0 generator cost curves generator costs are typically represented by up to four different curves –...

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Generator Cost Curves

Generator costs are typically represented by up to four different curves

– input/output (I/O) curve– fuel-cost curve– heat-rate curve– incremental cost curve

For reference- 1 Btu (British thermal unit) = 1054 J- 1 MBtu = 1x106 Btu- 1 MBtu = 0.293 MWh- 3.41 Mbtu = 1 MWh

2

I/O Curve

The IO curve plots fuel input (in MBtu/hr) versus net MW output.

3

Fuel-cost Curve

The fuel-cost curve is the I/O curve scaled by fuel cost. Coal prices vary; around $1/Mbtu to $2/Mbtu

4

Heat-rate Curve

Plots the average number of MBtu/hr of fuel input needed per MW of output.

Heat-rate curve is the I/O curve scaled by MW

Best for most efficient coal units is around 9.0

5

Incremental (Marginal) cost Curve

Plots the incremental $/MWh as a function of MW. Found by differentiating the cost curve

6

Mathematical Formulation of Costs

Generator cost curves are usually not smooth. However the curves can usually be adequately approximated using piece-wise smooth, functions.

Two representations predominate– quadratic or cubic functions– piecewise linear functions

In 476 we'll assume a quadratic presentation

2( ) $/hr (fuel-cost)

( )( ) 2 $/MWh

i Gi i Gi Gi

i Gii Gi Gi

Gi

C P P P

dC PIC P P

dP

7

Coal

• Four Types of Coal• Anthracite (15,000 Btu/lb), Eastern Pennsylvania; used mostly for heating because of its high value and cost• Bituminous (10,500 to 15,000 Btu/lb), most plentiful in US, used extensively in electric power industry; mined in Eastern US including

Southern Illinois.• Subbitunminous (8300 to 11,500 Btu/lb), most plentiful in Western US (Power River Basin in Wyoming); used in electric power

industry• Lignite or brown coal (4000 to 8300 Btu/lb), used in electric power industry

• Coals differ in impurities such as sulfur content

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Coal Prices

Source: US EIA

At $50 per ton and 11,800 Btu/lb, Illinois coal costs $2.12/Mbtu.Transportation by rail is around $0.03/ton/mile

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Coal Usage Example

A 500 MW (net) generator is 35% efficient. It is being supplied with Western grade coal, which costs $1.70 per MBtu and has 9000 Btu per pound. What is the coal usage in lbs/hr? What is the cost?

At 35% efficiency required fuel input per hour is

500 MWh 1428 MWh 1 MBtu 4924 MBtuhr 0.35 hr 0.29 MWh hr

4924 MBtu 1 lb 547,111 lbshr 0.009MBtu hr

4924 MBtu $1.70Cost = 8370.8 $/hr or $16.74/MWh

hr MBtu

10

Wasting Coal Example

Assume a 100W lamp is left on by mistake for 8 hours, and that the electricity is supplied by the previous coal plant and that transmission/distribution losses are 20%. How much irreplaceable coal has he/she wasted?

With 20% losses, a 100W load on for 8 hrs requires

1 kWh of energy. With 35% gen. efficiency this requires

1 kWh 1 MWh 1 MBtu 1 lb1.09 lb

0.35 1000 kWh 0.29 MWh 0.009MBtu

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Incremental Cost Example

21 1 1 1

22 2 2 2

1 11 1 1

1

2 22 2 2

2

For a two generator system assume

( ) 1000 20 0.01 $ /

( ) 400 15 0.03 $ /

Then

( )( ) 20 0.02 $/MWh

( )( ) 15 0.06 $/MWh

G G G

G G G

GG G

G

GG G

G

C P P P hr

C P P P hr

dC PIC P P

dP

dC PIC P P

dP

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Incremental Cost Example, cont'd

G1 G2

21

22

1

2

If P 250 MW and P 150 MW Then

(250) 1000 20 250 0.01 250 $ 6625/hr

(150) 400 15 150 0.03 150 $6025/hr

Then

(250) 20 0.02 250 $ 25/MWh

(150) 15 0.06 150 $ 24/MWh

C

C

IC

IC

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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

m

i=1

Minimize C ( )

