Lossy DC Power Flow. John W. Simpson-Porco. Universidad de Chile Santiago, Chile. March 2, 2017

Transcription

1 Lossy DC Power Flow John W. Simpson-Porco Universidad de Chile Santiago, Chile March 2, 2017

2 Our 20th Century Bulk Power System A large-scale, nonlinear, hybrid, stochastic, distributed, cyber-physical... Active Control Passive Consumption Today, we will focus on the network. 1 / 18

3 Our 20th Century Bulk Power System A large-scale, nonlinear, hybrid, stochastic, distributed, cyber-physical... Active Control Passive Consumption Today, we will focus on the network. 1 / 18

4 Modern Large-Scale Power rids 2 / 18

5 Problems in Power System Operations Power Flow Analysis Optimal Power Flow [Molzahn et al.] Contingency Analysis Transient Stability [Rezaei et al.] [Overbye et al.] 3 / 18

6 Linearized Optimal Power Flow Idea: Optimally match supply and demand (with constraints) 4 / 18

7 Linearized Optimal Power Flow Idea: Optimally match supply and demand (with constraints) minimize f i (P i ) θ,p i N subject to P i = j B ij(θ i θ j ) i N L N, P min i p min ij P i Pi max i N, p ij pij max (i, j) E, 4 / 18

8 Linearized Optimal Power Flow Idea: Optimally match supply and demand (with constraints) minimize f i (P i ) θ,p i N subject to P i = j B ij(θ i θ j ) i N L N, P min i p min ij P i Pi max i N, p ij pij max (i, j) E, 4 / 18

9 Linearized Optimal Power Flow Idea: Optimally match supply and demand (with constraints) minimize f i (P i ) θ,p i N subject to P i = j B ij(θ i θ j ) i N L N, P min i p min ij P i Pi max i N, p ij pij max (i, j) E, LP or QP depending on objective function, easy to solve 4 / 18

10 Linearized Optimal Power Flow Idea: Optimally match supply and demand (with constraints) minimize f i (P i ) θ,p i N subject to P i = j B ij(θ i θ j ) i N L N, P min i p min ij P i Pi max i N, p ij pij max (i, j) E, LP or QP depending on objective function, easy to solve Big drawback: model does not have resistances (power losses are not modelled) 4 / 18

11 AC Power Flow Equations 1 Network: n buses N, m lines E, line admittances y ij = g ij + jb ij 2 Bus Variables: voltage V i e jθ i, power S i = P i + jq i 3 Circuit Laws: Kirchhoff & Ohm 4 Admittance Matrix: Y = + jb Y ij = y ij, Y ii = n j=1 y ij 5 / 18

12 AC Power Flow Equations 1 Network: n buses N, m lines E, line admittances y ij = g ij + jb ij 2 Bus Variables: voltage V i e jθ i, power S i = P i + jq i 3 Circuit Laws: Kirchhoff & Ohm 4 Admittance Matrix: Y = + jb Y ij = y ij, Y ii = n j=1 y ij 5 / 18

13 AC Power Flow Equations 1 Network: n buses N, m lines E, line admittances y ij = g ij + jb ij 2 Bus Variables: voltage V i e jθ i, power S i = P i + jq i 3 Circuit Laws: Kirchhoff & Ohm 4 Admittance Matrix: Y = + jb Y ij = y ij, Y ii = n j=1 y ij 5 / 18

14 AC Power Flow Equations 1 Network: n buses N, m lines E, line admittances y ij = g ij + jb ij 2 Bus Variables: voltage V i e jθ i, power S i = P i + jq i 3 Circuit Laws: Kirchhoff & Ohm 4 Admittance Matrix: Y = + jb Y ij = y ij, Y ii = n j=1 y ij P i = n j=1 V iv j B ij sin(θ i θ j ) + n j=1 V iv j ij cos(θ i θ j ) 5 / 18

15 Simplified Models of Active Power Flow Many applications require some simplifying assumptions 1 constant voltage magnitudes: V i 1 2 no resistances: ij 0 P i = n ij sin(θ i θ j ) j=1 Decoupled PF 3 small phase differences: θ i θ j << 1 P i n ij(θ i θ j ) j=1 DC PF = θ = B 1 P 6 / 18

