1. ECE 476
POWER SYSTEM ANALYSIS
Lecture 16
Economic Dispatch
Professor Tom Overbye
Department of Electrical and
Computer Engineering

2. Announcements



Be reading Chapter 12.4 and 12.5 for lectures 15 and 16
HW 6 is 6.50, 6.52, 6.59, 12.20; due October 20 in class (for
Problem 6.52 the case new is Example6_52)
Office hours are changed for today only to 2 to 3 pm.
2

3. 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
3

4. I/O Curve
 The
IO curve plots fuel input (in MBtu/hr) versus
net MW output.
4

5. 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
5

6. 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
6

7. Incremental (Marginal) cost Curve
 Plots
the incremental $/MWh as a function of MW.
 Found by differentiating the cost curve
7

8. 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
–
–
 In
quadratic or cubic functions
piecewise linear functions
476 we'll assume a quadratic presentation
2
Ci ( PGi ) = α i + β PGi + γ PGi
$/hr (fuel-cost)
dCi ( PGi )
ICi ( PGi ) =
= β + 2γ PGi $/MWh
dPGi
8

9. 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
9

10. 10

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

12. 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 MBtu
=
×
=
hr × 0.35
hr
0.29 MWh
hr
4924 MBtu
1 lb
547,111 lbs
×
=
hr
0.009MBtu
hr
4924 MBtu $1.70
Cost =
×
= 8370.8 $/hr or $16.74/MWh
hr
MBtu
12

13. 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 lb
×
×
×
= 1.09 lb
0.35 1000 kWh 0.29 MWh 0.009MBtu
13

14. Incremental Cost Example
For a two generator system assume
2
C1 ( PG1 ) = 1000 +20 PG1 + 0.01PG1
$ / hr
2
C2 ( PG 2 ) = 400 +15 PG 2 + 0.03PG 2
$ / hr
Then
dC1 ( PG1 )
IC1 ( PG1 ) =
= 20 + 0.02 PG1 $/MWh
dPG1
dC2 ( PG 2 )
IC2 ( PG 2 ) =
= 15 + 0.06 PG 2 $/MWh
dPG 2
14

15. Incremental Cost Example, cont'd
If PG1 = 250 MW and PG2 = 150 MW Then
C1 (250) = 1000 +20 × 250 + 0.01 × 2502 = $ 6625/hr
C2 (150) = 400 +15 × 150 + 0.03 × 1502
= $6025/hr
Then
IC1 (250) = 20 + 0.02 × 250 = $ 25/MWh
IC2 (150) = 15 + 0.06 × 150 = $ 24/MWh
15

16. 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
Minimize
m
CT @ Ci ( PGi )
∑
i =1
Such that
m
∑ PGi = PD + PLosses
i=1
Initially we'll
ignore generator
limits and the
losses
16

17. 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, ∇f (x) = 0
 The gradient generalizes the first derivative for
multi-variable problems:
 ∂f (x) ∂f (x)
∂f (x) 
∇f (x) @
,
,K,
∂x1 ∂x2
∂xn 


17

18. 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 f (x) s.t. g(x) = 0
We define the Lagrangian L(xλ ) =f ( ) +λ Tg ( )
,
x
x
Then a necessary condition for a minimum is the
∇L xλ xλ ) = 0 and ∇L ( , ) = 0
( ,
xλ
18

19. Economic Dispatch Lagrangian
For the economic dispatch we have a minimization
constrained with a single equality constraint
L(PG , λ ) =
m
m
i =1
i =1
∑ Ci ( PGi ) + λ ( PD − ∑ PGi )
(no losses)
The necessary conditions for a minimum are
∂L(PG , λ )
dCi ( PGi )
=
− λ = 0 (for i = 1 to m)
∂PGi
dPGi
m
PD − ∑ PGi = 0
i =1
19

20. Economic Dispatch Example
What is economic dispatch for a two generator
system PD = PG1 + PG 2 = 500 MW and
2
C1 ( PG1 ) = 1000 +20 PG1 + 0.01PG1
$ / hr
2
C2 ( PG 2 ) = 400 +15 PG 2 + 0.03PG 2
$ / hr
Using the Largrange multiplier method we know
dC1 ( PG1 )
−λ = 20 + 0.02 PG1 − λ
=0
dPG1
dC2 ( PG 2 )
−λ
dPG 2
= 15 + 0.06 PG 2 − λ
=0
500 − PG1 − PG 2 = 0
20

