Lec2 Economic DispatchI
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Transcript of Lec2 Economic DispatchI
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EPE507
Power System Operation
Dr. Amer Al-Hinai
Economic Dispatch
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Power System Operation 2
The Economic Dispatch Problem
Consider a system that consists of N thermal-generating units serving an aggregated electrical load, Pload
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Power System Operation 3
The Economic Dispatch Problem
Input to each unit: cost rate of fuel consumed, Fi Output of each unit: electrical power generated, Pi
Total cost rate, FT, is the sum of the individual unit costs
Essential constraint: the sum of the output powers must equal the load demand
The problem is to minimize FT
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Power System Operation 4
Retail 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 half of retail costs.
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Power System Operation 5
Power System Economic Operation
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 higher operating costs.
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Power System Operation 6
Power System Economic Operation
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.
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Power System Operation 7
Electric Fuel Prices
Source: EIA Electric Power Annual, 2006 (October 2007)
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Power System Operation 8
Natural Gas Prices: 1990s to 2008
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Power System Operation 9
Coal Prices: 2005 to 2008
There are four main types of coal:
Bituminous sub-bituminous
lignite anthracite.
Heat values range from a low of 8Mbtu per ton to a high of 31 Mbtuper ton.
For Illinois coal price per Mbtu is about 4/Mbtu.
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Power System Operation 10
Power System Economic Operation
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, specified portfolio of generators.
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Power System Operation 11
US Generation Mix (Energy)
circa 2006-2009
Gen Type US % Illinois % California % Texas %
Coal 48 48 1 36
Nuclear 19 48 15 15
Hydro 6 0.1 22 2
Gas 21 3 50 40
Petroleum 1 0.1 1
Other Renewable 3 0.4 12 (14 in 1990) 7
Source: http://www.eia.doe.gov and Public Utility Commission of Texas
Indiana is 94% coal, while Oregon is 71% hydro, Washington State is 76% hydro. Canada is about 60% hydro, France is also 80% nuclear, China is about 80% coal
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Power System Operation 12
Power System Economic Operation
The two main types of generating units are thermal and
hydro, with wind 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 thermal units, looking at short-
term optimization.
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Power System Operation 13
Generator 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.
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Power System Operation 14
Block Diagram of Thermal Unit
To optimize generation costs we need to develop costrelationships between net power out and operatingcosts.
Between 2-10% of power is used within the generatingplant; this is known as the auxiliary power.
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Power System Operation 15
Generator Cost Curves
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
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Power System Operation 16
I/O Curve
The IO curve plots fuel input (in MBtu/hr) versus net MW output.
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Power System Operation 17
Fuel-cost Curve
The fuel-cost curve is the I/O curve multiplied by fuel cost.
A typical cost for coal is $ 1.70/MBtu.
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Power System Operation 18
Heat-rate Curve
Plots the average number of MBtu/hr of fuel inputneeded per MW of output.
Heat-rate curve is the I/O curve divided by MW.
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Power System Operation 19
Incremental (Marginal) cost Curve
Plots the incremental $/MWh as a function of MW.
Found by differentiating the cost curve.
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Power System Operation 20
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 approximations predominate:
quadratic or cubic functions
piecewise linear functions
We'll assume a quadratic approximation:
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Power System Operation 21
The Economic Dispatch Problem
H: Btu per hour heat input to the unit (Btu/h)
F: Fuel cost times H is (/h) input to the unit for
fuel
The per hour operating cost rate of a unit will
include prorated operation
and maintenance costs.
The labor cost for the operating crew is included
as part of the operation
cost.
