ESE 680 -- Modern Electrical Energy Infrastructure
Lecture 3
Power Flow Modeling and Optimization
(by an amateur)
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ESE 680 -- Modern Electrical Energy Infrastructure
Sample power networks
IEEE 14-bus case
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ESE 680 -- Modern Electrical Energy Infrastructure
Sample power networks
A distribution network from the service
area of Southern California Edison
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ESE 680 -- Modern Electrical Energy Infrastructure
Ohm’s law
resistance =
V
R=
I
potential difference (voltage)
current
=
Extension to AC circuits:
impedance =
complex voltage
complex current
Z = R + jX
Resistance: opposition
to the passage of an
electric current
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Reactance: opposition
to a change in current
or voltage (e.g., due to
inductors or capacitors)
ESE 680 -- Modern Electrical Energy Infrastructure
Ohm’s law
Complex form of the law:
V = IZ
Phasor form of voltage, current, and impedance:
V = |V |ej(ωt+φV )
I = |I|ej(ωt+φI )
Z = |Z|ejθ
V = IZ
|V |ej(ωt+φV ) = |I||Z|ej(ωt+φI +θ)
⇔
Last equality hold for all t. Hence...
|V | = |I||Z|
φV = φI + θ
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Impedance
Admittance
Impedance: Z = R + jX
Admittance: Y = Z
−1
= G + jB
Conductance
Susceptance
Re-write Ohm’s law in terms of admittance:
I =YV
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ESE 680 -- Modern Electrical Energy Infrastructure
Kirchhoff’s Current Law (KCL)
Let I1 , . . . , IN be the currents
flowing into or out from a node/bus (with
a proper sign convention). Then,
N
�
The algebraic sum of
currents in a network of
conductors meeting at a
point is zero.
Ij = 0.
j=1
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ESE 680 -- Modern Electrical Energy Infrastructure
Admittance matrix
KCL at node 1:
Re-write with the convention that the admittance between two buses
not connected be zero:
Re-arrange:
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ESE 680 -- Modern Electrical Energy Infrastructure
Admittance matrix



I1
y1 + y12 + y13 + y14
 I2  
−y21

=
 I3  
−y31
I4
−y41
−y12
y2 + y21 + y23 + y24
−y32
−y42
−y13
−y23
y3 + y31 + y32 + y34
−y43


