ECOOMC OEATO OF OWE SYSTEMS TOUCTO Oe of the earliest applicatios of o-lie cetralized cotrol was to provide a cetral facility, to operate ecoomically, several geeratig plats supplyig the loads of the system. Moder itegrated systems have differet types of geeratig plats, such as coal fired thermal plats, hydel plats, uclear plats, oil ad atural gas uits etc. The capital ivestmet, operatio ad maiteace ts are differet for differet types of plats. The operatio ecoomics ca agai be subdivided ito two parts. i) roblem of ecoomic dispatch, which deals with determiig the power output of each plat to meet the specified load, such that the overall fuel t is miimized. ii) roblem of optimal power flow, which deals with miimum loss delivery, where i the power flow, is optimized to miimize losses i the system. this chapter we cosider the problem of ecoomic dispatch. urig operatio of the plat, a geerator may be i oe of the followig states: i) Base supply without regulatio: the output is a costat. ii) Base supply with regulatio: output power is regulated based o system load. iii) Automatic o-ecoomic regulatio: output level chages aroud a base settig as area cotrol error chages. iv) Automatic ecoomic regulatio: output level is adjusted, with the area load ad area cotrol error, while trackig a ecoomic settig. egardless of the uits operatig state, it has a cotributio to the ecoomic operatio, eve though its output is chaged for differet reasos. The factors ifluecig the t of geeratio are the geerator efficiecy, fuel t ad trasmissio losses. The most efficiet geerator may ot give miimum t, sice it may be located i a place where fuel t is high. Further, if the plat is located far from the load ceters, trasmissio losses may be high ad ruig the plat may become uecoomical. The ecoomic dispatch problem basically determies the geeratio of differet plats to miimize total operatig t.
Moder geeratig plats like uclear plats, geo-thermal plats etc, may require capital ivestmet of millios of rupees. The ecoomic dispatch is however determied i terms of fuel t per uit power geerated ad does ot iclude capital ivestmet, maiteace, depreciatio, start-up ad shut dow ts etc. EFOMACE CUES UT-OUTUT CUE This is the fudametal curve for a thermal plat ad is a plot of the iput i British thermal uits (Btu) per hour versus the power output of the plat i MW as show i Fig. Btu / hr (put) (output) MW Fig : put output curve HEAT ATE CUE The heat rate is the ratio of fuel iput i Btu to eergy output i Wh. t is the slope of the iput output curve at ay poit. The reciprocal of heat rate is called fuel efficiecy. The heat rate curve is a plot of heat rate versus output i MW. A typical plot is show i Fig. (Heat rate) Btu / kw-hr (output) MW
Fig. Heat rate curve. CEMETA FUE ATE CUE The icremetal fuel rate is equal to a small chage i iput divided by the correspodig chage i output. put cremetal fuel rate = Output The uit is agai Btu / Wh. A plot of icremetal fuel rate versus the output is show i Fig 3 cremetal fuel rate (output) MW Fig 3: cremetal fuel rate curve cremetal t curve The icremetal t is the product of icremetal fuel rate ad fuel t (s / Btu or $ / Btu). The curve i show i Fig. 4. The uit of the icremetal fuel t is s / MWh or $ /MWh. 3
s / Mwhr approximate liear t actual t (output) MW Fig 4: cremetal t curve geeral, the fuel t F i for a plat, is approximated as a quadratic fuctio of the geerated output i. F i = a i + b i i + c i i s / h The icremetal fuel t is give by i = bi + c i i s / MWh d i The icremetal fuel t is a measure of how tly it will be produce a icremet of power. The icremetal productio t, is made up of icremetal fuel t plus the icremetal t of labour, water, maiteace etc. which ca be take to be some percetage of the icremetal fuel t, istead of resortig to a rigorous mathematical model. The t curve ca be approximated by a liear curve. While there is egligible operatig t for a hydel plat, there is a limitatio o the power output possible. ay plat, all uits ormally operate betwee mi, the miimum loadig limit, below which it is techically ifeasible to operate a uit ad max, which is the maximum output limit. ECOOMC EEATO SCHEU EECT OSSES A EEATO MTS The simplest case of ecoomic dispatch is the case whe trasmissio losses are eglected. The model does ot cosider the system cofiguratio or lie impedaces. Sice losses are eglected, the total geeratio is equal to the total demad. 4
