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

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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

WF/WTo =

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

R 1.6

T 1.4

A 1.2

N 1.0

0.8

T/W

SL TO

0.6

~I ' ' ' I ' ' ' I ' ' I

Solution

Landing

\ Supe -

\\\

~ruise AAF

r Takeoff

40 60

WING LOADING

Fig. 3.E5

5g~

ce at

80 100 120

W /S

(lbf/ft 2)

Revised AAF constraint diagram.

effects of rounding off. Thus we see that one of the motivating forces for carrying

four significant figures is to minimize the impact of these small imperfections on

the overall results.

Mission phase 1-2: Warm-up and takeoff.

This phase consists of segments

A (Warm-up), B (Takeoff Acceleration), and C (Takeoff Rotation).

A) Warm-up.

2000 ft PA, 100°E 60 s, mil power.

From Case 9, Eq. (3.42)

With

0 = 1.0796

ad~ = 0.5390

then

UA -~- 1- Cl~/-O~(-~o)mt

3=0.9298

M0=0

00=1.0796 30=0.9298

TsL/Wro=l.25

fl=l.0 At=60s Cl=0.9h -a

1--IA "~" 0.9895

fl = 0.9895

MISSION ANALYSIS 83

B) Takeoff acceleration.

2000 ft PA, 100°F,

kro

= 1.2,/Zro = 0.05,

CDR =

0.07,

C L

max =

2.0,

Macce I

= 0.1,

max power.

From Case 4, Eqs. (3.21) and (3.22)

1-I" = exp [ -(C' + C2M)~/-O \ ~- u,l ~ [

With

Vro

evaluated from Eq. (2.21), the takeoff estimate of~ro from Chapter 2,

Sec. 2.4.3, and a evaluated at

Maccel,

and with

/~ = 0.9895 a = ll60ft/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

= 0.9006

TsL/Wro

= 1.25 u = 0.2725

C1 + C2M

= 1.627h -1

then

C) Takeoff rotation.

From Case 10, Eq. (3.44)

I-Ic = 1--(C1

"}-C2M)~IIo~-~(~o)tR

With

=3S

6 = 0.9298

TSL/Wro

= 1.25

then

-I8 = 0.9958

/~ = 0.9853

Mro/2000

ft PA, 100°F,

= 3 s, max power.

/3 = 0.9853 0 = 1.0796

6o = 0.9515

Olwe t

= 0.9003

-Ic = 0.9984

The weight fraction for mission phases 1-2 is the product of the weight fractions

for segments A, B, and C.

I"I : I-IA FIB I"Ic :

0.9837

/3 = 0.9837

Macce I = O. 1

Vro

= 211.1 ft/s

q = 45.601bf/ft 2

~ro = 0.3612

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

= 812.0ft/s

A(vZ/2go)

= 9555 ft

Ah = 0.0ft

AZ e ~---

9555 ft

/3 = 0.9837

Wro/S

= 64 lbf/ft 2

8 = 0.9298

M = 0.4410

CL = 0.2351

Ka =0.18, K2=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

TAs/W

= ll,270ft

(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

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.

HE = Ha Hb Hc = 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 of the single gross in-

terval calculations for the illustrative example in Sec. 3.4.1. The amount of fuel

consumed for this phase has changed by only 3%. We should therefore not con-

sider breaking this phase into even finer intervals. Summarizing, then, we have the

following results for Mission Phase 2-3:

As23 =

Asp + Ase

= 23.04 n miles

At23 =

Ato

+ AtE = 2.906 min.

H = HD HE = 0.9742

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 of

TSFC

from Sec. 3.4.1 for all type B legs in these calculations.

From Case 7, Eq. (3.29), with

COR = K2 = 0

{ ~ (C1-'[- C2McRIT)AS34 I

]7.. = exp

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 2210 16,210 0.1601 0.1962 20,170 0.4915 4.067 1.462 0.9927

b 14,000 2.254 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 1550 41,550 57,490 2.356 20.27

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= 0.9736

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

fl = 0.9331

Mission phase 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

+ C2McAP)W/O~At}

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

/~ = 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

AZe, TAs/W, At, As, CI/M + C2,

Interval

Mi Mf Mavg

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-IF = I--Ia Ub Uc =

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 Ass = 7.01 n miles.

G) Supersonic penetration.

1.5M/30 kft, AsF + Asc =100 n miles, no after-

burning.

From Case 5, Eq. (3.23), with

CoR = 0

with

then

Asa = 92.99 n miles

CL = 0.05731

= 0.02889

C1/M

-t-

C2 :

0.9 h -1

K1 : 0.27

CD/CL

: 0.5040

-Ic = 0.9382

Segments F and G together yield

U = U F 1--IG =

0.9231

/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

AMRAAMs), 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

WpE 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

nfD

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

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

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 }dry

ot = 0.3020 | CI +C2M = 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 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.E7.

Then

VIK = rio I-Ib I-If =

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