電力系統(tǒng)分析英文課件OptimalDispatch.pdf

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 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 typical 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 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 n f x xx 12 n fff f xxx 0 Stationary point where f is flat in all directions Local minimum a pure source in f vector field 5 f x y cos2x cos2y 2 6 Minimize f x y x2 y2 2 2 0 ff fxy xy 22 2 2 2 20 02 ff xx y H f y x y 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 original problem 12 n f x xx 12 0 kn gx xx 1 2 kK 1 K kk i Lfg 1 0 K k k i iii gLf xxx 0 k k L g 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 25 Lxyxy 2 216 0 L xx x 2 212 0 L yy y 22 8 6 250 L xy g x y x 8 2 y 6 2 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 n f x xx 12 0 kn gx xx 1 2 kK 0 k k L g 12 0 jn ux xx 1 2 jm 11 Km kkjj kj Lfgu 1 in 1 2 kK 0 j j L u 1 jm 0 jj u 0 j Kuhn Tucker KKT necessary condition 11 0 Km j k kj ij iiii u gLf xxxx 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 20 L xx x 2 212 0 L yy y 22 8 6 250 L xy 212xy f 1220u x yxy 0 0 0 jjj L u 1220 0 or 1220 0 L xy L xy 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 2 iiiiii CPP 2 i iiii i dC P dP 12 A real case Gen ID PRIOR FUELCO MBtu PMAX MW PMIN WM HEMIN MBtu hr X1 MW Y1 Btu 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 050100150200250300 0 2 4 6 8 10 12 14 16 18 20 P MW Lambda MWh 050100150200250300 0 1000 2000 3000 4000 5000 6000 P 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 1 g n ti i CC 2 1 n iiiii i PP 1 g n iD i PP 1 g n tDi i LCPP 0 0 i L P L 0 t i C P 2 ti iii ii CdC P PdP 1 g in 1 g n iD i PP 1 2 g n i i i D P 1 1 2 1 2 g g n i Di i n i i P 2 i i i P All plants must operate at equal incremental cost Solve Pi 14 Example 7 4 The fuel cost functions for three thermal plants are C1 C2 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 h 2 111 2 222 2 333 5005 30 004 4005 50 006 2005 80 009 CPP CPP CPP 1 1 2 1 2 g g n i Di i n i i P 5 35 55 8 800 0 0080 0120 018 111 0 0080 0120 018 8001443 0555 8 5 MWh 263 8889 1 2 3 8 55 3 400 0000 2 0 004 8 55 5 250 0000 2 0 006 8 55 8 150 0000 2 0 009 P P P 1 11 1 5 30 008 dC P dP 2 22 2 5 50 012 dC P dP 3 33 3 5 80 018 dC P dP 6682 5 h t C 2 i i i P Equal incremental cost 15 Solving by the N R Method For a general case 11 2 gg nn i iD ii i PP kkk D df fP d 1 g n iD i fPP i g P 1 g n k kkDi k iD kkk i PP PfP dfdfdP ddd 1 kkk 1 kk until 16 050100150200250300350400450 5 5 5 6 6 5 7 7 5 8 8 5 9 9 5 10 10 5 11 P MW MWh Apply the R N Method in Example 7 4 1 6 0 1 1 6 05 3 87 5000 2 0 004 P 2 k k i i i P 1 2 1 3 6 05 5 41 6667 2 0 006 6 05 8 11 1111 2 0 009 P P 1 800 87 541 666711 1111 659 7222 P 1 659 7222 111 2 0 004 2 0 006 2 0 009 659 7222 2 5 263 8888 2 1 2 2 2 3 2 8 55 3 400 0000 2 0 004 8 55 5 250 0000 2 0 006 8 55 8 150 0000 2 0 009 800 400250150 0 0 P P P P 1 2 kk k k i i PP dP d 2 6 02 58 5 6682 5 h t C 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 iiig PPPin 1 g n ti i CC 2 1 n iiiii i PP 1 g n iD i PP i i i i i i i i dC dP dC dP dC dP max min 11 gg nn tDiiiiiii ii LCPPPPPP 0 0 i L P L 0 i ii i C P 1 g n iD i PP 0 0 i i L L min max 1 2 iiig PPPin max min 0 0 0 0 iiii iiii PP PP min max 0 iii ii PPP max 0 iii PP min 0 iii PP Excluding the plants that reach their limits the other plants still