Chapter 4 ECONOMIC DISPATCH AND UNIT COMMITMENT


 Edwin Shields
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1 Chapter 4 ECOOMIC DISATCH AD UIT COMMITMET
2 ITRODUCTIO A power system has several power plants. Each power plant has several generatng unts. At any pont of tme, the total load n the system s met by the generatng unts n dfferent power plants. Economc dspatch control determnes the power output of each power plant, and power output of each generatng unt wthn a power plant, whch wll mnmze the overall cost of fuel needed to serve the system load. We study frst the most economcal dstrbuton of the output of a power plant between the generatng unts n that plant. The method we develop also apples to economc schedulng of plant outputs for a gven system load wthout consderng the transmsson loss. ext, we express the transmsson loss as a functon of output of the varous plants. Then, we determne how the output of each of the plants of a system s scheduled to acheve the total cost of generaton mnmum, smultaneously meetng the system load plus transmsson loss.
3 IUT OUTUT CURVE OF GEERATIG UIT ower plants consstng of several generatng unts are constructed nvestng huge amount of money. Fuel cost, staff salary, nterest and deprecaton charges and mantenance cost are some of the components of operatng cost. Fuel cost s the major porton of operatng cost and t can be controlled. Therefore, we shall consder the fuel cost alone for further consderaton.
4 To get dfferent output power, we need to vary the fuel nput. Fuel nput can be measured n Tonnes / hour or Mllons of Btu / hour. Knowng the cost of the fuel, n terms of Rs. / Tonne or Rs. / Mllons of Btu, nput to the generatng unt can be expressed as Rs / hour. Let C Rs / h be the nput cost to generate a power of MW n unt. Fg. shows a typcal nput output curve of a generatng unt. For each generatng unt there shall be a mnmum and a maxmum power generated as mn and max. Input C n Rs / h mn max n MW Output Fg. InputOutput curve of a generatng unt
5 If the nputoutput curve of unt s quadratc, we can wrte C α β γ Rs / h () A power plant may have several generator unts. If the nputoutput characterstc of dfferent generator unts are dentcal, then the generatng unts can be equally loaded. But generatng unts wll generally have dfferent nputoutput characterstc. Ths means that, for partcular nput cost, the generator power wll be dfferent for dfferent generatng unts n a plant. 3 ICREMETAL COST CURVE As we shall see, the crteron for dstrbuton of the load between any two unts s based on whether ncreasng the generaton of one unt, and decreasng the generaton of the other unt by the same amount results n an ncrease or decrease n total cost. Ths can be obtaned f we can calculate the change n nput cost ΔC for a small change n power Δ. Snce dc = d Δ ΔC we can wrte ΔC = dc d Δ
6 ΔC = dc d Δ Thus whle decdng the optmal schedulng, we are concerned wth dc, ICREMETAL COST (IC) whch s determned by the slopes of the nputoutput curves. Thus the ncremental cost curve s the plot of versus dc d. The dmenson of dc d s Rs / MWh. The unt that has the nput output relaton as C α β γ Rs / h () has ncremental cost (IC) as IC dc α β () d d Here α, β and γ are constants.
7 A typcal plot of IC versus power output s shown n Fg.. IC n Rs / MWh Lnear approxmaton Actual ncremental cost Fg. Incremental cost curve n MW
8 Ths fgure shows that ncremental cost s qute lnear wth respect to power output over an apprecable range. In analytcal work, the curve s usually approxmated by one or two straght lnes. The dashed lne n the fgure s a good representaton of the curve. We now have the background to understand the prncple of economc dspatch whch gudes dstrbuton of load among the generatng unts wthn a plant.
9 4 ECOOMICAL DIVISIO OF LAT LOAD BETWEE GEERATIG UITS I A LAT Varous generatng unts n a plant generally have dfferent nputoutput characterstcs. Suppose that the total load n a plant s suppled by two unts and that the dvson of load between these unts s such that the ncremental cost of one unt s hgher than that of the other unt. ow suppose some of the load s transferred from the unt wth hgher ncremental cost to the unt wth lower ncremental cost. Reducng the load on the unt wth hgher ncremental cost wll result n greater reducton of cost than the ncrease n cost for addng the same amount of load to the unt wth lower ncremental cost. The transfer of load from one to other can be contnued wth a reducton of total cost untl the ncremental costs of the two unts are equal. Ths s llustrated through the characterstcs shown n Fg. 3
10 IC IC IC Fg. 3 Two unts case Intally, IC > IC. Decrease the output power n unt by Δ and ncrease output power n unt by Δ. ow IC Δ > IC Δ. Thus there wll be more decrease n cost and less ncrease n cost brngng the total cost lesser. Ths change can be contnued untl IC = IC at whch the total cost wll be mnmum. Further reducton n and ncrease n wll n IC >IC callng for decrease n and ncrease n untl IC = IC. Thus the total cost wll be mnmum when the ICREMETAL COSTS ARE EQUAL.
