5.3 The Optimal Power Flow Problem

5.3 The optimal power flow problem

5.3.1

OPF Introduction The idea of minimizing the total generation cost under full consideration of the power flow equations goes back to the decade between 1960 and 1970. The mathematical formulation and first solutions of this optimization problem have been given by Carpentier (1962) and Tinney, Dommel (1967).

Looking back to the Lagrangian function which has been used in economic dispatch; L = ∑ Fi (PGi ) + λ(PL + Ploss − ∑ PGi ) i

i = 1KN

i

We can realize, that the power flow in the network has been reduced to one simple equality constraint:

PL + Ploss − ∑ PGi = 0 i

5.3 The optimal power flow problem

5.3.2

The idea of Carpentier was to replace the simplifying constraint “load plus losses equals generation” by the power flow equations for each node in the network. This formulation of the minimization problem including the power flow equations is called Optimal Power Flow (OPF). Important applications of the OPF today: • Calculation of the optimum generation pattern to achieve the minimum total cost of generation. • Calculation of the optimum generation pattern to minimize air pollution by thermal generating units. • Using the network losses as objective, the OPF algorithm can find the optimum reactive power injections of generators, optimum settings of transformer taps and switched capacitors.

5.3 The optimal power flow problem

5.3.3

OPF Example: Find the operating pattern with minimum total fuel costs if the total load of 800 MW is distributed on the nodes PL1 = 100 MW, PL2 = 100 MW, PL3 = 600 MW.

[] + 0.01P [ ] + 0.012 P [ ]

F1 (PG1 ) = 1000 + 12 PG1 + 0.008 PG21

F2 (PG2 ) = 1500 + 14 PG2

F3 (PG3 ) = 2000 + 16 PG3

2 G2

$ h

$ h

2 G3

$ h

Line impedances: Z12 = Z13 = Z 23 = (12.414 + j48.616 ) Ω

Line shunt admittances: GSik + jBS ik = 0

PG2 PL2 =100 MW V2 = 224 kV

PG1 PL1 =100 MW V1 = 224 kV; θ1=0

~

~ 1

L13

L23

~

Network voltage: V1 = V2 = V3 = 224 kV

2

L12

3

PG3 PL3 = 600 MW V3 = 224 kV

5.3 The optimal power flow problem

5.3.4

Solution: Objective function: F = 1000 + 12 PG1 + 0.008 PG21 + 1500 + 14 PG 2 + 0.01 PG22 + 2000 + 16 PG3 + 0.012 PG23

Equality constraints (power flow equations, active power): Vi2Gii + ∑ Vi Vα [Giα cos(Θi − Θ α ) + Biα sin(Θi − Θ α )] − PGi + PLi = 0

i = 1, 2, 3

α

(a12 + a13 ) − a12 cos(Θ1 − Θ 2 ) + b12 sin(Θ1 − Θ 2 ) − a13 cos(Θ1 − Θ3 ) + b13 sin(Θ1 − Θ3 ) − PG1 + 100 = 0 (a21 + a23 ) − a21 cos(Θ2 − Θ1 ) + b21 sin(Θ2 − Θ1 ) − a23 cos(Θ2 − Θ3 ) + b23 sin(Θ 2 − Θ3 ) − PG2 + 100 = 0 (a31 + a32 ) − a31 cos(Θ3 − Θ1 ) + b31 sin(Θ3 − Θ1 ) − a32 cos(Θ3 − Θ 2 ) + b32 sin(Θ3 − Θ2 ) − PG3 + 600 = 0 a12 = a 21 = a13 = a31 = a 23 = a32 = 247.4 MW b12 = b 21 = b13 = b31 = b 23 = b32 = 968.9 MW

5.3 The optimal power flow problem x1 x2 x3

Variables:

5.3.5

PG1 PG2 PG3

Θ x4 = 2 Θ3 x5 λ1 λ1 λ2 λ3

λ2 λ3

Lagrange function: L = 1000 + 12 x1 + 0.008 x12 + 1500 + 14 x 2 + 0.01x12 + 2000 + 16 x 3 + 0.012 x 32

+ λ1[(a12 + a13 ) − a12 cos(− x 4 ) + b12 sin(− x 4 ) − a13 cos(− x 5 ) + b13 sin(− x 5 ) − x1 + 100 ]

+ λ 2 [(a 21 + a 23 ) − a 21 cos(x 4 ) + b 21 sin(x 4 ) − a 23 cos(x 4 − x 5 ) + b 23 sin(x 4 − x 5 ) − x 2 + 100 ] + λ 3 [(a31 + a32 ) − a31 cos(x 5 ) + b 31 sin(x 5 ) − a32 cos(x 5 − x 4 ) + b32 sin(x 5 − x 4 ) − x 3 + 600 ]

5.3 The optimal power flow problem

5.3.6

Necessary conditions for extremum: ∂L =0: ∂x1

12 + 0.016 x1 − λ1 = 0

(1)

∂L =0: ∂x 2

14 + 0.02 x 2 − λ 2 = 0

(2)

∂L =0: ∂x 3

16 + 0.024 x 3 − λ 3 = 0

(2)

∂L = 0: ∂x 4

λ1[− a12 sin(− x 4 ) − b12 cos(− x 4 )]

