# Economic Dispatch

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```					Portland State University

SYSC-505 Power System Economic Dispatch

Professors: Tim Anderson & Herman Migliore

Student: Donald Ogilvie

Date: Aug 17, 2010
Page
Abstract                                                                2
Introduction                                                            2
Economic Dispatch Common Formulae                                       3
1.0     Cost of Generating Electrical Power                             4
1.1     Monitoring System efficiency via Turbine Heat Rate (THR) Test   7
1.2     Typical Power Plant showing heat rate calculation               11
1.3     Analyzing the THR process                                       12
1.3.1 Heat Rate Verification & Validation Using Calculus                13
1.3.2 Matlab Software Code & Fuel Cost Curve                            14
2.0     Power Flow and Economic Dispatch                                17
2.1     Economic Dispatch System One generator Concept                  18
2.2     Incremental Cost (IC)                                           24
2.2.1 System-2 (Two Generator Concept)                                  24
2.3     System-3 Three Generators Economic Dispatch Applications        32
2.3.1 Matlab Software Code for Three Generators Lambda (Cost)           36
2.3.3 Economic Dispatch with Generator Limits                           38
3.0     System Modeling                                                 41
3.1     Computer Modeling & Simulation                                  45
3.2     Computer simulation of a two bus 1-generator system             47
3.3     Two generator economic dispatch modeling                        48
4.0     Conclusion                                                      50
Appendix                                                                52
References                                                              55
List of Figures & Tables                                                Page
Figure 1.0     Input/Output (I/O) Curve                                 7
Figure 1.1     Calculating Heat Rate                                    11
Figure 1.2     THR Test                                                 12
Figure 1.4     Heat Rate vs. Power                                      13
Figure 1.5     Typical Operating Cost Curve                             15
Figure 2.1     Single Generator Concept                                 18
Figure 2.2     Power Flow System                                        20
Figure 2.3     Two Generator Concept                                    24
Figure 2.4     Gauss Seidel Method                                      29
Figure 2.5     Three Generator Economic Dispatch Application            37
Figure 3.1     Modeling Two Bus System                                  42
Figure 3.2     Bus Modeling                                             44
Figure 3.3     Modeling 7-Bus System                                    45
Figure 3.4     Modeling A Single Generator                              47
Figure 3.5     Modeling A Two Bus Power Flow System                     48
Table-1.1      Fuel Cost                                                4
Table-1.2      Cost of Electricity Generation                           5
Table-2.1      Voltage Iteration Summary                                22
Table-2.2      Voltage Iteration                                        30

1
Abstract

This paper describes several efficient and practical methods to formulate and analyze the economic
dispatch of a power system, while taking into consideration the constraint of the transmission line
and load. Some of the practical approaches used are Lagrange method, Matlab software code, Power-
world modeling software, and fuel cost curve sometimes referred to as the input and output curve
(I/O). Iterative techniques for power flow are represented by Gauss-Seidel methods.

Introduction

In economic dispatch practices there are many choices for setting the operating points of generators.
The main aim of the economic dispatch paper is to include variables that affect operational costs,
such as the generator distance from the load, type of fuel, load capacity and transmission line losses.
By including these variables one will be able to perform economic dispatch and inter-connect
generators to minimize operating cost and functions. The generator cost is typically represented by
four curves, namely: Input/Output (I/O), heat rate, fuel cost and incremental cost curve. Generator
cost curves are not smooth, and they are generally represented by quadratic functions and piecewise
linear functions. Each plant uses a quadratic cost function such as the Fuel Cost Curve:
Ci ( PGi )     PGi   PGi .
2

Where i = generator i, one of the number of units
Ci = operating cost of unit in \$/h
PGi = electrical power output of a generator i in per unit on a common power base
Alpha, Beta & Gamma are in units of dollars ( \$/h)
This fuel cost curve allows us to look at a wide range of economic dispatch practice such as total
operating cost of a system, incremental cost and minute by minute loading of a generator. Power flow
concept is a major component of economic dispatch and this paper briefly covers the basic concepts
that will help us to look at power and power losses from the view point of power dynamics. The
power system modeling and economic dispatch used in this paper helps to drive home the fact that
power grid system is fast becoming a computerized control system. For example, small incremental
changes can be simulated on one input parameter at a time while looking at all the affected output
parameters in the system. By taking this approach, we are aiming at a much higher operating
precision and less human error which can be caused by countless pages of mathematical calculations
done by hand.

2
Economic Dispatch Common Formulae

Fuel Cost Curve: Ci ( PGi )     PGi   PGi
2

Power Flow Constraint:  PGi   PD

Cost function: Ci ( PGi )     PGi   PGi
2

Ci ( PGi )
Compute Incremental cost: ICi   
PGi

Ci ( PG1 )       C ( P )               C ( P )
The final system operating points:                      * L1  i G 2 * L2 .........  i Gi * Li
PG1              PG 2                  PGi

Constraints:     G  PG  PL  0   in1 PGi  PL  0

Ignore losses: L  C   ( in1 PGi  P )

Ci
If Lambda is the same for each generator:             i              2 i PGi   i
PGi
Ignoring losses: Ci   i PGi   PGi   i = incremental cost per unit
2

Lagrange is: L  C   (in1 PGi  PL )

L   C            Ci
The Partial derivatives are:              (1)         0               i  1, 2,...., n
PGi PGi           PGi

Ci
Incremental Cost:       i            2 i PGi   i
PGi

i   i
Lagrange can be formulated as: Pi                  or λ = dCi/dPi = βi + 2γiPi
2 i

3
1.0     Cost of Generating Electrical Power

The three main sources that are used to determine the cost of electrical energy (MW) generation are
facility construction, ownership cost and operating costs. Please note that energy sources, such as
solar, wind, nuclear, and hydro are not included in the cost components. Since operating cost is the
most significant of the three main sources, the focus will be on the economics of the operation. The
operating cost is dominated by the fuel cost, although labour is also a key component. The goal of
power system economic dispatch is to maximize system efficiency and minimize system losses that
cannot be billed or pass on to customers. Table-1.1 and Table-1.2 below the average costs of fuel
such as coal, petroleum and natural gas. The cost of uranium is around \$0.65/MBTU or \$0.65 per
Million British Terminal Unit (MBTU).

Table 1.1 – Fuel Cost
Receipts, Average Cost, and Quality of Fossil Fuels for the Electric Power Industry,
1991 through 2002, obtained from
All
Coal                                Petroleum                   Natural Gas        Fossil
Fuels
Avg.                             R    Avg.             Average Average
Period Receipts        Average Cost                  Receipts Average Cost                Receipts                R
Sulfur                                Sulfur             Cost    Cost
(cents/                Percent             (cents/           Percent
(thousand       6                           (thousand       6  (dollars/         (thousand (cents/ (cents/
10     (dollars/ton) by                    10                 by                6       6
tons)                           Weight     barrels)          barrel) WeightR     Mcf)    10 Btu) 10 Btu)
Btu)                                        Btu)
1991      769,923     144.7           30.0      1.3 172,051 252.7          15.9       1.1 2,630,818    215.3   160.2
1992      775,963     141.2           29.4      1.3 147,825 251.4          15.9       1.2 2,637,678    232.8   158.9
1993      769,152     138.5           28.6      1.2 154,144 237.3          14.9       1.3 2,574,523      256   159.4
1994      831,929     135.5           28.0      1.2 149,258 242.3          15.2       1.2 2,863,904      223   152.5
1995      826,860     131.8           27.0      1.1     89,908 256.6       16.1       1.2 3,023,327    198.4   145.2
1996      862,701     128.9           26.5      1.1 113,678 302.6          18.9       1.3 2,604,663    264.1   151.8
1997      880,588     127.3           26.2      1.1 128,749        273    17.28       1.4 2,764,734      276     152
1998      929,448     125.2           25.6      1.1 181,276 202.1          12.7       1.5 2,922,957    238.1   143.5
1999      908,232     121.6           24.7      1.0 145,939 235.9          14.8       1.5 2,809,455    257.4   143.8
2000      790,274     120.0           24.3      0.9 108,272 417.9          26.3       1.3 2,629,986    430.2   173.5
R                                                                               R
2001      762,815    123.2            24.7      0.9 124,618 369.3          23.2       1.4 2,152,366 448.7        173
2002      884,287     125.5           25.5      0.9 120,851 334.3          20.8       1.6 5,607,737      356   151.5

Source: Energy Information Administration, Form EIA-423, “Monthly Cost and Quality of Fuel for
Electric Plants Report,”

4
Aside: Cost of Electricity Generation

Table- 1.2      Cost of Electricity Generation 1994 Compared to 2003

Technology 1         1994 Cost of Electricity                   Current Cost of Electricity
(cents/KWh)                                (2003) data, cent/KWh

Hydroelectric        0.31 to 4.4                                0.25 to 2.7
Nuclear              2.5                                        1.4 to 1.9
Coal                 1.9 to 2.3                                 1.8 to 20
Natural Gas          2.5 to 11.7                                5.2 to 15.9
Solar                16.4 to 30.5                               13.5 to 15.9
Wind                 7.6                                        4.6

Source: California Energy Commission:

http://www.energy.ca.gov/electricity/comparative_costs_vl.html

All values are in 2003 dollars. Nuclear costs do not include decommissioning cost, which is < 0.1
cents/KWh. Please see Appendix for more current energy cost.

