Transcript EE 369 POWER SYSTEM ANALYSIS Lecture 16 Economic Dispatch Tom Overbye and Ross Baldick.

EE 369
POWER SYSTEM ANALYSIS
Lecture 16
Economic Dispatch
Tom Overbye and Ross Baldick
1
Announcements
Read 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: 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
m
Minimize
CT
 Ci ( PGi )
i 1
Such that
m
 PGi  PD  PLosses
i 1
Initially we'll
ignore generator
limits and the
losses
3
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) 
 x , x , , x 
 1
2
n 
4
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 f (x) s.t. g(x)  0
We define the Lagrangian L(x,λ )  f (x)  λ T g(x)
Then a necessary condition for a minimum is the
L x (x,λ )  0
and
L λ (x,λ )  0
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Economic Dispatch Lagrangian
For the economic dispatch we have a minimization
constrained with a single equality constraint
L(PG ,  ) 
m
m
 Ci ( PGi )   ( PD   PGi )
i 1
(no losses)
i 1
The necessary conditions for a minimum are
dCi ( PGi )
L(PG ,  )
   0 (for i  1 to m)

dPGi
PGi
m
PD   PGi  0
i 1
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Economic Dispatch Example
What is economic dispatch for a two generator
system PD  PG1  PG 2  500 MW and
C1 ( PG1 )  1000 20 PG1  0.01PG21
$/h
C2 ( PG 2 )  400 15PG 2  0.03PG22
$/h
Using the Lagrange multiplier method we know:
dC1 ( PG1 )

dPG1
dC2 ( PG 2 )

dPG 2
 20  0.02 PG1  
 15  0.06 PG 2  
0
0
500  PG1  PG 2  0
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Economic Dispatch Example, cont’d
We therefore need to solve three linear equations
20  0.02 PG1  
0
15  0.06 PG 2  
0
500  PG1  PG 2  0
0
1  PG1   20 
0.02
 0
0.06 1  PG 2    15 


 

1 0      500 
 1
 PG1 
 312.5 MW 
 P    187.5 MW 
 G2 


  
 26.2 $/MWh 
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Lambda-Iteration Solution Method
The 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: PGi ( ).
– 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
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brackets the optimal value.
Lambda-Iteration Algorithm
Pick  L and  H such that
m
L
P
(

 Gi )  PD  0
i 1
m
H
P
(

 Gi )  PD  0
i 1
 H   L   Do
While
 M  ( H   L ) / 2
m
If
M
H
M
P
(

)

P

0
Then



 Gi
D
i 1
Else  L   M
End While
10
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
Consider a three generator system with
IC1 ( PG1 )  15  0.02 PG1
  $/MWh
IC2 ( PG 2 )  20  0.01PG 2

$/MWh
IC3 ( PG 3 )  18  0.025 PG 3

$/MWh
and with constraint PG1  PG 2  PG 3  1000MW
Rewriting generation as a function of  , PGi ( ),
we have
PG1 ( ) 
  15
0.02
  18
PG3 ( ) 
0.025
PG2 ( ) 
  20
0.01
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Lambda-Iteration Example, cont’d
Pick  L so
m
L
P
(

 Gi )  1000  0 and
i=1
m
H
P
(

 Gi )  1000  0
i=1
Try  L  20 then
m
 PGi (20)  1000

i 1
  15   20   18
0.02

0.01
Try  H  30 then

0.025
m
 1000  670 MW
 PGi (30)  1000
 1230 MW
i 1
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Lambda-Iteration Example, cont’d
Pick convergence tolerance   0.05 $/MWh
Then iterate since  H   L  0.05
 M  ( H   L ) / 2  25
m
Then since
H
P
(25)

1000

280
we
set

 25
 Gi
i 1
Since 25  20  0.05
 M  (25  20) / 2  22.5
m
L
P
(22.5)

1000


195
we
set

 22.5
 Gi
i 1
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Lambda-Iteration Example, cont’d
Continue iterating until  H   L  0.05
The solution value of  ,  , is 23.53 $/MWh
*
Once  * is known we can calculate the PGi
23.53  15
PG1 (23.5) 
 426 MW
0.02
23.53  20
PG 2 (23.5) 
 353 MW
0.01
23.53  18
PG 3 (23.5) 
 221 MW
0.025
15
Thirty Bus ED Example
Case is economically dispatched (without considering
the incremental impact of the system losses).
16
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
In the lambda-iteration method the limits are taken
into account when calculating PGi ( ) :
if calculated production for PGi  PGi ,max
then set PGi ( )  PGi ,max
if calculated production for PGi  PGi ,min
then set PGi ( )  PGi ,min
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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  1000
 450 MW (compared to  670MW)
m
 PGi (30)  1000
i 1
 300  500  480  1000  280 MW
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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 cause  to either increase
or remain the same.
Final solution is:
PG1 (24.43)  300 MW (at maximum limit)
PG 2 (24.43)  443 MW
PG 3 (24.43)  257 MW
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Back of Envelope Values
 $/MWhr = fuelcost * heatrate + variable O&M
Typical incremental costs can be roughly
approximated:
– Typical heatrate for a coal plant is 10, modern
combustion turbine is 10, combined cycle plant is
7 to 8, older combustion turbine 15.
– Fuel costs ($/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
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(water) constrained.
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:
m


