82 AIRCRAFT ENGINE DESIGN 1.8
T H R 1.6 U S T 1.4 L O A 1.2 D
I N 1.0 G
0.8 T/W SL TO
0.6 20
~
~ I ' ' ' I ' ' ' I ' ' ISolution Landing ' /
\ \
MISSION ANALYSIS 83
B) Takeoff acceleration. 2 0 0 0 ft PA, 100°F, kro = 1 . 2 , / Z r o = 0.05, CDR = 0.07, C L max = 2.0, Macce I = 0.1, m a x power.
F r o m C a s e 4, Eqs. (3.21) and (3.22)
1-I" = exp [ -(C' + C2M)~/-O \ ~ - u,l ~ [
W i t h Vro e v a l u a t e d f r o m Eq. (2.21), the t a k e o f f e s t i m a t e o f ~ r o f r o m C h a p t e r 2, Sec. 2.4.3, and a e v a l u a t e d at Maccel, and w i t h
/~ = 0.9895 a = l l 6 0 f t / s /Zro = 0.05
Wro/S = 641bf/ft 2 CLmax = 2.0 kro = 1.2
Mro = 0.1819 0 = 1.0796 6 = 0.9298 0o = 1.0818 60 = 0.9363 Otwet = 0 . 9 0 0 6
TsL/Wro = 1.25 u = 0.2725 C1 + C2M = 1 . 6 2 7 h -1 then
C) Takeoff rotation.
F r o m C a s e 10, Eq. (3.44)
I - I c = 1--(C1 " } - C 2 M ) ~ I I o ~ - ~ ( ~ o ) t R W i t h
tR = 3 S 6 = 0.9298
TSL/Wro = 1.25
then
I
- I 8 = 0.9958 /~ = 0.9853Mro/2000 ft PA, 100°F, tR = 3 s, m a x power.
/3 = 0.9853 0 = 1.0796 6o = 0.9515 Olwe t = 0.9003
I
- I c = 0 . 9 9 8 4T h e w e i g h t fraction for m i s s i o n phases 1 - 2 is the p r o d u c t o f the w e i g h t fractions for s e g m e n t s A, B, and C.
I"I : I-IA F I B I " I c : 0.9837
12
/3 = 0.9837
Macce I = O. 1 Vro = 211.1 ft/s q = 45.601bf/ft 2
~ro = 0 . 3 6 1 2
Mro = 0.1819 00 = 1.0867
C1 --k C2M = 1.649 h -1
84 AIRCRAFT ENGINE DESIGN
Mission phase 2-3: Accelerate and climb. This phase consists of segments D (Horizontal Acceleration) and E (Climb and Acceleration).
D) Horizontal acceleration. Mro -+ Mct/2000 ft PA, 100°E mil power.
For this Mach number change of 0.1819 to 0.70, a single interval calculation will suffice. The Climb/Acceleration Weight Fraction Calculation Method of Sec. 3.4.1 is used with properties evaluated at the middle state point (M = 0.4410, 2000 ft PA, 100°F) and fl = fli.
With
(a/astd)i = 1.039
V/ = 211.1 ft/s
(a/a~td)f = 1.039 Vf = 812.0ft/s
A(vZ/2go) = 9555 ft A h = 0.0ft
A Z e ~--- 9555 ft /3 = 0.9837
Wro/S = 64 lbf/ft 2 8 = 0.9298
M = 0.4410 CL = 0.2351 Ka = 0 . 1 8 , K 2 = 0 . 0 CDo = 0.014 CD = 0.02395
CD/CL = 0.1019 0 = 1.0796 0o = 1.1215 8o = 1.0627 Otd~y = 0.5264
Tsz/Wro = 1.25 u = 0.1524
T A s / W = l l , 2 7 0 f t
(a/astd) = 1.039 V = 511.5 ft/s At = 32.97 s As = 2.775 n miles
C1/M + C2 = 2.340 h -1
then
H
e = 0.9935 fl = 0.9773 Note that Aso = 2.775 n miles and AtD = 32.97 S.E) Climbandacceleration. McL/2OOOftPA, 100°F--+ BCM/BCA, milpower.
The weight fraction of this mission segment is calculated along the minimum time-to-climb path depicted in Fig. 3.E2. In our illustrative example of Sec. 3.4.1, we found the weight fraction of this segment, using a single gross interval, to be 0.9812. Here we shall use the three successive integration intervals given in Table 3.E4 for our calculations, applying Eq. (3.20) to each. We use/3 =/3i and the values of the state properties (or, 0, Co/CL, and V) at the mid state point of each interval to obtain an appropriate average value of the required properties. The fli of each succeeding interval is obtained by multiplying the fli of the preceding interval by I-I. By comparing the final result with that of Sec. 3.4.1, we can determine whether it is necessary for sufficient accuracy to break the climb/acceleration calculations into still smaller intervals.