The data in Table-1.2 shows that a renewable energy source such as solar and wind are still very
expensive in contrast to all the other traditional methods of generating electricity. In short, the cost of
producing electricity by Solar or Wind Turbine system is not economically feasible to supply.
Therefore, it is yet to be determined if the initial implementation cost will drop low enough to make it
the source of choice for the future electricity market.

Major Components in Economic Dispatch
The three major components in economic dispatch are system efficiency, heat rate and power output
in megawatts. Moreover, in order to understand the concept of economic dispatch, we should start
with the cost of fuel. Table-1.1 and Table-1.2 show that the cost of fuel varies significantly with
different types of fuel.

5
Economic dispatch allows us to look at the cost of the fuel, labour cost, power generation cost and
transmission line losses. Therefore, the power generated will be PG = PL + PTL
Where PL = total system load
PTL = total system transmission losses per unit
Knowing PL + PTL can allow us to predict the total PG required for an area demand and various bus
and substation commitments. This information provides no insight in how the individual area
generator will share information on a common grid to meet the regional grid demand at various hours
such as peak and off-peak window. Furthermore, it is not economically feasible to run fossil fuel
systems when there is very low demand for power. Therefore, the process of load scheduling
becomes a major process in how a power system operates to stay economically feasible. In
minimizing cost without compromising system reliability and market security, one must take into
account one basic constraint of each generator which is Pmin ≤ PGi ≤ Pmax where the upper boundary
Pmax is directly related to upper rating of the generator. On the other hand, the lower boundary Pmin is
directly linked to thermal consideration that is required to maintain the boiler steam which drives the
turbine.
Power plants measure results in terms of efficiency, as seen in the equation below.

sec        watts
*106 *
P * T * 3600
                       hr        MW
BTU         lb           joules
12, 000     * 2000     *1054.85         * ytons
lb        ton           BTU

Some plants have an overall efficiency of 34% which is very good for this type of industry. Plant
efficiency varies as a function of the generator power level. The efficiency level shown in the above
equation is power output in megawatt hour (MWHR) divided by the energy input in British Thermal
Unit (BTU). The graph of MWHR versus power output, as seen in Figure 1.4 on page 14, is used to
monitor the health of the system. For example, if the heat rate is high then the system is running
inefficiently because a high heat rate will produce a low efficiency.

6
1.1 Monitoring System efficiency via Turbine Heat Rate (THR) Test:
The THR test was originally designed as a diagnostic type of test. This test provides a broad range of
data on the thermal efficiency and operating costs of the turbine steam cycle, and consequently of the
entire unit. Moreover, it is standard practice to run a THR test in order to determine the overall
efficiency of the system (since efficiency is the reciprocal of heat rate). Figure 1.0 below shows the
I/O curve for a 500 MW system after a major a boiler and turbine overhaul. The post overhaul test
was done over the entire load range. A minimum of four test points are required to monitor the I/O
curve for design compliance. This test is normally done at 25%, 50%, 75%, and 100% of the
maximum capacity (MC) rate. In cases where it is not feasible or possible to run tests in the 25%
range, the heat input from the latest I/O curve is added to the data points and used in curve fitting.
Test readings are taken over periods ranging from 1 hour to 2 hours of steady run.

I/O Equation (Post Overhaul)
5000

4500

Y = 0.001793*x. 2 + 8.05279*x + 399.3
4000
Heat Input (GJ/h)

3500     R2 = 1.0000

3000

2500

2000

1500

1000
100    150          200       250        300         350   400   450   500
Net Output MW
Figure 1.0 – Input. Output (I/O) Curve

In the real world, pre and post overhaul turbine heat rate tests are performed to monitor the system
efficiency at different levels of power output. In addition, heat rate tests are periodically run to look
at the unit efficiency throughout the year. The system condenser cooling process can allow an
additional 1.5% to 2% change in efficiency due to the change in temperature. Furthermore, 2%
change in efficiency can be the deciding factor that will determine if the generating plant will make
or lose money.

7
The Heat Rate (H) equation shown below shows that the efficiency is inversely proportional to the
heat rate. A typical coal fired plant heat rate (HR) is about 10.5 * 10^6 BTU/MW.h. The operating
cost of a generating plant or unit includes cost of fuel, labour, supplies, and maintenance is
commonly known as Ci = Fi Pi
Where Ci = operation cost of unit in \$/h

Fi = operation cost of unit in \$/MBTU

Pi = input unit power in MBTU/h

BTU       lb          1BTU
12, 000         2000     * ytons
H                    lb      ton         106 BTU
P *T

1        3600     3.41
H            *           
        1054.85    

The cost of fuel can be highly volatile and since Fi depends on the type of fuel such as coal, oil, gas
and nuclear – it is very hard to empirically determine a fixed price per MBTU. A rough average for
Fi in the power industry is about \$1.50 per MBTU. In the power industry, the fuel cost curve is the
standard that is the most accurate to analyze cost as it relates to heat rate. The cost is given as a
second order expression shown as: Fuel Cost Curve: Ci ( PGi )     PGi   PGi
2

Where i = generator i, one of the number of units
Ci = operating cost of unit in \$/h
PGi = electrical power output of a generator i in per unit on a common power base
Alpha, Beta & Gamma are in units of dollars ( \$/h)
The cost curve is commonly called the I/O curve. However, Fuel Cost Curve is the I/O Curve
multiplied by the cost of fuel. For example, if the cost fuel is \$1.90 per MBtu
then (A + B*MWn + C* MWn2) * 1.90 would be the Fuel cost curve. The concept of I/O will be
explained below, and coefficients A, B & C are given by design manufacture to determine the heat
rate.

8
Input & Output tests and calculations were used to provide system performance data, in the form of
input and output equation. I/O curves were primarily used in the calculation of the total incremental
cost. The following two methods were used to establish the I/O curve for a given unit.
(1) Direct I/O Method of Measurement: This is the most commonly used method because it only
requires measurements of fuel flow, calorific value of fuel samples taken during tests of net unit
output and reference operating conditions. The test is ideally suitable for gas and oil-fired units.
However, this test is not ideal for coal-fired units.
(2) Indirect I/O Method of Calculation: This method of calculating I/O curves are more complex than
the direct method. However, it is much more accurate for coal fired unit. I/O results are derived from
two sets of separate performance tests namely: Thermal Heat Rate Test (THR) and Steam Generator
efficiency test. Both tests require at least four test points, ranging from 25% to 100% of the
maximum capacity Rate (MCR) to calculate the heat inputs for given net outputs. Results are then
“curve-fitted” to yield the desired format of I/O equations, where the required input is expressed as a
function of the net load desired:
I = A + B*MWn + C* MWn2
Where I        Is Unit Heat Input (GJ/h)
A, B, C        Is the Coefficients of I/O Equation
MWn            Is Net Unit Output (MW)

Frequency of Updating of I/O Curves:
Currently, there is no set frequency for updating I/O curves applicable to all units. The main reason is
the large variation in Capacity Factors of individual power plants and units. Instead, I/O curves are
typically updated after major system overhauls, or when there are indications of measureable changes
in thermal efficiency of individual turbines, or steam generators. Generally, I/O tests have been “free
by-products” of diagnostic and optimization testing of turbines and boilers.