11 The same reasonng can be extended to a plant wth more than two generatng unts also. In ths case, f any two unts have dfferent ncremental costs, then n order to decrease the total cost of generaton, decrease the output power n unt havng hgher IC and ncrease the output power n unt havng lower IC. When ths process s contnued, a stage wll reach wheren ncremental costs of all the unts wll be equal. ow the total cost of generaton wll be mnmum. Thus the economcal dvson of load between unts wthn a plant s that all unts must operate at the same ncremental cost. ow we shall get the same result mathematcally.
12 Consder a plant havng number of generatng unts. Inputoutput curve of the unts are denoted as C ( ),C( )......, C ( ). Our problem s, for a gven load demand D, fnd the set of s whch mnmzes the cost functon C T = C ( ) C ( )..... C ( ). (3) subject to the constrants D (..... ) 0 (4) and mn =,,.., (5) max Omttng the nequalty constrants for the tme beng, the problem to be solved becomes Mnmze C C ( ) (6) T subject to D 0 Ths optmzng problem wth equalty constrant can be solved by the method of Lagrangan multplers. In ths method, the Lagrangan functon s formed by augmentng the equalty constrants to the objectve functon usng proper Lagrangan multplers. For ths case, Lagrangan functon s (7)
13 D L(,,......,,λ ) C ( ) λ ( ) (8) where λ s the Lagrangan multpler. ow ths Lagrangan functon has to be mnmzed wth no constrants on t. The necessary condtons for a mnmum are L L 0 =,,.., and 0. λ For a plant wth 3 unts L ( 3,,, λ ) = C ( ) C ( ) C 3 (3 ) λ(d 3 ) ecessary condtons for a mnmum are L C + λ (  ) = 0 L C + λ (  ) = 0 L 3 C λ (  ) = 0 L λ D 3 0
14 Generalzng the above, the necessary condtons are C λ =,,., and D 0 (9) (0) Here C s the change n producton cost n unt for a small change n generaton n unt. Snce change n generaton n unt wll affect the producton cost of ths unt ALOE, we can wrte C dc d Usng eqn.() n eqn. (9) we have () dc d λ =,,., ()
15 Thus the soluton for the problem Mnmze CT C ( ) subject to D 0 s obtaned when the followng equatons are satsfed. dc d λ =,,., (3) and D 0 (4) The above two condtons gve + number of equatons whch are to be solved for the + number of varables λ,. Equaton (3),,....., smply says that at the mnmum cost operatng pont, the ncremental cost for all the generatng unts must be equal. Ths condton s commonly known as EQUAL ICREMETAL COST RULE. Equaton (4) s known as OWER BALACE EQUATIO.
16 It s to be remembered that we have not yet consdered the nequalty constrants gven by mn =,,.., max Fortunately, f the soluton obtaned wthout consderng the nequalty constrants satsfes the nequalty constrants also, then the obtaned soluton wll be optmum. If for one or more generator unts, the nequalty constrants are not satsfed, the optmum strategy s obtaned by keepng these generator unts n ther nearest lmts and makng the other generator unts to supply the remanng power as per equal ncremental cost rule.
17 EXAMLE The cost characterstc of two unts n a plant are: C = K Rs./h C = K Rs. / h where and are power output n MW. Fnd the optmum load allocaton between the two unts, when the total load s 6.5 MW. What wll be the daly loss f the unts are loaded equally? SOLUTIO Incremental costs are: IC = Rs. / MW h IC = Rs. / MW h Usng the equal ncremental cost rule λ and λ λ 60 λ 0 Snce + = 6.5 we get e. λ [ ] 6.5.e..36 λ =
18 .36 λ = Ths gves λ = 0 Rs / MWh Knowng λ and λ Optmum load allocaton s MW and 00 MW When the unts are equally loaded, 8.5 MW and we have devated from the optmum value of = 6.5 MW and = 00 MW. Knowng C = K Rs./h C = K Rs. / h Daly loss can also be computed by calculatng the total cost C T, whch s C + C, for the two schedules. Thus, Daly loss = 4 x [ C (8.5,8.5) C (6.5,00)] T = 4 x [ K + K ( K + K )] = Rs T
19 EXAMLE A power plant has three unts wth the followng cost characterstcs: C C C Rs / h Rs / h Rs/h where s are the generatng powers n MW. The maxmum and mnmum loads allowable on each unt are 50 and 39 MW. Fnd the economc schedulng for a total load of ) 30 MW ) 00 MW SOLUTIO Knowng the cost characterstcs, ncremental cost characterstcs are obtaned as IC IC IC Rs / MWh 70 Rs / MWh 60 Rs / MWh Usng the equal ncremental cost rule = λ; = λ; = λ
