The document discusses economic operation of power systems. It covers characteristics of steam and hydro plants, economic load scheduling of thermal plants with and without considering transmission losses, penalty factors and loss coefficients. It also discusses hydrothermal scheduling. The document provides information on retail electricity prices and factors contributing to prices like generation, transmission and distribution costs. It describes different generation technologies in terms of capital and operating costs. The concepts of economic dispatch formulation, incremental costs, lambda iteration method and inclusion of transmission losses in economic dispatch are explained.

1. UNIT-IVUNIT-IV
Economic Operation of PowerEconomic Operation of Power
SystemsSystems
• Md Irshad Ahmad
• Irshad.ahmad@jit.edu.in
• Electrical Engineering Department
• Subject: : POWER STATION PRACTICE
• (NEE /NEN–702)

2. ContentsContents
• UNIT-IV:Economic Operation of Power
Systems:
• Characteristics of steam and hydro-
plants,Constraints in operation
• Economic load scheduling of thermal plants
Neglecting and considering transmission
Losses,
• Penalty factor, loss coefficients, Incremental
transmission loss. Hydrothermal
Scheduling.

3. Retail Electricity PricesRetail Electricity Prices
• There are many fixed and variable costs
associated with power systems, which
ultimately contribute to determining retail
electricity prices.
• The major variable operating cost is
associated with generation, primarily due
to fuel costs:
– Roughly 30% to 50% of retail costs.
• Retail prices also reflect the capital costs
of building the generation, transmission,
and distribution system as well as other 33

4. Power System EconomicPower System Economic
OperationOperation
• Different generation technologies vary in
the:
– capital costs necessary to build the generator
– fuel costs to actually produce electric power
• For example:
– nuclear and hydro have high capital costs and
low operating costs.
– Natural gas generators have low capital costs,
and (with gas available from fracking)
moderate operating costs. 44

5. Economic Dispatch:Economic Dispatch:
FormulationFormulation
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
T
1
1
Minimize C ( )
Such that
m
i Gi
i
m
Gi D Losses
i
C P
P P P
=
=
= +
∑
∑
@
Initially we'll
ignore generator
limits and the
losses
55

6. Unconstrained MinimizationUnconstrained 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 x
f x K@
( )∇ =f x 0
66

7. Minimization with EqualityMinimization with Equality
ConstraintConstraint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 ( ) s.t. ( )
We define the Lagrangian L( , ) ( ) ( )
Then a necessary condition for a minimum is the
L ( , ) 0 and L ( , ) 0
T
=
= +
∇ = ∇ =xλ
f x g x 0
xλ f x λ g x
xλ x λ 77

8. Economic Dispatch LagrangianEconomic Dispatch Lagrangian
G
1 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 Gi
i i
Gi
C P P P
dC
P
λ λ
λ
= =
= + −
∂
=
∂
∑ ∑P
P
1
( )
0 (for 1 to )
0
i Gi
Gi
m
D Gi
i
P
i m
dP
P P
λ
=
− = =
− =∑
88

9. Economic Dispatch ExampleEconomic Dispatch Example
1 2
2
1 1 1 1
2
2 2 2 2
1 1
1
What is economic dispatch for a two generator
system 500 MW and
( ) 1000 20 0.01 $/h
( ) 400 15 0.03 $/h
Using the Lagrange multiplier method we know:
( )
20 0.0
D G G
G G G
G G G
G
G
P P P
C P P P
C P P P
dC P
dP
λ
= + =
= + +
= + +
− = + 1
2 2
2
2
1 2
2 0
( )
15 0.06 0
500 0
G
G
G
G
G G
P
dC P
P
dP
P P
λ
λ λ
− =
− = + − =
− − =
99

10. Economic Dispatch Example,Economic Dispatch Example,
cont’dcont’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 0 500
312.5 MW
187.5 MW
26.2 $/MW
G
G
G G
G
G
G
G
P
P
P P
P
P
P
P
λ
λ
λ
λ
+ − =
+ − =
− − =
− −    
    − = −
    
− − −        
 
  =
 
   h
 
 
 
   1010

11. Economic dispatch example,Economic dispatch example,
cont’dcont’d
• At the solution, both generators have the same
marginal (or incremental) cost, and this common
marginal cost is equal to λ.
• Intuition behind solution:
– If marginal costs of generators were different, then by
decreasing production at higher marginal cost generator,
and increasing production at lower marginal cost
generator we could lower overall costs.
– Generalizes to any number of generators.
• If demand changes, then change in total costs can
be estimated from λ.
1111

12. Economic dispatch example,Economic dispatch example,
cont’dcont’d
• Another way to solve the equations is to:
– Rearrange the first two equations to solve for
PG1 and PG2 in terms of λ,
– Plug into third equation and solve for λ,
– Use the solved value of λ to evaluate PG1 and
PG2.
• This works even when relationship
between generation levels and λ is more
complicated:
– Equations are more complicated than linear 1212

