EE 369POWER SYSTEM ANALYSIS

Lecture 16Economic Dispatch

Tom Overbye and Ross Baldick

1

AnnouncementsRead Chapter 12, concentrating on sections

12.4 and 12.5.Read Chapter 7.Homework 12 is 6.59, 6.61, 12.19, 12.22,

12.24, 12.26, 12.28, 12.29, 7.1, 7.3, 7.4, 7.5, 7.6, 7.9, 7.12, 7.16; due Thursday, 12/3.

2

Economic Dispatch: FormulationThe goal of economic dispatch is to determine the

generation dispatch that minimizes the instantaneous operating cost, subject to the constraint that total generation = total load + losses

T1

1

Minimize C ( )

Such that

m

i Gii

m

Gi D Lossesi

C P

P P P

Initially we'll ignore generatorlimits and thelosses

3

Unconstrained MinimizationThis is a minimization problem with a single

equality constraintFor an unconstrained minimization a necessary

(but not sufficient) condition for a minimum is the gradient of the function must be zero,

The gradient generalizes the first derivative for multi-variable problems:

1 2

( ) ( ) ( )( ) , , ,nx x x

f x f x f xf x

( ) f x 0

4

Minimization with Equality ConstraintWhen the minimization is constrained with an equality

constraint we can solve the problem using the method of Lagrange Multipliers

Key idea is to represent a constrained minimization problem as an unconstrained problem.

That is, for the general problem minimize ( ) s.t. ( )

We define the Lagrangian L( , ) ( ) ( )Then a necessary condition for a minimum is the

L ( , ) 0 and L ( , ) 0

T

x λ

f x g x 0

x λ f x λ g x

x λ x λ 5

Economic Dispatch Lagrangian

G1 1

G

For the economic dispatch we have a minimization constrained with a single equality constraint

L( , ) ( ) ( ) (no losses)

The necessary conditions for a minimum areL ( , )

m m

i Gi D Gii i

Gi

C P P P

dCP

P

P

1

( ) 0 (for 1 to )

0

iGi

Gim

D Gii

P i mdP

P P

6

Economic Dispatch Example

1 22

1 1 1 12

2 2 2 2

11

1

What is economic dispatch for a two generator system 500 MW and

( ) 1000 20 0.01 $/h

( ) 400 15 0.03 $/hUsing the Lagrange multiplier method we know:

( ) 20 0.0

D G G

G G G

G G G

GG

P P P

C P P P

C P P P

dC PdP

1

22 2

2

1 2

2 0

( ) 15 0.06 0

500 0

G

G GG

G G

P

dC P PdP

P P

7

Economic Dispatch Example, cont’d

1

2

1 2

1

2

1

2

We therefore need to solve three linear equations20 0.02 015 0.06 0500 00.02 0 1 20

0 0.06 1 151 1 0 500

312.5 MW187.5 MW

26.2 $/MW

G

G

G G

G

G

G

G

PP

P PPP

PP

h

8

Economic dispatch example, cont’d• At the solution, both generators have the same marginal

(or incremental) cost, and this common marginal cost is equal to λ.

• Intuition behind solution:– If marginal costs of generators were different, then by

decreasing production at higher marginal cost generator, and increasing production at lower marginal cost generator we could lower overall costs.

– Generalizes to any number of generators.• If demand changes, then change in total costs can be

estimated from λ.

9

Economic dispatch example, cont’d

• Another way to solve the equations is to:– Rearrange the first two equations to solve for PG1 and PG2

in terms of λ,– Plug into third equation and solve for λ,– Use the solved value of λ to evaluate PG1 and PG2.

• This works even when relationship between generation levels and λ is more complicated:– Equations are more complicated than linear when there

are maximum and minimum generation limits or we consider losses.

10

Lambda-Iteration Solution Method• Discussion on previous page leads to “lambda-

iteration” method:– this method requires a unique mapping from a value of

lambda (marginal cost) to each generator’s MW output: – for any choice of lambda (common marginal cost), the

generators collectively produce a total MW output,– the method then starts with values of lambda below

and above the optimal value (corresponding to too little and too much total output), and then iteratively brackets the optimal value.

•11

( ).GiP

Lambda-Iteration AlgorithmL H

1 1H L

M H L

H M

1L M

Pick and such that

( ) 0 ( ) 0

While Do

( ) / 2

If ( ) 0 Then

Else End While

m mL H

Gi D Gi Di i

mM

Gi Di

P P P P

P P

12

Lambda-Iteration: Graphical ViewIn the graph shown below for each value of lambda there is a unique PGi for each generator. This relationship is the PGi() function.

