Mattingly J.D., Heiser W.H., Pratt D.T. Aircraft Engine Design

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70 AIRCRAFT ENGINE DESIGN

When WF and We are expressed in terms of Wro, Eq. (3.1) in the form

Wro = We + WE + Wpe + Wpp (3.48a)

leads to the desired relationship for Wro. For a mission with n junctions and

payload delivery at junction j(j < n), WF is expressed in terms of Wro by the

sum of the aircraft weight decrements between takeoff and junctionj and between

junctionj (after the delivery of Wpe) and landing thusly

which reduces to

The historical correlations of Figs. 3.1 and 3.2 and Sec. 3.3.1 relate We to Wro

via

F - -- -- f(Wro) (3.48c)

Wro

where f(Wro) is given by Eqs. (3.50-3.53). Combining Eqs. (3.48a), (3.48b), and

(3.48c) and solving for Wro yields

Wee + WPeI-- I

j n (3.49)

Wro

1---r

- F

This equation carries the same information found in Eq. (3.1a) and allows

the straightforward calculation of Wro. Of special note are the facts that Wpp

matters more than Wpe because it is carried the whole way, and that the original

form of Eq. (3.1a) is recovered when Wpe is zero. This approach also produces

useful results when payload is expended at multiple junctions in the mission.

An initial estimate of Wro must be made during this part of the design process

in order to place values upon the Wpe/Wro of the payload deliveries and the

allowable I'. If the initial estimate of Wro is different from the value given by

Eq. (3.46), then the I-I s computed after payload deliveries and/or the I" taken from

Fig. 3.1 or 3.2 or Sec. 3.3.1 will be in error and an iterative solution of Eq. (3.49)

will be necessary. Because both the l-I s and F vary slowly with the estimate of

Wro, this iteration process converges rapidly.

3.3 Aircraft Weight and Fuel Consumption Data

A preliminary estimate of the ratio of aircraft empty weight (WE) to takeoff

weight (Wro) is required to solve for the aircraft takeoff weight as indicated by

Eqs. (3.48c) and (3.49). In addition, the thrust specific fuel consumption (TSFC)

MISSION ANALYSIS 71

must be estimated for each phase/segment of the mission to determine its weight

fraction [see Eqs. (3.12) and (3.17) or (3.32)]. The following material is provided

to help make these preliminary estimates.

3.3.1 Empty Aircraft Weight Fraction (r)

As indicated by Figs. 3.1 and 3.2, there is a relationship between aircraft empty

weight (WE), aircraft takeoff weight (Wro), and the type of aircraft. Several

authors 3'4 have performed correlations of

Wro,

which are the bases of

the following results:

Cargo aircraft:

F : 1.26W~ -°'°8 (3.50)

Passenger aircraft:

F = 1.02

WTg "06

(3.51)

Fighter aircraft:

F = 2.34W~ -°'13 (3.52)

Twin

turboprop aircraft:

F = 0.96W~ -°'°5 (3.53)

3.3.2 Installed Thrust Specific Fuel Consumption (TSFC)

The installed thrust specific fuel consumption of an aircraft engine varies with

Mach number, altitude, type of engine, and throttle conditions. Estimates of

TSFC

can be obtained from engine manufacturer's published data or from the output

of performance (off-design) cycle analysis (see Chapter 5) compensated by in-

stallation losses. Another source is measurements of the airframe/engine system

obtained during flight test. The following models of

TSFC

(expected for ad-

vanced engines in the 2010 era and beyond) are provided to help these preliminary

estimates. The units of

TSFC are

1/hour (h-l).

High

bypass ratio

turbofan engine

(M < 0.9):

TSFC

= (0.45 + 0.54M0)v/-O (3.54)

Low bypass

ratio mixed turbofan engine:

TSFC

= (0.9 + 0.30M0)V'-0 mil power (3.55a)

TSFC

= (1.6 + 0.27M0)C'0 max power (3.55b)

Turbojet engine:

TSFC

= (1.1 + 0.30M0)~/-O mil power (3.56a)

TSFC

= (1.5 + 0.23M0)x/-0 max power (3.56b)

Turboprop engine:

TSFC

= (0.18 + 0.8M0)~¢/-0 (3.57)

72 AIRCRAFT ENGINE DESIGN

Subsonic Cruise Climb

~...~10 Escape ~ ~7

11 F -- ~ ~ Combat

D cen ~ Supersonic

12 P es d Deliver Expendables \ Spenetratio n

I Oesoen~ ~ I ~

I And ,~ I.