Following this approach, the Climb/Acceleration Weight Fraction Calculation Method of Sec. 3.4.1 was used to obtain the results given in Table 3.E5.
H E = H a H b H c = 0.9806
MISSION ANALYSIS 85 Table 3.E4 Segment F,---schedule and intervals
Integration interval
Climb schedule (state points)
Altitude, fl Mach number a b c
2,000 (100 °F) 0.70
9,000 0.775
16,000 0.85
23,000 0.875
30,000 0.90
36,000 0.90
42,000 0.90
Initial Mid
Final Initial Mid
Final Initial Mid Final
This result differs insignificantly from the earlier result o f the single gross in- terval calculations for the illustrative example in Sec. 3.4.1. The amount o f fuel consumed for this phase has changed b y only 3%. We should therefore not con- sider breaking this phase into even finer intervals. Summarizing, then, we have the following results for M i s s i o n Phase 2 - 3 :
As23 = Asp + Ase = 23.04 n miles At23 = Ato + AtE = 2.906 min.
H = H D H E = 0.9742 23
fi = 0.9584
Mission phase 3-4: Subsonic cruise climb. BCM/BCA, As23 -}- As34 = 150 n miles, mil power.
This is a type B (Ps = 0, T = D + R) mission leg for which full thrust is not usually applied. Even so, we shall use a conservative full thrust value o f TSFC
from Sec. 3.4.1 for all type B legs in these calculations.
F r o m Case 7, Eq. (3.29), with COR = K2 = 0
{ ~ (C1-'[- C2McRIT)AS34 I
]7.. = exp 34 MCRIT as----~d I
Table 3.E5 Segment E----climb/acceleration results
Ah, AV2/2gc, AZe, TAs/W, At, As, C]/M + C2, Interval ft ft ft CD/C L lg ft min n miles h- 1 I-I a 14,000 2 2 1 0 16,210 0.1601 0.1962 20,170 0.4915 4.067 1.462 0.9927 b 14,000 2 . 2 5 4 14,000 0.1384 0.2696 19,170 0.6913 6.119 1.329 0.9937 c 12,000 -662.4 11,340 0.1111 0.3755 18,150 1.1728 10.086 1.3 0.9941
40,000 1 5 5 0 41,550 57,490 2.356 20.27
86 with
then
AIRCRAFT ENGINE DESIGN
As34 = 127.0n miles K1 = 0.18 Coo = 0.016 Mcmr = 0.9 C1 -+- C2McRIT ---- 1.1700 h -1
I
- I = 0.97363 4
fl = 0.9331
Note should be taken of the fact that fuel consumption could be further reduced for this phase by capitalizing on the range factor as described in Sec. 3.2.7, Case 7.
Consequently, the preceding result may regarded as conservative.
with
then
Mission phase 4-5: Descend. BCM/BCA --~ MCAp/30 kfc - I = 1.0
4 5
fl = 0.9331
Mission p h a s e 5-6: Combat air patrol. 30 kft, 20 min., mil power.
From Case 8, Eq. (3.40), with CoR = K2 = 0
I-'I = exp{-(C1 + C 2 M c A P ) W / O ~ A t }
5 6
At = 1200s 0 = 0.7941 8 = 0.2976
K1 = 0.18 Coo = 0.014 C~ = 0.2789
MCA P = 0.6970 C1 + C2M = 1.1091h -1
- - I = 0.9675
5 6
/~ = 0.9027
At the end of combat air patrol MCA P = 0.6856 from Case 8, Eq. (3.38).
Note should be taken of the fact that fuel consumption could be further reduced for this phase by capitalizing on the endurance factor as described in Sec. 3.2.8, Case 8. Consequently, the preceding result may regarded as conservative.
Mission phase 6-7: Supersonic penetration. This phase consists of seg- ments F (Horizontal Acceleration) and G (Supersonic Penetration).
MISSION ANALYSIS
Table 3.E6 Segment F--horizontal acceleration
87
AZe, TAs/W, At, As, CI/M + C2,
Interval Mi Mf Mavg ft Co/CL u ft min n miles h -I
a 0.6856 0.95 0.8178 6,652 0.1085 0.1697 8,011 0.2567 2.062 2.227 0.9956
b 0.95 1.23 1.0900 9,389 0.2537 0.2907 13,240 0.2332 2.496 1.738 0.9943
c 1.23 1.5 1.3650 11,340 0.4192 0.3552 17,590 0.1828 2.452 1.442 0.9937
- - E 27,380 - - 38,840 - - 0.6727 7.010 - - - -
F) Horizontal acceleration. MCAP --+ 1.5M/30 kft, max power.
