Intro to Practical Fluid Flow Episode 10 pps

... test TiO 2 slurry. 17.3% in batch test Extended Wilhelm-Naide equation 10 1 10 2 10 –7 3456789 10 –2 10 –6 10 –4 10 –5 10 –3 10 –1 Figure 6.8 Measured settling velocities for unflocculated slurries. ... point of inflection = 109 .5kg/m 3 Figure 6.18 Simulated batch settling curve for a compressible sediment generated using the FLUIDS toolbox 188 Introduction to Practical...

Ngày tải lên: 13/08/2014, 16:21

... (angle)  1   (p=180) rad liter L 1 L  10 À3 m 3 metric ton t or tonne 1 t  100 0 kg bar bar 1 bar  0:1 Mpa  100 kPa  10 5 Pa 4 Introduction to Practical Fluid Flow //SYS21///INTEGRAS/B&H/IPF/FINAL_13-09-02/0750648856-CH000-PRELIMS.3D ... Data King, R.P. Introduction to practical fluid flow 1 Fluid dynamics I Title 620.1 H 064 Library of Congress Cataloguing...

Ngày tải lên: 13/08/2014, 16:21

... data Drag coefficient from the Karamanev equation 10 –2 10 –1 10 1 10 1 10 2 10 2 10 0 10 0 10 3 10 3 10 4 10 5 10 7 10 6 10 8 10 10 10 9 10 11 10 12 10 13 Figure 3.3 Drag coefficient for solid spheres ... from equation 3.8 Dimensionless group = /Φ 1pD Re C 10 –2 10 –1 10 1 10 1 10 2 10 2 10 0 10 0 10 3 10 3 10 4 10 5 10 7 10...

Ngày tải lên: 13/08/2014, 16:21

... average velocity to be known. The use of Q* for practical problem solving is described in the following illustrative example. 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 12 10 -3 10 -2 Dimensionless flowrate ... number Re Friction factor f Friction factor for Newtonian fluids 10 –3 10 3 10 4 10 5 10 6 10 7 e/D = 0.0500 e/D = 0.0250 e/D = 0. 0100 e/D = 0...

Ngày tải lên: 13/08/2014, 16:21

... 10 À3 m 3 =s h Pu  10: 15 m "  g f h Pu Q P  9:81 Â 100 0 Â 10: 15 Â 8:8325 Â 10 À3 2:983 Â 10 3  0:295 38 Introduction to Practical Fluid Flow //SYS21///INTEGRAS/B&H/IPF/FINAL_13-09-02/0750648856-CH02.3D ... multiply by  ÁP Pu ÀW Â 100 0  3:751 MPa h Pu  ÁP Pu g  382:4 m of water To convert to kW, multiply by mass flowrate Power  3:751 10 3 Â 6:5 Â...

Ngày tải lên: 13/08/2014, 16:21

... is referred to the FLUIDS 74 Introduction to Practical Fluid Flow //SYS21///INTEGRAS/B&H/IPF/FINAL_13-09-02/0750648856-CH04.3D ± 82 ± [81±116/36] 23.9.2002 4:56PM which is added to the pressure ... particle and fluid m/s.  p Volume of particle m 3 . V Ã T Dimensionless terminal settling velocity.  f Viscosity of fluid Pa s.  f Density of fluid kg/m 3 . 10 0 10 1...

Ngày tải lên: 13/08/2014, 16:21

... by C i  Solid flowrate type i Total flowrate   " V w À  Ti q i " V 4:70 The two important types of problem are solved as follows. 110 Introduction to Practical Fluid Flow //SYS21///INTEGRAS/B&H/IPF/FINAL_13-09-02/0750648856-CH04.3D ... the FLUIDS toolbox as shown in the 102 Introduction to Practical Fluid Flow //SYS21///INTEGRAS/B&H/IPF/FINAL_13-0...

Ngày tải lên: 13/08/2014, 16:21

... known Dimensionless flowrate * Q Friction factor f Friction factor for Bingham plastics Hedstrom number 10 –1 10 –2 10 –3 10 3 10 4 10 4 10 5 10 5 10 6 10 6 10 7 10 7 10 8 10 9 10 10 10 1 1 10 3 10 8 Figure ... having toknow the slurry velocity or flowrate. Reynolds number Re B 10 0 Friction factor f sl 10 –1 10 –2 10 –3 10 3 10 4 10 5 10 6...

Ngày tải lên: 13/08/2014, 16:21

... suspension He=4.7 10 4 Thorium oxide suspension He=7.1 10 3 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 Laminar-turbulent transition 10 0 10 1– 10 2– 10 3– 10 2 10 3 10 4 10 5 10 6 10 7 Reynolds number ... factor H = 0.1 – 0.2 H –= 0.3 0.6 H –= 1.0 1.5 H –= 2.0 3.0 H –= 4.0 8.0 H –= 8.0 10. 0 10 1 10 3 10 2 10 5 10 4 10 6 10 –3 10 –1 10 –...

Ngày tải lên: 13/08/2014, 16:21

... 650(C/700) + 4x10 (C/700) + 10 (C/700) vV 1.58 4 5.02 5 9.3 T Calculated settling velocity. / = 1 + 40.6(C /100 0) + 960(C /100 0) vV T 0.341 2.73 10 1 10 2 10 3 10 1 10 0 10 –1 10 –2 10 –3 10 –4 10 –5 Settling ... Â 10 À3  2 23 1=3 0:05  97:67 1 10 100 100 0 Dimensionless pipe diameter * D Slurry friction factor 10 –3 10 –2 10 –1 10 0 H = 0.15 0.45 1.25 2...

Ngày tải lên: 13/08/2014, 16:21