Jovan, I agree it's not fast and efficient, as it involves inverting the admittance matrix. However, I do not see why not Klein's impedance distance could be used in power networks. I mean, the fact that some (ok, most) injections are expressed as constant power does not invalidate the fact that it's an electric circuit governed by Kirchoff laws.
Incidentally, we have sometimes used the path of greatest admittance between two given nodes as an heuristic measure of "closeness" (actually, the net impedance of such path). It all depends what you want to use these distances for. -- Jose L. Marin Gridquant España SL Grupo AIA On Tue, Feb 17, 2015 at 4:35 PM, Jovan Ilic <jovan.i...@gmail.com> wrote: > > Paul, > > I would not call calculating Zbus "fast and efficient". Also, using > resistance distance > might make sense in standard electric circuits but it does not make sense > in power > networks with constant powers. > > As far as I know there is not a very good, theoretically sound, way of > calculating electrical > distance in power systems. I would love to be corrected on this one. > > Jovan > > > > On Tue, Feb 17, 2015 at 10:20 AM, Paul Cuffe <paul.cu...@ucd.ie> wrote: > >> Hi Hans, >> >> There is indeed a fast and efficient way to calculate this, though you >> don't encounter it often in the power systems literature. >> >> You can use the Klein resistance distance, as defined here: >> http://link.springer.com/article/10.1007/BF01164627 >> >> Once you have inverted your Ybus matrix to get the Zbus, you can >> calculate the Thevenin impedance between any two nodes, i and j, as follows: >> >> >> >> Of course, the reciprocal of the Zthev impedance value will give the >> effective admittance between any two nodes. >> >> Hope this helps, >> >> Paul >> >> >> On 17/02/2015 15:06, Barrios, Hans wrote: >> >> Hello everybody, >> >> >> >> I was wondering if somebody had already the following issue: >> >> I would like to create a “full version” of the Y-matrix, i.e. a matrix >> where (as long as there is only one synchronous grid) the admittance >> between each bus is given, even if the bus are not connected directly by >> one branch. >> >> If I am not missing anything, the Admittance between each bus should be a >> simple calculation of parallel an series admittances. >> >> But I was wondering, if anyone knows a fast and efficient way I can used >> to calculate this also for big grid structures. >> >> >> >> Thank you in advance for your contributions! >> >> >> >> Best regards >> >> Hans >> >> >> >> >> >> *Hans Barrios Büchel, M.Sc.* >> >> >> >> Institut für Hochspannungstechnik / Institute for High Voltage Technology >> >> - Nachhaltige Übertragungssysteme / Sustainable Transmission Systems >> - Wissenschaftlicher Mitarbeiter / Research Assistant >> >> >> >> RWTH Aachen University >> >> Schinkelstraße 2, Raum SG 115.1 >> >> 52056 Aachen >> >> Germany >> >> >> >> Tel. +49 241 80-94959 >> >> Fax. +49 241 80-92135 >> >> >> >> Mail. barr...@ifht.rwth-aachen.de >> >> Web. www.ifht.rwth-aachen.de >> >> >> >> >> -- >> Dr. Paul Cuffe, >> Senior Researcher, >> Electricity Research Centre, >> University College Dublin. >> >> Phone: +353-1-716 1743 >> >> >
