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

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286 AIRCRAFT ENGINE DESIGN

the especially devastating situation where resonance occurs because the upstream

or downstream disturbance has an organized pattern (caused, for example, by

the pressure fields and/or wakes of support struts, airfoil rows, or fuel injectors)

whose apparent or "blade passing" frequency coincides with one of the lower

natural frequencies of some airfoil. This condition can only be endured for very

short periods of time. 15

Buffeting can lead to an enormous accumulation of stress cycles during the life

of the engine because the natural frequency of parts is of the order of 1-10 kHz.

Thus, a single hour of excitation adds approximately 10 million cycles, and only

100 h adds about one billion cycles. HCF has become the leading cause of failures

in fighter engines and is presently the subject of intense investigation.

4) Thermal differential stress and low cycle fatigue (LCF) are important. Tem-

perature gradients, particularly in cooled turbine airfoils, disks, and burner liners,

can give rise to surprisingly high stresses as the material counteracts uneven local

expansion and contraction. These can be amplified during transients as the engine

is moving from one power setting or turbine inlet temperature to another. These

transient thermal differential stresses are the primary cause of thermal fatigue or

LCF, which, for obvious reasons, can rapidly consume the life of hot parts in high

temperature fighter engines. Recognition of the importance of the LCF during

the 1970s changed the entire approach of the engine community to design and

acceptance or qualification testing.15

Thermal differential stress is one example of the larger classes of strain-induced

stresses. Note should be taken of the fact that many engine parts can be geomet-

rically constrained by their neighbors, and that their stress analysis must take this

into account.

5) There are many sources of local stress concentrations, which can much more

than double the elastic stress level or even cause plastic flow and permanent de-

formation to take place. These include holes, slots, inside and outside corners,

machine marks, and, the most feared of all, cracks. Crack initiation and growth or

propagation is a leading determinant of engine life, and the recent development

and application of practical fracture mechanics design tools and test procedures is

one of the outstanding accomplishments of the aircraft engine community. 16

6) Foreign object damage (FOD) and domestic object damage (DOD) must be

guarded against if the probability of occurrence is deemed sufficiently high and

the consequences sufficiently severe.

There is an insidious, often disastrous, interaction between FOD and HCE

Cracks initiated by FOD are then propagated by HCF with the result that life is

significantly reduced. For this reason, FOD is known as the "finger of death."

A modern remedy for this problem is to apply surface treatments that reduce

or eliminate these interactions by preventing the defects from growing (see

Appendix N).

7) The first stage fan blades must withstand a variety of bird strikes, which may be

considered a soft-body relative of FOD. This has prevented the use of lightweight,

nonmetallic materials for this application for more than three decades. Bird strike

testing is a staple of the engine qualification process and is one of the most exciting

parts because it is dramatic and the cost of failure is great.

8) During strenuous maneuvers and hard landings, other forces and moments,

both inertial and gyroscopic, are generated within the engine. These can damage

DESIGN: ROTATING TURBOMACHINERY 287

the tips of rotating blades, their outer air seals, and any rotating seals, especially

when clearances are small, because they are asymmetric.

9) Torsional stresses are inevitable when power is transferred by shaft torque

from turbines to compressors. Although the magnitudes of these stresses are not

large, they cannot be ignored.

10) There are a number of material composition or chemistry effects that also

consume engine life, the most notable being erosion, corrosion, and creep. These

are likely to be important in applications where the engine remains at high temper-

ature and pressure for extended periods of time, such as supersonic cruise vehicles.

Each of these technical areas must be tended by true experts if surprises are to

be avoided. One must respect the organizations that are able to weld these talents

together and produce durable engines. No textbook could hope to capture the total

engine life design process, and so we must settle for less. In what follows, these

many phenomena shall be accounted for by using an "allowable working stress,"

which is the amount remaining for the principal tensile stresses alone.

Determining this remaining amount is itself no easy matter and is usually accom-

plished in engineering practice through the use of company materials specifications

that have been proven through time and are thoroughly documented. When such

resources are not available, the best approach is to use open literature references,

the Aerospace Structural Metals Handbook 17 being a good example for general

engine design. A cursory review of Ref. 17 will quickly reveal both the amazing

amount of information already in hand for construction metals and the many fac-

tors that must be considered before the design is final. To provide you with more

specific examples and guidance, the data for several materials typically found

in the failure critical parts of gas turbine engines are synopsized from Ref. 17 in

Appendix M. Failure critical parts, as defined in Appendix N, are those whose fail-

ure would threaten the integrity of the engine and/or aircraft and always include

the large rotating parts (for example, fan and cooled turbine rotor airfoils, and

cooled turbine rotor disks) and highly pressurized parts (for example, the burner

outer case).

