Hydrodynamic Lubrication 2009 Part 1 docx

Hydrodynamic Lubrication 2009 Part 1 docx

Hydrodynamic Lubrication 2009 Part 1 docx

... Rotor–Bearing Systems 11 1 5 .10 PreventionofOilWhip 11 3 References 11 4 6 Foil Bearings 11 9 6 .1 Basic Equations 12 1 6.2 Finite Element Solution of the Basic Equations 12 2 6.2 .1 Reynolds’ Equation 12 2 6.2.2 ... from <XNLR+RUL +\GURG\QDPLF/XEULFDWLRQ :LWK)LJXUHV ABC <XNLR+RUL +\GURG\QDPLF/XEULFDWLRQ Contents 1 Friction, Wear, and Lubrication 1 1 .1 Frictio...

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Hydrodynamic Lubrication 2009 Part 3 docx

Hydrodynamic Lubrication 2009 Part 3 docx

... (J 1 will be used later): J 3 =  dφ (1 + κ cos φ) 3 = 1 (1 − κ 2 ) 5/2  ϕ − 2κ sin ϕ + κ 2 ϕ 2 + κ 2 4 sin 2ϕ  (3 .11 ) J 2 =  dφ (1 + κ cos φ) 2 = 1 (1 − κ 2 ) 3/2  ϕ − κ sin ϕ  (3 .12 ) J 1 =  dφ 1 ... cos 2 φ = 1, as follows: cos φ = cos ϕ − κ 1 − κ cos ϕ , sin φ = (1 − κ 2 ) 1/ 2 sin ϕ 1 − κ cos ϕ (3.9) From these relations, we have the following: dφ = (1...

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Hydrodynamic Lubrication 2009 Part 8 docx

Hydrodynamic Lubrication 2009 Part 8 docx

... follows: K hi = 1 x i +1 − x i  1 1 11  (6.29) W i = p i +1 − p i T 1 x i +1 − x i  1 6 x i +1 2 + 1 6 x i +1 x i − 1 3 x i 2 , 1 3 x i +1 2 − 1 6 x i +1 x i − 1 6 x i 2  +  p i x i +1 − p i +1 x i T − x i +1 − ... h i +1 x i ) 3 (x i +1 − x i )   1 1 11  (6. 21) V i = U (x i +1 − x i ) 2  1 2 (h i +1 − h i )(x i +1 2 − x i 2 ) +(h i x i +1...

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Hydrodynamic Lubrication 2009 Part 12 docx

Hydrodynamic Lubrication 2009 Part 12 docx

... March, “A Thermohydrodynamic Analysis of Journal Bearings”, Proc. I. Mech. E., Vol. 18 1, Part 3O, 19 66 -19 67, pp. 11 7 - 12 6. 6. R.G. Woolacott, W.L. Cooke, “Thermal Aspects of Hydrodynamic Journal ... Vol. 21, No. 8, August 19 76, pp. 519 - 526. 13 . T. Suganami, A.Z. Szeri, “A Thermohydrodynamic Analysis of Journal Bearings”, Trans. ASME, Journal of Lubrication Technology, V...

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Hydrodynamic Lubrication 2009 Part 2 ppsx

Hydrodynamic Lubrication 2009 Part 2 ppsx

... hydrodynamic lubrication was first clarified experimentally by British railroad engineer Beauchamp Tower (18 45 – 19 04) in 18 83 [1] [2]. Based on Tower’s experiments, Osborn Reynolds (18 42 – 19 12), the ... 0 (2 .13 ) 20 2 Foundations of Hydrodynamic Lubrication Fig. 2.9a-c. Inclined pad and journal bearings (1) . a inclined pad bearing, b journal bearing, c simplified journal bear...

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Hydrodynamic Lubrication 2009 Part 4 pps

Hydrodynamic Lubrication 2009 Part 4 pps

... a 0 + a 1 p i +1, j + a 2 p i 1, j + a 3 p i, j +1 + a 4 p i, j 1 (3.72) (i = 1, 2, ···, m − 1; j = 1, 2, ···, n − 1) where p i, j is the pressure at the nodal point (i, j) and a 0 , a 1 , a 2 , ... be: P =  h 2 h 1 pdx=  xp  h 2 h 1 −  h 2 h 1 x dp dx dx = −  h 2 h 1 x dp dx dx (4 .14 ) = 6µU  h 2 h 1 x  1 h 2 − h m h 3  dx = 6µUB 2 h 2 2  h 2 h 1 ¯x ⎛ ⎜ ⎜ ⎜...

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Hydrodynamic Lubrication 2009 Part 5 potx

Hydrodynamic Lubrication 2009 Part 5 potx

... A 1 , ···will be as follows: A 0 = B 0 1 ω 2 ω 1 4  P 1 mc  2 A 1 = B 1 1 ωω 1 4  P 1 mc  2 − B 2 1 ωω 1 2  P 1 mc  A 2 = 2B 0 1 ω 2 ω 1 2  P 1 mc  2 + B 3 1 ω 1 4  P 1 mc  2 − B 4 1 ω 1 2  P 1 mc  + ... B 4 1 ω 1 2  P 1 mc  + 1 A 3 = 2B 1 1 ωω 1 2  P 1 mc  2 − B 2 1 ω  P 1 mc  A 4 = B 0 1 ω 2  P 1 mc  2 + 2B 3...

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Hydrodynamic Lubrication 2009 Part 6 pot

Hydrodynamic Lubrication 2009 Part 6 pot

... considering nonlinearity in X 1 and X 2 of second order only, as: p = p 0 + p 1 X 1 + p 2 X 2 + p 3 X 3 + p 4 X 4 + p 11 X 1 2 + p 12 X 1 X 2 + p 22 X 2 2 + p 13 X 1 X 3 + p 14 X 1 X 4 + p 23 X 2 X 3 + ... µ (mm) (kg) (kg·mm) (kg·mm) (rpm) (rpm) (mm) (mm) (mm) (mPa·s) A 10 00 10 10 5 18 93 3383 15 30 0.06 20 B 680 10 10 5 3376 5445 15 30 0.06 20 l, shaft length;...

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Hydrodynamic Lubrication 2009 Part 7 pot

Hydrodynamic Lubrication 2009 Part 7 pot

... an Unstable Domain 10 1 M dX 3 dt = − d 11 X 1 − d 12 X 2 − d 13 X 3 − d 14 X 4 − d 11 1 X 2 1 − d 11 2 X 1 X 2 − d 12 2 X 2 2 − d 11 3 X 1 X 3 − d 11 4 X 1 X 4 − d 12 3 X 2 X 3 − d 12 4 X 2 X 4 (5.94) M dX 4 dt = ... − d 11 X 1 − d 12 X 2 − d 13 X 3 − d 14 X 4 (5.99) M dX 4 dt = − d 21 X 1 − d 22 X 2 − d 23 X 3 − d 24 X 4 (5 .10 0) where d 11 , d 12...

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Hydrodynamic Lubrication 2009 Part 9 pot

Hydrodynamic Lubrication 2009 Part 9 pot

... follows.  r i+ 1 2  h i+ 1 2  3 p i +1 − p i ∆r i = 12 µ i  k =1  r k+ 1 2   h k+ 1 2  −  h  k+ 1 2  ∆t ∆r i (7.44) where h i+ 1 2 = 1 2 (h i (t) + h i +1 (t)) h  i+ 1 2 = 1 2  h i (t − ∆t) + h i +1 (t ... pressure be p 1 and the flow velocity be  1 just inside the gap between the disks, then Bernoulli’s equation p 1 + 1 2 ρ 1 2 = 0 + 0 (7 .13 )...

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