Engineering Tribology Episode 2 Part 11 doc

Engineering Tribology Episode 2 Part 11 doc

Engineering Tribology Episode 2 Part 11 doc

... Elsevier, 20 00. 45 G.W. Stachowiak and P. Podsiadlo, Surface Characterization of Wear Particles, Wear, Vol. 22 5 -22 9, 1999, pp. 117 1 -118 5. 46 P. Podsiadlo and G.W. Stachowiak, 3-D Imaging of Wear Particles ... body, three-body abrasive and erosive wear [109 ,110 ,113 ]. This is illustrated in Figure 11. 15 where TEAM LRN 496 ENGINEERING TRIBOLOGY 123 1 1 23 4 2 3 4 Very br...

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Engineering Tribology Episode 2 Part 1 doc

Engineering Tribology Episode 2 Part 1 doc

... parameter ‘G’ of the Vogelpohl equation, i.e.: = h* 1.5 ∂ 2 M v ∂x* 2 + ( R L ) 2 ∂ 2 M v ∂y* 2 = FM v + G FM v + ∂h* ∂x* + 2w* (5.100) Damping coefficients are computed in a similar manner ... from downstream of upstream groove TEAM LRN 24 4 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...

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Engineering Tribology Episode 2 Part 4 docx

Engineering Tribology Episode 2 Part 4 docx

... shown in Figure 7 .20 [7]. 100 20 0 500 1000 20 00 5000 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 20 0 500 1000 20 00 5000 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 ... 0.6 023 ln R y R x () = 1. 527 7 + 0.6 023 ln 0.03 0. 02 () · Contact Area Dimensions = 6 × 1.338...

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Engineering Tribology Episode 2 Part 8 docx

Engineering Tribology Episode 2 Part 8 docx

... Vapor Phase Lubrication of Oleic Acid and TCP, Wear , Vol. 21 4, 1998, pp. 20 7 -21 1. TEAM LRN 4 02 ENGINEERING TRIBOLOGY 0 0.1 0 .2 0.3 µ 0 100 20 0 300 Temperature [°C] FIGURE 8.54 Experimental friction ... London, 1963, pp. 70-80. 37 A. Dyson, Scuffing, A Review, Tribology International, Vol. 8, 1975, Part 1: pp. 77-87, Part 2: pp. 117 - 122 . 38 H. Blok, Les Temperature...

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Longman Toeic Intermediate Episode 2 Part 11 doc

Longman Toeic Intermediate Episode 2 Part 11 doc

... 42 195 165 43 20 0 170 44 21 0 175 45 21 5 180 46 22 0 190 47 23 0 195 48 24 0 20 0 49 24 5 21 0 50 25 0 21 5 # CORRECT LISTENING READING 51 25 5 22 0 52 26 0 22 5 53 27 0 23 0 54 27 5 ... 21 80 35 22 85 40 23 90 45 24 95 50 25 100 60 26 110 65 27 115 70 28 120 80 29 125 85 30 130 90 31 135 95 32 140 100 33 145 110 34 150 115 35...

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Engineering Tribology Episode 1 Part 4 doc

Engineering Tribology Episode 1 Part 4 doc

... stability [°C] 135 21 0 23 0 24 0 25 0 23 0 28 0 430 430 370 370 Kinematic viscosity [cSt] at -20 °C 170 193 16 85 115  20 0 850 1000 0°C 75 75 16 38 47 100 25 0 25 00 8000 440 40°C ... 180 185 20 0 26 0 24 0 29 0 none none Pour point [°C] -57 -60 - 62 -57 -65 -70 -70 -7 +4 -30 -67 Oxidative stability [°C] 24 0 29 0...

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Engineering Tribology Episode 1 Part 5 doc

Engineering Tribology Episode 1 Part 5 doc

... sulfonate 0 0.05 0.10 0.15 0 .20  0 .25  0.30 0.35 0 40 80 120  160 20 0 24 0 20  60 100 140 180 22 0 Coefficient of friction Tem p erature [°C] FIGURE 3 .24 Effect of dispersants on ... grease classification [28 ]. 000 445 - 475 00 400 - 430 0 355 - 385 1 310 - 340 2 26 5 - 29 5 3 22 0 - 25 0 4 175 - 20 5 5 130 - 160 6 85 -...

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Engineering Tribology Episode 2 Part 2 pdf

Engineering Tribology Episode 2 Part 2 pdf

... i.e.: H t = H f + H p TEAM LRN 26 2 ENGINEERING TRIBOLOGY and substituting gives: T = ⌠ ⌡ 0 2 ⌠ ⌡ 0 R 0 η r 3 dθdr + 2 n h r ⌠ ⌡ 0 2 ⌠ ⌡ R 0 η r 3 dθdr 2 n h R Assuming constant viscosity ... LRN 27 4 ENGINEERING TRIBOLOGY Equating flow through the orifice to the total lubricant flow through the bearing (6.19): πd 2 2 p s − p r 2 ( ( 1 /2 C d = p r h 3 η B Rearr...

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Engineering Tribology Episode 2 Part 3 pps

Engineering Tribology Episode 2 Part 3 pps

... −υ B 2 E B =+ 1 − 0.3 2 2.1 × 10 11 [] 1 2 1 − 0.3 2 2.1 × 10 11 ⇒ E' = 2. 308 × 10 11 [Pa] · Contact Area Dimensions a = 3WR' E' () 1/3 = 3 × 5 × (3 × 10 −3 ) 2. 308 × 10 11 () 1/3 = ... p max = 3W 2 a 2 = 3 × 5 2 (6.88 × 10 −5 ) 2 = 504.4 [MPa] p average = W πa 2 = 5 π(6.88 × 10 −5 ) 2 = 336 .2 [MPa] · Maximum Deflection δ= 1.0397 W...

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