The Request for Proposal (RFP) mission for the Air-to-Air Fighter (AAF) of Chapter 1 will be flown on paper in this example. The purpose initially is to find the weight fraction (Wf/Wi) of each mission phase and then a takeoff weight (Wro) estimate. Our ultimate goal is to use this estimate of Wro and the value of thrust loading (TsL/Wro) selected in the Chapter 2 example outlined in Sec. 2.4 to determine the AAF installed thrust requirement (TsL). The wing loading (WTo/S) selected and WTO will also give us our first estimate of the AAF wing area. The mis- sion to be flown is depicted in Fig. 3.El and is given with more detail in Table 3.El.
3.4.1 Weight Fraction Analysis
In addition to the design tools of this chapter, certain other information is re- quired to complete the weight fraction calculations. These supportive materials are the AAF preliminary data contained in Figs. 2.El and 2.E5, Sec. 3.3, and the standard atmosphere tables in Appendix B. From Eq. (3.55), the thrust specific fuel consumption (TSFC) in units of 1/hour is (1.6 + 0.27 M0)V'-O for maxi- mum power and (0.9 + 0.3 M0)~/-ff for military power and lower. Figure 2.E5 was used to determine the minimum time-to-climb path and the result is presented as Fig. 3.E2.
To illustrate the calculation procedure, let us determine the weight fraction of segment E in Mission Phase 2-3 (Table 3.El). First, please note that we shall round off to four significant digits all numbers recorded for weight fraction calculations in this chapter. However, the recorded numbers are obtained from calculations done by the AEDsys computer program. At times, therefore, you may not reproduce exactly our numbers because of a discrepancy in the significant digits of the input numbers and those used in serial internal calculations.
MISSION ANALYSIS Table 3 . E l AAF mission
73
Mission
phases Segments Condition
1-2 Warm-up and takeoff A--Warm-up B--Acceleration C--Rotation
2-3 Acceleration and climb D--Acceleration E--Climb/acceleration 3-4 Subsonic cruise climb 4-5 Descend
5-6 Combat air patrol 6-7 Supersonic penetration
F--Acceleration G--Penetration 7-8
8-9
Combat
H - - F i r e AMRAAMs I - - T u m 1
J--Turn 2 K--Acceleration
L - - F i r e AIM-9Ls & ½ ammo Escape dash
9-10 Minimum time climb 10-11 Subsonic cruise climb 11-12 Descend
12-13 Loiter
13-14 Descend and land
2000 ft PA, 100°F 60 s, mil power
kro = 1.2, IZro = 0.05, max power Mro, tR ---- 3 s, max power Minimum time-to-climb path
Mro to McL/2000 ft PA, 100°F, mil power Mcc/2000 ft PA, 100°F to BCM/BCA,
mil power
BCM/BCA, As23 + As34 = 150 n miles BCM/BCA to Mcae/30k ft
30 kft, 20 min 30,000 ft
McaP to 1.5M/30k ft, max power 1.5M, ASF + Asc = 100 n miles,
no afterburning 30,000 ft 652 lbf
1.6M, one 360 deg 5g sustained turn, with afterburning
0.9M, two 360 deg 5g sustained turn, with afterburning
0.8 to 1.6M, At < 50 s, max power 657 lbf
1.5M/30 kft, As89 = 25 n miles, no afterburning
1.5M/30 kft to BCM/BCA BCM/BCA, As10-11 = 150 n miles BCM/BCA to Mloiter/lO kft Mlo,er/10 kft, 20 min
Mloiter[lO kft to 2000 ft PA, 100°F
One may question the w i s d o m o f carrying calculations to four significant figures that are based upon numbers often estimated to be correct to a few percent. The underlying reason is that the 1--Is are generally the difference between unity and a quantity o f the order of 0.01, which is calculated from the estimated n u m b e r s - - often the exponents o f Eqs. (3.14) and (3.16). Therefore, in order to be certain that the small quantity is correct to two significant figures, it is necessary to maintain four places in I-I. This should also be understood to emphasize the fact that the weight fraction o f fuel consumed (1 - l-I) for most flight legs is quite small, but the cumulative effect is what matters.
