hydrodynamic lubrication - mechanical engineering lubrication • fluid lubricant: ... friction vs...
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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
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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
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Hydrodyamic Lubrication
• Review Navier Stokes Equations
• Derive Reynold s Equation
• Apply Reynolds equation to bearing
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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
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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
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• 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
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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)
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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)
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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
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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
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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)
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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)
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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)
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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
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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)
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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)
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Homogeneous & ParticularSolutions
Complete Solution
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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,
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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)
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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)
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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
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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
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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
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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
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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
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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
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• 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
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• 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
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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
]
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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
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• 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
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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
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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