9
Calculating Thermal Heat Rate (THR)
Thermal heat rate is required by a turbine-generator system to produce power at the generator
terminal. Heat rate is measured in kilojoules per kilowatt (KJ/KWh). The lower the heat rate, the
more efficient the turbine cycle, the more cost effective it is to operate, and the longer life cycle
expectancy. The turbine efficiency is the reciprocal value of the THR, which is the cycle Eff.          =
100 * (3600/THR) where 3600 KJ/h = 1KW. Figure 1.1 shows the operating points for the
calculation.
There are several different approaches to measure and analyze THR. However, the method used for
coal, gas and nuclear plants is shown in the equation below.
THR was calculated using the following equation:
GJ/h           =               W1 (H1 – h1) + Wr ( Hr – hr)
KWH                                  P + PB

Where:         W1 = Steam to HP Turbine Stop Valve             Kg/h
Wr = Steam Flow to Re-heater                    Kg/h
H1 = Enthalpy of Steam Supplied to the High Pressure Turbine Stop Valve (KW/Kg)
hr = Enthalpy of Steam Supplied to the Intermediate Pressure (IP)Turbine before
Interceptor Valve (KW/Kg)
h1 = Enthalpy at the Feed-water at HP outlet (KW/Kg)
hr = Enthalpy of Steam at HP Turbine Exhaust (KW/Kg)
P = Net Generated Output                                (KW)
PB = Equivalent Electrical Output from Boiler Feed Pump Turbine (KW)

In addition, other methods are used to measure THR, such as the Input and Output (I/O) method and
the condensate flow measurement test. The I/O method can be linked to a computerized software
that is programmed to do live trend mathematical analysis of the THR through a non-linear equation
such as:
Input (GJ/h) = A + B * MWnet + C*MWnet^2.
The result of I/O is generally analyzed in terms of the R-Square value.

10
1.2 Typical Power Plant showing the required data for heat rate calculation

Figure-1.1    Calculating Heat Rate

GJ/h          =              W1 (H1 – h1) + Wr ( Hr – hr)
KWH                                P + PB

1250.446(2790.6 – 750.8) + 59.93(2790.6 -1135.0) * 3600)
935300
Thermal Heat-Rate    =                      10199 KJ/KWh

This method of calculating heat rate is common to Group-1 (Coal Fire), Group-2 (Gas Fire, and
Group-3 Nuclear. Flow diagrams are often used to navigate the decision process. See Figure 1.2 and
Figure 1.3 below.

11
1.3   Analyzing the THR Process

Figure 1.2 THR Test

Method

12
1.3.1        Heat Rate Verification & Validation Using Calculus

Minute by minute loading also known as Incremental Heat Rate (IHR) is used to analyze the station
Operation Efficiency Factor from the ratio of
Actual Net Unit Heat Rate (KJ/KWh) or MWHR/MBTU to the Reference Net Unit Heat Rate
(KJ/KWh). The graph in Figure-1.4 shows the typical behaviour pattern for heat rate.

Figure 1.4 Heat Rate Vs Power (MW)

Actual Net Unit Heat Rate (KJ/KWh)
This can be verified from the equation shown below:
dI
IHR         b  2CW  3dW 2
dW
Substituting W for L

dI  ( IHR) dL

 II 1 dI  I 2  I1   ll1 ( IHR)dL
2                     2

The minute by minute loading allows us to observe and analyze the heat rate that is required between
selected points shown below. By using calculus one can verify and validate the heat rate used
between selected points with reference to the manufacturer design heat rate. A five point test will be
used to check and analyze how much deviation is present after a major unit overhaul. For example, a
generating unit rated at 610 MW could experience a planned maintenance system reliability test to

13
see if the heat rate is within the range of the manufacturer design specifications. See the 5-point test
below.
5-Point calculated steps are:
 610 [421.3  8.0l  0.00192l 2 ]dl =
450                                                         832756 BTU/hr

 350 [421.3  8.0l  0.00192l 2 ]dl =
610
1.22577E6 BTU/hr

 610 [421.3  8.0l  0.00192l 2 ]dl =
250                                                         1.52534E6 BTU/hr

100 [421.3  8.0l  0.00192l 2 ]dl =
610
1.80789E6 BTU/hr

 0 [421.3  8.0l  0.00192l 2 ]dl =
450
950683 BTU/hr

The numbers shown in Table-1.1 reflect only the cost of fuel to operate the generator(s), and they do
not reflect the actual cost of generating electricity because the actual outputs from the process of
producing electrical energy is subjected to a substantial loss of about 66%. Therefore, any power
plant that is running above 34% is considered to be very efficient.

1.3.2    Matlab Software Code & Fuel Cost Curve

Matlab software is an indispensable tool that is often used to solve power system problems. By
choosing four points in the system power output and the associated cost related to each points, the
result is the input and output curve shown below in Figure-1.5.

EDU>> p=[40 60 80 100]';
c=[650 850 930 1200]';
X=[ones(size(p)) p p.^2];
a = X\c
T = (0:1:100)';
Y = [ones(size(T)) T T.^2]* a;
plot(T,Y,'-',T,Y,'o'),grid on

a=
494.5000
2.5250
0.0438

Where “a” = the quadratic equation that represent the cost curve shown in C1 below.

14
C1 = 494.50 + 2.5250P1 + 0.0438 P12 (\$/MWhR)

Typical operating cost curve
1200

1100

1000
Operating cost in \$/h

900

800

700

600

500

400
0     10    20     30      40      50      60          70   80   90   100
Output Power (MW)
Figure 1.5 Typical Operating Cost Curve (Power MW)

In addition, the code creates the Quadratic equation for the cost curve, this curve is critical to
the lifecycle of the system, because all future curves will be taken with reference to the date
and time when this curve was generated. Below is a practical example that will allow one to
understand how to go from the fuel cost curve to the cost of generating power by including
input and output parameters.

The fuel cost per hour can also be shown as:

Fuel Cost Curve: Ci ( PGi )     PGi   PGi . For example, the cost curve at 100 MW is
2

(494.50 +2.525PG + 0.0876P2G) = 494.50 + 2.525 * 100 + 0.0876 * 100^2) = \$1623/hr

15
Ci ( PGi )
In addition, one can see how to implement the Incremental cost: ICi                  =
PGi
2.525 + 0.1752 * 100 = 20.045/MW

Another example, if a generating unit fuel input in millions of BTU/h is expressed as a function of
the output PG in megawatts as 0.032P2G + 5.8PG + 100), then as a System Engineer your duty is to
determine the following:
(a) The incremental cost equation in dollars per megawatt hour as function of the generating
power (PG) based on the fuel cost of \$2.00 per million BTU
(b) The cost of fuel per megawatt hour if the (PG) = 150 MW.
(c) The additional fuel cost per hour to raise the output of the unit from 150 MW to 151 MW.
Solution:
(a) The fuel cost curve in dollars per MWh is
FC          = (0.032P2G + 5.8PG + 100) * 2
= 0.064 P2G + 11.6PG + 200\$/MWh
Ci ( PGi )
Incremental cost: ICi                 = 0.128 PG + 11.6 \$/MWh
PGi
(b) The cost of fuel when PG = 150 MW is:

(0.064(150)2 +11.6 PG (150) + 200)/150 = 22.6 \$/MWh
(c) The Approximate ICi that is required to raise the power by 1MW from 150 MW to 151 MW is =
0.128 (150) + 11.6 = \$30.8 \$/MWh

Therefore, by using the Incremental Cost (CI) equation, one can formulate the equations into
computer code to automate the process to minimize potential error when making hard and explicit
decisions.

16
2.0           Power Flow and Economic Dispatch

This section covers several mock-up scenarios, examples and concepts that deal with a single
generator, two generator and three generator concepts and applications. In order to continue
exploring economic dispatch, the basics of power flow will be covered since it is directly linked to
the topic of economic dispatch. The study of power flow is an indispensable process in planning and
designing the future expansion of power systems as well as determining the best operating system.
The general information obtained from the study of power flow is the voltage magnitude and phase
angle at each bus in conjunction with real and real and reactive power flowing in the line.

There are three types of buses: load, voltage-controlled and slack. Each non-generator bus is called
the load bus where both Pgi and qi are at zero and the real and reactive bus Pdi and Qdi are power
and reactive bust respectively, their associated loads drawn are negatively input into the system.
The voltage controlled bus is any bus in the system at which the voltage magnitude is kept constant
and controlled. In short, the control is done with the generator excitation unit. Therefore, for each bus
where there is a generator, the generator must meet the condition of the power flow network to
maintain the credibility of the grid. This means that a generator must stay within a very tight window
of variance such as 5% tolerance in term of voltage swing. At the slack bus, the voltage angle of the
slack serves as reference for all the other buses, and mismatches are not defined. P and Q are not
scheduled at the slack bus, and the power loss at each bus in the system is given as:
N                           N                               N
PLoss   Pi                       Pgi                          P      di
i 1   i                     i 1                            i 1

Re al power loss              Total Generation                 Total load
Maximum and minimum generated reactive power is ignored at the slack bus.
Figure-2.1 shows a single generator system with a general concept of power flow that leads us into
power system economic dispatch. The examples shown will also briefly overview the topic of per-
unit system and how it relates to power system.

