How to optimize a power grid
Austin Tuttle
My background
Education
2014 - B.S in Physics and Applied Math (University of the Pacific)
2019 - PhD in Applied Math (UMN)
Internships What I learned
Current Job
“Power systems developer” in “DMS”(Distribution management systems) at OSI(Open Systems International)
Write software that assists in operation of the distribution grid.
2019 - Open Systems International (Private ~1000 employees)
2020 - Emerson (Public 87000 employees…)
2021 - AspenTech (Public 3000 employees)
Intro to power systems
Greater Picture
Grid structure:
Generation
Transmission
Distribution(ME)
3 phase power
Power is distributed along 3 “phases” (parallel lines)
Two phase does exist, those are 90 degrees out of phase.
Can translate between the two with transformers.
Grid Components and behavior
Lines Carry current
Switches Cutoff current/provide current
Transformers Step up/down voltage for optimal delivery
Regulators Maintain voltage balance
Capacitors Balance reactive power
Loads Eat power
Generators Create power(within the distribution grid itself)
Distribution:Substation
Distribution: Substation
Substation
Loads/Distribution transformer
Capacitors
Hierarchy structure
Looped structures
Interesting grid things
Reclosers
ATOs(automatic throw over)
Different connection type transformers(usually 3->3 phases, but some can transfer 1->2)
Delta-Wye transformer diagram
Steady State Calculations
Topology:
Load Estimation:
Power Flow:
Violations:
Basic Physics
Kirchoff's laws:
Current law: Sum of currents in/out of nodes is 0 (We use this one with I = YV to build matrices)
Voltage law: Sum of voltage differences around a loop is 0
Alternating current -> complex “phasor” form:
(Power = Voltage * Current)
(volt-ampere=Watts + Var), real and reactive power
Ohm’s Law DC to AC
Z = impedance
Y = admittance
Basics of Power Flow: Y-blocks
1
2
3
y12
y13
y23
y1
y2
y3
Basics of Power Flow: Y-blocks
1
2
3
4
5
6
yi can be zero if there’s no “shunt”.
This happens for lines. Causes Y-blocks for lines to be singular
Much larger than rest
Basics of Power Flow: Ohm’s Law
Vi=voltage at node i
Ii = current flow into node i
Y is very sparse, symmetric, complex, diagonally dominant
Y v=i
Basics of Power Flow: Current Injection
Sk=Pk+jQk=VkI*k
Power injected at a node is given by(P is power, Q is reactive power)
In a power system, current is injected from endpoints. So we compute:
Ik=S*k/Vk
Basics of Power Flow: Iteration
Assign all loads a power injection: S
With an initial voltage guess: V
Ik=S*k/Vk
Yv=i
Update voltage, recalculate Ik using it.
Let in be the current at step n, vn the voltage.
In=S*/Vn
Yvn+1=in
For step n:
Stop when converged
Solve for v
Basics of Power Flow: Complexity
Matrix: Examples
95,000 x 95,000: 700,000 nonzeros
15,633 x 15,633: 81,000 nonzeros
Matrix: Low Fill in
95,000 x 95,000: 1,250,000 nonzeros
15,633 x 15,633: 117,000 nonzeros
Comments
Managing a power grid in real time
Problem #1
Faults
Faults: How to find them and fix them
Example: Tree falls on a line.
Downstream: power is knocked out
Upstream: current surges as power goes to ground. Breaker trips
Breaker outages more customers
Where’s that tree?
Let’s show a simple example
Source
Closed switch
Open switch
Customer
Example
Source
Closed switch
Open switch
Customer
Trip
Measured fault current
Fault
Source
Closed switch
Open switch
Customer
Trip
Calculated
Location
Via simulating fault current
Source
Closed switch
Open switch
Customer
Isolate
Close
Open
Open
Source
Closed switch
Open switch
Customer
Restore #1
Causes overload
Close
Source
Closed switch
Open switch
Customer
Restore #2
Close
Causes overload
Source
Closed switch
Open switch
Customer
Restore #2
Close
Close
Share Capacity
Problem #2
Dealing with hot days
Var Control
Air conditioners, for example, consume reactive power reducing voltage. This can cause large amounts of “losses” to appear in the network.