During segment E, the A A F climbs and accelerates at military power to BCM/
BCA along the m i n i m u m time-to-climb path o f Fig. 3.E2 via the following climb schedule o f Table 3.E2.
74 AIRCRAFT ENGINE DESIGN
70
60
5 0 -
4 0 -
A l t _
3 0 -
(kft) -
2 0 -
1 0 -
0
Ps ( m i n = 0, m a x - 600, incr 50 fl/sec) Standard Day A t m o s p h e r e
t i i I i i I i t /
i i i l i i I I l i r I I I
100 500 1000 1500
Velocity (ft/sec)
M i n i m u m time-to-climb path on Ps chart.
Fig. 3.E2
l i l t
2000 W / S = 64
- T / W = 1 . 2 5
_ B e t a = 0.97 g's = 1.00
T m x = 1.0
- T R - 1.07
- A B o f f
_ E n g i n e #2 Aircraft #5 - M i n t i m e
The mid state point is chosen arbitrarily but j u d i c i o u s l y in order to obtain rea- sonable average values. The reader, as ever, is free to vary the mid point selection in order to determine the sensitivity of the results to this choice.
Equation (3.33) o f Case 7, with CDR = 0, yields the BCA P 2~(Wro/S) ~/-~--,v
~BCA - - P--~td - - ~ V ~ D O / ~ 1 } / r s t d jr1 CRIT
where Wro/S = 64 lbf/ft 2 and, from Fig. 2 . E l a , M c m r = 0.9, CDO = 0.016, and K1 = 0.18. F r o m weight fraction calculations preceding segment E, (fli)E = 0.9773, which we can use as an estimate for fl = (fli)e so that
•BCA = 0.1749 and h ~ 42,000 ft Table 3.E2 Climb schedule
Altitude, ft Mach number Computation interval 2,000 (100°F) 0.70 Initial state point
9,000 0.775
16,000 0.85
23,000 0.875 Mid state point
30,000 0.90
36,000 0.90
42,000 0.90 Final state point
MISSION ANALYSIS 75 Proceeding, then, with the weight fraction calculation for segment E, we have from Case 3, Eq. (3.20), with CDR = 0
where
I-'Ie - Wf - exp (C1/M + C2) [ A(h + V2-/u 2g°) ] -1
c_ _(WTo3
u = c L o~ \ Y Z s ~ J
For this illustrative calculation we select the single gross interval having the initial, middle, and final state points listed with Table 3.E2. An appropriate average value of u is found using fl = fli and values CD/CL and ot at the mid state point.
The initial, final, and middle state point values of h and M from Table 3.E2 and fl = fli are used in conjunction with the following Climb/Acceleration Weight Fraction Calculation Method to find l i e .
Climb/acceleration weight fraction calculation method Given:
Initial state point quantities: hi, M i , ( a / astd ) i , fl i Final state point quantities: h f, My, (a /asta) f
Middle state point quantities: h, M, 0, 8, 00, 80, (a/asta), K1, K2, CDO Other quantities: astd, go, WTo/ S, y, Pstd, TSL/ WTO, Cl, C2
Equations:
Vi = (a /astd)iastdMi Vf : (a/astd)fastdM f
\ 2g0 J 2go
o (W o3
U
T A s A Z e
W 1 - u V = (a/aste)astaM
A h = h i - h~ ,xt -- ~ - V -d \ - ~ s ~ )
,,Ze= +,',(
\ 2goJ , , , =2~(Wro/ PstdSM 2 ]---[ {
CL -- l . e = exp
CD K1 C2 't- K2CL q- CDO
CL CL
If max power,
Ol = Olwet(O0, TR, 60)
(C1/M -[- C2) astd
x [A h -1 +
76 AIRCRAFT ENGINE DESIGN If mil power,
With
ot = Otary(00,
TR, 30)
(a/astd)i
= 1.039 V/ = 811.7 ft/s(a/asta)f
= 0.8671Vf
= 870.9 ft/s(AV2/2go)
= 1549 ft Ah = 40,000 ft AZe = 4 1 , 5 5 0 f t /3 = 0.9773Wro/S
= 64 lbf/ft 2 3 = 0.4051M = 0.88 CL = 0.1346 K1 = 0.18, K2 = 0.0
Coo
= 0.016 CD = 0.01926CD/CL
= 0.1431 0 = 0.8420 00 = 0.9724 30 = 0.6706 Otary = 0.4024TsL/WT'o
= 1.25 u = 0.2780TAs/W
= 57,550ft(a/astd)
= 0.9176 V = 901.2ft/s At = 2.068 m i n As = 18.39 n milesCt/M
+ C2 = 1.323h -1then
F I E = 0.9812
This weight fraction of 0.9812 means that fuel in the amount (1 - 0.9812)
flWro
is consumed as 57,550 ft of weight specific thrust work is produced. The dissipated nonmechanical energy fraction of this work is u = 0.2780, while the fraction invested in airplane mechanical energy (Aze) is (1 -- u) = 0.7220, of which 96.27% is potential energy and 3.72% is kinetic energy. The climb time and ground distance covered are 2.068 m i n and 18.39 n miles, respectively.At the end of segment
E,
/3f = /3t FIE = 0.9589. Now a new mean value of /3 for segment E can be estimated and the calculation of F i e iterated. When this is done using the average value of (0.9773 + 0.9589)/2 = 0.9681, we obtain the same value for F i e , justifying the use of/3i for~3avg.
