Engineering Tribology Episode 2 Part 8 docx

Engineering Tribology Episode 2 Part 8 docx

Engineering Tribology Episode 2 Part 8 docx

... Phase Lubrication of Oleic Acid and TCP, Wear , Vol. 21 4, 19 98, 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 characteristic ... Mech. Engrs. Publ., 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...

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Engineering Tribology Episode 2 Part 4 docx

Engineering Tribology Episode 2 Part 4 docx

... 0.6 023 ln R y R x () = 1. 527 7 + 0.6 023 ln 0.03 0. 02 () · Contact Area Dimensions = 6 × 1.3 380 2 × 1.39 82 × 50 × 0.0 12 π (2. 3 08 × 10 11 ) () 1/3 = 2. 32 × 10 −4 [m]a = 6k 2 εWR' πE' () 1/3 ... p max = 3W 2 ab = 3 × 50 2 (2. 32 × 10 −4 ) × (1.75 × 10 −4 ) = 588 .0 [MPa] p average = W πab = 50 π (2. 32 × 10 −4 ) × (1.75 × 10 −4 ) = 3 92. 0 [MPa] · Maximu...

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Handbook of Corrosion Engineering Episode 2 Part 8 docx

Handbook of Corrosion Engineering Episode 2 Part 8 docx

... and 10 .2] : 3 Fe 2 O 3 ϩ 4 Cl Ϫ ϩ 6 H ϩ ϩ 2 e Ϫ → 2 FeCl 2( aq) ϩ 3 H 2 O (10 .2) Fe 3 O 4 ϩ 6 Cl Ϫ ϩ 8 H ϩ ϩ 2 e Ϫ → 3 FeCl 2( aq) ϩ 4 H 2 O (10.3) 2 H ϩ 2 e Ϫ → H 2 (10.4) Fe ϩ 2 Cl Ϫ → FeCl 2( aq) ϩ ... 1.55 18. 0 0 25 0 35 0. 62 7 .2 60 1000 143 0.1 52 1.76 90 5000 22 3 0.097 1.13 94 07651 62_ Ch10_Roberge 9/1/99 6:15 Page 84 7 84 6 Chapter Ten -405 -80...

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Mechanism Design - Enumeration of Kinema Episode 2 Part 8 docx

Mechanism Design - Enumeration of Kinema Episode 2 Part 8 docx

... 8 Automotive Mechanisms 8. 1 Introduction 8 .2 Variable-Stroke Engine Mechanisms 8 .2. 1 Functional Requirements 8 .2. 2 Structural Characteristics 8 .2. 3 Enumeration of VS-Engine Mechanisms 8. 3 ... journal and © 20 01 by CRC Press LLC 8. 3.3 Enumeration of C-V Shaft Couplings 8. 4 Automatic Transmission Mechanisms 8. 4.1 Functional Requirements 8. 4 .2 Structural Charact...

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Engineering Tribology Episode 1 Part 8 pptx

Engineering Tribology Episode 1 Part 8 pptx

... −ε 2 ) 2 2ε ⌠ ⌡ 0 π dθ = (1 +εcosθ) 3 sin 2 θ 2( 1 −ε 2 ) 3 /2 π 4 L 2 () − y 2 ⌠ ⌡ − L 2 L dy = 6 L 3 2 Substituting yields: W 1 = − c 2 (1 −ε 2 ) 2 UηL 3 ε 2 (4.105) W 2 = 4c 2 (1 ... = ⌠ ⌡ 0 π dθdy Rc 2 (1 + εcosθ) 3 3UηεRsinθcosθ 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 ⌠ ⌡ 0 π dθ c 2 3Uηε 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 (1 + εcosθ) 3 sinθco...

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Engineering Tribology Episode 2 Part 1 doc

Engineering Tribology Episode 2 Part 1 doc

... equation (5 .87 ) the value of flow ‘Q’ is: Q = 0.5 × 6 .8 × 0 .2 × 10 × 0.0004 = 2. 72 × 10 -3 [m 3 /s] = 2. 72 [litres/s] TEAM LRN 24 0 ENGINEERING TRIBOLOGY Computer Program for the Analysis of Grooved ... ∂p*/∂y* can be approximated by: ∂p* ∂x* ≈ P* i+1,j − P* i,j δx* (5. 82 ) TEAM LRN 24 8 ENGINEERING TRIBOLOGY K* = Kc W (5. 92) where: K* is the non-dimensional...

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Engineering Tribology Episode 2 Part 2 pdf

Engineering Tribology Episode 2 Part 2 pdf

... of these bearings are shown in Figure 6 .2. TEAM LRN 26 8 ENGINEERING TRIBOLOGY 1 2 5 10 20  B 0 0 .2 0.4 0.6 0 .8 1.0 C/B 0.4 0.5 0.6 0.7 0 .8 0.9 1.0 A A H B C B FIGURE 6.5 ... 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 Assumin...

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