13. Lambda-Iteration SolutionLambda-Iteration Solution
MethodMethod
• Discussion on previous page leads to
“lambda-iteration” method:
– this method requires a unique mapping from a
value of lambda (marginal cost) to each
generator’s MW output:
– for any choice of lambda (common 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 brackets the optimal• 1313
( ).Gi
P λ

14. Lambda-Iteration AlgorithmLambda-Iteration Algorithm
L H
1 1
H L
M H L
H M
1
L M
Pick and such that
( ) 0 ( ) 0
While Do
( )/2
If ( ) 0 Then
Else
End While
m m
L H
Gi D Gi D
i i
m
M
Gi D
i
P P P P
P P
λ λ
λ λ
λ λ ε
λ λ λ
λ λ λ
λ λ
= =
=
− < − >
− >
= +
− > =
=
∑ ∑
∑
1414

15. Lambda-Iteration: GraphicalLambda-Iteration: Graphical
ViewView
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.
1515

16. Lambda-Iteration ExampleLambda-Iteration Example
1 1 1
2 2 2
3 3 3
1 2 3
Consider a three generator system with
( ) 15 0.02 $/MWh
( ) 20 0.01 $/MWh
( ) 18 0.025 $/MWh
and with constraint 1000MW
Rewriting generation as a function of , (
G G
G G
G G
G G G
Gi
IC P P
IC P P
IC P P
P P P
P
λ
λ
λ
λ
= + =
= + =
= + =
+ + =
G1 G2
G3
),
we have
15 20
P ( ) P ( )
0.02 0.01
18
P ( )
0.025
λ
λ λ
λ λ
λ
λ
− −
= =
−
= 1616

17. Lambda-Iteration Example,Lambda-Iteration Example,
cont’dcont’dm
Gi
i=1
m
Gi
i=1
1
H
1
Pick so P ( ) 1000 0 and
P ( ) 1000 0
Try 20 then (20) 1000
15 20 18
1000 670 MW
0.02 0.01 0.025
Try 30 then (30) 1000 1230 MW
L L
H
m
L
Gi
i
m
Gi
i
P
P
λ λ
λ
λ
λ λ λ
λ
=
=
− <
− >
= − =
− − −
+ + − = −
= − =
∑
∑
∑
∑
1717

18. Lambda-Iteration Example,Lambda-Iteration Example,
cont’dcont’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
m
H
Gi
i
M
m
L
Gi
i
P
P
ε
λ λ
λ λ λ
λ
λ
λ
=
=
=
− >
= + =
− = =
− >
= + =
− = − =
∑
∑ 2.5
1818

19. Lambda-Iteration Example,Lambda-Iteration Example,
cont’dcont’d
H
*
*
1
2
3
Continue iterating until 0.05
The solution value of , , is 23.53 $/MWh
Once is known we can calculate the
23.53 15
(23.5) 426 MW
0.02
23.53 20
(23.5) 353 MW
0.01
23.53 18
(23.5)
0.025
L
Gi
G
G
G
P
P
P
P
λ λ
λ λ
λ
− <
−
= =
−
= =
−
= 221 MW=
1919

20. Thirty Bus ED ExampleThirty Bus ED Example
Case is economically dispatched (without considering
the incremental impact of the system losses).
2020

21. Generator MW LimitsGenerator 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.
2121

22. Lambda-Iteration with GenLambda-Iteration with Gen
LimitsLimits
,max
,max
In the lambda-iteration method the limits are taken
into account when calculating ( ) :
if calculated production for
then set ( )
if calculated production for
Gi
Gi Gi
Gi Gi
P
P P
P P
λ
λ
>
=
,min
,minthen set ( )
Gi Gi
Gi Gi
P P
P Pλ
<
=
2222

23. Lambda-Iteration Gen LimitLambda-Iteration Gen Limit
ExampleExample
G1 G2
G3
1 2 3
1
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) 10
m
Gi G G G
i
P P P P
=
≤ ≤ ≤ ≤
≤ ≤
− = + + −∑
1
00
250 100 200 1000
450 MW (compared to 670MW)
(30) 1000 300 500 480 1000 280 MW
m
Gi
i
P
=
= + + −
= − −
− = + + − =∑
2323

24. Lambda-Iteration LimitLambda-Iteration Limit
Example,cont’dExample,cont’dAgain 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 caus
λ
1
2
3
e to either increase
or remain the same.
Final solution is:
(24.43) 300 MW (at maximum limit)
(24.43) 443 MW
(24.43) 257 MW
G
G
G
P
P
P
λ
=
=
=
2424

