MDoT Sample Exam Paper

8/10/2019 MDoT Sample Exam Paper

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School of Engineering

Department of Power and Propulsion

MSc. in Thermal Power

MECHANICAL DESIGN OFTURBOMACHINERY

InstructionsAll questions carry equal marks

Candidates should answer no more than THREE questionsAll working is to be shown

This examination paper contains a cover sheet plus 6 pages of questions.An Examination Data Sheet is provided.

Closed Book Duration: 3 HoursDuration: 3 hoursDate: January 2014

Time: 09:30 12:30Venue:


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Question 1Part A (10 Marks)

a. Draw the Mohrs Circle for the following loading conditions (2 Marks):

- Simple Tension- Pure Shear

b. What will be the maximum shear stress acting on a cylinder that containsgas under pressure so that the walls are subjected to a longitudinal andhoop stress as shown in the diagram below

Explain clearly any assumption made (4 Marks)

c. Why single crystal materials have better resistance to creep? (1 Marks)

d. Draw a general creep curve and explain the shape of the curve and thedifferent regions. Why does the rate of creed reduce at the primary region?(3 Marks)

Part B (10 Marks)

You are asked to determine the creep life of a turbine blade that is uncooled,

shrouded and manufactured from the nickel alloy, Nimonic 115. The machine ofwhich the turbine blade is a component, is used for a whole range of operating

t

pdh

2

t

pdL

4


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conditions. However, to simplify the analysis, these conditions have beengrouped under three main headings: A,Band C. Condition A is at the maximumoperating speed, while conditions B and C are respectively at 95% and 90% ofthe maximum operating speed. The sketch below is a simple, one-dimensional,representation of the stress and temperature variation along the blade at

condition A. You can assume that the temperature is directly proportional to thespeed of the turbomachine and that the stress arises solely from the rotationalinertia of the shaft (ie from centrifugal action).

TDS

3/4

A

Mid

1/4

0 100 200 300 400 500 900 1000 1100 1200

Stress Mpa Temp (K)

It is assessed that the machine will spend 10% of its life at condition A, 30% at

conditionB and the balance (60%)at conditionC.

Determine the creep life of the blade and calculate where along the blade will theminimum life occur. You can reasonably assume that this minimum life will occur ata position between 50% blade height and the blade tip.

Quite often, the industry will use Larson-Miller data in the form of the coefficients of apolynomial which fits the curve reasonably well. It is therefore required that you usethe following polynomial to find the required Larson-Miller parameters

LMP = a0(log)0 + a1(log)

1 + a2(log)2 + a3(log)

3

The coefficients a0,a1,a2and a3can be found by assuming the following values of theLMP at particular values of stress:

Stress (MPa) 100 300 400 500

LMP 28 25.1 24.2 23.3

What are the coefficients of the LMP?(3 Marks)

How long would you be prepared to allow the blade to remain in service (4)Marks)?


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Is this an acceptable life for (a) an industrial gas turbine (b) an aero gasturbine (1 Marks)?

How could the creep life of the blade be improved(2 Marks)?


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Question 2

Part A (10 Marks)

a. A rotating shaft is subject to pure shear. Under what type of loading youexpect the shaft to fail if it is made from

- A ductile material- A brittle material

Draw a diagram showing the plane of failure in the two cases(4 Marks)

b. Explain briefly the types of stress that will be applied on a shaft that isconnected to an axial turbine.(2 Marks)

c. Explain briefly what are the stresses on the root of a turbomachine blade(2Marks)

d. What is the ratio of stress in the case of a rotating bending fatigue test? Whatare the limitations of test data that are derived from a rotating bending test? (1Marks)

e. How can the surface finish of the material affect its fatigue characteristics? (1Mark)

Part B (10 Marks)

As a preliminary design exercise you are asked to carry out a simple cumulativefatigue analysis on a compressor rotor blade. The LCFcycle arises from centrifugal(rotary inertial) action and leads to a direct stress, the maximum value of whichexists in the blade root with a magnitude of220MPa. The HCF minor cycles arisefrom the passing frequency of the rotor blades relative to the stator blades. The gas-

bending stress that, of course, varies as the blade passes behind a stator, has amaximum value of120MPaalso in the blade root.

The blades are to be designed for30000 major(start-stop) cycles with an averageduration of4 hours.

If the compressor shaft rotates at 10000 rpm and there are 107 stator blades in therow just upstream of the rotor blades being investigated, justify the use of theendurance limit data for this exercise. Do we need to consider modifying these data?