Such that

m

i Gii

Gi D Losses

C P

P P P

Initially we'll ignore generatorlimits and thelosses

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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

( ) ( ) ( )( ) , , ,

nx x x

f x f x f xf x

( ) f x 0

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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 modify a constrained minimization problem to be 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

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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 Gii i

Gi

C P P P

dCP

P

P

1

( )0 (for i 1 to m)

0

i Gi

Gi

m

D Gii

PdP

P P

17

Economic Dispatch Example

D 1 2

21 1 1 1

22 2 2 2

1 1

1

What is economic dispatch for a two generator

system P 500 MW and

( ) 1000 20 0.01 $/

( ) 400 15 0.03 $/

Using the Largrange multiplier method we know

( )20 0

G G

G G G

G G G

G

G

P P

C P P P hr

C P P P hr

dC PdP

1

2 22

2

1 2

.02 0

( )15 0.06 0

500 0

G

GG

G

G G

P

dC PP

dP

P P

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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 1 500

312.5 MW

187.5 MW

26.2 $/MWh

G

G

G G

G

G

G

G

P

P

P P

P

P

P

P

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Lambda-Iteration Solution Method

The direct solution only works well if the incremental cost curves are linear and no generators are at their limits

A more general method is known as the lambda-iteration– the method requires that there be a unique mapping

between a value of lambda and each generator’s MW output

– the method then starts with values of lambda below and above the optimal value, and then iteratively brackets the optimal value

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Lambda-Iteration Algorithm

L H

m mL H

Gi Gii=1 i=1

H L

M H L

mM H M

Gii=1

L M

Pick and such that

P ( ) 0 P ( ) 0

While Do

( ) / 2

If P ( ) 0 Then

Else

End While

D D

D

P P

P

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Lambda-Iteration: Graphical View

In 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

Gi

Consider a three generator system with

( ) 15 0.02 $/MWh

( ) 20 0.01 $/MWh

( ) 18 0.025 $/MWh

and with constraint 1000MW

Rewriting as a function of , P ( ), we have

G G

G G

G G

G G G

IC P P

IC P P

IC P P

P P P

G1 G2

G3

15 20P ( ) P ( )

0.02 0.0118

P ( )0.025

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Lambda-Iteration Example, cont’d

m

Gii=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

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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

mH

Gii

M

mL

Gii

P

P

2.5

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Lambda-Iteration Example, cont’d

H

*

*Gi

G1

G2

G3

Continue iterating until 0.05

The solution value of , , is 23.53 $/MWh

Once is known we can calculate the P

23.53 15P (23.5) 426 MW

0.0223.53 20

P (23.5) 353 MW0.01

23.53 18P (23.5)

0.025

L

221 MW

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Lambda-Iteration Solution Method

The direct solution only works well if the incremental cost curves are linear and no generators are at their limits

A more general method is known as the lambda-iteration– the method requires that there be a unique mapping

between a value of lambda and each generator’s MW output

– the method then starts with values of lambda below and above the optimal value, and then iteratively brackets the optimal value

27

Generator MW Limits

Generators have limits on the minimum and maximum amount of power they can produce

Often times the minimum limit is not zero. This represents a limit on the generator’s operation with the desired fuel type

Because of varying system economics usually many generators in a system are operated at their maximum MW limits.

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Lambda-Iteration with Gen Limits

Gi

Gi ,max Gi ,max

Gi ,min Gi ,min

In the lambda-iteration method the limits are taken

into account when calculating P ( ) :

if P ( ) then P ( )

if P ( ) then P ( )

Gi Gi

Gi Gi

P P

P P

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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) 100m

Gi G G Gi

P P P P

1

0

250 100 200 450 MW (compared to -670MW)

(30) 1000 300 500 480 1000 280 MWm

Gii

P

30

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). The presence of limits will alwa

G1

G2

G3

ys

cause to either increase or remain the same.

