2/20/2013. Topics. Power Flow Part 1 Text: Power Transmission. Power Transmission. Power Transmission. Power Transmission
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1 /0/0 Topcs Power Flow Part Text: 0-0. Power Trassso Revsted Power Flow Equatos Power Flow Proble Stateet ECEGR 45 Power Systes Power Trassso Power Trassso Recall that for a short trassso le, the power fro to, S ca be wrtte as: V Z Z Z S V I V V V V V S j e V V V e r jx r jx j j e V V V V V V cos js r jx r jx r jx r jx S Substtutg ad separatg real ad agary parts g jb r jx S V V V cos js r jx r jx g jb cos js g jb V V V P g V g V V cos b V V s Q b V g V V s b V V cos S 4 Power Trassso Power Trassso Real, Iagary Power fro to ust equal the power output by the geerator P g V g V V cos b V V s P Q b V g V V s b V V cos Q S P + jq S What f aother s added? S = S + S p Equatos becoe: P g V g V V cos b V V s g V g V V cos b V V s p p p p p p p Q b V g V V s b V V cos b V g V V s b V V cos p p p p p p p p P + jq S S p 5 6
2 /0/0 Power Trassso What f the trassso le fro to p s edu or log legth? Let the total le shut adttace be y shut Note that sce t s capactace oly, p the real power s ot affected P jq P g V g V V cos b V V s g V g V V cos b V V s p p p p p p p y Q V b V g V V s b V V cos shut b V g V V s b V V cos p p p p p p p S S p Power Trassso What f there was a shut adttace added to? Note ths s dfferece fro a load, sce the load s odeled ters of power Let the adttace be y = g + jb P V g g V g V V cos b V V s g V g V V cos b V V s p p p p p p p y shut Q V b V b V g V V s b V V cos b V g V V s b V V cos p p p p p p p P jq p S S p 7 8 Power Trassso Power Flow Equatos We ca wrte slar expressos for the power out of es ad p ou ca see that as we add ore es ad trassso les, the equatos becoe qute log Let s see f we ca splfy use works wth curret so we wll start out by cosderg curret Total curret eterg the trassso syste at s equal to the curret fro the geerator(s) at us the curret to load at I IG ID Ik,,..., k I s the total curret leavg through trassso les or shuts (ot loads) Usg : I k k k yelds I V V,,..., Bus I G I D I I p 9 0 Power Flow Equatos Now for power S VI Va substtuto S V k Vk k Geeral case k k k S V V,..., I V k k k Note: k refers to a eleet Bus Recall Power Flow Equatos k k k S V V j V V e k k va substtuto ad Euler s Idetty j jk S V e k Vk e k k k k V V e e j jk k k V V e G jb jk k k k k Gk jbk V V cos js G jb k k k k k locato
3 /0/0 Power Flow Equatos S V Vk cosk jsk Gk jbk k Splt to real ad reactve parts P V Vk Gk cosk Bk sk k Q V Vk Gk sk Bk cosk k Sped soe te becog falar wth these equatos These equatos atch the equatos derved for our sple ad cases Power Flow Equatos To recap: for each, we ca fd the et power jected (geerato us load) by P V Vk Gk cosk Bk sk k Q V Vk Gk sk Bk cos k k G k ad B k values correspod to the eleet k the, ad NOT the prtve adttace of le -k If we have voltage agtude ad agle values at each, the we ca easly detere all the power flows ad power jectos 4 Voltage at s.0 Voltage at s 0.97 p.u. at -7 Seres le pedace s j0. p.u. B = 0.0 p.u. costruct Ths s ot eleet, but s the total shut capactve reactace of le -. If gve a proble stateet, assue B s defed as such. Voltage at s.0 Voltage at s 0.97 p.u. at -7 Seres le pedace s j0. p.u. B = 0.0 p.u. costruct 0.99 j j j j Voltage at s.0 Voltage at s What s the real power jected to? 0.99 j j j j9.85 Voltage at s.0 Voltage at s What s the real power jected to? P V V G cos B s 0.99 j j j j9.85 k k k k k k P. (. cos. s ).. (. cos. s ). 7 8
4 /0/0 Voltage at s Voltage at s What s the reactve power jected to? 0.99 j j j j9.85 Voltage at s Voltage at s What s the reactve power jected to? Q V V G s B cos k k k k k k Q s cos 7 Q 0. p.u j j j j Power Flow Equatos I ths exaple, the power jected to the s also the power trastted dow the power le To detere the power out of the geerator, we ust kow the power to the load at If P D =.0 p.u. the the real power fro the geerator at s.0 p.u. follow a slar procedure for reactve power We ca use the power flow equatos to show the P- ad Q- V relatoshps The relatoshp ca be quatfed by fd the sestvty, or partal dervatve of P w.r.t. ad V Fro our exaple: P P V G V V G cos B s P 0 V V G s B cos P V V G V G cos B s Power Flow Equatos Evaluatg each dervatve P s cos P cos s 7 V 4.04 We expect a larger chage of real power f the dfferece voltage agles chage, rather tha a voltage agtude chage Now try a three syste Gve: V 0 V. 05 V Fd the power out of each geerator ad to the load j9.98 j0 j0 j0 j9.98 j0 j0 j0 j