The stress concentration effect of the radius between the blade and the root platformis contained in the factor Kt = 1.25 and the notch sensitivity for the same radius inthe blade material is given in the Sines and Waisman report as q = 0.78.


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There are, of course, several methods for determining the cumulative damagearising from amplitudes such as those given above. However, it is suggested thatyou consider the technique where the damage caused by one stress amplitude istransferred from one RM Diagram (Goodman Diagram) to another, where the

damage produced by the second stress amplitude is introduced, sometimes knownas theDouble Goodman Diagram Technique.

Do you think that this is an adequate preliminary design? (7 Marks)Does it satisfy monotonic and cyclic criteria? (2 Marks)If not, what would you suggest should be done? (1 Mark)

Material Properties. IMI 318A (Ti 6/4) Titanium.

Ultimate Tensile Strength = 950Mpa.0.2% Proof Strength = 700Mpa.

Density = 4440kg/m3

Elastic (Youngs) Modulus = 113Gpa.Poissons Ratio = 0.28

Endurance (Fatigue) Limit (107) cycles = 410MPa.Fracture Toughness = 121.4 Mpa.m0.5

Walker Constant m = 0.5


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Question 3

As a simple model, a turbomachine blade can be considered to be a rectangularcantilever 5 mm thick, 40 mm wide (chord) and 200 mm long. It is one of 67blades of a particular rotor stage and is manufactured by a material which has the

properties given below at this operating condition. The blade is set on a disc of500mm diameter with a stagger angle of 300. Assume that this stagger angle isconstant and that there is no blade pre-twist.

The turbomachine has a maximum speed of7000 rpm and the mass flow rate is 35kg/s.

The average fluid properties at the maximum speed are as follows:

Rotor inlet axial fluid velocity 300 m/sRotor inlet tangential fluid velocity 450 m/s

Rotor inlet static fluid pressure 150 kPaRotor exit axial fluid velocity 320 m/sRotor exit tangential fluid velocity -220 m/sRotor exit static fluid pressure 100 kPa

a. Is the blade part of the compressor or the turbine assembly?

b. Determine the centrifugal and gas bending stresses in the blade.

c. Combine the centrifugal and gas bending stresses to indicate which are thecritical points on the blade.

d. At what angle would it be necessary to lean the blade to counteract themaximum gas bending stress in the tangential direction?

e. Are the stress values acceptable for this particular blade material?

f. If you were asked to carry a fatigue analysis which stress you would consideras the mean stress on the blade and which as the alternating?

Note: In normal circumstances, the blade would be sub-divided into four or

more sections. In this question, treat the blade as a whole, that is, as a singlesection.

Material Properties:

Density 8030 kg/m3

Elastic (Youngs) Modulus 206 GPaUltimate Tensile Strength 1090 MPa0.2% Proof Strength 720 MPa


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Question 4

Part A (7 Marks)

a. Explain the concept of the free hoop radius (5 Marks)

b. Draw the stress distribution for a rotating constant thickness disc (no bladesattached) (1 Mark)

c. Draw the stress distribution for a rotating hollow constant thickness Disc withno blades attached (1Mark)

Part B (13 Marks)The shape of a gas turbine compressor disc is approximated by the two ringelements shown in the diagram below (where all dimensions are inmetres):

Blade

0.03

B

0.4

A

0.20.05

0.1

Engine Centre-Line

The disc is manufactured from titanium which has the following material properties:

Elastic Modulus E 109 GPaPoissons Ratio 0.3Density 4540 kg/m3

0.1% Proof Stress 0.1% 600 MPa

The hoop stress (h)at the bore of the disc is known to be 400 MPa. The centre ofgravity of the blade is known to be at a radius of0.45 mand there are 71bladeseach of mass0.15kgon the rim of the disc. Use the Discretised Ring Sum andDifference technique to estimate the rotational speed of the disc.


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There is no need to account for the thermal load in the disc (ie the operatingtemperature of all parts of the disc is constant).

Sketch the stress distribution in the disc(10 Marks).How would you expect this stress to be modified if the disk was subjected to a

temperature distributing with the highest temperature being at the rim. You dontneed to carry any calculations just comment on the diagram of the stress distribution.(3 Marks)


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Question 1

Solution

Part A.

a

Tension

Shear

b. Maximum shear will be equal to half the max principal minus the minimumprincipal that is equal to 0 because the thickness of the cylinder is assumed to beveru small: 0.5 (hoop stress 0)


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c. at low temperatures the boundaries of a crystal are stronger than the body of thecrystal. At high temperatures the boundaries lose their strength.

f. There is an elastic deformation, followed by the primary region where the

dislocations move freely but then their movement is restricted by otherdislocations, foreign atoms and the edges if the crystals. Then is thesecondary creep region with the minimum creep rate that is constant followedby the tertiary creep region and fracture.