17
2.1         Economic Dispatch System One Generator Concept

Figure-2.1        Single Generator Concept

(1)      Assume the generator below voltage is set at 10% above its rate value, and the load
connected to bus D is equal to 5 MW at a pf 0.9 lagging. Compute the per unit model for
the system shown below.
(2)      Find the load voltage at Bus-D

(3)      Find the active and reactive power supplied by Bus-A

(4)      Find the active and reactive losses

(5)      Find the economic loss from the active power loss

(6)      Use NPV to determine if this project is viable for investment

Solution:

Per-Unit System
In many engineering applications, it is useful to scale, or normalize quantities and dimensions. This
practice is commonly used in the power system industry, and the standard used is called per-unit
system. It is common for impedance value to convert from an old base to a new base, and in order to
do so we must take into consideration the other components in the system such as Voltage (V),
Current (I) and Power (W).
The formulae below will give a brief overview of the concept. It is very common practice to express
per-unit (pu) values as percentage such as 1pu =100%.

18
A typical conversion, is:
Per-unit value = (percent value/100).

(1) Solution
actual value in SI units
Per  unit value 
base value identical SI units

old base value
New pu value  old pu  value (                  )
newbase value

(actual im pedance, ) * (base KVA)
Per  unit impedance 
(base voltage, KV ) 2 *1000

base KVgiven          base KVAnew
Per  unit Z new  per  unit Z given (                      )2 (                 )
base KVnew           base KVAgiven

Vbase                 V 2base
Z base                                
I base                 Sbase

S base
I base             
Vbase

Transmission line in power system operates in units of kilovolt (KV), megawatts, kilovolt-amperes,
megavolt-amperes and ohms. These quantities are often expressed as per-unit value with respect to a
given base or reference base. For example, the base voltage is given as 120 KV in a system where
other voltages such as 108 KV, 120 KV and 126 KV become 0.90, 1.00 and 1.05 respectively. On the
other hand pu values can also be expressed as 90%, 100% and 1.05%. The pu value of any given
quantity is defined as the ratio of value expressed as a decimal or percentage.
Figure-2.2 below shows a typical generating system with the basic component that will allow one to
understand the basic concept of power flow. Economic dispatch is an integral part of power flow,
where one can look at economic dispatch without including the power flow components.

19
Figure-2.2   Power Flow System

20
13.2                           69
Sb  10 MVA, chooseVb, A  132(            )  13.2 KV ,Vb, D  132( )  66 KV
132                           138

1322
Zb, BC          1742.4
10

10               Z       10  100i
Z AB  0.10i( )  0.2i, Z BC  line             0.006  0.06i
5               Zb, BC 1742.4

6910
X T 2  0.08(        )  0.087i             Z CD  X T 2  0.087i
6910

5                          
Sload         cos 1 0.9  5.5625.8 MVA  5 MW  2.42 MVar
0.9



(2)         Ztotal  Z AB  Z BC  Z CD  0.2i  0.006  0.06i  0.087i  0.006  0.347i

1                
Y              2.8889  0.05  2.8796i
Ztotal

21
(3)                   Solution Below

GAUSS SEIDEL METHOD:

 Y1 1 Y1 2                          0.05 2.87960i 0.05 2.87960i
YBUS                                                                      
 Y2 1 Y2 2                          0.05 2.87960i 0.05 2.87960i

1    S
V2 (i  1)        [ 2   Y21V1 ]
Y 22 V2 (i)

v2 
1                                                   
 0.500 0.2419i [ ( 0.05 2.87960i]  1.1
                                   )       
( 0.05  2.87960i ( 1.0  0i)
)                                             

v2  1.013 0.172i

v2a 
1                                                       
 0.500 0.2419i [ ( 0.05 2.87960i]  1.1
                                   )       
( 0.05  2.87960i ( 1.013 0.172i
)                 )                              

v2a  0.988 0.151i

v2b 
1                                                        
 0.500 0.2419i [ ( 0.05  2.87960i]  1.1  0.988 0.157i
                                    )       
( 0.05  2.87960i ( 0.988 0.151i
)                 )                                

v2c 
1                                                        
 0.500 0.2419i [ ( 0.05  2.87960i]  1.1  0.987 0.156i
                                    )       
( 0.05  2.87960i ( 0.988 0.157i
)                 )                                  

Table 2.1 Voltage Iteration Summary
Iteration # V2
0           1.013- 0.172i
1           0.988- 0.151i
2           0.988- 0.157i
3           0.987- 0.1562i

Where V2 converges at 0.987-0.156i

22
(4) Solution

S1 = V1(Y11 *V1 * + Y12 *V2*)

= 0.5017 + j0.3492

Active power P1 = 0.5017 x 10 = 5.017MW
Reactive power Q1 = 0.3492 x 10 = 3.492MVar

Active power Ploss = P1 - P2 = 0.5017 - 5 = 0.017MW
Reactive loss Qloss = Q1 – Q2 = 3.492 – 2.42 = 1.072 MVar

(5) Solution Economic loss based on active power

Active power Ploss = P1 - P2 = 0.5017 - 5 = 0.017MW
= 0.17 MW
Assuming the cost of generating electricity is: \$25 per MWh

Therefore an annual economic loss would be: 0.017 MW * \$25 *24hr * 365days = \$3723

In addition, if the shareholder initial investment cost is \$10 M and there was no power loss, then the
\$3723 would be cash flow.
Now we can evaluate the economic dispatch process in terms of NPV
1  (1  rp ) n
NPV  [                  ]R  Investment Cost
rp
Where rP = % Return for the project
R = Cash Flow
N = Number of years

 1  ( 1.05  5
)
` NPV                    3723 10  106  9.984 106
 0.05 

The NPV Decision Rule is:
Accept a project if its NPV > 0
Reject a project if its NPV < 0
Indifferent where NPV = 0

23
2.2 Incremental Cost (IC)

Incremental cost is the slope of the fuel cost curve, and the unit of IC is in dollars per megawatt hour
(MWh). IC tells us how much it will cost to operate a generator to produce an additional 1MW of
power.
Let us assume a case where there is two generators no line losses, no generator limit and IC1 > IC2.
This means that for an additional 1MW, generator-1 operating cost is more than generating-2
operating cost. If our objective is to minimize operation cost, which always the business case, then it
is reasonable to reduce power output at generator-1 and, in return, increase the output of generator-2.
Then the optimal condition would be that the operation cost from generator-1 and generator-2 should
be the same. See example below in Figure-2.3.

2.2.1    System-2      Two Generator Concept

Figure-2.3     Two Generator Concept

24
Two Generator fuel cost curve equations are used to set Generator-1 & Generator -2 economic

Optimal Power Condition without line loss         =

Two generators with the following cost curve.
C1 (PG1) = 500 + 45PG1 + 0.01P2G1                       C2 (PG2) =2000 + 43PG2 + 0.003 P2G2
C1  PG1                                               C2 PG 2 
IC1                   45  0.02 PG1                   IC 2                 43  0.006 PG 2
PG1                                                     PG 2

PG1 + PG2 = PG2 = 500

IC1 = IC2 = 45 + 0.02 PG1 = 43 + 0.006 (500 - PG1)

PG1 = 38.4615 MW                    PG2 = 461.539 MW

PG1 + PG2 = 38.4615 + 461.539 = 500 MW

IC1 = 45 + 0.02 PG1 = 45 + 0.02 * 38.4615 = 45.769

IC2 = 43  0.006 PG 2 = 43 + 0.006 * 461.539 = 45.769

IC1 = C2 = \$45.769/MWh

25
Penalty Factor and Power Loss

On the other hand, if we have a scenario where the fuel cost curve is given as:
C1(PG1) = 600 + 15 * PG1 + 0.05 * (PG1)2
C2(PG2) = 800 + 20 * PG2 + 0.03 * (PG2)2
Assume the penalty factor for the first generator (slack bus) is set to1, and the second generator (PG2)
 Plosses
partial derivative of the losses is given as            0.05 .
 PG 2
The Systems Engineer is required to calculate the dispatch values of PG1 and PG2 for
load + Losses of 1000 MW.
Solution:
1
Li 
Power Loss is                     P
1  loss
PGi
2C1                            2C2
 15  0.1P1                     20  0.06 P2
2 PG1                          2 PG 2

2C1                           2C2
L1                           L2  
2 PG1                         2 PG 2

1       1                             1         1
L1                                  L2                        1.0526
2 PLoss 1  0                         2 PLoss 1  0.05
1                                    1
2 PG1                                 2 PG 2

  15
  15  0.01  PG1 
0.1

             0.95  20
(20  0.06 PG 2 )1.0526    PG 2  (              20) 
1.0526                0.06
0.06

15  0.1PG1  1.0526(20  0.06(1000  PG1 ))

14.25  0.095 PG1  20  0.06(1000  PG1 )

PG1  424.194 MW               PG 2  575.806 MW

26
By using the same cost curve given in the first scenario shown above, one can analyze the system to
show the potential cost saving that is possible without losses. See the tabulated results below.