20 = λ; = λ; = λ Case ) Total load = 30 MW Snce = 30 we have λ 5.0 λ 70.0 λ e. λ [ ] e..43 λ = Ths gves λ = RM / MWh Thus = ( ) /.0 = MW = ( ) /.0 = MW 3 = ( ) /.4 = MW All ' s le wthn maxmum and mnmum lmts. Therefore, economc schedulng s = MW; = MW; 3 = MW
21 Case ) Total load = 00 MW Snce = 00 we have λ [ ] 00.e..49 λ = Ths gves λ = 300 Rs / MWh Thus = ( ) /.0 = 85 MW = ( ) /.0 = 5 MW 3 = ( ) /.4 = 00 MW It s noted that mn. Therefore s set at the mn. value of 39 MW. Then MW. Ths power has to be scheduled between unts and 3. Therefore λ [ ] 6.e..749 λ = Ths gves λ = 86 Rs / MWh Thus = ( 865 ) /.0 = 7 MW 3 = ( ) /.4 = 90 MW and 3 are wthn the lmts. Therefore economc schedulng s 7MW; 39 MW; 3 90 MW
22 EXAMLE 3 Incremental cost of two unts n a plant are: IC Rs / MWh IC Rs / MWh where and are power output n MW. Assume that both the unts are operatng at all tmes. Total load vares from 50 to 50 MW and the mnmum and maxmum loads on each unt are 0 and 5 MW respectvely. Fnd the ncremental cost and optmal allocaton of loads between the unts for varous total loads and furnsh the results n a graphcal form. SOLUTIO For lower loads, IC of unt s hgher and hence t s loaded to mnmum value.e. = 0 MW. Total mnmum load beng 50 MW, when = 0 MW, must be equal to 30 MW. Thus ntally = 0 MW, IC = 76 Rs / MWh, = 30 MW and IC = 47 Rs / MWh. As the load ncreased from 50 MW, load on unt wll be ncreased untl ts IC.e. IC reaches a value of 76 Rs / MWh. When IC = 76 Rs / MWh load on unt s = ( 76 0 ) / 0.9 = 6. MW. Untl that pont s reached, shall reman at 0 MW and the plant IC,.e. λ s determned by unt.
23 When the plant IC, λ s ncreased beyond 76 Rs / MWh, unt loads are calculated as = ( λ  60 ) / 0.8 MW = ( λ  0 ) / 0.9 MW Then the load allocaton wll be as shown below. lant IC λ Load on unt Load on unt Total load Rs/MWh MW MW + MW
24 Load on unt reaches the maxmum value of 5 MW, when λ = (0.9 x 5) + 0 = 3.5 Rs / Mwh. When the plant IC, λ ncreases further, shall reman at 5 MW and the load on unt alone ncreases and ts value s computed as = ( λ  60 ) / 0.8 MW. Such load allocatons are shown below. lant IC λ Load on unt Load on unt Total load Rs / MWh MW MW + MW The results are shown n graphcal form n Fg. 4
25 / MW Total load D Fg. 4 Load allocaton for varous plant load
26 Alternatve way of explanaton for ths problem s shown n Fg. 5 IC Rs/MWh , 76 30, 47 6., , , 60 5, / MW Fg. 5 Load allocaton between two unts
27 5 TRASMISSIO LOSS Generally, n a power system, several plants are stuated at dfferent places. They are nterconnected by long transmsson lnes. The entre system load along wth transmsson loss shall be met by the power plants n the system. Transmsson loss depends on ) lne parameters ) bus voltages and ) power flow. Determnaton of transmsson loss requres complex computatons. However, wth reasonable approxmatons, for a power system wth number of power plants, transmsson loss can be represented as L B B B B B B B B B (5) where respectvely. are the powers suppled by the plants,,.,,,...,
28 L B B B B B B B B B (5) From eq.(5) L = n n n n n n n n n B B B = n n n B + n n n B +. + n n n B = n B n m m n m = n B n m m n m Thus L can be wrtten as L = n B n m m n m (6)
29 L = m n m B mn n (6) When the powers are n MW, the B mn coeffcents are of dmenson / MW. If powers are n perunt, then B mn coeffcents are also n perunt. Loss coeffcent matrx of a power system shall be determned before hand and made avalable for economc dspatch. For a two plant system, the expresson for the transmsson loss s L B B B B B B B B = B B B B
30 Snce B mn coeffcent matrx s symmetrc, for two plant system L B B B (7) In later calculatons we need the Incremental Transmsson Loss ( ITL ), L. For two plant system L B B (8) L B B (9) Ths can be generalzed as L m n B mn n m =,,., (0)
31 6 ECOOMIC DIVISIO OF SYSTEM LOAD BETWEE VARIOUS LATS I THE OWER SYSTEM It s to be noted that dfferent plants n a power system wll have dfferent cost characterstcs. Consder a power system havng number of plants. Inputoutput characterstcs of the plants are denoted as C ( ),C ( ),......, C ( ). Our problem s for a gven system load demand D, fnd the set of plant generaton whch mnmzes the cost functon,,...., C T = C ( ) C ( )..... C ( ) () subject to the constrants D (..... ) 0 () L and mn I =,,., (3) max Inequalty constrants are omtted for the tme beng.
