Hydrodynamic Lubrication

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1 ME 383S Bryant February 15, Hydrodynamic Lubrication Fluid Lubricant: liquid or gas (gas bearing) Mechanism: Pressures separate surfaces o Normal loads on bodies o Convergent profile between surfaces o Tangential motion between surfaces o Viscous effects generate shear stresses o Pressures equilibrate shear stresses o Surfaces lift apart

2 ME 383S Bryant February 15, 2005 Stribeck Curve 2 Hydrodynamic lubrication: full film formed, surfaces do not contact Friction vs stribeck number ηn/p η: dynamic viscosity, N: speed, P: pressure

3 Hydrodyamic Lubrication Review Navier Stokes Equations Derive Reynold s Equation Apply Reynolds equation to bearing 1

4 Navier Stokes Equations Indicial notation: x 1 = x, x 2 = y, x 3 = z Continuity Equation ρ t + 3 k=1 (ρu k ) x k = 0 (1) Momentum Equations ρ u j t + 3 k=1 + u j ρu k = P + λ x k x j x j 3 k=1 x k [ η ( uk x j + u j x k 3 k=1 )] u k x k + ρf j (2) Density ρ, Viscosity η, Bulk viscosity λ Unknowns: Flow velocities u j, Pressure p 2

5 Assumptions: Normal Lubrication Newtonian fluid (constitutive law) σ ij = pδ ij + λ u ( k x δ k ij + η ui x + u ) j j x i fluid stresses σ ij, velocities u i, { 1 if i = j Kroenecker delta δ ij = 0 otherwise quasi-steady flow: / t = 0 no slip between fluid particles & surfaces negligible fluid inertia (small Reynold s number) very thin film 3

6 Consequences: negligible variations in pressure p, temperature T, & fluid properties (density ρ, dynamic viscosity η) across film thickness 0 y h(x) effects of curvatures of bearing surfaces on flows negligible laminar flows Additional Assumption: Incompressible flow 4

7 Simplified Equations Continuity Equation (ρu x ) x + (ρu y) y + (ρu z) z = 0 (3) Momentum Equations η 2 u x y 2 = p x, (4) 2 u y 0, (5) y2 η 2 u z y 2 = p z (6) 5

8 Integrate Momentum Equations F y F x h 1 y x V 1 U 2 U 1 h o inclined pad V 2 B Integrate with respect to y Determine constants of integration from boundary velocities (U 1,V 1,W 1 ) and (U 2,V 2,W 2 ) 6

9 Flow Velocities & Continuity Equation Flow velocities u x = 1 2η u z = 1 2η p x y(y h)+u 1 + y h (U 2 U 1 ) u y =(V 2 V 1 ) y h + V 1 p z y(y h)+w 1 + y h (W 2 W 1 ) Steady State Continuity Equation (ρu x ) x + (ρu y) y + (ρu z) z =0 7

10 Reynold s Equation Derivation Substitute velocities into continuity equation, then integrate across film thickness 0 y h(x, z): h(x,z) y=0 h(x,z) + y=0 h(x,z) y=0 { ρ x z y { ρ [ 1 { p 2η x y(y h)+u 1 + y h (U 2 U 1 ) [ ρ (V 2 V 1 ) y ]} h + V 1 dy+ [ 1 2η p z y(y h)+w 1 + y h (W 2 W 1 ) =0 ]} ]} dy dy First and third terms require Leibnitz s rule: d dx b(x) b(x) f(y, x)dy = a(x) a(x) +f [b(x),x] db dx f(y, x) dy x f da [a(x),x] dx 8

11 Reynold s Equation x ρh3 η p + x z ρh3 η p =12ρ(V z 2 V 1 ) +6(U 1 U 2 ) (ρh) x +6ρh (U 1 + U 2 ) x +6(W 1 W 2 ) (ρh) z +6ρh (W 1 + W 2 ) 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: (U 1, V 1, W 1 ), (U 2, V 2, W 2 ) 9