Also the8BCA
calculation can be repeated now u s i n g / 3 f , and we find again that h ~ 42,000 ft.This calculation illustrates how a mission leg weight fraction is determined. The results of such calculations for all mission phases are contained in Table 3.E3 and Fig. 3.E3. The detailed computations leading to the data shown there are given in the Sec. 3.4.4 (Complete A A F Weight Fraction Computations).
--'[ = 0.7152 W F / W T o = 0.2787
1 14
1-I = 0.8585 1-I = 0.9164
7 14 8 14
MISSION ANALYSIS
Table 3.E3 Summary of results---mission analysis---24,000 lbf WTO 77
Mission Mission fl = W/Wro Fuel used, Payload,
phase segments End of leg I41[/Wi = 1-I lbf lbf
if
1-2 A Warm-up 0.9895 0.9895 252
1-2 B Takeoff acceleration 0.9853 0.9958 100
1-2 C Takeoff rotation 0.9837 0.9984 39
2-3 D Acceleration 0.9773 0.9935 155
2-3 E Climb/acceleration 0.9584 0.9806 453
3 - - 4 Subsonic cruise climb 0.9331 0.9736 607
4-5 Descend 0.9331 1.0000 0
5-6 Combat air patrol 0.9027 0.9675 729
6-7 F Acceleration 0.8880 0.9837 354
6-7 G Supersonic penetration 0.8331 0.9382 1317
7-8 H Fire AMRAAMs 0.8060 0 652
7-8 1 1.6M/5g turn 0.7860 0.9753 478
7-8 J 0.9M/5g turns 0.7682 0.9774 427
7-8 K Acceleration 0.7550 0.9828 317
7-8 L Fire AIM-9Ls and 0.7276 0 657
1 amino
8-9 Escape dash 0.7127 0.9795 358
9-10 Minimum time climb 0.7106 0.9970 51
10-11 Subsonic cruise climb 0.6891 0.9698 516
11-12 Descend 0.6891 1.0000 0
12-13 Loiter 0.6668 0.9677 535
13-14 Descend and land 0.6668 1.0000 0
End Remove permanent 0.6106 1348
payload
Total 6688 2657
3.4.2 Determination of WTO, TSL, and S
The time has come to compute the takeoff weight of the AAF. To do this, information obtained from the RFP, Table 3.E3, and Fig. 3.E4 will be inserted into the equation
Wpp -~- WpEI I'-I "~- WpE2 E
7 14 8 14
Wro
F [ - F
1 14
which can be easily recognized as the analytical extension of Eq. (3.49) of Sec. 3.2.12 to this multiple payload release mission.
The quantities appearing on the right-hand side of this equation are evaluated thusly:
78 AIRCRAFT ENGINE DESIGN 1.00
0.95
0.90
0.85
W/W 0.80
TO
0.75
0.70
0.65
0.60
0.80 0.75 0.70
0.65
We /Wro
0.600.55 0.50 0.45 0.40 10
1 2 3 4 5 6 7 7H 7L 8 9 10 11 12 13 14 Fig. 3.E3 Fraction of takeoff weight vs mission phase.
