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

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152 AIRCRAFT ENGINE DESIGN

Given

ht2.5i,

the subroutine FAIR will give the reduced pressure Pr

t2.5i.

Thus the

low-pressure compressor pressure ratio is calculated using Eq. (4.9b) written as

er t2.5i

7'i'cL -- --

(5.6d)

Prt2

For a calorically perfect gas, Eqs. (5.6a) and (5.6b) become

yrf =

Of(rf -

1)}~

(5.6b-CPG)

and Eqs. (5.6c) and (5.6d) become

zrcL = {1 +

rIcL(72cL --

1)}~c =7~

(5.6d-CPG)

High-pressure compressor temperature ratio

(rcH). The high-pressure

spool power balance of Eq. (4.21a) can be rearranged to yield

/ r

1 + (1 -

rtH)rlmH

(1 -- fl -- el -- e2)(1

+ f) ~ Lrr'CcLrlmPH.] thoho

"gcH

1 - el(1 -

"CtH)rlmH

(5.7)

so that

where

Ptl6 Pt2~ f Tf f

Pt6 Pt27rcL712cHYlYb71JtHT(tL 2TcL71JcHT@JrtHTftL

Ptl6

_ ~ f et6

(5.9)

P16

712cLT~cHTgbTt'tHTftL P6

For a calorically perfect gas, h0 =

cpcTo

in Eq. (5.7).

High-pressure compressor pressure ratio (Zrc,).

From the definition of

high-pressure compressor efficiency, Eq. (4.9c),

ht3i

= h/2.5{1 +

OcH(rcH --

1)} (5.8a)

Given

ht3i,

the subroutine FAIR will give the reduced pressure

Pr t3i.

Thus the

low-pressure compressor pressure ratio is calculated using Eq. (4.9c) written as

er t3i

7rcH - (5.8b)

Pr t2.5

For a calorically perfect gas, Eqs. (5.8a) and (5.8b) become

7rcH = {1 +

tlcH(rcn --

1)}3 (5.8b-CPG)

Mach number at station

16 (M16). The definition of total pressure yields

MI6 once

Pt16/P16

is evaluated. Since Pt6 = P6, then

P,6 Pt16 Pt6

P16

et6 P6

ENGINE SELECTION: PERFORMANCE CYCLE ANALYSIS 153

For the flow to be in the proper direction,

Ptl6

> P16.

With

Ptl6/P16

and

Ttl6

known, then the compressible flow subroutine RGCOMP yields M16. For the flow

to be subsonic, M16 < 1.0.

For a calorically perfect gas, Eq. (5.9) becomes

Pt16

ff~f Pt6 ffrf (l_4_~]t-lM~X~ yt''7-1

P16

7rcL37JcHTfbTftHTftL P6 7~'cLY'gcHY'(bTftHY'ftL

\]2

(5.9a-CPG)

and the Mach number at station 16 is given by

M16 ---- ~

\-~-t6] - 1

(5.9b-CPG)

Mixer bypass

ratio (o~').

the mass flow parameter,

From the definition of the mixer bypass ratio and

at ~/16

Ptl6

A16

MFP16 /

Tt6

(4.8f)

-- I~17 -- Pt6 "~6 ~ VTtl6

y't'f A16

MFP16 / Tt6

13/I

(5.10)

T(cL~cHTrbTrtH~tL

MFP6 VTtl6

For a calorically perfect gas, Eq. (5.10) is unchanged. The mixer bypass ratio is

iterated, as shown in Fig. 5.3a, until successive values are within 0.001.

Engine bypass ratio

(or). From the definitions of the engine and mixerbypass

ratios,

O/ ..... O/t{(1 -- fl -- S1 -- ~'2)(1

f) +

El -~- 5"2}

rhc rn6 rhc

Ot = Oft{(1 -- fl -- 81 -- 82)(1

+ f) + el + e2}

Equation (5.11) is unchanged for a calorically perfect gas.