Final solution is

P (24.43) 300 MW

P (24.43) 443 MW

P (24.43) 257 MW

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Back of Envelope Values

Often times incremental costs can be approximated by a constant value:

– $/MWhr = fuelcost * heatrate + variable O&M– 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

1.5 for coal, 4 for natural gas, 0.5 for nuclear, probably 10 for fuel oil.

– Hydro, solar and wind costs tend to be quite low, but for this sources the fuel is free but limited

32

Economic Load Dispatch including transmission Losses

The incremental production cost of a given plant over a limited range is given by:

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The optimal load dispatch problem including transmission losses is defined as:

Where PL is the total system loss which is assumed to be a function of generation.

Introducing the Lagrangian multiplier , the auxiliary function is given by:

The partial derivative of this expression when equated to zero, will give the condition for optimum load dispatch. i.e.

34

35

36

37

It is to note here that powers used on the R.H.S. of this eqn(8) during first iteration correspond to the values as calculated in step 2.

In case any of the generations violates the limit the generation of that generator is fixed at the limit violated.

Step 4. Check if the difference in power at all generators between two successive iterations is less than a tol. Value. If not, go back to step 3.

Step 5. Calculate the losses using the relation;

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39

Prob. A sample power system has two generators with the incremental cost curves as;

The loss coefficients in 1/MW units are ; B11 = 0.0015, B12 = -0.0005, B22= 0.0025.

Considering lemda = 2.6, if the losses are coordinated, find the economic load schedule, losses and generation cost.

40

41

42

Generally load is given and the problem is to find economic load dispatch, losses and cost of the generation.

For this we have to find out first the value of Lemda

Let the load is 160 MW and the losses are coordinated rather than included.

43

44

45

46

47

48

Prob. The fuel input in Btu per hour of two generators operating in a plant are given by:

The maximum and minimum loading limits on the generators are 100 MW and 10 MW respectively. Determine the minimum cost of the generation when the following loads are fed by the plant. The cost of fuel is SR 2.00 per million Btu.

49

50

51

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:

G1 1

L( , ) ( ) ( ( ) ) m m

i Gi D L G Gii i

C P P P P P

P

52

Impact of Transmission Losses

G1 1

G

This small change 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

i Gi

Gi

C P P P P P

dC PP d

P

P

1

( )(1 ) 0

( ) 0

L G

Gi Gi

m

D L G Gii

P PP P

P P P P

53

Impact of Transmission Losses

thi

i

Solving each equation for we get

( ) ( )(1 0

( )1

( )1

Define the penalty factor L for the i generator

1L

( )1

i Gi L G

Gi Gi

i Gi

GiL G

Gi

L G

Gi

dC P P PdP P

dC PdPP P

P

P PP

The penalty factorat the slack bus isalways unity!

54

Impact of Transmission Losses

1 1 1 2 2 2

i Gi

The condition for optimal dispatch with losses is then

( ) ( ) ( )

1Since L if increasing P increases

( )1

( )the losses then 0 1.0

This makes generator

G G m m Gm

L G

Gi

L Gi

Gi

L IC P L IC P L IC P

P PP

P PL

P

i

i appear to be more expensive

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

appear less expensive.

55

Calculation of Penalty Factors

i

Gi

Unfortunately, the analytic calculation of L is

somewhat involved. The problem is a small change

in the generation at P impacts the flows and hence

the losses throughout the entire system. However,

Gi

using a power flow you can approximate this function

by making a small change to P and then seeing how

the losses change:

( ) ( ) 1( )

1

L G L Gi

L GGi Gi

Gi

P P P PL

P PP PP

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Two Bus Penalty Factor Example

2

2 2

( ) ( ) 0.370.0387 0.037

10

0.9627 0.9643

L G L G

G Gi

P P P P MWP P MW

L L

57

Thirty Bus ED Example

Because of the penalty factors the generator incremental costs are no longer identical.

58

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 - 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)

Economic dispatch ignores the transmission system limitations