电力系统分析英文课件OptimalDispatch.pdf

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1、1 Instructor:Kai Sun Fall 2014 ECE 421/599 Electric Energy Systems 7 Optimal Dispatch of Generation 2 Background In a practical power system,the costs of generating and delivering electricity from power plants are different(due to fuel costs and distances to load centers)Under normal conditions,the

2、system generation capacity is more than the total load demand and losses.Thus,there is room to schedule generation within capacity limits Minimizing a cost function that represents,e.g.Operating costs Transmission losses System reliability impacts This is called Optimal Power Flow(OPF)problem A typi

3、cal problem is the Economic Dispatch(ED)of real power generation 3 Introduction of Nonlinear Function Optimization Unconstrained parameter optimization Constrained parameter optimization Equality constraints Inequality constraints 4 Unconstrained parameter optimization Minimize cost function 1.Solve

4、 all local minima satisfying two conditions(necessary&sufficient)Condition-1:Gradient vector Condition-2:Hessian matrix H is positive definite 2.Find the global minimum from all local minima 12(,)nf x xx12(,)nffffxxx=0Stationary point (where f is flat in all directions)Local minimum(a pure source in

5、 f vector field)5 f(x,y)=(cos2x+cos2y)2 6 Minimize f(x,y)=x2+y2(,)(2,2)0fffxyxy=222222002ffxx yHfy xy=0,0 xy=x y 0 f 7 Parameter Optimization with Equality Constraints Minimize Subject to Introduce Lagrange Multipliers 1 K Necessary conditions for the local minima of L(also necessary for the origina

6、l problem)12(,)nf x xx12(,)0kngx xx=1,2,kK=1KkkiLfg=+10KkkiiiigLfxxx=+=0kkLg=8 f Minimize f(x,y)=x2+y2 Subject to (x-8)2+(y-6)2=25 Solutions(from the N-R method):=1,x=4 and y=3 (f=25)=3,x=12 and y=9 (f=225)x y 0 8 6 2222(8)(6)25Lxyxy=+2(216)0Lxxx=+=2(212)0Lyyy=+=22(8)(6)250Lxy=+=g(x,y)=(x-8)2+(y-6)2

7、-25=0 9 Parameter Optimization with Inequality Constraints Minimize Subject to:Introduce Lagrange Multipliers 1 K and 1 m Necessary conditions for the local minima of L 12(,)nf x xx12(,)0kngx xx=1,2,kK=0kkLg=12(,)0jnux xx1,2,jm=11KmkkjjkjLfgu=+1,in=1,2,kK=0jjLu=1,jm=0jju=0jKuhn-Tucker(KKT)necessary

8、condition 110KmjkkjijiiiiugLfxxxx=+=10 Minimize f(x,y)=x2+y2 Subject to (x-8)2+(y-6)2=25 g(x,y)=(x-8)2+(y-6)2-25=0 Solutions:=0,=3,x=12 and y=9 (f=225)=5.6,=-0.2,x=5 and y=2 (f=29)=12,=-1.8,x=3 and y=6 (f=33)x y 0 8 6 2222(8)(6)25(122)Lxyxyxy=+2(216)20Lxxx=+=2(212)0Lyyy=+=22(8)(6)250Lxy=+=212xy+f(,)

9、1220u x yxy=0,0,0jjjLu=1220 0 or 1220 0LxyLxy=11 Operating Cost of a Thermal Plant Fuel-cost curve of a generator(represented by a quadratic function of real power)Incremental fuel-cost curve:2iiiiiiCPP=+2iiiiiidCPdP=+12 A real case Gen ID PRIOR FUELCO($/MBtu)PMAX(MW)PMIN(WM)HEMIN(MBtu/hr)X1(MW)Y1(B

10、tu/kWh)X2(MW)Y2(Btu/kWh)X3(Btu/kWh)Y3(Btu/kWh)A 1 0 1.91 230 65 532 65 8760 176 9507 260 10072 B 2 0 0.539 106 50 425 50 8501 75 9198 106 10341 Btu/h=Xi Yi 1000$/h=Btu/h$/MBtu/1000,000$/MWh =Yi/1000$/MBtu 05010015020025030002468101214161820P(MW)Lambda($/MWh)050100150200250300010002000300040005000600

11、0P(MW)Cost($/h)13 ED Neglecting Losses and No Generator Limits If transmission line losses are neglected,minimize the total production cost:subject to Apply the Lagrange multiplier method(ng+1 unknowns to solve):1gntiiCC=21niiiiiiPP=+1gniDiPP=1()gntDiiLCPP=+00iLPL=0tiCP=2tiiiiiiCdCPPdP=+=1,gin=1gniD

