Maintenance Fundamentals Episode 2 part 9 pps

Maintenance Fundamentals Episode 2 part 9 pps

Maintenance Fundamentals Episode 2 part 9 pps

... properly. Table 17 .2 Common Failure Modes of Rotary-Type, Positive-Displacement Pumps Source: Integrated Systems, Inc. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:12pm page 3 62 3 62 Maintenance ... Traps Source: Integrated Systems, Inc. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 7:39pm page 3 72 3 72 Maintenance Fundamentals Table 17...

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Maintenance Fundamentals Episode 2 part 4 ppsx

Maintenance Fundamentals Episode 2 part 4 ppsx

... In part, the increased level of vibration is caused by the impact as each piston reaches top dead center and Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 7:42pm page 26 2 26 2 Maintenance ... mechanical wear or Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 5:56pm page 27 2 27 2 Maintenance Fundamentals flow generated when the ball opening is...

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Maintenance Fundamentals Episode 2 part 6 ppsx

Maintenance Fundamentals Episode 2 part 6 ppsx

... Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:02pm page 29 9 29 9 Table 15.3 Common Failure Modes of Centrifugal Fans Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:02pm ... Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 5:58pm page 29 8 29 8 Maintenance Fundamentals 0.4 0.6 0.8 1.0 1 .2 1.4 1.6 1.8 CFM, THOUSANDS STATIC PRESSUR...

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Maintenance Fundamentals Episode 2 part 7 ppsx

Maintenance Fundamentals Episode 2 part 7 ppsx

... vendor Figure 17.1 Centrifugal pump. Figure 17 .2 Single and double volute. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:12pm page 3 32 3 32 Maintenance Fundamentals 16 DUST COLLECTORS The ... finer particles) following eddy current Figure 16 .2 Flow pattern through a typical cyclone separator. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:...

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Maintenance Fundamentals Episode 2 part 8 pps

Maintenance Fundamentals Episode 2 part 8 pps

... fixed volume of fluid for each Figure 17 .21 Lobe-type pump. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:12pm page 3 52 3 52 Maintenance Fundamentals Protection Positive-displacement ... by the characteristic curve for the pump. Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6:12pm page 3 42 3 42 Maintenance Fundamentals Impellers Impellers...

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Maintenance Fundamentals Episode 2 part 10 ppsx

Maintenance Fundamentals Episode 2 part 10 ppsx

... the material with a standard force (usually Keith Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6 :28 pm page 3 92 3 92 Maintenance Fundamentals 16. Breakdown Cost Component (period) Total breakdown ... Mobley /Maintenance Fundamentals Final Proof 15.6 .20 04 6 :28 pm page 391 Glossary 391 Other findings are that planned maintenance increases the technical life of t...

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Machinery Components Maintenance And Repair Episode 2 Part 9 pps

Machinery Components Maintenance And Repair Episode 2 Part 9 pps

... disassemble rotors, naturally the parts should be carefully marked as taken apart so that identical parts can be replaced in the proper 516 Machinery Component Maintenance and Repair 8. No chrome ... Component Maintenance and Repair * Source: Koppers Company, Inc., Power Transmission Division, Baltimore, Maryland 21 203. Reprinted by permission. Abrasive Wear Abrasive wear occurs when ha...

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Dimensioning and Tolerancing Handbook Episode 2 Part 9 pps

Dimensioning and Tolerancing Handbook Episode 2 Part 9 pps

... Model: ( ) ( ) ( ) 2 13 2 12 2 11 2 eScSaST ASM δδδ ++= ( ) ( )( )( ) ( )( )( ) ) 790 9 32( )4500 82( 2 ) 790 9 32( 2 2 222 627 221 018 696 45008 646 92 0 14 92 2 7 790 93 627 22 )0004)(548310()646 92 ( )017453( ./. ./ da ... 0.0005 0.00 12 0.600-0 .99 9 0.046 821 58 0.5654 92 0.0006 0.0015 1.000-1. 499 0.0 420 49 92 0.6 021 191 0.0008 0.0 02 1.500 -2. 799 0.048 096 84 0....

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Engineering Mechanics - Statics Episode 2 Part 9 ppsx

Engineering Mechanics - Statics Episode 2 Part 9 ppsx

... F H w 2h a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = tan θ max () wa 2F H = cos θ max () 2F H 4 F H 2 wa() 2 + = T max F H cos θ max () = F H 2 wa() 2 4 += wa 2 a 2 16h 2 1+= Guess h 1 m= Given T max wa 2 a 2 16h 2 1+= h Find h()= h 7. 09 m= Problem ... y−()= Σ M = 0; Mwy y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wa a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wby− 0= My() wby 1 2 wy 2 − 1 2 wb 2 − 1 2...

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