(4.8a)

(5.11)

Mixer enthalpy ratio

(rM). Equations (G.3) and (G.4) yield

"C M

1 + Ot'(Trrf/'CZ'rmlVtHVm2"rtL)

1 +or'

For a calorically perfect gas, Eq. (5.12) becomes

1 +

Olt('~rTf/'f)~Tml'gtHTm2"CtL)

(5.12)

rM = (5.12-CPG)

1 + ot'cpc/Cpt

154 AIRCRAFT ENGINE DESIGN

Mach number at station 6A

(M6A).

to the ideal mixer, we have

-1- Y6AM2A

__ R~

M6A V Y6

rearranged into

From the momentum equation applied

q- >'6M62) +

A16/A6(1 -t-

Y16M26)

M6(1 + or')

(4.29)

-- 1

+ Y6AM2A

m6a

V Y6A Constant

(5.13)

where

Rf-R~6T6 (1 +

y6M 2) -}- A16/A6(1 -1- F16M216)

Constant

"1 --V 76 M6(l+ot')

For a given value of the total temperature and fuel/air ratio at station 6A, Eq. (5.13)

is solved by functional iteration in combination with the isentropic temperature

ratio

(Tt6A/ T6A).

For a calorically perfect gas, the Mach number at station 6A (M6A) can be solved

for directly using the following system of equations:

ok(M6,

~6)~-

M6211 +

(Y6 --

1)M2/2].

(1+ y6M2) 2 '

~b(M16, Y16)~-"

M2611 +

(YI6 --

1)M26/2]

(1 + Y16M26) 2 (5.13a-CPG)

R6 + ot1R16

Cp6 A

R6A

= ; ~6A -- (5.13b-CPG)

1 + or'

Cp6A -- g6a

, 1

y6R6a

*---- (l+ot)/I - +~'./'~ -_/

/L~/O(M6, y6)

~ Y~-~-~6 ~ ~(~6, ~6)

Y6AR~ rM

(5.13c-CPG)

M6A

= (1 -- 2y6A) + ~i--- 2(Yaa + 1)~ (5.13d-CPG)

Mixer total pressure ratio

(TrM).

Equation (G.6) gives

T~A6

MFP(M6, Tt6,

f6)

7rMideal

---- (1 + t~')V

Tt6 A6A MFP(M6A, Tt6A,

f6A)

(5.14a)

ENGINE SELECTION: PERFORMANCE CYCLE ANALYSIS 155

The mixer total pressure ratio is the product of the frictional loss (7rM

max)

and the

mixing loss (rrM

ideal)

Y/'M = 7rM maxTrM

ideal

(5.14b)

For a calorically perfect gas, Eq. (1.3) is used to calculate the mass flow param-

eters in Eq. (5.14a), and Eq. (5.14b) is unchanged.

Mach number at station

8 (Ms). The Mach number at station 8 depends on

the pressure ratio

Pt9/Po

with the afterbumer off (dry). This ratio must be equal

to or greater than that corresponding to Mach one for the flow to be choked at

station 8. We write

[,,91

Po J dry = Zrr 2I'dYt'cL2TcHT(bYTtH~tLY'I'M 7~AB dry2Tn

(5.15)

The compressible flow subroutine RGCOMP is input this total to static pressure

ratio in combination with the fuel/air ratio (f6A) and the Mach number at station

9 (M9) is output for matched static pressures (P9 = P0) and afterburner off. If

/149 >_ 1, then M8 = 1 else M8 =/149.

For a calorically perfect gas, Eq. (5.15) is unchanged, and the Mach number at

station 9 (M9) is solved directly from Eq. (1.2) or

M9 ....

Y6A--1 LPo.ldry

If M9 >_ 1, then Ms = 1 else M8 = M9.

Mach number at station

6 (M6). M6 can be obtained from the mass flow

parameter

(MFP6)

once the latter is determined. Writing the ratio of mass flow

rates at station 8 to station 6 for nonafterbuming operation gives

th__88 -= Pt8dry A8dry

MFP8 T~t6 or'

th6 Pt6 A6 MFP6 V Tt8 = 1 +

Solving for the mass flow parameter at station 6 yields

A8dry MFP8 J Tt6

MFP6 = 7(MTgAB dry A6 1 Jr-

Tt6A

(5.16)

With

MFP6

known, the compressible flow subroutine RGCOMP yields/146 < 1.

The Mach number at station 6 is iterated, as shown in Fig. 5.3b, until successive

values are within 0.0005.