13

Lambda-Iteration Example

1 1 1

2 2 2

3 3 3

1 2 3

Consider a three generator system with( ) 15 0.02 $/MWh( ) 20 0.01 $/MWh( ) 18 0.025 $/MWh

and with constraint 1000MWRewriting generation as a function of , (

G G

G G

G G

G G G

Gi

IC P PIC P PIC P P

P P PP

1 2

3

), we have

15 20( ) ( )0.02 0.01

18( )0.025

G G

G

P P

P

14

Lambda-Iteration Example, cont’dm

i=1m

i=1

1

H

1

Pick so ( ) 1000 0 and

( ) 1000 0

Try 20 then (20) 1000

15 20 18 1000 670 MW0.02 0.01 0.025

Try 30 then (30) 1000 1230 MW

L LGi

HGi

mL

Gii

m

Gii

P

P

P

P

15

Lambda-Iteration Example, cont’d

1

1

Pick convergence tolerance 0.05 $/MWh

Then iterate since 0.05

( ) / 2 25

Then since (25) 1000 280 we set 25

Since 25 20 0.05

(25 20) / 2 22.5

(22.5) 1000 195 we set 2

H L

M H L

mH

Gii

M

mL

Gii

P

P

2.516

Lambda-Iteration Example, cont’dH

*

*

1

2

3

Continue iterating until 0.05

The solution value of , , is 23.53 $/MWh

Once is known we can calculate the 23.53 15(23.5) 426 MW

0.0223.53 20(23.5) 353 MW

0.0123.53 18(23.5)

0.025

L

Gi

G

G

G

P

P

P

P

221 MW

17

Thirty Bus ED ExampleCase is economically dispatched (without considering the incremental impact of the system losses).

18

Generator MW LimitsGenerators have limits on the minimum and

maximum amount of power they can produce

Typically the minimum limit is not zero. Because of varying system economics

usually many generators in a system are operated at their maximum MW limits:Baseload generators are at their maximum

limits except during the off-peak. 19

Lambda-Iteration with Gen Limits

,max

,max

In the lambda-iteration method the limits are takeninto account when calculating ( ) :

if calculated production for

then set ( )

if calculated production for

Gi

Gi Gi

Gi Gi

PP P

P P

,min

,min

then set ( )Gi Gi

Gi Gi

P P

P P

20

Lambda-Iteration Gen Limit Example

1 2

3

1 2 31

In the previous three generator example assumethe same cost characteristics but also with limits

0 300 MW 100 500 MW200 600 MW

With limits we get:

(20) 1000 (20) (20) (20) 10

G G

G

m

Gi G G Gi

P PP

P P P P

1

00

250 100 200 1000450 MW (compared to 670MW)

(30) 1000 300 500 480 1000 280 MWm

GiiP

21

Lambda-Iteration Limit Example,cont’dAgain we continue iterating until the convergencecondition is satisfied. With limits the final solution of , is 24.43 $/MWh (compared to 23.53 $/MWh without limits). Maximum limits will always caus

1

2

3

e to either increase or remain the same.Final solution is:

(24.43) 300 MW (at maximum limit)(24.43) 443 MW(24.43) 257 MW

G

G

G

PPP

22

Back of Envelope Values$/MWhr = fuelcost * heatrate + variable O&MTypical incremental costs can be roughly approximated:– Typical heatrate for a coal plant is 10, modern combustion

turbine is 10, combined cycle plant is 6 to 8, older combustion turbine 15.

– Fuel costs ($/MBtu) are quite variable, with current values around 2 for coal, 3 to 5 for natural gas, 0.5 for nuclear, probably 10 for fuel oil.

– Hydro costs tend to be quite low, but are fuel (water) constrained

– Wind and solar costs are zero.

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Inclusion of Transmission LossesThe losses on the transmission system are a

function of the generation dispatch. In general, using generators closer to the load

results in lower lossesThis impact on losses should be included when

doing the economic dispatchLosses can be included by slightly rewriting the

Lagrangian to include losses PL:

G1 1

L( , ) ( ) ( ) m m

i Gi D L G Gii iC P P P P P

P

24

Impact of Transmission Losses

G1 1

G

The inclusion of losses then impacts the necessaryconditions for an optimal economic dispatch:

L( , ) ( ) ( ) .

The necessary conditions for a minimum are now:

L (

m m

i Gi D L G Gii i

Gi

C P P P P P

P

P

P

1

, ) ( ) 1 ( ) 0

( ) 0

i LGi G

Gi Gim

D L G Gii

dC PP PdP P

P P P P

25

Impact of Transmission Losses

th

Solving for , we get: ( ) 1 ( ) 0

1 ( )1 ( )

Define the penalty factor for the generator(don't confuse with Lagrangian L!!!)

1

1 ( )

i LGi G

Gi Gi

iGi

GiLG

Gi

i

iL

GGi

dC PP PdP PdC PdPP P

P

L i

LP PP

The penalty factorat the slack bus isalways unity!

