Load Calculation & Selection of Bearing for Low Pressure Compressor (LPC) for Al -31FP Engine

Load Calculation & Selection of Bearing for Low Pressure Compressor (LPC) for Al -31FP Engine

Load Calculation & Selection of Bearing for Low Pressure Compressor (LPC) For Al -31FP Engine

Narayan Parida1*, Siva Sankara Raju R 2, K. C. Ratp

1* (M.Tech, MSD student, Department of Mechanical Engineering, GIET, Odisha, INDIA )

1. (Department of Mechanical Engineering, GIET, Odisha, INDIA )

2. (Department of Mechanical Engineering, GIET, Odisha, INDIA)

ABSTRACT

This paper presents based on the inputs obtained from a working engine AL-31FP. The AL-31FP Engine is used to power the Su-30 MKI aircraft. It is a twin spool afterburning turbofan engine with mixing main flow and by-pass flow downstream turbine, with thrust vectoring nozzle. Its compressor is an axial flow compressor with a pressure ratio of 23 and is made by integrating the low pressure compressor and the high pressure compressor. The low pressure compressor has 4 stages with one stage of inlet guide vanes which has a pressure ratio of 3.5. The high pressure compressor has 9 stages with 3 stages of variable stator vanes which has a pressure ratio of 6.6. Its engine consists of an annular combustion chamber with 3 igniters and 28 number of duplex fuel nozzles.

There are six bearings used in AL-31 FP in main transmission line. For low-pressure compressor two bearings are used i.e. front support roller bearing and rear support ball bearing. There are various forces acting on bearings such as radial forces, axial forces. Therefore estimation of these forces is necessary. From these forces equivalent bearing load can be calculated. After taking suitable bearing life and load factor dynamic load carrying capacity can be found out. Finally after knowing dynamic load carrying capacity and bore dimensions suitable bearing can be selected from the catalogue.

Key words- AL-31FP, 31-Series, FP-Thrust vector, TBO-Time before overhaul, dynamic loads.

1 INTRODUCTION

Al-31 FP( Arif & Lulca 31series Fersa Thrust Vector ) engine is afterburning turbofan engine with mixing of bypass and main ducts flows downstream the turbine with thrust vectoring nozzle.

Compressor is an axial, 2 spools and 13 stage compressor. The compressor consists of-

1. 4 stage low-pressure compressor (LPC) with variable inlet guide vane(IGV)

2. 9 stage high pressure compressor (HPC) with 3 variable stator vanes (Inlet and first 2 stages guide vanes)

3. Intermediate casing

Combustion chamber is annular Engine turbine is an axial, 2 stage turbine. It consists of

1. Single stage high pressure turbine (HPT) with cooled disc

2. Single stage low pressure turbine (LPT) with air cooled disc

3. Turbine support

Afterburner with flows mixing at the afterburner transition section inlet.Thrust vectoring nozzle

• It consists of tilting unit and jet nozzles.

Fig 1-High purpolsion with low thrust fig 2-low purpolsion with high thrust

Fig 3: Turbofan Engines

A turbofan engine may be considered a cross between turbojet engines& turboprop engines. In this engine air from the fan section of the forward blades is carried outside to the rear of the engines through ducting. The effect of turbofan engine design is to greatly increase the power-weight ratio of the engine& to improve the thrust specific fuel consumption. Turbofan engine may be high-by pass or low bypass engines. The ratio of the amount of air that by passes the core of the engine to the amount of air that passes through the core is called the by pass ratio. Many different types of turbofan engines are in use & can be found on air craft from small business jets to large transport type aircraft

The rotor blades increase the air velocity. When air velocity increases, the ram pressure of air passing through a rotor stage also increases. This increase in velocity and pressure is somewhat but not entirely nullified by diffusion. When air is forced past the thick sections of the rotor blades static pressure also increases. The larger area at the rear of the blades (due to its airfoil shape) acts as a diffuser