For a calorically perfect gas, Eq. (5.16) is unchanged, and the Mach number at

station 6 (M6) is solved for using Eq. (1.3).

Engine mass flow rate

(rho).

Conservation of mass and the definition of the

bypass ratio (a) yield

rh 4,

rh0

o/)/~/C

°t)(1 -- fl -- el --

e2)(1 -t- f)

156 AIRCRAFT ENGINE DESIGN

From Eq. (1.3)

~/4 t --

Pt4,A4'MFP4, Pt4A4,MFP4,

where, by assumption,

Pt4

= Pt4,

and

Zt4 = Zt4,.

Combining the preceding two

equations and denoting station 4' as 4 gives

(1 +

ol)PoTgrY'gdT~'cLT"gcH~Tfb

A 4

rho = (1 ~~1 ~ e2--~+f-)

vt-T-~t4 MFP4

(5.17)

For a calorically perfect gas, Eq. (5.17) is unchanged.

Mach number at station

9 (Mg). The Mach number at station 9 depends on

the pressure ratio

Pt9/P9.

We write

Pt9 Po

~9 ~-- "-~9YgrTgd2"[cL37:cHT(b2TtHTgtL71:MT"(ABJ'gn

(5.18)

The compressible flow subroutine RGCOMP is input this total to static pressure

ratio in combination with the fuel/air ratio at the afterburner exit (fT) and the Mach

number at station 9 (M9) is output.

For a calorically perfect gas, Eq. (5.18) is unchanged, and the Mach number at

station 9 (M9) is solved directly from Eq. (1.2), or

M9= T9j -1

5.2.6 Iterative Solution Scheme

Because there are 24 dependent variables and 24 equations, there are many

different ways that the off-design cycle analysis equations can be ordered to obtain

a solution. Our experience convinced us that the mixer bypass ratio (u'), the core

entrance Mach number to the mixer (M6), and the engine mass flow rate (#to) are

the preferred iteration variables. The iterative solution scheme shown in Figs. 5.2

and 5.3 has been custom tailored to this application so that it will converge on a

solution for a wide range of input values. Note that M6 is the second dependent

variable that is iterated and successful solution of the off-design performance

depends on its convergence. Referring to the functional relationships listed in

Sec. 5.2.5, note that values of or', M6, and rn0 are estimated to begin a solution

of the off-design equations. Making reasonable initial estimates can significantly

reduce the number of iterations required for a solution. Conversely, sufficiently

inaccurate initial estimates can prevent convergence altogether. Because M6 varies

only slightly over the off-design range of the engine and rh0 has a small influence

in the calculation of rcL in Eq. (5.5b), each has a secondary influence on the

iteration scheme. The reference point values of M6 and rh0 can therefore serve as

ENGINE SELECTION: PERFORMANCE CYCLE ANALYSIS 157

satisfactory initial estimates for these two terms, or

M6 = M6R (5.19a)

rno = #10g (5.19b)

However, the mixer bypass ratio

13/

c~' = (5.19c)

-- fl -- el -- e2)(1 -1- f) q-

gl -? S2

varies considerably over the expected flight envelope because it is proportional to

the engine bypass ratio oe (see Figs. 4.8 and 4.11). We have found that a reasonable

initial estimate of or' can be obtained using Eq. (5.19c) with o~ estimated by

To rr

ot ~ otR---- (5.19d)

TOR 72rR

where the altitude and Mach number effects on ot are accounted for by To and rr,

respectively. When using the AEDsys software for repeating calculations where

only one off-design independent variable is changed, the initial values of or', M6,

and rn0 for each subsequent calculation are taken from the solution already

converged.

Certain features of the programmed iteration methods used in the AEDsys per-

formance computations deserve highlighting, as follows:

1) If either the pressure ratio at the splitter plate (Pt 16//)16 ) or the Mach number

in the fan stream

(M16) is out

of limits, the core entrance Mach number to the

mixer (M6) is incremented by 0.01. If

M16 is too low, M6

is increased.

2) If the new value of the mixer bypass ratio (e() calculated by Eq. (5.10) is not

within 0.001 of the preceding value, then Newtonian iteration is used to converge

to a solution.