16 Simplified Models of Active Power Flow Many applications require some simplifying assumptions 1 constant voltage magnitudes: V i 1 2 no resistances: ij 0 P i = n ij sin(θ i θ j ) j=1 Decoupled PF 3 small phase differences: θ i θ j << 1 P i n ij(θ i θ j ) j=1 DC PF = θ = B 1 P 6 / 18

17 Simplified Models of Active Power Flow Many applications require some simplifying assumptions 1 constant voltage magnitudes: V i 1 2 no resistances: ij 0 P i = n ij sin(θ i θ j ) j=1 Decoupled PF 3 small phase differences: θ i θ j << 1 P i n ij(θ i θ j ) j=1 DC PF = θ = B 1 P 6 / 18

18 Simplified Models of Active Power Flow Many applications require some simplifying assumptions 1 constant voltage magnitudes: V i 1 2 no resistances: ij 0 P i = n ij sin(θ i θ j ) j=1 Decoupled PF 3 small phase differences: θ i θ j << 1 P i n ij(θ i θ j ) j=1 DC PF = θ = B 1 P Our goal is to include resistances (remove assumption #2) (we will not even need assumption #3) 6 / 18

19 Lossy DC Power Flow: Two-Bus Case y = g jb 1 2 P 1 P 2 P 1 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) P 2 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) slack/ref bus Let ψ = sin(θ 1 θ 2 ). Then cos(θ 1 θ 2 ) = 1 ψ 2 now isolate ψ P 1 = bψ + g g 1 ψ 2 7 / 18

20 Lossy DC Power Flow: Two-Bus Case y = g jb 1 2 P 1 P 2 P 1 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) P 2 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) slack/ref bus Let ψ = sin(θ 1 θ 2 ). Then cos(θ 1 θ 2 ) = 1 ψ 2 now isolate ψ P 1 = bψ + g g 1 ψ 2 7 / 18

21 Lossy DC Power Flow: Two-Bus Case y = g jb 1 2 P 1 P 2 P 1 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) P 2 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) slack/ref bus Let ψ = sin(θ 1 θ 2 ). Then cos(θ 1 θ 2 ) = 1 ψ 2 now isolate ψ P 1 = bψ + g g 1 ψ 2 7 / 18

22 Lossy DC Power Flow: Two-Bus Case y = g jb 1 2 P 1 P 2 P 1 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) P 2 = b sin(θ 1 θ 2 ) + g g cos(θ 1 θ 2 ) slack/ref bus Let ψ = sin(θ 1 θ 2 ). Then cos(θ 1 θ 2 ) = 1 ψ 2 now isolate ψ P 1 = bψ + g g 1 ψ 2 7 / 18

23 Lossy DC Power Flow: Two-Bus Case ψ = 1 b ( P 1 + g ) 1 ψ 2 g This is a fixed-point equation ψ = f (ψ) Interpret as an iteration. Start with initial guess ψ 0 and update: ψ k+1 = f (ψ k ) = 1 ( ) P 1 + g 1 ψ 2k b g }{{} :=P 1,k After ψ converges, simply compute θ 1 θ 2 = arcsin(ψ) 8 / 18

24 Lossy DC Power Flow: Two-Bus Case ψ = 1 b ( P 1 + g ) 1 ψ 2 g This is a fixed-point equation ψ = f (ψ) Interpret as an iteration. Start with initial guess ψ 0 and update: ψ k+1 = f (ψ k ) = 1 ( ) P 1 + g 1 ψ 2k b g }{{} :=P 1,k After ψ converges, simply compute θ 1 θ 2 = arcsin(ψ) 8 / 18

25 Lossy DC Power Flow: Two-Bus Case ψ = 1 b ( P 1 + g ) 1 ψ 2 g This is a fixed-point equation ψ = f (ψ) Interpret as an iteration. Start with initial guess ψ 0 and update: ψ k+1 = f (ψ k ) = 1 ( ) P 1 + g 1 ψ 2k b g }{{} :=P 1,k After ψ converges, simply compute θ 1 θ 2 = arcsin(ψ) 8 / 18

26 Lossy DC Power Flow: Two-Bus Case ψ = 1 b ( P 1 + g ) 1 ψ 2 g This is a fixed-point equation ψ = f (ψ) Interpret as an iteration. Start with initial guess ψ 0 and update: ψ k+1 = f (ψ k ) = 1 ( ) P 1 + g 1 ψ 2k b g }{{} :=P 1,k After ψ converges, simply compute θ 1 θ 2 = arcsin(ψ) 8 / 18

