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 > 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]] 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 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]] 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 with the "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 & 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 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 iteration. >>> 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 >>> >>> >> >> > >