operate at equal incremental cost P1 P2 P3 P4 18 Example 7 6 Consider generator limits in MW for Example 7 4 let PD 975MW 1 2 3 250450 150350 100225 P P P 1 1 1 5 30 008 dC P dP 2 2 2 5 50 012 dC P dP 3 3 3 5 80 018 dC P dP 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 1 6 05 3 87 5000 2 0 004 P 2 k k i i i P 1 2 1 3 6 05 5 41 6667 2 0 006 6 05 8 11 1111 2 0 009 P P 1 975 87 541 6667 11 1111 834 7222P 1 2 kk k k i i PP dP d 1 834 7222834 7222 3 1632 111 263 8888 2 0 004 2 0 006 2 0 009 2 6 03 16329 1632 2 1 2 2 2 3 9 16325 3 2 0 004 9 16325 5 305 2632 2 0 006 9 16325 482 89 8 186 8421 2 0 009 47P P P 1 2 3 250450 150350 100225 P P P 2 32 894732 8947 0 2368 11 132 8889 2 0 006 2 0 009 3 9 16320 23689 4 3 1 3 2 3 3 3 9 45 5 325 2 0 006 9 45 8 200 2 0 009 975 450325200 0 45 0 0 P P P P 2 975 305 2632186 8421 32 90 58 447P 1 max 1 so let4 50PP 20 Solve the optimal dispatch for PD 550MW 1 2 3 250450 150350 100225 P P P 1 1 1 5 30 008 dC P dP 2 2 2 5 50 012 dC P dP 3 3 3 5 80 018 dC P dP Solution 1 P1 280MW P2 170MW P3 100MW 7 54 MWh Ct 4676 h 2 111 2 222 2 333 5005 30 004 4005 50 006 2005 80 009 CPP CPP CPP 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 0MW P3 150MW 8 5 MWh Ct 4532 h 1 2 3 or 0 or 0 or 0 P P P 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 incremental 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 2 2 2 22 cos cos L P PIRR V R P V cos P I V P R jX I PD V P1 P2 PD R1 jX1 R2 jX2 R3 jX3 I1 I2 I3 V1 V2 222 1122123 22 12 12 1122 2 12 3 1122 cos cos coscos cos cos L PIRIRIIR PP RR VV PP R VV 2 L PBP 22 11 11212222L PB PB P PB P I1 I2 I3 23 More general loss formulas Quadratic form Kron s loss formula Bij B0i and B00 are Loss coefficients or B coefficients 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 000 111 ggg L nnn iijjii iji PPB PB PB 11 gg L nn iijj ij PPB P 24 Economic Dispatch Including Losses Cost function Subject to 000 111 ggg L nnn iijjii iji PPB PB PB 2 11 gg nn tiiiiii ii CCPP 1 g n iD i L PPP min max 1 iiig PPPin 0 1 2 g n L ijji j i P B PB P 2 i iii i dC P dP Incremental fuel cost Incremental transmission loss 25 Using Lagrange Multipliers max max min min 1111 gggg nnnn iDiiiiiLii iiii LCPPPPPPP 0 L 01 0 L i i i d P P P C d i i L i C dPP dP 00 111 0 1 gggg nnn Liij n iDDji jii i i PPBBPPPPB P 1 i i g i dC i P Ln d 1 1 i L i L P P Penalty factor 0 1 2 g n L ijji j i P B PB P 2 i iii i dC P dP 0 1 2 1 2 g n ijjiiii j g B PBPin 000 111 ggg L nnn iijjii iji PPB PB PB 2 11 gg nn tiiiiii ii CCPP 0 i L P where 26 1 1 11121 01 1 2 2 2212212 02 0 12 1 1 1 2 1 g g g ggg g n n n ng ng nnn n BBBB P PBBB B P BBBB Solve and Pi i 1 ng for the optimal dispatch 0 1 1 1 2 g n ii iiiijji j ji B PB PB 0 1 22 g n iiiijji j PB PB 000 1111 gggg nnnn iDLDiijjii iiji PPPPPB PB PB Check inequality constraints If PiPi max let Pi Pi max Remove that Pi from the equations 0 1 2 2 iijiijj i iii BB P P B 1 g n iDL i fPPP Use the N R Method kkkk DL df fPP d 1 g kk k k n k i i PP df P d kkk DL PPPf 1 g k n k i i Pdf d 27 1 1 1 1 0 1 2 2 iijiij iii kkk j k k i P PBB B 1 g k k k n i i P P 000 111 1 gggg kkkkk ij nnnn Diji jii i i i PPPBBBPPP 1 kkk 0 0 Initial and i P 0 1 2 2 iij iijj i iii BB P P B kkk DL PPPf 1 g kk k k n k i i PP df P d 0 1 2 1 1 2 2 k g n iiiiiijiij i i k j i i k i BBB B P P Until P k 28 Example 7 7 2 111 2 222 2 333 2007 00 008 1806 30 009 1406 80 00