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

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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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.

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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)

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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

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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.

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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

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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 #

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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

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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

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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

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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 (*)

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Thanks to Pratt & Whitney Engines

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

Questions (?)

TAMU Turbomachinery Research Consortium

Acknowledgements

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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