This acceleration calculation is divided into the three intervals shown in Table 3.E6. The initial, final, and average Mach number of each interval and h = 30,000 ft are used in the Climb/Acceleration Weight Fraction Calculation Method of Sec. 3.4.1. The calculated results are given in Table 3.E6.
Then
I - I F = I - - I a U b U c = 0.9837 /3 = 0.8880
It is interesting to note from Table 3.E6 that the total mission weight specific work is 38,840 ft with 70.49% used to increase the mechanical kinetic energy of the airplane. The remaining 29.51% is dissipated into nonmechanical energy of the airplane/atmosphere system. A single gross interval calculation for this segment gives higher values of total time and total ground distance by 1.16% and 4.22%, respectively, and lower values of weight fraction and specific thrust work by 0.024% and 0.33%, respectively. Also note that A s s = 7.01 n miles.
G) Supersonic penetration. 1.5M/30 kft, AsF + A s c = 1 0 0 n miles, no after- burning.
From Case 5, Eq. (3.23), with CoR = 0
with
then
A s a = 92.99 n miles CL = 0.05731 CD = 0.02889
C1/M -t- C 2 : 0.9 h -1 K1 : 0.27
CD/CL : 0.5040
- I c = 0.9382 Segments F and G together yield
U = U F 1--IG = 0.9231
6 7
/3 = 0.8331
6 = 0.2976 Coo = 0.028
88 AIRCRAFT ENGINE DESIGN
Mission phase 7-8: Combat. This phase consists of segments H (Fire A M R A A M s ) , I (Combat Turn 1), J (Combat Turn 2), K (Horizontal Accelera- tion), and L (Fire A1M-9Ls and ½ ammo).
H) Fire AMRAAMs. 652 lbf.
From Eq. (3.47), and since fl Wro = W7 here,
t ~ WTO -- WpE1 WpE1
- 1 - - -
Wro ~ Wro
with
then
W p E 1 = 652 lbf
= 0.8331
Wro = 24,0001bf
fl WTO -- WpE1
-- 0.9674 and fl = 0.8060
Wro
I) Combat turn 1. 1.6M/30 kft, one 360 deg 5g sustained turn, with after- burning.
From Case 6, Eq. (3.25), with CDR = 0
n f D
I-II=exp{_(CI+C2M)~/-~(___~_L ) 2zrNV ]
g0 /
with
0 = 0.7941 [ 0.7821 wet } ot = 0.4444 dry K1 = 0.290
nCo/CL
Co/CL = 0.1883 Olreq = ~ ~ - 0.6070
CI + C2M = 1.693 h -1 linearly interpolated then
3 = 0.2976
I
CI + CzM = 1.380
Coo = 0.028
0o = 1.201 CL = 0.2286 Co = 0.04305
%AB = 48.15 V = 1592ft/s
--II = 0.9753
J) Combat turn 2.
burning.
From Case 6, Eq. (3.25), with Cog = 0
1--i:=exp{_(Cl+C2M)v/-~(n~_~D L / go n2~/~--I
3o = 1.265
fl = 0.7860
0.9M/30 kft, two 360 deg 5g sustained tums, with after-
MISSION ANALYSIS
Table 3.E7 Segment J--horizontal acceleration
89
Aze, TAs/W, At, As, C1/M + C2,
Interval Mi Mf Mavg ft CD/CL u ft min n miles h-I ]7
a 0.8 1.06 0.930 7,438 0.1589 0.1878 9,158 0.1950 1.781 1.990 0.9955 b 1.06 1.33 1.195 9,925 0.3755 0.3205 14,610 0.1748 2.052 1.609 0.9942 c 1.33 1.6 1.465 12,170 0.5579 0.4086 20,580 0.1723 2.480 1.362 0.9931
E 29,530 44,350 0.5421 6.313
with
0 = 0.7941 3 = 0.2976 00 = 0.9228
{ 0"5033 wet/dry _ { 1.843 wet } d r y
ot = 0.3020 | CI + C 2 M = 1.170 Cc = 0.7045
K1 = 0.18 Coo = 0.016 Co = 0.1053
nCD/CL
Co/ CL = 0.1495 Otreq = P ~---"-/'~U-~,, -- 0.4700 %AB = 83.46 ISL/ vvTO
C1 + C2M = 1.732 h -l linearly interpolated V = 895.4 ft/s then
- ' L = 0.9774
30 = 0.5033
fl = 0.7682
0.8 --+ 1.6M/30 kft, max power.