5 /0/0 j9.98 j0 j0 start at = j0 j9.98 j0 P V Vk Gk cosk Bk s k k j0 j0 j9.98 for k = V 0 V V G cos B s 0 V. 05 for k = V V V G cos B s s for k = V V G cos B s s0.65 total real power: =.9 p.u. ow = P V Vk Gk cosk Bk sk k for k = V V G cos B s.05 0s for k = V V G cos B s 0 for k = V V G cos B s s7.5 total real power: = p.u. 5 6 ow = P V Vk Gk cosk Bk sk k for k = V V G cos B s.95 0s for k = V V G cos B s s 7.5 for k = V V G cos B s 0 total real power: = p.u. start at = Q V Vk Gk sk Bk cos k k for k = V G s B cos ( 9.98) for k = V V G s B cos.05 0cos for k = V V G s B cos cos total reactve power: = 0.65 p.u. 7 8 Followg the sae procedure we ca fd reactve power for ad We ca also fd the power flow o each le If we have voltage agle ad agtude, we ca copletely descrbe the load ad geerato, as well as power flows losses (real ad reactve) ca be coputed A very coo stake s to assue that the power flowg fro to k s foud by substtutg = ad k= to: P V Vk Gk cos k Bk s k k Ths s NOT the case, ad oly would yeld the correct aswer f the trassso les are lossless To fd the power flowg o the dvdual les, you ca use S V I V I0 V kvk k where I 0 s the curret through the shuts coected to, f ay 9 0 5
6 /0/0 Power Flow A slar stateet ca be ade for the agary power flowg fro to The value correspodg to = ad k=, geeral, has o physcal eag, rather t s sply a eleet a equato I geeral, whe calculatg the power flowg fro to, the power flowg to the shuts fro a p-odeled trassso le are cluded If the coplex power flowg fro to all other es s sued ad added to the power flowg to ay o-trassso le related shuts at, the the resultg power s equal to S Power Flow Proble As our dscusso, we geerally do ot kow the voltage forato at each Ofte, we kow ad ca cotrol the voltage agtude ad the real power put at es wth geerators If a oly has a load, we oly kow the real ad reactve power dead Solvg for voltage agtudes ad agles s kow as the power flow proble Power Flow Proble Solvg the power flow equatos for V ad s very dffcult For exaple, gve V 0 P. Q Fd V ad Two ukows, so we eed two equatos: P. V G V G cos B s Q.0 V B V G s B cos 4 Power Flow Proble Proble s olear, so t s dffcult to solve P. V G V G cos B s Q.0 V B V G s B cos Iage a slar proble o a three syste (equatos are coupled) Clearly, fdg the exact soluto s gog to take soe work Luckly, we do t eed a exact soluto. Oly a soluto that s close to the aswer If the exact soluto was V = , we ght be satsfed wth a soluto that s V = Power Flow Proble Use soe uercal ethods to approxate the soluto to the power flow proble We frst eed to have a better defto of the proble 5 6 6
7 /0/0 Power Flow Proble: Terology Power Flow Proble: Slack Bus Geerator Bus: a wth a coected geerator voltage agtude ca be cotrolled by adjustg the excter curret real power ca be cotrolled by adjustg power to the pre over Load Bus: a wthout a geerator real ad reactve power are kow Sce we do ot kow how uch power s flowg o each le, we caot kow advace the real power out of each geerator ad load. Why? losses! The aout of losses are ot kow beforehad Eve a lossless syste, the total power geerated ust equal the power at each load. Wth a gve syste load, all but geerator real power ca be specfed The geerator wthout ts real power specfed s kow as the slack 7 8 Power Flow Proble: Terology Geerator Bus: a wth a coected geerator voltage agtude ca be cotrolled by adjustg the excter curret real power ca be cotrolled by adjustg power to the pre over Load Bus: a wthout a geerator real ad reactve power are kow Slack Bus: a geerator wthout ts real power specfed voltage agtude s cotrolled voltage agle s set as the referece usually ubered Oly oe slack each syste Power Flow Proble For each type, there are two varables that ca be specfed (cotrolled) We ca also call geerator : PV load : PQ There are certa cases whch a geerator ay ot be PV, or a load ot PQ, for exaple: certa wd turbes load es wth cotrolled capactor baks 9 40 Power Flow Proble we are terested solvg olear equatos usg uercal ethods uercal ethods are teratve ad provde solutos at are exact to wth a tolerace We wll use Newto-Raphso to solve the proble 4 7
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