Lec2 Economic DispatchI

EPE507

Power System Operation

Dr. Amer Al-Hinai

Economic Dispatch

Power System Operation 2

The Economic Dispatch Problem

Consider a system that consists of N thermal-generating units serving an aggregated electrical load, Pload



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


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.


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.



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.


Electric Fuel Prices

Source: EIA Electric Power Annual, 2006 (October 2007)


Natural Gas Prices: 1990s to 2008


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.



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.


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



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.


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.


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.


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


I/O Curve

The IO curve plots fuel input (in MBtu/hr) versus net MW output.


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.


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.


Incremental (Marginal) cost Curve

Plots the incremental $/MWh as a function of MW.

Found by differentiating the cost curve.


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:



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.


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


N

1i

iiT PFF

N

1i

iLoad 0PP

N

1i

iLoad

N

1i

iiT PP.PFF


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



Fuel costs

coal: $ 3.30 / MBtu

oil: $ 3.00 / MBtu


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:


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


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


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


ED: Network Losses

Consider a similar system, which now has a transmission network that connects the generating

units to the load


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


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


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


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

Date post:	09-Mar-2016
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Lec2 Economic DispatchI

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