The creep rupture strength of a material is the maximum tensile stress it can

withstand without failing during a specified period of time at a given temperature.

As explained in Appendix M, a sensible approach for the designer is to limit stresses

to the appropriate creep rupture strength, or some fraction of that for safety. Thus,

for the purposes of exploration and the examples that follow, we will use typical

engine materials properties as summarized in Table 8.5 and Figs. 8.15 and 8.16

Table 8.5 Material types

Density

Material no. Type slug/ft 3 kg/m 3

1 Aluminum alloy 5.29 2726

2 Titanium alloy 9.08 4680

3 Wrought nickel alloy 16.0 8246

4 High-strength nickel alloy 17.0 8760

5 Single-crystal superalloy 17.0 8760

288 AIRCRAFT ENGINE DESIGN

lOO~

2oi

Fig. 8.15

1 I I I

0 400 800 1200

4 5

1600 2000

Temperature (°F)

Allowable stress vs temperature for typical engine materials.

1.0

0.8

o. 0.6

0.4

0.2

0.1 I

0 400

800 1200 1600 2000

Temperature (°F)

Fig. 8.16

Allowable strength-to-weight vs temperature for typical engine materials.

DESIGN: ROTATING TURBOMACHINERY 289

(Ref. 11). These were derived by taking 80% of the allowable 0.2% creep, 1000-h

tensile stress for aluminum alloy and 50% of the allowable 1% creep, 1000-h

tensile stress for the other alloys. Please note that, as the ensuing derivations will

demonstrate, strength-to-weight or specific strength is the most important property

for rotating parts. The typical properties also show that there is a temperature

beyond which every class of material rapidly loses capability.

It is therefore fortunate that the Larson-Miller parameter provides a convenient,

reliable procedure for extending the data for materials that have been only partially

characterized. According to this method, which is based on the assumption that

creep is thermally activated, the rupture life of a material at a given stress level

will vary with temperature in such a way that the Larson-Miller parameter

T(C +

log t) (8.58)

remains constant, where the material property C is a dimensionless constant. The

value of the material constant may be determined from data acquired at two test

conditions by applying Eq. (8.58) twice to obtain

C ---- (T2 log t2 - T1 log

h)/(Tl -

T2)

For example, for the Aluminum 2124 of Fig. M.2 (where rupture at 20 ksi occurs

at both 500°F/15 h and 400°F/950 h), C = 14.3. Also, for the Rene 80 of Fig. M. 13

(where rupture at 60 ksi occurs at both 1600°F/25 h and 1400°F/8000 h), C = 21.9.

You should try different combinations for these materials and for other materials

of Appendix M to increase your familiarity with and confidence for this procedure.

Representative values of C are also found in Table 8.6.

Equation (8.58) can also be used to demonstrate the surprising influence of tem-

perature on creep life. For example, a part having C = 20 that is designed to fail

after 2000 h at a temperature of 1400°F will last only 1130 h if the temperature

is raised merely 20°E This gives rise to the rule of thumb that the creep life of a

hot part is halved each time its temperature increases 20°E Fortunately, it is dou-

bled when the temperature decreases an equal amount. However, because creep

Table 8.6 Larson-Miller constants for selected

materials (Ref. 15)

Material Constant

Alloy C

Low carbon steel 18

Carbon moly steel 19

18-8

stainless steel 18

8-8 Mo stainless steel 17

2-1/4 Cr - 1 Mo steel 23

S-590 alloy 20

Haynes Stelite No. 34 20

Titanium D9 20

Cr-Mo-Ti-B steel 22

290 AIRCRAFT ENGINE DESIGN

Tip r"---~

Hub

Rim

w, -.~ [

rm = rh + ~--"- Wr----~

Disk /

1" t

Center line

Fig. 8.17

I ~ Was

• II

Shaft

Turbomachinery rotor nomenclature.

can cause failures in localized regions of high distress, the temperature distribu-

tion of critical parts must be accurately known for the structural analysis to be

reliable.

Creep will be a limiting factor for many likely future engine applications. These

include long-range supersonic transports, business jets, and attack aircraft, and

long-endurance unpiloted air vehicles. You may therefore find the Larson-Miller

parameter very handy in the years to come.

We now turn to the development of several important structural design tools.

