Advanced Engineering Dynamics 2011 Part 5 pptx
Advanced Engineering Dynamics 2010 Part 5 pdf
... Therefore (5. 5) Gm sin0 d0 dV = - 2r Using the cosine rule - 2Rx cos0 (5. 6) y = x2 + R~ and differentiating gives (5. 7) 2 2r dr = 2Rx sin0 d0 Substituting equation (5. 7) into ... m2(r2& + 2wi2) or (5. 15) o = 1 ~(m,r~w) = z 1 g(m2r:w) rl Substituting equation (5. 12) into equation (5. 14) gives lF121 = p2s + CLj: (5. 16) Now...
Ngày tải lên: 10/08/2014, 20:20
Computational Fluid Dynamics 2011 Part 4 pptx
... y8f w4 h16" alt="" 0 .5% xanthan gum 1 .5% xanthan gum Parameter Experiment CFD Experiment CFD f VV f VV Impeller Experiment CFD Experiment CFD Computational Fluid Dynamics 110 The difficulty ... annulling the viscosity and its effects, we get Euler’s fluid flow equation announced in 1 750 . Computational Fluid Dynamics 112 Fig. 3. Solutions of the same problem with a RAN...
Ngày tải lên: 21/06/2014, 18:20
Computational Fluid Dynamics 2011 Part 7 pptx
... comfortable. 0.0 1.0 2.0 3.0 4.0 5. 0 6.0 7.0 8.0 9.0 10.0 0123 456 v (m/s) D(vs) (a) 0 .51 5 0 .52 0 0 .52 5 0 .53 0 0 .53 5 0 .54 0 0 .54 5 0 .55 0 0 .55 5 0 .56 0 0 .56 5 0 153 0 45 a(º) D(vs) (b) Fig. 12. The ... air speed to wind speed against wind angle ranging 0° and 1 35 (CFD results for three cases) 0 50 100 150 200 250 300 350 400 450 50 0 0 20 40 60 80 100...
Ngày tải lên: 21/06/2014, 18:20
Computational Fluid Dynamics 2011 Part 11 pptx
... 6-DT, 6-PTD, 6-PDU T= 0.3–0. 15 m D = 0.1 75 0 .58 m C = 0.5T– 0.167T W/D = 0. 25 0.4 Particle density = 152 0 kg/m 3 Solid loading = 0 50 wt % Dp =100–2000 μm 0. 45 0.1 0.1 0.11 0.31 l js 1.16 gΔ fγ ... 0.1–0 .58 m Impeller type = 6-DT D = 0. 05 0.29 m C = 0.01 25- 0.07 25 m Dp = 50 50 00 μm Particle density=1049–3720 kg/m 3 Solid concentration = 0.1–2 vol. % 0.34 0.1...
Ngày tải lên: 21/06/2014, 18:20