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波状栓塞流是气力输送中的一种典型流型.为了发展更为准确和细致的栓塞流数学模型,本文引入动理学理论建立气固两相波状栓塞流动的数学模型,对管道内的波状栓塞流动进行了三维数值研究,再现了流动过程中栓塞和堆积层的形成、运动过程,得到了压力降、空气速度、栓塞长度和堆积层厚度等重要的流动参数;栓塞长度、堆积层体厚度随空气表观速度的变化等一些细微流动特征得到了体现,表明动理学理论能够反映出栓塞流这种非连续高浓度气-固流动的物理本质,具有很好的适用性.本文的研究为栓塞流数值模型的发展引入了一种新的数学工具和思路.

参考文献

[1] Pan R .Material Properties and Flow Modes in Pneumatic Conveying[J].Powder Technology,1999,104(02):157-163.
[2] Tsuji Y;Tanaka T;Ishida T .Lagrangian Numericl Simulation of Plug Flow of Cohesionless Particles in a Horizongtal Pipe[J].Powder Technology,1992,71:239-250.
[3] Tomita Y;Tateishi K .Pneumatics Slug Conveying in a Horizontal Pipeline[J].Powder Technology,1997,94(03):229-233.
[4] Levy A. .Two-fluid approach for plug fow simulations in horizontal pneumatic conveying[J].Powder Technology,2000(3):263-272.
[5] Ding Jianmin;Gidspow D .A Bubbling Fluidization Model Using Kinetic Theory of Granular Flow[J].,1990,36(04):523-538.
[6] Yonghao Zhang;Jason M. Reese .Particle-gas turbulence interactions in a kinetic theory approach to granular flows[J].International Journal of Multiphase Flow,2001(11):1945-1964.
[7] Gidaspow D.Hydrodynamics of Circulating Fluidized Beds,Kinetic Theory Approach,Fluidization Ⅶ[A].,1992:75-82.
[8] Wen C Y;Yu Y H .Mechanics of Fluidization[J].Chemical Engineering Progress Symposium Series,1966,62:100-111.
[9] Ergun S .Fluid Flow Through Packed Columns[J].Chemical Engineering Progress,1952,48(02):89-94.
[10] Chapman S;Cowling T G.The Mathematical Theory of Non-Uniform Gases[M].New York:Cambridge University Press,1970
[11] 刘大有;王伯懿.非均匀气体的数学理论[M].北京:科学出版社,1985
[12] Levy A et al.A Comparison of Analytical and Numerical Models with Experimental Data for Gas-Sold Flow Through a Straight Pipe at Differet Inclinations[J].Powder Technology,1997,93(03):253-260.
[13] Tomita Y;Jotaki T;Hayashi H .Wavelike Motion of Particulate Slugs in a Horizontal Pneumatic Pipeline[J].International Journal of Multiphase Flow,1981,7(02):151-166.
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