27 Lossy DC Power Flow: Two-Bus Case ψ = 1 b ( P 1 + g ) 1 ψ 2 g This is a fixed-point equation ψ = f (ψ) Interpret as an iteration. Start with initial guess ψ 0 and update: ψ k+1 = f (ψ k ) = 1 ( ) P 1 + g 1 ψ 2k b g }{{} :=P 1,k After ψ converges, simply compute θ 1 θ 2 = arcsin(ψ) 8 / 18

28 Convergence for Two-Bus Case Relative Error j3[k]! 3j= MDCPF k = 1 k = 2 k = g=b 9 / 18

29 Can we show this converges? Yes! Theorem: Convergence of Lossy DC Power Flow The Lossy DC Power Flow iteration ψ k+1 = f (ψ k ) converges exponentially to a power flow solution θ if ( ) 2 P1 ( g ) ( ) P < 1. b b b The power flow solution θ is the unique solution contained in the normal operating regime. P 1 /b is the DC PF solution ρ = g/b is the R/X ratio of the network Can we extend this idea to networks? 10 / 18

30 Can we show this converges? Yes! Theorem: Convergence of Lossy DC Power Flow The Lossy DC Power Flow iteration ψ k+1 = f (ψ k ) converges exponentially to a power flow solution θ if ( ) 2 P1 ( g ) ( ) P < 1. b b b The power flow solution θ is the unique solution contained in the normal operating regime. P 1 /b is the DC PF solution ρ = g/b is the R/X ratio of the network Can we extend this idea to networks? 10 / 18

31 raph and circuit matrices very convenient to write power flow in matrix/vector notation introduce incidence matrix A R n m 1 if bus k at sending-end of edge e A ke = 1 if bus k at recieving-end of edge e 0 otherwise e e 1 e 3 e A = ker(a) = {0} if graph has no cycles 11 / 18

32 raph and circuit matrices very convenient to write power flow in matrix/vector notation introduce incidence matrix A R n m 1 if bus k at sending-end of edge e A ke = 1 if bus k at recieving-end of edge e 0 otherwise e e 1 e 3 e A = ker(a) = {0} if graph has no cycles 11 / 18

33 Modified DC Power Flow with vector notation D B = diag(b ij ), P = (P 1,..., P n ) T, θ = (θ 1,..., θ n ) T decoupled power flow becomes P i = n ij sin(θ i θ j ) j=1 P = AD B sin(a T θ) candidate solution: The Modified DC PF ψ = sin(a T θ), ψ = A T B 1 P 12 / 18

34 Modified DC Power Flow with vector notation D B = diag(b ij ), P = (P 1,..., P n ) T, θ = (θ 1,..., θ n ) T decoupled power flow becomes P i = n ij sin(θ i θ j ) j=1 P = AD B sin(a T θ) candidate solution: The Modified DC PF ψ = sin(a T θ), ψ = A T B 1 P 12 / 18

35 Modified DC Power Flow with vector notation D B = diag(b ij ), P = (P 1,..., P n ) T, θ = (θ 1,..., θ n ) T decoupled power flow becomes P i = n ij sin(θ i θ j ) j=1 P = AD B sin(a T θ) candidate solution: The Modified DC PF ψ = sin(a T θ), ψ = A T B 1 P 12 / 18

36 Lossy DC Power Flow for Networks P i = j i B ij sin(θ ij ) + ii + j i ij cos(θ i θ j ), diag = ( 11, 22,..., nn ) T D = diag( ij ) Vectorized Lossy Power Flow P = AD B sin(a T θ) + diag A D cos(a T θ) substitute ψ = sin(a T θ) like before P = AD B ψ + diag A D 1 ψ 2 now isolate ψ! 13 / 18

37 Lossy DC Power Flow for Networks P i = j i B ij sin(θ ij ) + ii + j i ij cos(θ i θ j ), diag = ( 11, 22,..., nn ) T D = diag( ij ) Vectorized Lossy Power Flow P = AD B sin(a T θ) + diag A D cos(a T θ) substitute ψ = sin(a T θ) like before P = AD B ψ + diag A D 1 ψ 2 now isolate ψ! 13 / 18