These analyses will be based on the nomenclature and terminology found in

Figs. 8.17 and 8.18. It is most convenient to start at the outer radius and work

inward because the stress is known to be zero at the blade tips and because each

succeeding part must restrain or support all of the material beyond its own radial

location.

8.2.3.2 Rotor airfoil centrifugal stress trc.

Because each cross-sectional

area of the rotor airfoil must restrain the centrifugal force on all of the material

beyond its own radial location (see Figs. 8.17 and 8.18), the hub or base of the

airfoil must experience the greatest force. The total centrifugal force acting on

f~h t

Fc -= pW2 Ab r dr

(8.59)

DESIGN: ROTATING TURBOMACHINERY 291

/ Centrifugal

Rotor airfoil I

cross-sectio~l / [

area ~ l

Hub ~ --

Fig. 8.18 Rotor airfoil centrifugal stress nomenclature.

so that the principal tensile stress is

•c Ah -- pc°2 Ab r

-- ~ dr (8.60)

The airfoil cross-sectional area usually tapers down or diminishes with increasing

radius, which, according to Eq. (8.60), has the effect of reducing

at.

If the taper is

"linear," then

and Eq. (8.60) becomes

(8.62)

where A is the usual flowpath throughflow or annulus area 7r(r 2 - r~). Although

Eq. (8.62) can be directly integrated for any particular case, a useful and slightly

conservative result is obtained by taking r =

(rh + rt)/2,which

leads to the de-

sired relationship

( A,)

-- P°92A

1 + (8.63)

o-c 4~ Z

which can be employed directly in any situation to calculate the rotor airfoil cen-

trifugal stress.

At/Ah

is known as the taper ratio and is usually in the range of

0.8-1.0. For example, after choosing the values of rotor airfoil materials shown in

Table 8.7, Eq. (8.63) yields

= 17,900 and 14,900 psi for the compressor and

turbine, respectively.

292 AIRCRAFT ENGINE DESIGN

Table 8.7 Example values of rotor airfoil materials

Item Compressor Turbine

p, slug/ft 3 9.0 15.0

co, rad/s 1000 1000

A, ft 2 2.0 1.0

At/Ah

0.8 0.8

Moreover, since

AN 2

=- Ao)2(30/Tr) 2, then Eq. (8.63) can be rearranged into

AN 2

= 3600 ac (8.64)

yr(1 +

At/Ah) p

Thus, for a taper ratio

(At/Ah)

of 0.8, a strength-to-weight ratio

(ac/p)

3 ksi/(slug/fl 3) corresponds to an

AN 2

value of 4 x l01° in.2-rpm 2.

It is important for you to know that the accepted practice is to use Eq. (8.64) to

calculate the allowable

AN 2

using material properties such as those of Fig. 8.16

and to compare that with the value actually required by the engine rotating parts.

This quantity, referred to simply as "AN2, '' is thereby transformed into a surro-

gate for the allowable airfoil material specific strength. Put simply, if the required

AN 2

exceeds the allowable

AN 2,

a superior material must be found or the flowpath

design must be changed. In fact, designers often express their design capabilities

to each other in terms of

AN 2

and usually omit the units and the l0 l° factor in

conversations as a matter of convenience. For these reasons, the COMPR and

TURBN computer programs present

AN 2

for each stage so you may use it for

airfoil material selection. You may find Fig. 8.19 helpful for converting specific

strength into

AN 2.

It should be obvious by now that anything that reduces the amount of material

beyond the hub or base of the rotating airfoil will reduce the centrifugal stress

there or, conversely, increase the allowable

AN 2

of the airfoil. This explains the

great attraction of hollow fan blades, which, although extraordinarily expensive to

manufacture, are an essential ingredient of modem engines.

8.2.3.3 Rim web thickness

Wdr.

The rotating airfoils are inserted into

slots in an otherwise solid annulus of material known as a rim (see Fig. 8.17),

which maintains their circular motion. The airfoil hub tensile stress a, is treated

as though "smeared out" over the outer rim surface, so that

acNbAh

6blades -- --

(8.65)

2rr rh Wr

where Nb is the number of blades on the "wheel."

