Hydrodynamic Lubrication - Mechanical Engineering Lubrication • Fluid Lubricant: ... Friction vs...

ME 383S Bryant February 15, 20051

Hydrodynamic Lubrication

• Fluid Lubricant: liquid or gas (gas bearing)• Mechanism: Pressures separate surfaces

o Normal loads on bodieso Convergent profile between surfaceso Tangential motion between surfaceso Viscous effects generate shear stresseso Pressures equilibrate shear stresseso Surfaces “lift” apart

ME 383S Bryant February 15, 20052

Stribeck Curve Hydrodynamic lubrication: full film formed,

surfaces do not contact

Friction vs stribeck number ηN/P

η: dynamic viscosity, N: speed, P: pressure

Hydrodyamic Lubrication

• Review Navier Stokes Equations

• Derive Reynold s Equation

• Apply Reynolds equation to bearing

1

Navier Stokes Equations

Indicial notation: x1 = x, x2 = y, x3 = z

• Continuity Equation

∂ρ

∂t+

3∑k=1

∂(ρuk)

∂xk= 0 (1)

• Momentum Equations

ρ∂uj

∂t+

3∑k=1

ρuk∂uj

∂xk= −∂P

∂xj+ λ

∂

∂xj

3∑k=1

∂uk

∂xk

+3∑

k=1

∂

∂xk

[η

(∂uk

∂xj+

∂uj

∂xk

)]+ ρfj(2)

• Density ρ, Viscosity η, Bulk viscosity λ

• Unknowns: Flow velocities uj, Pressure p

2

Assumptions: NormalLubrication

• Newtonian fluid (constitutive law)

σij = −pδij + λ∂uk∂xk

δij + η

(∂ui∂xj

+∂uj∂xi

)fluid stresses σij, velocities ui,

Kroenecker delta δij =

1 if i = j0 otherwise

• quasi-steady flow: ∂/∂t = 0

• no slip between fluid particles & surfaces

• negligible fluid inertia (small Reynold s num-ber)

• very thin film

3

• Consequences:

negligible variations in pressure p, tem-

perature T , & fluid properties (density

ρ, dynamic viscosity η) across film thick-

ness 0 ≤ y ≤ h(x)

effects of curvatures of bearing surfaces

on flows negligible

laminar flows

• Additional Assumption: Incompressible flow

4

Simplified Equations

• Continuity Equation

∂(ρux)

∂x+

∂(ρuy)

∂y+

∂(ρuz)

∂z= 0 (3)

• Momentum Equations

η∂2ux

∂y2=

∂p

∂x, (4)

∂2uy

∂y2≈ 0, (5)

η∂2uz

∂y2=

∂p

∂z(6)

5

y

x

B

U1 hoh1

inclinedpad

Fx

V1

Fy

V2

U2

Integrate MomentumEquations

• Integrate with respect to y

• Determine constants of integration from

boundary velocities (U1, V1, W1) and (U2, V2, W2)

6

Flow Velocities & ContinuityEquation

• Flow velocities

ux =1

2η

∂p

∂xy(y − h) + U1 +

y

h(U2 − U1)

uy = (V2 − V1)y

h+ V1

uz =1

2η

∂p

∂zy(y − h) + W1 +

y

h(W2 − W1)

• Steady State Continuity Equation

∂(ρux)

∂x+

∂(ρuy)

∂y+

∂(ρuz)

∂z= 0

7

Reynold s Equation Derivation

• Substitute velocities into continuity equa-tion, then integrate across film thickness0 ≤ y ≤ h(x, z):

∫ h(x,z)

y=0

∂

∂x

ρ

[1

2η

∂p

∂xy(y − h)+U1+

y

h(U2 − U1)

]dy

+∫ h(x,z)

y=0

∂

∂y

ρ

[(V2 − V1)

y

h+ V1

]dy+∫ h(x,z)

y=0

∂

∂z

ρ

[1

2η

∂p

∂zy(y − h)+W1+

y

h(W2 − W1)

]dy

= 0

• First and third terms require Leibnitz s rule:

d

dx

∫ b(x)

a(x)f(y, x)dy =

∫ b(x)

a(x)

∂f(y, x)

∂xdy

+f [b(x), x] dbdx − f [a(x), x] da

dx

8

Reynold s Equation

∂

∂x

ρh3

η

∂p

∂x

+∂

∂z

ρh3

η

∂p

∂z

= 12ρ(V2 − V1)

+6(U1 − U2)∂(ρh)

∂x+ 6ρh

∂(U1 + U2)

∂x

+6(W1 − W2)∂(ρh)

∂z+ 6ρh

∂(W1 + W2)

∂z(7)

