Dear Prof. Ray, With reference to the same thread, I want to clear one doubt. I have imposed reactive power cost on conventional generators into my model, renewable generators delivers only active power with zero active power cost. Owing to this fact when I perform runopf successfully with active and reactive power demands, LAM_P and LAM_Q values for some buses goes negative. I know that LMP prices comprised of congestion, energy cost and losses together. But when I compare my case with other cases like case89pegase or case1354pegase, I noticed that in both cases susceptance values are zero and LMP_P values are always positive in opf results. Also without implementing reactive power cost in my model it also give all positive values for LMP_P. So in my case, lines susceptance which generates reactive power and to compensate such reactive power, generators absorb it and this is why few generators delivers negative (means absorbs) reactive power (in QG column). And due to implementation of reactive power cost into the model this negative generation of reactive power by generators causes LMP_P values negative or is there any other technical reason? For your information i have used linear cost function for active power generation and exponential cost for reactive power. Also I have attached diagram showing reactive power losses (Qf+Qt) in branches in pure reactive power grid model concern to our last discussion.
Regards Mirish. On Wed, Mar 2, 2016 at 12:48 AM, Mirish Thakur <[email protected]> wrote: > 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. >> >> >> >> >
