Economic Dispatch

Lecture No. 3

CHAPTER THREE

Economic Dispatch of Thermal Units 3.1 INTRODUCTION

The power system engineer is faced with the challenging task of planning and

successfully operating one of the most complex systems of today's civilization. The

efficient planning and optimum economic operation of power system has always

occupied an important position in the electric power industry.

A typical electric power system comprises of three main elements. Firstly,

there are consumers, whose requirements of electrical energy have to be served by

electrical power system. Secondly, there has to be means by which these requirements

are served i.e., the generating plants. Finally, there is transmission and distribution

network, which transports the power from producers to the consumers.

The prime objective of a power system is to transfer electrical energy from the

generating stations to the consumers with;

Maximum safety of personal and equipment.

Maximum continuity (security reliability and stability)

Maximum quality (frequency and voltage with in limits)

Minimum cost (optimum utilization of resources).

Electrical energy can not be stored economically. However, it can be stored as

potential energy in hydro systems and pumped storage schemes, but this represents a

small fraction of the installed capacity of most industrialized nations. The larger

portion of generating plant is thermal. For cold turbo alternator set 6 to 8 hours are

Power system Operational Planning Economic Dispatch

37

Lecture No. 3

required for preparation and readiness for synchronization. A hot unit can be

synchronized within 15 minutes and fully loaded in 30-60 minutes. It is therefore

essential that the demand must be met as and when it occurs. The present day load

demand handling is a challenging task for Power System Operation Engineer. In this

context system operation necessitates the following for power system:

it must be highly interconnected

it must be automated

it must have Operational Planning

In the early days the power system consisted of isolated stations and their

individual loads. But at present the power systems are highly interconnected in which

several generating stations run in parallel and feed a high voltage network which then

supplies a set of consuming centers. Operating an automated electric power system is

an extremely complex task.

The objective of power system control is to provide a secure supply at a

minimum cost. Figure 3.1 illustrates the operation and data flow in a modern power

system on the assumption of a fully automated power system based on real-time

digital control. Although such an extreme degree of automation has not yet been

implemented, the activities in the boxes are performed by most utilities.

In some cases computation is performed off-line, in others on-line, the degree

of human supervision or intervention, varying considerably from utility to utility.

There are three stages in system control, namely generator scheduling or unit

commitment, security analysis and economic dispatch.

• Generator scheduling involves the hour-by- hour ordering of generator

units on off the system to match the anticipated load and to allow a safety

margin.

• With a given power system topology and number of generators on the

bars, security analysis assesses the system response to a set of

contingencies and provides a set of constraints that should not be violated


38

Lecture No. 3

if the system is to remain in secure state.

• Economic dispatch orders the minute-to-minute loading of the connected

generating plant so that the cost of generation is a minimum with due

respect to the satisfaction of the security and other engineering

constraints.

These three control functions require reliable knowledge of the system

configuration, i.e., circuit breaker and isolator position and system actual P and Q

flows. This data is collected from thousands of metering devices and transmitted to

control centers, usually over hired telephone lines. It is statistically inevitable that

owing to the numbers of devices involved, interference over communication lines

unreliable data will be present. State estimation is a mathematical algorithm that

provides reliable database out of an unreliable set of information.

In traditional power control centers where all activities are channeled through

human operator, the experienced control engineer looking at wall mimic diagram of

power system takes in multiplicity of data. He mentally assesses their compatibility

with a degree of confidence and he can pinpoint grossly corrupted piece of

information. Human beings are good state estimators.


39

Lecture No. 3

Figure 3.1 Power System Control Activities

3.2 OPERATION PLANNING OF POWER SYSTEM

The operational planning of the power system involves the best utilization of

available energy resources subjected to various constraints to transfer electrical energy

from generating stations to the consumers with maximum safety of personal/equipment

without interruption of supply at minimum cost. In modern complex and highly

interconnected power systems, the operational planning involves the steps


40

Lecture No. 3

o Load Forecasting,

o Unit Commitment,

o Economic Dispatch

o Hydrothermal Coordination

o Voltage Control

o Frequency Control

TIMES SCALES INVOLVED FOR ACTIVITIES IN PLANNING

OPERATION AND CONTROL

Various activities that are combined under the broader area of power system

operation and control do not have the same time scale. For example at one extreme is

the time taken to build a hydropower station (up to 10 years), while on the other

extreme is the time interval between detection and interruption of a fault on power

system (up to 80 m sec). Time scales involved for various activities in power system

planning, operation and control are summarized as follows:

• YEARS

System expansion planning, construction, maintenance scheduling and planned

outages.

• MONTHS

Preliminary load forecasting, generation estimation and contingency planning

• DAYS

Short term load forecasting, reserve assessment and generation scheduling.

• HOURS

Unit commitment, preliminary economic dispatch and contingency analysis.

• MINUTES

Economic dispatch, power interchanges, frequency control and security

assessment.

• SECONDS


41

Lecture No. 3

Protection and C.B operation, automatic voltage and frequency control.

Time scale particular to system operation point of view may be stated as:

• Unit Commitment – hours to days to week

• Economic Dispatch – minutes to hours

• Security Analysis – every few minutes and on demand

• System equipment – milliseconds to seconds (Automatic voltage control i.e., tap changers and excitation control and generator set governor control)

SPINNING RESERVE

The total generating capacity required to be available on the bars is always

larger than the anticipated load. The difference between these quantities is called

“system spare” or “spinning reserve”.

The spinning reserve (s. r.) for a given generating unit can be defined as the

extra amount of the active power that can be obtained from that unit with in a

specified interval of time (a few minutes) by loading it at its maximum rate through

governor action. The spinning reserve of the power system will be available to

makeup the outage of any generating unit or to meet unexpected increase in demand.

