CONTROL VALVE HANDBOOK Episode 1 Part 5 ppt

... 14 .12 50 2 254 286 10 .00 11 . 25 292 3 75 11 .50 14 . 75 318 400 12 .50 15 . 75 65 2 1/ 2 292 311 11 .50 12 . 25 292 3 75 11 .50 14 . 75 318 400 12 .50 15 . 75 80 3 318 337 12 .50 13 . 25 318 460 12 .50 18 .12 3 81 498 15 .00 ... 292 10 . 75 11 .50 308 318 12 .12 12 .50 25 1 273 292 10 . 75 11 .50 273 292 10 . 75 11 .50 308 318 12 .12 12 .50...

Ngày tải lên: 07/08/2014, 12:22

... section 5. 8 .5. Figure 5 - 15 Conformance to limits of size for a width-type feature Figure 5 -16 Size limit boundaries control circularity at each cross section 5- 24 Chapter Five 5. 6.2 .1 Modifier ... Rule #1 mandates a ∅ .50 2 boundary of perfect form at MMC (envelope) Figure 5- 21 Rule #1 specifies a boundary of perfect form at MMC Geometric Dimensioning and Tolerancing...

Ngày tải lên: 21/07/2014, 15:20

... Positioning 19 - 8 19 .11 Backup Roll Design 19 - 12 19 .12 Automatic Control 19 - 12 19 .13 Deckling 19 - 14 Air Entrapment behind Deckling 19 .14 Die Cleanup 19 - 14 19 .1 Introduction ... Regulation? 17 .1 Introduction 17 - 1 17.2 Code of Federal Regulations 17 - 1 17.3 Title 29 (Labor) 17 - 1 17.4 Protection 17 - 2 17...

Ngày tải lên: 21/07/2014, 15:20

... fluid—superheated steam c. Service conditions— w = 12 5, 000 lb/h P 1 = 50 0 psig = 51 4 .7 psia P 2 = 250 psig = 264.7 psia DP = 250 psi x = DP/P 1 = 250 / 51 4 .7 = 0.49 T 1 = 50 0_F g 1 = 1. 0434 lb/ft 3 (from Proper- ties ... appropriate equation. C v + q N 1 F P P 1 *P 2 Ǹ G f + 800 ( 1. 0 )( 0.90 ) 25 0 .5 Ǹ + 12 5. 7 Chapter 5. Control Valve Selection 11...

Ngày tải lên: 07/08/2014, 12:22

... T D , LBFSIN. Composition Bearings 60 Degrees 70 Degrees 2 3 4 6 8 1/ 2 3/4 3/4 1 1 -1/ 4 0 . 15 0 .10 0 .10 1. 80 1. 80 80 280 380 50 0 750 0 .11 0 . 15 1. 10 1. 10 3.80 0.60 3.80 18 .0 36.0 60.0 51 5 212 0 212 0 414 0 9820 10 12 14 16 18 20 1- 1/4 1- 1/2 1- 3/4 2 2 -1/ 8 2 -1/ 2 1. 80 4.00 42 60 60 97 12 50 3000 2400 2800 2800 52 00 3.80 11 .0 75 10 5...

Ngày tải lên: 07/08/2014, 12:22