32 The problem to be solved becomes Mnmze C C ( ) (4) T subject to 0 D L (5) For ths case the Lagrangan functon s L(,,....,, λ) = C ( ) λ ( D ) (6) L Ths Lagrangan functon has to be mnmzed wth no constrant on t. The necessary condtons for a mnmum are L 0 =,,.., L and 0. λ
33 L(.,,...,, λ) = ( λ ( ) C ) L D (6) For a system wth plants L ( λ,, ) = ) λ( ) ( C ) ( C L D L C + λ ( L  ) = 0 L C + λ ( L  ) = 0 0 λ L L D
34 Generalzng ths, for system wth plants, the necessary condtons for a mnmum are C λ [ L ] 0 =,,.., (7) and 0 D L (8) As dscussed n earler case, we can wrte C dc d (9) Usng eqn. (9) n eqn. (7) we have dc d L λ λ =,,.., (30) and 0 D L (3)
35 dc d L λ λ =,,.., (30) and 0 D L These equatons can be solved for the plant generatons.,,...., (3) As shown n the next secton the value of λ n eqn. (30) s the ICREMETAL COST OF RECEIVED OWER. The eqns. descrbed n eqn. (30) are commonly known as COORDIATIO EQUATIOS as they lnk the ncremental cost of plant dc d, ncremental cost of receved power λ and the ncremental transmsson loss ( ITL ) L. Equaton (3) s the OWER BALACE EQUATIO.
36 Coordnaton equatons dc d L λ λ =,,.., (30) can also be wrtten as IC λ ITL λ =,,.., (3) The number of coordnate equatons together wth the power balance equaton are to be solved for the plant loads economc schedule. to obtan the,,....,
37 7 ICREMETAL COST OF RECEIVED OWER The value of λ n the coordnaton equatons s the ncremental cost of receved power. Ths can be proved as follows: The coordnaton equatons C λ L λ can be wrtten as C λ [ L ] (33).e. Δ C Δ Δ L Δ λ.e. ΔC Δ Δ L λ (34) Snce D L D L.e. Δ D Δ Δ (35) L Usng eqn,(35) n eqn.(34) gves Δ C Δ D C = λ D (36) Thus λ s the ncremental cost of receved power.
38 8 EALTY FACTORS To have a better feel about the coordnaton equatons, let us rewrte the dc L same as λ [ ] =,,.., (37) d Thus [ L ] dc d λ =,,.., (38) The above equaton s often wrtten as dc L = λ =,,.., (39) d where, L whch s called the EALTY FACTOR of plant, s gven by L = L =,,.., (40) The results of eqn. (39) means that mnmum fuel cost s obtaned when the ncremental cost of each plant multpled by ts penalty factor s the same for the plants n the power system.
39 9 OTIMUM SCHEDULIG OF SYSTEM LOAD BETWEE LATS  SOLUTIO ROCEDURE To determne the optmum schedulng of system load between plants, the data requred are ) system load, ) ncremental cost characterstcs of the plants and ) loss coeffcent matrx. The teratve soluton procedure s: Step For the frst teraton, choose sutable ntal value of λ. Whle fndng ths, one way s to assume that the transmsson losses are zero and the plants are loaded equally. Step Knowng C α β γ.e. IC = α + β substtute the value of λ nto the coordnaton equatons dc d L λ λ =,,..,.e. ( α β ) λ Bmn n n λ =,,.., The above set of lnear smultaneous equatons are to be solved for the values ' s.
