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 <[email protected]> wrote: > The following runs for me with a few changes. > > On Fri, May 6, 2016 at 1:25 PM, Mohammad Kazemi <[email protected]> > 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 > _______________________________________________ > fipy mailing list > [email protected] > http://www.ctcms.nist.gov/fipy > [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] > -- Mohammad Kazemi West Virginia University Department of Petroleum and Natural Gas Engineering
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