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Impact of Transmission Losses

1 1 1 2 2 2

The condition for optimal dispatch with losses is then( ) ( ) ( )

1 . So, if increasing increases1 ( )

the losses then ( ) 0 1.0

This makes generator

G G m m Gm

i GiL

GGi

LG i

Gi

L IC P L IC P L IC P

L PP PP

P P LP

appear to be more expensive(i.e., it is penalized). Likewise 1.0 makes a generatorappear less expensive.

i

iL

27

Calculation of Penalty FactorsUnfortunately, the analytic calculation of is somewhat involved. The problem is a small changein the generation at impacts the flows and hencethe losses throughout the entire system. However,

i

Gi

L

P

using a power flow you can approximate this function by making a small change to and then seeing howthe losses change:

1( )1

Gi

L LG i

LGi Gi

Gi

P

P PP L PP PP

28

Two Bus Penalty Factor Example

2 2

2 2

0.37( ) 0.0387 0.03710

0.9627 0.9643

L LG

G G

P P MWPP P MWL L

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Thirty Bus ED ExampleNow consider losses.Because of the penalty factors the generator incremental costs are no longer identical.

30

Area Supply Curve

0 100 200 300 400Total Area Generation (MW)

0.00

2.50

5.00

7.50

10.00

The area supply curve shows the cost to produce thenext MW of electricity, assuming area is economicallydispatched

Supplycurve forthirty bussystem

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Economic Dispatch - SummaryEconomic dispatch determines the best way to

minimize the current generator operating costs.The lambda-iteration method is a good approach for

solving the economic dispatch problem:– generator limits are easily handled,– penalty factors are used to consider the impact of losses.

Economic dispatch is not concerned with determining which units to turn on/off (this is the unit commitment problem).

Basic form of economic dispatch ignores the transmission system limitations.

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Security Constrained EDor Optimal Power Flow

Transmission constraints often limit ability to use lower cost power.

Such limits require deviations from what would otherwise be minimum cost dispatch in order to maintain system “security.”

Need to solve or approximate power flow in order to consider transmission constraints.

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Security Constrained EDor Optimal Power Flow

The goal of a security constrained ED or optimal power flow (OPF) is to determine the “best” way to instantaneously operate a power system, considering transmission limits.

Usually “best” = minimizing operating cost, while keeping flows on transmission below limits.

In three bus case the generation at bus 3 must be limited to avoid overloading the line from bus 3 to bus 2.

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Security Constrained Dispatch

Bus 2 Bus 1

Bus 3Home Area

Scheduled Transactions

357 MW179 MVR

194 MW

448 MW 19 MVR

232 MVR

179 MW 89 MVR

1.00 PU

-22 MW 4 MVR

22 MW -4 MVR

-142 MW 49 MVR

145 MW-37 MVR

124 MW-33 MVR

-122 MW 41 MVR

1.00 PU

1.00 PU

0 MW 37 MVR100%

100%100 MW

OFF AGCAVR ON

AGC ONAVR ON

100.0 MW

Need to dispatch to keep line from bus 3 to bus 2 from overloading

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Multi-Area OperationIn multi-area system, “rules” have been established

regarding transactions on tie-lines:– In Eastern interconnection, in principle, up to “nominal”

thermal interconnection capacity,– In Western interconnection there are more complicated

rulesThe actual power that flows through the entire network

depends on the impedance of the transmission lines, and ultimately determine what are acceptable patterns of dispatch:Can result in need to “curtail” transactions that otherwise

satisfy rules.Economically uncompensated flow through other areas

is known as “parallel path” or “loop flows.” Since ERCOT is one area, all of the flows on AC lines are

inside ERCOT and there is no uncompensated flow on AC lines.

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Seven Bus Case: One-line

Top Area Cost

Left Area Cost Right Area Cost

1

2

3 4

5

6 7

106 MW

168 MW

200 MW 201 MW

110 MW 40 MVR

80 MW 30 MVR

130 MW 40 MVR

40 MW 20 MVR

1.00 PU

1.01 PU

1.04 PU1.04 PU

1.04 PU

0.99 PU1.05 PU

62 MW

-61 MW

44 MW -42 MW -31 MW 31 MW

38 MW

-37 MW

79 MW -77 MW

-32 MW

32 MW-14 MW

-39 MW

40 MW-20 MW 20 MW

40 MW

-40 MW

94 MW

200 MW 0 MVR

200 MW 0 MVR

20 MW -20 MW

AGC ON

AGC ON

AGC ON

AGC ON

AGC ON

8029 $/MWH

4715 $/MWH 4189 $/MWH

Case Hourly Cost 16933 $/MWH

System hasthree areas

Left areahas onebus

Right area has onebus

Top areahas fivebuses

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No net interchangebetweenAny areas.

Seven Bus Case: Area View

System has40 MW of“Loop Flow”

Actualflowbetweenareas

Loop flow can result in higher losses

Area Losses

Area Losses Area Losses

Top

Left Right

-40.1 MW 0.0 MW

0.0 MW

0.0 MW

40.1 MW

40.1 MW

7.09 MW

0.33 MW 0.65 MW

Scheduledflow

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Seven Bus - Loop Flow?

Area Losses

Area Losses Area Losses

Top

Left Right

-4.8 MW 0.0 MW

100.0 MW

0.0 MW

104.8 MW

4.8 MW

9.44 MW

-0.00 MW 4.34 MW

100 MW Transactionbetween Left and Right

Transaction has actually decreasedthe loop flow

Note thatTop’s Losses haveincreasedfrom 7.09MW to9.44 MW

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