Proceedings of ASME Turbo Expo 2015: Turbine …rotorlab.tamu.edu/tribgroup/2015 TRC San Andres/2015...

1

Luis San AndrésMast-Childs Chair Professor, Fellow ASMETexas A&M University

Orbit-Model Force Coefficients for Fluid Film Bearings: a Step beyond Linearization

Sung-Hwa JeungGraduate Research Assistant, Student Member ASME Texas A&M University

Proceedings of ASME Turbo Expo 2015: Turbine Technical Conference and Exposition, June 15-19, 2015, Montreal, Canada

Paper GT2015-43487

Accepted for journal

publication

2

BackgroundRotordynamics models bearings and seals reactionforces F={FX, FY}T with linearized force coefficients(stiffness K, damping C, and inertia M).

( )t eF =F -K z -Cz -Mz : rotor (journal) displacements about static position (e).

z={x,y}T

, , ,; ;

e e e

X Y XXY YX XX

x y x y x y

F F FK C M

y x x

Linearized force coefficients denote changes in reaction force toinfinitesimally small amplitude motions about an equilibriumposition:

0

lim eY Y

x

F F

x

( ) 0lim

( )e

e

X X

y y e

F F

y y

0lim eX X

x

F F

x

Y

X

3

A concern:

Linearized force coefficients (K, C,M) are often inadequate to produceaccurate reaction forces for rotormotions of a sizeable magnitude.Squeeze film dampers (SFDs) are acase in point.

In practice rotor-bearing systems do not show infinitesimally smallmotions, hence

The ever present questions are:• How good are the linearized force coefficients?• Can they be used for rotor motions of large amplitude, like when

crossing a critical speed? • How can one obtain better (more accurate) coefficients? Should

one use higher order terms?

Y

X

4

Choy et al. (1991) Calculate higher-order force coefficients derived from a Taylor-series expansion.

( )

2 32 3

2 3

2 32 3

2 3

2

2 32 3

2 3

1 1 .....2! 3!

1 1 .....2! 3!

.....

1 1 .....2! 3!

t e

X X XX X

e e e

X X X

e e e

X

e

X X X

e e e

F F FF F x x xx x x

F F Fy y yy y y

F x yx y

F F Fx x xx x x

2 32 3

2 3

2

1 1 .....2! 3!

..... ....

X X X

e e e

X

e

F F Fy y yy y y

F x yx y

How many nonlinear terms are needed?

Past literature

Issues:What is the threshold

between small and large amplitude motions?

How to do it fast (in a computer)?

Past literature El-Shafei and Eranki (1994)Characterize a bearing (nonlinear) reaction force response to deliver best

estimations for the bearing stiffness and damping force coefficients.

Sawicki and Rao (2001)

Müller-Karger and Granados (1997)Nonlinearity of force coefficients depends on the size and shape of the orbital path.

Czolczynski (1999)Estimates force coefficients

from orbits (harmonic) motions.

Non-linear analyses

None of the methods benchmarked (to date) to experimental data.

5

6

In an analysis, how are the true linearized force coefficients obtained?

Y

X

7

Extended Reynolds eqn. for laminar thin film flows

3 2 2

21 2 2 1 2h P h h h hP

t t

c : clearance

eX, eY: journal eccentricity (varies with time)

P : film pressure

h : film thickness

: viscosity

Wo X

Y

eXo

eY

eo

X

clearancecircle

YStatic load

Journal center

Wo X

Y

eXo

eY

eo

X

clearancecircle

YStatic load

Journal center

o

Wo X

Y

eXo

eY

eo

X

clearancecircle

YStatic load

Journal center

Wo X

Y

eXo

eY

eo

X

clearancecircle

YStatic load

Journal center

o

Film thickness: 0 cos sin i th h X Y e

Pressure:

Let journal move with small (~0) amplitude displacements(∆X, ∆Y) with whirl frequency ωabout an equilibrium position:

cos sinX Yh c e e

0 + + .. i tX YX YP P P X P X P Y P Y e

0 + + ...e e e e

P P P PP P X X Y Y

X X Y Y

,...X Ye e

P PP P

X Y

i t

i t

X X e

Y Y e

8

Substitute h and P into modified Reynolds equation to obtain

30 0

0 0( )12 2h hP P

L

20

0( ) 1 Re2 4s

h hP i i h h P

L

Zeroth-order:

First-order:

Pressure fields & forces

00

cossin

o

o

LX

Y

FP Rd dz

F

Static (equilibrium) forces:

0 0

0 0

cos cosRe ; Im

sin sin

cos cosRe ; Im

sin sin

L LXX XX

X XXY XY

L LXY XY

Y YYY YX

K CP Rd dz P Rd dz

K C

K CP Rd dz P Rd dz

K C

True Linearized Force coefficients:

Limitation: journal motion of infinitesimally small amplitude about an equilibrium position.

F0+W=0balance of static forces

9

Orbit analysis

Method follows an experimental identification procedure.

Estimates (numerically) force coefficients accurate over a wide frequency range. In particular for rotor whirl motions of large amplitude and statically off-centered.

Y

X

10

Journal center kinetics and forcesJournal motion (X vsY) bearing reaction forces (FX vs FY).

Dots denote discrete points along the orbital path during a full period of whirl motion

Journal motion is neither circular nor small in amplitude.

Forces (N)

X

YJournal center orbit (µm)

X

Y

ω

Specify X(t),Y (t)

Solve unsteady Reynolds equation find pressure P( X,Y,dX/dt, dY/dt)

Integrate pressure field on journal surface

find ForcesFX, FY

Continue to complete whole orbit path.

11

The orbit analysis methodFor given whirl orbit with given shape and over specified frequency range:

• Calculate bearing reaction forces.

• Conduct Fourier analysis.• Estimate force coefficients.

0

0

( ) ( ) ( )

( ) ( ) ( )

cos( ) sin( )

sin( ) cos( )X t X X t Y t

Y t Y X t Y t

e e a a

e e a a

( ) cos( )X t Xa r t ( ) sin( )Y t Ya r t

,X Yr r : amplitudes of motion along X,Y axes.

Journal motion with frequency (ω)

X/c

Y/c Clearance (c)

4

0.3Xr c

0.4Yr c

12

Bearing reaction forces

The bearing reaction force: F = Fstatic + Fdyn(t).

The dynamic portion of the fluid film reaction force will be modeled in a linearized form as

dynF K z Cz+Mz

where z is a vector of dynamic displacements and are matrices of stiffness, viscous damping and inertia force coefficients

cos( ) sin( )X Y

Y X

i t i tr i r

i r re e

1z z

In the frequency domain, the journal dynamic motion is:

( , , )K C M

13

Fourier analysisThe time varying part of the reaction force is periodic with fundamental period T=2/.Using Fourier series decomposition,

2 31 ....i t i t i te e e dyn II IIIF F F F

To first order effects (fundamental frequency)

1i te dynF F 1 1F H z

, i.e. reaction force is proportional to motion with fundamental frequency.

is a matrix of complex stiffness

14

Procedure to obtain multiple solutions

Forward () and backward (-) whirl orbits ensure linear independence of two forces needed for identification. z1 F1 z2 F2

X

Y

+ωX

Y

-ω

15

Y

X

FY

FX

1z1 1F11

FY

FX

2z1 2F12

3, 4,….……N-1,

ω

Forward whirl motions

2 22 21i te dynF F

2 21 1F H z

FY

FX

2z1 2F1

Fourier analysis:

FX

Nz1 NF1N

FY

16

Y

X

FY

FX

1z2 1F21

FY

FX

2z2 2F22

-ω

Backward whirl motions

2 22 22i te dynF F

2 22 2F H z

FY

FX

2z2 2F2

Fourier analysis:

3, 4,….……N-1,

FX

Nz2 NF2N

FY

17

Procedure to obtain complex stiffnesses

To find :

1 2k k k k

k 1 2F F H z z

Solve algebraic equations at each frequency (k):

XX XYk

YX YY k

H H=H H

H

1,.....k N H H H z1 F1 z2 F2

X

Y

+ωX

Y

-ω

18

Stack Hk for a set of frequencies (k= 1,2,….N). A simple linear curve fit delivers:

2Re( )

Im( )k

k

k

k

H K - M

H C

Derived K,C,M coefficients

The orbit analysis procedure numerically replicates experimental procedures to identify force coefficients.