25. Back of Envelope ValuesBack of Envelope Values
$/MWhr = fuelcost * heatrate + variable O&M
Typical incremental costs can be roughly
approximated:
– Typical heatrate for a coal plant is 10, modern
combustion turbine is 10, combined cycle plant
is 6 to 8, older combustion turbine 15.
– Fuel costs ($/MBtu) are quite variable, with
current values around 2 for coal, 3 to 5 for
natural gas, 0.5 for nuclear, probably 10 for fuel
oil.
– Hydro costs tend to be quite low, but are fuel
(water) constrained 2525

26. Inclusion of TransmissionInclusion of Transmission
LossesLosses
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 to include
G
1 1
L( , ) ( ) ( )
m m
i Gi D L G Gi
i i
C P P P P Pλ λ
= =
 
= + + − ÷
 
∑ ∑P
2626

27. Impact of Transmission LossesImpact of Transmission Losses
G
1 1
G
The inclusion of losses 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 Gi
i i
C P P P P Pλ λ
λ
= =
 
= + + − ÷
 
∂
∂
∑ ∑P
P
1
( ) ( )
1 0
( ) 0
i Gi L G
Gi Gi Gi
m
D L G Gi
i
dC P P P
P dP P
P P P P
λ
=
 ∂
= − − = ÷∂ 
+ − =∑
2727

28. Impact of Transmission LossesImpact of Transmission Losses
th
( ) ( )
Solving for , we get: 1 0
( )1
( )
1
Define the penalty factor for the generator
(don't confuse with Lagrangian L!!!)
1
( )
1
i Gi L G
Gi Gi
i Gi
GiL G
Gi
i
i
L G
Gi
dC P P P
dP P
dC P
dPP P
P
L i
L
P P
P
λ λ
λ
 ∂
− − = ÷∂ 
=
 ∂
− ÷∂ 
=
 ∂
− ∂

÷

The penalty factor
at the slack bus is
always unity!
2828

29. Impact of Transmission LossesImpact of Transmission Losses
1 1 1 2 2 2
The condition for optimal dispatch with losses is then
( ) ( ) ( )
1
. So, if increasing increases
( )
1
( )
the losses then 0 1.0
This makes generator
G G m m Gm
i Gi
L G
Gi
L G
i
Gi
L IC P L IC P L IC P
L P
P P
P
P P
L
P
λ= = =
=
 ∂
− ÷∂ 
∂
> ⇒ >
∂
appear to be more expensive
(i.e., it is penalized). Likewise 1.0 makes a generator
appear less expensive.
i
i
L <
2929

30. Calculation of Penalty FactorsCalculation of Penalty Factors
Unfortunately, the analytic calculation of is
somewhat involved. The problem is a small change
in the generation at impacts the flows and hence
the losses throughout the entire system. However,
i
Gi
L
P
using a power flow you can approximate this function
by making a small change to and then seeing how
the losses change:
( ) ( ) 1
( )
1
Gi
L G L G
i
L GGi Gi
Gi
P
P P P P
L
P PP P
P
∂ ∆
≈ ≈
∆∂ ∆ −
∆
3030

31. Two Bus Penalty FactorTwo Bus Penalty Factor
ExampleExample
2 2
2 2
( ) ( ) 0.37
0.0387 0.037
10
0.9627 0.9643
L G L G
G G
P P P P MW
P P MW
L L
∂ ∆ −
= − = = −
∂ ∆
= ≈
3131

32. Thirty Bus ED ExampleThirty Bus ED Example
Now consider losses.
Because of the penalty factors the generator incremental
costs are no longer identical.
3232

33. Area Supply CurveArea Supply Curve
0 100 200 300 400
Total AreaGeneration(MW)
0.00
2.50
5.00
7.50
10.00
The area supply curve shows the cost to produce the
next MW of electricity, assuming area is economically
dispatched
Supply
curve for
thirty bus
system
3333

34. Economic Dispatch - SummaryEconomic 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.
3434

35. Security Constrained EDSecurity Constrained ED
or Optimal Power Flowor 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.”
Need to solve or approximate power flow in
order to consider transmission constraints.
3535

36. Security Constrained EDSecurity Constrained ED
or Optimal Power Flowor 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 3636

37. Security Constrained DispatchSecurity Constrained Dispatch
Bus 2 Bus 1
Bus 3Home Area
Scheduled Transactions
357 MW
179 MVR
194 MW
448 MW
19 MVR
232 MVR
179 MW
89 MVR
1.00 PU
-22 MW
4 MVR
22 MW
-4 MVR
-142 MW
49 MVR
145 MW
-37 MVR
124 MW
-33 MVR
-122 MW
41 MVR
1.00 PU
1.00 PU
0 MW
37 MVR100%
100%
100 MW
OFF AGC
AVR ON
AGC ON
AVR ON
100.0 MW
Need to dispatch to keep line
from bus 3 to bus 2 from overloading
3737