K) Horizontal acceleration.
This acceleration calculation is divided into the three intervals shown in Table 3.E7. The initial, final, and average Mach number o f each interval and h = 30,000 ft are used in the Climb/Acceleration Weight Fraction Calculation Method of Sec. 3.4.1. The calculated results are given in Table 3.E7.
Then
V I K = r i o I - I b I - I f = 0.9828 fl = 0.7550
The weight specific thrust work of segment K is 44,350 ft with 66.58% going to increase the kinetic energy and 33.43% dissipated into nonmechanical energy by the drag forces.
A single interval calculation for this segment gives essentially the same weight fraction value but lower values of time, distance, and specific thrust work by 4.44%, 3.26%, and 1.78%, respectively.
L) Fire AIM-9Ls and ~ ammunition. 657 lbf.
From Eq. (3.47), and since Wj = fl Wro here, fl Wro - Wpz2 Wpe2
3 Wro 3 Wro
90 with
AIRCRAFT ENGINE DESIGN
WpE2 = 657 lbf /~ = 0.7550 WTo = 24,0001bf then
Wro - W p ~
- 0.9637 and Wro
For mission phase 7-8 we have
- I 0.8734
7 8
fl = 0.7276
= 0.7276
Mission phase 8-9: Escape dash. 1.5M/30 kft, As89 = 25 n miles, no afterbuming.
The computational procedure is identical with that of segment G. Here the distance is smaller (25 n miles vs 92.99 n miles) and the fraction of takeoff weight (fl) is less (0.7276 vs 0.8880), making CL smaller (0.04696 vs 0.05731) and C D / C L greater (0.6090 vs 0.5041). The net effect is a higher weight fraction for this leg (0.9795 vs 0.9382) and less fuel consumed, as
- I 0.9795
8 9
fl = 0.7127
Mission phase 9-10: Minimum time climb. 1.5M/30 kft --~ BCM/BCA, mil power.
In this leg kinetic energy is exchanged for potential energy as the airplane climbs from 30,000 ft to the Case 7 48,000 ft (BCA) and the Mach number is reduced from 1.5 to 0.9 (BCM). Using the initial and final values of h and V in the following listing, we find that the energy height diminishes by 4800 ft. We assume, therefore, in accordance with the RFP, a constant energy height maneuver for the weight fraction calculation of this phase.
From Case 11, Eq. (3.42), with CDR = 0
]--I = exp { - ( C , - q - C e M ) ~ / C O ( ~ L ) A t ]
9 10
with (assuming the vertical speed is 0.5 of Vmia) "
A h V Zmia
A t -- 0.5Vmi~a 2g0 Zei - - havg
MISSION ANALYSIS 91 and
Then,
Ot = 0.7941
= 1492fffs hi = 3 0 , 0 0 0 f l Zei = 6 4 , 6 0 0 h h f = 4 8 , 0 0 0 h
Of = 0.7519 V f = 871.2fffs z ~ = 5 9 , 8 0 0 h
AZe = - - 4 8 0 0 h h a ~ = 3 9 , 0 0 0 h
Vm~ = 1284fffs A t = 28.04 s Omm = 0.7519 Mmid = 1.326
$mid= 0.1949 CLmia = 0.08986
K l m i d = 0 . 2 3 8 7 K2mid = 0 C oomid = 0 . 0 2 8
Comia = 0.2993 (Co/CL)mia = 0.3330 C1 + C2Mmid = 1.298 h -1 A s = 5.131 n miles
H
= 0.99719 10
fl = 0.7106
This is a conservative estimate in as much a s AZe = - - 4 8 0 0 ft < 0.
Mission phase 10-11: Subsonic cruise climb. BCM/BCA, A s 1 0 _ l l = 150 n miles, mil power.
Refer to the outbound subsonic cruise climb leg data, mission phase 3-4. The weight fraction o f this leg is lower (0.9698 vs 0.9736) due only to a higher cruise distance (144.9 vs 127.0 n miles). The starting BCA is higher (48,000 ft vs 42,000 ft) because of the airplane's lower weight (0.7106 Wro vs 0.9584 Wro).
Then,
I
- ' [ = 0.9698 10 11fl = 0.6891
Mission phase 11-12: Descend. BCM/BCA ---* Mtoiter/lO lift.
--I = 1.O 11 12
~ = 0.6891
Mission phase 12-13: Loiter. Mzoit, r/ l O kft, 20 min., mil power.
