04 Unit Commitment
-
Upload
mohsinaman -
Category
Documents
-
view
218 -
download
0
Transcript of 04 Unit Commitment
-
8/8/2019 04 Unit Commitment
1/19
-
8/8/2019 04 Unit Commitment
2/19
Unit Commitment
l Given load profile(e.g. values of the load for each hour of a day)
l Given set of units availablel When should each unit be started, stopped and how
much should it generate to meet the load at minimumcost?
G G GLoad Profile
? ? ?
TypicalSummer AndWinter Demands
-
8/8/2019 04 Unit Commitment
3/19
Why try to optimise?
l Energy traded through the Electricity Pool of England andWales: ~ 7 billion per year
l 0.1% cost reduction through better scheduling:~ 7 million
A Simple Example
l Unit 1: P Min = 250 MW, P Max = 600 MW C 1 = 510.0 + 7.9 P 1 + 0.00172 P 12 /h
l Unit 2: P Min = 200 MW, P Max = 400 MW C 2 = 310.0 + 7.85 P 2 + 0.00194 P 22 /h
l Unit 3: P Min = 150 MW, P Max = 500 MW C 3 = 78.0 + 9.56 P 3 + 0.00694 P 32 /h
l What combination of units 1, 2 and 3 will produce 550MW at minimum cost?
l How much should each unit in that combinationgenerate?
-
8/8/2019 04 Unit Commitment
4/19
Cost of the various combinations
1 2 3 P min P max P 1 P 2 P 3 C total Off Off Off 0 0 Infeasible
Off Off On 150 500 Infeasible
Off On Off 200 400 Infeasible
Off On On 350 900 0 400 150 5418
On Off Off 250 600 550 0 0 5389
On Off On 400 1100 400 0 150 5613On On Off 450 1000 295 255 0 5471
On On On 600 1500 Infeasible 5617
l Far too few units committed:Cant meet the demand
l Not enough units committed:Some units operate above optimum
l Too many units committed:Some units below optimum
l Far too many units committed:Minimum generation exceeds demand
l No-load cost affects choice of optimal combination
Observations on the example:
-
8/8/2019 04 Unit Commitment
5/19
Another Example
Load
Time
12 18 24
500
1000
l Optimal generationschedule for a loadprofile
l Decompose the profileinto a set of period
l Assume load is constantover each period
l For each time period,
which units should becommitted to generate atminimum cost during thatperiod?
Optimal combination for each hour
Load Unit 1 Unit 2 Unit 3
1100 On On On
1000 On On Off
900 On On Off
800 On On Off700 On On Off
600 On Off Off
500 On Off Off
-
8/8/2019 04 Unit Commitment
6/19
Matching the combinations to the load
Load
Time1260 18 24
Unit 1
Unit 2
Unit 3
Issues
l Must consider all constraintsn Unit constraintsn System constraints
l Some constraints create a link between the periodsl Starting up a generating unit costs money in addition to
the running cost considered in economic dispatchl Curse of dimensionality
-
8/8/2019 04 Unit Commitment
7/19
Unit Constraints
l Constraints that affect each unitindividually:
n Maximum generating capacityn Minimum stable generationn Flexibilityn Minimum up timen Minimum down time
n Ramp rate
Flexible Plants
l Power output can be adjusted (within limits)l Examples:
n Coal-firedn Oil-firedn Open cycle gas turbinesn Combined cycle gas turbinesn Hydro plants with storage
l Status and power output can be optimised
Thermal units
-
8/8/2019 04 Unit Commitment
8/19
Inflexible Plants
l Power output cannot be adjusted for technical or commercial reasons
l Examples:n Nuclear n Run-of-the-river hydron Renewables (wind, solar,)n
Combined heat and power (CHP, cogeneration)l Output treated as given when optimising
Notations
X i ( t ) : Status of unit i at period t
X i ( t ) = 1: Unit i is on during period t
X i ( t ) = 0 : Unit i is off during period t
Pi ( t ) : Power produced by unit i during period t
-
8/8/2019 04 Unit Commitment
9/19
Unit Constraints
l Minimum up timen Once a unit is running it may not be shut down
immediately:
l Minimum down timen Once a unit is shut down, it may not be started
immediately
If X i (t ) = 1 and t iup < t i
up ,min then X i (t + 1) = 1
If X i (t ) = 0 and t idown
< t idown ,min then X i (t + 1) = 0
Unit Constraints
l Maximum ramp ratesn To avoid damaging the turbine, the electrical output of a unit
cannot change by more than a certain amount over a periodof time:
Pi (t + 1) Pi (t ) Piup ,max
P i (t ) P i (t + 1) P idown ,max
Maximum ramp up rate constraint:
Maximum ramp down rate constraint:
-
8/8/2019 04 Unit Commitment
10/19
System Constraints
l Constraints that affect more than one unitn Load/generation balancen Reserve generation capacityn Crew constraintsn Emission constraintsn Network constraints
System Constraints: Load/generation balance
P i (t )iC ( t ) = L (t )
C (t ) = i X i (t ) = 1{ }: Set of units committed at time t
-
8/8/2019 04 Unit Commitment
11/19
System Constraint: Reserve Capacity
l Unanticipated loss of a generating unit or aninterconnection causes unacceptable frequency drop if not corrected
l Need to increase production from other units to keepfrequency drop within acceptable limits
l Rapid increase in production only possible if committedunits are not all operating at their maximum capacity
P imax
iC ( t ) L (t )+ R (t )
R (t ): Reserve requirement at time t
How much reserve?
