Download Notes for Smart Grid and Power Supply (markets, etc.)

EE 295 - Smarter Electric Energy Systems
(i.e., Smart Grid)
Section 2: Smart Grid and Power Supply
Paul Hines
September 16, 2015
1
Introduction
See the slides for topics that we did from the slides.
2
Power flow
See notes from Section 1 of this class. Key part repeated here:
If we assume that θf t is small, then sin θf t ∼
= θf t and we get:
1
Pf t ∼
θf t
=
xf t
(1)
If we furthermore assume that Qf t = 0 (not a particularly good assumption), then the
current magnitude and the power are equal, |If t | = Pf t , and we can use eq. (1) to roughly
simulate power flows in a power system.
In order to solve for the flows Pf t in simulation, we put eq. (1) into matrix form as
follows. Let A denote the line-to-node incidence matrix with 1 and -1 in each row indicating
the endpoints of each line, θ be the vector of voltage phase angles, be a diagonal matrix of
line reactances, and Pflow be a vector of active power flows along transmission lines. Then,
we can solve for the vector of power flows Pflow given that we know the vector of voltage
phase angles θ as shown in the following:
A> θ = XPflow
Pflow = X−1 A> θ
(2)
(3)
In order to solve for θ, we use information about the sources (generators) and sinks (loads)
to build a vector of net injected powers (generation minus load), P. Given P, we can solve
the following to find θ:
P = APflow = AX−1 A> θ = Bθ
(4)
The matrix B is known as the bus susceptance matrix, and has the properties of a weighted
graph Laplacian matrix describing the network of transmission lines, where the link weights
are the susceptances −bf t = 1/xf t .
1
3
Optimal Power Flow and its friends
3.1
Notation
Let’s say that you have
• ng generators, located at buses in the set G, each of which is producing Pg of power
• nd loads, located at buses in the set D, each of which is consuming Pd of power.
– We’ll assume for notational simplicity that there is one load per bus and that
they are at different buses D ∩ G = ∅. These formulations can be extended to the
other cases, but
• Each generator g has a cost function for cg (Pg ) ($) for producing electricity.
• Each load has some cost function cd (Pd0 − Pd ) describing the cost of not consuming
the full amount of load at bus d.
3.2
Economic dispatch
The economic dispatch problem is as follows:
X
X
cd (Pd0 − Pd )
min
cg (Pg ) +
Pg ,Pd
s.t.
g∈G
d∈D
Pg ≤ Pg ≤ Pg , ∀g ∈ G
0 ≤ Pd ≤ Pd , ∀d ∈ D
X
X
Pg =
Pd + Ploss
g∈G
d∈D
For the moment let’s ignore the generator limits and losses, and solve for the first order
optimality conditions. To do so we can write the Lagrangian for the problem:
X
X
X
X
cg (Pg ) +
cd (Pd0 − Pd ) − λ(
Pg −
Pd )
g∈G
g∈G
d∈D
d∈D
We find the optimality conditions by solving for the point at which all the partial derivatives
of this are equal to zero.
dcg
= λ
dPg
dcd
= λ
dPd
X
X
Pg −
Pd = 0
g∈G
d∈D
This tells us that if we find an optimal solution, there will also be an optimal price λ, that
tells us the cost of having demand at this level. This price is known as the system marginal
price, or often “the system lambda.”
2
3.3
Now add transmission line constraints
The above ignores transmission system constraints, which are very important to our problem.
To do so we need not just a system-wide balance between supply and demand, but also a
nodal one. We can use the DC power flow to give us this.
X
X
cg (Pg ) +
min
cd (Pd0 − Pd )
Pg ,Pd
s.t.
g∈G
d∈D
Pg ≤ Pg ≤ Pg , ∀g ∈ G
Bθ = PG − PD
1
Fij ≤ Fij = (θi − θj ) ≤ Fij
X
θr = 0
0 ≤ Pd ≤ Pd , ∀d ∈ D
X
X
Pd
Pg =
g∈G
d∈D
When you do this you get a new set of LeGrange multipliers, but this time associated with
the nodal equality constraint
Bθ = PG − PD
These nodal prices are called “Location-Based Marginal Prices” or the “Locational Marginal
Prices.” You can see these prices by looking at the ISO New England web site, for example:
http://www.iso-ne.com/markets/hrly_data/hourlyLMP.do
3.4
Security constrained OPF
3.5
Unit commitment
Add up/down constraints
3.6
Security constrained unit commitment
3.7
Stochastic unit commitment
3