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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
Power system Operational Planning Economic Dispatch
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.
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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.
Power system Operational Planning Economic Dispatch
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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.
Power system Operational Planning Economic Dispatch
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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
Power system Operational Planning Economic Dispatch
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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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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 =′
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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 ++=
Power system Operational Planning Economic Dispatch
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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).
Power system Operational Planning Economic Dispatch
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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.,
Power system Operational Planning Economic Dispatch
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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
Power system Operational Planning Economic Dispatch
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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 =′=′
Power system Operational Planning Economic Dispatch
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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
Power system Operational Planning Economic Dispatch
58
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
Power system Operational Planning Economic Dispatch
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.
Power system Operational Planning Economic Dispatch
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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 $/h are given by
F1 (P1) = 500 + 5.3 P1 + 0.004 P12
F2 (P2) = 400 + 5.5 P2 + 0.006 P22
F3 (P3) = 200 + 5.8 P3 + 0.009 P32
If the total load to be supplied PD = 800 MW, then find the optimal
dispatch and total cost by the following methods:
1. Analytical method
2. Graphical demonstration
3. Iterative technique using gradient method
Given that line losses and generation limit have been neglected.
Solution:
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Lecture No. 3
0.0181
0.0121
0.0081
0.0185.8
0.0125.5
0.0085.3800
32c1
22c1
12c1
32c3b
22c2b
12c1b
DP
N
1i 22c1
N
1i i2cib
DPλ
++
+++=
++
+++=
∑=
∑=
+=
8.5$/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 =+=
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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
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Lecture No. 3
λ-ITERATION
METHOD
Power system Operational Planning Economic Dispatch
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
Power system Operational Planning Economic Dispatch
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 =−
=
Power system Operational Planning Economic Dispatch
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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
Input - output curve:
Power system Operational Planning Economic Dispatch
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Lecture No. 3
400P100 00194.085.7310 12
222 ≤≤++=⎟⎠⎞
⎜⎝⎛ PP
hMBtuH
Unit 3: coal fired steam unit
Input - output curve:
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)
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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
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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
−=
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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)
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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
Power system Operational Planning Economic Dispatch
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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
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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)
Power system Operational Planning Economic Dispatch
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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
Power system Operational Planning Economic Dispatch
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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
$/Mwh of each unit ad plot against output in Mw.
F1 (P1) = H1 x 1.5 = 120 + 12P1 + 0.036 P12
F1 (P2) = H2 x 1.5 = 180 + a P2 + 0.06 P22
11
1 P 072.012dPdF
+=
22
2 P 12.09dPdF
+=
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Lecture No. 3
P1& P2 (Mw) dF1/dP1 ($/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
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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
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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
Power system Operational Planning Economic Dispatch
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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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