L(PG ,  )   Ci ( PGi )    PD  PL ( PG )   PGi 


i 1
i 1
m
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Impact of Transmission Losses
The inclusion of losses then impacts the necessary
conditions for an optimal economic dispatch:
m


L(PG ,  )   Ci ( PGi )    PD  PL ( PG )   PGi  .


i 1
i 1
The necessary conditions for a minimum are now:
m
L(PG ,  )
PGi
 PL ( PG ) 
dCi ( PGi )

  1 
0

dPGi
PGi 

m
PD  PL ( PG )   PGi  0
i 1
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Impact of Transmission Losses
 PL ( PG ) 
dCi ( PGi )
Solving for  , we get:
  1 
0

dPGi
PGi 

dCi ( PGi )
1
 
 PL ( PG )  dPGi
 1  P


Gi 
Define the penalty factor Li for the i th generator
(don't confuse with Lagrangian L!!!)
The penalty factor
1
Li 
at the slack bus is
 PL ( PG ) 
always unity!
 1  P


Gi 
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Impact of Transmission Losses
The condition for optimal dispatch with losses is then
L1 IC1 ( PG1 )  L2 IC2 ( PG 2 )  Lm ICm ( PGm )  
1
Li 
. So, if increasing PGi increases
 PL ( PG ) 
 1  P


Gi 
PL ( PG )
the losses then
 0  Li  1.0
PGi
This makes generator i appear to be more expensive
(i.e., it is penalized). Likewise Li  1.0 makes a generator
appear less expensive.
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Calculation of Penalty Factors
Unfortunately, the analytic calculation of Li is
somewhat involved. The problem is a small change
in the generation at PGi impacts the flows and hence
the losses throughout the entire system. However,
using a power flow you can approximate this function
by making a small change to PGi and then seeing how
the losses change:
PL ( PG ) PL ( PG )

PGi
PGi
1
Li 
PL ( PG )
1
PGi
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Two Bus Penalty Factor Example
PL ( PG )
 0.0387
PG 2
PL ( PG ) 0.37 MW

 0.037
PG 2
10 MW
L2  0.9627
L2  0.9643
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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
The area supply curve shows the cost to produce the
next MW of electricity, assuming area is economically
dispatched
10.00
7.50
Supply
curve for
thirty bus
system
5.00
2.50
0.00
0
100
200
Total Area Generation (MW)
300
400
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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.
30
Security Constrained ED
or 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.”
31
Security Constrained ED
or 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
-22 MW
4 MVR
22 MW
-4 MVR
Bus 1
1.00 PU
357 MW
179 MVR
1.00 PU
0 MW
37 MVR
100%
194 MW OFF AGC -142 MW
49 MVR
232 MVR AVR ON
145 MW 100%
-37 MVR
Home Area
Bus 3
Scheduled Transactions
100.0 MW
-122 MW
41 MVR
100 MW
124 MW
-33 MVR
1.00 PU
179 MW
89 MVR
448 MW AGC ON
19 MVR AVR ON
Need to dispatch to keep line
from bus 3 to bus 2 from overloading
33
Multi-Area Operation
 In 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 rules
 Actual 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.
34
Seven Bus Case: One-line
System has
three areas
44 MW
-42 MW
-31 MW
0.99 PU
3
1.05 PU
1
106 MW -37 MW
AGC ON
62 MW
79 MW
2
40 MW
20 MVR
Top Area Cost
8029 $/MWH
1.00 PU
-32 MW
Case Hourly Cost
16933 $/MWH
32 MW
80 MW
30 MVR
4
110 MW
40 MVR
38 MW
-61 MW
1.04 PU
31 MW
-77 MW
5
-39 MW
40 MW
94 MW
AGC ON
-14 MW
1.01 PU
130 MW
40 MVR
168 MW AGC ON
-40 MW
1.04 PU
6
Left area
has one
bus
20 MW
-20 MW
40 MW
1.04 PU
20 MW
200 MW
0 MVR Left Area Cost
4189 $/MWH
200 MW AGC ON
-20 MW
Top area
has five
buses
7
No net
interchange
between
Any areas.
200 MW
Right Area Cost
0 MVR
4715 $/MWH
201 MW AGC ON
Right area has one
bus
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Seven Bus Case: Area View
Top
40.1 MW
0.0 MW
Area Losses
7.09 MW
-40.1 MW
0.0 MW
System has
40 MW of
“Loop Flow”
Left
Area Losses
0.33 MW
Right
40.1 MW
0.0 MW
Actual
flow
between
areas
Scheduled
flow
Area Losses
0.65 MW
Loop flow can result in higher losses
36
Seven Bus - Loop Flow?
Top
4.8 MW
0.0 MW
-4.8 MW
0.0 MW
Left
Area Losses
-0.00 MW
100 MW Transaction
between Left and Right
Area Losses
9.44 MW
Right
104.8 MW
100.0 MW
Note that
Top’s
Losses have
increased
from
7.09MW to
9.44 MW
Area Losses
4.34 MW
Transaction has
actually decreased
the loop flow
37