ECE 476
Power System Analysis
Lecture 16: Economic Dispatch,
Optimal Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Chapter 12.4 and 12.5, Chapter 7
• HW 7 is due now
• HW 8 is 12.19, 12.20, 12.26, 12.28; due October 29 in
class (no quiz)
1
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
L(PG ,  )
dCi ( PGi )

   0 (for i  1 to m)
PGi
dPGi
m
PD   PGi  0
i 1
2
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
$ / hr
C2 ( PG 2 )  400 15 PG 2  0.03PG22
$ / hr
Using the Largrange multiplier method we know
dC1 ( PG1 )

dPG1
 20  0.02 PG1  
0
dC2 ( PG 2 )

dPG 2
 15  0.06 PG 2  
0
500  PG1  PG 2  0
3
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
 1
     500 
 PG1 
 312.5 MW 
 P    187.5 MW 
 G2 


  
 26.2 $/MWh 
4
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
5
Lambda-Iteration Algorithm
Pick  L and  H such that
m
L
P
(

 Gi )  PD  0
H
P
(

 Gi )  PD  0
i=1
i=1
While
m
 H   L   Do
 M  ( H   L ) / 2
m
If
M
H
M
P
(

)

P

0
Then



 Gi
D
i=1
Else  L   M
End While
6
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.
7
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  1000 MW
Rewriting as a function of  , PGi ( ), we have
PG1 ( ) 
  15
0.02
  18
PG3 ( ) 
0.025
PG2 ( ) 
  20
0.01
8
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
i 1
 1230 MW
9
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
10
Lambda-Iteration Example, cont’d
Continue iterating until     0.05
H
L
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
PG2 (23.5) 
 353 MW
0.01
23.53  18
PG3 (23.5) 
 221 MW
0.025
11
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.
12
Lambda-Iteration with Gen Limits
In the lambda-iteration method the limits are taken
into account when calculating PGi ( ) :
if PGi ( )  PGi ,max then PGi ( )  PGi ,max
if PGi ( )  PGi ,min then PGi ( )  PGi ,min
13
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
14
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 always
cause  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
15
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
16
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:
L(PG ,  ) 
m
m
i 1
i 1
 Ci ( PGi )   ( PD  PL ( PG )   PGi )
17
Impact of Transmission Losses
This small change then impacts the necessary
conditions for an optimal economic dispatch
L(PG ,  ) 
m
m
i 1
i 1
 Ci ( PGi )   ( PD  PL ( PG )   PGi )
The necessary conditions for a minimum are now
L(PG ,  )
PGi
dCi ( PGi )
PL ( PG )

  (1 
)0
dPGi
PGi
m
PD  PL ( PG )   PGi  0
i 1
18
Impact of Transmission Losses
Solving each equation for  we get
dCi ( PGi )
PL ( PG )
  (1 
0
dPGi
PGi
dCi ( PGi )
1
 
 PL ( PG )  dPGi
1  P


Gi 
Define the penalty factor Li for the i th generator
1
Li 
 PL ( PG ) 
1  P


Gi 
The penalty factor
at the slack bus is
always unity!
19
Impact of Transmission Losses
The condition for optimal dispatch with losses is then
L1IC1 ( PG1 )  L2 IC2 ( PG 2 )  Lm ICm ( PGm )  
1
Since Li 
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.
20
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
21
Two Bus Penalty Factor Example
PL ( PG )
 0.0387
PG 2
L2  0.9627
PL ( PG ) 0.37 MW

 0.037
PGi
10MW
L2  0.9643
22
Thirty Bus ED Example
• Case is economically dispatched without considering
the incremental impact of the system losses
23
Thirty Bus ED Example, cont
• Because of the penalty factors the generator
incremental costs are no longer identical.
24
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
25
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
26
Optimal Power Flow
• The goal of an optimal power flow (OPF) is to
determine the “best” way to instantaneously operate
a power system.
• Usually “best” = minimizing operating cost.
• OPF considers the impact of the transmission
system
• OPF is used as basis for real-time pricing in major
US electricity markets such as MISO and PJM.
• ECE 476 introduces the OPF problem and provides
some demonstrations.
27
Electricity Markets
• Over last fifteen years electricity markets have
moved from bilateral contracts between utilities to
also include spot markets (day ahead and realtime).
• Electricity (MWh) is now being treated as a
commodity (like corn, coffee, natural gas) with the
size of the market transmission system dependent.
• Tools of commodity trading are being widely
adopted (options, forwards, hedges, swaps).
28
Electricity Futures Example
Source: Wall Street Journal Online, 10/21/2015
29
Historical Variation in Oct 2015 Price
Price has dropped, following the drop in natural gas prices
Source: Wall Street Journal Online, 10/21/2015
“Ideal” Power Market
• Ideal power market is analogous to a lake.
Generators supply energy to lake and loads remove
energy.
• Ideal power market has no transmission constraints
• Single marginal cost associated with enforcing
constraint that supply = demand
–
–
buy from the least cost unit that is not at a limit
this price is the marginal cost
• This solution is identical to the economic dispatch
problem solution
31
Two Bus Economic Dispatch
Example
Total Hourly Cost : 8459 $/hr
Area Lambda : 13.02
Bus A
Bus B
300.0 MW
199.6 MW
AGC ON
300.0 MW
400.4 MW
AGC ON
32
Market Marginal (Incremental) Cost
Below are some graphs associated with this two bus
system. The graph on left shows the marginal cost for each
of the generators. The graph on the right shows the
system supply curve, assuming the system is optimally
dispatched.
16.00
16.00
15.00
15.00
14.00
14.00
13.00
13.00
12.00
12.00
0
175
350
525
Generator Power (MW)
700
0
350
Current generator operating point
700
1050
Total Area Generation (MW)
1400
33
Real Power Markets
• Different operating regions impose constraints -total demand in region must equal total supply
• Transmission system imposes constraints on the
market
• Marginal costs become localized
• Requires solution by an optimal power flow
34
Optimal Power Flow (OPF)
• OPF functionally combines the power flow with
economic dispatch
• Minimize cost function, such as operating cost,
taking into account realistic equality and inequality
constraints
• Equality constraints
–
–
–
bus real and reactive power balance
generator voltage setpoints
area MW interchange
35
OPF, cont’d
• Inequality constraints
–
–
–
–
transmission line/transformer/interface flow limits
generator MW limits
generator reactive power capability curves
bus voltage magnitudes (not yet implemented in
Simulator OPF)
• Available Controls
–
–
generator MW outputs
transformer taps and phase angles
36