Part B.

Use the given LMPs to set up 4 simultaneous equations:

a0+2a1+4a2+8a3= 28 (1)a0+2.477a1+6.136a2+15.2a3=25.1 (2)a0+2.602a1+6.771a2+17.618a3=24.2 (3)a0+2.699a1+7.284a2+19.66a3=23.3 (4)

Solution of these equations yields:

a0 = 169.625, a1= -176.393, a2= 74.313 and a3= -10.762

Hence: LMP = 169.625 176.393(log)1+ 74.313(log)2 10.762(log)3

The investigation is restricted to positions on the blade between 50% and 100% ofthe blade height.

Temperature is directly proportional to speed andStress is proportional to speed2

For condition A, the stresses and temperatures can be extrapolated from the graphsusing similar triangles. For example, at 60% blade height:

(400 A60)/(60 50) = 100/25 hence A60= 360 MPa

and B60 = 360 x (0.95)2= 324.9 MPa and C60= 360 x (0.9)

2 = 291.6 MPa

Also (TA60-950)/60 = (1150 950)/100 hence TA60= 1070K

and TB60= 1070 x 0.95 = 1016.5K and TC60= 1070 x 0.9 = 963K

Therefore the following table of results can be drawn:


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Ht% A B C

T P T P T P

506070

8090

100

400360320

260180100

105010701090

111011301150

24.2024.5624.92

25.4726.3428.00

361324.9288.8

234.7162.590.3

997.51016.51035.5

1054.51073.51092.5

24.5524.8725.20

25.7226.5928.39

324291.6259.2

210.6145.8

81

945963981

99910171035

24.8825.1825.48

25.9826.8628.85

The life to rupture time is calculated for each blade height using the LMP andMiners Law. Thus for the 50% blade height:

Condition A: 1050/1000(logtr+ 20) = 24.20, hence tr= 1116 hours

Condition B: 997.5/1000(logtr+20) = 24.55, hence tr= 40796.8 hours

Condition C: 945/1000(logtr+ 20) = 24.88, hence tr= 2137701 hours

Using Miners Law: 1/tr50= 0.1/1116 + 0.3/40796.8 + 0.6/2137701

Hence tr50= 10284 hours

Spreadsheet calculations yield the following results for all blade heights:

Ht% trA trB trC trTotal50

60708090

100

6575

1116

895.33726.06891.25

2054.4122288.81

804.69658.63

40797

29538.0921853.4124821.6158573.02

967587.09

25317.2619035.83

2137701

1396624.54945644.431009964.542569998.5

74494063.51

1142873.58793335.98

10284

817865758007.5

18507.9208131.3

7318.25940

Initially, it was found that the minimum life occurred at around 70% of blade height.Blade height positions at 65% and 75% heights were also checked and it was found

that the minimum life occurred at approximately 75% of the blade height. This lifewas 5940 hours which is a time to rupture.

Applying a factor of safety of 0.67, the allowable life would be about 3960 hours.This would be far too short for an industrial gas turbine where lives of the order of50000 to 100000 hours are not uncommon. The best of the civil aero gas turbinesare obtaining high pressure turbine lives of the order of 10000 to 20000 hours andtherefore our blade would also fall short of this expectation. However, for a militarytraining aircraft, the life would certainly be adequate and would be excellent for a fastjet engine where life expectancy is of the order of 1000 hours.


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There are various ways in which the creep life of the blade could be improved. Theeasiest and the cheapest way would be to employ a material with better creepproperties. However, you could also consider the following:

1. Introduce blade cooling to reduce the metal temperature. This would be a very

expensive option and would entail considerable re-design effort. Certainly, bladecooling should have been considered at the early design stages and not as aretrospective repair scheme.

2. Reduce blade stresses and temperatures. This option would certainly improvethe creep life but would reduce the overall performance of the machine and thereforeprobably increase the life cycle costs.

3. Change the operating regime of the engine so that it spends more time atconditions B and C and less at condition A which is the most life consuming. Thismay not be possible since the requirement for the machine is generally dictated by

the operating conditions.


