.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:
() 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,
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
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 0 5 3 576 5 50 75 00 55 Increase load by MW at each bus, compensated by gen ncrease at bus 00.8 8 3 576 4 49 76 0 56 3 3 3 ( 0.) 3. 0.9375 Increase load by MW at each bus, compensated by gen ncrease at bus 3 0. 5 3 579 4 5 76 0 56 Fg. 3 3 3 ( 0.).8 3.074 4
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
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
Fg. But why do we actually call them penalty factors? Consder the crteron for optmalty n the EDC wth losses: C ( ),... m (7) 7
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( )=45+0.0. 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 λ=0.98(45+0.0 )=44.+0.96 λ=.0(45+0.0 )=45+0.0 λ=.0(45+0.0 3 )=45.9+0.004 3 uttng these three equatons nto matrx form results n: 300 45.9 45 44. 0 0.004 0 0 0 0.0 0 0 0 0.096 3 Solvng n Matlab yelds: 46.9875 53.3 99.37 47.3 3 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.
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
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
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
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
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
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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[] 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. 48-43.