michael 2007-9-4 11:13
How does FlexPDE fill holes?
[size=2]Hello, I had a question about how FlexPDE is doing things. e)St^n(eJniM
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Say I have a 2-d tube (like in the sample problem "viscous") with a hole in the middle of it. I can solve the stokes equations with flow from, say, left to right. There is Poiseuille flow at the left and right ends, viscous flow that passes around the middle hole, and solid boundaries on the top and bottom. I run FlexPDE to get the flow for this system. I export u, v, and p to a data file. I then comment out the sphere so that I should have just Poiseuille flow through the newly vacant tube. I import the data file to get what I call u0, v0, and p0 everywhere except where the hole is. Yet it seems (from a little investigation of my own) that u0, v0, and p0 are interpolated in the once empty hole region. What sort of interpolation is being used to find the values of u0, v0, and p0 where they are undefined (undefined according to the data file)? 2] F6}:k#P h$zV
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I've attached two files, a data file, and a pde file. To see what I'm talking about, you can comment and uncomment the "exclude" portion of the code (the hole). [/size]*i%y8z7{ k
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[code] { VISCOUS.PDE }m4k6Eu3F["X ~|
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This example shows the application of FlexPDE to problems in
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viscous flow.._B.~$F;Q6x
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The Navier-Stokes equation for steady incompressible flow in two
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cartesian dimensions is
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dens*(dt(U) + U*dx(U) + V*dy(U)) = visc*del2(U) - dx(P) + dens*Fx
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dens*(dt(V) + U*dx(V) + V*dy(V)) = visc*del2(V) - dy(P) + dens*Fy
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together with the continuity equationph,\,H gd0zV
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div[U,V] = 0
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where7a}1r'@1e'zj
U and V are the X- and Y- components of the flow velocity0[u8R(~ M4Z a)S-Z
P is the fluid pressurekQ?9r}C$O S6_
dens is the fluid density3gf(w;x b%{
visc is the fluid viscosityi4D'RRE!@e\&e
Fx and Fy are the X- and Y- components of the body force.
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In principle, the third equation enforces incompressible mass conservation,B%x h x9F3Ss y
but the equation contains no reference to the pressure, which is nominally
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the remaining variable to be determined.0d%x${ m-i_a.Q
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In practice, we choose to solve a "slightly compressible" system by defining a(w`r6P.P7~(?
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hypothetical equation of state
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p(dens) = p0 + L*(dens-dens0)a`"Rp`h D1pW
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where p0 and dens0 represent a reference density and pressure, and L is a large9fH$i!??8c6Q
number representing a strong response of pressure to changes of density. L is4X5N Q w/J0h5s'dVY-r C
chosen large enough to enforce the near-incompressibility of the fluid, yet not~6\^7r
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so large as to erase the other terms of the equation in the finite precision8S9RI1W,{m
of the computer arithmetic.d] k6co[
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The compressible form of the continuity equation isy_/j6F/?
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dt(dens) + div(dens*U) = 0 B c1JeJ
pHj+KI9d1w p
which, together with the equation of state yields
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dt(p) = -L*dens0*div(U)K IhK;Ls/\]3H u
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In steady state, we can replace the dt(p) by -div(grad(p))6oG i%z^&F(f
[see Help | Tech Notes | Smoothing Operators in PDEs"],$G)p:X4C Q:h1de
resulting in the final pressure equation: q9X,r$[ T;Zp7t
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div(grad(p)) = M*div(U)&?(QEd\Y
cp Z
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Here M has the dimensions of density/time or viscosity/distance^2.
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The problem posed here shows flow in a 2D channel blocked by a bar of square*| S/Yp9tK\6r| v4E
cross-section. We have chosen a "convenient" value of M, one that gives goodO\m0ks
accuracy in reasonable time. The user can alter this value to find one
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which is satisfactory for his application.7pcE3B I)u
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We have included three elevation plots of X-velocity, at the inlet, center/r[G/Cm4z _T/E
and outlet of the channel. The integrals presented on these plots show theEy_7~ vil
consistency of mass transport across the channel.