40 Step 3 Compute the transmsson loss L from L = [ ] [ B ] [ t ] where [ ] = [.... ] and [ B ] s the loss coeffcent matrx. Step 4 Compare wth D + L to check the power balance. If the power balance s satsfed wthn a specfed tolerance, then the present soluton s the optmal soluton; otherwse update the value of λ. Frst tme updatng can be done judcously. Value of λ s ncreased by about 5% Value of λ s decreased by about 5% f f D + L. D + L In the subsequent teratons, usng lnear nterpolaton, value of λ can be updated as
41 k k L D k k k k k k ] [ λ λ λ λ (4) Here k, k and k+ are the prevous teraton count, present teratve count and the next teraton count respectvely. Step 5 Return to Step and contnue the calculatons of Steps, 3 and 4 untl the power balance equaton s satsfed wth desred accuracy. The above procedure s now llustrated through an example. k k k λ k λ k λ D + L k
42 EXAMLE 4 Consder a power system wth two plants havng ncremental cost as IC.0 00 Rs / MWh IC.0 50 Rs / MWh Loss coeffcent matrx s gven by B = Fnd the optmum schedulng for a system load of 00 MW. SOLUTIO Assume that there s no transmsson loss and the plants are loaded equally. Then MWh. 50MW. Intal value of λ = (.0 x 50 ) + 00 = 50 Rs / Coordnaton equatons dc d λ L λ ( 0.00 ( ) 50 ) 50.e and = ; On solvng 00 = MW and = MW
43 L = MW = + = MW and MW D L Snce < D L, λ value should be ncreased. It s ncreased by 4 %. ew value of λ = 50 x.04 = 60 Rs / MWh. Coordnaton equatons: ( 0.00 ( ) ) e and =.48 ; On solvng 0 = MW and = MW
44 L = MW = MW ; MW D L D L Knowng two values of λ and the correspondng total generaton powers, new value of λ s computed as λ k λ k λ k k λ k k [ D k L k ] λ ( ) 63 Rs / MWh Wth ths new value of λ, coordnaton procedure has to be repeated. equatons are formed and the
45 The followng table shows the results obtaned. λ L + D + L Optmum schedule s 5.40MW MW For ths transmsson loss s MW
46 0 BASE OIT AD ARTICIATIO FACTORS The system load wll keep changng n a cyclc manner. It wll be hgher durng day tme and early evenng when ndustral loads are hgh. However durng nght and early mornng the system load wll be much less. The optmal generatng schedulng need to be solved for dfferent load condtons because load demand D keeps changng. When load changes are small, t s possble to move from one optmal schedule to another usng ARTICIATIG FACTORS. We start wth a known optmal generaton schedule, 0, 0,, 0, for a partcular load D. Ths schedule s taken as BASE OIT and the correspondng ncremental cost s λ 0. Let there be a small ncrease n load of Δ D. To meet wth ths ncreased load, generatons are to be ncreased as Δ, Δ,., Δ. Correspondngly ncremental cost ncreases by Δλ.
47 Knowng that for th unt, C = α + β + γ, ncremental cost s IC = α + β = λ (4) Small change n ncremental cost and correspondng change n generaton are related as Δλ = α (43) Δ Thus Δ = Δλ α for =,,, (44) Total change n generatons s equal to the change n load. Therefore Δ = Δ D.e. Δ D = Δλ α (45) From the above two equatons Δ Δ D α α = k for =,,, (46) The rato Δ s known as the ARTICIATIO FACTOR of generator, Δ D represented as k. Once all the k s, are calculated from eq.(46), the change n generatons are gven by Δ = k Δ D for =,,, (47)
48 EXAMLE 5 Incremental cost of three unts n a plant are: IC = Rs / MWh; IC = Rs / MWh; and IC 3 = Rs / MWh where, and 3 are power output n MW. Fnd the optmum load allocaton when the total load s 4.5 MW. Usng artcpatng Factors, determne the optmum schedulng when the load ncreases to 50 MW. Soluton Usng the equal ncremental cost rule λ ; λ ; = λ λ 60 λ 0 λ 0 Snce = 4.5 we get e λ [ ] 4.5.e. 3.6 λ = Ths gves λ = 0 Rs / MWh Optmum load allocaton s MW ; 00 MW ; 3 80 MW
49 artcpaton Factors are: k = k = k 3 = = 3.6 = = = = 3.6 = 0.53 Change n load Δ D = = 7.5 MW Change n generatons are: Δ = x 7.5 =.9655 MW Δ = x 7.5 =.6363 MW Δ 3 = 0.53 x 7.5 =.898 MW Thus optmum schedule s: = MW; = MW; 3 = MW
50 Example 6 A power plant has two unts wth the followng cost characterstcs: C = Rs / hour C = Rs / hour where and are the generatng powers n MW. The daly load cycle s as follows: 6:00 A.M. to 6:00.M. 50 MW 6:00.M. to 6:00 A.M. 50 MW The cost of takng ether unt off the lne and returnng to servce after hours s Rs Maxmum generaton of each unt s 00 MW. Consderng 4 hour perod from 6:00 A.M. one mornng to 6:00 A.M. the next mornng
51 a. Would t be economcal to keep both unts n servce for ths 4 hour perod or remove one unt from servce for hour perod from 6:00.M. one evenng to 6:00 A.M. the next mornng? b. Compute the economc schedule for the peak load and off peak load condtons. c. Calculate the optmum operatng cost per day. d. If operatng one unt durng off peak load s decded, up to what cost of takng one unt off and returnng to servce after hours, ths decson s acceptable? e. If the cost of takng one unt off and returnng to servce after hours exceeds the value calculated n d, what must be done durng off peak perod?