+ λ 2 [a 21 sin(x 4 ) + b 21 cos(x 4 ) + a 23 sin(x 4 − x 5 ) + b 23 cos(x 4 − x 5 )] + λ 3 [− a32 sin(x 5 − x 4 ) − b32 cos(x 5 − x 4 )] = 0

∂L =0: ∂x 5

(4)

λ1[− a13 sin(− x 5 ) − b13 cos(− x 5 )]

+ λ 2 [− a 23 sin(x 4 − x 5 ) − b 23 cos(x 4 − x 5 )]

+ λ 3 [a31 sin(x 5 ) + b31 cos(x 5 ) + a32 sin(x 5 − x 4 ) + b32 cos(x 5 − x 4 )] = 0

(5)

5.3 The optimal power flow problem

∂L = 0: ∂λ1

h1 (x1, x 4 , x 5 ) = 0

(a12 + a13 ) − a12 cos(− x 4 ) + b12 sin(− x 4 ) − a13 cos(− x 5 ) + b13 sin(− x 5 ) − x1 + 100 = 0

∂L = 0: ∂λ 2

(6)

h2 (x 2 , x 4 , x 5 ) = 0

(a21 + a23 ) − a 21 cos(x 4 ) + b21 sin(x 4 ) − a 23 cos(x 4 − x 5 ) + b23 sin(x 4 − x 5 ) − x 2 + 100 = 0 ∂L =0: ∂λ 3

5.3.7

(7)

h3 (x 3 , x 4 , x 5 ) = 0

(a31 + a32 ) − a31 cos(x 5 ) + b31 sin(x 5 ) − a32 cos(x 5 − x 4 ) + b32 sin(x 5 − x 4 ) − x 3 + 600 = 0

(8)

5.3 The optimal power flow problem

∇L

G

GT

0

2

5.3.8

0.016

0

0

0

0

−1

0

0

0

0.020

0

0

0

0

−1

0

0

0

0.024

0

0

0

0

−1

0

0

0

0

0

0

∂h2 ∂X 4 ∂h2 ∂X 5

∂h3 ∂X 4 ∂h3 ∂X5

0

0

0

0

0

0

−1

0

0

0

0

0

0

−1

∂ 2L ∂X5 ∂X 4 ∂ 2L ∂X 5 ∂X 5 ∂h1 ∂X 5 ∂h2 ∂X 5 ∂h3 ∂X 5

∂h1 ∂X 4 ∂h1 ∂X 5

−1

∂ 2L ∂X 4 ∂X 4 ∂ 2L ∂X 4 ∂X 5 ∂h1 ∂X 4 ∂h2 ∂X 4 ∂h3 ∂X 4

0

0

0

=

5.3 The optimal power flow problem

5.3.9

12 + 0.016 x1 − λ1 14 + 0.020 x 2 − λ 2 ∇L =

Vector of right hand side:

h

λ 1[

λ 1[

16 + 0.024 x 3 − λ 3

] + λ2 [ ] + λ3 [ ] + λ2 [ ] + λ3 [ h1 (x1, x 4 , x 5 ) h2 (x 2 , x 4 , x 5 ) h3 (x 3 , x 4 , x 5 )

] ]

5.3 The optimal power flow problem

x1 = 390 .6

Result of iterative procedure:

5.3.10

PG1 = 390 .6 MW

x 2 = 237 .6

PG 2 = 237.6 MW

x 3 = 195 .9

PG3 = 195 .9 MW

x 4 = −0.0521

Θ 2 = −2.987 °

x 5 = −0.2424

Θ 3 = −13.889°

λ1 = 18.250

$ λ1 = 18.250 MWh

λ 2 = 18.752

$ λ 2 = 18.752 MWh

λ 3 = 20.701

$ λ 3 = 20.701 MWh

F = 17892 .96 $h

5.3 The optimal power flow problem

PG1=390.6 MW V1 = 224 kV ; θ1=0 PL1 =100 MW λ1=18.250 $/MWh

5.3.11 PG2=237.6 MW V2 = 224 kV ; θ2=-2.987° PL2 =100 MW λ2=18.752 $/MWh

~

~ 1

50.82

2

50.15

~

22 5. 35

17 8. 79

18 7. 72

23 9. 81

3

Ploss=24.06 MW F = 17892.96 $/h

PG3=195.9 MW V3 = 224 kV ; θ3=-13.889° PL3 =600 MW λ3=20.701 $/MWh

5.3 The optimal power flow problem

5.3.12

Locational Marginal Price (LMP) Cost to supply the next unit of energy in the most economic way at a particular location in the network

∆PLi

∆PGi (result of OPF calculation)

LMP

1

∆PL1=1

∆PG1 = 0.471; ∆PG2 = 0.256; ∆PG3 = 0.234

∆F $ = 18.250 MWh = λ1 ∆PL1

2

∆PL2=1

∆PG1 = 0.320; ∆PG2 = 0.414; ∆PG3 = 0.249

∆F $ = 18.752 MWh = λ2 ∆PL 2

3

∆PL3=1

∆PG1 = 0.351; ∆PG2 = 0.248; ∆PG3 = 0.420

∆F $ = 20.701 MWh = λ3 ∆PL 3

Node