5.4 Security Constrained Optimal Power Flow
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5.4 Security constrained optimal power flow
5.4.1
The example of a power system discussed in chapter 5.3 is optimal with respect to economy, but it is operated disregarding any security constraints. Considering the outage of line L13, carrying a power flow of 239.81 MW, will result in a post contingency flow of more than 400 MW in line L23. A similar overload of line L13 will be the result of the outage of line L23. Only the outage of line L12 will not lead to any overload situation in line L13 or L23 provided that each of the 3 lines can carry a maximum flow of 300 MW.
Therefore, as security constraints, we have to limit the flows in line L13 and L23. In both lines the limit is fixed to 150 MW in the pre-contingency situation.
5.4 Security constrained optimal power flow
5.4.2
Security constraints: P13 = a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) ≤ 150
P23 = a 23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) ≤ 150
Introduction of slack variables x6, x7 to transform the inequality constraints to equality constraints: a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 26 − 150 = 0
a23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150 = 0
To stay within the same solution procedure as in chapter 5.3, we have to introduce two additional terms in the Lagrange function with λ4 and λ5 as Lagrange multipliers.
5.4 Security constrained optimal power flow
5.4.3
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 ]
[ + λ [a (1 − cos(x
+ λ 4 a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 26 − 150 5
23
4
]
− x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150
]
5.4 Security constrained optimal power flow Necessary conditions for extremum: ∂L =0: ∂x1 ∂L =0: ∂x 2 ∂L =0: ∂x 3
5.4.4
(1) (2)
Identical with eq. (1) to (5) of chapter 5.3
(3)
∂L = 0: ∂x 4
(4)
∂L =0: ∂x 5
(5)
∂L =0: ∂x 6
2λ 4 x 6 = 0
∂L =0: ∂x 7
2λ 5 x 7 = 0
(6) Complimentary rule (7)
5.4 Security constrained optimal power flow
5.4.5
∂L = 0: ∂λ1
h1 (x1, x 4 , x 5 ) = 0
∂L = 0: ∂λ 2
h2 (x 2 , x 4 , x 5 ) = 0
∂L =0: ∂λ 3
h3 (x 3 , x 4 , x 5 ) = 0
(10)
∂L = 0: ∂λ 4
h 4 (x 5 , x 6 ) = 0
(11)
∂L =0: ∂λ 5
(8) Identical with eq. (6) to (8) of chapter 5.3
(9)
a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 62 − 150 = 0
h5 (x 4 , x 5 , x 7 ) = 0
a 23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150 = 0
(12)
5.4 Security constrained optimal power flow
∇ 2L
G
GT
0
5.4.6
0.016
0
0
0
0
0
0
−1
0
0
0
0
0
0.020
0
0
0
0
0
0
−1
0
0
0
0
0
0.024
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
∂ 2L ∂x 5 ∂x 4 ∂ 2L ∂x 5 ∂x 5
0
0
∂ 2L ∂x 4 ∂x 4 ∂ 2L ∂x 4 ∂x 5
0
0
∂h1 ∂x 4 ∂h1 ∂x 5
∂h2 ∂x 4 ∂h2 ∂x 5
∂h3 ∂x 4 ∂h3 ∂x 5
∂h 4 ∂x 5
∂h5 ∂x 4 ∂h5 ∂x 5
0
0
0
0
0
2λ 4
0
0
0
0
2x 6
0
0
0
0
0
0
0
2λ 5
0
0
0
0
2x 7
−1
0
0
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
0
0
0
2x 6
0
0
0
0
0
0
0
0
0
∂h5 ∂x 4
∂h1 ∂x 5 ∂h2 ∂x 5 ∂h3 ∂x 5 ∂h 4 ∂x 5 ∂h5 ∂x 5
0
0
∂h1 ∂x 4 ∂h2 ∂x 4 ∂h3 ∂x 4
0
2x 7
0
0
0
0
0
=
0
5.4 Security constrained optimal power flow
5.4.7
12 + 0.016 x1 − λ1 14 + 0.020 x 2 − λ 2 λ1[
∇L Vector of right
λ1[ =
hand side:
h
15 + 0.024 x 3 − λ 3 ] + λ2 [ ] + λ3 [
] + λ2 [
] + λ3 [
2λ 4 x 6 2λ 5 x 7 h1 (x1, x 2 , x 5 )
h2 (x 2 , x 4 , x 5 ) h3 (x 3 , x 4 , x 5 )
h4 (x 5 , x 6 ) h5 (x 4 , x 5 , x 7 )
] ]
5.4 Security constrained optimal power flow
PG1 = 251 .59 MW
x1 = 251 .59
x 2 = 246 .78
PG 2 = 246.78 MW
x 4 = −0.0016
Θ 2 = −0.094 °
PG3 = 312 .98 MW
x 3 = 312 .98
Θ 3 = −8.734°
x 5 = −0.1524 Result of iterative procedure:
5.4.8
x6 = 0
x 7 = 1.28
λ1 = 16.025
slack var. = 0 slack var. ≠ 0
λ 3 = 23.512 λ5 = 0
x 72 = 1.64 MW
$ λ1 = 16.025 MWh
$ λ 2 = 18.936 MWh
λ 2 = 18.936
λ 4 = 8.530
x6 = 0
$ λ 3 = 23.512 MWh
binding constr. non bind. constr.
