optimal dispatch

8/19/2019 Optimal Dispatch

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Economic Dispatch

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Announcements

Read 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.

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Economic Dispatch: Formulation

The 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 Gi

i

m

Gi D Losses

i

C P

P P P

Initially we'll

ignore generator

limits and the

losses3


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Unconstrained Minimization

This is a minimization problem with a single equality constraint

For 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

( ) ( ) ( )( ) , , ,

n x x x

f x f x f xf x

( ) f x 0

4


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Minimization with Equality Constraint

When 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 theL ( , ) 0 and L ( , ) 0

T

x λ

f x g x 0

x λ f x λ g x

x λ x λ 5


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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 Gi

i i

Gi

C P P P

dC

P

P

P

1

( ) 0 (for 1 to )

0

iGi

Gi

m

D Gi

i

P i m

dP

P P

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Economic Dispatch Example

1 2

21 1 1 1

22 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 $/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 P

dP

1

22 2

2

1 2

2 0

( ) 15 0.06 0

500 0

G

G G

G

G G

P

dC P P

dP

P P

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conom c spatc xamp e, 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 500312.5 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


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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


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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


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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

( ).Gi

P


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Lambda‐Iteration Algorithm

L 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 m L H

Gi D Gi D

i i

m M

Gi D

i

P P P P

P P

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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.

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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 P

IC P P

IC P P

P P P

P

1 2

3

),

we have

15 20( ) ( )

0.02 0.01

18( )0.025

G G

G

P P

P

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Lambda‐Iteration Example, cont’dm

i=1

m

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 L

Gi

H

Gi

m L

Gi

i

m

Gii

P

P

P

P

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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

m

H Gi

i

M

m L

Gi

i

P

P

2.516


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Lambda‐Iteration Example, cont’d

H

*

*

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 MW0.02

23.53 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).

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Generator MW Limits

Generators 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.

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Lambda‐Iteration with Gen Limits

,max

,max

In the lambda-iteration method the limits are taken

into account when calculating ( ) :

if calculated production forthen 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

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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 MW

200 600 MW

With limits we get:

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

G G

G

m

Gi G G G

i

P P

P

P P P P

1

00

250 100 200 1000450 MW (compared to 670MW)

(30) 1000 300 500 480 1000 280 MWm

Gi

i

P

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Lambda‐Iteration Limit Example,cont’d

Again 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&M

Typical 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. 23


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Inclusion of Transmission Losses

The losses on the transmission system are a

function of

the

generation

dispatch.

In general, using generators closer to the

load results in lower losses

This impact on losses should be included

when doing the economic dispatch

Losses can be included by slightly rewriting

the Lagrangian

to include losses PL

:

G1 1

L( , ) ( ) ( )m m

i Gi D L G Gi

i i

C P P P P P

P24


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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 Gi

i i

Gi

C P P P P P

P

P

P

1

, ) ( ) 1 ( ) 0

( ) 0

i LGi G

Gi Gi

m

D L G Gi

i

dC PP P

dP P

P P P P

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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

Gi LG

Gi

i

i

LG

Gi

dC PP P

dP P

dC P

dPPP

P

L i

LP

PP

The penalty factor

at the slack bus is

always unity!26


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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 Gi

LG

Gi

LG i

Gi

L IC P L IC P L IC P

L PP

P

P

PP L

P

appear to be more expensive

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

appear less expensive.

i

i

L

27


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Calculation of Penalty Factors

Unfortunately, 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 LG i

LGi Gi

Gi

P

P PP L

PP P

P

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Two Bus Penalty Factor Example

2 2

2 2

0.37( ) 0.0387 0.037

10

0.9627 0.9643

L LG

G G

P P MW P

P P MW

L L

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Thirty Bus ED Example

Now consider losses.Because of the penalty factors the generator incremental

costs are no longer identical.

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Area Supply Curve

0 100 200 300 400

Total Area Generation (MW)

0.00

2.50

5.00

7.50

10.00

The area supply curve shows the cost to produce the

next MW of electricity, assuming area is economically

dispatched

Supplycurve for

thirty bus

system

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Economic Dispatch ‐

Summary

Economic 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 ED

or Optimal Power FlowTransmission 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 ED

or Optimal Power FlowThe 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 34


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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 MVR 100%

100%

100 MW OFF AGC AVR ON

AGC ON

AVR ON

100.0 MW

Need to dispatch to keep line

from bus 3 to bus 2 from overloading35


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Multi‐Area Operation

In 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 rules

The 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.36


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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 has

three areas

Left area

has one

bus Right area has onebus

Top area

has fivebuses

37

No net

interchange

between

Any areas.


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Seven Bus Case: Area View

System has

40 MW of

“Loop Flow”

Actualflow

between

areas

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

Scheduled

flow

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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 Transaction

between Left and Right

Transaction has

actually decreasedthe loop flow

Note thatTop’s

Losses have

increasedfrom

7.09MW to

9.44 MW

39

optimal dispatch

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