Because of the coupling between active and reactive power, costs on one affect the nodal prices for the other. For example, even with zero cost on reactive power, the price of reactive power at a particular bus may be positive or negative, depending on the sensitivity of the real power dispatch to reactive demand at that bus. So, yes, they are coupled and when you put a cost on one it affects prices of both.
If you want to see exactly *how* the negative prices on active power arises in your case, I suggest perturbing the active power demand at the bus of interest by a small amount and observing how all of the active and reactive dispatches change. Presumably, increasing the active demand will allow for a redispatch of reactive power whose cost decreases by more than the cost of the required increase in active power. I’ve found that doing an explicit perturbation and sensitivity analysis like this can shed light on these sorts of things. I once used it to understand a case in which a nodal price was much higher than any of the generator costs in the system, which can legitimately happen. Ray > On Mar 11, 2016, at 9:54 AM, Mirish Thakur <[email protected]> wrote: > > 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] > <mailto:[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] > <mailto:[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] >> <mailto:[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 >> <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 >> <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. >> > > > > <reactivepowerloss.jpg>
