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Old   August 2, 2007, 05:10
Default CFD: Using Taylor series
  #1
charlie ryan
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Hi

I've been looking at a fluid dynamics paper and have a specific question. Here's the link to the paper; http://arxiv.org/PS_cache/physics/pd.../0110081v1.pdf "Drop Formation in a One-Dimensional Approximation of the Navier-Stokes Equation" It involves fluid falling from a pipe, and breaking up into drops due to surface tension. It starts with general equations [N-S, continuity, BC's], and creates a 1D computational model. So it is CFD, just pretty imple computationally! What im really interested in is their simplication of the navier stokes equations, continuity etc, by the use of a power series.

Looking at the paper, i don't understand how they got equations 9, 10 and 11. Looking at Eq. 9, it seems that it is a power series. Now i understand that a power series looks like; f(x)=a + bx + cx^2 + dx^3... So in equation 9, where is v_11*r?! They say by symmetry we get equation 9 - what does this mean? Equation 11 seems to follow the same pattern - again, why? And where in earth do they get equation 10 from - what do they differentiate to get it? I understand that this is quite a specific question, but pls, any help is greatly appreciated! If the link doesn't work please tell me!

Thanks Charlie

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Old   August 2, 2007, 05:35
Default Re: CFD: Using Taylor series
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Richard
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I don't know where equations 9 and 11 come from, though presumably if they are right they have the same origin. But equation 10 comes from substituting equation 9 into equation 3.
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Old   August 2, 2007, 08:49
Default Re: CFD: Using Taylor series
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Praveen. C
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I dont know the full answer but

first assume

v_z(z,r) = v_0 + v_1*r + ...

Note that

v_0(z) = v_z(z,r=0)

v_1(z) = d(v_z)/dr(z,r=0)

Now v_z has axisymmetry, just draw z-axis downwards, r-axis horizontal think how the profile of v_z should look like near r=0. Then you will realize that v_z must have a smooth extremum at r=0, otherwise it will not have slope continuity at r=0. Hence v_1=0.
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Old   August 2, 2007, 09:11
Default Re: CFD: Using Taylor series
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ag
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It looks to me like the key assumption in establishing the form of (9) is not symmetry but rather the relative sizes in the two dimensions r and z. They specifically state that they are going to focus on very thin columns relative to the elongation, indicating that gradients in r will be much larger than those in z. Then symmetry is used (as noted by Praveen) to further reduce the series. This is just a first impression based on a very cursory look at the article.
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Old   August 3, 2007, 04:56
Default Re: CFD: Using Taylor series
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charlie ryan
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Hi

Thanks for your help - very enlightening! Richard i think your right - to get 10 you do just substitute 9 into Eq. 3. So basically they get rid of v_1 by saying that the gradient at at r=0 is zero - i.e. the velocity profile is smooth across the r axis. One other Q - would there be r^3 terms in Eq. (9) and (11)? Thanks very much for your help

Charlie

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