Because I want to have PF=0.9 or 0.95 for all generators. As you know, it is very important
On Mon, Mar 14, 2011 at 16:36, Ray Zimmerman <[email protected]> wrote: > You need a row of A for each generator, each one with QPratio in the column > corresponding to Pg(i) and -1 in the column corresponding to Qg(i) ... > > mpc.A = sparse([1:ng; 1:ng], [2*nb+(1:ng); 2*nb+ng+(1:ng)], > [QPratio*ones(ng,1); -ones(ng,1)], 1:ng, 2*nb+2*ng); > > As I said, this is likely to make the OPF infeasible, so I'm not sure why > you'd want to do it. > > -- > Ray Zimmerman > Senior Research Associate > 211 Warren Hall, Cornell University, Ithaca, NY 14853 > phone: (607) 255-9645 > > > > > On Mar 14, 2011, at 11:17 AM, Roberto Carvalini wrote: > > Thank you very much. > > Is it possible for you to give an example, for example case9, for all > generators? I will be appreciate if you do it. > > Best Regards > > R. Carvalini > > On Mon, Mar 14, 2011 at 16:08, Ray Zimmerman <[email protected]> wrote: > >> The *A* matrix is from equation (5.25) in the User's >> Manual<http://www.pserc.cornell.edu/matpower/manual.pdf>. >> As I mentioned, the coefficients need to go in the columns corresponding to >> Pg(1) and Qg(1). So, Pg(1) is in the first column following the voltage >> angles (1st nb columns) and magnitudes (next nb columns), so it's in column >> 2*nb+1. For Qg(1) it's the first column after the Pg columns, so 2*nb (for >> the voltage angles and magnitudes) plus ng columns for Pg. >> >> Yes, if you wanted to do this for all generators, you need to include a >> row in *A* for each. Be careful though, this will likely over-constrain >> the problem and make it infeasible. If all generators have a fixed power >> factor, then balancing the real power output to match the load will mean a >> specific reactive power output as well, which will not necessarily match the >> reactive power load. >> >> -- >> Ray Zimmerman >> Senior Research Associate >> 211 Warren Hall, Cornell University, Ithaca, NY 14853 >> phone: <%28607%29%20255-9645>(607) 255-9645 >> >> >> >> On Mar 11, 2011, at 11:40 AM, Roberto Carvalini wrote: >> >> Thank you very much. >> >> I have read the manual but I didn't understand by which formula you >> computed this matrix (highlighted) >> >> mpc.A = sparse([1; 1], *[2*nb+1; 2*nb+ng+1]*, [QPratio; -1], 1, >> 2*nb+2*ng); >> >> and if want to define the same PF for all generators, must I repeat this >> for all or it depends on the type buses (PV or Slack)? >> >> Regards >> >> R. Carvalini >> >> On Fri, Mar 11, 2011 at 16:27, Ray Zimmerman <[email protected]> wrote: >> >>> Dirk's suggestion is fine for a simple power flow. But if you want to run >>> an OPF, where the generator dispatch is to be determined by the >>> optimization, but subject to a constant power factor constraint, then you >>> will need to supply an additional user-defined linear constraint. For >>> example, if you want to run case9 with a constant power factor constraint of >>> 0.95 on the first generator, you could do it like this ... >>> >>> define_constants; >>> mpc = loadcase('case9'); >>> nb = size(mpc.bus, 1); >>> ng = size(mpc.gen, 1); >>> pf = 0.95; >>> QPratio = sqrt(1/pf^2 -1); >>> %% add constraint that QPratio * Pg(1) - Qg(1) = 0 >>> mpc.A = sparse([1; 1], [2*nb+1; 2*nb+ng+1], [QPratio; -1], 1, 2*nb+2*ng); >>> mpc.l = 0; >>> mpc.u = 0; >>> r = runopf(mpc); >>> resulting_pf = r.gen(1, PG) / sqrt(r.gen(1, PG)^2 + r.gen(1, QG)^2) >>> >>> The important part here is defining the A matrix properly, where the >>> columns correspond to voltage angles, voltage magnitudes, generator real >>> injections, and generator reactive injections (2*nb + 2*ng columns in all). >>> So the coefficients for our constraint co in the columns corresponding to >>> Pg(1) and Qg(1). >>> >>> -- >>> Ray Zimmerman >>> Senior Research Associate >>> 211 Warren Hall, Cornell University, Ithaca, NY 14853 >>> phone: <%28607%29%20255-9645> <%28607%29%20255-9645>(607) 255-9645 >>> >>> >>> >>> On Mar 11, 2011, at 9:50 AM, Dirk Van Hertem wrote: >>> >>> You model your generator as a load, so it is not represented in your gen >>> matrix. Of course, if you want model your generator in a unit commitment, >>> this may not be trivial... >>> >>> Dirk >>> >>> On 03/11/2011 12:32 PM, Roberto Carvalini wrote: >>> >>> Thank you >>> >>> How can I add? my problem is this >>> >>> Best Wishes >>> >>> Roberto >>> >>> >>> On Fri, Mar 11, 2011 at 12:22, Dirk Van Hertem >>> <[email protected]>wrote: >>> >>>> You can just define them as a PQ bus with negative power... (negative >>>> load) >>>> >>>> Dirk >>>> >>>> >>>> On 03/11/2011 12:11 PM, Roberto Carvalini wrote: >>>> >>>>> Hi >>>>> >>>>> How can I define constant power factor in MAPOWER for example, for >>>>> generators PF=0.9? >>>>> >>>>> Please help me >>>>> >>>>> Regards >>>>> >>>>> Roberto >>>>> >>>>> >>>>> >>>> >>>> -- >>>> Dirk Van Hertem [email protected] >>>> Electrical Engineering Department http://www.esat.kuleuven.be/electa >>>> K.U. Leuven, ESAT-ELECTA GSM: +32-(O)498-61.74.98 >>>> 10, Kasteelpark Arenberg, B-3001 Heverlee fax: >>>> <%2B32-16-32.19.85> <%2B32-16-32.19.85> <%2B32-16-32.19.85> >>>> +32-16-32.19.85 >>>> >>>> >>>> >>> >>> >>> -- >>> Dirk Van Hertem [email protected] >>> Electrical Engineering Department http://www.esat.kuleuven.be/electa >>> K.U. Leuven, ESAT-ELECTA GSM: +32-(O)498-61.74.98 >>> 10, Kasteelpark Arenberg, B-3001 Heverlee fax: <%2B32-16-32.19.85> >>> <%2B32-16-32.19.85>+32-16-32.19.85 >>> >>> >>> >>> >> >> >> >> > > > > > Best Wishes Roberto
