PARTICLE-LADEN FLOW - ERCOFTAC SERIES Phần 2 docx

... 11.8 180 20 0 22 0 24 0 26 0 SOLVE ER 2 campaign 27 /01 /20 00 N 2 O (ppbv) (a) 10.6 11 11.4 11.8 180 20 0 22 0 24 0 26 0 recons. D=1 m 2 s −1 (b) 10.6 11 11.4 11.8 180 20 0 22 0 24 0 26 0 recons. D=10 −1 m 2 s −1 (c) 10.6 ... 11.4 11.8 180 20 0 22 0 24 0 26 0 recons. D=10 2 m 2 s −1 (d) 13 .2 13.6 14 14.4 80 100 120 140 160 SOLVE ER 2 campaign 27 /01 /20...

Ngày tải lên: 05/08/2014, 14:20

... turbulent-like flows 367 0 5 10 15 20 25 30 0 50 100 150 20 0 25 0 300 350 t/τ k Nc/sec St~0.14 St~0.91 St ~2. 29 St~9.17 Fig. 2. Effects of St on the encounter (collision) rates. 0 5 10 15 20 25 30 ... ESF. 623 9. References [1] Kiørboe, T. (1997) Scientia Marina 61:14 1-1 58. [2] Wang, L-P. and M.R. Maxey, (1993) J. Fluid Mech. 25 6 :2 7-6 8. [3] Fung, JCH (1990) Kinematic S...

Ngày tải lên: 05/08/2014, 14:20

... 20 0 −5 0 5 10 15 20 25 30 FA5010 −50 25 0 25 50 −5 0 5 10 15 20 25 30 FA5010 20 0 −100 0 100 20 0 −5 0 5 10 15 20 25 30 MA5010 z (mm) sand flux (mm/s) 20 0 −100 0 100 20 0 −5 0 5 10 15 20 25 30 MA5010 sand ... ‘on-shore’, maximum ‘offshore’ and total mean flux profiles for two grain sizes, i.e., 0 .28 (MA5010) and 0.15 mm (FA5010). 20 0 −100 0 100 20 0 −5 0 5 10 15 20 25...

Ngày tải lên: 05/08/2014, 14:20

... 1000 = 0 = 1 = 10 = 100 = 1000 2 10 -2 10 -1 10 0 10 1 t - t 0 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω Fig. 2. Single Particle Dispersion in horizontal ... Herring N 2 = 1 = 0 = 1 = 10 = 100 = 1000 2 10 -2 10 -1 10 0 10 1 10 -4 10 -3 10 -2 10 -1 10 0 10 1 t -...

Ngày tải lên: 05/08/2014, 14:20

... 145 x z 0 4 8 12 16 20 24 28 32 -0 .5 -0 .25 0 0 .25 0.5 (a) γ =0. 92 x z 0 4 8 12 16 20 24 28 32 -0 .5 -0 .25 0 0 .25 0.5 (b) γ =0.7 x z 0 4 8 12 16 20 24 28 32 -0 .5 -0 .25 0 0 .25 0.5 (c) γ =0 .2 Fig. 3. ... turbulence 125 -0 .1 0 0.1 0 .2 0.3 0.4 〈ω 1 3 〉/〈ω 3 2 〉 3 /2 C1 C2 C3 C4 -0 .1 0 0.1 0 .2 0.3 0.4 〈ω 1 ω 3 2 〉/〈ω 3 2...

Ngày tải lên: 05/08/2014, 14:20

... two-fluid equations of the dispersed-phase, results in: ∇·(α 2 U 2 )=0 (1) for the average mass balance and: ∇·(α 2 U 2 U 2 )=−∇ ·  α 2  u  2 u  2  − α 2  τ 2  U rel (2) for the momentum-balance. ... Berlin. [26 ] Minier J-P., Peirano E., Chibarro S. (20 03) Monte Carlo Methods and Appl., Vol.9, No. 2, pp. 9 3-1 33. [27 ] Maclnnes J. M., Bracco F. V. (19 92)...

Ngày tải lên: 05/08/2014, 14:20

... J 0 =4πR 2 K(R)dn(r)/dr| r=R ,inthe negative r-direction, towards the spherical surface. In normalized units: 22 4 Guido Boffetta 10 4 10 2 10 0 10 -2 10 -4 10 -6 10 2 10 1 10 0 10 -1 S p (τ) τ/τ η 3 2 1 10 2 10 1 10 0 Fig. ... turbulence 22 7 1 10 -2 10 -4 10 -6 10 -8 0 5 10 15 20 25 30 35 40 P(a)a rms a/a rms Fig. 4. Acceleration pdf in lin-log...

Ngày tải lên: 05/08/2014, 14:20

... 8. 100 120 160 20 0 24 0 300 1 2 3 4 6 8 10 ∆/η (1/s 2 ), ( − ) 100 120 160 20 0 24 0 300 10 20 30 50 70 100 ∆/η (1/s 2 ), ( − ) s 2 [1/s 2 ] ω 2 [1/s 2 ] s 2 ⋅t 2 * ω 2 ⋅t 2 * s 2 [1/s 2 ] ω 2 [1/s 2 ] s 2 ⋅t 2 * ω 2 ⋅t 2 * r −4/3 r −4/3 a) b) > 12 ... near-isotropic turbulence Phys. Fluids, 15(4), 86 0-8 80. 28 0 M. Guala et al. 0 500 1000...

Ngày tải lên: 05/08/2014, 14:20

... 25 δ S . 0 500 1000 1500 x -2 5 0 25 50 y 0 10 20 30 40 z a) 0 500 1000 1500 x -2 5 0 25 y 0 10 20 30 40 z b) Fig. 2. Trajectories of particles along an oscillatory cycle, released at a fixed ho- rizontal ... the −6 −4 2 0 2 4 6 10 −6 10 −4 10 2 10 0 u / <u 2 > 1 /2 Probability Density (a) −6 −4 2 0 2 4 6 10 −6 10 −4 10 2 10 0 u / <u 2 > 1 /2 Probabi...

Ngày tải lên: 05/08/2014, 14:20

... Large-Eddy Simulation, held at the Munich University of Technology, August 27 29 , 20 03. 20 04 ISBN 1-4 02 0 -2 03 2- 5 10. E. Lamballais, R. Friedrich, B.J. Geurts and O. Métais (eds.): Direct and Large-Eddy ... quality and emissions through 20 03. EPA-454-R-0 4-0 02. US Envir- onmental Protection Agency [22 ] Wiscombe WJ (1980) Improved Mie scattering algorithms. Appl Opt 19:...

Ngày tải lên: 05/08/2014, 14:20