21. Economic Dispatch Example, cont’d
We therefore need to solve three linear equations
20 + 0.02 PG1 − λ
=0
15 + 0.06 PG 2 − λ
=0
500 − PG1 − PG 2 = 0
0
−1  PG1   −20 
0.02
 0
0.06 −1  PG 2  =  −15 


 

−1
 −1
  λ   −500


 

 PG1 
 312.5 MW 
 P  =  187.5 MW 
 G2 


 λ 
 26.2 $/MWh 




21

22. 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 lambdaiteration
–
–
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
22

23. Lambda-Iteration Algorithm
Pick λ L and λ H such that
m
PGi (λ L ) − PD < 0
∑
i=1
While
m
PGi (λ H ) − PD > 0
∑
i=1
λ H − λ L > ε Do
λ M = (λ H + λ L ) / 2
If
m
PGi (λ M ) − PD > 0 Then λ H = λ M
∑
i=1
Else λ L = λ M
End While
23

24. 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(λ) function.
24

25. Lambda-Iteration Example
Consider a three generator system with
IC1 ( PG1 ) = 15 + 0.02 PG1
= λ $/MWh
IC2 ( PG 2 ) = 20 + 0.01PG 2
=λ
$/MWh
IC3 ( PG 3 ) = 18 + 0.025 PG 3
=λ
$/MWh
and with constraint PG1 + PG 2 + PG 3 = 1000 MW
Rewriting as a function of λ , PGi (λ ), we have
λ − 15
PG1 (λ ) =
0.02
λ − 18
PG3 (λ ) =
0.025
λ − 20
PG2 (λ ) =
0.01
25

26. Lambda-Iteration Example, cont’d
Pick λ L so
m
PGi (λ L ) − 1000 < 0 and
∑
i=1
m
PGi (λ H ) − 1000 > 0
∑
i=1
Try λ
L
= 20 then
m
∑ PGi (20) − 1000
=
i =1
λ − 15 λ − 20 λ − 18
+
+
− 1000 = −670 MW
0.02
0.01 0.025
Try λ H = 30 then
m
∑ PGi (30) − 1000
= 1230 MW
i =1
26

27. Lambda-Iteration Example, cont’d
Pick convergence tolerance ε = 0.05 $/MWh
Then iterate since λ H − λ L > 0.05
λ M = (λ H + λ L ) / 2 = 25
Then since
m
PGi (25) − 1000 = 280 we set λ H = 25
∑
i =1
Since 25 − 20 > 0.05
λ M = (25 + 20) / 2 = 22.5
m
PGi (22.5) − 1000 = −195 we set λ L = 22.5
∑
i =1
27

28. Lambda-Iteration Example, cont’d
Continue iterating until λ H − λ L < 0.05
*
The solution value of λ , λ , is 23.53 $/MWh
Once λ * is known we can calculate the PGi
23.53 − 15
PG1 (23.5) =
= 426 MW
0.02
23.53 − 20
PG2 (23.5) =
= 353 MW
0.01
23.53 − 18
PG3 (23.5) =
= 221 MW
0.025
28

29. 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 lambdaiteration
–
–
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
29

30. 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.
30

31. Lambda-Iteration with Gen Limits
In the lambda-iteration method the limits are taken
into account when calculating PGi (λ ) :
if PGi (λ ) > PGi ,max then PGi (λ ) = PGi ,max
if PGi (λ ) < PGi ,min then PGi (λ ) = PGi ,min
31

32. Lambda-Iteration Gen Limit Example
In the previous three generator example assume
the same cost characteristics but also with limits
0 ≤ PG1 ≤ 300 MW
100 ≤ PG2 ≤ 500 MW
200 ≤ PG3 ≤ 600 MW
With limits we get
m
∑ PGi (20) − 1000
i =1
= PG1 (20) + PG 2 (20) + PG 3 (20) − 1000
= 250 + 100 + 200 = −450 MW (compared to -670MW)
m
∑ PGi (30) − 1000
i =1
= 300 + 500 + 480 − 1000 = 280 MW
32

33. Lambda-Iteration Limit Example,cont’d
Again we continue iterating until the convergence
condition is satisfied. With limits the final solution
of λ , is 24.43 $/MWh (compared to 23.53 $/MWh
without limits). The presence of limits will always
cause λ to either increase or remain the same.
Final solution is
PG1 (24.43) = 300 MW
PG2 (24.43) = 443 MW
PG3 (24.43) = 257 MW
33

34. 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
34