$ λ 4 = 8.530 MWh
λ5 = 0
F = 18272 .66 $h
5.4 Security constrained optimal power flow
PG1=251.59 MW V1 = 224 kV ; θ1=0 PL1 =100 MW λ1=16.025 $/MWh
5.4.9
~
~ 1
1.59
2
1.59
~
14 4. 26
14 2. 76
14 8. 37
15 0. 00
3
Ploss=11.35 MW F = 18272.66 $/h
PG3=312.98 MW V3 = 224 kV ; θ3=-8.732° PL3 =600 MW λ3=23.512 $/MWh
PG2=246.78 MW V2 = 224 kV ; θ2=-0.094° PL2 =100 MW λ2=18.936 $/MWh
5.4 Security constrained optimal power flow
5.4.10
Locational Marginal Price (LMP)
∆PLi
∆PGi (result of OPF calculation)
LMP
1
∆PL1=1
∆PG1 = 0.886; ∆PG2 = 0.230; ∆PG3 = -0.108
∆F $ = 16.025 MWh = λ1 ∆PL1
2
∆PL2=1
∆PG1 = 0.288; ∆PG2 = 0.417; ∆PG3 = 0.274
∆F $ = 18.936 MWh = λ2 ∆PL 2
3
∆PL3=1
∆PG1 = -0.162; ∆PG2 = 0.328; ∆PG3 = 0.846
∆F $ = 23.512 MWh = λ3 ∆PL 3
Node
5.4.1
The example of a power system discussed in chapter 5.3 is optimal with respect to economy, but it is operated disregarding any security constraints. Considering the outage of line L13, carrying a power flow of 239.81 MW, will result in a post contingency flow of more than 400 MW in line L23. A similar overload of line L13 will be the result of the outage of line L23. Only the outage of line L12 will not lead to any overload situation in line L13 or L23 provided that each of the 3 lines can carry a maximum flow of 300 MW.
Therefore, as security constraints, we have to limit the flows in line L13 and L23. In both lines the limit is fixed to 150 MW in the pre-contingency situation.
5.4 Security constrained optimal power flow
5.4.2
Security constraints: P13 = a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) ≤ 150
P23 = a 23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) ≤ 150
Introduction of slack variables x6, x7 to transform the inequality constraints to equality constraints: a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 26 − 150 = 0
a23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150 = 0
To stay within the same solution procedure as in chapter 5.3, we have to introduce two additional terms in the Lagrange function with λ4 and λ5 as Lagrange multipliers.