, ,K C M : stiffness, damping and mass coefficientsvalid over a frequency range and for specified whirl amplitude (and shape) sets.

19

Bearing forces - compare Damper B (cB = 127 μm)

Orbit model reproduces best periodic force at highest frequency.

(Numerical) nonlinear force: integration of pressure field (solution of Reynolds eqn.)

Orbit analysis:

First Fourier component: 1i te F

21 i 1F K M C z

50% off-centered orbit (20% c)

20

Comparisons of experimental results to predictions from models (orbit and linearized)

Orbit model

TrueLinear

Note in the following slides:

( , , )K C M

Y

X

( , , )K C M

21

SFD Test Rig

Static loader

Shaker in X direction

Shaker in Y direction

Top view

SFD test bearing

2 electro magnetic-shakers (2 kN ~ 550 lbf)Static loader at 45°Customizable SFD test bearing.

22

Example: Squeeze Film Damper

San Andrés, L., and Jeung, S.-H., 2015, “Experimental Performance of an Open Ends, Centrally Grooved, Squeeze Film Damper Operating with Large Amplitude Orbital Motions,” ASME J. Eng. Gas Turbines Power, 137(3).

L/D=0.4, 2 x 25.4 mm lands

L/D=0.2, 25.4 mm lands

Damper A Damper BJournal diameter, D 127 mm 127 mmRadial clearance, cA, cB 0.251 mm 0.129 mmFilm land length, L 25.4 mm 25.4 mmCentral Groove axial length, LG 12.7 mm none

depth, dG 9.5 mmLubricant ISO VG 2

Density, r 785 kg/m3 799 kg/m3

Viscosity m at TS 0.0029 Pa.s 0.0025 Pa.s

Frequency range (Hz) 10-100Whirl amplitude, r (μm) 20 - 178 6.- 76Static eccentricity, es (μm) 0 - 191 0 - 63

23

Evaluate SFD force coefficients fromMax. clearance (c)

X Displacement [μm]

YD

ispl

acem

ent [μm

]Tests conducted Excitation frequency 10 – 100 Hz

circular orbits: amplitude (r) grows.Y

X

with offset or static eccentricity (es) – 45o from

X-Y axes

Operating condition Damper A0.251 mm

Damper B0.129 mm

Whirl amplitude r (μm) 20 – 178 6 – 76

Static eccentricity es (μm) 0 - 191 0 – 63

Max. squeeze film Reynolds No. (Res)

10.5 3.2

2

ScRe Squeeze film

Reynolds #

24

Complex stiffness vs. frequency

Ima(H)Real(H)

Model

force coefficients

Goodness of Fit (R2)

MXX [kg] CXX [kN.s/m] Re (XX) Im (XX)

Fit to test data 6.5 5.8 0.99 0.99

Linear model 7.0 6.0 0.95 0.99

Orbit model 7.2 6.5 0.93 0.94

Damper B (cB = 127 μm)

Centered orbit (30% c)

Orbit model True Linear

25

Complex stiffness vs. frequency

Ima(H)Real(H)

Model

force coefficients

Goodness of Fit (R2)

MXX [kg] CXX [kN.s/m] Re (XX) Im (XX)

Fit to test data 6.1 6.1 0.99 0.99

Linear model 7.0 6.0 0.98 0.99

Orbit model 6.1 6.1 0.98 0.99

Damper B (cB = 127 μm)

Centered orbit (50% c)

Orbit model True Linear

26

Complex stiffness vs. frequency

Ima(H)Real(H)

Model

force coefficients

Goodness of Fit (R2)

MXX [kg] CXX [kN.s/m] Re (XX) Im (XX)

Fit to test data 6.4 7.8 0.93 0.97

Linear model 8.2 9.2 0.73 0.85

Orbit model 5.8 8.6 0.92 0.93

Damper B (cB = 127 μm)

40% off-centered orbit (30% c)

Orbit model True Linear

orbit model ~ test data

27

SFD effective force coefficients

2X eff XX XY XXK K C M

XYX eff XX XY

KC C M

For circular orbits (only), dynamic forces reduce to

radial eff

tangential eff

F K r

F C r

r

-Fradial

-Ftangential

r

X

Y

28

Orbit modelpredictions agree

well with experimental

results.

