35 h-1 Cdo = 0.018
aSL = 1116 ft/s Mcrit = 0.9
Then:
9141.0
9678.034
Mission Phase 5-6: Combat Air Patrol
30k ft , 20 min.
Assume: (K2 = 0)
tKCC DO 156 4exp
Where: Δt = 1200 sec θ = 0.7940
K1 = 0.18 C = 1.35 h-1
Cdo = 0.014
Then:
8780.0
9605.056
Mission Phase 4-5: Descend
BCM/BCA -> Mcap/30k ft
9141.0
0.145
Example Mission Analysis
Page 88 starts here
• ragini
Mission Phase 8-9 Continued
98
9769.0
6757.0
Mission Phase 9-10: Minimum Time Climb
109
exp tC
Cc
L
D
BCM/BCA = 1.5M/30kftAirplane climbs from 30,00 ft to 50,000 ft and reduces speed from 1.5 to 0.9 MaFrom Eq 3.42:
With (assuming vertical speed = 0.7 Vavg)
avgV
ht
7.0
mideo
mid hzg
Vi
2
2
Mission Phase 9-10: Minimum Time Climb
Given information:hi = 30,00 ftMi = 1.5(a/aSL)i= 0.8911Vi = 1492 ft/sZei = 64,600 ftHf = 50,000 fth = 20,000 ftMf = 0.9
(a/aSL)f= 0.8671Vf = 870.9 ft/sVavg = 1181 ft/st = 24.19 sHmid = 40,000 ftVmid = 1258 ft/s(a/aSL)mid= 0.8671Mmid = 1.3WTO/S = 64 lb/ft2
= 0.6757= 0.1858CL = 0.093K1 = 0.23K2 = 0CD0 = 0.023CD/CL = 0.2687= 0.7519C = 1.35 h-1
109
9979.0
Calculation from given information
6743.0
Mission Phase 10-11: Subsonic Cruise Climb
1110
9620.
BCM/BCA, s10-11 = 150 n.mi.
6487.0
Mission Phase 11-12: DescendBCM/BCA = Mloiter/10kft
6487.0
1211
1
Mission Phase 12-13: Loiter
1312
9573.
BCM/BCA = Mloiter/10kft, 20 min
6210.0
Mission Phase 13-14: Descend and Land
Mloiter/10kft --> 2000 ft PA, 100oF
6210.0
1413
1
Page 94-95
• Kc and someone else
Case X: Minimum Fuel-to-Climb Path
NOTE: AAF is not required to fly a climb path for minimum fuel burn.
Fuel Consumed Specific Work (fs) is calculated with Eq. (3.8) at military power using:
145.1
135.1
97.0
/64
2.1
01
01
2
MforhC
MforhC
ftlbSW
WT
TO
TOSL
Case X: Minimum Fuel-to-Climb Path
Contours of constant fs and constant energy heights (ze) are shown in the velocity-altitude space (figure 3.E5).
From energy height ze1(h1,V1) to ze2(h2,V2), the min fuel-to-climb path has a maximum value of fs at each ze.
ze = 12 kft @ SL to ze = 100 kft @ 57 kft is also shown in figure 3.E5
Case X: Minimum Fuel-to-Climb Path
By comparing figures 3.E2 and 3.E5:
• Ps = 0 and fs = 0 are the same [from Eq (3.8)]
• Due to the step change in C across M = 1, each fs contour has a discontinuity– Not including fs = 0
• Minimum time-to-climb path occurs at about 50 fps greater than subsonic minimum fuel-to-climb path
•Minimum time-to-climb paths reach transonic and then supersonic speeds at a lower altitude than minimum fuel-to-climb paths
By dividing Eq. (3.9) by W1 = βWTO and given WF1-2 = W1-W2 & 1Π2 = W2/W1
The aircraft weight fraction (1Π2) results in:
In figure 3.E5 (ze1 = 12 kft, ze2 = 64 kft):
Case X: Minimum Fuel-to-Climb Path
2
1
21
211
1e
e
z
z s
e
TO
SLF
f
dz
W
T
W
W
2
1
121
e
e
z
z s
e
TO
SL
f
dz
W
T
9685.021
Note: No equation number were given in the text