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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> 
> 
> 
> -- 
> Mohammad Kazemi
> West Virginia University
> Department of Petroleum and Natural Gas Engineering
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