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Power System Operation 22
The mathematical statement of the problem is a constrained optimization with the following functions:
objective function:
equality constraint:
Note that any transmission losses are neglected and any operating limits are not explicitly stated when formulating this problem
Problem may be solved using the Lagrange function
The Economic Dispatch Problem
N
1i
iiT PFF
N
1i
iLoad 0PP
N
1i
iLoad
N
1i
iiT PP.PFF
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Power System Operation 23
The Lagrange function establishes the necessary
conditions for finding an extreme of an objective function
with constraints
Taking the first derivatives of the Lagrange function with
respect to the independent variables allows us to find the
extreme value when the derivatives are set to zero
There are (NF + N) derivatives, one for each independent variable
and one for each equality constraint
The derivatives of the Lagrange function with respect to the
Lagrange multiplier merely gives back the constraint equation
The NF partial derivatives result in 0
P
PF
P
L
i
ii
i
The Economic Dispatch Problem
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Power System Operation 24
Fuel costs
coal: $ 3.30 / MBtu
oil: $ 3.00 / MBtu
The Economic Dispatch Problem
Unit # 1: coal-fired steam unit: H1 = 510 + 7.20P1 + 0.00142P12 MBtu/h
Unit # 2: oil-fired steam unit : H2 = 310 + 7.85P2 + 0.00194P22 MBtu/h
Unit # 3: oil-fired steam unit : H3 = 78 + 7.97P3 + 0.00482P32 MBtu/h
Example # 1
Determine the economic operating point for the three generating units when delivering a total of 850 MW
Input-output curves:
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Power System Operation 25
ED: Inequality Constraints In addition to the cost function and the equality constraint
each generation unit must satisfy two inequalities
The power output must be greater than or equal to the minimum power permitted:
Minimum heat generation for stable fuel burning and temperature
The power output must be less than or equal to the maximum power permitted:
Maximum shaft torque without permanent deformation
Maximum stator currents without overheating the conductor
min,ii PP
max,ii PP
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Power System Operation 26
Then the necessary conditions are expanded slightly
max,ii
i
i
min,ii
i
i
max,iimin,ii
i
i
PPdP
dF
PPdP
dF
PPP:PdP
dF
ED: Inequality Constraints
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Power System Operation 27
ED: Inequality Constraints
Example # 2
Reconsider the example # 1 with the following generator limits and the price of coal decreased to $2.70 / MBtu
Generator limits:
Unit # 1: 150 P1 600 MW
Unit # 2: 100 P2 400 MW
Unit # 3: 50 P3 200 MW
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Power System Operation 28
ED: Network Losses
Consider a similar system, which now has a transmission network that connects the generating
units to the load
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Power System Operation 29
ED: Network Losses
The economic dispatch problem is slightly more complicated
The constraint equation must include the network losses, Ploss
The objective function, FT is the same as before
The constraint equation must be expanded as:
N
1i
iLoosLoad 0PPP
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Power System Operation 30
The same math procedure is followed to establish the necessary conditions for a minimum-cost operating solution Lagrange function and its derivatives w.r.t. the input power:
ED: Network Losses
N
1i
iLoosLoad
N
1i
iiT PPP.PFF
i
loos
i
i
i
loos
i
i
i dP
dP
dP
dF0
dP
dP1
dP
dF
dP
d
The transmission network loss is a function of the impedances and the currents flowing in the network
For convenience, the currents may be considered functions of the input and load powers
It is more difficult to solve this set of equations
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Power System Operation 31
step 1 pick starting values for Pi that sum to the load
step 2 calculate Ploss/Pi and the total losses Ploss
step 3 calculate that causes Pi to sum to Pload & Ploss
step 4 compare Pi of step 3 to the values used in step 2; if there is significant change to any value, go back to step 2, otherwise, the procedure is done
ED: Iterative MethodPick Pi,
load
N
1i
i PP
Calculate
i
loss
P
P
Calculate totallosses Ploss
Solve for
and Pinew
is
Pinew - Pi ?
Done
Yes
No
Pi = Pinew
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Power System Operation 32
Example # 3
Repeat the first example, but include a
simplified loss expression for the transmission
network
ED: Network Losses
Ploss = 0.00003 P12 + 0.00009 P2
2 + 0.00012 P32 MW