52 Soluton To meet the peak load of 50 MW, both the unts are to be operated. However, durng 6:00 pm to 6:00 am, load s 50 MW and there s a choce ) both the unts are operatng ) one unt (ether or to be decded) s operatng ) When both the unts are operatng IC = Rs / MWh IC = Rs / MWh λ Usng equal IC rule λ 50 = 50;.5 λ = and λ = Therefore = MW; = MW Then C T = C ( = ) + C ( = ) = Rs / h For hour perod, cost of operaton = Rs
53 ) Cost of operaton for hours = Rs ) If unt s operatng, C Ι = 50 = 3500 Rs / h If unt s operatng, C Ι = 50 = 3000 Rs / h Between unts and, t s economcal to operate unt. If only one unt s operatng durng offpeak perod, cost towards takng out and connectng t back also must be taken. Therefore, for offpeak perod (wth unt alone operatng) cost of operaton = (3000 x ) = Rs Between the two choces () and (), choce () s cheaper. Therefore, durng 6:00 pm to 6:00 am, t s better to operate unt alone.
54 b. Durng the peak perod, D = 50 MW. Wth equal IC rule λ Therefore = 86. MW; Thus, economc schedule s: λ 50 = 50;.5 λ = and λ = = MW Durng 6:00 am to 6:00 pm = 86. MW; = MW Durng 6:00 pm to 6:00 am = 0; = 50 MW c. Cost of operaton = [ C Ι = C Ι = ] for peak perod = x = Rs Cost of operaton for offpeak perod = Rs 6000 Therefore, optmal operatng cost per day = Rs
55 d. If both the unts are operatng durng offpeak perod, cost of operaton = Rs 7833 If unt alone s operatng durng = Rs (3000 x ) + x offpeak perod, cost of operaton = Rs x For Crtcal value of x: x = 7833 x = Rs 5833 Therefore, untl the cost of takng one unt off and returnng t to servce after hours, s less than Rs 5833, operatng unt alone durng the offpeak perod s acceptable. e. If the cost of takng one unt off and returnng t to servce exceeds Rs.5833, then both the unts are to be operated all through the day.
56 UIT COMMITMET Economc dspatch gves the optmum schedule correspondng to one partcular load on the system. The total load n the power system vares throughout the day and reaches dfferent peak value from one day to another. Dfferent combnaton of generators, are to be connected n the system to meet the varyng load. When the load ncreases, the utlty has to decde n advance the sequence n whch the generator unts are to be brought n. Smlarly, when the load decreases, the operatng engneer need to know n advance the sequence n whch the generatng unts are to be shut down. The problem of fndng the order n whch the unts are to be brought n and the order n whch the unts are to be shut down over a perod of tme, say one day, so the total operatng cost nvolved on that day s mnmum, s known as Unt Commtment (UC) problem. Thus UC problem s economc dspatch over a day. The perod consdered may a week, month or a year.
57 But why s ths problem n the operaton of electrc power system? Why not just smply commt enough unts to cover the maxmum system load and leave them runnng? ote that to commt means a generatng unt s to be turned on ; that s, brng the unt up to speed, synchronze t to the system and make t to delver power to the network. Commt enough unts and leave them on lne s one soluton. However, t s qute expensve to run too many generatng unts when the load s not large enough. As seen n prevous example, a great deal of money can be saved by turnng unts off (decommtng them) when they are not needed. Example 7 The followng are data pertanng to three unts n a plant. Unt : Mn. = 50 MW; Max. = 600 MW C = Rs / h Unt : Mn. = 00 MW; Max. = 400 MW C = Rs / h Unt 3: Mn. = 50 MW; Max. = 00 MW C 3 = Rs / h What unt or combnaton of unts should be used to supply a load of 550 MW most economcally?
58 Soluton To solve ths problem, smply try all combnaton of three unts. Some combnatons wll be nfeasble f the sum of all maxmum MW for the unts commtted s less than the load or f the sum of all mnmum MW for the unts commtted s greater than the load. For each feasble combnaton, unts wll be dspatched usng equal ncremental cost rule studed earler. The results are presented n the Table below.
59 Unt Mn Max Unt Unt Unt 3 Mn. Gen Max. Gen 3 Total cost Off Off Off 0 0 Infeasble On Off Off Off On Off Infeasble Off Off On Infeasble On On Off Off On On On Off On On On On ote that the least expensve way of meetng the load s not wth all the three unts runnng, or any combnaton nvolvng two unts. Rather t s economcal to run unt one alone.
