Hydrodynamic Lubrication 2009 Part 4 pps

... [ 24] :  r = 1 2µ ∂p ∂r z(z − h) + ρrω 2 µ  hz 4 − z 2 2 + z 3 3h − z 4 12h 2  (4. 37) G c = ρrω 2 µ h 3 40 (4. 38) If G c = 0 is assumed in Eq. 4. 34, Reynolds’ equation ignoring the centrifugal force ... exit clearance h 0 of the pad (see Fig. 4. 7). ¯r = r − R ∆R , ¯ θ = θ θ 0 , ¯ h = h h 0 , ¯p = ph 0 2 ωµR 2 (4. 43) 4. 1 Infinitely Long Plane Pad Bearings 49 4. 1.1 Basic...

Ngày tải lên: 11/08/2014, 08:21

... is 2 Foundations of Hydrodynamic Lubrication The essence of hydrodynamic lubrication was first clarified experimentally by British railroad engineer Beauchamp Tower (1 845 – 19 04) in 1883 [1][2]. ... lubrication. Recently developed theories of elastohy- drodynamic lubrication, thermohydrodynamic lubrication, turbulent hydrodynamic lubrication, and others are regarded as extension...

Ngày tải lên: 11/08/2014, 08:21

... Basic Equations for Thermohydrodynamic Lubrication Hydrodynamic lubrication that takes heat generation and temperature rise into con- sideration is called thermohydrodynamic lubrication, orTHL. To ... 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...

Ngày tải lên: 11/08/2014, 08:21

... 43 9 - 44 4. 30. T. Kato and Y. Hori, “Taylor Vortices in a Journal Bearing” (in Japanese), Trans. JSME, Vol. 49 , No. 44 5, September 1983, pp. 1510 - 1520. 31. T. Kato and Y. Hori, “Turbulent Lubrication ... Lubrication Theory Using the Frictional Law (First Report, Derivation of Turbulent Coefficient and Lubrication Equation)” (in Japanese), Trans. JSME, Vol. 44 , No. 382, June 19...

Ngày tải lên: 11/08/2014, 08:21

... [ 24] :  r = 1 2µ ∂p ∂r z(z − h) + ρrω 2 µ  hz 4 − z 2 2 + z 3 3h − z 4 12h 2  (4. 37) G c = ρrω 2 µ h 3 40 (4. 38) If G c = 0 is assumed in Eq. 4. 34, Reynolds’ equation ignoring the centrifugal force ... µU  h 2 h 1  3h m h 2 − 4 h  dx (4. 21) = µUB h 2  h 2 h 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 3 h m h 2 − 4 h ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4. 22) In the case of an infinitely long plane pad, substituting...

Ngày tải lên: 11/08/2014, 08:21

... 1982. VIII Contents 4 Fundamentals of Thrust Bearings 47 4. 1 InfinitelyLongPlanePadBearings 48 4. 1.1 BasicFormulae 49 4. 1.2 Basic Characteristics . 49 4. 2 Finite Length Plane Pad Bearings 54 4.3 Sector ... Bearings 55 4. 3.1 Reynolds’ Equation in Cylindrical Coordinates . 55 4. 3.2 Numerical Solution of a Sector Pad 57 4. 4 Additional Topics . . 58 4. 4.1 Influence of Deforma...

Ngày tải lên: 11/08/2014, 08:21

... Sommerfeld’s condition. From Eq. 3 .42 and Eq. 3 .43 , the bearing load P 1 is obtained as: P 1 = µUL  R c  2 6κ {4 2 + π 2 (1 − κ 2 )} 1/2 (2 + κ 2 )(1 − κ 2 ) . (3 .44 ) This can be rewritten as follows ... : 1 S = 6πκ {4 2 + π 2 (1 − κ 2 )} 1/2 (2 + κ 2 )(1 − κ 2 ) (3 .45 ) where S = (µN/p m )(R/c) 2 as before, with N = U/(2πR) and p m = P 1 /(2RL). Further, dividing Eq. 3 .43 by Eq. 3...

Ngày tải lên: 11/08/2014, 08:21

... (π 2 − 4) κ 0 2 (5.38) C xx = − 2(2 + κ 0 2 ) κ 0  1 − κ 0 2  π 2 − (π 2 − 4) κ 0 2  π 2 − 8 π(2 + κ 0 2 )  (5.39) C xy =+ 4  π 2 − (π 2 − 4) κ 0 2 (5 .40 ) K yx =+ π  1 − κ 0 2  π 2 − (π 2 − 4) κ 0 2  1 κ 0 − 2κ 0 2 ... jour- nal center (x j , y j ) from the equation of motion Eqs. 5 .47 and 5 .48 . Thus, substitution of Eqs. 5 .49 and 5.50 into Eqs. 5 .47 and 5 .48 , respectiv...

Ngày tải lên: 11/08/2014, 08:21

... 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 + p 24 X 2 X 4 (5. 84) where X 3 = dX 1 dt , X 4 = dX 2 dt (5.85) In the above ... and p 1 , p 2 , ···, p 24 may be referred to as dynamic pressure coefficients. These coefficients are defined as: p 1 = ∂p ∂X 1  0 , p 2 = ∂p ∂X 2  0 , ··· p 24 = ∂ 2 p ∂X...

Ngày tải lên: 11/08/2014, 08:21

... 36.51 d ij j = 1 = 2 = 3 = 4 i = 1 3 .42 55 1.2681 5.9031 -2.1 143 = 2 -2 .44 68 0.9712 -2.1 143 2.5770 = 11 = 12 = 22 = 13 = 14 = 23 = 24 23.2595 -0. 147 8 -8.9769 -9.7937 1.99 34 -3.0892 -2 .43 69 -7.7506 -0.8730 ... d 13 X 3 − d 14 X 4 − d 111 X 2 1 − d 112 X 1 X 2 − d 122 X 2 2 − d 113 X 1 X 3 − d 1 14 X 1 X 4 − d 123 X 2 X 3 − d 1 24 X 2 X 4 (5. 94) M dX 4 dt = − d 21...

Ngày tải lên: 11/08/2014, 08:21