Engineering Tribology 2011 Part 8 pptx
Engineering Tribology 2011 Part 8 pptx
... 10 −3 1 −30 × 10 −3 ) 1 = 0.2 = 16.67 83 .33 From Figure 7.12, for k 0 = 0.2: k 1 = 1.17, k 2 = 0 .88 and k 3 = 1. 98 and from Figure 7.13 where k 2 /k 1 = 0 .88 /1.17 = 0.75, the other constants ... 1.17 3 × 50 × 0.012 2.3 08 × 10 11 () 1/3 = 2.32 × 10 −4 [m] b = k 2 3WR' E' () 1/3 = 0 .88 3 × 50 × 0.012 2.3 08 × 10 11 () 1/3 = 1.75 × 10 −4 [m] 310 ENGINEERING TRIBO...
Ngày tải lên: 12/08/2014, 03:20
Manufacturing Handbook of Best Practices 2011 Part 8 pptx
... result FIGURE 8. 10 Process-based layout. Lathe Lathe Drill Drill Mill Mill Cell B Cell A Shared X-ray SL3003Ch08Frame Page 187 Friday, November 9, 2001 1:52 PM © 2002 by CRC Press LLC 184 The Manufacturing ... justify on the SL3003Ch08Frame Page 182 Friday, November 9, 2001 1:52 PM © 2002 by CRC Press LLC 202 The Manufacturing Handbook of Best Practices APPENDIX 8. 6 APPENDIX 8. 7 When I...
Ngày tải lên: 11/08/2014, 14:20
Vorticity and Vortex Dynamics 2011 Part 8 pptx
... while f(η)=e −η [ln η +ei(η) −0 .80 7] + 2ei(η) −2ei(2η). For sufficiently large z, the leading terms of (8. 32) is W − w ∼ Γ 2 8 z ln Wz ν e −η . (8. 33) The solutions (8. 30), (8. 31a), and (8. 32) have been independent ... (η)= ∞ η (1 −e −ξ ) 2 ξ 2 dξ (8. 31b) = (1 −e −η ) 2 η + 2ei(η) − 2ei(2η), (8. 31c) where ei(η)= ∞ η (e −ξ /ξ)dξ is the exponential integral. Then, substituting (...
Ngày tải lên: 11/08/2014, 17:22
Thumb for Mechanical Engineers 2011 Part 8 pptx
... 1 58 304 1 48 286 1 40 271 133 537 483 03 4 38 81 6 402 7 48 370 689 343 639 320 594 299 555 280 521 263 490 242 462 2 28 437 214 41 1 201 385 189 363 1 78 616 ... 143 274 130 249 1 18 2 28 109 21 1 101 197 95 184 88 173 83 163 78 154 74 146 70 1 38 67 437 38 1 OD 336 626 302 561 273 5 08 249 462 2...
Ngày tải lên: 11/08/2014, 21:21