Handbook of Lubrication Episode 2 Part 1 pdf
Handbook of Lubrication Episode 2 Part 1 pdf
... Patent 2, 11 3,353, 19 38. 10 . Opitz, H. and Koenig, W., Ind. Anzieger, 87, 46 (Part I), 26 , 19 65; Ind. Anzeiger. 87, 845 (Part II); 43, 19 65; Ind. Anmger, 87, 10 33 (Part III), 51, 19 65. 11 . Sata, ... of work removed volume of work consumed 3 52 CRC Handbook of Lubrication 335-356 4 /11 /06 12 : 34 PM Page 3 52 Copyright © 19 83 CRC Press LLC Copyright © 19 8...
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Handbook of Lubrication Episode 2 Part 1 pot
... Patent 2, 11 3,353, 19 38. 10 . Opitz, H. and Koenig, W., Ind. Anzieger, 87, 46 (Part I), 26 , 19 65; Ind. Anzeiger. 87, 845 (Part II); 43, 19 65; Ind. Anmger, 87, 10 33 (Part III), 51, 19 65. 11 . Sata, ... Deterioration of Cutting Fluids, Tech. Pap. MR 72- 22 6, American Society of Mechanical Engineers, New York, 19 72. 10 . Suskind, R., Occupational skin problems, J. Oc...
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Handbook of Lubrication Episode 2 Part 9 pdf
... Lubr. Eng., 32, 16 6, 19 75. 3. Ku, P. M., Gear failure modes — importance of lubrication and mechanics, ASLE Trans., T19, 23 9, 19 76. 4. Wedeven, L. D., What is EHD?, Lubr. Eng., 31, 29 1, 19 75. 5. ... Eng., 32, 8, 19 76. 8. Standard AGMA 25 0.04, Lubrication of Industrial Enclosed Gear Drives, American Gear Manufacturers Association, Arlington, Va., 19 74. 9. Standard AG...
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Handbook of Lubrication Episode 2 Part 12 pdf
... 76, 19 78. 36. Desforges, C. D., Laser heat treatment, Tribol. Int., 11 (2) , 13 9, 19 78. 37. Elliot, T. L., Surface hardening, Tribol. Int., 11 (2) , 12 1 , 19 78. 38. ASM Committee Case hardening of ... Ohio, 19 78, 364. 41. Wolfe, G. F., Effect of surface coatings on the load-carrying capacity of steel, Lubr. Eng., 19 , 28 , 19 63. 42. Tilly, G. P., Sand erosion of me...
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Handbook of Lubrication Episode 2 Part 2 pot
... Lubr. Eng., 32, 24 9, 19 76. 12 . Bennett, E. O., The deterioration of metal cutting fluids, Prog. Ind. Microbiol., 13 , 12 1 , 19 74. 13 . Tant, C. O. and Bennett, E. O., The isolation of pathogenic ... that hold as much as 11 ,400 kg (25 ,000 lb) of grease. 3 92 CRC Handbook of Lubrication FIGURE 13 . Typical gear spray system. 379-393 4 /10 /06 4: 51 PM Page 3 92 Cop...
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Handbook of Lubrication Episode 2 Part 2 pps
... reservoir layout. 395- 411 4 /10 /06 4:54 PM Page 398 Copyright © 19 83 CRC Press LLC Viscosity Watts rating SSU cSt in. 2 cm 2 300 65 8 1 .25 300 11 60 65 25 0 6 1. 0 11 60 and above 25 0 and above 4 0.75 PUMPS The ... Bacterial., 24 , 12 5 , 19 61. 46. Bennett, E. O., The Effect of Water Hardness on the Deterioration of Cutting Fluids, Tech. Pap. No. MR 72- 226 , Society...
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Handbook of Lubrication Episode 2 Part 3 ppt
... µm. 395- 411 4 /10 /06 4:54 PM Page 4 02 Copyright © 19 83 CRC Press LLC 422 CRC Handbook of Lubrication Table 3 (continued) JOURNAL BEARING PERFORMANCE INCLUDING INERTIA AND TURBULENCE (L/D = 1. 0, ββ = 16 0°) Copyright ... which gives: ΔΤ ϭ H/[ ρ c(Q Ϫ Q s /2) ] (1) For typical petroleum oils, ρc = 1 12 lb F /in. 2 F (13 9 N/cm 2 C) and for water ρc = 327 lb F /in. 2...
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Handbook of Lubrication Episode 2 Part 3 ppt
... operating. Normally below 50 µm. 395- 411 4 /10 /06 4:54 PM Page 4 02 Copyright © 19 83 CRC Press LLC Design Principles 395- 411 4 /10 /06 4:55 PM Page 411 Copyright © 19 83 CRC Press LLC Radiator— Oil through ... which gives: ΔΤ ϭ H/[ ρ c(Q Ϫ Q s /2) ] (1) For typical petroleum oils, ρc = 1 12 lb F /in. 2 F (13 9 N/cm 2 C) and for water ρc = 327 lb F /in. 2 F (406 N/cm 2 C)...
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Handbook of Lubrication Episode 2 Part 4 ppt
... factor C W,L = 0 .1 32; from Equation 11 , using C W,T = 1 for laminar flow: W = 0 .1 32 × 1 × 4.5 × 10 –8 × 12 9 6 × (2. 75) 2 × 2. 75/(0.00 016 ) 2 = 62. 5 lb for load capacity (P = W/BL = 8 .26 psi). Similarly ... 20 0 psi) ρ = 7.77 × 10 –5 lb sec 2 /in. 4 = µ/ ρ = 2. 32 × 10 2 in. 2 /sec Calculating the Reynolds and bearing characteristic numbers: Re =...
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Handbook of Lubrication Episode 2 Part 4 pps
... Equation 1 with pc = 1 12 as typical for petroleum oils. ΔT = 1. 71 × 10 4 / [1 12 ( 4.08 – 1. 66 /2) ] = 47.0 F This temperature rise implies an inlet temperature T i of T i = T S – ΔT /2 = 13 5 – 47.0 /2 = 11 1.5 ... psi) ρ = 7.77 × 10 –5 lb sec 2 /in. 4 = µ/ ρ = 2. 32 × 10 2 in. 2 /sec Calculating the Reynolds and bearing characteristic numbers: Re = R ω C...
Ngày tải lên: 05/08/2014, 09:20