The spinning reserve should be at least equal to the rating of the largest unit on the

bars. The system characteristics will determine the percentage interval after which the

spinning reserve must be available if excessive drop in frequency is to be averted.

The specified post – outage time and maximum loading rate of a generating

unit will fix its ceiling of spinning reserve. The maximum loading rate of a turbo

alternator unit is determined empirically and is dictated thermal considerations.

Typical maximum loading rate of

Turbo alternator : 2-5 MW /min (Ramping)

Gas turbine : 30 MW /min

First response reserve capacity can be provided by hydro or pump storage and

by gas turbine. Such plant can be started up automatically when frequency falls below

critical value.


42

Lecture No. 3

The decision of how much reserve capacity the system should carry depends on

diverse factors such as type of generating plant, unit sizes, and degree of desirable

reliability and security. It usually amounts to less than 10% of the load.

Typical spinning reserve on CEGB

Day time s. r of 1000 MW

Drops at. Night to 680 MW

Of 1000 MW, 640 MW is provided by partly loaded sets on the bars capable of

supplying the demands within 5 minutes and sustaining this output. The remaining

360 MW is available from the F testing pump storage scheme there is also a standing

reserve of 500 MW provided by gas turbines not in synchronization with the system

but able to supply the demand within 5 minutes. Finally there is a standby reserve of

800 MWof gas turbine plant not included in standby reserve capacity capable of

achieving the demand within hours.

3.3 ECONOMIC DISPATCH

Having solved the unit commitment problem and having ensured through

security analysis that present system is in a secure state then the efforts are made to

adjust the loading on the individual generators to achieve minimum production cost

on minute-to-minute basis. This loading of generators subjected to minimum cost is

in essence the economic dispatch problem and can be defined as a computational

process of allocating generation levels to the generating units in the mix so that the

system load may be supplied entirely and most economically.

Load dispatching is essentially an online activity and is normally

associated with an online forecasting / prediction system. The economic dispatch

calculations are performed every few minutes, which must ensure that all the

committed units, sharing in the economic dispatch calculations, are operating in

such a way that the overall system operation cost is minimum and the recognized

system constraints are satisfied.


43

Lecture No. 3

3.3.1 POWER SYSTEM VARIABLES FOR ECONOMIC DISPATCH

On a bus bar there are four variables namely P, Q, V and δ. Out of the four

two are specified and two can be determined from load flow analysis. In economic

dispatch problem, further degrees of freedom are necessary, in other words some

elbow room is available within which variables can be adjusted for the minimization

of the operating cost.

For the economic dispatch analysis, we can classify the variables in the power

system into three categories:

1. Control or decision variables U

These are the variables over which we have complete control, within specified

limits. Typical control variables are the active power injection and voltage magnitude

at generation buses.

2. State or dependent variables X

We have no direct control over these, as their value is not known until the

completion of load flow study. Typical dependent variables may be voltage magnitude

and angles at load buses.

3. Output variables Z

These are function of other variables at least one of which is state variable. The

output variables are determined after the completion of load flow study. Typical

output variables are the P and Q injected at slack bus, the complex power flows over

the transmission lines, the cost of power generation, the network losses etc.

3.3.2 VARIOUS CONSTRAINTS IN POWER SYSTEM

The present modern power system has to operate under various operational and

network constraints. Broadly speaking there are two types of constraints:

Equality Constraints

Inequality Constraints

EQUALITY CONSTRAINTS


44

Lecture No. 3

Equality or network constraints are basic load flow equations under steady

state condition. There is a balance between active and reactive power, therefore

( )

( ) 0QQQ

0PPP

n

1ildig,

ld

n

1iig,

=+−

=+−

∑

∑

=

= (3.1)

Where Pl and Ql are network losses. In terms of and state variables mathematically can

be written in short hand form

g ( X , U ) = 0 (3.2)

INEQUALITY CONSTRAINTS

Inequality constraints are of two types

Hard type

Soft type

Hard type are those which are definite and specific like the tapping range of

on-load tap changing transformer, where as soft type are those which have some

flexibility associated with them like the nodal voltages and phase angles between

nodal voltages.

Generator Constraints

The KVA loading on a generator is given by 22 QP + and this should not exceed a

pre-specified value Cp because of temperature rise conditions i.e.,

Pi2 + Qi

2 ≤ Ci2

The maximum active power generation of a source is limited by thermal consideration

and minimum power generation is limited by flame instability of the boiler. If the

power output of a generator for optimum operation of the system is less than a pre-

specified value Pmin, the unit is not put on the bar because it is not possible to generate

that low value of the power from that unit. Hence generator power Pi can not be


45

Lecture No. 3

outside the range stated by the inequality

Pg, min. ≤ Pg, i ≤ Pg, max.

Voltage Constraints

Voltage magnitude and phase angle at various nodes should vary within certain limits

i.e.,

Vi, min. ≤ Vi ≤Vi, max.

δi, min. ≤δi ≤ δi, max

Running Spare Capacity Constraints

These constraints are required to meet

Forced outages of one or more generators on the system

Un expected load on the system

The total generation should be such that in addition to meeting the load and losses, a

minimum spare capacity should be available i.e.,

G ≥ Pi + Pso

Where G is total generation and Pso is some prespecified power. A well planned

system is one in which spare capacity Pso is minimum.

Transformer Tap Settings

In case of auto transformer we have

0 ≤ t < 1.0

For two winding transformer, if tapings are provided on secondary side, then we have

0 ≤ t ≤ n

Where n is transformation ratio

Phase shift limits of phase shifting transformer

θi,min. ≤ θi ≤ θi, max.

Transmission Line Constraints


46

Lecture No. 3

P and Q flows through the line are limited by the thermal capability of the circuit and

is expressed as:

C i ≤ Ci, max.

Where Ci, max. is the maximum loading capacity of the ith line.