5.4 Security constrained optimal power flow
5.4.3
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 ]
[ + λ [a (1 − cos(x
+ λ 4 a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 26 − 150 5
23
4
]
− x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150
]
5.4 Security constrained optimal power flow Necessary conditions for extremum: ∂L =0: ∂x1 ∂L =0: ∂x 2 ∂L =0: ∂x 3
5.4.4
(1) (2)
Identical with eq. (1) to (5) of chapter 5.3
(3)
∂L = 0: ∂x 4
(4)
∂L =0: ∂x 5
(5)
∂L =0: ∂x 6
2λ 4 x 6 = 0
∂L =0: ∂x 7
2λ 5 x 7 = 0
(6) Complimentary rule (7)
5.4 Security constrained optimal power flow
5.4.5
∂L = 0: ∂λ1
h1 (x1, x 4 , x 5 ) = 0
∂L = 0: ∂λ 2
h2 (x 2 , x 4 , x 5 ) = 0
∂L =0: ∂λ 3
h3 (x 3 , x 4 , x 5 ) = 0
(10)
∂L = 0: ∂λ 4
h 4 (x 5 , x 6 ) = 0
(11)
∂L =0: ∂λ 5
(8) Identical with eq. (6) to (8) of chapter 5.3
(9)
a13 (1 − cos(− x 5 )) + b13 sin(− x 5 ) + x 62 − 150 = 0
h5 (x 4 , x 5 , x 7 ) = 0
a 23 (1 − cos(x 4 − x 5 )) + b 23 sin(x 4 − x 5 ) + x 72 − 150 = 0
(12)
5.4 Security constrained optimal power flow
∇ 2L
G
GT
0
5.4.6
0.016
0
0
0
0
0
0
−1
0
0
0
0
0
0.020
0
0
0
0
0
0
−1
0
0
0
0
0
0.024
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
∂ 2L ∂x 5 ∂x 4 ∂ 2L ∂x 5 ∂x 5
0
0
∂ 2L ∂x 4 ∂x 4 ∂ 2L ∂x 4 ∂x 5
0
0
∂h1 ∂x 4 ∂h1 ∂x 5
∂h2 ∂x 4 ∂h2 ∂x 5
∂h3 ∂x 4 ∂h3 ∂x 5
∂h 4 ∂x 5
∂h5 ∂x 4 ∂h5 ∂x 5
0
0
0
0
0
2λ 4
0
0
0
0
2x 6
0
0
0
0
0
0
0
2λ 5
0
0
0
0
2x 7
−1
0
0
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
0
0
−1
0
0
0
0
0
0
0
0
0
0
0
2x 6
0
0
0
0
0
0
0
0
0
∂h5 ∂x 4
∂h1 ∂x 5 ∂h2 ∂x 5 ∂h3 ∂x 5 ∂h 4 ∂x 5 ∂h5 ∂x 5
0
0
∂h1 ∂x 4 ∂h2 ∂x 4 ∂h3 ∂x 4
0
2x 7
0
0
0
0
0
=
0
5.4 Security constrained optimal power flow
5.4.7
12 + 0.016 x1 − λ1 14 + 0.020 x 2 − λ 2 λ1[
∇L Vector of right
λ1[ =
hand side:
h
15 + 0.024 x 3 − λ 3 ] + λ2 [ ] + λ3 [
] + λ2 [
] + λ3 [
2λ 4 x 6 2λ 5 x 7 h1 (x1, x 2 , x 5 )
h2 (x 2 , x 4 , x 5 ) h3 (x 3 , x 4 , x 5 )
h4 (x 5 , x 6 ) h5 (x 4 , x 5 , x 7 )
] ]
5.4 Security constrained optimal power flow
PG1 = 251 .59 MW
x1 = 251 .59
x 2 = 246 .78
PG 2 = 246.78 MW
x 4 = −0.0016
Θ 2 = −0.094 °
PG3 = 312 .98 MW
x 3 = 312 .98
Θ 3 = −8.734°
x 5 = −0.1524 Result of iterative procedure:
5.4.8
x6 = 0
x 7 = 1.28
λ1 = 16.025
slack var. = 0 slack var. ≠ 0
λ 3 = 23.512 λ5 = 0
x 72 = 1.64 MW
$ λ1 = 16.025 MWh
$ λ 2 = 18.936 MWh
λ 2 = 18.936
λ 4 = 8.530
x6 = 0
$ λ 3 = 23.512 MWh
binding constr. non bind. constr.
$ λ 4 = 8.530 MWh
λ5 = 0
F = 18272 .66 $h
5.4 Security constrained optimal power flow
PG1=251.59 MW V1 = 224 kV ; θ1=0 PL1 =100 MW λ1=16.025 $/MWh
5.4.9
~
~ 1
1.59
2
1.59
~
14 4. 26
14 2. 76
14 8. 37
15 0. 00
3
Ploss=11.35 MW F = 18272.66 $/h
PG3=312.98 MW V3 = 224 kV ; θ3=-8.732° PL3 =600 MW λ3=23.512 $/MWh
PG2=246.78 MW V2 = 224 kV ; θ2=-0.094° PL2 =100 MW λ2=18.936 $/MWh
5.4 Security constrained optimal power flow
5.4.10
Locational Marginal Price (LMP)
∆PLi
∆PGi (result of OPF calculation)
LMP
1
∆PL1=1
∆PG1 = 0.886; ∆PG2 = 0.230; ∆PG3 = -0.108
∆F $ = 16.025 MWh = λ1 ∆PL1
2
∆PL2=1
∆PG1 = 0.288; ∆PG2 = 0.417; ∆PG3 = 0.274
∆F $ = 18.936 MWh = λ2 ∆PL 2
3
∆PL3=1
∆PG1 = -0.162; ∆PG2 = 0.328; ∆PG3 = 0.846
∆F $ = 23.512 MWh = λ3 ∆PL 3
Node
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