Thank you very much Leon! Best regards
Citando Leon Thurner <[email protected]>: > That is not correct, the impedance has to be in series, not in parallel. The > admittance are indeed in parallel, but remember that a parallel connection > of admittances is calculated as Y = Y1 + Y2. The equivalents are then > simply: > > > > Z = Z1 + Z2 (series impedances) > > Y = Ysh1 + Ysh2 (parallel admittance) > > > > The resulting line won’t be exactly the same, since you are moving some of > the admittance from the middle of the line to the end, but pretty close. > > > > Best regards > > Leon > > > > Von: [email protected] > [mailto:[email protected]] Im Auftrag von Saranya A > Gesendet: Donnerstag, 6. Juli 2017 06:25 > An: MATPOWER discussion forum <[email protected]> > Betreff: Re: Series branches PI model equivalent > > > > The single branch equivalent will be: > > Z= Z1 || Z2 and Ysh= Ysh1|| Ysh2 > > > > On Thu, Jul 6, 2017 at 3:59 AM, André Silva <[email protected] > <mailto:[email protected]> > wrote: > > Hi everyone, > > > Can anyone please clarify this for me? > > Lets suppose that we have 2 line represented by their equivalent PI model. > > We would have something like (everything already in per unit): > > Line1, characterized by Z1 and Ysh1 > > Line2, characterized by Z2 and Ysh2 > > Z1 and Z2 are the series impedances of each line. > > Ysh1 and Ysh2 are shunt admittances of each line (corresponding to > MATPOWER's 'b' parameter - branch charging, not yet divided by half!). > > If one wishes to obtain a single branch which is eletrically equivalent to > both branches parameters, what should one do? > > First add Z1 and Z2, right? And then calculate the parallel between Ysh1 > and Ysh2? > > Would this logic be correct for obtaining the same Power Flow results for > the two branches or the single equivalent branch scenarios? > > > Thank you! > > > Best Regards > > > > ---------------------------------------------------------------- This message was sent using IMP, the Internet Messaging Program.
