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