EDC Introduction

Transcription

1 .0 Introducton EDC3 In the last set of notes (EDC), we saw how to use penalty factors n solvng the EDC problem wth losses. In ths set of notes, we want to address two closely related ssues. What are, exactly, penalty factors? How to obtan the penalty factors n practce?.0 What are penalty factors? Recall the defnton: (,, m) () In order to gan ntutve nsght nto what s a penalty factor, let s replace the numerator and denomnator of the partal dervatve n () wth the approxmaton of Δ /Δ, so:

2 () Multplyng top and bottom by Δ, we get: (3) What s Δ? It s a small change n generaton. But that cannot be all, because f you mae a change n generaton, then there must be a change n njecton at, at least, one other bus. et s assume that a compensatng change s dstrbuted throughout all other load buses accordng to a fxed percentage for each bus. By dong so, we are embracng the so-called conformng load assumpton, whch ndcates that all loads change proportonally. Therefore Δ =Δ D. But ths wll also cause a change n losses of Δ, whch wll be offset by a compensatng change n swng bus generaton Δ. So,

3 D (4) where we see generaton changes are on the left and load & loss changes are on the rght. Solvng for Δ -Δ (because t s n the denomnator of (3)), we get D (5) Substtutng (5) nto (3), we obtan: D (6a) Recognze that Δ n (6a) reflects the losses, we have D (6b) So from (6b), we extract the followng nterpretaton of the penalty factor: It s the amount of generaton at unt necessary to supply Δ D, as a percentage of Δ D -Δ. Ths depends on how the load s changed (whch s why we use the conformng load assumpton). If the change ncreases losses (Δ >0), then >. If the change decreases losses(δ <0), then <. 3

4 An example wll llustrate the sgnfcance of (6a) & (6b). Consder Fg.. Observe that the flows gven on the crcuts are nto bus (the flows along the lne out of buses and 3, respectvely, are hgher). Basecase Increase load by MW at each bus, compensated by gen ncrease at bus ( 0.) Increase load by MW at each bus, compensated by gen ncrease at bus Fg ( 0.)

5 One observes that <. Ths s because a load change compensated by a gen change at bus decreases the losses as ndcated by the fact that the bus generaton decreased by 0. MW. On the other hand, 3 >. Ths s because a load change compensated by a gen change at bus 3 ncreases the losses as ndcated by the fact that the bus generaton ncreases by 0.. MW. Why does the bus generaton reduce losses whereas the bus 3 generaton ncreases losses? Answer: Because ncreasng bus tends to reduce lne flows, whereas ncreasng bus 3 tends to ncrease lne flows. So we see that n general, generators on the recevng end of flows wll tend to have lower penalty factors (below.0); generators on the sendng end of flows wll tend to have hgher penalty factors (above.0). 5

6 Because transmsson systems are n fact relatvely effcent, wth reasonably small losses n the crcuts, the amount of generaton necessary to supply a load change tends to be very close to that load change. Therefore penalty factors tend to be relatvely close to.0. A lst of typcal penalty factors for the power system n Northern Calforna s llustrated n Fg.. enerators mared to the rght are unts n the San Francsco Bay Area, whch s a relatvely hgh mport area for the Northern Calforna system. Most of the penalty factors for these unts are below.0. Unts havng penalty factors>. are manly unts close to the Oregon border (a long way from the SF load center), such that they tend to add to the north-to-south flow that results from the northwest hydro beng sold nto the Calforna load centers. 6

7 Fg. But why do we actually call them penalty factors? Consder the crteron for optmalty n the EDC wth losses: C ( ),... m (7) 7

