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 6.43, 6.48, 6.59, 6.61,

12.19, 12.22, 12.20, 12.24, 12.26, 12.28, 12.29; due Tuesday Nov. 25.

Homework 13 is 12.21, 12.25, 12.27, 7.1, 7.3, 7.4, 7.5, 7.6, 7.9, 7.12, 7.16; due Thursday, December 4.

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 are

L( , )

m m

i Gi D Gii i

Gi

C P P P

dCP

P

P

1

( )0 (for 1 to )

0

i Gi

Gi

m

D Gii

Pi m

dP

P P

6

Economic Dispatch Example

1 2

21 1 1 1

22 2 2 2

1 1

1

What is economic dispatch for a two generator

system 500 MW and

( ) 1000 20 0.01 $/h

( ) 400 15 0.03 $/h

Using the Lagrange multiplier method we know:

( )20 0.0

D G G

G G G

G G G

G

G

P P P

C P P P

C P P P

dC PdP

1

2 22

2

1 2

2 0

( )15 0.06 0

500 0

G

GG

G

G G

P

dC PP

dP

P P

7

Economic Dispatch Example, cont’d

1

2

1 2

1

2

1

2

We therefore need to solve three linear equations

20 0.02 0

15 0.06 0

500 0

0.02 0 1 20

0 0.06 1 15

1 1 0 500

312.5 MW

187.5 MW

26.2 $/MW

G

G

G G

G

G

G

G

P

P

P P

P

P

P

P

h

8

Lambda-Iteration Solution MethodThe direct solution using Lagrange multipliers only

works if no generators are at their limits.Another method is known as lambda-iteration

– the method requires that there to be a unique mapping from a value of lambda (marginal cost) to each generator’s MW output:

– for any choice of lambda (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.

( ).GiP

9

Lambda-Iteration AlgorithmL H

1 1

H L

M H L

H M

1

L 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

10

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.

11

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 1000MW

Rewriting generation as a function of , (

G G

G G

G G

G G G

Gi

IC P P

IC P P

IC P P

P P P

P

G1 G2

G3

),

we have

15 20P ( ) P ( )

0.02 0.0118

P ( )0.025

12

Lambda-Iteration Example, cont’dm

Gi

i=1

m

Gii=1

1

H

1

Pick so P ( ) 1000 0 and

P ( ) 1000 0

Try 20 then (20) 1000

15 20 181000 670 MW

0.02 0.01 0.025

Try 30 then (30) 1000 1230 MW

L L

H

mL

Gii

m

Gii

P

P

13

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

2.514

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 MW0.01

23.53 18(23.5)

0.025

L

Gi

G

G

G

P

P

P

P

221 MW

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

16

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. 17

Lambda-Iteration with Gen Limits

,max

,max

In the lambda-iteration method the limits are taken

into account when calculating ( ) :

if calculated production for

then set ( )

if calculated production for

Gi

Gi Gi

Gi Gi

P

P P

P P

,min

,min

then set ( )Gi Gi

Gi Gi

P P

P P

18

Lambda-Iteration Gen Limit Example

G1 G2

G3

1 2 31

In the previous three generator example assume

the same cost characteristics but also with limits

0 P 300 MW 100 P 500 MW

200 P 600 MW

With limits we get:

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

Gi G G Gi

P P P P

1

00

250 100 200 1000

450 MW (compared to 670MW)

(30) 1000 300 500 480 1000 280 MWm

Gii

P

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Lambda-Iteration Limit Example,cont’dAgain we continue iterating until the convergence

condition 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

P

P

P

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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 7 to 8, older combustion turbine 15.

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

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

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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:

G1 1

L( , ) ( ) ( ) m m

i Gi D L G Gii i

C P P P P P

P

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Impact of Transmission Losses

G1 1

G

The inclusion of losses then impacts the necessary

conditions for an optimal economic dispatch:

L( , ) ( ) ( ) .

The necessary conditions for a minimum are now:

L( , )

m m

i Gi D L G Gii i

C P P P P P

P

P

1

( ) ( )1 0

( ) 0

i Gi L G

Gi Gi Gi

m

D L G Gii

dC P P PP dP P

P P P P

23

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 Gi L G

Gi Gi

i Gi

GiL G

Gi

i

iL G

Gi

dC P P PdP P

dC 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 increases

( )1

( )the losses then 0 1.0

This makes generator

G G m m Gm

i GiL G

Gi

L Gi

Gi

L IC P L IC P L IC P

L PP PP

P PL

P

appear to be more expensive

(i.e., it is penalized). Likewise 1.0 makes a generator

appear less expensive. i

i

L

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Calculation of Penalty FactorsUnfortunately, the analytic calculation of is

somewhat involved. The problem is a small change

in the generation at impacts the flows and hence

the 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 how

the losses change:

( ) ( ) 1( )

1

Gi

L G L Gi

L GGi Gi

Gi

P

P P P PL

P PP PP

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Two Bus Penalty Factor Example

2 2

2 2

( ) ( ) 0.370.0387 0.037

10

0.9627 0.9643

L G L G

G G

P P P P MWP P MW

L L

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

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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.”

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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 MW

179 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 MWOFF 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, areas with direct

interconnections can transact according to rules:– In Eastern interconnection, in principle, up to

“nominal” thermal interconnection capacity,– In Western interconnection there are more

complicated rulesActual power flows through the entire network

according to the impedance of the transmission lines, and ultimately determine what are acceptable patterns of dispatch.

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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