New Formulations of the Optimal Power Flow Problem

Transcription

1 New Formulations of the Optimal Power Flow Problem Prof. Daniel Kirschen The University of Manchester 2010 D. Kirschen and The University of Manchester 1

2 2010 D. Kirschen and The University of Manchester 2

3 Outline A bit of background The power flow problem The optimal power flow problem (OPF) The security-constrained OPF (SCOPF) The worst-case problem 2010 D. Kirschen and The University of Manchester 3

4 What is a power system? Generators Power Transmission Network Loads 2010 D. Kirschen and The University of Manchester 4

5 What is running a power system about? Greed Minimum cost Maximum profit 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 5

6 What is running a power system about? Fear Avoid outages and blackouts 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 6

7 What is running a power system about? Green Accommodate renewables 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 7

8 Balancing conflicting aspirations Environmental impact Cost Reliability 2010 D. Kirschen and The University of Manchester 8

9 The Power Flow Problem 2010 D. Kirschen and The University of Manchester 9

10 State variables Voltage at every node (a.k.a. bus ) of the network Because we are dealing with ac, voltages are represented by phasors, i.e. complex numbers in polar representation: Voltage magnitude at each bus: Voltage angle at each bus:! k V k 2010 D. Kirschen and The University of Manchester 10

11 Other variables Active and reactive power consumed at each bus: P L L k,q k a.k.a. the load at each bus Active and reactive power produced by renewable generators: P k W,Q k W Assumed known in deterministic problems In practice, they are stochastic variables 2010 D. Kirschen and The University of Manchester 11

12 What is reactive power? Reactive power Active power 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 12

13 Injections Bus k P k,q k P k G,Q k G P k W,Q k W P k L,Q k L G W P k = P k G + P k W! P k L Q k = Q k G + Q k W! Q k L There is usually only one P and Q component at each bus 2010 D. Kirschen and The University of Manchester 13

14 Injections V k!" k Bus k P k,q k Two of these four variables are specified at each bus: Load bus: P k,q k Generator bus: Pk,V k Reference bus: V k,! k 2010 D. Kirschen and The University of Manchester 14

15 Line flows To bus i P ki,q ki P kj,q kj To bus j V k!" k Bus k P k,q k The line flows depend on the bus voltage magnitude and angle as well as the network parameters G ki, B ki (real and imaginary part of the network admittance matrix) 2010 D. Kirschen and The University of Manchester 15

16 Power flow equations To bus i P ki,q ki P kj,q kj To bus j V k!" k Bus k P k,q k Write active and reactive power balance at each bus: N " i=1 P k = V k V i [G ki cos! ki + B ki sin! ki ] N " i=1 Q k = V k V i [G ki sin! ki # B ki cos! ki ] k = 1,!N with:! ki =! k "! i, N : number of nodes in the network 2010 D. Kirschen and The University of Manchester 16

17 The power flow problem Given the injections and the generator voltages, Solve the power flow equations to find the voltage magnitude and angle at each bus and hence the flow in each branch N " P k = V k V i [G ki cos! ki + B ki sin! ki ] i=1 N " Q k = V k V i [G ki sin! ki # B ki cos! ki ] i=1 k = 1,!N Typical values of N: GB transmission network: N~1,500 Continental European network (UCTE): N~13,000 However, the equations are highly sparse! 2010 D. Kirschen and The University of Manchester 17

18 Applications of the power flow problem Check the state of the network for an actual or postulated set of injections for an actual or postulated network configuration Are all the line flows within limits? Are all the voltage magnitudes within limits? 2010 D. Kirschen and The University of Manchester 18

19 Linear approximation N " P k = V k V i [G ki cos! ki + B ki sin! ki ] i=1 N " Q k = V k V i [G ki sin! ki # B ki cos! ki ] i=1 N " P k = B ki! ki i=1 Ignores reactive power Assumes that all voltage magnitudes are nominal Useful when concerned with line flows only 2010 D. Kirschen and The University of Manchester 19

20 The Optimal Power Flow Problem (OPF) 2010 D. Kirschen and The University of Manchester 20

21 Control variables Control variables which have a cost: Active power production of thermal generating units: P i G Control variables that do not have a cost: Magnitude of voltage at the generating units: Tap ratio of the transformers: t ij V i G 2010 D. Kirschen and The University of Manchester 21

22 Possible objective functions Minimise the cost of producing power with conventional generating units: g! i=1 min C i (P i G ) Minimise deviations of the control variables from a given operating point (e.g. the outcome of a market): g # i=1 ( ) min c i +!P i G+ " c i "!P i G" 2010 D. Kirschen and The University of Manchester 22

23 Equality constraints Power balance at each node bus, i.e. power flow equations N " P k = V k V i [G ki cos! ki + B ki sin! ki ] i=1 N " Q k = V k V i [G ki sin! ki # B ki cos! ki ] i=1 k = 1,!N 2010 D. Kirschen and The University of Manchester 23

24 Inequality constraints Upper limit on the power flowing though every branch of the network Upper and lower limit on the voltage at every node of the network Upper and lower limits on the control variables Active and reactive power output of the generators Voltage settings of the generators Position of the transformer taps and other control devices 2010 D. Kirschen and The University of Manchester 24

