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

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APPENDIX K: TURBOPROP ENGINE CYCLE ANALYSIS 603

Outputs

Overall performance:

Component behavior:

F, P, rho, S, Se, fo, rTe, 0TH, 00, Cc, Cprop

V9 / ao, Pt9 / P9, P9 / Po, T9 / To

Tf c , YgtH , 7"f tL ~ "f'c ~ "gtH , "~tL , 7j)~

rml, rm2, f, g8, g9

Equations

Preliminary computations:

FAIR (1, 0, To, ho, Pro, ~bo, cpo, Ro, Fo, ao)

Vo = Moao

hto =

ho + V~

2gc

FAIR (2, 0, Tto,

hto, Prto, C~to, CptO, Rto, )'to, ato)

rr = hto/ ho

rCr = P.o/ Pro

Tfd = ~dmax

tiptop = l O Morlpropmax

rlprop = lTpropmax

rlprop= (1

M°30"7 )

ht2 = hto

ert2 = Prto

Set initial values:

~tH ~ 9TtHR,

T( c = 7"gcR ,

~gml ~ "gmlg,

M4=l,

~] prop max

YgtL ~ TgtLR, ~gtH ~ 15tHR,

~c = 7JcR, fflo ~ l~loR

rm2 =

75m2R , f = f R

M4.5

= 1, M9 = Ms =

MsR

Mo<I

Mo _< 0.1

0.1 < M0 < 0.7

0.7 < Mo < 0.85

"CtL ~ ~tLR

High-pressure and low-pressure turbines:

FAIR(l, f,

Tt4 , ht4, Prt4, ill)t4, Cpt4, Rt4, Yt4, at4)

ht4.5 = ht4"Cml'CtHTm2

f4.5 =/(1 -/3 - el - e2)/(1 -/3)

FAIR (2, f4.5,

Tt4.5i, ht4.5, Prt4.5,

~bt4.5,

Cpt4.5, Rt4.5,

Yt4.5, at4.5)

ht5 = ht4.5"CtL

FAIR (2, f4.5, Ttsi, hts, Prts, qStS, Opt5, Rts, YtS, ats)

1 ht3

= ht2"cc

FAIR (2, 0, Tt3, ht3, Prt3, fbt3, Cpt3, Rt3, Yt3, at3)

FAIR (1, f, Tt4, ht4, Prt4, ~t4, Cpt4, Rt4, Yt4, at4)

f4.5 = f(1 -- fl -- el -- e2)/(1 -- t)

TURBC(Tt4, f, A4/A4.5, M4, M4.5, OtH, Tt4.5i, Tt3,

t ~,

el, e2, zrttt, ~tH, Tt4.5)

604 AIRCRAFT ENGINE DESIGN

TURB (Tt4.5 , f4.5, A4.5/A8, M4.5, M8, rhL, Tt5i, 2"f'tL, 7"~n, Z'tL, Tt5)

VX = ht4/ ho

Compressor and engine mass flow:

vc = 1 + Omn{(1 -- fl -- el -- e2)(1 + f) + el} rXrml(1 -- rtu)

ht3 = htzrc

ht3i = ht2{l + rlcH(rcH --

1)}

FAIR (2, 0, Tt3, ht3, Prt3, flPt3, Cpt3, Rt3, Yt3, at3)

FAIR

(2,

0, Zt3i, ht3i, Prt3i, ~)t3i, Cpt3i, Rt3i, )/t3i, at3i)

Jrc = Pr,3i / Pr,2

2 fiemp=f

FAIR(l,

f, Tt4,

ht4, Prt4, fl/)t4, Cpt4, Rt4, Yt4, at4)

ht4 - ht3

f--

heROb

- ht4

if If

ftemp[

> 0.000l goto 2

P,9

-- ~ 7rr Tfd 2"fc 71"b ~tH 7rdtL 2"fn

RGCOMPR (3,

Tts, M9 ....

f4.5,

TtT9, Pt9/ Po, MFP9)

M9error = [M9 - M9new[

if M9error > 0.0l then

Mg = Mg.ew

goto

end if

if M9> 1 thenM8=l elseM8=M9

RGCOMPR (1,

Tt4, M4, f, TtT, PtP, MFP4)

1 + fR PoTrrTrdTrc MFP4

l'l'lOnew : #lOR

1 +-----f

{PoTrrTfd7"(c} R MFPaR l Tt4

• = l~tOnew _~ #to

mOerr°r [ ~tOR

if rh0e

.... >

0.001 then rh0

= ghOnew

goto 1

Tt9

T~5

RGCOMPR (1, Tt9, M9, f4.5,

TtT, Pt9/P9, MFP9)

1,9/t"9

Po/P9 -- --

P,9 / Po

T9 = Tt9/TtT9

FAIR (1, f4.5, T9, h9, er9, ¢9, cp9, R9, )/9, a9)

