UNIT - IV
CONTENTS
Economic Dispatch
Unit Commitment
Economic Dispatch
Thermal power plant
H=Btu per hour heat input to the unit
F=Fuel cost times H is the Rs per hour input to the unit for fuel
Input output curve for a steam turbine generator
Hydro electric power plant
System Variables
Control variables (Pg and Qg)
Disturbance Variables (PD and QD)
State variables (V and )
PROBLEM OF OPTIMUM DISPATCH
FORMULATION
Scheduling is the process of allocation of generation
among different generating units.
Economic scheduling is a cost effective mode of
allocation of generation among the different units in
such a way that the overall cost of generation
should be minimum. This can also be termed as
an optimal dispatch.
Incremental Fuel Cost curve
Incremental Fuel Cost
Incremental Heat rate
Incremental efficiency
input
output
F
PG
dCi
dPGi
dPG
output
input
dPC
Classical methods for economic operation of system plants
Base loading to capacity
Base loading to most efficient load
Proportional loading to capacity
Merit Order Dispatch
If the incremental generation costs are substantially
constant over the range of operation, then without
considering
reserve
and
transmission
line
limitations, the most economic way of scheduling
generation to load each unit in the system to its rated
capacity in the order of highest incremental
efficiency.
Optimization problem
Objective function
The objective function is to minimize the overall
cost of the production of power generation.
Total cost C= C1+C2+C3+..Cn
i.e.,
n
C Ci
i 1
n
C Ci ( PGi )
i 1
i.e., C C1 ( PG1 ) C2 ( PG 2 ) C3 ( PG 3 ) ....Cn ( PGn )
Constraint Equation
1. Equality Constraints
The sum of real-power generation of all
various units must always be equal to the total
real-power demand on the system.
n
PD PGi
i 1
2. Inequality Constraints
ECONOMIC DISPATCH PROBLEM (Solution by Direct Method)
Analytical Solution for (Losses are neglected with no generator
limits)
Problem 2
The fuel cost of two units are given by,
F1 1.6 25PG1 0.1PG1 Rs/Hr
2
F2 2.1 32PG1 0.1PG1 Rs/H
2
If the total demand on the generators is 250 MW,
find the economic load scheduling of the two units.
Problem 2
A power plant consists of two 200 MW units, whose
input cost data given by,
F1 0.004P1 2P1 80 Rs/Hr
2
F2 0.006P2 1.5P2 100 Rs/Hr
2
If the total load of 250 MW, what should be the
division of load between two units for most economic
generation?
Problem 3
A power plant consists of two 200 MW units, whose
input cost data given by,
F1 0.05P1 23.5P1 700 Rs/Hr
2
F2 0.2P2 20P2 850 Rs/Hr
2
F3 0.9P3 18P3 960 Rs/Hr
2
Maximum and minimum loading allowable on each
unit are: 150 MW and 40 MW. Find the optimal
scheduling for a load of 275 MW.
Problem 4
A power plant consists of two 100 MW units, whose input cost data given
by,
F1 0.05P1 20P1 800 Rs/Hr
2
F2 0.06P2 15P2 1000 Rs/Hr
2
If the total received power is 150 MW,
(i) What would be the division of each load between the units for the
most economic operation?
(ii) Find the savings per hour realized from economic allocation of load
between the units in comparison with their sharing the output equally,
when the load is 150 MW.
Problem 5
Determine the economic scheduling of he three generating units in
a power system to meet the load of 925 MW. The operating limits
and cost function is given below
Operating limits
250 MW PG1 450 MW
200 MW PG1 350 MW
125 MW PG1 225 MW
F1 (P1 ) 0.0045P1 5.2P1 580 Rs/Hr
2
Cost Function is
F2 (P2 ) 0.0056P2 4.5P2 640 Rs/Hr
2
F3 (P3 ) 0.0079P3 5.8P3 820 Rs/Hr
2
Base point and Participation Factors
1
Pi
Fi "
N
1
PD
i 1 Fi "
Pnew ,i
Pi
Pbase ,i (
) PD
PD
Problem 6
The input output curve characteristics of three units are
F1 0.0016P1 5.46P1 940 Rs/Hr
2
F2 0.0019P2 5.35P2 820 Rs/Hr
2
F
0
.