Refer to data for the combat air patrol loiter leg, mission phase 5 - 6 . C o m p a r e d to that leg the weight fraction o f this leg is about the same (0.9677 vs 0.9675) because the offsetting effects o f the dimensionless temperature 0 (0.9312 vs 0.7941) and M a c h number (0.3940 vs 0.6970). Because o f the lower vehicle weight and altitude
9 2 A I R C R A F T E N G I N E D E S I G N
ofthisleg, theloiterMach n u m b e r i s m u c h l e s s t h a n t h a t o f t h e C A P l e g . T h e n ,
H
= 0 . 9 6 7 712 13
fl = 0 . 6 6 6 8
Mission phase 13-14: Descend and land. Mloiter/lO kft --> 2000 ft PA, 100°F.
H
= I . 013 14
fl = 0 . 6 6 6 8
The results of all of the AAF weight fraction computations in the preceding are summarized in Table 3.E3 and Fig. 3.E3.
Minimum fuel-to-climb path. Although the AAF is not required by the RFP to fly a minimum fuel climb, this section is included to show the application of the material on minimum fuel path included as a note in Sec. 3.2 and for comparison with the minimum time path of Fig. 3.E2. Values of fuel consumed specific work ( f i ) were calculated for the AAF at military power using Eq. (3.8) for the following data:
TSL/WTO : 1.25 WTo/S = 6 4 l b f / f t 2 /3 = 0 . 9 7 q -}- C2M : 0.9 + 0 . 3 M The results of these calculations are presented in Fig. 3.E6 as contours of constant fuel consumed specific work (fi) plotted in the velocity-altitude space with con- tours of constant energy height (Ze). The minimum fuel-to-climb path from energy
7O i
6 0 -
5 0 -
4 0 - - Alt
3 0 - (kft) -
2 0 -
fs (min = 0, max = 2800, incr 200 kft) Standard Day Atmosphere
1 0 -
I I I I I I I ~
100 500
Fig. 3.E6
i i I I I I / I i i /
• I i i i l I I I i
1000 1500
Velocity (fl/sec)
Minimum fuel-to-climb path on fs chart.
2000 W/S = 64
"TAV - 1.25 _ B e t a 0.97 g's - 1.00
Tmx = 1.0
"TR = 1.07 - A B off
Engine #2 TSFC #2
"Aircraft #5 Min fuel
MISSION ANALYSIS 93
height Zel(hl, V1) to energy height Ze2(h2, V2) corresponds to a path having a max- imum value of fs at each Ze. The resulting minimum fuel-to-climb path from a Ze of 12 kft at sea level (where 111 = 879 ft/s) to a Ze of 95 kft at 57 kft altitude (where
V2 = 1564 ft/s) is also shown on Fig. 3.E6.
Based on the comparison of Figs. 3.E2 and 3.E6, the following comments can be made:
1) The contours Ps = 0 and f~ = 0 are the same. This follows from the definition of f~ given in Eq. (3.8).
2) The minimum fuel-to-climb path is similar but different from the minimum time-to-climb path for evident reasons [see Eq. (3.8)].
3) The minimum fuel-to-climb path stays subsonic to a higher altitude (about 37 kft) than that of the minimum time-to-climb (about 15 kft).
You may find it useful to know that the weight fraction for a type A flight path can be expressed conveniently in terms of Tslff Wvo, fs, Ze, and/3 by dividing Eq. (3.9) by the aircraft weight at the start W 1 ( = • WTO), noting that WFI_2 = W1 - W2, and using the definition of 1-[1 2( = W2/WO. Making these substitutions gives
Ze2
w 12 _1_1-I_ raze
W1 1 2 fl Wro fs
Zel
The resulting equation for the aircraft weight fraction 1-[1 2 is TSL f Ze2 dZe I - I = l f l ~ o Z
1 2 Zel
Application of this equation to the subsonic portion of the minimum fuel-to- climb path of Fig. 3.E6 (Zel = 12 kft, Ze2 = 52 kft) where fs is approximately 2800 kft gives the following estimate for the weight fraction:
-I 0.9815 12
References
1"2002 Aerospace Source Book:' Aviation Week and Space Technology, 14 Jan. 2002.
2jane's All the World Aircraft, 2000-2001, Jane's Yearbooks, Franklin Watts, Inc., New York, 2000.
3Raymer, D. P., Aircraft Design: A Conceptual Approach, 3rd ed., AIAA Education Series, AIAA, Reston, VA, 2000.
4jonas, F., Aircraft Design Lecture Notes, U.S. Air Force Academy, Colorado Springs, CO, 1984.