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
