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EE 220UNIT COMMITMENT

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Load Demand Cycles

Human activity follows cycles

Systems supplying services will also experienceusage cycles transportation, communication, andelectric power systemsElectric power consumption follows a daily, weekly,and seasonal cycles

High power usage during the day and eveninghours industrial & commercial operations andlighting loadsLower usage on the weekends

Higher usage during the summer

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Example: Demand ProfileJune 10, 2010

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Load cycles create economic problems forpower generation it is quite expensive tocontinuously run all generation, which is needed

to meet the peak power demands

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Unit Commitment

Definition: Commitment means to turn on a givengeneration unit

Have the prime mover operating the unit atsynchronous speedSynchronize and connect the unit to the network

Economics

Savings are gained by de-committing some of the

generation units when they are not needed tomeet the current load demand

The engineering problem is committing enoughunits to meet current and future load demandswhile minimizing starting and operating costs

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Example

Consider the cost of operating three generationunits

What combination of units is best to supply a 550MW load?

Unit 1 F 1(P1) = 561 + 7.92 P 1 + 0.001562 P 12 R/hr 150 P1 600 MW

Unit 2 F 2(P2) = 310 + 7.85 P 2 + 0.00194 P 22 R/hr 100 P2 400 MW

Unit 3 F 3(P3) = 93.6 + 9.56 P 3 + 0.005784 P 32 R/hr 50 P3 200 MW

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Using unit combination

Unit

1

Unit

2

Unit

3

Max.

Gen.

Min.

Gen.P

1P

2P

3F

1F

2F

3F

T

Off Off Off 0 0 Infeasible

Off Off On 200 50 Infeasible

Off On Off 400 100 Infeasible

Off On On 600 150 0 400 150 0 3760 1658 5418

On Off Off 600 150 550 0 0 5389 0 0 5389

On Off On 800 200 500 0 50 4911 0 586 5497

On On Off 1000 250 295 255 0 3030 2440 0 5471

On On On 1200 300 267 233 50 2787 2244 586 5617

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The least expensive way to supply the generation is not with allthree units running or with any combination involving two units

The optimal commitment is to only run unit # 1, the mosteconomic unit

By only running the most economic unit, the load can be suppliedby that unit operating closer to its best efficiency.

If another unit is committed, both unit # 1 and the other unit will

be loaded further from their best efficiency points, resulting in ahigher net cost

But from reliability considerations, more units should be committed

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UC with Demand

Pattern/CycleConsider the load demand with a simple peak-valley pattern

450 MW min. load

Total Loading

3 PM 9 PM 3 PM

1150 MW peak load

3 AM 9 AM

Time of Day

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In order to optimize the operation of the system

Units must be shut down as load goes down

Then the units must be recommitted as load goes back upSimple approach to the solution is a simple priority list schemewith the following rule

Rules

When the load is above 1000 MW, run all three unitsLoading between 600 MW and 1000 MW, run units#1 and #2

Loading below 600 MW, only run unit #1

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Generator Commitment

600 MW

Total Loading

3 PM 9 PM 3 PM

200 MW

3 AM 9 AM

Time of Day

400 MW

1000 MW

1200 MW

Unit #1

Unit #2

Unit #3

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Unit Commitment

ConstraintsPrimary constraints

Enough units are committed to supply the load and the losseseconomically.

Additional Constraints in Unit Commitment

Spinning reserveThermal unit constraints

Minimum up timeMinimum down time

Crew constraintsHydro-constraintsMust runFuel constraints

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Unit Commitment Solution

Priority-List Methods

Consists of a simple shut-down rule

Obtain by an exhaustive enumeration of all unitcombinations at each load level

Or obtained by noting the full-load averageproduction cost of each unit = net heat rate at full-load fuel cost.

Various enhancements can be made to the priority-

list scheme by the grouping of units to ensure thatvarious constraints are met

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Typical shut-down rules

At each hour when load is dropping, determine whether dropping

the next unit on the list leaves sufficient generation to supply theload plus the spinning-reserve requirements

If the supply is not sufficient, keep the unit committed

Determine the number of hours before the unit is needed again

If the time is less than the minimum shut-down time for the unit, keep it

committedPerform a cost comparison

The sum of the hourly production costs for the next number of hourswith the next unit to be dropped being committed

And the sum of the restart costs for the next unit based on the

minimum cost of cooling the unit or banking the unit

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Example

Construct a priority list for the units in the firstexample using the same cost equations

Getting the Full-Load Average Cost

Unit 1 F1(P1) = 561 + 7.92 P 1 + 0.001562 P 1 R/hr 150 P1 600 MWUnit 2 F2(P2) = 310 + 7.85 P 2 + 0.00194 P 2 R/hr 100 P2 400 MWUnit 3 F3(P3) = 93.6 + 9.56 P 3 + 0.005784 P 3 R/hr 50 P3 200 MW

Unit 1 F1(600)/600 = 9.7922Unit 2 F2(400)/400 = 9.4010Unit 3 F3(200)/200 = 11.1848

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A strict priority order for these units:

[ 2, 1, 3]

The commitment schemeIgnoring minimum up/down times and start-up costs

Combination Min MW Max MW

2 + 1 + 3 300 1200

2 + 1 250 10002 100 400

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DYNAMIC PROGRAMMING

SOLUTIONChief advantage over enumeration schemes isthe reduction in the dimensionality of the problem

In a strict priority order scheme, there are only Ncombinations to try for an N-unit system`

A strict priority list would result in a theoreticallycorrect dispatch and commitment only if

The no-load costs are zero

Unit input-output characteristics are linear There are no limits, constraints, or restrictions

Start-up costs are a fixed amount

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The following assumptions are made in this implementation ofthe DP approach

A state consists of an array of unitsWith specified units operating and the rest de-committed (off-line)

A feasible state is one in which the committed units can supply therequired load and meets the minimum capacity for each period

Start-up costs are independent of the off-time or down-timeI.e., it is a fixed amount with respect to time

No unit shutting-down costs

A strict priority order will be used within each interval

A specified minimum amount of capacity must be operatingwithin each interval

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Recursive algorithm to compute the minimum cost in hour Kwith combination I:

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State (K,I) = I th combination in hour K

A strategy is the transition or path from one state at a given hourto a state at the next hour

X = number of states to search in each period

N = number of strategies or paths to save at each step

These variables allow control of the computational effort

For complete enumeration, the maximum value of X or N is 2 N - 1

For a simple priority-list ordering, the upper bound on X is n, the number ofunits

Reducing N means that information is discarded about the highestcost schedules at each interval and saving only the lowest Npaths or strategies

There is no assurance that the theoretical optimum will be found

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Example

Fuel Cost = 2.0 R/MBtu for each unit

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State Unit 1 Unit 2 Unit 3 Unit 4 Max. Net Capacityfor Combination

15 1 1 1 1 69014 1 1 1 0 63013 0 1 1 1 61012 0 1 1 0 55011 1 0 1 1 44010 1 1 0 1 390

9 1 0 1 0 3808 0 0 1 1 3607 1 1 0 0 3306 0 1 0 1 3105 0 0 1 0 3004 0 1 0 0 2503 1 0 0 1 140

2 1 0 0 0 801 0 0 0 1 600 0 0 0 0 0

Allowable states: {5,12,14,15}

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Strict Priority

Unit Pmax(MW)

Pmin(MW)

No-LoadCost (R/h)

IncrementalCost

(R/MWh)

Cost at FullLoad

Priority

1 80 25 213 20.88 23.5425 32 250 60 585.62 18 20.34248 23 300 75 684.74 17.46 19.74246667 14 60 20 252 23.8 28 4

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