• Describes flow through convergent channel

• Left side: tangential & out of plane flows

• Film thickness h = h(x, z)

• Pressure p = p(x, z)

• Boundary velocities on surfaces:(U1, V1, W1), (U2, V2, W2)

9

y

x

B

U1 = U hoh1

inclinedpad

V1 = - V

FxW

Inclined Pad Bearing

• Normal load W , velocity V = −dhodt

• Tangential force Fx, velocity U

• Film thickness:

h(x, z) = h(x) = ho + (h1 − ho)(1 − x/B)

10

Particular (Long Bearing)Solution

• Assumptions

Long bearing (BL 1) =⇒ ∂/∂z = 0

No out of plane motions: W1 = W2 = 0

Relative velocity:

U = U1 − U2, V = V2 − V1

Rigid pad/Stiff bearing:∂(U1+U2)

∂x ≈ 0

incompressible & iso-viscous

• Apply to Reynold s Equation:

1

η

d

dx

(h3dp

dx

)= 12V + 6U

dh

dx(8)

11

Long Bearing Solution

• Integrate:

dp

dx= 12η

V x

h3+ 6η

U

h2+ C1

• Integrate:

p(x) = 12ηV∫

xdx

h3+ 6ηU

∫dx

h2+ C1x + C2

• Film thickness:

h(x) = ho + (h1 − ho)(1 − x/B)

• Pressure boundary conditions:

p(0) = p(B) = pa

12

Long Bearing Solution

pp =6 η n U x

(1 − x

B

)h2 (2ho + n)

−12B η V x

(1 − x

B

)h2 (2ho + n)

+ pa

• 1st term = load support

• 2nd term = ``squeezefilm effect

• pa: ambient pressure

• n = h1 − ho

• pp solution of Reynold s eqn (8)

• pp = pp(x) independent of z & satisfies (7)

with W1 = W2 =∂(U1+U2)

∂x = 0=⇒ particular solution of (7)

13

Homogeneous Solution

• Homogeneous Reynold s Equation

set all excitations to zero

U1 = V1 = W1 = U2 = V2 = W2 = 0

film thickness h = h(x)

∂

∂x

(h3∂p

∂x

)+ h3 ∂

∂z

(∂p

∂z

)= 0

• Separable solution: let ph = X(x)Z(z)

14

Homogeneous & ParticularSolutions

Complete Solution

15

Complete Solution

• Sum homogeneous & particular solutions

• Apply boundary conditions: velocities &

pressures

• Special cases:

long bearing: Lz/B >> 1,

ph = ph(x) & p = pp(x)

short bearing: Lz/B << 1,

16

Bearing Forces

• Forces: integrate over areas

W = −∫Ab

− [p(x, z) − pa] + 2η

∂uy

∂y

y=0

dxdz

Fx = −∫Ab

η

∂ux

∂y+

∂uy

∂x

y=0

dxdz

Fz = −∫Ab

η

∂uz

∂y+

∂uy

∂z

y=0

dxdz

where

ux =1

2η

∂p

∂xy(y − h) + U1 +

y

h(U2 − U1),

uy = (V2 − V1)y

h+ V1, uz =

1

2η

∂p

∂zy(y − h)

17

Apply to Inclined Pad

• Normal force

W (ho, U, V ) = (LB)2[2BR1U −

(4B2R1 + R3

)V

]

• Tangential force

Fx(ho, U, V ) = (LB)2[(R1 + R2

)U − 2BR1V

]

• where B = B/n,

R1(ho) =3 η

n L B

[− 2n

2ho + n+ log(1 + n/ho)

]

R2(ho) =R3(ho)

2=

η

n L Blog(1 + n/ho)

18

1Sf : U

R: ℜ2

UTF: LB 1 0

R: ℜ1

TF: n/2B

R: ℜ3

TF: 1/LBFx

VW

Sf : VP

τ σyy

tangential motions

Coquette Flow:power losses

Squeeze film:power losses

Wedge effect:lift + losses

normal motions

R2

U R1

R3

V

LB 1/LBn/2B

Fx

+

-

+

-

W

``BondGraph Equation

• Resistance field =⇒ W & Fx equations

• Bond graph & bearing equivalent circuit

19

Gear: Ig, mg

Load: Ws

Shaft: ksr, ksa

Tiltedpads

Bearingplate: Ip, mp

Te

Bearing: b1, b2

Thrust Bearing with M Pads

• In bearing bond graph, (Tb, ωb) replaces (Fx, U)