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************************************************************** }
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title 'Viscous flow in 2D channel, Re < 0.1'
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variables1{ ]z PvuA H
u(0.1)
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v(0.01)
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p(1)0D#sxF9M0s,B fL
(kg.m*e TM#S$M
definitions
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Lx = 5 Ly = 1.5
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Gx = 0 Gy = 0
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p0 = 1FH#j:m m.h sm
r
speed2 = u^2+v^2
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speed = sqrt(speed2)
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dens = 14q nkqr `J [
visc = 17v5X6W `)dv?
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vxx = (p0/(2*visc*Lx))*(Ly-y)^2 { open-channel x-velocity }
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rball=0.5 \q b;pi;Q#C0e
cut = 0.05 { cut the corners of the obstruction }'kzMu3oN-EK
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penalty = 100*visc/rball^2 { "equation of state" }1M0R9ye;VJ
transfer('data_in.dat',u0,v0,ps)+b!S1{Z%q!g
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initial values;k
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u = 0.5*vxx v = 0 p = p0*x/Lx$@5K L ao+x*w}
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equations
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u: div(grad(u)) - dx(p)/visc = dens*(u*dx(u) + v*dy(u))/viscZ)h
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v: div(grad(v)) - dy(p)/visc = dens*(u*dx(v) + v*dy(v))/visc
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p: div(grad(p)) = penalty*(dx(u)+dy(v))
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Boundaries
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region 1SSS p]w?/J
start(0,0)
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value(u) = 0 value(v) = 0 load(p) = 0 line to (Lx,0)UN*m*?LS
load(u) = 0 value(p) = p0 line to (Lx,Ly)
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value(u) = 0 value(v) = 0 load(p) =0 line to (0,Ly) E+x){Tt'W%MA p
r
load(u) = 0 value(p) = 0 line to finish
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start(Lx/2+rball,Ly/2)
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arc (center = Lx/2,Ly/2) angle = 360}R,`:b#oY5S
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monitorsG)A#I#W @!U~ W\6K$P\
contour(speed)
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plots k;z4? k2W{ Rw
grid(x,y)
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contour(u)Q @3^/?Z
contour(v)/E Q}8pMX8E.Y
contour(speed) painted
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contour(u0)
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contour(v0)$k h4X5C.pw
contour(sqrt(u0^2+v0^2))
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vector(u,v) as "flow"s$IO4Q[J
contour(p) as "Pressure" painted
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contour(dx(u)+dy(v)) as "Continuity Error"G;c1D[.C2]4C-fZ#V
contour(p) zoom(Lx/2,0,1,1) as "Pressure"!S/y7b-IZ N B
elevation(u) from (0,0) to (0,Ly)vm:IS
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elevation(u) from (Lx/2,0) to (Lx/2,Ly)
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elevation(u) from (Lx,0) to (Lx,Ly)(FC@!C0B"uX
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transfer(u,v,p) file = "data_in.dat"vz
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end 236852924
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FlexPDE Version 5.0.17 20:02:48 Jun 23 2007
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viscous 2.pde 00:29:06 7/8/07
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Title: uBbD
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FlexPDE v500 Transfer File7Hk ZFO:Yh
Viscous flow in 2D channel, Re < 0.1ayOrR(HKf a Ge
2 DIMENSIONS ll][ ?b"n
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[/code]
michael 2007-9-4 11:13
Whenever you import a TRANSFER, even if it is for a domain identical to the exported one, there are numerical roundoff conditions that can make border nodes appear to be outside the imported mesh. In order to handle these events cleanly, FlexPDE uses the following procedure:
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If a lookup point is outside the lookup mesh, find a point on the border of the lookup mesh that is close to the lookup point, and return that value.
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When you have a transfer mesh with a hole in it, and ask to look up values in the hole, you are essentially asking "what is the value at a point where I have no data?". There is no reasonable answer to this question.
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A
You must either provide data in the hole (by including the hole in the exporting problem and forcing the flow velocity to zero), or provide a switch in the importing problem so that you don't try to look up values where there are none.