Fluid Flow at Small Reynolds Number: Numerical Applications by Tayfour El-Bashir

By Tayfour El-Bashir

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Similarly non zero values of the moment requires a ln(r) term. Obviously the presence of such terms violates the boundary condition that f = 0 as r → ∞, and it is only when all such terms are absent that one will achieve the required solution. This is why at Re = 0 it is impossible to obtain a convergent solution for an arbitrary value of β, unlike the situation at Re = 0. However, for Re = 0 one can think of the lift, drag and moment as functions of β and only when these quantities acquire the value of zero will β achieve its required value, namely β = c, and produce a convergent solution.

31) 2π r[r 0 where r, Ψ and ω are all non-dimensional. 33) reduce to those given by Tang (1990), except for a reversal of the sign in CL and the omission of a factor 2 in the definition of both CL and CD . CHAPTER 2. 34) becomes, in dimensional form, on substituting the appropriate constitutive equation M = −μ 2π r2 [ 0 1 ∂2Ψ ∂2ψ 1 ∂ψ − 2 − ] dϑ. 36) results in the non dimensional expression CM = 2 Re 2π r2 [ω + 0 2 ∂ψ 2 ∂ 2Ψ ] dϑ. 38) and the boundary conditions on the cylinder reduce this expression to CM = 2 Re 2π 0 [ω − 2ω0 ] dϑ.

This matching process leads to the obvious linkage between the coefficients of the terms in Oseen’s expansion in the outer region, closely related to Filon’s expansion, with those from the in the inner region. Hence, the appropriate value of A can be obtained, as already indicated from the solution in the outer region. Any similar term in a Stokes expansion, such as a Brln(r) cos(ϑ) term in the stream function expansion, has its coefficient similarly related to the lift. Solutions of the Stokes equation represented by the form f (r) sin(ϑ) or g(r) sin(ϑ) fail to produce any moment contribution, but it can easily be seen that the solution of ∇4 Ψ = 0, which has a non-zero moment arises from the ln(r/a) term.

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