Static load and dynamic forced performance of rigid rotorsupported on short length journal bearings (includesthermal effects) MEEN 626. Luis San Andres (c)

Dr. Luis San Andres, UT/2000

Disk 2M

Journal bearing

X

Y

2Fo

disk

Clearance circle

Ωt

e

Static load

u Rigid shaft

Figure 5.1. Rigid rotor supported on journal bearings.(u) imbalance, (e) journal eccentricity

Updated 09/04/09

ORIGIN 1:=

rotor properties

bearing

2MKrot

bearing

2MKrot 2MKrot

DATA for rotorWT 500 lbf⋅:= Rotor weight

Load per bearingWWT

2:=

W 1.112 103× newton=

1/2 rotor massMWg

:= M 250 lb=

Rotor stiffness on each side of disk (at midspan)kshaft 40 106⋅Nm

⋅:=

Rotor_sagW

kshaft:= Rotor sag at midspan:

kshaft 2.284 105×lbfin

=Rotor_sag 1.095 10 3−× in=

rotor properties

bearing geometry

BEARING GEOMETRY

D 6 in⋅:= journal diameter RD2

:= journal radius

Static shaft deflection due to rotorweight = % of clearance

bearing lengthL 2 in⋅:=

c 0.003 in⋅:= radial clearance Rotor_sagc

0.365=LD

0.333=

bearing geometry

RPM_max 10000:= MAXIMUM & design speeds Nmax 50:=# of cases for analysis

RPMdesign 7200:=

a 0.2 c⋅:= Amplitude of imbalance on rotor diskMechanical energy convected by lubricant.κ 0.8:=Thermal model conditionsHeat carry over - thermal mixing coefficient.λ 0.70:=

Tsupply 50K:= Supply Oil Temperature Take deg-K as deg-C

Lubricant properties

PROPERTIES OF LUBRICANT MOBIL velocite No 10 (ISO VG 22)

μsupply 0.0143N s⋅

m2⋅:= Lubricant viscosity at Tsupply in Pa-sec.

α 0.0281K

⋅:= Alpha coefficient for viscosity equation.

ρ 862kg

m3⋅:= Density in kg/m3.

Cp 1880J

kg K⋅⋅:= Specific Heat in kJ/(kg C).

Lubricant properties

Calculations

CALCULATIONS =======FIND journal eccentricity as rotor speed increases for given load

DrRPM_max

Nmax( ):= step in rpm n 1 Nmax..:= Tsupply 50 K=

rpmn Dr n( )⋅:= rotor running speed(rpm)

Ωn rpmnπ

30⋅

radsec

⋅:= rotor speed(rad/sec) krpmrpm1000

:=

itmax 10:= Max iterations for thermal loop VISCOSITY formula

μ T( ) μsupply eα− T Tsupply−( )⋅

⋅:=

Journal eccentricity ratio and attitude angle for STATIC equilibrium

Find journal eccentricity and operating temperatures

εguess 0.8:= Ttrial Tsupply 20K+:=ITERATIVE LOOP -Given load, find operatingeccentricity & find temperature rise

SteadyJB ε rpm, Tguess, ( ) it 0←

x ε←

N rpm1

60 sec⋅⋅←

Ω rpmπ

30 sec⋅⋅←

Teff Tguess←

Tguess Teff 1.1⋅←

it it 1+←

μ μsupply eα− Teff Tsupply−( )⋅

⋅←

σ πμ N⋅ L⋅ D⋅

WRc

⎛⎜⎝⎞⎟⎠

2⋅

⎡⎢⎣

⎤⎥⎦

⋅LD

⎛⎜⎝⎞⎟⎠

2⋅←

ε root1 x2−( )2

x π2 1 x2−( )⋅ 16 x2⋅+⋅

σ− x, ⎡⎢⎢⎣

⎤⎥⎥⎦

←

Power Ω( )2μ R3⋅ L⋅

π

c⋅⎛⎜⎝

⎞⎟⎠⋅

1

1 ε2−( ) .5

⋅←

Q0 c R⋅ L⋅ 1 ε+( )( )⋅Ω

2⋅←

Qπ c R⋅ L⋅ 1 ε−( )( )⋅Ω

2⋅←

Qe c R⋅ L⋅ ε⋅ Ω⋅←

δTπ

κ Power⋅( )