3) Functional iteration is used for both core entrance Mach number to the mixer

(M6) and the mass flow rate (rn0).

5.2.7

Variation in Engine

Speed

As is shown in Sec. 8.2.1, the change in total enthalpy across a fan or compressor

is proportional to the shaft rotational speed (N) squared. For the low-pressure

compressor, we can therefore write

ht2.5 - ht2 = KIN 2

which can be rewritten, using referencing, as

NL _ / ht2.5 - ht2 / ht2 "CcL -- 1

Nt~ V ht2"5R

-htzR

V ht2R rcU¢ -- 1

_~ht0 teL--1 ,.~00

rcL--1

h toR vcz~ -- 1 )90R ~cLtf-- -1

(5.20a)

158 AIRCRAFT ENGINE DESIGN

Likewise, for the high-pressure compressor, we have

[

NIt / ht3

- ht2.5 [ ht2.5

"rcH -- 1

V V

NoR ht3R --ht2.5R ht2.5R

rcl~R - 1

~ hto rcL rcH -- 1 ~ / O0 tel rcH -- 1

-- ht0~ rcLR rc/4R- 1 -- V00R rcLa rc,qR -- 1 (5.20b)

For a calorically perfect gas, Eqs. (5.20a) and (5.20b) are unchanged.

5.2.8 Software Implementation of Performance Calculations

The performance computer program embedded in the Engine Test portion of

the AEDsys program uses the system of equations listed in Appendix I, which

is based on the solution technique outlined in Secs. 5.2.4-5.2.6 and portrayed in

Figs. 5.3a and 5.3b. The performance program is intended to be used in conjunction

with the reference point program ONX discussed in Chapter 4.

The reference point of the engine whose performance behavior at off-design is to

be investigated is initially obtained using the computer program ONX. The inputs

and outputs of the reference point analysis are the source of the reference values

(subscript R) used in the performance computer program. The inputs listed next

for the performance analysis, therefore, include those required for the reference

point analysis as well as those that specify and are unique to the performance point

being analyzed (e.g., flight conditions and limits).

Sample printouts of the reference point (ONX) and off-design calculations

(Engine Test portion of AEDsys) for a typical performance analysis with and

without afterburning are given in printout samples A, B, and C. AEDsys will auto-

matically transfer all input values (reference values) from ONX for performance

computations other than the Performance Choices and Engine Control Limits. The

output consists of overall engine performance parameters, component behavior in-

formation, and a large selection of internal variables of interest. In fact, sufficient

output data are provided to allow the hand calculation of any desired quantity not

included in the printout.

The Engine Test portion of the AEDsys program is a powerful learning and

design tool. The AEDsys program will automatically scale the thrust and mass

flow of the input reference engine when required to match the thrust loading set

in the Mission Analysis portion. The program uses the thrust scale factor

(TSF)

represent this scaling.

TSF

is calculated by determining the input engine's thrust

at sea level, static conditions

(FsL),

and dividing this by the required sea level,

static thrust

(Tsc

req)

TSF -= FsL/TsL req.

Note that a thrust scale factor

(TSF)

0.9372 is printed on printout samples B and C, and the reference mass flow rate

has been reduced to 187.45 lbm/s from the 200 lbm/s (Design Point of sample

printout A).

As shown in Fig. 5.3b, the software checks to see if a control limit has been

exceeded. It uses an iterative procedure to reduce the throttle (Tt4) until the most

constraining limit is just met.

ENGINE SELECTION: PERFORMANCE CYCLE ANALYSIS 159

Inputs

Performance choices:

Flight parameters:

Throttle setting:

Exhaust nozzle setting:

Design constants:

rr:

r/:

Others:

Reference condition:

Flight parameters:

Component behavior:

Others:

Engine control limits:

Mo, To, Po

Zt4, Tt7

P0/e9

T(dmax, 7~b, 7t'Mmax, T(ABR, 7g n

Of, tlcL, TIcH, T]tH, 17tL, Ob, TIroL, 17mH,

FImPL, OmPH

A4, A4.5, A6, A16,

A6A, A8wo/AB

fl, el, e2, hpR, ProL, Pron

MOR, TOR,

POR

7~fR, 712cLR~ 7rcHR, YgtHR, T(tLR

75fR, 75cLR, TcHR, TtHR, "gtLR

M6R, M16R, M6AR, MSR

rmlR, rm2R, fR, fABR, M9R, M19R, OtR, O{ R

FR, /nOR, SR

7(cmax, Tt3max,

Pt3max, %Nc, %NH

Outputs

Overall performance:

