Hydrodynamic Lubrication 2011 Part 1 pptx

Hydrodynamic Lubrication 2011 Part 1 pptx

Hydrodynamic Lubrication 2011 Part 1 pptx

... 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 ... Friction,Wear,andLubrication—Tribology 1 1.2 VariousFormsofLubrication 2 1. 2 .1 SolidFriction 4 1. 2.2 Hydrodynamic Lubrication . . . 6 1. 3 Meanings of Trib...

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Hydrodynamic Lubrication 2011 Part 5 pptx

Hydrodynamic Lubrication 2011 Part 5 pptx

... 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 2011 Part 7 pptx

Hydrodynamic Lubrication 2011 Part 7 pptx

... 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 − ... gradient: dp i (x) dx = R i · P (6 .12 ) where R i is: R i =  0, 0, ···, 1 x i +1 − x i , 1 x i +1 − x i , ···, 0, 0  (6 .13 ) 12 2 6 Fo...

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Recent Advances in Biomedical Engineering 2011 Part 1 pptx

Recent Advances in Biomedical Engineering 2011 Part 1 pptx

... interpolation, using the generic expression (Sörnmo & Laguna, 2006): 0 1 2 1 0 1 , 0 1 , 2 1 ( ) , , ; ( ), 1, , 1; ( ) ( ), 1, , 1. K K K n n n n K K x n n n n f n n n n x n f n n n n          ... matrix (Figure 10 ). This leads us to a two-dimensional real random field, x s [n,m], where: n = 0, 1, …, N 1; m = 0, 1, …, M 1; s   ; and   1, 2,  ...

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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 2011 Part 2 doc

Hydrodynamic Lubrication 2011 Part 2 doc

... Theory of Hydrodynamic Lubrication 21 1. If U 1 = U and U j = U,RHS= 6µ  0 + 2U ∂h ∂x  = 12 µU ∂h ∂x 2. If U 1 = 0 and U j = U, RHS = 6µ  −U ∂h ∂x + 2U ∂h ∂x  = 6µU ∂h ∂x 3. If U 1 = −U and ... follows: u = U 1 , w = W 1 at y = 0 u = U 2 , w = W 2 at y = h  (2.9) Then the fluid velocities will be as follows: u = − 1 2µ ∂p ∂x y(h − y) +  1 − y h  U 1 + y h U 2  (2...

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Hydrodynamic Lubrication 2011 Part 3 pps

Hydrodynamic Lubrication 2011 Part 3 pps

... side are expanded into partial fractions as follows: cos φ (1 + κ cos φ) 2 = 1 κ  1 (1 + κ cos φ) − 1 (1 + κ cos φ) 2  cos φ (1 + κ cos φ) 3 = 1 κ  1 (1 + κ cos φ) 2 − 1 (1 + κ cos φ) 3  Integrating ... 45 ∂ ∂x  h 3 ∂p ∂x  = h 3 i +1/ 2, j p i +1, j − p i, j ∆x − h 3 i 1/ 2, j p i, j − p i 1, j ∆x ∆x (3.69) ∂ ∂z  h 3 ∂p ∂z  = h 3 i, j +1/ 2 p i, j +1 − p i, j ∆z −...

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Hydrodynamic Lubrication 2011 Part 4 ppsx

Hydrodynamic Lubrication 2011 Part 4 ppsx

... φ 1 + κ cos φ dφ = − 1 κ ln 1 + κ cos φ 1 + κ (5 .18 ) J 2 s (φ) =  φ 0 sin φ (1 + κ cos φ) 2 dφ = 1 1 + κ 1 − cos φ 1 + κ cos φ (5 .19 ) J 3 s (φ) =  φ 0 sin φ (1 + κ cos φ) 3 dφ = 1 2κ  1 (1 ... =  φ 0 dφ (1 + κ cos φ) 2 = 1 1 − κ 2  J 1 (φ) − κ sin φ 1 + κ cos φ  (5 .13 ) J 3 (φ) =  φ 0 dφ (α + κ cos φ) 3 = 1 2 (1 − κ 2 ) 2 ×  (2 + κ 2 )J 1 (φ) −...

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Hydrodynamic Lubrication 2011 Part 8 pps

Hydrodynamic Lubrication 2011 Part 8 pps

... 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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Hydrodynamic Lubrication 2011 Part 9 pot

Hydrodynamic Lubrication 2011 Part 9 pot

... b 2 (s) three-element 80.5 1. 68 10 −2 0.0 2.40 10 −2 0.0 four-element 80.5 2.63 10 −2 0.0 3.63 10 −2 6.94 10 −5 five-element 85 .1 3.46 10 −2 6.22 10 −5 4 .19 10 −2 1. 30 10 −4 Coefficients of the constitutive ... 16 - 17 , 19 79, Suita, pp. 16 - 18 . 12 . Y. Hori, T. Kato and H. Narumiya, “Rubber Surface Squeeze Film”, Trans. ASME, Jour- nal of Lubrication Technology, Vol....

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