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Research Papers: Fundamental Issues and Canonical Flows

Slip Flow in Eccentric Annuli

[+] Author and Article Information
R. S. Alassar

Department of Mathematics and Statistics,
King Fahd University of Petroleum and Minerals,
Dhahran 31261, Saudi Arabia

Contributed by the Fluids Engineering Division of ASME for publication in the JOURNAL OF FLUIDS ENGINEERING. Manuscript received May 13, 2016; final manuscript received October 10, 2016; published online January 20, 2017. Assoc. Editor: Daniel Maynes.

J. Fluids Eng 139(4), 041201 (Jan 20, 2017) (7 pages) Paper No: FE-16-1304; doi: 10.1115/1.4035115 History: Received May 13, 2016; Revised October 10, 2016

A solution of the problem of Poiseuille slip flow through an eccentric cylindrical annulus is obtained in bipolar coordinates. The solution is in excellent agreement with the two published limiting cases of slip flow through concentric annuli and no-slip flow through eccentric annuli. It is shown that for a fixed aspect ratio, fully eccentric channels sustain the maximum average velocity (flow rate) under the same pressure gradient and slip conditions. For a given channel geometry, the average velocity varies linearly with Knudsen number except for small aspect ratio. It is also shown that the extrema of the friction factor Reynolds number product is determined by how this product is defined or scaled.

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Figures

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Fig. 1

Problem configuration

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Fig. 2

Dimensionless velocity distribution for r1/r2=1/2, H/r2=1/4, (a) Kn=0, (b) Kn=0.01, and (c) Kn=0.1

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Fig. 3

Dimensionless velocity at θ=0,  and  π for r1/r2=1/4, (a) H/r2=0, (b) H/r2=1/2. Curves from top to bottom: Kn=0.1,   0.09,   0.08,   ⋯  ,   0.0 .

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Fig. 4

Average dimensionless velocity for r1/r2=1/2

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Fig. 5

Average dimensionless velocity for (a) H/r2=1/4 and (b) H/r2=0

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Fig. 6

Average dimensionless velocity for (a) H/r2=1/2 and (b) H/r2=0

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Fig. 7

fRe for r1/r2=1/2, (a) based on Dh and (b) based on A

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Fig. 8

fReA for (a) H/r2=1/4 and (b) H/r2=0

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Fig. 9

fReDh for (a) H/r2=1/4 and (b) H/r2=0

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Fig. 10

σ for r1/r2=1/2, Kn=0.05

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Fig. 11

σ for r1/r2=1/2, e=0.5

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Fig. 12

σ¯ for r1/r2=1/2

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Fig. 13

σ¯ for H/r2=1/4

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