Component behavior:

F,/no, S, fo, rip, TITH, 110,

V9/Olo,

Pt9 / Pg, eg / Po, Tg / To

T( f , 7"(cL , YfcH~ 7[tH~ 2"l'tL

"t'f ~ TcL , "6cH , "CtH , "CtL ,

rml, rm2,

fAB

M6, M16,

M6A, M8, M9

Sample Printout A

On-Design Calcs (ONX V5.00) Date: 10/01/2002 6:00:00 AM

File: C:\Program Files\AEDsys\AAF Base Line Engine.ref

Turbofan Engine with Afterbuming

using Modified Specific Heat (MSH) Model

Mach No

Alt (ft)

TO (R)

P0 (psia)

Density

(Slug/ft^3)

Cp c

Cp t

Gamma c

Gamma t

Input Data

= 1.451 Alpha

= 36000 Pi f/Pi cL

= 390.50 Pi d (max) =

= 3.306 Pi b --

= .0007102 Pin =

Efficiency

= 0.2400 Btu/lbm-R Burner =

-- 0.2950 Btu/lbm-R Mech Hi Pr =

= 1.4000 Mech Lo Pr =

= 1.3000 Fan/LP Comp =

Tt4 max = 3200.0 R

= -001.000

= 3.900/3.900

0.960

0.950

0.970

HP Comp =

0.999

0.995

0.890/0.890

(ef/ecL)

0.900 (ecH)

160 AIRCRAFT ENGINE DESIGN

h--fuel = 18400 Btu/lbm

CTO Low = 0.0000

CTO High = 0.0152

Cooling Air #1 = 5.000%

Cooling Air #2 = 5.000%

P0/P9 = 1.0000

** Afterburner **

Tt7 max = 3600.0 R

Cp AB = 0.2950 Btu/lbm-R

Gamma AB = 1.3000

*** Mixer ***

************************** RESULTS

liP Turbine = 0.890 (etH)

LP Turbine = 0.900 (etL)

Pwr Mech Elf L = 1.000

Pwr Mech Eft H = 1.000

Bleed Air = 1.000%

Pi AB = 0.950

Eta A/B = 0.990

Pi Mixer max = 0.970

**************************

Tau r = 1.421 a0 (ft/sec) = 968.8

Pi r = 3.421 V0 (ft/sec) = 1405.7

Pi d = 0.935 Mass Flow = 200.0 lbrn/sec

TauL = 10.073 Area Zero = 6.227 sqft

PTO Low = 0.00 KW Area Zero* = 5.440 sqft

PTO High = 300.61 KW

Ptl6/P0 = 12.481 Ttl6/T0 = 2.1997

Pt6/P0 = 12.428 Tt6/T0 = 5.5576

Alpha = 0.4487

Pi c = 20.000 Tau ml = 0.9673

Pi f = 3.9000 Tau m2 = 0.9731

Tau f = 1.5479 Tau M = 0.8404

Eta f = 0.8674 Pi M = 0.9637

Pi cL = 3.900 Tau cL = 1.5479

Eta cL = 0.8674

PicH = 5.1282 M6 = 0.4000

Tau cH = 1.6803 M16 = 0.3940

EtacH = 0.8751 M6A = 0.4188

PitH = 0.4231 A16/A6 = 0.2715

Tau tH = 0.8381 Gamma M = 1.3250

Eta tH = 0.8995 CP M = 0.2782

Pi tL = 0.4831 Eta tL = 0.9074

Tau tL = 0.8598

Without AB With AB

Pt9/P9 = 11.327 Pt9/P9 = 11.036

f = 0.03069 f = 0.03069

f AB = 0.03352

F/mdot = 62.493 lbf/(lbm/s) F/mdot = 110.829

lbf/(lbm/s)

