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
... (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
... 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
... 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 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
... 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
... 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
... 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
... 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
... 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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