The mathematical short hand notation that encompasses all the inequality constraints

can be written as:

( ) 0XU,h ≤ (3.3)

3.3.3 Economic Dispatch – Mathematical Formulation

In general the economic dispatch problem can be formulated in mathematical

terms as:

Minimize a scalar objective function

f ( X , U )

Subject to gi ( X , U ) = 0

hk ( X , U ) ≥ 0

Where

X = state (dependent) variables

U = control (independent) variables

gi( X , U ) = power flow equations ( non- linear )

hk( X , U ) = inequalities ( limits )

VARIABLES

The system state or dependent variables X are generally

⎩⎨⎧= nodes PQon voltagenodal V,

nodes PV and PQon angle voltage,θX

The control or independent variables U are generally

⎪⎩

⎪⎨⎧

=nodesslack on V,

node PQon QP,nodes PVon V P,

mer tapt transforU

s

θ


47

Lecture No. 3

Equality Constraints

The equality constraints gi ( X , U ) = 0 are the nodal load flow equations i.e.,

YV = I

Where

I = S*/V*

For analysis and computation purposes it is more convenient to express the

problem by a single relation and to separate the complex quantities into their co-

ordinate components. Usually the problem can be decomposed into rectangular (real

and imaginary) or polar (magnitude and angle) co-ordinates.

Simple arithmetic operations are faster in rectangular system than in polar form

which involves trigonometrically expressions. The polar form however uses V and θ

explicitly which offers opportunity to exploit the natural physical operation of the

system into P- θ and Q-V dependency.

The power flow equations in polar form are usually expressed in terms of

power mismatch at each node i.e.,

( ) 0SinθBCosθGVVPPΔPk

ikikikikkidigii =+−−= ∑

( ) 0CosθBSinθGVVQQΔQ ikikikikk

kidigii =−−−= ∑

Where

θik = θi – θk

g = generation

d = demand

G = Network Conductance

B = Network Susceptance

K = node number directly connected to node i usually

G << B, θik << 20◦ and Vi = 1 ± 10%

Load flow problem is non-linear and sparse and can be solved by Newton or

approximate Newton methods. In this method, Jacobin is formed in each iteration and


48

Lecture No. 3

equations solved to correct the unknown V and θ. Mathematically we can write as:

⎥⎦⎤

⎢⎣⎡ΔΔ⎥⎦

⎤⎢⎣⎡=⎥⎦

⎤⎢⎣⎡ θ

V NL H

JΔPΔQ

3.3.4 Solution Methods

Economic dispatch is a constrained optimization problem. The solution of

optimization problems is province of a branch of mathematics called mathematical

programming. The label is some what confusing as the work programming has nothing

to do with computer programming , although of course , computer are invariably used

for the solution of such problems.

A wide variety of optimization techniques has been applied for solving

Economic Dispatch problems. The techniques can be classified as

The approaches used for ED problem may be listed as:

Merit Order Approach (old method)

Equal Incremental Cost Criterion (widely used)

Linear Programming (Easy Constraint Handling)

Dynamic Programming

Non-Linear Programming (Exact Methods)

a) Ist Order Gradient Based Techniques

(Dommel & Tinney Method)

b) Second Order Method (Optimal Power Flow ) 1 Non-linear Programming (NLP) Objective function and constraints are non-linear.

2 Quadratic Programming (QP) Special form of non-linear programming

Objective function-quadratic

Constraints are linear

3 Newton–Based Solution

Necessary conditions of optimality commonly

referred to as Khun-Tucker conditions are

obtained

In general these are non-linear equations

requiring iterative methods of solution. The


49

Lecture No. 3

Newton method is favoured for its quadratic

characteristics.

4 Linear Programming Linear programming treats problems with

constraints and objective function formulated in

linear forms with non-negative variables.

Simplex and Revise Simplex methods are known

to be quite effective for solving LP problems.

3.5 ECONOMIC DISPATCH OF THERMAL UNITS-------METHOD OF

SOLUTIONS

3.5.1 Merit Order Approach

In this method it is assumed that incremental cost of all generators is constant

over the full range or over successive discrete portions within the range. The economic

way of loading the machine is in the order of highest incremental efficiency. This

method therefore needs forming of a table which could be looked into for any load

condition and does not need any complicated calculations.

Assuming linear cost function, the optimal economic dispatch can be described

as:

Minimize ( )∑=

=N

1iiiT PFF

Subject to ∑=

−=N

1iiD PPΦ

And 0PP im

i ≤−

0PP Mii ≤−

The Augmented cost function is given by

∑∑ ∑∑== ==

−+−+−+=N

1i

Mii

Mi

N

1ii

mi

N

1i

mii

N

1iDiacost )P(Pμ)P(Pμ)Pλ(PFF

1. If for some i the inequality constraints are inactive then

λF =′


50

Lecture No. 3

2. If for some i Mii PP =

then λF ≤′

3. If some i mii PP =

then λF ≥′

Thus it can be concluded that if ith generator is between its limits, the

remaining generators are either at its upper or lower limit depending on whether their

incremental cost F is larger or smaller than F.

This analysis leads to the following strategy for optimum dispatch. A table is

made up of the available generating units with their corresponding values of the

incremental cost in ascending order. For an example

Unit dF/dt $/Mwh

2 dF2/dt

5 dF5/dt

7 dF7/dt

13 dF13/dt

10 dF10/dt

1 dF1/dt

For a given load if unit 13 is operating between its limits, units 2,5 and 7 are at

their upper limit and units 10,1 at their lower limit. In practice it is arranged that not

all units below 13 are at their lower limit, but perhaps only 10 and 1, so that spinning

reserve is ensured.

The above table is known as the merit order list of the units, the method is

known as “Order of Merit” method.

3.5.2 Equal Incremental Cost Criterion (Neglecting Transmission Loss)

Let there be a system of thermal generating units connected to a single bus-bar

serving a received electrical load Pload. It is required to run the machines so that cost of


51

Lecture No. 3

generation is minimum subject to total generation are equal to total demand when

transmission losses have been neglected.

