I don't have any special knowledge about that system. The data says that some buses have a nominal voltage of 138kV, others 230kV. I suppose you could assume that any branch connecting the two voltage levels is a transformer. You will have to convert any input impedance data to p.u. manually.
-- Ray Zimmerman Senior Research Associate B30 Warren Hall, Cornell University, Ithaca, NY 14853 phone: (607) 255-9645 On Jun 10, 2013, at 10:06 AM, Alexandra Kapetanaki <[email protected]> wrote: > Dear Ray, > > Thank you for the clarification. > It helped me a lot. > I would like to change the electrical properties of the branches and use the > properties of specific conductors. > A long as, Matpower uses the r,x,b in p.u. can you please inform me what is > the Vbase of the case IEEE RTS 24 bus system(138KV or 230 KV)? > Is there any way to just insert the real values of my values in Matpower and > define the values in p.u.? > > Thank you in advance, > Kind Regards, > 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 Ray Zimmerman > [[email protected]] > Sent: 07 June 2013 16:30 > To: MATPOWER discussion forum > Subject: Re: Dispatchable load > > You specified the $1000 value for the dispatchable loads in the wrong column. > Should be 5, not 6. > > -- > Ray Zimmerman > Senior Research Associate > B30 Warren Hall, Cornell University, Ithaca, NY 14853 > phone: (607) 255-9645 > > > > > > On Jun 6, 2013, at 5:27 PM, Alexandra Kapetanaki > <[email protected]> wrote: > >> Dear Ray, >> >> The Matpower converges in both cases. >> Is there any possibility the problem derives from voltage constraints? >> Please find attached a simplified .m file with only the extra generators in >> order to reproduce the problem. >> In the case without the extra generators(in 24 buses) the total load >> dispatched . >> Please, can you please help me so as to understand if I have to take voltage >> control actions? >> >> Thank very much, >> 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 Shri >> [[email protected]] >> Sent: 06 June 2013 21:41 >> To: MATPOWER discussion forum >> Subject: Re: Dispatchable load >> >> Perhaps MATPOWER should (i) avoid calling printpf if the PF/OPF does not >> converge or (ii) print the 'Converged/Not converged' line after printpf. The >> problem is that 'converged/Not converged' line is not glaringly visible. So >> even if the OPF does not converge printpf produces output which might fool >> some users to think that the system has actually converged. This is >> certainly an issue when running large systems because users have to scroll >> 3-4 screens up to check whether the PF/OPF has converged when the output is >> directed to stdout. >> >> Shri >> On Jun 6, 2013, at 1:38 PM, Ray Zimmerman 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 >>>> 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 >>>>>> >>>>>> >>>>> >>>>> >>>> >>>> >>> >> >> <testlosses.m><case24_ieee_rts_min_load_losess.m> >