Spang and Brown[1999][1] were reviewed the basics of controlling an engine while satisfying numerous constraints. In the published paper by Maggiore[2003][2] the concept so developed were applied to estimator design for jet engine thrust, stall margins, and an unmeasurable state. Boyce [2006][3] focuses on bearings and seals. The bearings in gas and steam turbines provide support and positioning for the rotating components. Montazeri-Gh et al.[2012][4] presented an actuator-based hardware-in-the-loop (HIL) simulation for testing of a jet engine fuel control unit (FCU). Mlynarek and Van Orden [2012][5] were demonstrated that the disturbance of asbestos-containing gaskets. H.K.D.H. Bhadeshia [2012][6] was given a report on steels used in the manufacture of some of the most technologically important engineering components, the rolling bearings. The jet engine maintenance process is complex, expensive and time-consuming. It often requires engine disassembly or baroscopic examinations. In order to accelerate the process and reduce the down time of an engine Adamczuk and Seume [2013][7] . Progress in the design and structural analysis of commercial jet engine fan blades is reviewed and presented. The article published by Amoo [2013][8] was motivated by the key role fan blades play in the performance of advanced gas turbine jet engines.

1. COMPRESSOR DETAIL

   The compressor is axial, 2-spool, 13 stages is made of titanium & steel alloys.It consists of 4 stages LPC with adjustable flaps of one inlet guide vane assembly and 9 stages of HP stator with 3 adjustable guide vane units (inlet and first two stages) and the intermediate casing.

   Total Number of blades:2443 In which Rotor blades: 981

   Stator blades: 1462

   Fig 4: AL31FP ENGINES

   The Compressor comprises:

   • 4 Stage Low Pressure Compressor (FAN) with Inlet Guide Vanes

   • 9 Stage High Pressure Compressor (GAS GENERATOR) with Three Swiveling Guide Vanes Intermediate Casing.

     Assumptions :

     Axial velocity is constant in LP compressor

   • Polytropic efficiency of compressor = 0.87

   • Polytropic efficiency of LPC = 0.89

   • Polytropic efficiency of 1st 2nd 3rd & 4th stage = 0.9

   • Degree of reaction of 2nd stage = 0.65

   • Degree of reaction of 3rd & 4th stage = 0.50

   • Pressure of air at the front side of LPC disc is as same as that of after 4th stage of LPC

   • Pressure of air at the rear side of LPC disc is as same as that of after 4th stage of LPC

   • Combustion is constant pressure process

   • expansion ratio = 7

   • polytropic efficiency of LPT stage = 0.9

   • Area before LPT disc is pressure vessel having uniform thickness

   • Average life of ball & roller bearings = 3000 hrs

   • Dynamic load carrying capacity of existing roller bearing C= 75 k N

2.2 COMPRESSOR CHARACTERISTIC

Fig: compressor p-v diagram

Moving from A-B, Pressure increases & Density. But the mass flow decreases slightly. It results in increase in axial velocity & increase in incidence, and stalling will happen in last stage. Speed reduced from A-C, Mass flow falls off more rapidly than the speed, and the effect is to decrease the axial velocity at inlet and cause theincidence on first stage blade to increase.

1. FORCES ACTING ON LP COMPRESSOR FRONT & REAR SUPPORT BEARINGS

   There are various forces acting on LPC bearings i.e. front support roller bearing & rear support ball bearing. These forces can be divided into 2 types radial forces and axial forces.