S = 1.0862 (lbm/hr)/lbf S = 1.6938

(lbm/hr)/lbf

T9/T0 = 2.5755 T9/T0 = 5.2970

V9/V0 = 2.402 V9/V0 = 3.384

M9/M0 = 1.542 M9/M0 = 1.531

A9/A0 = 1.080 A9/A0 = 1.625

A9/A8 = 2.261 A9/A8 = 2.272

Thrust = 12499 lbf Thrust = 22166 lbf

Thermal Eff = 55.43% Thermal Eft = 45.25%

Propulsive Eft = 59.14% Propulsive Eft = 46.26%

ENGINE SELECTION: PERFORMANCE CYCLE ANALYSIS 161

Sample Printout B

Turbofan with AB--Dual Spool AEDsys (Ver. 3.00)

Engine File: C:\Program Files\AEDsys\AAF Base Line Engine.ref

Date: 10/01/2002 6:00:00 AM

Input Constants

Pidmax = 0.9600 Pi b = 0.9500 Eta b = 0.9990

cp c = 0.2400 cp t = 0.2950 Gam c = 1.4000

Pi AB = 0.9500 Eta AB = 0.9900 cp AB = 0.2950

EtacL = 0.8674 EtacH = 0.8751 EtatH = 0.8995

Eta mL = 0.9950 Eta mH = 0.9950 Eta PL = 1.0000

Eta f = 0.8674 PTO L = 0.0KW PTO H = 281.9KW

Bleed = 1.00% Cool 1 = 5.00% Cool 2 = 5.00%

Control Limits: Tt4 ----- 3200.0 Pi c = 20.00

** Thrust Scale Factor = 0.9372

Parameter Reference**

Mach Number @ 0 1.4510

Temperature @ 0 390.50

Pressure @ 0 3.3063

Altitude @ 0 36000

Total Temp @ 4 3200.00

Total Temp @ 7 3600.00

Pi r/Tau r 3.4211/1.4211

Pi d 0.9354

Pi f/Tau f 3.9000/1.5479

Pi cL/Tau cL 3.9000/1.5479

Pi cH/Tau cH 5.1282/1.6803

Tau ml 0.9673

Pi tH/Tau tH 0.4231/0.8381

Tau m2 0.9731

Pi tL/Tau tL 0.4831/0.8598

Control Limit

LP Spool RPM (% of Reference Pt) 100.00

HP Spool RPM (% of Reference Pt) 100.00

Mach Number @ 6 0.4000

Mach Number @ 16 0.3940

Mach Number @ 6A 0.4188

Gamma @ 6A 1.3250

cp @ 6A 0.2782

Pt 16/Pt6 1.0042

Pi M/Tau M 0.9637/0.8404

Alpha 0.449

Pt9/P9 11.0362

P0/P9 1.0000

Mach Number @ 9 2.2217

Mass Flow Rate @ 0 187.45

Corr Mass Flow @ 0 251.91

Flow Area @ 0 5.836

Flow Area* @ 0 5.099

Flow Area @ 9 9.481

MB--Fuel/Air Ratio (f) 0.03070

AB--Fuel/Air Ratio (fAB) 0.03353

Overall Fuel/Air Ratio (fo) 0.05216

Specific Thrust (F/m0) 110.83

Thrust Spec Fuel Consumption (S) 1.6941

Thrust (F) 20775

Fuel Flow Rate 35195

Propulsive Efficiency (%) 46.26

Thermal Efficiency (%) 45.25

Overall Efficiency (%) 20.93

Pin = 0.9700

Gam t = 1.3000

Gam AB = 1.3000

Eta tL = 0.9074

Eta PH = 1.0000

hPR = 18400

Test**

1.8000

390.00

2.7299

40000

3200.00

3600.00

5.7458/1.6480

0.9067

3.0054/1.4259

4.7208/1.6377

0.9673

0.4231/0.8381

0.9731

0.5023/0.8667

Tt4

94.88

100.00

0.3835

0.4559

0.4187

1.3282

0.2762

1.0492

0.9735/0.8268

0.530

13.3874

1.0000

2.3377

188.72

196.83

5.734

3.985

10.736

0.02975

0.03371

0.05080

104.69

1.7468

19757

34513

53.42

47.10

25.16