EEL 6266 Power System Operation and Control. Chapter 5 Unit Commitment

Transcription

1 EEL 6266 Power Sysem Operaon and Conrol Chaper 5 Un Commmen

2 Dynamc programmng chef advanage over enumeraon schemes s he reducon n he dmensonaly of he problem n a src prory order scheme, here are only N combnaons o ry for an N un sysem a src prory ls would resul n a heorecally correc dspach and commmen only f he no-load coss are zero un npu-oupu characerscs are lnear here are no oher lms, consrans, or resrcons sar-up coss are a fxed amoun 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 2

3 Dynamc programmng he followng assumpons are made n hs mplemenaon of he DP approach a sae consss of an array of uns wh specfed uns operang and he res decommed (off-lne) a feasble sae s one n whch he commed uns can supply he requred load and mees he mnmum capacy for each perod sar-up coss are ndependen of he off-lne or down-me.e., s a fxed amoun w.r.. me no un shung-down coss a src prory order wll be used whn each nerval a specfed mnmum amoun of capacy mus be operang whn each nerval 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 3

4 The forward DP approach runs forward n me from he nal hour o he fnal hour he problem could run from he fnal hour back o he nal hour he forward approach can handle a un s sar-up coss ha are a funcon of he me has been off-lne (emperaure dependen) he forward approach can readly accoun for he sysem s hsory nal condons are easer o specfed when gong forward he mnmum cos funcon for hour K wh combnaon I: F cos ( K I ) = mn[ P ( K, I ) + S ( K, L : K, I ) + F ( K, L) ], cos cos cos { L} F cos (K, I) = leas oal cos o arrve a sae (K, I) P cos (K, I) = producon cos for sae (K, I) S cos (K, L: K, I) = ranson cos from sae (K, L) o (K, I) 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 4

5 The forward DP approach sae (K, I) s he I h commmen combnaon n hour K a sraegy s he ranson or pah from one sae a a gven hour o a sae a he nex hour X s defned as he number of saes o search each perod N s defned as he number of sraeges o be saved a each sep hese varable allow conrol of he compuaonal effor for complee enumeraon, he maxmum value of X or N s 2 N for a smple prory-ls orderng, he upper bound on X s n, he number of uns reducng N means ha nformaon s dscarded abou he hghes cos schedules a each nerval and savng only he lowes N pahs or sraeges here s no assurance ha he heorecal opmum wll be found 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 5

6 The forward DP approach resrced search pahs N = 3 X = 5 N X X N X Inerval K Inerval K Inerval K , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 6

7 Example consder a sysem wh 4 uns o serve an 8 hour load paern Incremenal No-load Full-load Mn. Tme Un Pmax Pmn Hea Rae Cos Ave. Cos (h) (MW) (MW) (Bu / kwh) ($ / h) ($ / mwh) Up Down Inal Condon Sar-up Coss Un off(-) / on(+) Ho Cold Cold sar (h) ($) ($) (h) Hour Load (MW) , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 7

8 Example o smplfy he generaor cos funcon, a sragh lne ncremenal curve s used he uns n hs example have lnear F(P) funcons: df F( P) = Fno load + P dp he uns mus operae whn her lms F(P) F no-load P mn P max P No-load Incremenal Un Pmax Pmn Cos Cos (MW) (MW) ($ / h) ($ / MWh) , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 8

9 Case : Src prory-ls orderng he only saes examned each hour conss of he lsed four: sae 5: un 3, sae : sae 4: , sae 5: all four all possble commmens sar from sae (nal condon) mnmum un up and down mes are gnored n hour : P F cos = cos Sae Un saus Capacy MW MW MW MW possble saes ha mee load demand (450 MW):, 4, & 5 (,5) = F ( 25) + F2 ( 05) + F3 ( 300) + F ( 20) ( 25) ( 05) ( 300) ( 20) (,5) = Pcos (,5) + Scos( 0, :,5) = ( ) + [( 350) + ( 0.02) ] = Economc Dspach Eq. 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 9 = DP Sae Transon Eq.