ρ Cp⋅ Q0Qe

2−

⎛⎜⎝

⎞⎟⎠

⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

1 λQπ

Q0⋅+

←

δTin λQπ

Q0

⎛⎜⎝

⎞⎟⎠

⋅ δTπ⋅←

Tπ Tsupply δTπ+←

Tin Tsupply δTin+←

Tguess Teff←

Teff 0.5 Tin Tπ+( )⋅←

x ε←

it itmax<( ) Teff Tguess− 0.1K>( )∧while

:=

Speed in rev/sec

Speed in rad/sec

Guess (set) Effective temperature

effective viscosity

σ = short length bearingSommerfeld #

Root solver: find eccentricity ε forspecified Sommerfeld #

Power loss

Qo = inlet flow rate, θ=0

Qπ = flow rate at min film θ=π(start of cavitation zone

Exit flow (side flow)

Temperature rise at θ=π

Thermal energy equations areprobably incorrect. Please checkthem with those in handout

Temperature rise at inlet plane

Max temperature at θ=π

δT T TS−=

Effective temperature

Supply flow rate

Qsupply Q0 λ Qπ⋅−←

εTπ

KTin

KTeff

KPowerwatt

Qsupplys

m3⋅ σ

⎛⎜⎝

⎞⎟⎠

return

εn TπnTinn

Teff nPowern Qsupplyn

σn( ) SteadyJB .1 rpmn, Ttrial, ( ):=

convert to physical units

Teff Teff 1⋅ K:= Tin Tin 1⋅ K:=Power Power watt⋅:= Tπ Tπ 1⋅ K:=

Qsupply Qsupply 60000⋅:= LPMQsidesnc R⋅ L⋅ εn⋅ Ωn⋅ 60000⋅:=

effective viscosity: deg-C (actual)μ_n μsupply e

α− Teff nTsupply−( )⋅

⋅:=

Journal Attitude angleϕn atan

π 1 εn( )2−⋅

4 εn⋅

⎡⎢⎣

⎤⎥⎦

180π

⋅:=

Calculations

STATIC LOAD PERFORMANCE: journal cccentricity and attitude angle

0 2 4 6 8 100

0.10.20.30.40.50.60.70.80.9

1

rotor speed (krpm)

jour

nal e

ccen

trici

ty e

/c

0 2 4 6 8 100

102030405060708090

rotor speed (kRPM)

Atti

tude

ang

le (d

eg)

ε W 250 lbf=φc 3 10 3−× in=

POWER LOSS (kW) on each bearing

0 2 4 6 8 100

2.385

4.769

7.154

9.539

rotor speed (kRPM)

Pow

er (k

W) p

er b

earin

g

WL D⋅

20.833 psi=

L 2 in=D 6 in=

0 2 4 6 8 100

1

2

3

4

5

QsupplyQsides

rotor speed (krpm)

Bea

ring

inle

t flo

w ra

te (L

PM)

LUBRICANT FLOW RATE (LPM)

FILM temperatures: effective, inlet to film, exit at 180deg Tsupply 50 K=

0 2 4 6 8 1050

55

60

65

70

T (pi)T inlet filmT effec

Exit Temperature vs. Rotor Speed

Rotor Speed (krpm)

Exit

Tem

pera

ture

(C)

Tπ

Tin

Teff

krpm

*

OTHER BEARING OPERATING PARAMETERS

Minimum Film Thickness h2nc 1 εn−( )⋅⎡⎣ ⎤⎦ 106⋅:=

0 2 4 6 8 100

16

32

48

64

80Minimum Film Thickness vs. Speed

Rotor Speed (krpm)

Film

Thi

ckne

ss (u

m)

h2

krpm

c 0.076 mm=

Effective viscosity (C-Poise)

0 2 4 6 8 100

5

10

Viscosity vs. Rotor Speed

Rotor Speed (krpm)

Effe

ctiv

e vi

scos

ity (c

-Poi

se)

μ_ 1000⋅

krpm

μ Tsupply( ) 0.014 Ns

m2⋅=

7Friction Coefficientμfn

PowernΩn R⋅ W⋅

:=

0 2 4 6 8 100

0.05

0.1

0.15Coefficient of Friction vs. Rotor Speed

Rotor Speed (krpm)

Coe

ffic

ient

of F

rictio

n

μf

krpm

Reynolds NumberRen ρ Ωn⋅ R⋅

cμ_n

⋅:=

0 2 4 6 8 100

200

400

600

800

Rotor Speed (krpm)