   1. Radial forces

      1. Due to dynamic unbalance state of rotor

      2. Due to gyroscopic effect

      3. Due to weight of rotor

   2. Axial forces

      1. Axial force due to action of gases on LPC rotor

      2. Axial force due to action of gases on LPT rotor

2. FORCE ACTING DUE TO UNBALANCE STATE OF ROTOR

   Centrifugal force = m r 2

   Centrifugal force on 1st stage =(8.86 × 103 × 102 × 1151.922) = 123.82 N

   Centrifugal force on 4th stage = 5.98× 103 × 102 × 1151.922 =80 N

3. FORCE ACTING DUE TO GYROSCOPIC EFFECT

   Inertial forces and moment appear in the engine due to change in value or direction of speed of flight of the aircraft on which the engine is mounted. These forces are transmitted to the body through the bearing in the form of radial forces. The direction of these forces changes in time with change of angle of rotation of the rotor. This causes appearance of gyroscopic moment of the rotor and inertial overloads.

   The gyroscopic moment is given by M= J sin

   Max. Velocity of SU-30 = 2290 km/hr = 636.11 m/s & Max k = 9

   = 9×9.8/636.11

   = 0.1387 rad/s

   Mass moment of inertia of rotor = 9 kg-2 N = 11000 rpm

   = 1151.92 rad/s

   Taking = 75

   The gyroscopic moment= J sin (75)= Max. Velocity of SU-30 = 2290 km/hr

   = 636.11 m/s

   & Max k = 9

   = 9×9.8/636.11

   = 0.1387 rad/s

   Mass moment of inertia of rotor = 9 kg-2 N = 11000 rpm

   = 1151.92 rad/s

   Taking = 75

   The gyroscopic moment= J sin (75) =1388.9 N-m

   Force acting due to gyroscopic effect = 1437.94/0.480=2.9 kN

4. STATIC FORCE ACTING DUE TO WEIGHT OF ROTOR

   In static condition force due to weight of rotor is acting on bearings Considering mass of discs and rotor blades

   Mass of 1st stage disc =12.743 kg = 124.88 N

   Mass of 1st stage blades = 0.28 kg (37) = 10.36 N

   Mass of 1st stage disc & blades = (124.88+10.36) = 135.24 N Mass of 2nd stage disc & blades = (122.5+6.975) = 129.475 N

   Mass of 3rd stage disc & blades = (187.18+5.415) = 192.595 N

   Mass of 4th stage disc & blades = (91.14+3.913) = 95.053 N

   & Mass of shaft = 1.1 kg For calculating reactions at bearings

   Total radial load acting on roller bearing = 3518.2 N = 3.52 kN

   Total radial load acting on ball bearing = 3477 N = 3.47 kN

5. FORCES DUE TO ACTION OF GASES

   Forces produced from static pressure of a gas are determined by the production of pressure of gas on the corresponding areas. The action of static gas pressure causes appearance of stresses of tension or compression, bending and torsion and axial forces. This may be in that case when various static pressures act on a part from both sides when there is area difference.

   = 1 2 +3

   Axial force Q1 is the resultant from forces appearing on each stage

   Q1=

   Q1=

   4

   =1

   = (2 2 – 1 1 ) + C

   Assuming axial velocity is constant in compressor

   mC=0

   Pressure after downstream of compressor = 33.5±0.5 = 34 bar

   P02 = 34 bar

   Taking compression ratio = 23

   r = P02 = 23

   P01

   = 1.478 bar

   Now, for ramming effect

   Assuming it as isentropic process

   P0a P01

   T0a /(1)

   =

   T01

   = 324.246 K (stagnation temperature before 1st stage of LPC) Now, for calculating

   P01

   P02

   T01 c /(1)

   =

   T02

   Taking polytropic efficiency of compressor = 0.87 & assuming it as an isentropic process

   = 908.527 K (stagnation temperature after (3th stage of LPC)

   Assuming

   polytropic efficiency of LPC = 0.89

   5.173 (T )

   5.173 (T )

   0.891.44/(1.441)

   = 02 LPC

   1.478 324.25

   () = 484.775 K (stagnation temperature after 4th stage of LPC) Stagnation temperature after 1st stage = 364.68 K

   Temperature

   Temperature

   Stagnation temperature after 2nd stage = 404.81 K Stagnation temperature after 3rd stage = 444.94 K Stagnation temperature after 4th stage = 484.78 K