Cosider a system with g umber of geeratig plats supplyig the total demad. f F i is the t of plat i i s/h, the mathematical formulatio of the problem of ecoomic schedulig ca be stated as follows: Miimize Such that where g F i i F T = g i i F T = total t. i = geeratio of plat i. = total demad. This is a costraied optimizatio problem, which ca be solved by agrage s method. AAE METHO FO SOUTO OF ECOOMC SCHEUE The problem is restated below: Miimize g F T F i i Such that i 0 g i The augmeted t fuctio is give by g F T i i The miimum is obtaied whe 0 i ad 0 i F T i 0 g i i 0 5
The secod equatio is simply the origial costrait of the problem. The t of a plat F i depeds oly o its ow output i, hece Usig the above, We ca write F T i F i i Fi i i d i i d i ; i =. g b i + c i i = i =. g The above equatio is called the co-ordiatio equatio. Simply stated, for ecoomic geeratio schedulig to meet a particular load demad, whe trasmissio losses are eglected ad geeratio limits are ot imposed, all plats must operate at equal icremetal productio ts, subject to the costrait that the total geeratio be equal to the demad. From we have We kow i a loss less system i bi c i Substitutig (8.6) we get g i i g bi i ci A aalytical solutio of λ is obtaied from (8.7) as g i g i c b i c i i t ca be see that is depedet o the demad ad the coefficiets of the t fuctio. Example. The fuel ts of two uits are give by 6
F =.5 + 0 + 0. s/h F =.9 + 30 + 0. s/h, are i MW. Fid the optimal schedule eglectig losses, whe the demad is 00 MW. Solutio: d d 0 0. 30 0. s / MWh s / MWh For ecoomic schedule Solvig we get, d d 00 MW 0 + 0. = 30 + 0. (00 - ) = 5 MW = 75 MW = 0 + 0. (5) = 45 s / MWh Example The fuel t i $ / h for two 800 MW plats is give by F = 400 + 6.0 + 0.004 F = 500 + b + c where, are i MW (a) The icremetal t of power, is $8 / MWh whe total demad is 550MW. etermie optimal geeratio schedule eglectig losses. (b) The icremetal t of power is $0/MWh whe total demad is 300 MW. etermie optimal schedule eglectig losses. (c) From (a) ad (b) fid the coefficiets b ad c. Solutio: b 8.0 6.0 a) 50 c 0.004 MW 550 50 300 MW 7
b 0 6 b) 500 C 0.004 MW 300 500 800 MW c) From From b c 8.0 b (a) 300 c 0.0 b (b) 800 c Solvig we get b = 6.8 c = 0.00 ECOOMC SCHEUE CU MTS O EEATO (EECT OSSES) The power output of ay geerator has a maximum value depedet o the ratig of the geerator. t also has a miimum limit set by stable boiler operatio. The ecoomic dispatch problem ow is to schedule geeratio to miimize t, subject to the equality costrait. g i i ad the iequality costrait i (mi) i i (max) ; i =, g The procedure followed is same as before i.e. the plats are operated with equal icremetal fuel ts, till their limits are ot violated. As soo as a plat reaches the limit (maximum or miimum) its output is fixed at that poit ad is maitaied a costat. The other plats are operated at equal icremetal ts. Example 3 cremetal fuel ts i $ / MWh for two uits are give below: d 0.0.0 $ / MWh 8