This problem may be solved by using “Equal Incremental Cost Criterion”.

Working philosophy of their criterion is as: “When the incremental costs of all the

machines are equal, and then cost of generation would be minimum subject to equality

constraints”.

The economic dispatch problem mathematically may be defined as:

Minimize (3.5) ( )∑=

=N

1iiiT PFF

Subject to or (3.6) ∑=

=N

1iiD PP 0PPΦ

n

1iiload =−= ∑

=

Where

FT = F1 + F2 + ……FN is the total fuel input to the system

Fi = Fuel input to ith unit

Pi = The generation of ith unit

This is the constrained optimization problem with equality constraints only.

Three methods can be applied for locating the optimum of the objective function.

These methods are:

Direct substitution

Solution by constraint variation

The method of Lagrange multipliers

However, in this analysis method of Lagrange multiplier will be used. The

working philosophy of this method is that constrained problem can be converted into

an unconstrained problem by forming the Lagrange, or augmented function. Optimum

is obtained by using necessary conditions.

Case-1

No Generation Limits i.e. ( ) (max.iimin.i PPP )≤≤ Neglected.

The augmented unconstrained cost function is given by


52

Lecture No. 3

⎟⎠

⎞⎜⎝

⎛−+=

+=

∑=

N

1iiDT

T

PPλ.F

λ.ΦFL (3.7)

The necessary conditions for constrained local minima of L are the following:

0PL

i

=∂∂ (3.8)

0λL=

∂∂

(3.9)

C-I

First condition gives

( ) 010λ.PF

PL

i

T

i

=−+∂∂

=∂∂

or

λPF0λ

PF

i

T

i

T =∂∂

⇒=−∂∂

Q N21T FFFF −−−−−−−++=

then

λdPdF

PF

i

T

i

T ==∂∂

and therefore the condition for optimum dispatch is

λdpdF

i

T = (3.10)

or

λP2cb iii =+ (3.10)

where 2

iiiiiT PcPbaF ++=


53

Lecture No. 3

C-II

Second condition results in

∑=

=−=∂∂ N

1iiD 0PP

λL

or

∑=

=N

1iDi PP (3.12)

In summary,

“When losses are neglected with no generator limits, for most economical

operation, all plants must operate at equal incremental production cost while

satisfying the equality constraint given by equation (3.12 ).”

i

ii 2c

bλ.P −= (3.13)

The relations given by equation (3.13) are known as the co-ordination

equations. They are function of λ. An analytical solution for λ is given by

substituting the value of Pi in equation (3.12), i.e.,

D

N

1i i

i P2c

bλ.=

−∑=

(3.14)

D

N

1i

N

1i i

i

i

P 2cb

2c1 .λ =−∑ ∑

= =

∑

∑

=

=

+= N

1i i

N

1i i

iD

2c1

2cbP

λ (3.15)

Optimal schedule of generation is obtained by substituting the value of λ from

eq. (3.15) into eq. (3.13).


54

Lecture No. 3

Solution of Equation (3.13) Iteratively

In an iterative search technique, starting with two values of λ, a better value of λ

is obtained by extrapolation, and process is continued until ∆Pi is within a specified

accuracy. However rapid solution is obtained by the use of gradient method. Equation

(3.14) can be written as

( ) DPλ.f = Expanding left hand side of the above equation in Taylor’s series about an

operating point λ (k) and neglecting higher order terms results in

( )( ) ( ) ( )D

kk

k PΔλ.dλλ.dfλ.f =⎟

⎠⎞

⎜⎝⎛+

( ) ( )( )

( ) ( )k

kDk

dλλ.dfλ.fP

Δλ.

⎟⎠⎞

⎜⎝⎛

−=

( )( )

( ) ( )

( )

∑=

⎟⎠⎞

⎜⎝⎛

=

i

k

k

kk.

2c1

ΔP

dλλ.df

ΔPΔλ (3.16)

and therefore ( ) ( ) (kk1k Δλ.λ.λ. +=+ ) (3.17)

where

( ) ( )∑=

−=N

1i

kiD

k PPΔP (3.18)

The process is continued until ∆PP

(k) is less than a pre-specified accuracy. Case II

Generation limits i.e., Pi(min) ≤ Pi ≤ Pi(max.) Included

The power output of any generator should not exceed its rating nor should it be

below that necessary for stable boiler operation. Thus generators are restricted to be

within given maximum and minimum limits. The problem is to find the real power

generation for each unit such that objective function (i.e., total production cost) is

minimum, subject to equality constraints and the inequality constraints i.e.,


55

Lecture No. 3

Minimize: ( )∑=

=N

1iiiT PFF

Subject to: ∑=

−=N

1iiD PPΦ

And: for i =1,2,…,N ( )

( ) ⎪⎭

⎪⎬⎫

≤−

≤−

0PP0PP

imin.i

max.ii

This is constrained optimization problem both with equality and inequality

constraints. The optimization problem with equality and inequality constraints is

handled well by Lagrange multiplier method. However, Khun-Tucker conditions

complement the Lagrange conditions to include inequality constraints as additional

terms.

For establishing optimal solution for this problem, we consider the case of two

machines and would be generalized for N machines. The problem may be stated as:

Minimize: ( ) ( )2211T PFPFF +=

Subject to: ( ) 21D21 PPPP,PΦ −−=

⎪⎩

⎪⎨⎧ ≤+−=

≤−−=⇒≤≤

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

01P1P1P1g

01P1P1P2gPPP 111

⎪⎩

⎪⎨⎧ ≤+−=

≤−−=⇒≤≤

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

0112302224

PPP 222

PPPg

PPPg

Lagrange function becomes

( ) ( ) ( ) ( ) ( ) (( ) ( ) ( ) ( ) ( )( ) ( )224223

11211121D2211

24423312211121T

PPμ.PPμ.

