Minimizing load shedding in electricity networks bypreventive islanding

Waqquas Bukhsh, Andreas Grothey, Ken McKinnon,Jacek Gondzio, Paul Trodden

School of MathematicsUniversity of Edinburgh

FERC25th June 2012

Background

Project structure

◮ EPSRC funded project led by Janusz Bialek at Durham Universitywith groups at Edinburgh and Southampton.

Group Responsibilities:

◮ Durham: Electrical Engineering: System dynamics

◮ Southampton: Mathematics: Laplacian based graph partitioning

◮ Edinburgh: OR and Optimization: Steady state optimization

USA Aug 2003

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Background

Goal

◮ Develop a tool that can create a Firewall to isolate the troubledarea from the rest of the network.

◮ Leave network in a stable steady state.

◮ Could be used for off line analysis to prepare responses to faultsin different areas or, if fast enough, to react in real time.

Sectioning “uncertain” parts of network

◮ Network has “uncertain” buses/lines/generators ? ? ?

??

? ?

Sectioning “uncertain” parts of network

Form a Firewall

◮ to: Achieve a new “safe” steady state thatisolates all “uncertain” buses/lines/generators

◮ by: Cutting lines & Shedding loads & Adjusting generation

◮ goal: Minimize the load shed

?

??

?

Sectioning “uncertain” parts of network

◮ Section 0 – uncertain parts, Section 1 – certain parts

◮ All load in section 0 is at risk, with chance β of surviving

?

??

?

Island 1 Island 2

Section 0 Section 1Section 1

Island 3 Island 4

Sectioning with uncertainty: β = 0.5Before

~

~3

75

14

13

6

1

10

9

4

1112

2?

◮ 259 MW total load

◮ 0.0 MW load shed

◮ 129.5 expected loss

After

~

~3

75

14

13

6

1

10

9

4

1112

2?

◮ 0.0 MW shed in Section 1

◮ 35.9 MW shed in Section 0

◮ 127.8 MW at risk in Section 0

◮ 99.8 MW expected loss

Growth of section 0 in 24 bus network as β → 1

0.00 ≤ β ≤ 1.00~ ~~

~

~~ ~ ~

~ ~~

2118 22

16 19

17

12

15 14 13

9 10

1 2 7

4

11

8

2320

65

24

3?

?

?

Sectioning Constraints

◮ Uncertain buses, B0, and lines, L0

◮ Optimization decides what else in 0 & 1 and which line to cut

◮ γb = section for bus b

◮ ρbb′ = 0 if line bb′ is cut, and 1 otherwise

γ = 1 γ = 0 γ = 1

?

??

?

Island 1 Island 2

Section 0 Section 1Section 1

Island 3 Island 4

Sectioning Constraints

ρbb′ ≤ 1+ γb − γb′ bb′ ∈ L \L0 No lines between sections

ρbb′ ≤ 1− γb + γb′ bb′ ∈ L \L0 No lines between sections

ρbb′ ≤ 1− γb bb′ ∈ L0 No unsafe lines in Section 1

ρbb′ ≤ 1− γb′ bb′ ∈ L0 No unsafe lines in Section 1

γb = 0 b ∈ B0 All unsafe buses in Section 0

γ = 1 γ = 0 γ = 1

?

??

?

Island 1 Island 2

Section 0 Section 1Section 1

Island 3 Island 4

AC power flow model: Notation

3

~2

21

~1 ,( )L

12L12p q ,( )L

21L21p q

,( )L13

L13p q ,( )L

23L23p q

,( )L31

L31p q ,( )L

32L32p q

1v 1θ

3v 3θ

2v 2θ

,( )G1

G1p q ,( )G

2G2p q

,( )D2

D3p qα3

,( )D2

D2p qα2,( )D

1D1p qα1

d

b

~g

demand

generator

bus

1 2

3

vb, θb voltage and phase angle at bus bpL

bb′ ,qLbb′ real and reactive power flows into line (b,b′) from bus b

pGg ,qG

g real and reactive power output of generator gPD

d ,QDd real and reactive power demand at load d

αd proportion of load d connected — αd ∈ [0,1]

Kirchhoff Current Law (KCL): conservation of flow at buses

Gg(p G

g,q )

,qbb’L(pL )bb’

αd ),QDDd(P d

~g

bv bθb

d

∑g∈Gb

pGg = ∑

d∈Db

αdPDd + ∑

b′∈Bb

pLbb′ Real balance

∑g∈Gb

qGg = ∑

d∈Db

αd QDd + ∑

b′∈Bb

qLbb′ Reactive balance

Kirchhoff Voltage Law (KVL): flow-voltage relations on lines

G Bbb’ bb’

L(pLbb’ bb’

)^ ^,q L(pLb’b b’b

)^ ^,qb

bv bθb’

b’v b’θ

δbb′ = θb −θb′

p̂Lbb′ = Gbb′vb(vb − vb′ cosδbb′)−Bbb′vbvb′ sinδbb′

q̂Lbb′ = Bbb′vb(vb′ cosδbb′ − vb)−Gbb′vbvb′ sinδbb′ − v2

b Cbb′

◮ Because of power losses in the lines

p̂Lbb′ 6= −p̂L

b′b and q̂Lbb′ 6= −q̂L

b′b.