10 Case n hour : mnmum a sae (9208) n hour 2: K P cos S cos F cos possble saes ha mee load demand (530 MW):, 4, & 5 P 2,5 = F 25 + F 85 + F F 20 F cos = cos ( ) ( ) 2( ) 3( ) ( ) ( 25) ( 85) ( 300) ( 20) ( 2, 5) = P ( 2,5) + S (, L : 2,5) = cos ( 30) + mn = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 0 cos = 30 DP Sae Transon Eq.

11 Case : DP dagram sae un saus capacy hour 0 hour 450 MW hour MW hour MW hour MW hour MW hour MW hour MW hour MW MW MW MW MW oal cos: 73,439 prory order ls, up-mes and down-mes negleced 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol

12 Case 2: complee enumeraon ( possbles) forunaely, mos are no feasble because hey do no supply suffcen capacy n hs case, he rue opmal commmen s found he only dfference n he wo rajecores occurs n hour 3 s less expensve o urn on he less effcen peakng un #4 for hree hours han o sar up he more effcen un # for ha same me perod only mnor mprovemen o he oal cos case : 73,439 case 2: 73, , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol

13 Case 2: DP dagram sae un saus capacy hour 0 hour 450 MW hour MW hour MW hour MW MW MW MW MW MW 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 3

14 Lagrange Relaxaon dual varables and dual opmzaon consder he classcal consraned opmzaon problem prmal problem: mnmze f(x,,x n ), subjec o ω(x,,x n ) = 0 he Lagrangan funcon: L(x,,x n ) = f(x,,x n ) + ω(x,,x n ) defne a dual funcon q( ) = mn L( x, x2,) x, x2 hen he dual problem s o fnd q = max q ( ) ( ) 0 he soluon nvolves wo separae opmzaon problems n he case of convex funcons hs procedure s guaraneed o solve he problem 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 4

15 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 5 Example mnmze f(x,x 2 ) = 0.25x + x 2 2 subjec o ω(x,x 2 ) = 5 x x 2 he Lagrangan funcon: he dual funcon: he dual problem: ( ) ( ) ( ) max = = = = q q q ( ) ( ) ( ) 5 2 & 2,, mn , 2 + = = = = q x x x x L q x x ( ) ( ) ,, x x x x x x L + + =

16 Ierave form of Lagrange relaxaon mehod he opmzaon may conan non-lnear or non-convex funcons erave process based on ncremenal mprovemens of s requred o solve he problem selec a arbrary sarng solve he dual problem such ha q() becomes larger d updae usng a graden adjusmen: = + q d fnd closeness o he soluon by comparng he gap beween he prmal funcon and he dual funcon prmal funcon: J * = mn L * relave dualy gap: J q * zero q * ( ) ( ) α, n pracce he gap never reaches 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 6

17 Lagrange relaxaon for un commmen loadng consran P load N = P un lms U = 0 = KT U P mn P U P mn = K N and = KT un mnmum up-me and down me consrans he objecve funcon T N [ ( ) ] F ( P + Ssar up, U = F P, U ) = = 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 7

18 Formaon of he Lagrange funcon n a smlar way o he economc dspach problem L ( P, U, ) = F( P, U ) + Pload T = = un commmen requres ha he mnmzaon of he Lagrange funcon subjec o all he consrans he cos funcon and he un consrans are each separaed over he se of uns wha s done wh one un does no affec he cos of runnng anoher un as far as he cos funcon, un lms, and he up-me and down-me consrans are concerned he loadng consran s a couplng consran across all he uns he Lagrange relaxaon procedure solves he un commmen by emporarly gnorng he couplng consran N P U 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 8

19 The dual procedure aemps o reach he consraned opmum by maxmzng he Lagrangan wh respec o he Lagrange mulpler ( ) ( ) q( ) = mn L( P, U, ) q = max q where done n wo basc seps Sep : fnd a value for each whch moves q() owards a larger value Sep 2: assumng ha found n Sep s fxed, fnd he mnmum of L by adjusng he values of P and U mnmzng L L = = T N [ F ( P ) + S, ] = = N T T { [ F ( P ) + S ] U P U } +, = = U + T Pload = T P, U = = = N load P P U 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 9