12 Inclined Pad Bearing W F x h 1 y x U 1 = U V 1 = - V h o inclined pad B Normal load W, velocity V = dh o dt Tangential force F x, velocity U Film thickness: h(x, z) =h(x) =h o +(h 1 h o )(1 x/b) 10

13 Particular (Long Bearing) Solution Assumptions Long bearing ( B L 1) = / z =0 No out of plane motions: W 1 = W 2 =0 Relative velocity: U = U 1 U 2, V = V 2 V 1 Rigid pad/stiff bearing: (U 1 +U 2 ) x 0 incompressible & iso-viscous Apply to Reynold s Equation: 1 d η dx ( h 3dp dx ) =12V +6U dh dx (8) 11

14 Long Bearing Solution Integrate: dp dx =12ηVx h 3 +6η U h 2 + C 1 Integrate: p(x) =12ηV xdx h 3 +6ηU dx h 2 + C 1 x + C 2 Film thickness: h(x) =h o +(h 1 h o )(1 x/b) Pressure boundary conditions: p(0) = p(b) =p a 12

15 Long Bearing Solution p p = 6 ηnux ( 1 B x h 2 (2 h o + n) ) 12 BηV x ( 1 B x h 2 (2 h o + n) ) + p a 1st term = load support 2nd term = ``squeezefilm effect p a : ambient pressure n = h 1 h o p p solution of Reynold s eqn (8) p p = p p (x) independent of z & satisfies (7) with W 1 = W 2 = (U 1 +U 2 ) x =0 = particular solution of (7) 13

16 Homogeneous Solution Homogeneous Reynold s Equation set all excitations to zero U 1 = V 1 = W 1 = U 2 = V 2 = W 2 =0 film thickness h = h(x) x ( h 3 p x ) + h 3 z ( p z ) =0 Separable solution: let p h = X(x)Z(z) 14

17 Homogeneous & Particular Solutions Complete Solution 15

18 Complete Solution Sum homogeneous & particular solutions Apply boundary conditions: pressures velocities & Special cases: long bearing: L z /B >> 1, p h = p h (x) &p = p p (x) short bearing: L z /B << 1, 16

19 Bearing Forces Forces: integrate over areas W = F x = F z = A b { η A b η A b [p(x, z) p a ] +2η u y y { ux y + u } y dxdz x y=0 { uz y + u } y dxdz z y=0 } y=0 dxdz where u x = 1 p 2η x y(y h)+u 1 + y h (U 2 U 1 ), u y =(V 2 V 1 ) y h + V 1,u z = 1 p y(y h) 2η z 17

20 Apply to Inclined Pad Normal force W (h o,u,v)=(lb) 2 [ 2 BR 1 U ( 4 B 2 R 1 + R 3 ) V ] Tangential force F x (h o,u,v)=(lb) 2 [( R 1 + R 2 ) U 2 BR1 V ] where B = B/n, R 1 (h o )= 3 η nlb [ R 2 (h o )= R 3 (h o) 2 2 n 2 h o + n + log(1 + n/h o) = η nlb log(1 + n/h o) ] 18

21 ``BondGraph Equation Resistance field = W & F x equations Coquette Flow: power losses Wedge effect: lift + losses Squeeze film: power losses R: R 2 R: R 1 R: R 3 P F x τ Sf : U TF: LB 1 0 U tangential motions TF: n/2b 1 σ yy TF: 1/LB W V S f : V normal motions + R 2 R 3 + F x U R 1 V W - LB n/2b 1/LB - Bond graph & bearing equivalent circuit 19

22 Thrust Bearing with M Pads Load: W s Gear: I g, m g Shaft: k sr, k sa Bearing: b 1, b 2 T e Bearing plate: I p, m p Tilted pads In bearing bond graph, (T b,ω b ) replaces (F x,u) ω b = MU/R T b = RMF x 20