38 Lossy DC Power Flow for Networks P i = j i B ij sin(θ ij ) + ii + j i ij cos(θ i θ j ), diag = ( 11, 22,..., nn ) T D = diag( ij ) Vectorized Lossy Power Flow P = AD B sin(a T θ) + diag A D cos(a T θ) substitute ψ = sin(a T θ) like before P = AD B ψ + diag A D 1 ψ 2 now isolate ψ! 13 / 18

39 Lossy DC Power Flow for Networks P i = j i B ij sin(θ ij ) + ii + j i ij cos(θ i θ j ), diag = ( 11, 22,..., nn ) T D = diag( ij ) Vectorized Lossy Power Flow P = AD B sin(a T θ) + diag A D cos(a T θ) substitute ψ = sin(a T θ) like before P = AD B ψ + diag A D 1 ψ 2 now isolate ψ! 13 / 18

40 Lossy DC Power Flow for Networks contd. We obtain a fixed-point equation ( ) ψ = f (ψ) := A T B 1 P diag + A D 1 ψ 2 which we interpret as an iteration Lossy DC Power Flow Iteration ( ) ψ k+1 = f (ψ k ) := A T B 1 P diag + A D 1 ψk 2 phase angles calculated as A T θ k = arcsin(ψ k ) 14 / 18

41 Lossy DC Power Flow for Networks contd. We obtain a fixed-point equation ( ) ψ = f (ψ) := A T B 1 P diag + A D 1 ψ 2 which we interpret as an iteration Lossy DC Power Flow Iteration ( ) ψ k+1 = f (ψ k ) := A T B 1 P diag + A D 1 ψk 2 phase angles calculated as A T θ k = arcsin(ψ k ) 14 / 18

42 Lossy DC Power Flow for Networks contd. We obtain a fixed-point equation ( ) ψ = f (ψ) := A T B 1 P diag + A D 1 ψ 2 which we interpret as an iteration Lossy DC Power Flow Iteration ( ) ψ k+1 = f (ψ k ) := A T B 1 P diag + A D 1 ψk 2 phase angles calculated as A T θ k = arcsin(ψ k ) 14 / 18

43 Convergence of Lossy DC Power Flow Theorem The Lossy DCPF iteration ψ k+1 = f (ψ k ) converges exponentially to a power flow solution θ if the network has no cycles and ψ DC 2 + 2ρ ψ DC < 1 The power flow solution θ is the unique solution contained in the normal operating regime ψ DC = A T B 1 P is the DC PF solution ρ = D 1 B A 1 A D measures R/X ratio Proof based on contraction mapping principle 15 / 18

44 Simulations on MATPOWER Test Cases Table: Base Case Testing of Lossy Modified DC Power Flow Test System Error (deg) Error (deg) Error (deg) DCPF k = 1 k = 2 New England RTS 96 (2 area) bus system RTS 96 (3 area) bus system bus system Polish 2383wp PEASE PEASE PEASE 13, / 18

45 Conclusions Lossy DC Power Flow extends the DC PF 1 Iterative approximation 2 Convergence conditions 3 Successful numerical tests Future work: 1 Analysis for meshed networks 2 Application to stability/contingency 3 Application to market LMPs 17 / 18

46 Conclusions Lossy DC Power Flow extends the DC PF 1 Iterative approximation 2 Convergence conditions 3 Successful numerical tests Future work: 1 Analysis for meshed networks 2 Application to stability/contingency 3 Application to market LMPs 17 / 18

47 Questions 18 / 18

48 appendix

49 Algorithmic Implementation Algorithm 1: Lossy Modified/Unmodified DCPF Inputs: grid data, approximation order k N Outputs: voltage phase angles θ r [k] ψ[0] = 0 m for l 1 to k do P r [l 1] P r diag + A r D 1m (ψ[l 1]) 2 δ r [l] Solve(L B δ r [l] = P r [l 1]) ψ[l] A T r δ r [l] if Lossy Modified DCPF then θ r [k] Solve(A T r θ r [k] = arcsin(ψ[k])) return L-MDCPF voltage angles θ r [k] else θ r [k] δ r [k] return L-DCPF voltage angles θ r [k]