Rey

nold

s Num

ber

Ren

Ren 1 εn+( )⋅

Ren 1 ε−( )n⎡⎣ ⎤⎦⋅

krpmn

Force coefficients evaluation

Short length journal bearing: rotordynamic force coefficients at equilibrium position

fron

4 σn⋅ εn( )2⋅

1 εn( )2−⎡⎣ ⎤⎦2

:= fton

π σn⋅ εn⋅

1 εn( )2−⎡⎣ ⎤⎦1.5

:=

Dimensional Stiffness (Fo=W)Dimensionless Stiffness

kxxn

fron

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fron( )2 1+ 2 εn( )2⋅+⎡

⎣⎤⎦

⋅:=Fon

μ_n Ωn⋅Lc

⎛⎜⎝⎞⎟⎠

2⋅ L⋅ R⋅

4 σn⋅:=

kyyn

fron

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fton( )2 1+ εn( )2−⎡

⎣⎤⎦

⋅:=Kxxn

kxxn

Fon

c⋅:=

kyxn

fton

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fron( )2 1− εn( )2+⎡

⎣⎤⎦

⋅:= Kyynkyyn

Fon

c⋅:=

Kyxnkyxn

Fon

c⋅:=

kxyn

fton

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fron( )2 1+ 2 εn( )2⋅+⎡

⎣⎤⎦

⋅:=

Kxynkxyn

Fon

c⋅:=

MATRIX of STIFFNESSES Kbn

Kxxn

Kyxn

Kxyn

Kyyn

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

Dimensionless DampingDimensional DampingCoefficients

cxxn

2 fton⋅

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fron( )2 2 εn( )2+⎡⎣ ⎤⎦⋅ 1+ εn( )2−⎡

⎣⎤⎦

⋅:=

Cxxncxxn

Fon

c Ωn⋅⋅:=

cyyn

2 fton⋅

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fton( )2 2 εn( )2+⎡⎣ ⎤⎦⋅ 1− εn( )2+⎡

⎣⎤⎦

⋅:=Cyyn

cyyn

Fon

c Ωn⋅⋅:=

cxyn

2 fron⋅

εn 1 εn( )2−⎡⎣ ⎤⎦⋅fton( )2 2 εn( )2+⎡⎣ ⎤⎦⋅ 1− εn( )2+⎡

⎣⎤⎦

⋅:=Cxyn

cxyn

Fon

c Ωn⋅⋅:=

cyxncxyn

:= CyxnCxyn

:=

MATRIX OF DAMPING COEFFS Cbn

Cxxn

Cyxn

Cxyn

Cyyn

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

Force coefficients evaluation

STIFFNESS AND DAMPING COEFFICIENTS

0 2 4 6 8 105− 107×

0

5 107×

1 108×

1.5 108×

2 108×

KxxKyyKxy-Kyx

Rotor Speed (krpm)

Stiff

ness

(N/m

)

0 2 4 6 8 100

10

20

30

CxxCyyCxyCyx

Rotor Speed (krpm)

Dam

ping

(N.s/

m)

Equivalent bearing stiffness for rigid rotorkeqn

kxxncyyn

⋅ kyyncxxn

⋅+ cyxnkxyn

⋅− cxynkyxn

⋅−

cxxncyyn

+:=

0 2 4 6 8 102.2 107×

2.3 107×

2.4 107×

2.5 107×

2.6 107×

2.7 107×

2.8 107×

rotor speed (krpm)

Equi

v. S

tiffn

ess (

N/m

)

Whirl frequency ratioWFRn

keqnkxxn

−( ) keqnkyyn

−( )⋅ kxynkyxn

⋅−

cxxncyyn

⋅ cxyncyxn

⋅−:=

0 2.5 5 7.5 100.35

0.4

0.45

0.5

0.55

Rotor Speed (krpm)

whi

rl fr

eque

ncy

ratio

Threshold speed of instability(krpm)

Ωtn

keqnFon

⋅

M c⋅

30π

⎛⎜⎝⎞⎟⎠

WFRn⋅:=

0 1.25 2.5 3.75 5 6.25 7.5 8.75 100

2.882

5.765

8.647

11.53Rigid Rotor

rotor speed (krpm)

Thre

shol

d sp

eed

(krp

m)

Ωt1000

krpm

krpm

Critical Rotor Mass for rigid rotor

Recall the rotor mass: 2 M⋅ 226.796 kg=Mcn

keqnFon

⋅

c Ωn( )2⋅ WFRn( )2⋅:=

McRn2 Mcn

⋅:=

0 1.25 2.5 3.75 5 6.25 7.5 8.75 100.1

1

10

100

rotor speed (krpm)

Mcr

titic

al/R

otor

Mas

s

McR2 M⋅

krpm

Effect of shaft flexibility on threshold speed of instability:Threshold speed of instability calculated including rotor sag at midspan.

