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<p style="margin-top:0;margin-bottom:0">Dear Zhai,</p>
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<p style="margin-top:0;margin-bottom:0">ffx will have no impact on the flow if you set p54=-1, p55=1.</p>
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<p style="margin-top:0;margin-bottom:0">The code fixes the flow rate by adding an auxiliary solution,</p>
<p style="margin-top:0;margin-bottom:0">(U0,p0) such that <U+a*U0> = Ubar, where a is a constant</p>
<p style="margin-top:0;margin-bottom:0">determined by the equation the requirement that <U+a*U0>=Ubar.</p>
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<p style="margin-top:0;margin-bottom:0">The velocity pressure pair (U0,p0) is the solution to the linear</p>
<p style="margin-top:0;margin-bottom:0">unsteady Stokes problem:</p>
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<p style="margin-top:0;margin-bottom:0"> (1/dt) U0 - 1/Re \nabla^2 U0 = -grad p0 + F0</p>
<p style="margin-top:0;margin-bottom:0"> div U0 = 0</p>
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<p style="margin-top:0;margin-bottom:0">where F0=[1 , 0 , 0 ] is a unit forcing in the x direction.</p>
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<p style="margin-top:0;margin-bottom:0">If you add your own ffx, then the net forcing will be:</p>
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<p style="margin-top:0;margin-bottom:0"> ffx + a*1</p>
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<p style="margin-top:0;margin-bottom:0">If you don't, it will be "a*1 = a".</p>
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<p style="margin-top:0;margin-bottom:0">Note that if you add your own forcing, then "a" will be smaller</p>
<p style="margin-top:0;margin-bottom:0">because the preliminary solution, U, will be closer to the target</p>
<p style="margin-top:0;margin-bottom:0">Ubar value, assuming your ffx value is reasonably close to the</p>
<p style="margin-top:0;margin-bottom:0">actual mean pressure drop in your system.</p>
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<p style="margin-top:0;margin-bottom:0">Note that because these equations are _linear_ it really doesn't</p>
<p style="margin-top:0;margin-bottom:0">matter how you satisfy the condition <U+a*U0> = Ubar.</p>
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<p style="margin-top:0;margin-bottom:0">Also, note that, in the above, I've been using "U" to represent the</p>
<p style="margin-top:0;margin-bottom:0">solution computed by Nek5000 _prior_ to the addition of the auxiliary solution U0.</p>
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<p style="margin-top:0;margin-bottom:0">The real solution that is returned at the end of the tilmestep is:</p>
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<p style="margin-top:0;margin-bottom:0"> U := U + a*U0</p>
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<p style="margin-top:0;margin-bottom:0"> p := p + a*p0</p>
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<p style="margin-top:0;margin-bottom:0">(Of course, like U, p and p0 are periodic --- so, the "true" pressure is p + a linear term corresponding to the mean pressure drop...)</p>
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<p style="margin-top:0;margin-bottom:0">The reason I prefer p54/p55 to adjusting ffx via a feedback loop is that the feedback loop introduces its own timescale -- i.e., the response time that you find in any feedback system. Since we already have multiple
timescales in a turbulent flow, I prefer to not add an additional one that is not well characterized.</p>
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<p style="margin-top:0;margin-bottom:0">hth,</p>
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<p style="margin-top:0;margin-bottom:0">Paul</p>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" style="font-size:11pt" color="#000000"><b>From:</b> Nek5000-users <nek5000-users-bounces@lists.mcs.anl.gov> on behalf of nek5000-users@lists.mcs.anl.gov <nek5000-users@lists.mcs.anl.gov><br>
<b>Sent:</b> Monday, March 26, 2018 10:39:52 AM<br>
<b>To:</b> nek5000-users@lists.mcs.anl.gov<br>
<b>Subject:</b> [Nek5000-users] Further to ffx, ffy and ffz</font>
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<div dir="ltr">Dear Paul and Phillip,
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<div>Thanks for your reply. </div>
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<div>Paul, I am sorry say that I made an mistake for my question last time. </div>
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<div>I intended to ask if I set </div>
<div>P54=-1</div>
<div>P55=1</div>
<div>ffx: non-zero (x is streamwise)</div>
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<div>Will the fluid be changed by the effect of non-zero ffx which I am thinking can be considered as the effect of the particles acting on the fluid while I am doing the lagrangian paticle tracking.</div>
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<div>By the way, can I fix the pressure gradient as (Re_t/Re_b)^2 for the pipe as well?</div>
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<div>Kind regards,</div>
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<div>Zhai</div>
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