60 Example 8 Daly load curve to be met by a plant havng three unts s shown below. 00 MW 500 MW noon 4 pm 8 pm am 6 am noon Data pertanng to the three unts are the same n prevous example. Startng from the load of 00 MW, takng steps of 50 MW fnd the shutdown rule.
61 Soluton For each load startng from 00 MW to 500 MW n steps of 50 MW, we smply use a bruteforce technque wheren all combnatons of unts wll be tred as n prevous example. The results obtaned are shown below. Load Optmum combnaton Unt Unt Unt 3 00 On On On 50 On On On 00 On On On 050 On On On 000 On On Off 950 On On Off 900 On On Off 850 On On Off 800 On On Off 750 On On Off 700 On On Off 650 On On Off 600 On Off Off 550 On Off Off 500 On Off Off
62 Load Optmum combnaton Unt Unt Unt 3 00 On On On 50 On On On 00 On On On 050 On On On 000 On On Off 950 On On Off 900 On On Off 850 On On Off 800 On On Off 750 On On Off 700 On On Off 650 On On Off 600 On Off Off 550 On Off Off 500 On Off Off The shutdown rule s qute smple. When load s above 000 MW, run all three unts; more than 600 MW and less than 000 MW, run unts and ; below 600 MW, run only unt.
63 The above shutdown rule s qute smple; but t fals to take the economy over a day. In a power plant wth unts, for each load step, (neglectng the number of nfeasble solutons) economc dspatch problem s to solved for ( ) tmes. Durng a day, f there are M load steps, (snce each combnaton n one load step can go wth each combnaton of another load step) to arrve at the economy over a day, n ths bruteforce technque, economc dspatch problem s to be solved for ( ) M. Ths number wll be too large for practcal case. UC problem become much more complcated when we need to consder power system havng several plants each plant havng several generatng unts and the system load to be served has several load steps. So far, we have only obeyed one smple constrant: Enough unts wll be connected to supply the load. There are several other constrants to be satsfed n practcal UC problem.
64 COSTRAITS O UC ROBLEM Some of the constrants that are to be met wth whle solvng UC problem are lsted below.. Spnnng reserve: There may be sudden ncrease n load, more than what was predcted. Further there may be a stuaton that one generatng unt may have to be shut down because of fault n generator or any of ts auxlares. Some system capacty has to be kept as spnnng reserve ) to meet an unexpected ncrease n demand and ) to ensure power supply n the event of any generatng unt sufferng a forced outage.. Mnmum up tme: When a thermal unt s brought n, t cannot be turned off mmedately. Once t s commtted, t has to be n the system for a specfed mnmum up tme. 3. Mnmum down tme: When a thermal unt s decommtted, t cannot be turned on mmedately. It has to reman decommtted for a specfed mnmum down tme.
65 4. Crew constrant: A plant always has two or more generatng unts. It may not be possble to turn on more than one generatng unt at the same tme due to nonavalablty of operatng personnel. 5. Transton cost: Whenever the status of one unt s changed some transton cost s nvolved and ths has to be taken nto account. 6. Hydro constrants: Most of the systems have hydroelectrc unts also. The operaton of hydro unts, depend on the avalablty of water. Moreover, hydroprojects are multpurpose projects. Irrgaton requrements also determne the operaton of hydro plants.
66 7. uclear constrant: If a nuclear plant s part of the system, another constrant s added. A nuclear plant has to be operated as a base load plant only. 8. Must run unt: Sometme t s a must to run one or two unts from the consderaton of voltage support and system stablty. 9. Fuel supply constrant: Some plants cannot be operated due to defcent fuel supply. 0. Transmsson lne lmtaton: Reserve must be spread around the power system to avod transmsson system lmtaton, often called bottlng of reserves.
67 RIORITY LIST METHOD In ths method the full load average producton cost of each unt s calculated frst. Usng ths, prorty lst s prepared. Full load average producton of a unt roducton cost correspondng Full load to full load Example 9 The followng are data pertanng to three unts n a plant. Unt : Max. = 600 MW C = Rs / h Unt : Max. = 400 MW C = Rs / h Unt 3: Max. = 00 MW Obtan the prorty lst C 3 = Rs / h
68 Soluton Full load average producton of a unt x x Full load average producton of a unt x x Full load average producton of a unt x x A strct prorty order for these unts, based on the average producton cost, would order them as follows: Unt Rs. / h Max. MW
69 The shutdown scheme would (gnorng mn. up / down tme, start up costs etc.) smply use the followng combnatons. Combnaton Load D MW D < 00 MW MW D < 000 MW D < 400 MW ote that such a scheme would not gve the same shut down sequence descrbed n Example 7 wheren unt was shut down at 600 MW leavng unt. Wth the prorty lst scheme both unts would be held on untl load reached 400 MW, then unt would be dropped.