PPμ.PPμ.PPPλ.PFPF

PgμPgμPgμPgμP,Pλ.ΦFμλ,P,L

−+−+

−+−+−−++=

+ )++++=

−+

−+

The necessary conditions for an optimum for the point Po, λo, µo are:

Condition 1


56

Lecture No. 3

( ) 0μ,λ,PPL

i

=°°°∂∂

( ) 0μ.μ.λPF 2111 =−−−′

( ) 0μ.μ.λPF 4322 =−−−′

Condition 2

( ) 0PΦ i =°

0PPP 21D =−−

Condition 3

( ) 0Pgi ≤°

0PP 11 ≤− +

0PP 11 ≤−−

0PP 22 ≤− +

0PP 22 ≤−−

Condition 4

( ) 0Pgμ. i =°° 0μ.i ≥°

( ) 0μ. 0PPμ. 1111 ≥=− +

( ) 0μ. 0PPμ. 2112 ≥=−−

( ) 0μ. 0PPμ. 3223 ≥=− +

( ) 0μ. 0PPμ. 4224 ≥=−−

Case-I

If optimum solution occurs at values for P1 and P2 that are not at either upper

limit or lower limit’ then all µ values are equal to zero and

( ) ( ) λPFPF 2211 =′=′


57

Lecture No. 3

That is, the incremental cost associated with each variable is equal and this

value is exactly the λ.

Case- II

Now suppose that the optimum solution requires that P1 be at its upper limit

( )0PP i.e., 11 =− + and that P2 is not at its upper or lower limit. Then

0μ.1 ≥ and 0μ.μ.μ. 432 ===

From condition-1

( ) ( ) λPFμ.λPF 11111 ≤′⇒−=′

( ) λPF 22 =′

Therefore, the incremental cost associated with the variable that is at is upper

limit will always be less than or equal to λ, whereas the variable that is not at

limit will exactly equal to λ.

Case- III

Now suppose that optimum solution requires that P1 be at its lower

limit ( )0PPi.e., 11 =−− and that P2 is not at its upper and lower limit. Then

0μ.2 ≥ and 0μ.μ.μ 431 ===

From Condition I, we get:

( ) ( ) 0PFμ.λPF 11211 ≥′⇒+=′⇒

( ) λPF 22 =′

Therefore, the cost associated with a variable at its lower limit will be greater

than or equal to λ, whereas, again incremental cost associated with variable

that is not at limit will exactly equal to λ.

Case- IV

If the optimum solution requires that both P1, P2 are at limit and equality

constraint can be met, then λ and non-zeroµ values are indeterminate. Let


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Lecture No. 3

0PP 11 =− +

0PP 22 =− +

Then 0μ.1 ≥ 0μ.3 ≥ 0μ.μ. 42 ==

Condition-1 would give

( ) 111 μ.λPF −=′

( ) 222 μ.λPF −=′

Specific values for λ, µ1 and µ2 would be undetermined.

For the general problem of N variables

Minimize: ( ) ( ) ( ) ( )NN22i1iT PF...PFPFPF +++=

Subject to: 0P...PPL N21 =−−−−

And: N1,2,...,1for 0iPiP

0PP ii

=⎪⎭

⎪⎬⎫≤+−

≤−−

Let the optimum lie at Pi = Pi0, i= 1,2,…,N and assume that at least one xi is

not at limit. Then,

if ( ) λPF then PP and PP iiiiii =°′>°<° −+

if ( ) λPF then PP iiii ≤°′=° +

if ( ) λPF then PP iiii ≥°′=° −

3.5.3 λ ITERATION METHOD

The incremental production cost of a given plant over a limited range is given

by

iiii

i PbadPdF

+= (1)

Where bi = slope of incremental production cost curve


59

Lecture No. 3

ai = y-intercept of incremental production cost curve

The necessary condition for optimum schedule is as:

λdPdF

i

i = (2)

Subject to equality and inequality constraints i.e.,

∑=

−=N

1iiD PPΦ (3)

( )

( )N1,2,3,...,fori

0PP0PP

imin.i

max.ii=

⎪⎭

⎪⎬⎫

≤−

≤−

For optimal schedule equations (2) and (3) can be solved simultaneously. As

inequality constraints have also to be taken into account, the following iterative

method known as λ iteration method may be used for solution.

1. Assume suitable value λo This value should be more than the largest intercept

of the incremental cost characteristic of various generators.


60

Lecture No. 3

2. Compute the individual generations P1, P2,…,PN corresponding to the

incremental cost of production from equation 2. In case generation at any bus

is violated during that iteration and the remaining load is distributed among the

remaining generators.

3. Check if the equality

D

N

1ii PP =∑

=

is satisfied.

4. If not, make a second guess λ′ and repeat the above steps.

3.6 PROBLEM

PROBLEM #1 This problem has been taken as standard test system for testing the algorithms in IEEE Transaction papers.

The fuel costs functions for three thermal plants in $/h are given by

F1 (P1) = 500 + 5.3 P1 + 0.004 P12

F2 (P2) = 400 + 5.5 P2 + 0.006 P22

F3 (P3) = 200 + 5.8 P3 + 0.009 P32

If the total load to be supplied PD = 800 MW, then find the optimal

dispatch and total cost by the following methods:

1. Analytical method

2. Graphical demonstration

3. Iterative technique using gradient method

Given that line losses and generation limit have been neglected.