23 Thrust Bearing Bond Graphs R: MR 2 R: MR 1 R: MR 3 T S f : ω b b TF: R ωb P τ TF: LB 1 0 TF: n/2b 1 σ yy TF: 1/LB V W Sf : V R: R 2 R: R 1 R: R 3 Sf : Tb ωb ω b TF: R 1 TF: LB 1 0 TF: n/2b 1 TF: 1/LB 1 W V S f : V R: R 2 R: R 1 R: R 3 TF: LB 1 0 TF: n/2b 1 TF: 1/LB Load: W s Gear: I g, m g Shaft: k sr, k sa Bearing: b 1, b 2 T e Bearing plate: I p, m p Tilted pads 21

24 Rayleigh Step Bearing W F x α B y x U 1 = U h o V 1 = -V h 1 Ri Ro step pad n s B B Easier to manufacture step replaces incline Film thickness: h(x) = { h1, 0 x<b s h o, B s <x B 22

25 Step Bearing Pressures Triangular Pressures p(x) p a step pad B s B Maximum pressure at step x = B s : p max p a =6η (B B s) B s (nu + BV) (B B s ) h B s h 3 o 23

26 Eccentric Journal Bearings φ θ W Bearing 2R Shaft 2R b z x e ω y L y ω b 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 center relative to bearing center Clearance: c = R b R n = e/c 24

27 Problem: shaft at (e, φ) orbits bearing Coordinate system attached to load W Journal rotates at relative ω ω b Bearing rotates at relative ω b Could be piston rod-crankshaft bearing 25

28 Surface velocities: U 1 = Rω + de dt V 1 = de dt U 2 = R b ω b R dφ dt V 2 = eω b sin θ sin θ edφ dt cos θ edφ dt sin θ cos θ Rdφ dt Reynold s equation, right side: 6η {[ R b ω b R (ω ω b ) + e φ cos θ e sin θ +h ( e φcos θ+ esin ) θ x +2R (ω ω b ) h x +2e φ sin θ +2 e cos θ } ] h x 26

29 Reynold s Equation Ω=ω + ω b 2 φ Approximations: c/r 1, e/r 1 Assumptions: L/2R is large (generally > 4) Similar procedure gives: θ ρh3 p + R 2 η θ z [ =6R 2 2 (ρh) t ρh3 η p z +Ω (ρh) θ ] 27

30 p(θ) p a =6η R2 Solve Pressures c 2 [ n 2+n 2Ω sin θ n cos θ 1 (1+ncos θ) (1+ncos θ) ] Suppose n= 0. For π θ<2π, p p a < 0 = subambient pressures & cavitation in liquids p - p a 6ηΩR 2 /c 2. n = 0 n = 0.5 θ 28

31 Cavitation = lubricant vapors & air bubbles Pressure solution invalid, options: Solve with cavitation algorithm: Define: φ = ρ ρ c, p = p c + gβ ln φ ρ c : vapor density, p c : vapor pressure dp = g β ρ dρ = gβ φ dφ Step function g = g(φ) = { 0, φ < 1 1, φ 1 d dx ( ρh 3 12η ) dp dx = d dx ( ρc βh 3 ) 12η gdφ dx Yields ``ModifiedReynold s Equation in unknown φ Approximation: when p<p a, set p = p a in integrals for force & torque 29

32 Forces & Torque Pressure integrals = forces & torque: W e = 12πηL R3 c 2 W φ =12Lπη R3 c 2 T f =4πηLΩ R3 c n ( 1 n 2 ) 3/2 nω ( 2+n 2 ) 1 n 2 1+2n 2 ( 2+n 2 ) 1 n 2 T f shaft rotation ω circumferential motions W. φ e φ R ė TF: φ W e radial motions Yields 3-port resistive field: W Y.. Y Z W Z TF transforms (e, φ) to cartesian system 30

33 Rotor dynamics in bond graph 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

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