We can utilize grid components to minimize this power loss.
Example
Source
Capacitors
Transformer
Regulator
Regulator
Generator
Load
Regulator
Capacitor
Capacitor
Load
Loads
Control Variables
Source
Reactive Power: Q
Downstream Voltage: Vd
Q
Vd
Q
Vd
Vd
Real Power: P
Complex Power: P+jQ
P+jQ
P+jQ
State Space
Source
Capacitors: S:{0,1}
{0,1}
{0,1}
Transformer:
t:{ t=Real, -16<t<16}
Regulator:
r:{-8<r<8}
r:{ -12<r<12}
r:{ -12<r<12}
Generator:
g:{ g=Real, 0<g<max}
Loads:
l:{ Outaged/Not}*
Not a very good choice….
Problem Statement
Minimize: F(Cap States, Tap Positions, Generators)
Subject to:
Performed periodically, as demands/conditions change
Algorithm(s)
Break up into separate parts
Voltage Optimization:
Minimize F(tap positions)
Var Optimization:
Minimize F(cap states)
Gen Optimization:
Minimize F(gen power)
Volt/Var Optimization
Power Flow Optimization
3rd option:
Feeder reconfiguration
Switch changes to move loads
From one feeder to another
Problem #3
Faults and Real-time Data
Problems of Timing
General problem statement:
Customer sees a fault. When recovering from it the estimated loading is way off causing an underloading condition that is not safe.
During the fault, the “last energized” value recorded on loads appears off.
Why….
5 kW
5 kW
5 kW
2 kW
2 kW
2 kW
20 kW
15 kW
6 kW
21 kW
20 kW
41 kW
Not measured but inferred from above
Same with all the loads
Nominal
50 kW
0 kW
0 kW
0 kW
2 kW
2 kW
2 kW
20 kW
0 kW
6 kW
6 kW
26 kW
Nominal
50 kW
0 kW?
2 kW
2 kW
2 kW
20 kW
0 kW
6 kW
6 kW
26 kW
Nominal
50 kW
0 kW?
0 kW?
Or maybe 10 kw shared?
From
And this one is wrong?
15 kW
19.7 kW
2 kW
2 kW
2 kW
20 kW
26.3 kW
0 kW
6 kW
6 kW
26 kW
Nominal
50 kW
50 kW - 6 = 46 remain
Proportions were:
Bottom = 20/(41-6)->26.3
Top = 15/(41-6)->19.7
15 kW
19.7 kW
2 kW
2 kW
2 kW
20 kW
26.3 kW
0 kW
6 kW
6 kW
26 kW
Fault still exists, we reopen
Nominal
50 kW
Save “last energized” value
With last energized we can tell if a tie can
Take the load
But 19.7 is not a good value.
We shouldn’t have saved it, we should have
Kept 15!
15 kW
19.7 kW
2 kW
2 kW
2 kW
20 kW
26.3 kW
0 kW
6 kW
6 kW
26 kW
Problem:
Nominal
50 kW
15 kW
19.7 kW
2 kW
2 kW
2 kW
20 kW
26.3 kW
0 kW
6 kW
6 kW
26 kW
Problem:
Nominal
50 kW
State: Closed
State: Closed
State: Closed
State: Closed
State: Closed
6 kW
These are “SCADA points”
Status values: On/Off, Open/Closed
Analog values: 0,1,1.1,-1 etc
They are measurements that get “scanned” and updated. We receive those updates when they change values.
15 kW
19.7 kW
2 kW
2 kW
2 kW
20 kW
26.3 kW
Analog:
10 sec
Everything has a timestamp
Nominal
50 kW
State: 20 sec
State: 0 sec
State: 0 sec
State: 0 sec
Analog:
12 sec
Analog:
10 sec
Analog:
4 sec
Misc Thoughts
Unexpected Problems
What a PhD brings
Misc advice
Demo?
Questions?
End
Problem #4
Investigating reliability and planning
Test future planned changes to the grid
Increased loading
New grid components
How in sync are fault protection devices?
Study historical data
Test future planned changes to the grid
These tests can involve perform thousands of tests on similar systems, how can we minimize the time of this?
Small Problem 1
Coloring