C1(PG1) = 600 + 15 * PG1 + 0.05 * (PG1)2
C2(PG2) = 800 + 20 * PG2 + 0.03 * (PG2)2

2C1                        2C2
 15  0.1P1                 20  0.06 P2
2 PG1                      2 PG 2

IC1 = IC2 = 15 + 0.1 PG1 = 20 + 0.06 (1000- PG1)
PG1= 406.25 MW                PG2 = (1000 - PG1) = 593.75 MW

The cost associated with 1C1 & 1C2 are:
1C1 = 15 + 0.1 * 406.25 = \$55.625/MW
1C2 = 20 + 0.06 * 593.75 = \$55.625/MW

In contrast, the cost associated power loss in scenario-1is shown below:
1C1 = 15 + 0.1 * 424.194 = \$57.419/MWh
1C2 = 20 + 0.06 * 575.806 = \$54.548/MWh
1C1 cost per megawatt shows an increase of \$1.8 Per MWh when losses at generator-2 were set to
5%.

Another Method used to set the incremental cost (IC) is the Lagrange Multiplier (2.2.3). This method
uses the matrix approach that will result in a \$-value for Lambda. With this \$-value the aim is to get
IC1 for generator one equal to IC2 for generator two even when the two generators may not be the
same. This approach offers the most optimum cost benefit when (IC1 = IC2).

27
Two Generators Alternate Solution Using Lagrange Multiplier

System PD = PG1 + PG2 = 1000MW

And C1(Pg1) = 1500 + 20PG1 + 0.01P2g1

C2(Pg2) = 500 + 15Pg2 + 0.03P2g2

Using the Lagrange multiplier method, we know:

dc1(Pg1) – λ = 20 + 0.02Pg1 – λ = 0

dc2(Pg2) – λ = 15 + 0.06Pg2 – λ = 0

1000 – PG1 – PG2 = 0

Solve three linear equations:

20 + 0.02Pg1 – λ = 0

15 + 0.06 Pg2 – λ = 0

1000 – Pg1 – Pg2 = 0

Solution to the Matrix yields: Pg1 = 687.5 MW

Pg2 = 312.5 MW

λ = \$33.75/MWh

Comparing the results with previous method used, yield similar results
IC1 = IC2 = 20 + 0.02Pg1 = 15 + 0.06(1000- Pg1),
Pg1 = 687.5
Pg2 = 1000- 687.5) = 312.5
Cost = 20 + 0.02 * 687.5 = 15 + 0.06 * 312.50 = \$ 33.75

28
Two Generator (GEN) Power Flow & Economic Dispatch System Example

The example shown in Figure 2-4 below covers the general concept of how we go from power flow
to economic dispatch

Figure-2.4                    Gauss-Seidel-Method

Bus 1 is slack bus: V =1@ 0° degree
Schedule power at Bus 2 is 1.2 pu
Schedule power at Bus 3 is 1.5 pu

Calculate:

1.        YBus model
2.        Bus voltage using Gauss-Seidel Method
3.        Power mismatch at Bus 2 & 3
4.        Power supply by the swing bus
5.        Power loss of transmission lines
6.        Assume there was no loss, what would be the additional revenue for this system over a 5-
year period @ 5% and \$20/MW

Solution

(1)        Y
11        
   Y
12
Y
13        Y
22


   Y
12
Y
23   

Y
33


   Y
23
Y
13   

                

 Y11 Y12 Y13 
               14 4 10 
Y     Y Y Y    4 9 5 
BUS  21     22  23             
 Y31 Y32 Y33   10 5 15 
             

29
1 S k      k 1               n
(2)    Vk (i  1)     [        YknVn (i  1)   YknVn (i)]
Ykk Vk (i) n1              n  k 1

1    S
V2 (i  1)        [ 2   Y21V1 (i  1)  Y23V3 (i)]
Y 22 V2 (i)

1    S3
V3 (i  1)      [         Y31V1 (i  1)  Y32V2 (i  1)]
Y 33 V3 (i)

Table 2.2 Voltage Iteration
Iteration                          Voltage (V2)                     Voltage (V3)
0                                  1                                1
1                                  1.1333                           0.9
2                                  1.0621                           0.9333
3                                  1.0885                           0.9224
4                                  1.0784                           0.9180
5                                  1.0781                           0.917

V2 = 1.078 p.u., V3 = 0.917 pu

S2 = V2 (Y21 * V1* Y22 * V2 * + Y23 * V3 *)

= 1.078(-4x1 + 9 x1.078-5.0917)

=      1.2041

S3 = V3 (Y31 * V1* Y23 * V2 * + Y33 * V3 *)

= 0.917(-10 x 1 - 5 x1.078 + 15 x.0917)

= 1.4993

30
Power mismatch at Bus 2

∆S2 = S2cal – S2sch = 1.2041 – 1.2 = 0.041
∆P2 = P2cal – P2 = 1.2041 – 1.2 = 0.041
∆Q2 = Q2cal – Q2sch = 0 – 0 = 0

Power mismatch at Bus 3

∆S3 = S3cal – S3sch = 1.4993 – (-1.5) = 0.0007
∆P3 = P3cal – P3 = 1.4993 – (-1.5) = 0.0007
∆Q3 = Q3cal – Q3sch = 0 – 0 = 0

(4) Power supply by swing bus

S1 = V1 (Y11* V1 * Y12 * V2 * +Y13 * V3 *)
∆P2 = 1 ∙ (14 ∙1-4 ∙ 1.078-10 ∙ 0.917)
= 0.518

For a resistive power system, P1=S1=0.518, Q1=0.

(5) Power loss of transmission lines:
S loss = P loss = S1 + S2 + S3
= 0.518 + 1.2041 – 1.4993
= 0.223
(6)
1  (1  rp ) n
NPV  [                 ]R  Investment Cost
rp
Where rP = %Return for the project
R = Cash Flow
N = Number of years
Given: 5% interest rate, \$20/MW and a power loss of 0.233 pu (22.3 MW)

Cash flow = 20 * 22.3 * 24 * 365 = 3,906,960

 1  ( 1.05  5
)
NPV
                3906960 0  1.692 107
 0.05 

NPV = \$16.92 M (If no power loss in the system)

31
2.3 System-3 Three Generators Economic Dispatch Applications

In economic dispatch systems where there are more than 2-generating units, two pieces of data will
be available for us to work with. They are total load and IC curves of each unit. By using the Lambda
Iteration Method, we can find Lambda which is the cost that will produce optimum condition. This
approach is carried out by using an iterative procedure that will allow us to:
(1) pick an initial value for Lambda
(2) find the corresponding output power each generating unit.
(3) Check if output power is less required load
In addition, if the total power is less than the load, then we would increase Lambda and go step two,
Else, condition met (stop).
Lambda Iteration Method & Example

Pick  L and  H Such that
m                               m

P
i 1
Gi   ( L )  P  0
D       P
i 1
( H )  P  0
D
Gi

While  H   L   Do
( H   L )
M 
2
m
If     P
i 1
Gi   ( M )  P  0 Then  H   M
D

Else  L   M

Assuming a three generator system with the following:

IC1 (PG1) = 15 + 0.02PG1 = λ \$/MWh
IC2 (PG2) = 20 + 0.01PG1 = λ \$/MWh
IC3 (PG3) = 18 + 0.025PG1 = λ \$/MWh

and with a constraint PG1 + PG2 + PG3 = 1000 MW.
Rearranging the equations as a function of λ, PGi (λ), to allow
PG1 (λ) =   15             PG2 (λ) =   20
0.02                     0.01
PG3 (λ) =   18
0.025

32
Now the task is to find what value of Lambda that will allow us to meet the system demand (1000
MW).
m
Pick  L so that            P
i 1
Gi   ( L )  1000  0 and

m
Pick  H so that            P
i 1
( H )  1000  0
Gi

m
Try  L  18 Then  P (18)  1000 
i 1         Gi

  15             20               18
                                        1000  650
0.02             0.01              0.025

m
Try  H  32 Then  P (32)  1000 
i 1        Gi

  15             20               18
                                        1000  1610
0.02             0.01              0.025

Assume convergence tolerance   0.05\$ / MWh
The iterate since |  L   H |                             0.05
 
L         H
18  32
M                                        25
2                    2

m
Then since           P
i 1
(25)  1000  280 set  H  25
Gi

and         |25-18| > 0.05

L  H                18  25
M                                        21.5
2                    2

m

P
i 1
(21.5)  1000   385 set  L  21.5
Gi

Continue the iteration until |  H   L |  0.05
The convergence value of lambda (   ) is 23.53 \$/MWh
Upon knowing   we can calculate the PGi

33
23.53  15
P 1 (23.5) 
G                         426.5 MW
0.02

23.53  20
(23.5)                 353 MW
0.01

23.53  18
P 3 (23.5) 
G                         221.2 MW
0.025

P 1  P 2  P 3  1000 MW
G     G     G

The results are within the constraint of PG1  PG 2    P 3  1000 MW
G
given at the out set.
This method starts with a Lambda valve below and above the optimal value shown here.
The main disadvantages of this approach in comparison to the direct method are: (a) more
challenging (b) more time consuming. The main advantage in using this method is that it works for
both linear and non linear application. While on the other hand, the direct method is only accurate for
linear applications.