12、iPP=12gniiiDP=11212ggniDiiniiP=+=2iiiP=All plants must operate at equal incremental cost Solve Pi 14 Example 7.4 The fuel-cost functions for three thermal plants are C1C2 in$/h.P1,P2 and P3 are in MW.PD=800MW.Neglecting line losses and generator limits,find the optimal dispatch and the total cost in

13、$/h 2111222223335005.30.0044005.50.0062005.80.009CPPCPPCPP=+=+=+11212ggniDiiniiP=+=5.35.55.88000.0080.0120.0181110.0080.0120.018+=+8001443.05558.5$/MWh263.8889+=1238.55.3400.00002(0.004)8.55.5250.00002(0.006)8.55.8150.00002(0.009)PPP=11115.30.008dCPdP=+22225.50.012dCPdP=+33335.80.018dCPdP=+6682.5$/h

14、tC=2iiiP=Equal incremental cost 15 Solving by the N-R Method For a general case:112ggnniiDiiiPP=()()()()()()kkkDdffPd+1()gniDifPP=()ig P=()()()()1()()()()()()()()()gnkkkDikiDkkkiPPPfPdfdfdPddd=(1)()()kkk+=+(1)()|kk+until 16 05010015020025030035040045055.566.577.588.599.51010.511P(MW)$/MWhApply the R

15、-N Method in Example 7.4(1)6.0=(1)16.05.387.50002(0.004)P=()()2kkiiiP=(1)2(1)36.05.541.66672(0.006)6.05.811.11112(0.009)PP=(1)800(87.541.666711.1111)659.7222P=+=(1)659.72221112(0.004)2(0.006)2(0.009)659.7222 2.5263.8888=+=(2)1(2)2(2)3(2)8.55.3400.00002(0.004)8.55.5250.00002(0.006)8.55.8150.00002(0.0

16、09)800(400250150)0.0PPPP=+=()()()()1()2kkkkiiPPdPd=(2)6.02.58.5=+=6682.5$/htC=17 ED Neglecting Losses but Including Generator Limits Considering the maximum(by rating)and minimum(for stability)generation limits,Minimize Subject to (min)(max)1,2,iiigPPPin=，1gntiiCC=21niiiiiiPP=+1gniDiPP=iiiiiiiidCdPd

17、CdPdCdP=+(max)(min)11()()()ggnntDiiiiiiiiiLCPPPPPP=+00iLPL=0iiiiCP+=1gniDiPP=00iiLL(min)(max)1,2,iiigPPPin=,(max)(min)()0,0()0 0iiiiiiiiPPPP=(min)(max)(0)iiiiiPPP=(max)(0)iiiPP=(min)(0)iiiPP=Excluding the plants that reach their limits,the other plants still operate at equal incremental cost P1 P2 P

18、3 P4 18 Example 7.6 Consider generator limits(in MW)for Example 7.4,let PD=975MW 123250450150350100225PPP1115.30.008dCPdP=+=2225.50.012dCPdP=+=3335.80.018dCPdP=+=PD=800MW PD=906MW PD=975MW Solution:P1=450MW P2=325MW P3=200MW=9.4$/MWh Ct=8236.25$/h 19(1)6.0=(1)16.05.387.50002(0.004)P=()()2kkiiiP=(1)2

19、(1)36.05.541.66672(0.006)6.05.811.11112(0.009)PP=(1)975(87.541.6667 11.1111)834.7222P=+=()()()()1()2kkkkiiPPdPd=(1)834.7222834.72223.1632111263.88882(0.004)2(0.006)2(0.009)=+(2)6.03.16329.1632=+=(2)1(2)2(2)39.16325.32(0.004)9.16325.5305.26322(0.006)9.16325.482.898186.84212(0.009)47PPP=12325045015035

20、0100225PPP(2)32.894732.89470.236811132.88892(0.006)2(0.009)=+(3)9.16320.23689.4=+=(3)1(3)2(3)3(3)9.45.53252(0.006)9.45.82002(0.009)975(450325200)04500.PPPP=+=(2)975(305.2632186.8421)32 90.58 447P=+=1(max)1,so let4 50PP20 Solve the optimal dispatch for PD=550MW 123250450150350100225PPP1115.30.008dCPd

21、P=+=2225.50.012dCPdP=+=3335.80.018dCPdP=+=Solution 1:P1=280MW P2=170MW P3=100MW=7.54$/MWh Ct=4676$/h 2111222223335005.30.0044005.50.0062005.80.009CPPCPPCPP=+=+=+Solution 2:P1=340MW P2=210MW P3=0MW=8.02$/MWh Ct=4584$/h Solution 3:P1=0MW P2=340MW P3=210MW=9.58$/MWh Ct=47785$/h Solution 4:P1=400MW P2=0

22、MW P3=150MW=8.5$/MWh Ct=4532$/h 123or 0or 0or 0PPP=Unit Commitment Problem(mixed-integer optimization)P3=0 P1=0 P2=0 21 Transmission Loss When transmission distances are very small and load density in the system is very high Transmission losses may be neglected All plants operate at equal incrementa