Disconnecting lines (pLbb′ case)

bv b’θ

Lbb’

(p Lbb’,q ) L

b’b(p L

b’b,q )Lb’b

(p Lb’b,q )^^L

bb’(p L

bb’,q )^^

ρbb’

vb’ b’θb b’

When ρbb′ = 0, line is not connected, so pLbb′ = 0

When ρbb′ = 1, line is connected, so pLbb′ = p̂L

bb′

MILP Formulation

−ρbb′PLmaxbb′ ≤ pL

bb′ ≤ PLmaxbb′ ρbb′ ,

−(1−ρbb′)P̂Lmaxbb′ ≤ p̂L

bb′ −pLbb′ ≤ P̂Lmax

bb′ (1−ρbb′).

PLmaxbb′ is the maximum allowed real flow into the line from b

P̂Lmaxbb′ is the maximum possible value of p̂L

bb′ −pLbb′

Bounds on variables

pbb′ + pb′b ≤ Hbb′ Thermal limit on line

V−b ≤ vb ≤ V +

b Voltage limit at bus e.g. vb ∈ [0.95,1.05]

QG−g ≤ qG

g ≤ QG+g Reactive Power limits QG−

g ,QG+g

independent of current operating point

ζgPG−g ≤ pG

g ≤ ζgPG+g Real Power limits PG−

g ,PG+g

dependent on current operating point

ζg = 0 if generator is switched off

ζg = 1 if generator remains on

Objective

Definition

◮ Planned supply: PDd αd

◮ Supply probability: β in Section 0

◮ Objective: Maximize expected load supplied

Objective: max β ∑d∈S0

PDd αd + ∑

d∈S1

PDd αd

IP Formulation

◮ Split each αd into α0d for section 0 and α1d for Section 1

αd = α0d + α1d , 0 ≤ α1d ≤ γb

Objective: max∑d

PDd (βα0d + α1d)

Variables for average line flow and loss

L(pLbb’ bb’

)^ ^,q L(pLb’b b’b

)^ ^,qb

bv bθb’

b’v b’θ

◮ Define

pAvbb′ =

p̂bb′ − p̂b′b

2qAv

bb′ =q̂bb′ − q̂b′b

2pLoss

bb′ = p̂bb′ + p̂b′b qLossbb′ = q̂bb′ + q̂b′b

p̂bb′ = pAvbb′ +

pLossbb′

2q̂bb′ = qAv

bb′ +qLoss

bb′

2

p̂b′b = −pAvbb′ +

pLossbb′

2q̂b′b = −qAv

bb′ +qLoss

bb′

2

Approximations of Kirchhoff’s Laws

We consider 3 approximations to Kirchhoff’s Laws

◮ “DC” Constant Loss:

◮ Drop the reactive power constraints◮ Set all voltages to 1◮ Assume line loss pLoss

bb′ = its pre-islanding value

◮ AC Linear:◮ Include reactive power constraints◮ Linearize KVL

◮ AC PWL:◮ Include reactive power constraints◮ Linearize KVL◮ Include PWL approximation of cosine terms

Case: B = 5,G = 1,C = 0.5, |δ| <= 57◦, 0.95 ≤ v ≤ 1.05.

max “DC” AC Lin AC PWL

pAv =−Bδ 2.500 X X X

+ G(vb − vb′) 0.100 X X

+ G/2 ((vb −1)2 − (vb′ −1)2) 0.001

+ B(1− vbvb′)sinδ 0.245

+ B(δ− sinδ) 0.103

pLoss = 2G(1− cosδ) 0.245 X

+ G(vb − vb′)2 0.001

+ 2G(vbvb′ −1)(1− cosδ) 0.023

qAv =−Gδ 0.500 X X

− (B + C)(vb − vb′) 0.495 X X

− (B + C)/2 ((vb −1)2 − (vb′ −1)2) 0.001

+ G(δ− sinδ)) 0.021

+ G(1− vbvb′)sinδ) 0.051

qLoss =−C(vb − vb′ −1) 0.550 X X

−2B(1− cosδ) 1.224 X

−B(vb − vb′)2 0.050

−C((vb −1)2 +(vb′ −1)2)+ 0.000

+ 2B(vbvb′ −1)(1− cosδ) 0.125

Computational tests

Steps

1. Solve OPF: for pre-islanded state using the full AC equations.

2. Based on this solution solve the MILP problem to decide the bestislands and generator shut downs.

3. With the islands and generator shutdowns fixed, solve an optimalload shedding problem using the full AC equations.

Software and examples

◮ 1 and 3 use IPOPT (COIN-OR).

