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 >> >> > >