Solution:


61

Lecture No. 3

0.0181

0.0121

0.0081

0.0185.8

0.0125.5

0.0085.3800

32c1

22c1

12c1

32c3b

22c2b

12c1b

DP

N

1i 22c1

N

1i i2cib

DPλ

++

+++=

++

+++=

∑=

∑=

+=

8.5$/MWh263.8889

1443.0555800λ =+

=

The co-ordination equation is given by

2

22 2c

bλP −=

Substituting the values of λ, we have optimal dispatch,

400.002(0.004)

5.38.52c

bλP1

101 =

−=

−=

250.002(0.006)

5.68.5

22c2bλ0

2P =−

=−

=

150.002(0.009)

5.88.5

32c3bλ0

3P =−

=−

=

P10 + P2

0 + P30 = 400+250+150 = 800 MW = PD

2. Graphical Demonstration

The necessary conditions for optimal dispatch are.

λ0.008P5.3dPdF

11

1 =+=

λ0.012P5.5dPdF

22

2 =+=


62

Lecture No. 3

λ0.018P5.8dPdF

33

3 =+=

Subject to P1 + P2 + P3 = PD

To demonstrate the concept of equal incremental cost for optimal dispatch, we

plot the equal incremental cost of each plant on the same graph as shown in figure3.

To obtain a solution various values of λ could be tried until one is found which

produces ∑Pi = PD for each λ, if ∑Pi < PD, we increase λ otherwise if ∑Pi > PD we

reduce λ. Thus the horizontal dashed line move up or down until at optimum point λ0,

∑Pi = PD

For λ0 = 8.5 P10 = 400 P2

0 = 250 P30 = 150

Satisfying ∑Pi = PD

Figure 3.2 Plot of Incremental Costs


63

Lecture No. 3

λ-ITERATION

METHOD


64

Lecture No. 3

3. λ Iteration using gradient method

For the solution using iterative method, we assume starting value of λ(1) = 6.0

from the co-ordination equations.

i

ii 2c

bλP −=

We find Pi and then check ∑Pi = PD if this condition is not satisfied, we


65

Lecture No. 3

update λ by

kΔλ

(k)λ

1)(kλ +=

+

Where

22c

1

(k)ΔP(k)Δλ∑

=

ITERATION 1

0.6λ (1) =

50.87)004.0(23.50.6P(1)

1 =−

=

67.41)006.0(25.50.6P(1)

2 =−

=

11.11)009.0(28.50.6P(1)

3 =−

=

∑Pi(1) = 87.50 + 41.67 + 11.11 = 140.28

PD = 800 MW

ΔPP

(1) = 800-140.2777 = 659.72

∑Pi(1) ≠ PD we calculate

2.5

0.0191

0.121

00081

659.7222

2c1

2c1

2c1

ΔPΔλ

321

(1)(1) =

++=

++=

Thus new value of λ is

λ(2) = λ(1) +Δλ(1) = 6.0+2.5=8.5

ITERATION 2

00.400)004.0(23.55.8P(2)

1 =−

=


66

Lecture No. 3

00.250)006.0(25.55.8P(2)

2 =−

=

00.150)009.0(28.55.8P(2)

3 =−

=

Δ Pi(2) = PD - ∑Pi

(2) = 800 – (400+250+150) = 800 – 800 = 0

ΔPi(2) = 0, the equality constraint is met in two iteration.

Optimal Dispatch

P10 = 400 MW

P20 = 250 MW

P30 = 150 MW

λ0 = 8.5

Total fuel cost is given by

Ft = F1 (P1) + F2 (P2) + F3 (P3)

= {500 + 5.3(400) + 0.004(400)2} + {400+5.5(250) + 0.006(250)}

+ {200+5.8(150)+0.009 (150)2}

Ft = 6, 682.5 $/h.

PROBLEM 2

Let there be three units with following data:

Unit 1: Coal fired steam units

Input - output curve:

600P150 00142.02.7510 12

111 ≤≤++=⎟⎠⎞

⎜⎝⎛ PP

hMBtuH

Unit 2: coal fired steam unit



67

Lecture No. 3

400P100 00194.085.7310 12

222 ≤≤++=⎟⎠⎞

⎜⎝⎛ PP

hMBtuH

Unit 3: coal fired steam unit


MWPPh

MBtuH 200P50 00482.097.778 12

333 ≤≤++=⎟⎠⎞

⎜⎝⎛

It is required to determine the economic operating point for these three

units when delivering a total of 850 MW with reference to the following cost data:

Case-I

Unit 1: fuel cost = 1.1 $/MBtu

Unit 2: fuel cost = 1.0 $/MBtu

Unit 3: fuel cost = 1.0 $/Mbtu

Case-II

The fuel cost of Unit 1 decreased to 0.9 $/MBtu

Solution.

Case-I

Fuel costs are given by:

F1 (P1) = H1 (P1) x 1.1 = 561+7.92P1 + 0.001562 P12 $/h

F2 (P2) = H2 (P2) x 1.0 = 310+7.85P2 + 0.00194 P22 $/h

F3 (P3) = H3 (P3) x 1.0 = 78+7.97P3 + 0.00482 P32 $/h

If λ is the incremental cost, then condition for optimal dispatch are:

λ0.003124P7.92dPdF

1

1 =+= (1)


68

Lecture No. 3

λ 0.003887.85dPdF

2

2 =+=

λ 0.009647.97dPdF

3

3 =+=

and P1+P2+P3 = 850 (4)

003124.092.7

1−

=λP

00388.085.7

2−

=λP

00964.097.7

3−

=λP

Putting the values of P1, P2, & P3 in = (4), we have

+−

003124.092.7λ

00388.085.7−λ 850

00964.097.7

=−

+λ

λ = 9.148 $/MWh

Now

P10 = 393.2 Mw 150 ≤ P1 ≤ 600

P20 = 334.6 Mw 100 ≤ P2 ≤ 400

P30 = 122.2 Mw 50 ≤ P3 ≤ 250

∑ Pi0 = 393.2 + 334.6+122.2=850

All the generations are within their limits and equality constraint is also

satisfied.