−y14
V1
 V2 
−y24


 V3 
−y34
y4 + y41 + y42 + y43
V4
In general:
Note: Y is
symmetric.
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ESE 680 -- Modern Electrical Energy Infrastructure
Electric power
P = word done by electric charges moving through an electric potential
(voltage) difference per unit time
= (charge flow per unit time) * (voltage difference)
= IV
S
2
V
=I R=
R
2
Q
(by Ohm’s law)
P
Generalization to complex power:
S = V I∗
= P + jQ
reactive power
real power
•
•
When voltage and current are in phase, no
reactive power.
Capacitors generate and inductors consume
reactive power.
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ESE 680 -- Modern Electrical Energy Infrastructure
Reactive power voltage support
(mentioned in the last lecture in an even sloppier way)
R
Vr = Vs
R + jωL
1
Vr = Vs
(1 − ω 2 LC) +
jωL
R
|Vr |
v.s. capacitance
|Vs |
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ESE 680 -- Modern Electrical Energy Infrastructure
Power balance equations and “power flow problem”
Power injection into a bus k:
Sk =
We also know:
Ik =
N
�
∗
Vk I k
Yjk Vjk
j=1
Can re-write by eliminating the currents:
I k = Vk
N
�
∗ ∗
Yjk
Vjk
j=1
•
Variables for each bus:
• Voltage magnitude and phase
• Current magnitude and phase
• Real and reactive power
Generator bus: real power and voltage
magnitude specified
• Load bus: Real and reactive power
specified
•
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Power flow problem: Given some of the
variables, find the rest subject to...
• the network constraints above, and
• additional performance/safety/reliability
constraints.
ESE 680 -- Modern Electrical Energy Infrastructure
Typical (additional) constraints & objectives
Constraints:
• Physical limitations of the generators
• Transmission lines can only carry certain
amount of current
• Transformers have limited ratio or phase
shifting capability and they heat up
• Performance: bounds on voltage
magnitudes and angles
• Limits on reactive power sources
• Limited rate of change of almost
everything
• ...
Objectives:
• Total generation cost
• Losses (difference between total
generation and use)
• “Time” to reach acceptable network
configurations
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow (a formulation of it)
x : all variables, including
power injections, Sk
voltages, Vk
current, Ik
minimize
over
x
cost(x)
subject to
f (x) ≤ 0
g(x) ≤ 0
Sk = Vk Ik∗
for all k:
Ik =
N
�
j=1
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Yjk Vjk
eliminating
the current
Sk = V k
N
�
∗ ∗
Yjk
Vjk
j=1
ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow: X
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow: X
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow: X
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow: X
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow: X
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ESE 680 -- Modern Electrical Energy Infrastructure
Optimal power flow (a formulation of it)
cost(x)
minimize
over
x
Ax ≤ b
subject to
Cx = d
Sk = V k
N
�
pick linear,
quadratic,
convex,
etc.
affine
constraints
(due to
operation,
performan
ce, safety,
etc.)
network
for all k constraints
∗ ∗
Yjk
Vjk
j=1
How to “deal” with the network constraints
(usually the bottleneck)?
• approximate
• relax
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ESE 680 -- Modern Electrical Energy Infrastructure
Approximations
Re-write the bus power injections:
Case 1 (for transmission lines): resistance << reactance
→ Gkj ≈ 0
Case 2: θj − θk small for neighboring buses j and j
→ cos(θj − θk ) ≈ 1
sin(θj − θk ) ≈ θj − θk
Case 3: Voltage magnitude at every bus is equal or
close to some base voltage (normalized to 1)
All three simplifications together lead to
the so-called DC OPF.
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ESE 680 -- Modern Electrical Energy Infrastructure
Another formulation of optimal power flow
(for tree networks)
loss minimization
real and reactive power injections, real and reactive
power flows, losses, voltage magnitudes
underlying
physical laws
and network
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ESE 680 -- Modern Electrical Energy Infrastructure
Idea behind the solution technique
Nonconvex:
Quadratically-constrained,
quadratic program
minimize
subject to
c(x)
convex constraints on x
Convex relaxation:
Second order cone program
minimize
subject to
q(x) = 0
c(x)
convex constraints on x
q(x) ≤ 0
convex, quadratic function
search on the boundary
of the cone
When does the relaxation give solutions to
the original problem?
• In
practice, quite often.
search inside the cone
SDP relaxation: Bai et al. ’08
SOCP relaxation: Farivar et al. ’11,
Taylor et al. ’11
Exactness conditions: Lavaei et al.
’10, Zhang et al. ’11, Bose et al. ’11
Any sufficient exactness certificates?
• Only
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in special cases. One such certificate explained in the next slide.
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ESE 680 -- Modern Electrical Energy Infrastructure
A priori exactness certificate
Theorem: For tree networks, the solution to the relaxation
solves the original problem, if
•
[non-decreasing resistance-to-reactance ratio]
rij
rjk
≤
for each (i, j, k) s.t. (i, j), (j, k) ∈ E
xij
xjk
•
relatively
outdated
results
−1
yij
= rij + xij i
resistance
reactance
admittance
[unidirectional real power flow in the lossless case]
Pij ≥ 0 for each (i, j) ∈ E
Proof idea:
Involved algebraic
manipulations
↓
Lagrange multipliers for
q(x) ≤ 0 are strictly
positive at optimality
↓
Complementary
slackness to show q(x) =
0 at optimality
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ESE 680 -- Modern Electrical Energy Infrastructure
Exactness certificate as a design guide
tree network
•
non-decreasing resistance-toreactance ratio
•
unidirectional real power flow in the
lossless case
valid for most
distribution circuits
constrains the
placement of distributed
generation and storage
substation
47-bus distribution circuit
from SCE’s service area
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{
•
{
Exactness conditions:
relatively
outdated
results
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ESE 680 -- Modern Electrical Energy Infrastructure
Credits:
[1] http://www.eeh.ee.ethz.ch/uploads/tx_ethstudies/modelling_hs08_script_02.pdf
[2] http://people.ee.ethz.ch/~bacher/publications/thun_paper.pdf (very accessible
overview)
[3] http://home.eng.iastate.edu/~jdm/ee552/PowerFlowEquations.pdf
[4] J. L. Kirtley, “Electric Power Principles: Sources, Conversion, Distribution, and
Use,” Wiley, 2010.
[5] http://arxiv.org/pdf/1208.4076.pdf
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ESE 680 -- Modern Electrical Energy Infrastructure