38. Multi-Area OperationMulti-Area Operation
In multi-area system, “rules” have been established
regarding transactions on tie-lines:
– In Eastern interconnection, in principle, up to “nominal”
thermal interconnection capacity,
– In Western interconnection there are more complicated
rules
The actual power that flows through the entire
network depends on the impedance of the
transmission lines, and ultimately determine what
are acceptable patterns of dispatch:
Can result in need to “curtail” transactions that
otherwise satisfy rules.
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. 3838

39. Power System EconomicPower System Economic
OperationOperation
• Fuel cost to generate a MWh can vary
widely from technology to technology.
• For some types of units, such as hydro,
“fuel” costs are zero but the limit on total
available water gives it an implicit value.
• For thermal units it is much easier to
characterize costs.
• We will focus on minimizing the variable
operating costs (primarily fuel costs) to
meet demand. 3939

40. Power System EconomicPower System Economic
OperationOperation
• Power system loads are cyclical.
• Therefore the installed generation capacity
is usually much greater than the current
load.
• This means that there are typically many
ways we could meet the current load.
• Since different states have different mixes
of generation, we will consider how
generally to minimize the variable
operating costs given an arbitrary, 4040

41. Thermal versus OtherThermal versus Other
GenerationGenerationThe main types of generating units are
thermal and hydro, with wind and solar
rapidly growing.
For hydro the fuel (water) is free but there
may be many constraints on operation:
– fixed amounts of water available,
– reservoir levels must be managed and
coordinated,
– downstream flow rates for fish and navigation.
Hydro optimization is typically longer term
(many months or years).
We will concentrate on dispatchable
thermal units, looking at short-term 4141

42. Generator typesGenerator types
Traditionally utilities have had three broad
groups of generators:
– “Baseload” units: large coal/nuclear; almost
always on at max.
– “Midload,” ‘intermediate,” or “cycling” units: smaller
coal or gas that cycle on/off daily or weekly.
– “Peaker” units: combustion turbines used only for
several hours. during periods of high demand
4242

43. Block Diagram of Thermal UnitBlock Diagram of Thermal Unit
•To optimize generation costs we need to develop
cost relationships between net power out and
operating costs.
•Between 2-10% of power is used within the
generating plant; this is known as the auxiliary 4343

44. Thermal generator Cost CurvesThermal generator Cost Curves
Thermal generator costs are typically
represented by one or other of the
following four 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 = 0.29 MWh 4444

45. I/O CurveI/O Curve
The IO curve plots fuel input (in
MBtu/hr) versus net MW output.
4545

46. Fuel-cost CurveFuel-cost Curve
The fuel-cost curve is the I/O curve
multiplied by fuel cost.
A typical cost for coal is $ 1.70/MBtu.
4646

47. Heat-rate CurveHeat-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 divided by
MW.
Best heat-rate for most efficient coal
units is around 9.0
4747

48. Incremental (Marginal) costIncremental (Marginal) cost
CurveCurvePlots the incremental $/MWh as a
function of MW.
Found by differentiating the cost curve.
4848

49. Mathematical Formulation ofMathematical Formulation of
CostsCostsGenerator cost curves are usually not
smooth. However the curves can
usually be adequately approximated
using piece-wise smooth, functions.
Two approximations predominate:
– quadratic or cubic functions
– piecewise linear functions
We'll assume a quadratic
approximation: 2
( ) $/hr (fuel-cost)
( )
( ) 2 $/MWh
i Gi i i Gi i Gi
i Gi
i Gi i i Gi
Gi
C P P P
dC P
IC P P
dP
α β γ
β γ
= + +
= = +
4949

50. Coal Usage ExampleCoal Usage Example
•A 500 MW (net) generator is 35% efficient.
It is being supplied with coal costing $1.70
per MBtu and with heat content 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
= × =
×
× =
× =
5050

51. Wasting Coal ExampleWasting 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 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
× × × =
5151

52. Incremental Cost ExampleIncremental Cost Example
2
1 1 1 1
2
2 2 2 2
1 1
1 1 1
1
2 2
2 2 2
2
For a two generator system assume
( ) 1000 20 0.01 $/hr
( ) 400 15 0.03 $/hr
Then
( )
( ) 20 0.02 $/MWh
( )
( ) 15 0.06 $/MWh
G G G
G G G
G
G G
G
G
G G
G
C P P P
C P P P
dC P
IC P P
dP
dC P
IC P P
dP
= + +
= + +
= = +
= = +
5252

53. Incremental Cost Example,Incremental Cost Example,
cont'dcont'd
1 2
2
1
2
2
1
2
If 250 MW and 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
G GP P
C
C
IC
IC
= =
= + × + × =
= + × + × =
= + × =
= + × =
5353

54. 5454
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