I I
" • X-29A
I I I I I I
F-~2A
X ~ 1 6 ~ D F-15 Eq. (3.52)
¥ ~ L ~ F.~ooo, , ~ , ,
:~-4~---._._~
F-101B F/A-18E F-111 Ft
I I I I I I I I
80 90 100
20 30 40 50 60 70
W (1,000 lbfs)
TO
Fig. 3.E4 Weight fraction of fighter aircraft.
MISSION ANALYSIS 79 Wet, = 200 lbf Pilot plus equipment
270 Cannon
405 Ammunition feed system 275 Returning ammunition 198 Casings weight 1348 lbf
Wt,el = 652 lbf AMRAAMs Wpe2 = 382 lbf Sidewinder missiles
275 Spent ammunition 657 lbf
Wp = 2657 lbf
(Source: AAF RFP Sec. D.)
F I =
I-I"" I-I
= 0.85857 14 7 8 13 14
U = H " " 1-I = 0 . 9 1 6 4
8 14 8 9 13 14
1-I = F I F l = 07152
1 14 1 2 13 14
(Source: Table 3.E3.)
If we assume that Wro = 24,000 lbf, Eq. (3.52) gives I" = We/Wro = 0.6306 Thus
1348 + (652)(0.8585) + (657)(0.9164)
Wro = = 29,6701bf
0.7152 - 0.6306
This result evidently calls for another iteration, in which the small changes in the I-I s after junction 8 are neglected. If we assume that Wro = 28,000 lbf, Eq. (3.32) gives
Thus
WTO =
I" = We/Wro = 0.6217
1348 + (652)(0.8585) + (657)(0.9164)
= 26,840 lbf 0.7152 - 0.6217
Referring to Fig. 3.2, reproduced here as Fig. 3.E4, it can be seen that this conventional metal version of the AAF would be a little heavier than existing lightweight fighters (e.g., F-16). In the time period of the expected introduction and serial manufacture of the AAF, the available nonmetals that can be used with confidence will reduce I" for aircraft of this size by approximately 3 to 5%. With this new assumption in hand, Wro is recalculated for the assumed Wro of 24,000 lbf and a conservative estimate of 3% empty weight savings,
1348 + (652)(0.8585) + (657)(0.9164)
Wro = = 26,840 lbf
0.7152 - 0.97 x 0.6306
For an assumed Wro of 24,150 lbf, the resulting Wro is 24,1501bf.
80 AIRCRAFT ENGINE DESIGN
The conclusion, then, is that the AAF can have a practical (and affordable) size, but only if reliable nonmetallics having competitive strength, durability, and repara- bility become available according to schedule. Because the choices of TsL/Wro = 1.25 and Wro/S = 64.0 lbf/ft a have already been made, the description of the AAF at this stage of design, in dimensional terms to three significant figures, using W F / W T o = 0.2787, is
WTo = 24,000 lbf TsL = 30,000 lbf S = 375 ft 2 We = 26601bf WE = 14,650 lbf Wv = 6,6901bf
This information can be used to generate a number of perspectives on the na- ture and shape of the corresponding aircraft. For example, because afterbuming military engines ordinarily have TsL in the range of 15,000-35,000 lbf, the deci- sion may remain open as to whether the AAF is a single- or twin-engine design.
Further, if it is assumed that the wing is a delta planform with an aspect ratio of 2, the total wing span will be approximately 27 ft. Because the takeoff pay- load fraction (Hip~ WTo) of fighter aircraft is usually in the range of 0.05-0.10, the AAF value of 0.111 shows that it is quite a workhorse. For further refer- ence, We / Wzo for the largest cargo and transport aircraft is in the range of only 0.15-0.25.
3.4.3 First Reprise
This is the first opportunity to evaluate some of the key assumptions that have been made along the way. Hopefully, they will be found to be sufficiently accurate that the present results are acceptable. Otherwise, it may be necessary to reiterate the process up to this point with new assumptions.
