I'm afraid I don't understand what you are requesting. The form of the OPF is fully described in the MATPOWER User's Manual.
-- Ray Zimmerman Senior Research Associate B30 Warren Hall, Cornell University, Ithaca, NY 14853 phone: (607) 255-9645 On Jun 9, 2013, at 12:01 PM, 舒 <[email protected]> wrote: > Dear Professor > Is that OK send me an orginal form of OPF for the matpower?just OPF > > > > > schools of electrical engineering and automation ,Tsinghua University,China.. > > > At 2013-06-07 02:38:53,"Ray Zimmerman" <[email protected]> wrote: > Are you sure the OPF converged? > > -- > Ray Zimmerman > Senior Research Associate > B30 Warren Hall, Cornell University, Ithaca, NY 14853 > phone: (607) 255-9645 > > > > > > On Jun 5, 2013, at 12:35 PM, Alexandra Kapetanaki > <[email protected]> wrote: > >> Dear Ray, >> >> I run DC opf for the 24 bus IEEE network and I have negative virtual gen for >> the representation of the load and for an intact system with 2850 MW load >> there is 2850 MW dispatchable load. >> However, in the case I add extra virtual gen to all the 24 buses with zero >> output (also zero Pmin and Pmax) the dispatchable load is 1036MW instead of >> 2850MW. >> Basically, the network is the same eg generation capacity, load, >> transmission lines's capacity in both cases. >> The only difference is the extra generators with zero output in the second >> case. >> Can you please insight me why the load is not satisfied in the second case? >> >> Thank you in advance, >> Alexandra >> >> Alexandra Kapetanaki >> PhD Student >> Electrical Energy & Power Systems Group, School of Electrical & Electronic >> Engineering >> Ferranti Building (B18), The University of Manchester, M13 9PL, United >> Kingdom >> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179 >> < b>From: [email protected] >> [[email protected]] on behalf of Alexandra >> Kapetanaki [[email protected]] >> Sent: 04 June 2013 15:39 >> To: MATPOWER discussion forum >> Subject: RE: transmission losses-DC opf >> >> How can I exclude the Q losses in AC opf? >> Because the formula is applicable in a DC model without the effect of Q. >> >> Thank you in advance! >> >> Alexandra Kapetanaki >> PhD Student >> Electrical Energy & Power Systems Group, School of Electrical & Electronic >> Engineering >> Ferranti Building (B18), The University of Manchester, M13 9PL, United >> Kingdom >> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179 >> From: [email protected] >> [[email protected]! ornell.e du] on behalf of Ray Zimmerman >> [[email protected]] >> Sent: 04 June 2013 15:05 >> To: MATPOWER discussion forum >> Subject: Re: transmission losses-DC opf >> >> I'm not sure, but you could try your formula to estimate the losses for the >> AC model as well. That would give you an idea of how good your estimate is. >> >> -- >> Ray Zimmerman >> Senior Research Associate >> B30 Warren Hall, Cornell University, Ithaca, NY 14853 >> phone: (607) 255-9645 >> >> >> >> >> >> On Jun 4, 2013, at 7:10 AM, Alexandra Kapetanaki >> <[email protected]> wrote: >> >>> Dear Ray, >>> >>> Thank you for your support! >>> >>> I run DC OPF in the 24 bus IEEE network and my aim is to include a >>> transmission losses model in the system. >>> I followed the procedure you insisted to me and for the losses per line I >>> used the already proved equation Ploss = rij(Pk)^2 which is applicable in a >>> p.u. system. >>> So I divided the Pk of each line by baseMVA(=100) whereas the rij is >>> already in p.u.. >>> However, I compared the losses of the branches for an intact network(24bus) >>> by using AC opf to the corresponding losses by using DC opf and they seem >>> to be different as you can see below: >>> Losses for each branch_DC(real values)= >>> [4,32 >>> 6,306 >>> 2,4 >>> 2,21 >>> 2,114 >>> 1,173 >>> 1,318 >>> 5,486 >>> 2,31 >>> 2,4 >>> 5,67 >>> 12,4 >>> 6,18 >>> 3,27 >>> 6,791 >>> 3,83 >>> 14,3655 >>> 4,268 >>> 5,459 >>> 3,761 >>> 5,155 >>> 5,11 >>> 3,94 >>> 4,13] >>> >>> Losses for each branch_AC(real values)= >>> [0,0022 >>> 0,13 >>> 0,864 >>> 0,644 >>> 1,25 >>> 0,156 >>> 0,933 >>> 0,245 >>> 0,059 >>> 0,94 >>> 1,63 >>> 0,756 >>> 0,423 >>> 0,285 >>> 0,35 >>> 0,50 >>> 0,611 >>> 0,537 >>> 1,391 >>> 0,309 >>> 5,80 >>> 4,69 >>> 6,129 >>> 0,030 >>> 2,187 >>> 2,187 >>> 2,692 >>> 4,1436 >>> 0,230 >>> 0,833 >>> 2,697 >>> 0,191 >>> 0,191 >>> 0,100 >>> 0,100 >>> 0,313 >>> 0,31 >>> 1,829] >>> >>> I would like to let you know that Plosses_AC_opf were obtained by the >>> following command: >>> results_2(N).Plossesbranch_AC(br)=results.branch(br,14)+results.branch(br,16); >>> >>> >>> Why do you think the losses are considerable different between the AC,DC >>> models? >>> In DC the Q is not included so this is one reason of the deviation but is >>> that enough? >>> >>> Thank you ! in advan ce, >>> >>> Alexandra Kapetanaki >>> PhD Student >>> Electrical Energy & Power Systems Group, School of Electrical & Electronic >>> Engineering >>> Ferranti Building (B18), The University of Manchester, M13 9PL, United >>> Kingdom >>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179 >>> From: [email protected] >>> [[email protected]] on behalf of Ray Zimmerman >>> [[email protected]] >>> Sent: 28 May 2013 17:58 >>> To: MATPOWER discussion forum >>> Subject: Re: transmission losses-piece wise linear approach >>> >>> - Solve a DC OPF with no losses. >>> - Compute the loss for each branch, Ploss = rij(Pk)^2. >>> - Add half of the loss to the load at the buses connected by the branch. >>> - Re-solve the DC OPF. >>> - Recompute the loss for each branch based on the new flow. >>> - Adjust the loads at each end of the branch to reflect the change in >>> losses. >>> - Repeat, until the change from one iteration to the next is smaller than >>> some threshold. >>> >>> -- >>> Ray Zimmerman >>> Senior Research Associate >>> 419A Warren Hall, Cornell University, Ithaca, NY 14853 >>> phone: (607) 255-9645 >>> >>> >>> >>> >>> On May 28, 2013, at 12:25 PM, Alexandra Kapetanaki >>> <[email protected]> wrote: >>> >>>> Dear Ray, >>>> >>>> Thank you for your help! >>>> Can you please explain me more what do you mean wi! th the & >>>> nbsp;"adjusting the set of dummy loads used to represent losses based on >>>> flows in the previous iteration". >>>> For example, I want to represent the losses as dummy loads according to >>>> the following figure: >>>> >>>> >>>> <losses.png> >>>> >>>> >>>> >>>> >>>> and express the losses by using the quadratic equation : Ploss=rij(Pk)^2 . >>>> How can I practically implement that within "few iterations", as you >>>> recommend? >>>> >>>> Thank you once again for your support, >>>> >>>> >>>> Alexandra Kapetanaki >>>> PhD Student >>>> Electrical Energy! & P ower Systems Group, School of Electrical & >>>> Electronic Engineering >>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United >>>> Kingdom >>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179 >>>> From: [email protected] >>>> [[email protected]] on behalf of Ray Zimmerman >>>> [[email protected]] >>>> Sent: 28 May 2013 16:55 >>>> To: MATPOWER discussion forum >>>> Subject: Re: transmission losses-piece wise linear approach >>>> >>>> Dear Alexandra, >>>> >>>> Since the network equations in MATPOWER's DC OPF assume no losses, it >>>> seems you would have to introduce the losses as dummy (dispatchable) loads >>>> at the downstream end of each branch. You would have to add an additional >>>> set of constraints for each of these dummy loads constraining the >>>> consumption lie above the piecewise linear constraints you propose. These >>>> constraints could be added using the mechanism for user-defined >>>> constraints described in section 5.3.2 and chapter 6 of the User's Manual. >>>> >>>> Another approach to using MATPOWER to solve a "DC OPF with losses" is to >>>> simply run the DC OPF iteratively, each time adjusting the set of dummy >>>> loads used to represent losses based on flows in the previous! iterati on. >>>> I'm guessing it wouldn't require more than a few iterations to converge to >>>> a pretty good solution. This approach is a bit brute-force, but may be >>>> simpler to implement and allows you to use whatever function you like >>>> (e.g. a quadratic) to compute the losses. Just another idea. >>>> >>>> -- >>>> Ray Zimmerman >>>> Senior Research Associate >>>> 419A Warren Hall, Cornell University, Ithaca, NY 14853 >>>> phone: (607) 255-9645 >>>> >>>> >>>> >>>> >>>> On May 27, 2013, at 3:16 PM, Alexandra Kapetanaki >>>> <[email protected]> wrote: >>>> >>>> Dear Dr Ray, >>>> >>>> As the AC power flow requires high computation time for the losses to be >>>> calculated, a DC opf can be used in conjuction with a linear model for >>>> transmission losses. >>>> My aim is to adjust the piece wise linear approach of Matpower in order to >>>> accommodate the losses of a transmission line. >>>> More particularly, the losses can be expressed through the following >>>> equation: Ploss=rij(Pk)^2 [where rij the resistance of the line and Pk the >>>> power flow in the transmission line]. >>>> However, the above equation is a quadratic function but can be expressed >>>> with a piecewise linear model. The figure below shows a linear model >>>> consists of N line pieces >>>> <piecewiselinear.png> >>>> >>>> >>>> >>>> the equation of the nth line piece is: >>>> >>>> <bbbbbbbbbbbbbbb.png> >>>> >>>> >>>> >>>> >>>> >>>> >>>> -How can I modify the piece wise linear approach of Matpower with the view >>>> to include the losses model? >>>> >>>> >>>> Thank you in advance, >>>> >>>> Alexandra Kapetanaki >>>> PhD Student >>>> Electrical Energy & Power Systems Group, School of Electrical & Electronic >>>> Engineering >>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United >>>> Kingdom >>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179 >>> >> > > > > > > >