d 0.0.6 $ / MWh The limits o the plats are mi = 0 MW, max = 5 MW. Obtai the optimal schedule if the load varies from 50 50 MW. Solutio: The icremetal fuel ts of the two plats are evaluated at their lower limits ad upper limits of geeratio. At (mi) = 0 MW. (mi) (mi) = 0.0x 0+.0 =.$ / MWh d = 0.0 x 0 +.6 =.84 $ / MWh d At (Max) =5 Mw (max) (max) = 0.0 x 5 +.0 = 3.5 $ / MWh = 0.0 x 5 +.6 = 3. $ / MWh ow at light loads uit has a higher icremetal t ad hece will operate at its lower limit of 0 MW. itially, additioal load is take up by uit, till such time its icremetal fuel t becomes equal to.$ / MWh at = 50 MW. Beyod this, the two uits are operated with equal icremetal fuel ts. The cotributio of each uit to meet the demad is obtaied by assumig differet values of ; Whe = 3. $ / MWh, uit operates at its upper limit. Further loads are take up by uit. The computatios are show i Table Table lat output ad output of the two uits lat lat Output d $/MWh d $/MWh $/MWh MW MW MW..96.96 0 + 30 50... 0 + 50 70.4.4.4 40 66.7 06.7.6.6.6 60 83.3 43.3.8.8.8 80 00 80 3.0 3.0 3.0 00 6.7 6.7 3. 3. 3. 0 5* 35 9
3.5 3. 3.5 5* 5* 50 For a particular value of, ad are calculated usig (8.6). Fig 8.5 Shows plot of each uit output versus the total plat output. For ay particular load, the schedule for each uit for ecoomic dispatch ca be obtaied. Example 4. example 3, what is the savig i fuel t for the ecoomic schedule compared to the case where the load is shared equally. The load is 80 MW. Solutio: From Table it is see that for a load of 80 MW, the ecoomic schedule is = 80 MW ad = 00 MW. Whe load is shared equally = = 90 MW. Hece, the geeratio of uit icreases from 80 MW to 90 MW ad that of uit decreases from 00 MW to 90 MW, whe the load is shared equally. There is a icrease i t of uit sice icreases ad decrease i t of uit sice decreases. 0
90 crease i t of uit = d 80 d 90 = 0.0.0d 8. 5 80 ecrease i t of uit = 90 90 d 00 d $ / h = 0.0.6d 7. 4 00 $ / h Total icrease i t if load is shared equally = 8.5 7.4 =. $ / h Hece the savig i fuel t is. $ / h if coordiated ecoomic schedule is used. ECOOMC SATCH CU TASMSSO OSSES Whe trasmissio distaces are large, the trasmissio losses are a sigificat part of the geeratio ad have to be cosidered i the geeratio schedule for ecoomic operatio. The mathematical formulatio is ow stated as Miimize g F T F i i Such That where is the total loss. g i i The agrage fuctio is ow writte as g = F T i 0 i The miimum poit is obtaied whe i F T i i 0 ; i = g g i i 0 (Same as the costrait) Sice F T i i d i, (8.7) ca be writte as d i i i
i d i i The term is called the pealty factor of plat i, i. The coordiatio i equatios icludig losses are give by i d i i ; i =. g The miimum operatio t is obtaied whe the product of the icremetal fuel t ad the pealty factor of all uits is the same, whe losses are cosidered. A rigorous geeral expressio for the loss is give by = m m B m + B o + B oo where B m, B o, B oo called loss coefficiets, deped o the load compositio. The assumptio here is that the load varies liearly betwee maximum ad miimum values. A simpler expressio is = m m B m The expressio assumes that all load currets vary together as a costat complex fractio of the total load curret. Experieces with large systems has show that the loss of accuracy is ot sigificat if this approximatio is used. A average set of loss coefficiets may be used over the complete daily cycle i the coordiatio of icremetal productio ts ad icremetal trasmissio losses. geeral, B m = B m ad ca be expaded for a two plat system as = B + B + B
Example 5 A geerator is supplyig a load. A icremetal chage i load of 4 MW requires geeratio to be icreased by 6 MW. The icremetal t at the plat bus is s 30 / MWh. What is the icremetal t at the receivig ed? Solutio: = 30 d d 30 = MW oad = 6MW = 4MW Fig ; Oe lie diagram of example 5 = - = MW at receivig ed is give by d 6 30 45 s / MWh 4 or 30 45 s / MWh d 6 Example 6 a system with two plats, the icremetal fuel ts are give by d d 0.0 0 s / MWh 0.05.5 s / MWh The system is ruig uder optimal schedule with = = 00 MW. f = 0., fid the plat pealty factors ad. 3
Solutio: For ecoomic schedule, d i i i ; For plat, = 00 MW i 0. ; 0.0500.5. Solvig, 30 s / MWh =. 5 0. d =.48 = (0.0x00+0) = 30 i.48 = ; Solvig Example 7 = 0.3 A two bus system is show i Fig. 8.8 f 00 MW is trasmitted from plat to the load, a loss of 0 MW is icurred. System icremetal t is s 30 / MWh. Fid, ad power received by load if d 0.0 6.0 s / MWh d 0.04 0.0 s / MWh oad 4