25 Formulation of the OPF problem ( ) min u 0 f 0 x 0,u 0 ( ) = 0 ( )! 0 g x 0,u 0 h x 0,u 0 x 0 : vector of dependent (or state) variables u 0 : vector of independent (or control) variables Nothing extraordinary, except that we are dealing with a fairly large (but sparse) non-linear problem D. Kirschen and The University of Manchester 25

26 The Security Constrained Optimal Power Flow Problem (SCOPF) 2010 D. Kirschen and The University of Manchester 26

27 Bad things happen 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 27

28 Sudden changes in the system A line is disconnected because of an insulation failure or a lightning strike A generator is disconnected because of a mechanical problem A transformer blows up The system must keep going despite such events N-1 security criterion 2010 D. Kirschen and The University of Manchester 28

29 Security-constrained OPF How should the control variables be set to minimise the cost of running the system while ensuring that the operating constraints are satisfied in both the normal and all the contingency states? 2010 D. Kirschen and The University of Manchester 29

30 Formulation of the SCOPF problem min f u 0 x 0,u 0 k ( ) s.t. g k (x k,u k ) = 0 k = 0,..., N c h k (x k,u k )! 0 k = 0,..., N c u k " u 0! #u k max k = 1,..., N c k = 0 : normal conditions k = 1,..., N c : contingency conditions!u k max : vector of maximum allowed adjustments after contingency k has occured 2010 D. Kirschen and The University of Manchester 30

31 Preventive or corrective SCOPF min f u 0 x 0,u 0 k ( ) s.t. g k (x k,u k ) = 0 k = 0,..., N c h k (x k,u k )! 0 k = 0,..., N c u k " u 0! #u k max k = 1,..., N c Preventive SCOPF: no corrective actions are considered!u k max = 0 " u k = u 0 #k = 1, N c Corrective SCOPF: some corrective actions are allowed! k = 1, N c "u k max # D. Kirschen and The University of Manchester 31

32 Size of the SCOPF problem SCOPF is (N c +1) times larger than the OPF Pan-European transmission system model contains about 13,000 nodes, 20,000 branches and 2,000 generators Based on N-1 criterion, we should consider the outage of each branch and each generator as a contingency However: Not all contingencies are critical (but which ones?) Most contingencies affect only a part of the network (but what part of the network do we need to consider?) 2010 D. Kirschen and The University of Manchester 32

33 A few additional complications Some of the control variables are discrete: Transformer and phase shifter taps Capacitor and reactor banks Starting up of generating units There is only time for a limited number of corrective actions after a contingency 2010 D. Kirschen and The University of Manchester 33

34 The Worst-Case Problems 2010 D. Kirschen and The University of Manchester 34

35 Good things happen 2010 D. Kirschen and The University of Manchester Photo credit: FreeDigitalPhotos.net 35

36 but there is no free lunch! Wind generation and solar generation can only be predicted with limited accuracy When planning the operation of the system a day ahead, some of the injections are thus stochastic variables Power system operators do not like probabilistic approaches 2010 D. Kirschen and The University of Manchester 36

37 Formulation of the OPF with uncertainty!#"# $ min c T p! M 0 " p 0 market-based generation ( ) additional generation!##" ## $ + b *T nd 0 c 0 + p! 0 c T ( ) s.t. g 0 (x 0,u 0,p 0,b 0,p nd 0,s) = 0 h 0 (x 0,u 0,p 0,b 0,p nd 0,s) # 0 init u 0 " u 0 max # $u 0 p 0 " p M 0 max # $p 0 p min nd b T 0 # p nd 0 b T 0 # p nd max { } b 0 % 0,1 s min # s # s max b 0 T Deviations in cost-free controls Deviations in market generation Deviations in extra generation Decisions about extra generation Vector of uncertainties 2010 D. Kirschen and The University of Manchester 37

38 Worst-case OPF bi-level formulation max s s.t. c T ( p! M 0 " p ) 0 + b!t ( nd 0 c 0 + p! 0 c T ) s min # s # s max ( p! 0,u! 0,b! nd! 0,p ) 0 = arg min c T ( M p 0 " p ) 0 + b T ( 0 c 0 + p nd 0 c T ) s.t. g 0 (x 0,u 0,p 0,b 0,p 0 nd,s) = 0 h 0 (x 0,u 0,p 0,b 0,p 0 nd,s) # 0 u 0 " u 0 init p 0 " p 0 M # $u 0 max # $p 0 max p min nd b T 0 # p nd 0 b T 0 # p nd max { } b 0 % 0,1 b 0 T 2010 D. Kirschen and The University of Manchester 38

39 Worst-case SCOPF bi-level formulation max s s.t. c T ( p! M 0 " p ) 0 + b!t ( nd 0 c 0 + p! 0 c T ) s min # s # s max ( p! 0,p! k,u! 0,u! k,b! nd! 0,p ) 0 = arg min c T ( M p 0 " p ) 0 + b T ( 0 c 0 + p nd 0 c T ) s.t. g 0 (x 0,u 0,p 0,b 0,p 0 nd,s) = 0 h 0 (x 0,u 0,p 0,b 0,p 0 nd,s) # 0 g k (x k,u k,p k,b 0,p 0 nd,s) = 0 h k (x k,u k,p k,b 0,p 0 nd,s) # 0 p k " p 0 # $p k max u k " u 0 # $u k max p min nd b T 0 # p nd 0 b T 0 # p nd max { } b 0 % 0,1 b 0 T 2010 D. Kirschen and The University of Manchester 39