◮ 2 uses CPLEX 12.3 (using 8 cores)

◮ IEEE test cases up to 300 buses.

Computational Times (DC constant loss)Min, Median and Max

nB

t com

p(s

)

feasible

5%

1%

optimal

9 14 24 30 39 57 118 30010−2

10−1

1

101

102

103

104

MILP Suboptimality (DC constant loss)

How close is the MILP solution to the MILP optimal?

Much better than the bounds indicate:

Average amount above proven optimal solution

Feasible 5% gap 1% gap

9.12% 0.37% 0.04%

AC infeasibility (DC constant loss)

◮ Some AC solutions were infeasible because voltage limits wereviolated.

◮ Proportion where one or more methods gave an AC infeasibility:

nB 9 14 24 30 39 57 118 300

AC infeasible % 0% 15% 50% 27% 35% 17% 14% 28%

What can be done?

◮ Relaxing voltage limits in the contingency?0.9 ≤ v ≤ 1.1 “cures” some cases.

◮ Alter the tap setting post-islanding?

◮ Use AC PWL in MILP for islanding

AC feasibility problems

DC constant loss MILP

Voltage problemsv2 = 1.1461v6 = 0.8452

Disconnect static reactor at bus 6:v2 = 1.0709v6 = 0.9291

~ ~~

~

~~ ~ ~

~ ~~

2118 22

16 19

17

12

15 14 13

9 10

1 2 7

4

11

8

2320

5

24

3

?6

AC feasibility problem solved

AC PWL MILP

No Voltage problemsv2 = 1.02v6 = 0.9851v10 = 0.9984

~ ~~

~

~~ ~ ~

~ ~~

2118 22

16 19

17

12

15 14 13

9 10

1 2 7

4

11

8

2320

5

24

3

?6

Comparison of AC PWL and AC solutions24-bus network: angles, voltages & flows

Line l

δL l(d

eg)

PWL-ACAC-OLS

0 5 10 15 20 25 30−20

−15

−10

−5

0

5

10

(a) Angle differencesLine l

pL,

frl

(MW

)

PWL-ACAC-OLS

0 5 10 15 20 25 30−500

−400

−300

−200

−100

0

100

200

300

(b) Real flows

Bus b

v b(p

.u.)

PWL-ACAC-OLS

0 5 10 15 20 250.94

0.96

0.98

1

1.02

1.04

1.06

(c) VoltagesLine l

qL,

frl

(MV

Ar)

PWL-ACAC-OLS

0 5 10 15 20 25 30−150

−100

−50

0

50

100

(d) Reactive flows

AC Infeasible cases

24 Bus case. 24 single bus contingencies

Method Number AC Average Elapsedinfeasible Time (sec)

DC Constant 7 0.32AC Linear 3 2.33AC PWL 0 7.61

Results are similar for other IEEE cases

Dynamic instability

◮ Islanding creates a shock that could excite dynamic instability,and so prevent the system converging to the planned steadystate.

◮ Tests of previous 452 islanding solutions using 2nd order modelsshowed 14 to be transiently unstable.

◮ Penalise line and generator disconnections to reduce shock

Times with and without line and generator penalties

nB

t com

p(s

)

original penaltiesnew penalties

9 14 2430 39 57 118 30010−2

10−1

1

101

102

103

104

◮ All problems now transiently stable

◮ Average load supply decreased by 0.34%

Transient dynamics 24 Bus

Note different scales

Time (s)

δr g−

δr s,0

(deg

)

0 1 2 3 4 5 6−200

0

200

400

600

800

1000

1200

1400

Time (s)

δr g−

δr s,0

(deg

)

0 1 2 3 4 5 6 7 8−30

−20

−10

0

10

20

30

40

Low penalty - Unstable High penalty - StablePredicted load supplied Predicted load supplied

= 2707 MW = 2675 MW25 generators lose synchrony No loss of synchrony

— all in “safe” section

Current & Future

◮ Scale up: We can solve “2500” bus Polish networks with DC MILP

◮ Computation: Develop heuristics and aggregation methods

◮ Bus splitting: to give more flexibility in islanding.

◮ Dynamic stability: Incorporate more realistic indicators in MILP— using work being done at Durham.

Thank You

Questions?