Another method is to use the direct formula for Lambda which will yield similar results. This direct
method works well if the incremental cost curves are linear and all the generators are below their
limits. See formula below.

ngen
P
Demand
         
i
i1        2i
 
ngen


1
2 i
i1

For example, if generating station has a three generator system with the following,
IC1 (PG1) = 15 + 0.02 PG1 = λ           \$/MWh
IC2 (PG2) = 20 + 0.01 PG1 = λ           \$/MWh
IC3 (PG3) = 18 + 0.025 PG1 = λ          \$/MWh

With a constraint of: PG1 + PG2 + PG3 = 1000 MW

34
The solution using direct method is shown below

ngen
P
Demand
             
i           1000 
15

20

18
i1            2i                     0.02       0.01       0.025
 23.526
                                                   1          1          1
ngen                                                       

1                           0.02       0.01        0.025
2 i
i1
23.53 15
P        ( 23.53
)            P 
G1                           1         0.02                           P  426.5
1

P        ( 23.53
)                   23.53 20
G2                          P                                       P  353
2         0.01                            2
PG1 = 426.5 MW,                  PG2 = 253 MW,
23.53 18                       PG3 = 221.2 MW
P        ( 23.53
)            P 
G3                           3         0.025                          P  221.2
3

PG1 + PG2 + PG3 = 1000 MW

In addition, we can use software option to solve for lambda. This will take into account the complete
cost curve, and provide the option of setting the limit on each generator.
For example, if you choose the PG1 = 426.5 MW, PG2 = 253 MW and PG3 = 221.2 MW
with an operating cost of 1700, 1000, and 250 respectively, then Matlab code will allow you to find
the best value for Lambda. Bear in mind that the Lambda will not be the same since the total
operating cost is now factored in. See examples below.

35
2.3.1   Matlab Software Code for 3-Generator Lambda (Cost)

Code and Results

EDU>> % Example 3-Generator Station
gendata = [1700, 15, 0.02; 1000, 20, 0.01; 250, 18, 0.025];
power = [426.5, 353, 221.2];

% find lambda0
n = length (gendata);
lambda0 = 0;
for i = 1:n
lambda0 = lambda0 + gendata (i,2) + 2*gendata(i,3)*power(i);
end
lambda0 = lambda0 / 3
clear x0
x0 = power, x0(n+1) = lambda0
for k = 1 : 10
disp (k)
pgen = 0, cost = 0;
for i = 1:n
gradient(i) = gendata(i,2) + 2 * gendata (i,3) * x0(i)- x0(n+i);
pgen = pgen + x0(i);
cost = cost + gendata (i,1) + gendata (i,2) * x0(i) + gendata (i,3) * x0(i) * x0(i);
end
disp ([x0, Pgen, cost/1000])
x1 = x0 - gradient * alpha;
x0 = x1;
lambda0 =

29.3933

x0 =

426.5000 353.0000 221.2000

x0 =

426.5000 353.0000 221.2000 29.3933

36
Please see Matlab code examples for Lagrange Method and Kuhn-Tucker Conditions shown below.

Figure-2.5 Three Generator Economic Dispatch Application Using Lagrange Method & Matlab Code

EDU>> % Example 3-Generator Station
gendata = [1700, 23.76, 0.004686; 1000, 23.55, 0.00582; 250, 23.70, 0.01446];
power = [400, 200, 400];

% find lambda0
n = length (gendata);
lambda0 = 0;
for i = 1: n
lambda0 = lambda0 + gendata (i,2) + 2*gendata(i,3)*power(i);
end
lambda0 = lambda0 / 3
clear x0
x0 = power, x0(n+1) = lambda0
for k = 1 : 10
disp (k)
pgen = 0, cost = 0;
for i = 1:n
gradient(i) = gendata(i,2) + 2 * gendata (i,3) * x0(i)- x0(n+i);
pgen = pgen + x0(i);
cost = cost + gendata (i,1) + gendata (i,2) * x0(i) + gendata (i,3) * x0(i) * x0(i);

37
end
Disp ([x0, Pgen, cost/1000])
x1 = x0 - gradient * alpha;
x0 = x1;

lambda0 = 29.5516

x0 = 400 200 400

x0 = 400.0000 200.0000 400.0000 29.5516
Therefore the cost of generating power is \$29.55 per MW/hr

2.3.2 Economic Dispatch with Generator Limits

The power output of any generator must not exceed its rating nor drop below a given value for stable
boiler operation. The Aim is to find the real power generated for each plant to minimize cost,
subject to:
(1) Meeting load demand - equality constraints
(2) Constrained by generator limits – inequality constraints.
By using the Kuhn-Tucker Conditions we can formulate the equation that allows one work within
such limits. See equation below:
dCi/dPi = λ                  Pi(min) < Pi < Pi(max)
dCi/dPi ≤ λ               Pi = Pi(max)
dCi/dPi ≥ λ               Pi = Pi(min)

In this example, if one ignores the system losses, the optimal dispatch and the total cost in \$/hr for
three generators can be found from the following equations:

C1 = 1700 + 23.76 P1 + 0.00468P2 (\$/MWhr)
C2 = 1000 + 23.76 P2 + 0.00468P2 (\$/MWhr)
C3 = 250+ 23.76 P3 + 0.00468P2           (\$/MWhr)
200 ≤ P1 ≤ 450
150 ≤ P2 ≤ 350
100 ≤ P3 ≤ 200

38
Total Demand = 1000
Total cost: C1 = (1700 + 23.76 * 450) + (0.00468 * 4502 ) = 13339.7
+ C2 = 1000 + (23.55 * 350) + (0.0058 2* 3502) = 9955.45
+ C3 = 250+ (23.70 * 200) + (0.00468 * 2002) = 5568.4
Total Cost = 28,863.6 \$/hr

Setting Limits to Generators

In some applications the Systems Engineer will have to run the generators without limits and in other
applications limits must be enforced. An example using three generators will be used to explore the
concept of limits.

If the generator cost function of three generators are given as:

C1 = 472.8 + 7.47 P1 + 0.023 P12 (\$/h/MW) 100≤ P1 ≤ 400
C2 = 432.9 + 7.87 P2 + 0.020 P22 (\$/h/MW)      150≤ P2 ≤ 350
C3 = 463.9 + 7.56 P3 + 0.022 P32 (\$/h/MW)      120≤ P3 ≤ 400

(a) Find the optimal dispatch of the three generators without any limits
(b) What is the new dispatch if the generator limits are enforced?

39
Solution

ngen
P
Demand
              
i                            1800 
7.47

7.87

7.56
i1             2i                                      0.046          0.040          0.044
 33.555
                                                                           1              1              1
ngen                                                                                 

1                                             0.046       0.040          0.044
2 i
i1
33.555 7.47
P        ( 33.555
)                           P 
G1                                           1           0.046                             P  567.065
1

P        ( 33.555
)                                  33.555 7.87
G2                                          P                                            P  642.125
2           0.040                                 2

33.555 7.56
P        ( 33.555
)                           P 
G3                                           3           0.044                             P  590.795
3

Lower limits are enforced

P                350       Meets lower limit
G2

ngen
P
Demand
             
i                       350 
7.47
0.046

7.56
0.044
i1            2i                                                          15.387
                                                                           1              1

ngen                                                0.046           0.044

1
2 i
i1

P                                    15.387 7.47
G1                                                    172.109
0.046

15.387 7.56
P                                                     177.886
G3                                      0.044

40
3.0    System Modeling

This section explains how modeling is used in decision analysis as it relates to power system
economic dispatch. Modeling is the cog that drives the concept of decision making. It can be defined
as a process where the risk is measured and controlled by checks and balances such as planning,
assessment, action taken to mitigate risks, monitor the situation and report the findings.