l Protect the system against credible outagesl Deterministic criteria:
n Capacity of largest unit or interconnectionn Percentage of peak load
l Probabilistic criteria:n Takes into account the number and size of the
committed units as well as their outage rate
-
8/8/2019 04 Unit Commitment
12/19
Types of Reservel Spinning reserve
n Primary quick response for a short time
n Secondary slower response for a longer time
n High frequency ability to reduce output when frequency is high
l Scheduled or off-line reserven Unit that can start quickly (e.g. gas turbines)
l Other sources of reserven Pumped hydro plantsn Demand reduction
l Reserve must be spread around the network
Cost of Reserve
l Reserve has a cost even when it is not calledn More units scheduled than required
Units not operated at their maximum efficiency Extra start up costs
n Must build units capable of rapid responsen Cost of reserve proportionally larger in small systems
Important driver for the creation of interconnections between systems
-
8/8/2019 04 Unit Commitment
13/19
Crew Constraints
l It may not be possible to start more than one generatingunit at a time in a power station because of the number of people required to supervise the start-up
l Less of a problem than it use to be thanks to automation
Emission Constraints
l Amount of pollutants that generating units can emit maybe limited
l Pollutants:n SO 2, NO x
l Various forms:n Limit on each plant at each hour n Limit on plant over a year n Limit on a group of plants over a period
-
8/8/2019 04 Unit Commitment
14/19
Network Constraints
l Transmission network may have an effect on thecommitment of units
n Some units must run to provide voltage supportn The output of some units may be limited because their
output would exceed the transmission capacity of thenetwork
Cheap generatorsMay be constrained off
More expensive generator May be constrained on
A B
Start-up Costs
l Thermal units must be warmed up before they can bebrought on-line
l Warming up a unit costs moneyl Start-up cost depends on time unit has been off
SC i (t iOFF ) = i + i (1 e
t i
OFF
i )
t iOFF
i
i + i
-
8/8/2019 04 Unit Commitment
15/19
-
8/8/2019 04 Unit Commitment
16/19
Solving the Unit Commitment Problem
l Decision variables:n Status of each unit at each period:
n Output of each unit at each period:
l Combination of integer and continuous variables
X i (t ) 0,1{ } i , t
Pi ( t )
X i (t )
Pi (t ) 0, P i
min ; P imax{ } i , t
Optimisation with integer variables
l Continuous variablesn Can follow the gradientsn Any value within the feasible set is OK
l Discrete variablesn There is no gradientn Can only take a finite number of valuesn Must try combinations of discrete values
-
8/8/2019 04 Unit Commitment
17/19
How many combinations are there?
111
110
101
100
011
010
001
000
l Examplesn 3 units: 8 possible statesn N units: 2 N possible states
How many solutions are there anyway?
1 2 3 4 5 6T=
l Optimisation over a timehorizon divided into intervals
l A solution is a path linking onecombination at each interval
l How many such path arethere?
-
8/8/2019 04 Unit Commitment
18/19
How many solutions are there anyway?
1 2 3 4 5 6T=
l Optimisation over a timehorizon divided into intervals
l A solution is a path linking onecombination at each interval
l How many such path arethere?
l Answer:
2 N ( )2 N ( )K 2 N ( )= 2 N ( )T
The Curse of Dimensionality
l Example: 5 units, 24 hours
l Processing 10 9 combinations/second, this would take 1.9
10 19 years to solvel There are over 100 units in England and Wales...l Many of these combinations do not satisfy the constraints
2 N ( )T
= 2 5( )24
= 6.2 10 35 combinations
-
8/8/2019 04 Unit Commitment
19/19
How do you Beat the Curse?
Brute force approach wont work!
l Need to be smartl Try only a small subset of all combinationsl Cant guarantee optimality of the solutionl Try to get as close as possible within a reasonable
amount of time
Main Solution Techniques
l Priority list / heuristic approachl Dynamic programmingl Lagrangian relaxationl Mixed Integer Programming
l Characteristics of a good techniquen Solution close to the optimumn Low computing timen Ability to model constraints