Engineering Tribology 2E Episode 9 potx
... Institute for Technology and Mechanical Engineering, Moscow, 194 9. 2 H.M. Martin, Lubrication of Gear Teeth, Engineering, London, Vol. 102, 191 6, pp. 1 19- 121. 3 D. Dowson and G.R. Higginson, Elastohydrodynamic ... Journal of Lubrication Technology, Vol. 99 , 197 7, pp. 485-487. 13 F.T. Barwell, Bearing Systems, Principles and Practice, Oxford University Press, 197 9. 14 Engineering S...
Ngày tải lên: 13/08/2014, 16:21
Engineering Tribology 2E Episode 3 potx
... NLGI Spokesman, Vol. 19, 195 5, pp. 26-27 and 30-31. 31 N.A. Scarlett, Use of Grease in Rolling Bearings, Proc. Inst. Mech. Engrs., Vol. 182, Pt. 3A, 196 7- 196 8, pp. 585- 593 . 32 A.G. Papay, Oil-Soluble ... Application, Lubrication Engineering, Vol. 39, 198 3, pp. 4 19- 426. 33 P. Cann, H.A. Spikes and A. Cameron, Thick Film Formation by Zinc Dialkyldithiophosphates, ASLE Transactions, V...
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... combustion chamber. ( # Rolls-Royce Limited.) 390 Gas Turbine Engineering Handbook //INTEGRA/B&H/GTE/FINAL (26-10-01)/CHAPTER 10.3D ± 399 ± [370±408/ 39] 29. 10.2001 4:02PM The DLE injector has two ... Larson-Miller parameter T temperature, R t rupture time, hr 1200 1600 2000 2600 195 0 196 0 197 0 198 0 199 0 2000 2010 YEAR U 500 RENE 77 IN 733 GTD111 GTD 111 DS GTD 111 SC GTD...
Ngày tải lên: 13/08/2014, 09:20
Engineering Tribology 2E Episode 1 pot
... lubrication in gas bearings 1 89 4 .9 Reynolds equation for squeeze films 191 Pressure distribution 192 Load capacity 193 Squeeze time 194 Cavitation and squeeze films 195 Microscopic squeeze film ... the following expression: α = 1.216+4.143×(log 10 υ 0 ) 3.0627 +2.848×10 −4 ×b 5. 190 3 (log 10 υ 0 ) 1. 597 6 −3 .99 9× (log 10 υ 0 ) 3. 097 5 ρ 0.1162 (2.12) where: α is the pressure-v...
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Engineering Tribology 2E Episode 2 ppsx
... 1st World Congress in Bioengineering, San Diego, Vol. II, 199 0, pp. 2 69. 21 Z. Rymuza, Tribology of Miniature Systems, Elsevier, Tribology Series 13, 198 9. 40 ENGINEERING TRIBOLOGY Oil sample 300 ... Vol. 36, 192 9, pp. 618-6 19. 2 G.H.B. Davis, G.M. Lapeyrouse and E.W. Dean, Applying Viscosity Index to Solution of Lubricating Problems, Journal of Oil and Gas, Vol. 30, 193 2, p...
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Engineering Tribology 2E Episode 4 pptx
... 2) − () 6UηB dx ⌠ ⌡ 0 B = L W (4. 49) Again there are two variables in (4. 49) , ‘x’ and ‘h’, and one has to be replaced by the other before the integration can be performed. Substituting (4. 39) for ‘dx’, 152 ENGINEERING TRIBOLOGY Introducing ... extent of the pressure field Constants W* θ 1 θ 2 sec Cθ 90 ° 90 ° 1.1781 0 0 0° 90 ° 1.1781 0 0.020041 −27.05° 90 ° 1.1228...
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Engineering Tribology 2E Episode 5 pptx
... 0.881 0. 496 -0. 690 -0.348 - - - 7 W [N] 1.7575 ×10 4 -0. 793 2.033 2. 596 1.042 0.884 -1.510 - - 1.108 - 8 H [W] 2. 791 5× 10 3 -0.5 79 1.530 0.873 2.500 1.642 -0.225 - - 9 ε 1.0516× 10 2 0. 399 -1.040 ... (Part 1), JSLE Transactions, Vol. 33, 198 8, pp. 790 - 797 . 14 W. Lewicki, Theory of Hydrodynamic Lubrication in Parallel Sliding, Engineer, Vol. 200, 195 5, pp. 93 9 -94 1. 15 D.D....
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Engineering Tribology 2E Episode 6 pdf
... Reynolds equation and was introduced by Vogelpohl [1] in the 193 0's. The Vogelpohl parameter ‘M v ’ is defined as follows: 244 ENGINEERING TRIBOLOGY 1.5 2.5 0 0.1 Groove len g th/axial bearin g ... be parallel to ‘y*’ then φ = 0 and when the front is parallel to ‘x*’ then φ = π/2. In 202 ENGINEERING TRIBOLOGY extends the generality of a numerical solution. A basic disadvant...
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Engineering Tribology 2E Episode 7 doc
... Design, London, 196 7, Proc. Inst. Mech. Engrs., Vol. 182, Pt. 3A, 196 7- 196 8, pp. 100-115. 9 R. Bassani and B. Piccigallo, Hydrostatic Lubrication: Theory and Practice, Elsevier, Amsterdam, 199 2. 10 J.K. ... of Thermohydrodynamic Lubrication, Tribology Transactions, Vol. 32, 198 9, pp. 346-356. 9 C.M.Mc. Ettles, Transient Thermoelastic Effects in Fluid Film Bearings, Wear, Vol. 79,...
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Engineering Tribology 2E Episode 8 ppt
... × 10 11 () 1/3 = 5. 799 × 10 −5 [m] · Maximum and Average Contact Pressures p max = 3W 2πa 2 = 3 × 5 2π(5. 799 × 10 −5 ) 2 = 7 09. 9 [MPa] p average = W πa 2 = 5 π(5. 799 × 10 −5 ) 2 = 473.3 ... = 1.3380k = 1.03 39 R y R x () 0.03 0.02 () 0.636 = 1.03 39 0.636 · Simplified Elliptical Integrals = 1. 398 2ε = 1.0003 + 0. 596 8R x R y = 1.0003 + 0. 596 8 × 0.02 0.03 = 1.77 19 = 1....
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