23、l production cost to achieve optimal dispatch of generation However,in a large interconnected network Power is transmitted over long distances with low load density areas Transmission losses are a major factor affecting the optimal dispatch.22 Calculation of Transmission Losses 22222|cos|cosLPPIRRVR

24、PV=|cosPIV=P R+jX I PD V P1 P2 PD R1+jX1 R2+jX2 R3+jX3 I1 I2 I3 V1 V2 2221122123221212112221231122|cos|coscoscos|cos|cosLPIRIRIIRPPRRVVPPRVV=+=+2LPBP=2211 11212222LPB PB P PB P=+I1 I2 I3 23 More general loss formulas:Quadratic form Krons loss formula Bij,B0i and B00 are Loss coefficients or B-coeffi

25、cients.See Chapter 7.7 for details of calculating B-coefficients Changes with power flows but usually assumed constant and estimated for a power-flow base case 000111gggLnnniijjiiijiPPB PB PB11ggLnniijjijPPB P24 Economic Dispatch Including Losses Cost function Subject to 000111gggLnnniijjiiijiPPB PB

26、 PB211ggnntiiiiiiiiCCPP1gniDiLPPP(min)(max)1,iiigPPPin012gnLijjijiPB PBP2iiiiidCPdP=+Incremental fuel cost Incremental transmission loss 25 Using Lagrange Multipliers (max)(max)(min)(min)1111()()()ggggnnnniDiiiiiLiiiiiiLCPPPPPPP0L=(01)0LiiidPPPCd+=iiLiCdPPdP=+0011101ggggnnnLiijniDDjijiiiiPPBBPPPPB P

27、 1,iigidCiPLnd11iLiLPPPenalty factor:012gnLijjijiPB PBP2iiiiidCPdP=+012 1,2gnijjiiiijgB PBPin000111gggLnnniijjiiijiPPB PB PB211ggnntiiiiiiiiCCPP0iLP=where 26 111112101122221221202012111 21gggggg gnnnngngnnn nBBBBPPBBBBPBBBBSolve and Pi(i=1ng)for the optimal dispatch:011()(1)2gniiiiiijjijjiB PB PB012

28、2gniiiijjijPB PB0001111ggggnnnniDLDiijjiiiijiPPPPPB PB PBCheck inequality constraints:If PiPi(max),let Pi Pi(max)Remove that Pi from the equations 0(1)22()iijiijjiiiiBB PPB1()gniDLifPPPUse the N-R Method:()()()()()()()kkkkDLdffPPd()()()()()1()()gkkkknkiiPPdfPd()()()()kkkDLPPPf()()1()()gknkiiPdfd27(1

29、)(1)()1)(1)0(1)22()iijiijiiikkkjkkiPPBBB()()()1gkkkniiPP000111()()()()1)ggggkkkkkijnnnnDijijiiiiiPPPBBBPPP()()1)kkk(0)(0)Initial and iP0(1)22()iij iijjiiiiBB PPB()()()()kkkDLPPPf()()()()()1()()gkkkknkiiPPdfPd()()01(2)1(1)22()kgniiiiiijiijiikjiikiBBBBPPUntil|P(k)|28 Example 7.7 2111222223332007.00.00

30、8$/1806.30.009$/1406.80.007$/CPPhCPPhCPPh12310 MW85 MW10 MW80 MW10 MW70 MWPPP222(pu)1(pu)2(pu)3(pu)0.02180.02280.0179LPPPP2223122221230.0218()0.0228()0.0179()100 MW1001001000.0002180.0002280.000179MWLPPPPPPP(only Bii 0)Solve the optimal dispatch of three thermal plants in a power system.The total sy

31、stem load is 150MW.The base for per unit values is 100MVA.29(0)8.0=1)(1)()(2()iiikkkiiPB()112)1(2()kggnniiiiikiiiiiBPB12310 MW85 MW10 MW80 MW10 MW70 MWPPP(1)1(1)2(1)38.07.051.3136 MW2(0.0080.80.000218)8.06.378.5292 MW2(0.0080.80.000228)8.06.871.1575 MW2(0.0070.80.000179)PPP(1)2220.000218(51.3136)0.0

32、00228(78.5292)0.000179(71.1575)2.886LP(1)1502.8864 (51.313678.529271.1575)48.1139P(1)31()152.4924iiP(1)48.11390.31552152.4924(3)7.6789(4)1(4)2(4)37.67897.035.0907 MW2(0.0087.6790.000218)7.67896.364.1317 MW2(0.0097.6790.000228)7.67896.852.4767 MW2(0.0077.6790.000179)PPP(4)2220.000218(35.0907)0.000228(64.1317)0.000179(52.4767)1.699LP(4)0.0001P1592.65$/htC()()()1gkkkniiPP