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Question 2

Solution

Part A

a. The ductile material will fail under shear and the fracture will be parallel to thedirection of shear i.e on a plane that cuts at 90 degrees through the shaft. Thebrittle material will fail under direct stress and the fracture will be at 90degrees to the maximum direct stress therefore on a plane that cuts at 45degrees through the shaft.

b. There will be an axial stress applied on the shaft because of the pressuredifference across the turbine. This load will push the turbine in the direction ofthe flow, there will be a direct stress associated with the bending resultingfrom the weight of the turbine and will have its maximum value aon thesurface of the shaft at the position of the bearing that holds the shaft in place,

there will be torsion associated with the transmission of power that will resultin a shear stress.

c. A stress associated to the centrifugal load, a bending moment associated tothe pressure difference across the blade a bending moment associated to thechange in axial and tangential velocity of the gas.

d. R=-1. Needs to be corrected for mean stress, temperature, stressconcentration, type of load, corrosion, size, surface finish.

e. Failures originate at the outer surface and a smooth surface would meanreduction in irregularities/defects and stress concentrations

Part B

The combined cycle is as follows:

Stress Mpa B = 120MPa

cyclic = 60MPa

CF= 220MPa Mean= 280MPa

Time

Basic assumptions are that the flight cycle can be broken down into major cycleswith superimposed fluctuations (minor cycles); that the gas bending stress does

actually reduce to zero when the flow is obscured by a stator blade (unlikely but most


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severe, ie conservative) and probably most important that stresses are all elasticotherwise they cannot be added linearly.

For the endurance limit data:

Number of HCF cycles = 30000x4x60x10000x107 = 77.04 x 1010

cycles

This is obviously well above the endurance level of 107 cycles and justifies using theendurance limit. In fact it is so far above 107 that we must consider using a factor. Ifwe were dealing with a ferritic material, we could probably assume the S-N curvewas asymptotic to the 107 value, but this is not the case for titanium. In fact, somecompanies require their designers to use only half of the endurance limit value insuch situations.

For the cyclic strength at 30,000 cycles:

Log (0.9x950) = 2.932

Log30000

Log410 =2.6128

0 1 2 3 4 5 6 7 Log N

Log 30,000 = 4.4771

0798.037

6128.2932.2

4771.47

6128.2000,30

Log

Log +30,000= 2.8141

+30,000= 652 Mpa (3)

For the fatigue concentration factor at 30,000 cycles:

Ktat 107 cycles = 1.25


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Since 195.125.078.0111,78.01

1

)10( 7

xKqK

K

Ktf

t

f

Kf 1.195

Kf(30,000)1.0

0 1 2 3 4 5 6 7Log 30,000 = 4.4771 Log N

072.14

19.04771.11

37

1195.1

4771.4

1

)000,30(

)000,30(

xK

K

f

f

(2)

The major (LCF) R-M Diagram:

652 Mpa (30,000 cycles)

+cyc

1.072 x 280 Design Point2 (a)

0 140 1 950 Mpa

MPax

359652

9501.150140

650

950

1.150

1401

1

(2)

The combined Goodman Diagram:

410 (107 cycles)


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205 (0.5x107

value)B

B1.195x60 = 71.7Mpa A

0 359 Mpa 950

Plotted on graph paper, the safety factor for the reduced endurance limit (ie 410/2) is1.39 and for the full endurance limit is 1.82. Equivalent monotonic safety factor =

700/359 = 1.95 which is satisfactory. (2)

Conclusions

Obviously, the design point A lies within both the Goodman Diagram, for which thetechnique was originally devised, and for the reduced diagram which takes accountof the fact that the component suffers considerably more than 107 cycles. For theformer, from a plot on graph paper, I get a safety factor of 1.82, and for the lattercase the SF = 1.39. Since we have agreed that there should be a limiting safetyfactor of 1.5 for flying components, the design is a satisfactory one if we take the fullvalue of the endurance limit but must be considered a failure if apply a factor (say0.5) to the fatigue (endurance) limit, which we should. Hence, we must considersome alteration in the design, for example, a stronger material (perhaps best) orsome improvement in the design of the blade so as to reduce the stressconcentration factor. (2)


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SolutionQuestion 3The pressure falls and the blade is a turbine blade