Fig Oe lie diagram of example 7 Solutio; Sice the load is coected at bus, o loss is icurred whe plat two supplies the load. Therefore i (8.36) B = 0 ad B = 0 B ; B ; 0. 0 From data we have = 0 MW, if = 00 MW 0 = B (00) B = 0.00 MW - Coordiatio equatio with loss is d i i For plat i d (0.0 +6.0) + 30 ( x 0.00 x ) = 30 0.08 = 30-6.0. From which, = 75 MW For lat d 0.04 + 0.0 = 30 or = 50 MW oss = B = 0.00 x (75) = 30.65 MW = ( + ) = 394. 375 MW EATO OF TASMSSO OSS FOMUA A accurate method of obtaiig geeral loss coefficiets has bee preseted by ro. The method is elaborate ad a simpler approach is possible by makig the followig assumptios: (i) All load currets have same phase agle with respect to a commo referece (ii) The ratio X / is the same for all the etwork braches. 5
Cosider the simple case of two geeratig plats coected to a arbitrary umber of loads through a trasmissio etwork as show i Fig a (a) = = 0 (b) = 0 = (c) Fig Two plats coected to a umber of loads through a trasmissio etwork et s assume that the total load is supplied by oly geerator as show i Fig 8.9b. et the curret through a brach i the etwork be. We defie t is to be oted that = i this case. Similarly with oly plat supplyig the load curret, as show i Fig 8.9c, we defie 6
7 ad are called curret distributio factors ad their values deped o the impedaces of the lies ad the etwork coectio. They are idepedet of. Whe both geerators are supplyig the load, the by priciple of superpositio = + where, are the currets supplied by plats ad respectively, to meet the demad. Because of the assumptios made, ad have same phase agle, as do ad. Therefore, the curret distributio factors are real rather tha complex. et ad. where ad are phase agles of ad with respect to a commo referece. We ca write si si = si si si si = ow 3 ad 3 where, are three phase real power outputs of plat ad plat ;, are the lie to lie bus voltages of the plats ad, are the power factor agles. The total trasmissio loss i the system is give by = 3 where the summatio is take over all braches of the etwork ad is the brach resistace. Substitutig we get B B B where B
B B The loss coefficiets are called the B coefficiets ad have uit MW -. For a geeral system with plats the trasmissio loss is expressed as a compact form... p p q q p q, q p p q q k q B q p q p p B q p q q q q q B Coefficiets ca be treated as costats over the load cycle by computig them at average operatig coditios, without sigificat loss of accuracy. Example 8 Calculate the loss coefficiets i pu ad MW - o a base of 50MA for the etwork of Fig below. Correspodig data is give below. a =. j 0.4 pu b = 0.4 - j 0. pu c = 0.8 - j 0. pu d = 0.8 - j 0. pu e =. - j 0.3 pu Z a = 0.0 + j 0.08 pu Z b = 0.08 + j 0.3 pu Z c = 0.0 + j 0.08 pu Z d = 0.03 + j 0. pu Z e = 0.03 + j 0. pu 8
a ref =.0 0 0 b c a b c d d e e Solutio: Total load curret = d + e =.0 j 0.5 =.06-4.03 0 A = d = 0.8 - j 0. = 0.846-4.03 0 A oad oad Fig : Example 8 0.4;.0 0.4 0.6 f geerator, supplies the load the =. The curret distributio is show i Fig a. a b c 0.6 c = 0 = 0 d 0.4 e 0.6 a b a.0; 0.6; Fig a : eerator supplyig the total load 0 b C ; oad oad 0.4 d ; 0.6 e. Similarly the curret distributio whe oly geerator supplies the load is show i Fig b. 9
a = 0 a b 0.4 c d 0.4 e 0.6 oad oad Fig b: eerator supplyig the total load a =0; b = -0.4; c =.0; d = 0.4; e = 0.6 From Fig 8.0, = ref + Z a a = 0 0 + (. j 0.4) (0.0 + j0.08) =.06 4.78 0 =.056 + j 0.088 pu. = ref b Z b + c Z c =.0 0 0 (0.4 j 0.) (0.08 + j 0.3) + (0.8 j 0.) (0.0 + j 0.08) = 0.98 j 0.05 = 0.93-3.0 0 pu. Curret hase agles agle of (= a ) = ta - 0.4 8.43 0. 0. 0 agle of c ta 7. 3 0.8 0. 98 ower factor agles 0 0.78 8.43 3. ; 0.9 4 0 0 0 7.3 3.0 4.03 ; 0.998 B.06 0.90.0 0.0 0.6 0.08 0.4 0.03 0.6 0.03 = 0.0677 pu = 0.0677 x 0.354 0 MW - 50 Cos B 0
0.98.06 0.93 0.998 = - 0.00389 pu = 0.4 0.6 0.08 0.4 0.4 0.03 0.6 0.6 0.03 0.9 = - 0.0078 x 0 - MW - B 0.4 0.08.0 0.0 0.4 0.03 0.6 0.93 0.998 = 0.056pu 0. 0 MW - 0.03