Requisite decision modeling will be the focus as it is the answer to “what if” questions in decision-
making. Modeling is also used for analyzing policy, alternative solution and strategy that helps in
making the right decisions with a high degree of confidence.
Requisite decision modeling requires a wide range of testing and no new intuitions emerge about the
problem the model set out to solve. By using a wide range of testing, one can verify the sensitivity
points or parameters that will affect the model. Therefore, by knowing the sensitivity limits on these
parameters, one can make decisions with less risks and less uncertainty. In addition, a requisite model
will provide the modeler or user with trade-off options between long-term returns and volatility.

Modeling in this case is used to ultimately replicate the historical performance of similar projects. It
is also very important that a model replicate the data for the right reasons while capturing the
underlying structure of the project. In addition, modeling is used to promote inquiry, expose hidden
assumptions, motivate the widest range of empirical tests, challenge preconceptions and support
multiple viewpoints. Modeling the power system economic dispatch with power world software
helps one to get reliable, useful and effective models to serve as virtual worlds to aid learning and
policy design. Furthermore, simulation models are always necessary for learning about dynamically
complex systems.
Figure 3.1 shows a system with 250 MW load and shunt of 50 Mvar with a line condition of 25.92
MW, -8.64 Mvar and 27.32 MVA. By using incremental modeling techniques, one can achieve the
condition that will minimize line loss and in return boost cash/NPV.

41
25.92 MW
-8.64 Mvar        50.0 Mvar
250 MW
250 MW
27.32 MVA
30.0 Mvar

2

1

slack                                                                 100 MW
60 Mvar

152 MW

43 Mvar

Figure 3.1           Modeling Two Bus System

By applying Computerized Simulation using power world software to show power system and power
flow as it relates to economic dispatch, we obtain a wide range of power data. The above Figure-3.1
is used to generate the Ybus Sparse to give us the voltage magnitude and angle. The phase angles of
different substation or bus cannot be economically measured, but the voltage magnitude is constantly
monitored.

42
In order to effectively estimate the voltage magnitude and angle, one must choose a bus as reference,
and that bus will now become the standard by which all other bus angles will be measured. The
Matlab Software was used to generate voltage magnitude and angle from the above Figure-3.2. See
Ybus data below.

EDU>> j = sqrt (-1);
Ybus = sparse (3);
Ybus ( 1,    1) =      10.0000+ j*( -29.6750);
Ybus( 1,     2) = -10.0000+ j*( 30.0000);
Ybus( 2,     1) = -10.0000+ j*( 30.0000);
Ybus( 2,     2) =    10.0000+ j*( -29.4750);
Ybus( 3,     3) =      0.0000+ j*(   0.0000);
V( 1) = 1.000000 + j*( 0.000000);
V( 2) = 1.000000 + j*(-0.017278);
V( 3) = 0.000000 + j*( 0.000000);
EDU>>
The voltage angle of the slack bus shown in Figure-3.2 at bus-1 is used as reference to all the other
bus voltages. The angle of the slack bus is not important because the voltage angle difference is used
to determine the power (Pi) and the imaginary (Qi) at the load. The slack bus also known as the
swing bus is a fictitious concept that is created and defined in the problem formulations. This
process allows us to address the I2R losses that are not known before the load flow calculation.

43
0.00 MW
0.00 Mvar       50.0 Mvar
250 MW
250 MW
0.00 MVA
30.0 Mvar

2

1

slack                                                              152 MW
42 Mvar

100 MW

60 Mvar

Figure-3.2               Bus Modeling

By changing line conditions, we were able to eliminate the line loss during the simulation process.
Line loss is one of the key components of power system economic dispatch. For example, if we have
a scenario where the loss between buses resulted in 50 MW and the cost of generating power is \$50
per MWh, then there will be a loss of \$50 for every hour which results in \$438,000 per year.
Therefore, by using modeling techniques, one can exponentially minimize the hours spent on various
power flow data such as Ybus, voltage magnitude/angle mismatches and power loss.

44
3.1        Computer Modeling & Simulation

By using simulation process to design the system in Figure-3.3, one can learn about the system and
document the system behaviors from the view point of power flow and economic dispatch. The
approach used is to design a basic system with two generators with the per-unit voltage of the system
set 1.0 pu. A system of 1.0 pu is considered a perfect system. In power system design and planning
this method is known as a flat start.
After the flat start is achieved, the iteration technique is used to increment the system load while
documenting the system. Moreover, this process allows us to check the sensitivity of the system to
see when the system will collapse. The system collapses at bus 7 when the load of 220MW and 50
Mvar was applied.

1
2                  300 MW
A
99 MW                                                  A

slack
17 Mvar                                                Amps
MVA
30 Mvar

220 MW
A          30 Mvar
A
50 Mvar
MVA

Amps
System Callaspses @ 0.95 PU on Bus 7            4

A
7

Amps
A

99.45 MW                 A
MVA                                                                50 MW
5   0 Mvar
MVA
6
20 MW
500 MW
SYSE 505 Student: Ogilvie Donald                              100 Mvar

Figure-3.3                                   7-Bus System

45
The data shown below is from the simulation process in Figure-3.3 and it provides us with the Ybus
record for each bus. In order to get the power system economic dispatch, one must deal with the
following power flow conditions:

(1)          Ybus model
(2)          Bus voltage
(3)          Power mismatch at each bus
(4)          Power supplied by the swing bus
(5)          Power loss of transmission lines
(6)          Power system economic dispatch.

Therefore, since our objective is power system economic dispatch, we must meet all the power flow
conditions in order get to the system economics, and there is no better way than simulation modeling
to eliminate the hundreds of hours in crunching numbers. Furthermore, when we factor in human
fatigue, errors and the duplications, simulation modeling will no doubt be the future of all power
system planning and economic dispatch

YBus Records
Number               Name   Bus    1         Bus    2         Bus    3         Bus    4         Bus    5         Bus    6         Bus    7

1    1      5.88 - j23.52    -2.94 + j11.76                                                                       -2.94 + j11.76
2    2      -2.94 + j11.76   5.88 - j23.52    -2.94 + j11.76
3    3                       -2.94 + j11.76   5.88 - j23.52    0.00 + j0.00                      -2.94 + j11.76
4    4                                        0.00 + j0.00     2.94 - j11.76    -2.94 + j11.76
5    5                                                         -2.94 + j11.76   5.88 - j23.52    -2.94 + j11.76
6    6                                        -2.94 + j11.76                    -2.94 + j11.76   8.82 - j35.28    -2.94 + j11.76
7    7      -2.94 + j11.76                                                                       -2.94 + j11.76   5.88 - j23.52

46
3.2     Computer simulation of a two bus 1-generator system

The Ybus matrix, voltage magnitude and angle are shown below for Figure-3.4. The transmission
line data that was used in simulation process came from Figure-2.2.
Using Gauss Seidel method of iteration for the problem associated with Figure-2.2 resulted in V2 =
0.987-0.156i, while on the other hand computer simulation yields
V2 = 0 979-0.173i for Figure 3.5 resulted in a difference of 0.008 pu. The acceptable tolerance is plus
or minus 5% of 100% where 100% = 1.0 pu.

j = sqrt(-1);
Ybus = sparse(2);
Ybus(        1,              1) =    0.0498+ j*(        -2.5910);
Ybus(        1,              2) =    -0.0498+ j*(        2.8810);
Ybus(        2,              1) =    -0.0498+ j*(        2.8810);
Ybus(        2,               2) =   0.0498+ j*(        -2.5910);
V(      1) =         1.000000 + j*( 0.000000);
V(      2) =         0.979153 + j*(-0.173715);

50 MW
-24 Mv ar

2
1

sla c k

Figure 3.4           Modeling a Single Generator

47
3.3    Two generator economic dispatch system modeling

Figure-3.5 shows a typical 3-bus scenario for modeling.
By using computer based simulating we were able to capture all the key components needed to meet
our economic dispatch requirements. For example, the bus record number, the bus sensitivity and the
bus mismatches provided data such as sensitivity limits and upper and lower per-unit voltage limits.
In addition, we can clearly see that under Gen Marginal Loss Sensitivities that there is no megawatt
loss or system sensitivity issues. Under bus sensitivities, we can see the assigned limits for each bus
for both MW & Mvar.