Natural Frequency (rpm) of Flexible ShaftΩtf n

Ωtn

1 keqn

Rotor_sagc

⋅+⎛⎜⎝⎞⎟⎠

12

:=

ωnnWFRn Ωtf n

⋅:=

0 2 4 6 8 100

2.5

5

7.5

10

12.5

15

flex rotorrigid rotornat. freq rotsync line

Flexible Rotor

rotor speed (krpm)

Thre

shol

d sp

eed

(krp

m)

Ωtf1000

Ωt1000

ωn1000

krpm

krpm

kshaft 4 107×Nm

=

Rotor_sagc

0.365=

The thresholdspeed ofinstability is lowerfor the flexiblerotor than for therigid rotor model.

Synchronous response of flexible rotor due to imbalance

The equations of motion for both rotor and journal bearings are given below. Thecoordinates of rotor and disk motion have origin at the static equilibrium position.

k kshaft:=Let:ROTOR MASSLESS BEARINGS

md2 X⋅

dt2

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ k X x−( )⋅+ m a⋅ ω2⋅ cos ω t⋅( )⋅=

CI J, ( )

dxdt

dydt

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅ KI J, ( )x

y⎛⎜⎝

⎞⎟⎠

⋅+ k−x X−

y Y−⎛⎜⎝

⎞⎟⎠

⋅=

md2 Y⋅

dt2

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ k Y y−( )⋅+ m a⋅ ω2⋅ sin ω t⋅( )⋅=

a 1.524 10 5−× m= imbalance displacement ac

0.2=

The rotor disk (X,Y) and journal center displacements (x,y) are synchronous withthe imbalance excitation, i.e.

X Xc cos ω t⋅( )⋅ Xs sin ω t⋅( )⋅+= Y Yc cos ω t⋅( )⋅ Ys sin ω t⋅( )⋅+=

x xc cos ω t⋅( )⋅ xs sin ω t⋅( )⋅+= y yc cos ω t⋅( )⋅ ys sin ω t⋅( )⋅+=

Find the solution. Define UNIT matrix I21

0

0

1⎛⎜⎝

⎞⎟⎠

:=

Rotor (complex) displacements

Xn

Yn

⎛⎜⎝

⎞⎟⎠

Ωn( )2− M⋅ I2⋅ Cbni⋅ Ωn⋅( ) Kbn

+⎡⎣ ⎤⎦1− 1

kshaftI2⋅+⎡

⎢⎣

⎤⎥⎦

1−+

⎡⎢⎣

⎤⎥⎦

1−M a⋅ Ωn( )2⋅

i− M a⋅ Ωn( )2⋅⎡⎣ ⎤⎦⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅:=

Journal (complex) displacements

xn

yn

⎛⎜⎝

⎞⎟⎠

kshaft I2⋅ Cbni⋅ Ωn⋅( ) Kbn

+⎡⎣ ⎤⎦+⎡⎣ ⎤⎦

1− kshaft⋅Xn

Yn

⎛⎜⎝

⎞⎟⎠

⋅:=

Amplitudes of rotor and journal motion

XmnXn:= Ymn

Yn:=

xmnxn:= ymn

yn:=

show responses in dimensionless form (amplitude/clearance)

0 2 4 6 8 100

0.2

0.4

0.6

0.8

X/cY/c

Imbalance response Rotor center

rotor speed (krpm)

Rot

or c

ente

r (X

& Y

)

ROTORCENTER

ac

0.2=

0 2 4 6 8 100

0.1

0.2

0.3

X/cY/c

Imbalance response - journals

rotor speed (krpm)

Jour

nal d

isps

(X

& Y

)

JOURNALSCENTER

Exercise: Calculate the major and minor axes of the ellipses describing the (X,Y)motions. See Appendix A of Childs' Rotordynamics Book:.