Case-II

Cost of Coal for Unit 1 decreases to = 0.9 $/MBtu

The fuel cost of coal-fired plats is given by


69

Lecture No. 3

F1 (P) = H1 (P) x 0.9 = ( 510 + 7.2 P1 + 0.00142 P12) x 0.9

= 459 + 6.48 P1 + 0.0128 P12

F2 (P2) = 310 + 7.85 P2 + 0.00194 P22

F3 (P3) = 78 + 7.97 P3 + 0.00482 P32

11

1 0.0256P6.48dPdF

+=

22

2 0.00388P7.85dPdF

+=

33

3 0.00964P7.97dPdF

+=

The necessary conditions for an optimum dispatch are:

λ0.0256P6.48dPdF

11

1 =+= (1)

λ0.00388P7.85dPdF

22

2 =+= (2)

λ0.00964P7.97dPdF

33

3 =+= (3)

And

P1 + P2 + P3 = 850 MW (4)

From equations (1), (2) & (3), we have.

0.002566.48λP1

−=

0.003887.85λP2

−=

o.oo9647.97λP3

−=


70

Lecture No. 3

Substituting the value in equation (4), we get

λ = 8.284 $ / Mwh

Substituting the value of λ, we have optimal schedule as:

P1 = 704.6 MW

P2 = 111.8 MW

P3 = 32.6 MW

∑Pi = P1 + P2 + P3 = 849.0 Mw ≅ 850 Mw

This schedule satisfies the equality constraint, but unit 1 and unit 3 are not

within the limits. Unit 1 violate upper limit & unit 3 violate lower limit. To

solve this problem, we clamp the violated generations to their

corresponding limits and optimal set is obtained by satisfying necessary

conditions.

Unit 1 is fixed to Max limit i.e., P1 = 600 Mw

Unit 3 is fixed to Min limit i.e., P3 = 50 Mw

Q P1 + P2 + P3 = 850

P2 = 850- (P1+P3) = 850- (600+50) = 200 Mw

Now the schedule is as:

P1 = 600 Mw

P2 = 200 Mw

P3 = 50 Mw

The necessary conditions for this schedule are as:

MW 600(Max)PPλdPdFC 11

1

11 ==≤→ Q (Clamped)

λdPdFC

2

22 =→ Since P2 is within inequality 200 ≤ P2 ≤ 400 = 200Mw

50MW(Min)PPλdPdF

C 333

33 ==≥→ Q (Clamped)


71

Lecture No. 3

8.016$/MWh)0.0256(6006.48dPdF

6001

1 =+=

8.626$/MWh0)0.00388(207.85dPdF

2002

2 =+=

8.626$/MWh)0.00964(507.97dPdF

503

3 =+=

Condition 1

626.8 016.8dPdF

6001

1 <= True

Condition 2

λ = 8.626 $/MWh True

Condition 3

8.626 452.8dPdF

503

3 <= False

Condition 3 is violated, so for the optimal schedule, we keep unit 1 at its max

limit, allow unit 2 and unit 3 incremental costs equal to λ.

P1 = 600 Mw

λ0.00388P7.85dPdF

22

2 =+= (1)

λ0.00964P7.97dPdF

33

3 =+= (2)

P2 + P3 = 850 – P1 = 250 MW (3)

From equations (1) & (2), we have

8.576 λ

2500.00964

7.97λ0.00388

7.85λPP 32

=⇒

=−

+−

=+

Now the economic schedule for unit 2 and 3 corresponding to value of λ = 8.576

$/MWh


72

Lecture No. 3

P2 = 187.1 Mw

P3 = 62.90 Mw

8.016dPdF

6001

1 = < 8.576 True

8.576.dPdF

dPdF

3

3

2

2 == True

The figure represents the graph of the economic schedule.

PROBLEM # 3

Unit 1: H1 = 80 + 8 P1 + 0.024 P12 20 ≤ P1 ≤ 100

Unit 2: H2 = 120 + 6 P2 + 0.04 P22 20 ≤ P2 ≤ 100

Where

Hi = Fuel input to unit i in Mbtu/hour

Pi = Unit output in Mw.

1. Plot input/output characteristic for each unit.

2. Calculate net heat rate in Btu/Kwh and plot against output in Mw

3. Assume Fuel cost 1.5 $/Mbtu. Calculate the incremental production cost in $/Mwh

of each unit and plot against output in Mw.

Solution

1. Plot input/output characteristic for each unit.

UNIT 1 UNIT 2

P1 (Mw) H1 (Mbtu/h) P2 (Mw) H2 (Mbtu/h)

20 249.6 20 256.0

30 341.6 30 336.0

40 438.4 40 424.0

50 540.0 50 520.0

60 646.4 60 624.0


73

Lecture No. 3

70 757.6 70 736.0

80 873.6 80 856.0

90 994.4 90 984.0

100 1120.0 100 1120.0

0

200

400

600

800

1000

1200

20 30 40 50 60 70 80 90 100

20 ≤P1≤ 100 H1=8 P1+0.024 P1

2+80

H1

BTU/hr

P1 (MW)


74

Lecture No. 3

0

200

400

600

800

1000

1200

20 30 40 50 60 70 80 90 100

20 ≤P1≤ 100 H2=6 P2+0.04 P2

2+120

H2

BTU/hr

P2 (MW)

b) Calculate net heat rate in Btu/Kwh and plot against output in MW

UNIT 1 UNIT 2

P1 H1/P1 P2 H2/P2

20 249.6/20 = 12.48 20 256/20 = 12.80

30 341.6/30 = 11.38 30 336/30 = 11.20

40 438.4/40 = 10.96 40 424/40 = 10.60

50 540/50 = 10.8 50 520/50 = 10.40

60 646.4/60 = 10.77 60 624/60 = 10.40

70 757.6/70 = 10.82 70 736/70 = 10.50

80 873.6/80 = 10.92 80 856/80 = 10.70

90 994.4/90 = 11.04 90 984/90 = 10.93

100 1120/100 = 11.20 100 1120/100 = 11.20


75

Lecture No. 3

H1/P1 vs P1

0

5

10

15

0 20 40 60 80 100 120

P1

H1/P

1

H2/P2 vs P2

0

5

10

15

0 20 40 60 80 100 120

P2

H2/P

2

2. Assume Fuel cost 1.5 $/Mbtu. Calculate the incremental production cost in

$/Mwh of each unit ad plot against output in Mw.