20 separaon of he uns from one anoher; he nsde erm can now be solved ndependenly for each generang un T [ F ( ) + ] P S =, U P U he mnmum of he Lagrangan s found by solvng for he mnmum for each generang un over all me perods mn q N T ( ) [ = { ( ) + ] mn F P S U P U }, = = subjec o he up-me and down-me consrans and U P mn P U = KT hs s easly solved as a wosae dynamc programmng problem of one varable P mn U = S S S U = 0 = = 2 = 3 = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 20

21 Mnmzng he funcon wh respec o P a he U = 0 sae, he mnmzaon s rval and equals zero a he U = sae, he mnmzaon w.r.. P s: mn d dp [ F + P ] ( P ) [ ] F + P = F ( P ) + = 0 F ( P ) = ( P ) d dp here are hree cases o be consdered for P op and he lms f P op P mn hen mn [F (P ) P ] = F (P mn ) P mn f P mn P op P max hen mn [F (P ) P ] = F (P op ) P op f P max P op hen mn [F (P ) P ] = F (P max ) P max he wo-sae DP s solved o mnmze he cos of each un for U = 0 he mnmum s zero; herefore he only way o have a lower cos s o have [F (P ) P ] < 0 d dp 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 2

22 Adjusng mus be carefully adjused o maxmze q() varous echnques use a mxure of heursc sraeges and graden search mehods o acheve a rapd soluon for he un commmen problem s a vecor of s o be adjused each hour smple echnque graden componen: heursc componen: α = 0.0 when d d q α = when d d = + q d ( ) ( ) d q s posve s negave ( ) d d α q where N ( ) = Pload = P U 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 22

23 The relave dualy gap used as a measure of he closeness o he soluon for large real-sze power-sysems un-commmen calculaons, he dualy gap becomes que small as he dual opmzaon proceeds he larger he commmen problem, he smaller he gap he convergence s unsable a he end some uns are beng swched n and ou he process never comes o a defne end here s no guaranee ha when he dual soluon process sops, wll be a a feasble soluon * * he gap equaon: ( J q ) q * 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 23

24 Lagrange relaxaon algorhm usng dual opmzaon updae for all sar pck sarng for = T k = 0 loop for each un solve for he dual value q * ( ) wo-sae dynamc program wh T sages and solve for P and U gap > hreshold solve he economc dspach for each hour usng commed uns calculae he prmal value ( ) J * q * q * gap < hreshold make fnal adjusmens o schedule o acheve feasbly end 2002, 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 24

25 Example fnd he un commmen for a hree-generaor sysem over a four-hour segmen of me assume no sar-up cos and no mnmum up-me or down-me F F 2 F 2 ( P ) = P P 00 < P < 2 ( P2 ) = P P2 00 < P2 < 2 ( P ) = P P 50 < P < sar wh all values se o zero execue an economc dspach for each hour when here s suffcen generaon commed for ha hour f here s no enough generaon, arbrarly se he cos o 0,000 he prmal value J * s he oal generaon cos summed over all 3 P load ( MW ) , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 25

26 Ieraon : Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 0.0 J * = 40,000 (J * q * ) / q * = undefned Ieraon 2: dynamc programmng for un #3 =.7 F(P) P = P 3 = P mn 3 U 3 = P 32 = P 3 mn P 33 = P 3 max P 34 = P 3 mn U 3 = 0 = = 2 = 3 = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 26

27 Ieraon 2: Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 4,982 J * = 40,000 (J * q * ) / q * = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 27

28 Ieraon 3: Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 8,344 J * = 36,024 (J * q * ) / q * = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 28

29 Ieraon 4: Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 9,24 J * = 28,906 (J * q * ) / q * = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 29

30 Ieraon 5: Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 9,532 J * = 36,024 (J * q * ) / q * = , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 30

31 Ieraon 6: Hour u u 2 u 3 P P 2 P 3 dq()/d P P 2 P q() = 9,442 J * = 20,70 (J * q * ) / q * = Remarks he commmen schedule does no change sgnfcanly wh furher eraons however, he soluon s no sable (oscllaon of un 2) he dualy gap does reduce afer 0 eraons, he gap reduces o good soppng crera would be when he gap reaches , 2004 Florda Sae Unversy EEL 6266 Power Sysem Operaon and Conrol 3