CF load at RDS = mr2= 0.005 x 0.04 x 0.2 x 8030 x 0.35 x (7000/60 x2 x )2

= 60408.3 N

CF stress at RDS = 302 MPa

Bending moment from pressure gradient = Aannx dp/N x rmom

= 50 x 1000 x (0.452 0.252)/67 x 0.1 = 32.8 Nm

Bending moment from axial velocity component = mbladex dvaxial x rmom

= 35/67 x 20 x 0.1 = 1.045 Nm

Bending moment from tangential velocity component = mbladex dvtangx rmom

= 35/67 x 670 x 0.1 = 35 Nm

Bending moments acting at RDS:

Axial: 32.8 1.045 = 31.76 Nm (fore to aft)Tangential: 35 Nm (10)

Y XC

31.76 Nm D

A

X BMyy Y MXX

35 Nm

MYY= MAcos - MTsin = 31.76 x 0.8660 35 x 0.5 = 10.0 Nm

MXX= MAsin + MTcos = 31.76 x 0.5 + 35 x 0.8660 = 46.2 Nm

IYY=

483

106667.212

04.0005.0

mx

x


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IXX= 410

3

101667.412

005.004.0mx

x

Converting bending moments into stresses by means of bending theory:

The stresses due to MYY (10 Nm) will be:

MPax

xBYY 49.7

106667.2

02.0108

The stresses due to MXXwill be:

MPax

xBXX 2.277

101667.4

0025.02.4610

Considering the four corners of the blade at the RDS:

CF XX YY Total

A 302.0 277.2 7.49 586.7

B 302.0 -277.2 7.49 32.3

C 302.0 277.2 -7.49 571.7

D 302.0 -277.2 -7.49 17.3

Blade leaning:

CF CF x 0.1 sin = MT

MT or sin = 35/(60408.3 x 0.1) = 0.0058

= 0.330

0.1

Conclusions:

The safety factor for the blade = 720/587 = 1.23 which is too low. The blade istherefore a failed design. The CF stress is within reasonable limits but the gasbending stress takes the total beyond the required 65% of the proof stress.Solutions to the problem are to use an improved material or tolean the blade/

The major cycle should include the centrifugal stress plus half the gas bendingstress, the minor cycle should consider the equivalent stress of the major cycle andhalf the gas bending stress


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Question 4

Solution

Part A

a. The free hoop radius is the position on the disc at which a thin ring of equal

radius would exhibit the same strain as the disc material at the same

rotational speed. This is a useful position to make attachments to the disc

since there would be no conflicting strain between the disc and the

attachment if both were made from the same material

b.

c.


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Part BSolution

Using the formulae given for the discretised ring technique:

SiA= 400 MPa DiA= 400 MPa (1)

SOA= 400x106 1.3/2 x 45402(0.22-0.12) = 400x106 88.532 (2)

DOA= 400x106x0.25 0.7x4540/4x2x[0.14/0.22-0.22] = 100x106+29.7942

(2)Hence hOA= (SOA+ DOA)/2 = 250x10

6-29.3682

(1)rOA= (SOA DOA)/2 = 150x10

6-59.1622

(1)Also r= (150x10

6-59.1622)x[0.05/0.03-1] = 100x106-39.442

and h= 30x106-11.8322

(2)Hence: hiB= 280x10

6-41.22and riB= 250x106-98.6022

(2)Therefore: SiB= 530x10

6-139.8022 and DiB= 30x106+57.4022

(2)Using the discretised ring formulae again gives:

SOB= 530x106-139.8022-1.3/2x45402[0.42-0.22]=530x106-493.9222

(2)

DOB= 7.5x10

6

+14.3505

2

+0.7x4540/4

2

[0.15] = 7.5x10

6

+133.5255

2

(2)


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Hence: hOB= 268.75x106-180.198252

rOB= 261.25x106-313.723752

(2)Since there are 71 blades on the disc, then:

rOB = 71x0.15x0.45x2/(x0.8x0.03)= 63.5632

= 261.25x106-313.723752 (1)

Hence = 832.1 rads/sec = 132.4revs/sec = 7946 rpm (1)

However: hiB= 280x106-41.22and riB= 250x10

6-98.6022

Hence: hiB= 251.5 MPa riB= 181.7 MPa (1)

and: hOB= 268.75x106-180.198252 = 143.97 MPa

rOB = 71x0.15x0.45x2 = 44.01 MPa (1)

Stress distribution is therefore as follows:

400

251.5

181.7 143.97

44.01

Bore 0.2 Rim

(1)The temperature profile would introduce compressive stresses in the hoop directionat the rim

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MDoT Sample Exam Paper