F1 (P1) = H1 x 1.5 = 120 + 12P1 + 0.036 P12

F1 (P2) = H2 x 1.5 = 180 + a P2 + 0.06 P22

11

1 P 072.012dPdF

+=

22

2 P 12.09dPdF

+=


76

Lecture No. 3

P1& P2 (Mw) dF1/dP1 ($/Mwh) dF2/dP2 ($/Mwh)

20 13.44 11.4

30 14.16 12.6

40 14.88 13.0

50 15.6 15.0

60 16.32 16.2

70 17.04 17.4

80 17.76 18.6

90 18.48 19.8

100 19.20 21.0

PROBLEM # 4

Unit 1: Coal Fired Steam

H1 (P1) = 510 + 7.2 P1 + 0.00142 P12 150 ≤ P1 ≤ 600 Mw Fuel Cost = 1.1 Rs/Mbtu

Unit 2: Coal Fired Steam

H2 (P2) = 310 + 7.85 P2 + 0.00194 P22 100 ≤ P2 ≤ 400 Mw Fuel cost = 1.0 Rs/Mbtu

1. Determine Economic Schedule for load demand of 728 Mw.

2. Determine total fuel cost/hour.

3. Determine the specific cost at Economic Scheduled generation.

4. Determine the average specific cost of unit 1 & unit 2.

5. For the following capacity charges determine the total cost per unit for unit ≠ 1

& Unit # 2

Capacity Charges for Unit 1 = 6 times the av. sp. cost

Capacity Charges for Unit 2 = 4 times the av. sp. cost

Solution

F1 (P1) = H1 (P1) x 1.1 = 561 + 7.92 P1 + 0.001562 P12

F2 (P2) = H2 (P2) x 1.0 = 310 + 7.85 P2 + 0.00194 P12


77

Lecture No. 3

(3)728PP

(2)λP 0.003887.85dPdF

(1)λP 0.0031247.92dPdF

21

22

2

11

1

L

L

L

=+

=+=

=+=

For economic schedule the necessary condition is:

λdPdF

dPdF

2

2

1

1 ==

From =S (1) & (2), we have,

0.003887.85λP

0.0031247.92λP 21

−=

−=

Substituting in equation (3), we get,

λ = 9.1486 Rs/Mwh

Corresponding to this value of λ, the optimum values of P1 and P2 (satisfying

equality constraints are :

P1 = 393 MW

P2 = 335 MW

2. Fuel cost per hour

Rs/h7072.3 Rs/h 5.31578.3914 FFFFF 3352393121t

=

+=+=+=

3. The specific cost at Economic Scheduled generation.

Specific Cost of Unit 1.

Sp Cost1 = 3914.8/393 Rs./Mwh

= 9.96134 Rs./Mwh

= 0.996134 Paisa/Kwh

Specific Cost of Unit 2

Sp Cost2 = 3157.5/335

= 9.4253 Rs./Mwh


78

Lecture No. 3

= 0.9425 Paisa/Kwh

Specific Cost of System = 9.801466 Rs./Mwh = 0.97146 Paisa/Kwh.

4. Specific Cost of Unit 1

9.7922 05875.32/60 5875.32 PF 100%

9.80236 05293.28/54 5293.28 PF 90%

9.83851 04722.48/48 4722.48 PF 80%

9.911754 04162.94/42 4162.94 PF 70%

10.0406 03614.64/36 3614.64 PF 60%

10.2586 03077.58/30 3077.58 PF 50%

11.8943 0178.145/15 145.1784F 25%

561/0 561 F 0

/)(F Load Various F (Rs./Mwh)Cost Speicfic (Rs./h)Hour Per Cost Fuel Load %

60011

54011

48011

42011

36011

30011

15011

011

111011

==

==

==

==

==

==

==

==

=

=

=

=

=

=

=

=

=

=

P

P

PPP

α

Average Sp Cost of Unit 1 = 61.74618/6 = 10.29103 Rs./Mwh = 10.29103 Rs./Mwh (Average of Six readings 25% to 90%) Average Specific Cost of Unit 2.

Average Sp Cost of Unit 2 = 58.88669/6 = 9.814778 Rs./Mwh = 0.9814778 Rs./Kwh (Average of Six readings 25% to 90%)

Capacity Charges Evaluation

Unit 1

Rate of Capacity Charges = 6 x Av. Sp Cost

= 6 x 10.29103 = 61.74618 Rs./Mwh

= 6.174618 Ps./Kwh

Capacity Charges/h for the declared capacity of 600 Mw = 61.74618 x 600

= 37047.708 Rs./h

Unit 2

Rate of Capacity Charges = 4 x Av. Sp. Cost


79

Lecture No. 3

= 4 x 9.814778

= 39.259 Rs./h

Capacity Charges /h for the declared capacity of 400 Mw = 39.259 x 400

= 15703.6 Rs./hour.

Total Cost per Unit of Energy = Capacity (fixed) Charges + Av. Sp Cost

Unit 1 = 61.7461 + 10.29103 = 72.03721 Rs./Mwh

Unit 2 = 39.259 + 9.81477 =49.07389 Rs./Mwh

3.9 REFERENCE

[1]. A.J.Wood & Bruce F. Wollenberg, Power Generation , Operation & Control, John Wiley & Sons, 1996.


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Economic Dispatch