• ωb = MU/R Tb = RMFx

20

1

R: Mℜ2

TF: LB 1 0

R: Mℜ1

TF: n/2B

R: Mℜ3

TF: 1/LB VW Sf : V

Pτ σyySf : ωb ωb

TF: RTb

1

R: ℜ2

TF: LB 1 0

R: ℜ1

TF: n/2B

R: ℜ3

TF: 1/LBSf : ωb ωbTF: R

TbVW Sf : V1 1

1

R: ℜ2

TF: LB 1 0

R: ℜ1

TF: n/2B

R: ℜ3

TF: 1/LB

Gear: Ig, mg

Load: Ws

Shaft: ksr, ksa

Tiltedpads

Bearingplate: Ip, mp

Te

Bearing: b1, b2

Thrust Bearing Bond Graphs

21

B

y

x U1 = U ho

h1

steppad

V1 = -V

FxW

nsB

B

RoRi

α

Rayleigh Step Bearing

• Easier to manufacture step replaces incline

• Film thickness:

h(x) =

h1, 0 ≤ x < Bs

ho, Bs < x ≤ B

22

B

steppad

Bs

p(x)

pa

Step Bearing PressuresTriangular

• Pressures

• Maximum pressure at step x = Bs:

pmax − pa = 6η(B − Bs) Bs (nU + B V )

(B − Bs) h31 + Bs h

3o

23

Shaft

yz

W2R

2Rb

Bearing

L

φ

ωb

θ

e ωx

y

Eccentric Journal Bearings

• Film thickness:

h = h(θ) = c + e cos θ = c(1 + n cos θ)

Eccentricity: e Attitude angle: φ

Polar coordinates (e, φ) locates shaft cen-

ter relative to bearing center

Clearance: c = Rb − R

n = e/c

24

• Problem: shaft at (e, φ) orbits bearing

• Coordinate system attached to load W

• Journal rotates at relative ω − ωb

• Bearing rotates at relative ωb

• Could be piston rod-crankshaft bearing

25

• Surface velocities:

U1 = Rω + dedt sin θ − edφ

dt cos θ − Rdφdt

V1 = −dedt cos θ − edφ

dt sin θ

U2 = Rbωb − Rdφdt

V2 = −eωb sin θ

• Reynold s equation, right side:

6η

[−Rbωb − R (ω − ωb) + e

·φ cos θ− ·

e sin θ

]∂h∂x

+h∂

(−e

·φcos θ+

·esin θ

)∂x

+2R (ω − ωb)∂h∂x + 2e

·φ sin θ + 2

·e cos θ

26

Reynold s Equation

• Ω = ω + ωb − 2·φ

• Approximations: c/R 1, e/R 1

• Assumptions: L/2R is large (generally > 4)

• Similar procedure gives:

∂

∂θ

ρh3

η

∂p

∂θ

+ R2 ∂

∂z

ρh3

η

∂p

∂z

= 6R2

[2

∂(ρh)

∂t+ Ω

∂(ρh)

∂θ

]

27

θ

p - pa6ηΩR2/c2

n = 0.5n = 0.

Solve Pressures

p(θ) − pa = 6ηR2

c2

[n

2 + n2Ωsin θ− ·

n cos θ

]×

× 1

(1 + n cos θ)2+

1

(1 + n cos θ)

• Suppose

·n= 0. For π ≤ θ < 2π, p − pa < 0

=⇒ subambient pressures & cavitation inliquids

28

• Cavitation =⇒ lubricant vapors & air bub-bles

• Pressure solution invalid, options:

Solve with cavitation algorithm:

∗ Define: φ = ρρc

, p = pc + gβ lnφ

∗ ρc: vapor density, pc: vapor pressure

∗ dp = gβρdρ = gβ

φdφ

∗ Step function g = g(φ) =

0, φ < 11, φ ≥ 1

∗ ddx

(ρh3

12ηdpdx

)= d

dx

(ρcβh3

12ηgdφ

dx

)∗ Yields ``ModifiedReynold s Equation

in unknown φ

Approximation: when p < pa, set p = pa

in integrals for force & torque

29

e φ.Wφ

circumferentialmotions

e. We

radialmotions

R

ωTf

shaftrotation

TF: φ

WZWY

Y.

Z.

Forces & Torque

• Pressure integrals =⇒ forces & torque:

We = −12πηLR3

c2

·n(

1 − n2)3/2

Wφ = 12LπηR3

c2nΩ(

2 + n2) √

1 − n2

Tf = 4πηLΩR3

c

1 + 2n2(2 + n2

) √1 − n2

• Yields 3-port resistive field:

• TF transforms (e, φ) to cartesian system

30

Rotor dynamics in bondgraph form

• Shaft bending & torsion via FEM in bond

graphs

• Rotating coordinate in bond graph form

[Hubbard, 1979]

• Connect to rest of system via bond graph

• Possible: shaft whirl, bending, etc. excited

directly by system

31