Engineering Tribology 2E Episode 5 pptx
Engineering Tribology 2E Episode 5 pptx
... [N] 1. 757 5 ×10 4 -0.793 2.033 2 .59 6 1.042 0.884 -1 .51 0 - - 1.108 - 8 H [W] 2.79 15 10 3 -0 .57 9 1 .53 0 0.873 2 .50 0 1.642 -0.2 25 - - 9 ε 1. 051 6× 10 2 0.399 -1.040 1.372 -0 .53 9 -0. 458 0.7 65 - - 10 T max ... 10 1 υ 37.8°C −1.1 υ 93.3°C 2.46 L 2 .51 5 D 0 .56 3 N 0 .52 8 c −1.09 T S −0.383 () ε 1 − ε 2 1.3 85 or row 7: W = 1. 757 5 × 10 4 υ 37.8°C −0.793 υ 93.3°C 2.033 L 2...
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Engineering Tribology 2E Episode 4 pptx
... described below. HYDRODYNAMIC LUBRICATION 153 0 0.1 0.2 0.3 0.4 0 .5 0.6 0.7 0.8 0.9 1.0 0.01 0.02 0.03 0.04 0. 05 0.06 0.08 0.1 0.2 0.3 0.4 0 .5 0.6 0.8 1 2 3 4 5 6 8 10 20 30 40 50 60 80 100 L D = ∞ = 1 1 2 = 1 4 = 1 8 = 0.02 0.03 0.04 0. 05 0.06 0.08 0.1 0.2 0.3 0.4 0 .5 0.6 0.8 1 2 3 4 5 6 8 10 20 0.004 0.0 05 0.006 0.008 0.01 1 8 Ocvirk W/L Uη ∆ ... K) (4 .57 ) Substitut...
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Engineering Tribology 2E Episode 10 pptx
... concentrations reaching 5% [49]. 0 50 100 150 10 100 1000 10 4 10 5 10 6 Number of cycles to disruption Slip amplitude [µm] Vacuum evaporation Sputtering 100V bias 200V bias Ion Plating FIGURE 9. 15 Comparison ... December, 1987, pp. 52 -58 . 123 M.A. Keller and C.S. Saba, Catalytic Degradation of a Perfluoroalkylether in a Thermogravimetric Analyzer, Tribology Transactions, 19...
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Engineering Mathematics 4 Episode 5 pptx
... of mid-ordinate 15 ° 10 sin 15 ° D 2 .58 8 V 45 ° 10 sin 45 ° D 7.071 V 75 ° 10 sin 75 ° D 9. 659 V 1 05 ° 10 sin 1 05 ° D 9. 659 V 1 35 ° 10 sin 1 35 ° D 7.071 V 1 65 ° 10 sin 1 65 ° D 2 .58 8 V Sum of mid-ordinates ... speed/time is shown in Fig. 20.3. 30 25 Graph of speed/time 20 15 Speed (m/s) 10 5 0 12 3 Time (seconds) 45 6 2 .5 4.0 7.0 15. 0 5. 5 8. 75 10. 75...
Ngày tải lên: 13/08/2014, 08:21
Engineering Tribology 2E Episode 1 pot
... temperature of interest in [g/cm 3 ]. 1000 50 0 200 100 50 20 10 5 2 1 [ 10 Pa ] × -9 -1 C 0123 Viscosity [cP]η 0 0.001 0.002 0.0 05 0.01 0.02 0. 05 0.1 0.2 0 .5 1 10243 852 65 79 93 107 121 149 177 204 Viscosity ... 1983, pp. 53 -56 respectively. Figures 14.2 and 15. 2: Royal Society of London. From Proceedings of the Royal Society of London, Vol. 394, 1984, pp. 161-181 and Vol....
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Engineering Tribology 2E Episode 2 ppsx
... 000 2 50 0 3 000 3 50 0 4 000 35 40 45 50 60 70 80 90 2 2 .5 3 3 .5 4 4 .5 5 6 7 8 9 10 15 20 25 30 35 40 45 50 60 70 80 90 1.9 1.8 1.7 1.6 1 .5 1.4 1.3 1.2 25 30 35 40 45 50 60 70 80 90 100 150 200 250 300 350 400 450 30 35 40 45 50 60 70 80 90 100 150 200 250 300 350 400 100 120 FIGURE ... cS 100 150 200 250 300 350 400 450 50 0 600 700 800...
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Engineering Tribology 2E Episode 3 potx
... classification [28]. 000 4 45 - 4 75 00 400 - 430 0 355 - 3 85 1 310 - 340 2 2 65 - 2 95 3 220 - 250 4 1 75 - 2 05 5 130 - 160 6 85 - 1 15 NLGI grade Worked (60 strokes) penetration ... soap -40 Bentonite clay -30 Di-ester Lithium soap - 75 Di-ester Bentonite clay -55 Silicone Lithium soap -55 Dye - 75 Silica -50 Mineral oil Mineral oil...
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Engineering Tribology 2E Episode 6 pdf
... equations (5. 37), (5. 38) and (5. 39), i.e.: v = z η dz ⌠ ⌡ 0 h ⌠ ⌡ 0 h dz η dp dy z η ⌠ ⌡ z 0 dz − dz η ⌠ ⌡ 0 z ()( ) (5. 45) Substituting for ‘M’ equation (5. 45) simplifies to: v = M dp dy (5. 46) An ... defined as follows: 244 ENGINEERING TRIBOLOGY 1 .5 2 .5 0 0.1 Groove len g th/axial bearin g len g th Dimensionless load 2 0.2 0.3 0.4 0 .5 0.6 0.7 0.8 0.9 1 0...
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Engineering Tribology 2E Episode 7 doc
... ‘critical frequency’ as it is often called: A 1 A 3 A 5 2 (A 1 2 + A 2 A 5 2 − A 1 A 4 A 5 )(A 5 +γA 1 ) ω* c 2 = (5. 104) where: A 1 ,A 2 ,A 5 are the dimensionless stiffness and damping products; ω c * is ... surfaces: if the centre of curvature 254 ENGINEERING TRIBOLOGY 1 3 4 0 0.1 0.2 0.3 0.4 0 .5 0.6 Misalignment parameter t Dimensionless critical frequency 2 FIGURE...
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Engineering Tribology 2E Episode 8 ppt
... 100 200 50 0 1000 2000 50 00 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 2 × 10 6 5 × 10 6 10 7 200 50 0 1000 2000 50 00 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 100 50 20 10 Piezoviscous-elastic Piezoviscous-rigid Lubrication ... 0. 25 µW U A − U B Ka 0.1 < L < 5 T f a = 0 .5 NL = α π 4 or qa K T f a = 0. 25 µW U A − U B Ka α ranges...
Ngày tải lên: 13/08/2014, 16:21