   600

   500

   400

   300

   200

   100

   0

   Stagnation

   temperature

   600

   500

   400

   300

   200

   100

   0

   Stagnation

   temperature

   0 2 4 6

   NO.of Stages

   0 2 4 6

   NO.of Stages

   Velocity triangle for LPC For 1st stage

   At max rating IGVs are fully opened

   = 0

   =

   U

   Fig:Velocity triangle for 1st stage LPC

   U(Cw ) = Cp (T0)stage (= work done factor) for 1st stage = 0.98

   C

   = 1004 ×40.13 = 112m/s

   0.98× 365.7

   P03

   P01

   T03 stage /(1)

   =

   T01

   Assuming polytropic efficiency of 1st stage = 0.9

   P03

   1.478

   0.91.4/(1.41)

   364.68

   364.68

   =

   324.24

   = 2.062 bar (Stagnation pressure after 1st stage)

   P2

   P01

   = T2

   T01

   /(1)

   = 1.764 bar (Static pressure after 1st stage rotor)

   P1 T1 /(1)

   pressure before 1st stage rotor)

   For 2nd stage

   =

   P2 T2

   = 1.213 bar (Static

   By using ,

   (T0) = 40.13 K & = 0.93

   (T ) =

   U C ( tan tan )

   0 C

   a 1 2

   tan 1 tan2 = 0.6345 ..(a)

   using Degree of reaction (DOR) = Ca

   2

   ( tan 1

   + tan2 )

   1 = 57.84 & 2 =43.71

   U

   Using = tan + tan

   Ca 2

   = 45.08

   & = tan + tan

   Ca 1 1

   1 = 20.22

   (T1)2nd stage = (T01 )2nd stage

   C12

   2C

   = 344.96 K (Static temperature before 2nd

   stage rotor)

   (T2)2nd stage = (T02 )2nd stage

   C22

   2C

   = 370.06 K (Static temperature after 2nd

   stage rotor)

   P03

   2.062

   0.91.4/(1.41)

   404.81

   404.81

   =

   364.68

   P03 = 2.865 bar (Stagnation pressure after 2nd stage)

   P2

   P01

   = T2

   T01

   /(1)

   = 2.17 bar(Static pressure after 2nd stage rotor)

   P1 T1

   P1 T1

   /(1)

   = =1.697 bar.. (Static pressure before 2nd stage rotor)

   P2 T2

   For 3rd stage

   By using ,

   (T0) = 40.13 K & = 0.88

   (T ) =

   U C ( tan tan )

   0 C

   a 1 2

   tan 1 tan2 = 0.6705 ..(a) 1 = 52.74 & 2 = 32.8

   C12

   (T1)3rd stage = (T01 )3rd stage

   2C

   2

   2

   = 404.81 221.98

   21004

   (T1 )3rd stage = 380.27 K (Static temperature beforer 3rd stage rotor)

   (T2)3rd stage = (T02 )3ed stage

   C22

   2C

   = 397.65 K (Static temperature after 3rd

   stage rotor)

   /(1)

   /(1)

   P03 = T03 stage = 3.858 bar Stagnation pressure after 3rd stage)

   P01

   P2

   P01

   T01

   = T2

   T01

   /(1)

   = 2.692 bar (Static pressure after 3rd stage rotor)

   P1 380.27 1.4/(1.41)

   =

   2.692 397.65

   P1= 2.302 bar (Static pressure before 3rd stage rotor)

   For 4th stage

   (T0) = 40.13 K & = 0.83

   (T1 )4th stage = 420.82 K (Static temperature before 4th stage rotor)

   (T2 )4th stage = 436.516 K (Static temperature after 4th stage rotor)

   P03 = 5.056 bar (Stagnation pressure after 4th stage)

   p>P2= 3.609 bar (Static pressure after 4th stage rotor)