During the calculation of various other boundaries on the constraint diagram, it was necessary to assume values for the weight fraction (fl). In fact, /3 was taken to be 1.0 at takeoff, 0.56 at landing, and 0.78 everywhere else (an estimate of the aircraft maneuver weight). The maneuver weight (50% fuel, 2 AIM-9Ls, 1 ammunition) and its associated fl can be estimated now that the aircraft and
takeoff weight has been determined. The easiest method is to subtract the weight of 1 ammunition from the takeoff weight (Wro) to obtain _12 fuel, AMRAAMs, and
a maneuver weight of 19,730 lbf or/3 = 0.8221. Because this weight fraction is greater than the estimated 0.78 used in the constraint analysis of Chapter 2, the associated constraint lines will require a larger thrust loading for each wing loading. In addition the results of the mission analysis give a better estimate of the takeoff and landing weight fractions (see Table 3.E3). The new estimates for /3 resulting from the mission analysis and the estimate of the maneuver weight encourage a reconsideration of the legs that bound the solution space of Fig. 2.E3, as summarized next:
1) Takeoff, Mission Phase 1-2: Because the real/3 must be less than 1.0, the takeoff constraint boundary is conservative and is less critical than originally es- timated.
MISSION ANALYSIS 81 2) Combat Acceleration: The maneuver fl value of 0.8221 is greater than the initial estimate of 0.78, and therefore this constraint requires a larger thrust loading of 0.969 at a wing loading of 64 lbf/ft 2, more than originally estimated.
3) Supercruise, Mission Phase 6-7 and Maneuver Weight: Here the initial fl value for mission phase 6-7 of 0.8880 is higher than the initial estimate, and the required thrust loading increases to 0.934 at a wing loading of 64 lbf/ft 2, much less than the selected thrust loading of 1.25. The maneuver fl value requires a thrust loading of 0.930 at a wing loading of 64 lbf/ft 2, less than that of mission phase 6-7.
4) 5g turn at 0.9M/30 kft, Maneuver Weight: Here the maneuver fl value for turn is higher than the initial estimate, and the required thrust loading increases to 1.261 at a wing loading of 64 lbf/ft 2, slightly more than the selected thrust loading of 1.25. If the difference were larger, a reconsideration of the thrust loading would be appropriate, but this small change does not warrant a reselecfion. Chapter 6 will better define the required thrust loading.
5) 5g turn at 1.6M/30 kft, Maneuver Weight: Here the maneuver fi value for turn is higher than the initial estimate, and the required thrust loading increases to 0.931 at a wing loading of 64 lbf/ft 2.
6) Landing, Mission Phase 13-14: In the calculation of landing distance using a drag chute (Sec. 2.4, Mission Phase 13-14), it was assumed that the wing area (S) was 500 f12, whereas the presently derived value is 375 ft 2. Substituting the new value would have the effect of increasing the chute drag coefficient to 0.7131.
When combined with the new fl value of 0.6668, the
Wro/S
boundary is moved to 64 lbfffl 2.7) 1.8M/40 kft, Maneuver Weight: Here the maneuver fi value for turn is higher than the initial estimate, and the required thrust loading increases to 0.728 at a wing loading of 64 lbf/ft 2, much less than the selected thrust loading of 1.25.
A revised constraint, diagram is given in Fig. 3.E5. Note that the AAF design point is now on the landing constraint line, slightly below the 0.9M/30 kft 5g turn line, and above the other constraint lines. Further work in this textbook will focus on the engine and the design thrust loading
(TsL/Wro)
may change from its current design value of 1.25. However the wing loading(WTo/S)
is assumed to remain at 64 lbf/fl 2.3.4.4 Complete AAF Weight Fraction Computations
The RFP mission of the AAF is flown in this section to find the weight fraction of each mission segment and phase. These calculations are the source of the data given in Table 3.E3 and Fig. 3.E3. The same procedure is used in each weight fraction calculation. First, the applicable weight fraction equation from Sec. 3.2 is written. Next, data are given to determine the values of all quantities in the order they appear from right to left in the weight fraction equation. Finally, the weight fraction and the accumulated fl are calculated. Please note that for all mission legs we shall use
fli
for[~avg
and K2 = 0.Careful note should be taken of the fact that these hand calculations com- pletely duplicate the AEDsys Mission Analysis computations obtained by using the AAEAED file. Any minor differences are simply the result of the cumulative
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 ' /