V 9 = M9a 9

= f(1 -/3 - el

- E2)

CTOH

OmPH

APPENDIX K: TURBOPROP ENGINE CYCLE ANALYSIS 605

CrOL = (CroH)R thor hoR

r'no ho

thor hoR

CroH = (CroH)R--

rho ho

C~ = (Yo - 1)Mo [(1 + fo - ,B) V9 - Mo + (1 + fo - ~)

R9 T9/To (1 - Po/P9) I

x Ro V9/ao )/0

Cpr°p~pr°prig{ rImL(l-'}- f°--t~)~k~'CmlTtH~m2(1--TtL)--CToL}rimPL

CTOTAL = Cprop "JI- Cc

-- = CTOTALho

rho

P = rhoCToTALho

Sp----

CrorALho

F CrorALho

tho Vo

F = fnoCTorAtho Vo

S--

F/fno

/ [Cprop ~0~2 11 (V9~ 2

rlP = CTOTAL [

~]prop

-[-

(1 + fo -/~) \ao--/ - M°2

CTOTAL "4- CTOL "1- CTOH

ri TH

foheR/ ho

riO = ~rHOP

RGCOMPR (1, Tto, O, Mo, TtT, PtP, MFPo)

th o ~/C-T t o

Ao----

PtoMFPo

FAIR(l, f4.5, Tt4.5,

ht4.5, Prt4.5,

~t4.5,

Cpt4.5, Rt4.5,

)/t4.5, at4.5)

f4.5R

fRO --/~ -- el -- e2)/(1 --/~)

FAIR(l, fR, TtaR,

htaR, Prt4R, dPtnR, Cpt4R, Rt4R, )/taR, ataR)

ht4.5R ~ ht4R'CmlR'CtHR'gm2R

100/

ht4.5(1 --

"¢tL)

%NL

[ht4.5(1 - rtC)]R

~ horr(rc- 1)

%NH = 100 [horr(Zc - 1)]n

606 AIRCRAFT ENGINE DESIGN

If any of the control limits (rrc,

Tt3, Pt3,

etc.) are exceeded, then reduce

Tt4

and

go to 1. A Newtonian iteration scheme is used to rapidly converge on the value of

Tt4

that meets the most constraining limit.

Example Performance Results

Consider a turboprop engine designed for a Mach number of 0.8 at a standard day

altitude of 25 kft with a compressor pressure ratio of 30, a turbine enthalpy ratio of

0.6, and the other information given in the Example Design Point Results Section.

Engine controls limit the compressor pressure ratio to 30, the combustor exit

temperature (Tt4) to 3200°R, and the compressor exit temperature (Tt3) to 1600°R.

The performance of the engine is desired at other altitudes, Mach numbers, and

partial throttle. The variation of the compressor pressure ratio (Zrc) with changes in

flight Mach number and altitude is presented in Fig. K.6 for full throttle operation.

This is the same trend as observed before and corresponds to a 00 break of 0.933.

Figure K.7 shows the variation in full throttle uninstalled thrust with changes

in flight Mach number and altitude for an engine sized for a static sea level thrust

of 14,000 lbf. The variation of uninstalled thrust specific fuel consumption at full

throttle is shown in Fig. K.8. The sharp break at M0 = 0.1 in Figs. K.7 and K.8 is

due to the assumed

Oprop

of Fig. K.5. The partial throttle performance at an altitude

of 15 kft is presented in Fig. K.9 for selected flight Mach numbers. Note that S is

very low compared with other turbine engine cycles.

Fig. K.6

''I'''I ''I

Alt (kft)

0.2 0.4 0.6 0.8 1

Compressor pressure ratio variation at full throttle (standard day).

15000

10000

(lbf)

5000

Alt (kft)

0 0.2

0.4 0.6 0.8

Fig. K.7 Example turboprop's thrust at full throttle (standard day).

0.9

0.8

0.7

0.6

(1/11) 0.5

0.4

0.3

0.2

0.1

I I ' ' I

Alt

(]~ft) '

i i , I I I i i [ i i i i i , i

0 0.2 0.4 0.6 0.8

Fig. K.8 Example turboprop's thrust specific fuel consumption at full throttle

(standard day).

607

608 AIRCRAFT ENGINE DESIGN

1.0 ' ' ' ' I ' ' ' ' I I ' ' ' ' I ' ' ' '

0.9

0.8

..... M 0 = 0.8

0.7 ~

0.7

(l/h) ...-- 0.6

0.6

0.5

0.4 0.3

0.3 , L ~ ~ I ~ ~ , , I ~ i .... i ....

0 1000 2000 3000 4000 5000

F (]bf)

Fig. K.9 Partial throttle performance of example turboprop at 15 kft (standard day).

