Thanks for your response. Here are my equations:
eqpk:
[image: \frac{\partial P_k}{\partial t} = D \frac{\partial^2 P_k}{\partial
x^2}]
with IC and BCs:
[image: P_k (x,0) = Pi]
[image: P_k(0,t) = P_d(t)]
[image: P_k(L,t) = P_u(t)]
eqpf:
∂Pf/∂t = C1 ∂2 Pf/∂x2 - C2 (Pf-Pk)
with IC and BCs:
[image: P_f (x,0) = Pi]
[image: P_f(0,t) = P_d(t)]
[image: P_f(L,t) = P_u(t)]
eqpd:
[image: - \frac{\partial P_d}{\partial t} = C_3 (P_f - P_k)+C_4 (P_k-P_d)]
with IC:
[image: P_d (0) = Pi]
eqpu:
[image: \frac{\partial P_u}{\partial t} = C_3 (P_f - P_u)+C_4 (P_k-P_u)]
with IC:
[image: P_u (0) = Pi]
So [image: P_k] and [image: P_f] are functions of space and time while [image:
P_d] and [image: P_u] are only function of time. How should i define a
variable which is only a function of time? If I want to have schematic of
problem, it is as below.
The pressure in tanks [image: P_u] and [image: P_d] are uniform and only
changes with time.
I appreciate it,
On Mon, May 9, 2016 at 9:43 AM, Guyer, Jonathan E. Dr. (Fed) <
jonathan.gu...@nist.gov> wrote:
> By declaring Pd and Pu as CellVariables, you are declaring them to be
> functions of space. That's what a CellVariable is.
>
> Furthermore, eqpd and eqpu depend on Pf and Pk, so that necessitates that
> Pd and Pu are functions of space.
>
> Can you show us the math for the equations you are trying to solve?
>
> > On May 7, 2016, at 1:12 PM, Mohammad Kazemi <kazem...@gmail.com> wrote:
> >
> > Thank you. It is running now, but I have a question. Pf and Pk are
> functions of location (x) and time (t). However, Pd and Pu are only
> functions of time (they represent pressure at downstream and upstream
> tanks). When I solve this problem, I will get a vector for Pd and Pu (at
> last timestep) which varies with location. For example, for Pu,
> >
> > print Pu
> >
> > [ 7171802.05029459 7171719.44929532 7171632.68237858 7171538.95796536
> > 7171435.46595385 7171319.37283423 7171187.81897787 7171037.91833804
> > 7170866.76056175 7170671.41527712]
> >
> > Which shows the Pu values at 10 different mesh blocks. Is it related to
> how I define my cell variables?
> >
> > I appreciate your helps.
> >
> > Bests,
> >
> > On Fri, May 6, 2016 at 5:21 PM, Daniel Wheeler <
> daniel.wheel...@gmail.com> wrote:
> > The following runs for me with a few changes.
> >
> > On Fri, May 6, 2016 at 1:25 PM, Mohammad Kazemi <kazem...@gmail.com>
> wrote:
> > >
> > > Pk.constrain([Pd], mesh.facesLeft)
> > > Pk.constrain([Pu], mesh.facesRight)
> >
> > Pk.constrain(Pd.faceValue, mesh.facesLeft)
> > Pk.constrain(Pu.faceValue, mesh.facesRight)
> >
> > > Pf.constrain([Pd], mesh.facesLeft)
> > > Pf.constrain([Pu], mesh.facesRight)
> >
> > Pf.constrain(Pd.faceValue, mesh.facesLeft)
> > Pf.constrain(Pu.faceValue, mesh.facesRight)
> >
> > This makes sense since the Pf and Pk are scalars and [Pd] and [Pu] are
> > effectively vectors. Also the face value is required since the
> > constraint is on the faces.
> >
> > I hope this helps.
> >
> > --
> > Daniel Wheeler
> > _______________________________________________
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> > fipy@nist.gov
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> >
> >
> >
> > --
> > Mohammad Kazemi
> > West Virginia University
> > Department of Petroleum and Natural Gas Engineering
> > _______________________________________________
> > fipy mailing list
> > fipy@nist.gov
> > http://www.ctcms.nist.gov/fipy
> > [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
>
>
> _______________________________________________
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>
--
Mohammad Kazemi
West Virginia University
Department of Petroleum and Natural Gas Engineering
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