Transcript New Formulations of the Optimal Power Flow Problem

New Formulations of the Optimal Power
Flow Problem
Prof. Daniel Kirschen
The University of Manchester
© 2010 D. Kirschen and The University of Manchester
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© 2010 D. Kirschen and The University of Manchester
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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
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What is a power system?
Generators
Power
Transmission Network
Loads
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What is running a power system about?
Greed
Minimum cost
Maximum profit
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What is running a power system about?
Fear
Avoid outages and blackouts
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What is running a power system about?
Green
Accommodate renewables
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Balancing conflicting aspirations
Environmental impact
Cost
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Reliability
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The Power Flow Problem
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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: Vk
 Voltage angle at each bus: k
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Other variables
• Active and reactive power consumed at each
bus: PkL ,QkL
 a.k.a. the load at each bus
• Active and reactive power produced by
renewable generators: PkW ,QkW
• Assumed known in deterministic problems
• In practice, they are stochastic variables
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What is reactive power?
Reactive power
Active power
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Injections
Bus k
Pk ,Qk
PkG ,Q kG
PkL ,Q kL
PkW ,QkW
G
W
Pk  PkG  PkW  PkL
Qk  QkG  QkW  QkL
There is usually only one P and Q component at each bus
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Injections
Vk  k
Bus k
Pk ,Qk
Two of these four variables are specified at each bus:
• Load bus: Pk ,Qk
• Generator bus: Pk ,Vk
• Reference bus: Vk , k
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Line flows
To bus i
Pkj ,Qkj
Pki ,Q ki
Vk  k
To bus j
Bus k
Pk ,Qk
The line flows depend on the bus voltage magnitude
and angle as well as the network parameters Gki , Bki
(real and imaginary part of the network admittance matrix)
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Power flow equations
To bus i
Pk j ,Qk j
Pki ,Q ki
Vk  k
To bus j
Bus k
Pk ,Qk
Write active and reactive power balance at each bus:
N
Pk   Vk Vi [Gki cos  ki  Bki sin  ki ]
i1
N
Qk   Vk Vi [Gki sin  ki  Bki cos ki ]
k  1,L N
i1
with: ki  k  i, N : number of nodes in the network
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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
Pk   Vk Vi [Gki cos  ki  Bki sin  ki ]
i1
N
Qk   Vk Vi [Gki sin  ki  Bki cos ki ]
k  1,L N
i1
Typical values of N:
GB transmission network: N~1,500
Continental European network (UCTE): N~13,000
However, the equations are highly sparse!
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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?
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Linear approximation
N
Pk   Vk Vi [Gki cos  ki  Bki sin  ki ]
i1
N
Qk   Vk Vi [Gki sin  ki  Bki cos ki ]
N
Pk   Bki ki
i1
i1
• Ignores reactive power
• Assumes that all voltage magnitudes are nominal
• Useful when concerned with line flows only
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The Optimal Power Flow Problem
(OPF)
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Control variables
• Control variables which have a cost:
 Active power production of thermal generating units:Pi G
• Control variables that do not have a cost:
 Magnitude of voltage at the generating units: Vi G
 Tap ratio of the transformers: t ij
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Possible objective functions
• Minimise the cost of producing power with
conventional generating units:
g
min  Ci ( Pi G )
i1
• Minimise deviations of the control variables from
a given operating point (e.g. the outcome of a
market):
g

min  ci Pi G  ci Pi G

i1
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Equality constraints
• Power balance at each node bus, i.e. power flow
equations
N
Pk   Vk Vi [Gki cos  ki  Bki sin  ki ]
i1
N
Qk   Vk Vi [Gki sin  ki  Bki cos ki ]
k  1,L N
i1
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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
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Formulation of the OPF problem

min f0 x 0 ,u0
u0

 
h x ,u  0
g x 0 ,u0  0
0
0
x 0: vector of dependent (or state) variables
u0 : vector of independent (or control) variables
Nothing extraordinary, except that we are dealing
with a fairly large (but sparse) non-linear problem.
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The Security Constrained
Optimal Power Flow Problem
(SCOPF)
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Bad things happen…
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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
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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?
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Formulation of the SCOPF problem
min
f0 x 0 ,u0 
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  u0  u max
k
k  1,..., N c
uk
k  0: normal conditions
k  1,..., Nc : contingency conditions
u max
:
k
vector of maximum allowed adjustments after
contingency k has occured
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Preventive or corrective SCOPF
min
f0 x 0 ,u0 
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  u0  u max
k
k  1,..., N c
uk
Preventive SCOPF: no corrective actions are considered
u max
 0  u k  u0 k  1,K N c
k
Corrective SCOPF: some corrective actions are allowed
Êk  1,K N c ÊÊu max
0
k
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Size of the SCOPF problem
• SCOPF is (Nc+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?)
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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
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The Worst-Case Problems
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Good things happen…
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… 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
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Formulation of the OPF with uncertainty
market-based generation
min
s.t.
6 4 7 4 8
6 additional
4 4 7generation
4 48
nd  T
c T p0  p0M  b*T
c

p
c
0
0
0




g 0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
h0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
u0  u0init  u0max
Deviations in cost-free controls
p0  p0M  p0max
Deviations in market generation
nd
nd
p min
b0T  p0nd b0T  p max
b0T
Deviations in extra generation
b0 0,1
Decisions about extra generation
s min  s  s max
Vector of uncertainties
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Worst-case OPF bi-level formulation



nd  T
max c T p0  p0M  bT
c

p
c
0
0
0
s
s.t.

s min  s  s max
p ,u ,b ,p  

0

0

0
nd 
0
arg



min c T p0  p0M  b0T c0  p0nd c T
s.t.

g 0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
h0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
u0  u0init  u0max
p0  p0M  p0max
nd
nd
p min
b0T  p0nd b0T  p max
b0T
b0 0,1
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Worst-case SCOPF bi-level formulation



nd  T
max c T p0  p0M  bT
c

p
c
0
0
0
s
s.t.

s min  s  s max
p ,p ,u ,u ,b ,p  

0

k

0

k

0
nd 
0
arg



min c T p0  p0M  b0T c0  p0nd c T
s.t.

g 0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
h0 (x 0 ,u0 ,p0 ,b0 ,p0nd ,s)  0
g k (x k ,u k ,p k ,b0 ,p0nd ,s)  0
h k (x k ,u k ,p k ,b0 ,p0nd ,s)  0
p k  p0  p max
k
u k  u0  u max
k
nd
nd
p min
b0T  p0nd b0T  p max
b0T
b0 0,1
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