Thank you very much Dr. Ray and Dirk. Regards Mirish On Mon, Feb 29, 2016 at 4:14 PM, Ray Zimmerman <[email protected]> wrote:
> Hi Mirish and Dirk, > > I believe Dirk is close, but the signs are incorrect. It is in fact the > line charging capacitance (like a any other shunt capacitor used for > voltage support) that produces reactive power (negative reactive losses) > rather than consuming it. And the series component always consumes reactive > power rather than generating it. If you look at Fig 3-1 and do the math > carefully, the series reactive losses are: > > series_Q_losses = |Vf – Vt|^2 * 1/x > shunt_Q_losses = – (|Vf|^2 + |Vt|^2) * b/2 > > The total reactive losses in the line including both components can always > be found by adding Qf + Qt. > > So, yes, the reason for the overall negative reactive losses in the second > line is that the shunt components are generating more reactive power than > the series element is consuming. As Dirk said, it’s important to realize > that reactive power is generated by network elements such as transmission > lines and shunt capacitors, not just generators. > > Ray > > > > On Feb 28, 2016, at 1:03 PM, Dirk Toewe <[email protected]> wrote: > > Hello Mirish, > > *A branch can have negative reactive power losses* > The shunt charging susceptance is *positive* so it *cannot* * produce* > reactive power but it will always *consume* reactive power: > > 0 < Q_branch_shunt_loss = b/2 * |Vf|² + b/2 * |Vt|² = b/2 * ( > (V_magnitude_f)² + (V_magnitude_t)² ) > > so it can only be the inductance that produced reactive power. And indeed: > The series element susceptance is -1/X ( since 1/j = -j ) > > 0 > Q_series_loss = -1/x * | Vf - Vt |² > > so the series element actually *generates* reactive power (*negative* > loss). The second Line has a slightly higher X in relation to B (0.91% > against 0.87%) and therefore has a net reactive power production which is > even positive. Impedances can produce reactive power if their suscepance is > negative. Why can't they produce real power? Because the resistance R is > usally positive (There is few exceptions to that as well though: > https://en.wikipedia.org/wiki/Negative_resistance). You must free > yourself from the idea that reactive power is something only a generator > produces. > > I believe what also confuses you, is power flow and current flow. Current > flow and power flow must not have the same direction. If you for example > have the: > > - complex voltages Uf = -Ut > - and the complex currents If = -It > > (see Figure 3-1, http://www.pserc.cornell.edu//matpower/manual.pdf) > Then the currents If = -It point in the same direction but the power flows: > > Pf = Uf * conj(If) = -Ut * conj(-It) = Ut * conj(It) = Pt > > point in opposite directions. > > > *How to calculate branch losses *There seems to be a little bit of a > difference of opinions here. I a agree with your approach: To me, the loss > of a branch is: > > P_loss = Pf+Pt > Q_loss = Qf+Qt. > > The to equations a simple conservation of energy. And yes the conservation > of energy fortunately also extends to reactive powers. > > You have to be careful though: In the pretty printed output of the runpf() > method, the branch losses are *only *the series element losses (I^2 * Z). > That might cause some confusion. > > > I hope this helps and i hope my statements are correct, since I only work > on power grids for 5 months now. > > P.S.: Next time please also include the 'from' and 'to' bus entry. That > would make it easier for me to give you a practical example. > > Best regards > Dirk > > > On 28.02.2016 02:36, Mirish Thakur wrote: > > Hello Matpower friends, > > > I have a question over similar sign of flows at both nodes of a branch. I > have performed PF with PG=0, PD=0 BRANCH_R=0; means total reactive power > model (reactive power generation and reactive demand exist) and I got > successful convergence. When I observed reactive power flows in branches I > got some surprising results in branch field which I would like to share- > > [ 2 44 0 0.00103 0.1179 > 501.0822 0 0 0 0 1 > -360 360 0 2.53088 0 > -14.8590] > > > [ 2 44 0 0.001491 0.16406 > 501.0822 0 0 0 0 1 > -360 360 0 -2.53088 0 > -14.6211] > > > So in above two branches (2 to 44) I can understand first branch flow, but > in second one both flows (QF and QT) have same negative sign. I didn’t > understand why this happened? Is this due to more charging susceptance of > second line or what? Another question is how can I estimate MVAr losses in > these scenarios? By simply adding both flows in first transmission line > gives Q loss = -12.3281 MVAr or by using average power formula > > From = sqrt(result.branch(:, 14).^2 + result.branch(:, 15).^2); > > To = sqrt(result.branch(:, 16).^2 + result.branch(:, 17).^2); > > Losses = (To + from)./2 = 8.6949 MVAr; ? > > From my point of view simple addition will give me MVAr loss value but I’m > confused how to calculate MVAr loss in second line? Thanks in advance. > > > Regards > > Mirish. > > > >