Making the conventional assumption of uniform stress within the rim (at), the

force diagram of Fig. 8.20 may be used to determine the dimension

War

necessary

to balance the blade and rim centrifugal forces. Please note that Wr and

hr are

simply sensible initial choices, where

approximates the axial chord of the

airfoil and hr is similar in magnitude to Wr. It is very important to realize that it

will always be possible to design a rim large enough to "carry" the airfoils. The real

question is whether the size of the rim is practical from the standpoint of the space

DESIGN: ROTATING TURBOMACHINERY 293

Fig. 8.19

~ 0.8

~ A,um!~al'°Y[

S .... ~it~a.loy I

/~~u~b.nea.oys

I I I I I

l 2 3 4 5

~/p [ksi/(slug/ft3)]

AN 2 as a function of specific strength

and taper

ratio.

/ /

Fig. 8.20 Rim segment radial equilibrium nomenclature.

294 AIRCRAFT ENGINE DESIGN

Table 8.8 Example values of disk materials

Item Compressor Turbine

~blades

/ar

0.10 0.20

hr/rr

0.05 0.10

p(O)rr)2/ar

6.0 4.0

required, weight, and manufacturing cost. Thus, there is no absolute solution, and

the final choice must be based on experience and a sense of proportion.

The radial force balance leads to the equation for the minimum Wdr

6blades rh Wr dO + ,oo)2hr Wr rr + dO = ar rr Wdr dO + 2(7rh r Wr

sin

which, for an infinitesimal dO, becomes

r- (rr)P(Wrr)2 (hr'~2 ] hr

Wdr = ~btades

1+ +-- 1+ --1 -- (8.66)

err g tTr 2rr,J rr

which can be employed directly in any situation to calculate the rotor airfoil cen-

trifugal stress. If ar is sufficiently large, Eq. (8.66) clearly shows that

Wdr

can be

zero or less, which means that the rim is "self-supporting." Nevertheless, a token

disk will still be required in order to transfer torque to the shaft and, of course, to

keep the rim and airfoils in their correct axial and radial positions. For example,

after choosing the values of disk materials shown in Table 8.8, then Eq. (8.66)

yields

Wdr/Wr

= 0.37 for the compressor and 0.56 for the turbine.

8.2.3.4 Disk of uniform stress.

The disk supports and positions the rim

while connecting it to the shaft (see Fig. 8.17). Its thickness begins with the value

Wdr

at the inside edge of the rim and generally grows as the radius decreases

because of the accumulating centrifugal force that must be resisted. Just as was

discovered for the rim, a disk that will perform the required job can always be

found, but the size, weight, and/or cost may be excessive. Thus, the final design

choice usually involves trial-and-error and judgment.

It is impossible to overemphasize the importance of ensuring the structural

integrity of disks, particularly the large ones found in cooled high-pressure turbines.

Because they are very difficult to inspect and because the massive fragments that fly

loose when they disintegrate cannot be contained, they must not be allowed to fail.

The most efficient way to use available disk materials is to design the disk for

constant radial and circumferential stress. Because the rim and disk are one contin-

uous piece of material, the design stress would be the same throughout (at

= O'd).

Applying locally radial equilibrium to the infinitesimal element of the disk of

Fig. 8.21 leads to the equation

p(wrr)2Wddr dO = ad [(r -- d-~ ) (wd - d--~) dO

DESIGN: ROTATING TURBOMACHINERY 295

d~ Cen6a ~.,.~~i 0t~fugal /~

\dO / r-dr/2 r r + dr[2

///

Fig. 8.21 Disk element radial equilibrium nomenclature.

which becomes in the limit

dWdwd

+pO~2ad

d(~)=0

This equation may be integrated, starting from r =

and

Wd = Wdr,

tO yield the

desired result:

p(a~rr) 2

[ (r)211

exp / 1 - (8.67)

Wdr= I

The main feature of Eq. (8.67) is that the disk thickness grows exponentially in

proportion to (wrr)2, which is the square of the maximum or rim velocity of the

disk.

What does Eq. (8.67) look like? Figure 8.22 shows the disk thickness distribution

for typical values of the disk shape factor

(DSF):

DSF = p(ogrr)2 /2ad

(8.68)

Judging by the looks of these thickness distributions, the maximum allowable

value of

DSF

is not much more than two, so that

[Wrr]max ~

44ffd/P

(8.69)

which, for typical compressor disk values

ofad

= 30,000 psi and p = 9.0 slug/ft 3,

is about 1400 ft/s, whereas for typical turbine disk values of

= 20,000 psi and

p = 16.0 slug/ft 3 it is about 850 ft/s. Why does

Wa/Wdr

grow more slowly as r

approaches zero?

It is important for you to know that [ogrr]m~x, known to designers as the

"allowable wheel speed," is a surrogate for the allowable disk material specific

strength. For this reason, the COMPR and TURBN computer programs present