. . ~ ~J~

~.,.~ ~ "~ 14

and Takeoff

Fig. 3.E1 Mission profile by phases.

3.4 Example Mission Analysis

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.El AAF mission

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--Fire AMRAAMs

I--Tum 1

J--Turn 2

K--Acceleration

L--Fire 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

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

Mcae/30k

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

Mloiter/lO kft

Mlo,er/10 kft, 20 min

Mloiter[lO

kft to 2000 ft PA, 100°F

One may question the wisdom of 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 of the order of 0.01, which is calculated from the estimated numbers--

often the exponents of 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 of fuel consumed (1 - l-I) for most flight legs is quite small, but

the cumulative effect is what matters.

During segment E, the AAF climbs and accelerates at military power to

BCM/

BCA

along the minimum time-to-climb path of Fig. 3.E2 via the following climb

schedule of Table 3.E2.

74 AIRCRAFT ENGINE DESIGN

50-

40-

Alt

30-

(kft) -

20-

10-

Ps (min = 0, max - 600, incr 50 fl/sec) Standard Day Atmosphere

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)

Minimum time-to-climb path on

chart. Fig. 3.E2

lilt

2000

W/S = 64

- T/W = 1.25

_Beta = 0.97

g's = 1.00

Tmx = 1.0

- TR -

1.07

-AB off

Engine #2

Aircraft #5

-Min time

The mid state point is chosen arbitrarily but judiciously 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) of Case 7, with CDR = 0, yields the

BCA

P 2~(Wro/S) ~/-~--,v

~BCA -- P--~td -- ~V ~DO/ ~1

}/rstd jr1 CRIT

where

Wro/S =

64 lbf/ft 2 and, from Fig. 2.Ela,

Mcmr=

0.9,

CDO

= 0.016,

and K1 = 0.18. From 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 = cL o~ \YZs~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 lie.

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

TAs

AZ e

W 1-u

V = (a/aste)astaM

Ah = hi - h~ ,xt -- ~- V -d \-~s~ )

,,Ze= +,',(

,,, =

\ 2goJ

2~(Wro/ PstdSM 2 ]---[ {

CL -- l.e = exp

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

= 0.8671

= 870.9 ft/s

(AV2/2go)

= 1549 ft

Ah = 40,000 ft

AZe = 41,550ft

/3 = 0.9773

Wro/S

= 64 lbf/ft 2

3 = 0.4051

M = 0.88

CL = 0.1346

K1 = 0.18, K2 = 0.0

Coo

= 0.016

CD = 0.01926

CD/CL

= 0.1431

0 = 0.8420

00 = 0.9724

30 = 0.6706

Otary = 0.4024

TsL/WT'o

= 1.25

u = 0.2780

TAs/W

= 57,550ft

(a/astd)

= 0.9176

V = 901.2ft/s

At = 2.068 min

As = 18.39 n miles

Ct/M

+ C2 = 1.323h -1

then

FIE = 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 min and 18.39 n miles, respectively.

At the end of segment

/3f = /3t FIE =

0.9589. Now a new mean value of

/3 for segment E can be estimated and the calculation of Fie iterated. When this

is done using the average value of (0.9773 + 0.9589)/2 = 0.9681, we obtain the

same value for Fie, justifying the use of/3i for

~3avg.

Also the

8BCA

calculation can

be repeated now using/3f, 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 AAF Weight Fraction Computations).

--'[ = 0.7152

WF/WTo

= 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

Mission Mission fl =

W/Wro

Fuel used, Payload,

phase segments End of leg

I41[/Wi

= 1-I lbf lbf

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

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

0.75

0.70

0.65

0.60

0.80

0.75

0.70

0.65

We /Wro

0.60

0.55

0.50

0.45

0.40

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~16~D

F-15 Eq.

(3.52)

¥~L ~ F.~ooo, , ~,,

:~-4~------._._~

F-101B Ft

F/A-18E F-111

I I I I I I I I

80 90 100

20 30 40 50 60 70

W (1,000 lbfs)

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.)

FI =

I-I"" I-I

= 0.8585

7 14 7 8 13 14

U = H"" 1-I =0.9164

8 14 8 9 13 14

1-I = FI Fl = 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.