Appendix L

Propeller Design Tools

When turboprop cycles are being studied, it is important to have realistic design

tools for the propeller because most of the thrust is provided by the propeller and

because the behavior of propellers is far from simple. The emphasis here is on the

efficiency of conversion of mechanical or shaft power to thrust power because that

is the main function of the propeller.

The propeller accomplishes its purpose by spreading the mechanical energy

that would otherwise reside in the core jet exhaust flow over a much greater

flow of the oncoming air (Fig. L. 1). This greatly increases the propulsive efficiency

of the turboprop cycle and gives it the characteristics of a very high bypass ratio

turbofan engine. However, the absence of a diffuser to reduce the Mach number of

the approaching flow leads to supersonic relative speeds at the propeller blade tips

while the flight speed is still subsonic. This, in turn, leads to unacceptably high

propeller drag losses as the speed of sound is neared.

Propeller technology has been pursued by outstanding aerodynamicists for more

than a century, including W. J. M. Rankine, R. E. Froude, S. Drzewiecki, L. Prandtl,

A. Betz, H. Glauert, 1 S. Goldstein, and T. Theodorsen. 2'3 One of their legacies to

us is a number of extremely clever analyses with fascinating results. It is worth-

while understanding what they have accomplished, whether or not you care about

propellers.

Propeller Momentum or Actuator Disk Theory

This analysis is remarkable for the wealth of information brought forth by the

simple, but insightful, model shown in Fig. L.2.

The assumptions are as follows:

1) The flow is everywhere inviscid; in particular, the slipstream allows a discon-

tinuity between the freestream velocity and the velocity of the air that has passed

through the propeller disk.

2) The entire effect of the propeller is a uniform increase in the static pressure

across the "actuator disk" (which represents the smeared-out influence of the

propeller) of magnitude A

pp.

3) Because the propeller spreads the available mechanical power over a rel-

atively large quantity of air, the density of the air may be taken as constant

at P0.

4) Because the streamlines are parallel at sections far upstream and far down-

stream, the static pressure there equals P0. This is not true near the propeller disk

where there is considerable streamline curvature.

609

610 AIRCRAFT ENGINE DESIGN

propeller

lcuon gear

1~22

bleed air

31 3.1

44.1 4.4 4.5

8,9

Fig. L.1 Turboprop engine.

Freestream

~ Slipstream

Far upstream

(o)

"~- Ap

r <

Pp = Po

Pp = PO

v~_>Vo

P~ = Po

g=Po

Far downstream

Propeller

disk

(p)

(e)

Fig. L.2 Propeller momentum theory nomenclature.

APPENDIX L: PROPELLER DESIGN TOOLS 611

Conservation of Mass

Since

then

po VoAo

= po VpAp

VoAo

Ap--

Incompressible Bernoulli Equation

Since

and

then their difference is

po V~ po V~

Pp----Po+--

2 2

poVJ poV

P+ = Po+--

2 2

PO(- 2

P+ - P; "-- APp = ~ , V e - Vg)

Propeller Thrust

The thrust of the propeller can be obtained either from using Eq. (L.2)

ApAPp -- poAp (Vf - Vg)

or from

(L.1)

(L.2)

F = rhp(Ve - Vo) = PoVoAo(Ve - Vo)

(L.3)

Setting these equal yields

( )

Ap= I + Ve/Vo Ao <_Ao

(L.4)

which may be combined with Eq. (L. 1) to produce the very useful result:

Vp = (Ve +

V0)/2 (L.5)

In words, the flow velocity at the propeller actuator disk is the exact average of the

far upstream and far downstream values.

612 AIRCRAFT ENGINE DESIGN

Ideal Propulsive Efficiency 017,)

Using the definition

Propulsive power

r/~ = Rate of kinetic energy input

FVo

rl*p = poVoAo(V2_ V2)/2

and Eq. (L.3) gives

(L.6)

C~ -- poVZAp/2

This may be combined with Eqs. (L.3) and (L.4) to yield

(V~ 2

- 1 -- 4(1 - r/~)

c7 = \ Vo /

Ideal Thrust Coefficient (C~)

This quantity is a measure of the thrust per unit actuator disk area (or thrust

loading). As it grows larger, the required propeller diameter for a given thrust

grows smaller, in accordance with the definition

(L.8)

Ideal Power Coefficient (C~)

This quantity is a measure of the ability of the propeller to transfer power to the

airstream and is defined as

Rate of kinetic energy input

C*p = poV3Ap/2

po VoAo(V 2 - V2)/2

C*p = poV3ap/2

which becomes, using Eq. (L.4),

't }

C;=~ 1+~o °

[k, Vo]-1

C~. 4(1 -

0~)

(L.9)

-- < 1 (L.7)

r/~-- -- _

1 +

Ve/Vo Ao

This is a familiar result that confirms the earlier assertion that the more air the

propeller power is distributed amongst, the more thrust power is generated.