8 Ths says that all unts (or all regulatng unts) must be at a generaton level such that the product of ther ncremental cost and ther penalty factor must be equal to the system ncremental cost λ. et s do an experment to see what ths means. Consder that we have three dentcal unts such that ther ncremental cost-rate curves are dentcal, gven by IC( )= Now consder the three unts are so located such that unt has penalty factor of 0.98, unt has penalty factor of.0, and unt 3 has penalty factor of.0, and the demand s 300 MW. Wthout accountng for losses, ths problem would be very smple n that each unt would carry 00 MW. But wth losses, the problem s as follows: 8

9 9 λ=0.98( )= λ=.0( )= λ=.0( )= uttng these three equatons nto matrx form results n: Solvng n Matlab yelds: One notes that the unt wth the lower penalty (unt ) was turned up and the unt wth the hgher penalty (unt 3) was turned down. The reason for ths s that unt has a better effect on losses.

10 3.0 enalty factor calculaton There are several methods for penalty factor calculaton. We wll revew several of them n ths secton. Ths method s descrbed n []. Consder a power system wth total of n buses of whch bus s the swng bus, buses m are the V buses, and buses m+ n are the Q buses. Consder that losses must be equal to the dfference between the total system generaton and the total system demand: D (8) Recall the defnton for bus njectons, whch s D (9) Now sum the njectons over all buses to get: 0

11 D n D n n D n ) ( (0) Therefore, n () Now dfferentate wth respect to a partcular bus angle θ (where s any bus number except ) to obtan: n n m m,...,, () Assumpton to the above: All voltages are fxed at.0; ths releves us from accountng for varaton n power wth angle through the voltage magntude term. Otherwse, each term n () would appear as V V

12 Now let s assume that we have an expresson for losses as a functon of generaton, 3,, m,.e., = (, 3,, m ) (3) Then we can use the chan rule of dfferentaton to express that n m m,...,, ) ( ) ( (4) In (4), we assume that at generator buses, loads are constant, and / θ = / θ. Subtractng (4) from (), we obtan, for =,,n: n m m m m m n m m ) ( ) ( 0 from (4) ) ( ) ( from () Now brng the frst term to the left-handsde, for =,,n

13 3 Wrtng the above n m m m ) ( ) ( The above equaton, when wrtten for =,,n, can be expressed n matrx form as ) ( ) ( m n n n m n n m (5) The matrx on the left-hand sde s the transpose of the upper left-hand submatrx of the power flow Jacoban (we called t J θ ), and so codes are readly avalable to compute t. The elements of the rght-handsde vector may be found by dfferentatng the real power equaton for bus, whch s: ) sn( ) cos( N B V V (6) wth respect to each angle, resultng n

14 V V sn cos B The soluton vector contans the nverse of the penalty factors n the frst m- terms. 4.0 Usng loss formula The method of loss formula results n an approxmate expresson gven by T T 0 B 00 B B (7) where s the vector of generaton T (8) m Development of the coeffcent matrces n (7) has been done n several ways. The frst edton of the W&W text (986) presented a method developed by Meyer [] n Appendx B of chapter 4; t was removed from the second edton. I developed another method based on the wor of Kron, whch s partally artculated n the boo by El-Harawry and Chrstenson, and attached to the end of these notes. 4

15 Some mportant smlartes n the methods:. Both are dependent on the followng assumptons: Each bus can be clearly dstngushed as ether a load bus or a generaton bus. Reactve generaton vares lnearly wth generaton,.e., Q g =Q go +f g.. Both end up wth expressons for of the same form. 3. Both expressons for are dependent on the elements of the Z bus matrx. But there s one major dfference between the formulatons n that Kron s approach maes no assumpton regardng conformng loads. However, the method of W&W (Meyers) does,.e., n Meyer s approach, all loads must ncrease or decrease unformly. We assume that we have the so-called B- coeffcents n the example whch follows. 5

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21 [] A. Bergen and V. Vttal, ower System Analyss, rentce-hall, 000. [] W. Meyer, Effcent computer soluton for Kron and Kron-Early oss Formulas, roc of the 973 ICA conference, IEEE 73 CHO 740-, WR, pp

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