   P1= 3.175 bar (Static pressure before 4th stage rotor) Axial force on compressor is as follows = 1 2 +3

   Qst = (P2i A2i – P1i A1i) + C

   Qst 1 = (P21 A21 – P11 A11) + C =19.702 kN Qst 2 = (P22 A22 – P12 A12) + C= 12.573 kN Qst 3 = (P23 A23 – P13 A13) + C== 13.644 kN Qst 4 = (P24 A24 – P14 A14) + C= 16.471 kN

   Q1 = Qst 1 + Qst 2 + Qst 3 + Qst 4 = 19.702 + 12.573 + 13.644 + 16.471=62.39 kN

6. FORCES ACTING ON TURBINE ROTOR

   = 1 +2 3 4

   1 = P × A

   2 = P × A

   3 = m (2 1) 4 = P × A

   Force 1 acts on the left on annular surface of the disk

   Assuming it as pressure vessel with thickness 0.8 mm Using equation

   t= P ×d

   2 × t ×

   3 6

   3 6

   P = 2×0.8 ×10 ×269.5×10 ×0.89

   524 × 103

   P = 7.2 × 105 Pa

   Assuming 20% pressure losses & finding out required areas Q1 = 40.44 kN

   For calculating forces caused by difference in pressure of gas before and after the blades of turbine wheel.

1. BEARING SELECTION Frball = 3.47 kN Faball = 3.21 kN

   Now C= dynamic load carrying capacity

   C = P × (L10 ) 1/3 Load factor C=7.1 (2030.58) 1/3 1.4

   C= 98210 N = 98.21 k N

   For Front support Roller Bearing :

   Average life of roller bearing = 3000 hrs

   L10

   = 60×11000 ×3000

   106

   L10 = 2030.58 million rev

   Now C= dynamic load carrying capacity

   C = P × (L10 ) 3/10 Load factor C= 48512 N = 48.512 k N

2. CONCLUSION

In analytical design, by increasing numbers of ball, decreasing ball diameter, and changing the contact angle, the life in working hours can be increased. From the results of analytical design and analysis it can be seen that life of bearing is nearly same in both cases. It can be further increased by changing the

parameters, material and lubrication of the Bearing. In this paper the proposed work focused to calculate the axial and radial forces acting on front & rear bearing of LPCR and also calculated the bearings parameters, which are useful for bearing selection.

REFERENCES

1. H Austin Spang , Harold Brown, Control of jet engines, Control Engineering Practice, Volume 7, Issue 9, Pages 10431059, 1999.

2. Manfredi Maggiore, Raúl Ordóñez, Kevin M. Passino, Shrider Adibhatla, Estimator design in jet engine applications, Engineering Applications of Artificial Intelligence, Volume 16, Issues 78, Pages 579593, 2003.

3. Meherwan P. Boyce, 13 Bearings and Seals, Gas Turbine Engineering Handbook (Third Edition), Pages 501552, 2006.

4. M. Montazeri-Gh, M. Nasiri, M. Rajabi, M. Jamshidfard, Actuator-based hardware-in-the-loop testing of a jet engine fuel control unit in flight conditions, Simulation Modelling Practice and Theory, Volume 21, Issue 1, Pages 6577, 2012.

5. S.P. Mlynarek, D.R. Van Orden, Assessment of potential asbestos exposures from jet engine overhaul work, Regulatory Toxicology and Pharmacology, Volume 63, Issue 1, Pages 7883, 2012.

6. H.K.D.H. Bhadeshia, Steels for bearings, Progress in Materials Science, Volume 57, Issue 2, Pages 268435, 2012.

7. Rafael R. Adamczuk, Joerg R. Seume, Early Assessment of Defects and Damage in Jet Engines , Procedia CIRP, Volume 11, Pages 328333, 2013.

8. Leye M. Amoo, On the design and structural analysis of jet engine fan blade structures, Progress in Aerospace Sciences, Volume 60, Pages 111, 2013.