YBus
Records
Number           Name     Bus 1                     Bus 2                       Bus 3
1   1        1.85 - j24.80             -0.31 + j12.49              -1.54 + j12.31
2   2        -0.31 + j12.49            0.87 - j29.14               -0.55 + j16.65
3   3        -1.54 + j12.31            -0.55 + j16.65              2.09 - j28.96

1
2

150 MW
0 Mv ar

slack

51 MW                                          100 MW
0 Mv ar                  3                     7 Mv ar

Figure 3.5              Modeling a two Bus Power Flow System

48
Bus Sensitivities
Area          Area           P                 Q
Number           Name        Num           Name           Sensitivity       Sensitivity
1   1                    1    1                       0
-
2   2                    1    1              0.63929194
-
3   3                    1    1              0.36759728        0.01529269

Ctg Limit   Ctg Limit
Area                        Limit                     Volt        Limit Low     Limit High    Low PU      High PU
Number       Name       Name        Monitor         Group       PU Volt       (kV)        PU Volt       PU Volt       Volt        Volt
1       1          1           Yes             Default           1            138            0.9           1.1         0.9         1.1
2       2          1           Yes             Default           1            138            0.9           1.1         0.9         1.1
3       3          1           Yes             Default     0.99507        137.319            0.9           1.1         0.9         1.1

Bus Mismatches
Area             Mismatch        Mismatch            Mismatch
Number           Name    Name             MW              Mvar                MVA
1       1       1                            0                 0                 0
2       2       1                            0                 0                 0
3       3       1                            0                 0                 0

Gen Marginal Loss
Sensitivities
Loss MW             Penalty        Gen                  Max        Gen
Number                Name            ID   Status     Sens                Factor         MW        Min MW     MW         Mvar
1     1               1    Closed                     0              1     50.52         0      1000        0.42
2     2               1    Closed                     0              1       100         0      1000        7.49

49
4.0 Conclusion

This paper has given me the opportunity to research and develop new techniques to simulate and
model power systems. For example, Figure 3.3, the 7-Bus System, was designed and developed to
meet the requirements of the North American power grid system. The simulated Y-bus data is shown
on page 46. In addition, I was able to successfully set the per-unit voltage limits to trip the circuit
breaker. If the per-unit voltage varies by +/- 5%, then this action will remove the system from the
grid.

The focus of this economic dispatch paper is to compare the different ways in which to analyze
began with the different fuel sources and their associated cost, and how to minimize cost by making
the system more efficient, because power plant measures result in terms of efficiency. In addition,
efficiency is output divided by input and in this case our input is heat rate (MWHR/MBTU) and our
output in power is MW. By using the input and output concept one can conclude that the system of
generating power will maximize profit when the heat is maintained as low as possible while
maximizing your output requirements. In the power industry, fuel cost curve provides the best
accuracy to effectively analyze heat rate. In fact, power system economic dispatch is driven by the
fuel cost curve:
Ci ( PGi )     PGi   PGi and incremental cost which is a derivitive of the fuel cost.
2

Heat rate is widely used in the power industry to troubleshoot the system when the system efficiency
drops below optimum setting before or after major system repairs or overhaul. Furthermore after a
major overhaul, it is standard practice in this industry to apply minute by minute loading of the
system. This minute by minute loading process is shown here with the use of calculus as a source for
verification and validation. Matlab Software code was used to generate the fuel cost curve equation,
and graph the operational cost against the output power in MW.

This paper also focused on 1-generator, 2-generator operation and a 3-generator operation to look at
various methods used to determine the cost of power with and without power loss.
Although the focus of the paper is on economic dispatch, areas such as per-unit system and power
flow was briefly covered to allow one to complete the process from a typical scenarios such as
transformers, generators and z-line conditions. Gauss Seidel method of iteration was used to show

50
the overall concept of obtaining the bus voltage from the Ybus matrix and thus the economic
dispatch.

Moreover, from the power loss conditions this paper was able to obtain the equivalent cash flow
using NPV if the system was operating without power loss. Modeling and simulation process was
also used to show the effectiveness of using software as a tool to make quick, clear, accurate and
explicit decision when planning economic dispatch operation as a system engineer.

By using Powerworld software in modeling power system and economic dispatch, this paper shows
how to analyze the power loss and the per unit voltage limits to make the system economically
viable. For example, Figure 3.5 shows a complete power flow analysis and the data required to
perform economic dispatch. In addition, the analysis shows zero power loss and zero mismatches
which would result in a perfect system.
I would recommend modeling and simulation to be the first step in planning and assessing economic
dispatch. It is also an indispensable source for performing verification and validation of power
system economic dispatch.

51
Appendix

Total Electricity System Power

2009 Total California In-State Power Generation
In-State             Percent of                                             Total
Northwest    Southwest
Fuel Type          Generation        California In-State                                       System
Imports      Imports
(GWh)                 Power                                               Power

Coal                         3,735                  1.8%                N/A             N/A            N/A

Large Hydro                 25,094                 12.2%                N/A             N/A            N/A

Natural Gas               116,716                  56.7%                N/A             N/A            N/A

Nuclear                     31,509                 15.3%                N/A             N/A            N/A

Oil                              67                 0.0%                N/A             N/A            N/A

Other                             7                 0.0%                N/A             N/A            N/A

Renewables                  28,567                 13.9%                N/A             N/A            N/A

Biomass                      5,685                  2.8%                N/A             N/A            N/A

Geothermal                  12,907                  6.3%                N/A             N/A            N/A

Small Hydro                  4,181                  2.0%                N/A             N/A            N/A

Solar                          846                  0.4%                N/A             N/A            N/A

Wind                         4,949                  2.4%                N/A             N/A            N/A

Total                     205,695                 100.0%             19,929         71,201         296,827

Source:

EIA, QFER, and SB 105 Reporting Requirements

Note: Due to legislative changes required by Assembly Bill 162 (2009), the California Air Resources Board is
currently undertaking the task of identifying the fuel sources associated with all imported power entering into
California.

1.   In-state generation: Reported generation from units 1 MW and larger.

Previous years information (2008 Total System Power).

52
Total Electricity System Power

2008 Total System Power in Gigawatt Hours
Total
In-State          Northwest         Southwest                        Percent of Total
Fuel Type                                                                   System
Generation[1]       Imports[2]        Imports[2]                        System Power
Power

Coal*                         3,977             8,581            43,271          55,829              18.21%

Large Hydro                  21,040             9,334             3,359          33,733              11.00%

Natural Gas                122,216              2,939            15,060        140,215               45.74%

Nuclear                      32,482              747             11,039          44,268              14.44%

Renewables                   28,804             2,344             1,384          32,532              10.61%
Biomass                5,720              654                  3           6,377                2.08%

Geothermal                 12,907                 0              755           13,662                4.46%

Small Hydro                  3,729              674                  13          4,416                1.44%

Solar                 724                  0                 22            746                0.24%

Wind                 5,724             1,016              591            7,331                2.39%

Total                      208,519            23,945             74,113        306,577               100.0%

Source:

2008 Net System Power Report - Staff Report,
Publication number CEC-200-2009-010, to be considered for adoption July 15, 2009. (PDF file, 26 pages, 650
kb)

EIA, QFER, and SB 105 Reporting Requirements

*Note: In earlier years the in-state coal number included coal-fired power plants owned by California utilities located
out-of-state.

1.    In-state generation: Reported generation from units 1 MW and larger.
2.    Net electricity imports are based on metered power flows between California and out-of-state
balancing authorities.
The resource mix is based on utility power source disclosure claims, contract information, and
calculated estimates on the remaining balance of net imports.

Previous years information (2007 Total System Power).

53

http://greenecon.net/understanding-the-cost-of-solar-energy/energy_economics.html

http://greenecon.net/category/alternative-energy

http://energyalmanac.ca.gov/electricity/index.html#table

54
References

1. Florida State University 2002 and 2004 EEL 6266 Power System Economics and Control with
Matlab Software Code Lecture Notes

2. Tom Overbye, ECE 476, Power System Analysis Lecture 16 Economic Dispatch

3. Thomas J. Overbye, Powerworld Simulator Software

4. Ali. Keyhani, Power Flow Problem Tutorial and Course Notes and Matlab Software Codes

5. Robert T. Clement, Making Hard Decisions

6. Chanan Singh and Panida Jirutitijaroen Power System Control and Operation: Economic
Dispatch

7. Form E1A-423 Monthly Cost of Fuels for Electric Plants Report

8. Grainger Stevenson, Power System Analysis

9. Charles A. Gross Power System Analysis

10. http://greenecon.net/understanding-the-cost-of-solar-energy/energy_economics.html

11. http://greenecon.net/category/alternative-energy

12. http://energyalmanac.ca.gov/electricity/index.html#table

55

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