I think you need to take some time to understand the examples I already gave. If you understand them, the answer to your question should be obvious. Simply eliminate the row of A corresponding to the generator whose power factor you want to free up.
-- Ray Zimmerman Senior Research Associate 211 Warren Hall, Cornell University, Ithaca, NY 14853 phone: (607) 255-9645 On Mar 14, 2011, at 12:04 PM, Roberto Carvalini wrote: > Ok. Except generator 1, I want to have specified PF for other generators. > Could you please give me an example? I will be appreciate > > Also I did the last email command but it has error. > > Best Wishes > > Roberto > > > On Mon, Mar 14, 2011 at 17:01, Ray Zimmerman <[email protected]> wrote: > You have to leave at least one with a free power factor and large enough > reactive power range to balance the reactive power in the system. > > -- > Ray Zimmerman > Senior Research Associate > 211 Warren Hall, Cornell University, Ithaca, NY 14853 > phone: (607) 255-9645 > > > > On Mar 14, 2011, at 11:53 AM, Roberto Carvalini wrote: > >> How can I solve this problem? >> >> Best Wishes >> >> Roberto >> >> >> On Mon, Mar 14, 2011 at 16:44, Ray Zimmerman <[email protected]> wrote: >> But you are over-specifying the problem and it will not have a solution. You >> can't specify all of the power factors exactly and expect the power flow >> equations to be satisfied. >> >> -- >> Ray Zimmerman >> Senior Research Associate >> 211 Warren Hall, Cornell University, Ithaca, NY 14853 >> phone: (607) 255-9645 >> >> >> >> On Mar 14, 2011, at 11:39 AM, Roberto Carvalini wrote: >> >>> 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. 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: (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: (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: +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: +32-16-32.19.85 >>>>> >>>>> >>>>> >>>>> >>>>> >>>> >>>> >>>> >>>> >>>> >>> >>> >>> >>> >>> Best Wishes >>> >>> Roberto >> >> > >