0032
P
P3 method
99 Rs/Hr
3 5.65
The total load is 3600 MW. Use participation
factor
to calculate the dispatch for a
load is reduced to 550 MW
Problem 7
The input output curve characteristics of three units are
F1 0.0035P1 6.49P1 750 Rs/Hr
2
F2 0.0015P2 5.75P2 870 Rs/Hr
2
TheF
fuel
cost
of
unit
1
,unit
2
is 1.0
load is 800 MW. Use the
0.0010P3 and
unit
8.356
PRs./Mbtu.Total
3
3 620 Rs/Hr
participation factor method to calculate he dispatch for a load is increased to 880 MW.
ECONOMIC DISPATACH
Vs
UNIT COMMITMENT
ECONOMIC DISPATCH:
The economic dispatch problem assumes that there
are N units already connected to the system. The
purpose of economic dispatch problem is to find the
optimum operating policy for these N units.
UNIT COMMITMENT:
The unit commitment problem is more complex. We
may assume that we have N units available to us and
that we have a forecast of the demand to be served.
Problem 8
So far, We have obeyed one simple constraint: Enough units
will be committed to supply the load. Thus we could stop here
and state that the problem was solved.
Unfortunately, other constraints and other phenomena must
be taken into account in order to claim an optimum solution.
Constraints in Unit Commitment
Spinning reserve
Thermal constraints
Other Constraints
Spinning Reserve (or) Synchronized Reserve :
Spinning reserve is the term used to describe the total
amount of generation available from all units
synchronized(i.e., spinning) on the system, minus present
load and losses.
If one unit is list, there must be ample reserve on the other
units to make up for the loss in a specified time period.
Reserves finally, must be spread around the power system
to avoid transmission system limitations (often called
bottling of reserve) and to allow various parts of the
system to run as islands, should they become electrically
disconnected.
Thermal Unit Constraints
Thermal units usually require a crew to operate them,
especially when turned on and turned off. A thermal unit
can undergo only gradual temperature changes, and
this translates into a time period of some hours required
to bring the unit on-line.
Minimum up time:
Once the unit is running, it should not be turned off
immediately.
Minimum down time:
Once the unit is de-committed, there is a minimum time
before it can be recommitted
Crew constraints:
If a plant consists of two or more units, they cannot
both be turned on at the same time since there are
not enough crew members to attend both units while
starting up.
Start-Up Cost:
(curve)
Methods to solve Unit Commitment Problem
1.
Priority list method
2. Dynamic Programming method
3. Lagrange relation method
4. Integer programming method
5. Branch and Bound method
6. Evolutionary algorithms
What is the need for different methods
Procedure:
We must establish a loading pattern for M periods
We have N units to commit and dispatch
The M load levels and operating limits on the N units are
such that any one unit can supply the individual loads
and that any combination of units can also supply loads
C( N,2) ...C( N, N 1) C( N, N) 2 1
N
N!
C( N, j)
(
N
j
)!
j
!
j ! 1 2 3.... j
For the total period of M intervals, the maximum
number of possible combinations is (2 N 1) M , which
can become a horrid number to think about
Priority List methods
The simplest Unit Commitment solution method consists of
creating priority list of units.
It is a simple shun-down rule or priority list scheme.
The priority list can be obtained in a much simpler manner by
noting down the full load average production cost (FLAPC) of
each unit.
Full load average production cost is simply the neat heat rate at
he full load multiplied by the fuel cost.
FLAPC
Ci (PG i ) K.H i (PG i )
PG i
PG i
Assumptions in Priority List method
No load cost are zero
Unit input- output characteristics are linear between zero
output and full load.
There are no other constraints(Minimum up/Down time)
Start-up costs are fixed amount.
Assumptions in Dynamic Programming method
1.
A state consists of an array of units with specified units
operating(on-line) and the rest off-line
2. The startup cost of a unit is independent of the time if it has been
off-line.
3. There are no costs for shutting down a unit.
4. There is a strict order, and in each interval a specifies minimum
amount of capacity must be operating.