(es/cB=0)

Effective force coefficients Damper B (cB = 127 μm)

Circular centered orbits

Large amplitude motions.

29

Orbit modelpredictions agree

well with experimental results.

SFD effective force coefficients Damper B (cB = 127 μm)

(es/cB=0.4)Circular off-centered orbits

Large amplitude eccentric motions.

30

Coefficients from elliptical orbits

Y

X

Y

X

Y

X

Dynamic load measurements:Elliptical orbit amplitude (r) grows.rX : rY = 5 : 1 at static eccentricity eS/cB=0.1.

(rX, rY)=(0.1, 0.02) cB

(rX, rY)=(0.35, 0.07) cB

(rX, rY)=(0.6, 0.12) cB

Damper B (cB = 127 μm)

Note: True force coefficients can’t be found large motions.

31

Coefficients from elliptical orbits Damper B (cB = 127 μm)

Predicted and experimental results show CXX ≠ CYY and MXX ≠ MYY.Orbit-based force coefficients follow trend of test coefficients;but over predict MYY by ~42% at rX/cB=0.6.

32

SFD forces and energy dissipation

( )E dt TSFDz F

BCM SFD S S SF L M z C z K z a

SFD SFD SFD SFDF M z C z K z

Mechanical work

Experimentally derived force

Orbit model derived force

Over a full period of motion

: Applied Dynamic load

: Mass of bearing cartridge

: Rig structure force coefficients, ,S S SM C K LBCM

= Dissipated energyA way to characterize the goodness of the force coefficients.

33

Evaluation of dynamic forcesω=100 Hz

Y

X(a)

(c)

(b)

For small amplitude motions, experimental

force and that constructed from orbit

model are similar.

For a large amplitude orbit, the differences

are significant.

Damper A (cA = 253 μm)

small to large size orbits:

34Energy dissipation increases with increasing orbit amplitude.

E < 0 = energy dissipated by SFD

Mechanical energy dissipation ( )E dt TSFDz F

35

Energy difference (%)100 Hz

For all cases the difference is less than ~25%.

experimental

orbit_model

1diff

EE

E

Test damper A Test damper B

The measure gives validity to the force coefficients derived from an orbit analysis.

36

ConclusionThe orbit analysis numerically replicates an experimentalprocedure, it calculates forces as a function of (any)motion amplitude and identifies force coefficients (K,C,M)over a frequency range.

Comparison to test results from two distinct SFDsFor large amplitude whirl orbits (r/c>0.3),

• True linearized force coefficients correlate poorly with test data.

• Orbit model force coefficients predict well the experimentallyforce coefficients, reproduce measured forces, and dissipatesame amount of energy as that in the experiments.

37

Observation

One step beyond the conventional….

Orbit analysis model proposes a methodology to escape the limitations of linearized force coefficients while

avoiding time-consuming (transient response) numerical integration.

(*) In the near future, with ubiquitous super fast & super cheap computer power, every engineering process will be modeled with the greatest fidelity possible (number crunching rules! ).

for the time being (*)

38

Thanks to Pratt & Whitney Engines

Learn more at http://rotorlab.tamu.edu

Questions (?)

TAMU Turbomachinery Research Consortium

Acknowledgements

39

Other slidesY

X

Y

X

Y

X

40

Types of journal motionx= R

Y

X

e

h

R

e: amplitude of motion

whirl frequency

eXo

eYo

whirlingjournal

Film thickness

x= R

Y

X

2rX

rX, rY : amplitudes of motion

whirl frequency

eo

2rY

K, C, M (force coefficients)RBS stability analysis

Applications:

FX, FY (reaction forces)RBS imbalance response& transient load effects

(a) Small amplitude journal motions (b) Large amplitude journal motions