Hydrodynamic Lubrication 2009 Part 6 pot

... (kg·mm) (rpm) (rpm) (mm) (mm) (mm) (mPa·s) A 1000 10 10 5 1893 3383 15 30 0. 06 20 B 68 0 10 10 5 33 76 5445 15 30 0. 06 20 l, shaft length; m, disk mass; I p , I d , moment of inertia; N c , critical ... 8.490 ω/ω c = 6. 0 An actual example of a generator rotor with such parameter values is shown in Table 5.2. Table 5.2. An example of generator rotors Rotor Bearing M = 44 ton D = 66 2...

Ngày tải lên: 11/08/2014, 08:21

... example, Someya (1 963 )[17] (1 964 )[19] (1 965 ) [23] performed many detailed numerical computations of this kind of problem in 5.2 Oil Whip Theory 69 1 R j 2 d dψ  h 3 dp dψ  = 6 U 1 R j dh dψ + ... follows: A 0 , A 1 , A 2 , A 3 , A 4 , A 5 , A 6 > 0 (5.58)               A 1 A 3 A 5 00 A 0 A 2 A 4 A 6 0 0 A 1 A 3 A 5 0 0 A 0 A 2 A 4 A 6 00A 1 A 3 A 5     ...

Ngày tải lên: 11/08/2014, 08:21

... Japanese), Trans. JSME, Vol. 31, No. 229, Septem- ber 1 965 , pp. 1398 - 1408. 25. A. Cameron, “Principles of Lubrication , Longman, London, 1 966 26. F.K. Orcutt and E.B. Arwas, “The Steady State and ... 1972, pp. 66 5 - 66 7. 35. S. Iida, “A Study on the Vibration Characteristics of Tilting Pad Journal Bearings” (in Japanese), Trans. JSME, Vol. 40, No. 331, March 1974, pp. 875 - 884....

Ngày tải lên: 11/08/2014, 08:21

... called the squeeze film. O. Reynolds referred to the squeeze effect in his famous paper on lubrication (18 86) and stated that it was an important mechanism, together with the wedge effect, for the ... drop ∆p will be: 7.3 Sinusoidal Squeeze by a Rigid Surface (Experiments) 147 7 8 6 5 1 9 10 11 2 3 4 12 Fig. 7 .6. Pressure change at measuring points for an intense sinusoidal squeeze [9] 7.3...

Ngày tải lên: 11/08/2014, 08:21

... follows: ∂ ∂z  µ ∂ r ∂z  = ∂p ∂r − ρ θ 2 r (8 .63 ) ∂ ∂z  µ ∂ θ ∂z  = 1 r ∂p ∂θ (8 .64 ) ∂p ∂z = 0 (8 .65 ) The second term on the right-hand side of Eq. 8 .63 is an inertia term. Integrating Eq. 8 .64 under the boundary ... integration yields: r ∂ ∂r  h 0 ρ r dz +  h 0 ρ r dz + ∂ ∂θ  h 0 ρ θ dz = 0 (8 .69 ) Substituting Eqs. 8 .66 and 8 .67 into Eq. 8 .69 yields the following Re...

Ngày tải lên: 11/08/2014, 08:21

... ρν t ∂u ∂y (9 .67 ) −ρ   w  = ρν t ∂w ∂y (9 .68 ) Under these assumptions, a turbulent lubrication equation is derived from Eq. 9 .66 . First, disregarding the left-hand side of Eq. 9 .66 from assumption ... Vol. 6, 1 963 , pp. 249 - 253. 9. Chung-Wah Ng and C.H.T. Pan, “A Linearized Turbulent Lubrication Theory”, Journal of Basic Engineering, Trans. ASME, Series D, Vol. 87, September...

Ngày tải lên: 11/08/2014, 08:21

... – d m 1,7374 0,9 364 C – d m 109,09 1,32 36 C 0 – d m 12 ,60 3 1,7 564 Cylindrical roller bearings c  d  d – d m 0,4 167 1,0 966 D – d m 1,7114 0,9481 C – d m 69 ,121 1 ,66 75 C 0 – d m 22,552 ... 0,3984x 1,1179 R 2 = 0,99 96 0 400 800 1200 160 0 0 200 400 60 0 800 1000 1200 1400 160 0 d m [mm] d [mm] D - d m y = 1,7374x 0,9 364 R 2 = 0,9998 0 400 800 1200 160 0...

Ngày tải lên: 19/06/2014, 15:20

... 1 26 6.3 Characteristics of Foil Bearings 1 26 6.3.1 Single Cylinder Heads 127 6. 3.2 Double Cylinder Heads . . 128 6. 3.3 Comparison with Experiments 130 6. 4 Additional Topics . . . . . . 130 6. 4.1 ... Rise 161 8.1 Basic Equations for Thermohydrodynamic Lubrication 162 8.2 Generalized Reynolds’ Equation . 163 8.2.1 BalanceofForces 163 8.2.2 FlowVelocity 164 8.2.3 Continuity Eq...

Ngày tải lên: 11/08/2014, 08:21

... experiments and studied them theoretically. In the introductory part of his famous paper of 18 86 [3] on hydrodynamic lubrication, Reynolds wrote: Lubrication, or the action of oils and other viscous fluids ... Sternlicht, “Theory of Hydrodynamic Lubrication , McGraw-Hill Book Company, New York, 1 961 . 5. Norimune Soda,“Bearings”(in Japanese), Iwanami Zensho Series, Iwanami Shoten,...

Ngày tải lên: 11/08/2014, 08:21

... (µN/p m )(R/c) 2 , as: S  L D  2 = (1 − κ 2 ) 2 πκ{ 16 2 + π 2 (1 − κ 2 )} 1/2 (3 .62 ) Further, from Eq. 3.59 and Eq. 3 .60 , we can write the locus of the bearing center as: tan θ = π(1 − κ 2 ) 1/2 4κ (3 .63 ) This is shown in ... = p µω  c R  2 (3 .67 ) Then, Reynolds’ equation, Eq. 3 .66 can be nondimensionalized in the following form, where U = Rω: ∂ ∂ ¯x  ¯ h 3 ∂ ¯p ∂ ¯x  +  D L...

Ngày tải lên: 11/08/2014, 08:21