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Transcript of 0 Generator Cost Curves Generator costs are typically represented by up to four different curves –...
1
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
11
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
12
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
13
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
14
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
19
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
21
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.
22
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
23
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
24
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
25
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:
33
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;
38
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
56
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
59
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