70 Most prorty schemes are bult around a smple shut down algorthm that mght operate as follows: At each hour when the load s droppng, determne whether droppng the next unt on the prorty lst wll leave suffcent generaton to supply the load plus spnnng reserve requrements. If not, contnue operatng as s; f yes, go to next step. Determne the number of hours, H, before the unt wll be needed agan assumng the load s ncreasng some hours later. If H s less than the mnmum shut down tme for that unt, keep the commtment as t s and go to last step; f not, go to next step. Calculate the two costs. The frst s the sum of the hourly producton costs for the next H hours wth the unt up. Then recalculate the same sum for the unt down and add the start up cost. If there s suffcent savng from shuttng down the unt, t should be shut down; otherwse keep t on. Repeat the entre procedure for the next unt on the prorty lst. If t s also dropped, go to the next unt and so forth.
71 Questons on Economc Dspatch and Unt Commtment. What do you understand by Economc Dspatch problem?. For a power plant havng generator unts, derve the equal ncremental cost rule. 3. The cost characterstcs of three unts n a power plant are gven by C = 0.5 C = 0.6 C 3 = Rs / hour Rs / hour Rs / hour where, and 3 are generatng powers n MW. Maxmum and mnmum loads on each unt are 5 MW and 0 MW respectvely. Obtan the economc dspatch when the total load s 60 MW. What wll be the loss per hour f the unts are operated wth equal loadng?
72 4. The ncremental cost of two unts n a power statons are: dc d dc d = Rs / hour = Rs / hour a) Assumng contnuous runnng wth a load of 50 MW, calculate the savng per hour obtaned by usng most economcal dvson of load between the unts as compared to loadng each equally. The maxmum and mnmum operatonal loadngs of both the unts are 5 and 0 MW respectvely. b. What wll be the savng f the operatng lmts are 80 and 0 MW?
73 5. A power plant has two unts wth the followng cost characterstcs: C = Rs / hour C = Rs / hour where and are the generatng powers n MW. The daly load cycle s as follows: 6:00 A.M. to 6:00.M. 50 MW 6:00.M. to 6:00 A.M. 50 MW The cost of takng ether unt off the lne and returnng to servce after hours s Rs Consderng 4 hour perod from 6:00 A.M. one mornng to 6:00 A.M. the next mornng a. Would t be economcal to keep both unts n servce for ths 4 hour perod or remove one unt from servce for hour perod from 6:00.M. one evenng to 6:00 A.M. the next mornng? b. Compute the economc schedule for the peak load and off peak load condtons. c. Calculate the optmum operatng cost per day. d. If operatng one unt durng off peak load s decded, up to what cost of takng one unt off and returnng to servce after hours, ths decson s acceptable?
74 6. What do you understand by Loss coeffcents? 7. The transmsson loss coeffcents B mn, expressed n MW  of a power system network havng three plants are gven by B = Three plants supply powers of 00 MW, 00 MW and 300 MW respectvely nto the network. Calculate the transmsson loss and the ncremental transmsson losses of the plants. 8. Derve the coordnaton equaton for the power system havng number of power plants.
75 9. The fuel nput data for a three plant system are: f = Mllons of BTU / hour f = Mllons of BTU / hour f 3 = Mllons of BTU / hour where s are the generaton powers n MW. The fuel cost of the plants are Rs 50, Rs 30 and Rs 40 per Mllon of BTU for the plants, and 3 respectvely. The loss coeffcent matrx expressed n MW  s gven by B = The load on the system s 60 MW. Compute the power dspatch for λ = 0 Rs / MWh. Calculate the transmsson loss. Also determne the power dspatch wth the revsed value of λ takng 0 % change n ts value. Estmate the next new value of λ.
76 0. What are artcpatng Factors? Derve the expresson for artcpatng Factors.. Incremental cost of three unts n a plant are: IC IC IC Rs / MWh Rs / MWh Rs /MWh where, and 3 are power output n MW. Fnd the optmum load allocaton when the total load s 85 MW. Usng artcpatng Factors, determne the optmum schedulng when the load decreases to 75 MW.. What s Unt Commtment problem? Dstngush between Economc Dspatch and Unt Commtment problems.
77 3. Dscuss the constrans on Unt Commtment problem. 4. Explan what s rorty Lst method. ASWERS MW MW 9.45 MW Rs Rs.39 Rs It s economcal to operate unt alone durng the off peak perod. 86. MW MW 0 50 MW Rs Rs MW MW 5.88 MW 8.5 MW MW MW MW Rs / MWh. 3.5 MW; 30.0 MW;.5 MW MW; MW; MW
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