Chapter – 1 INTRODUCTION

Chapter – 1
INTRODUCTION
Power System Operation and Control
1. INTRODUCTION
Power System comprises group of generators connected to the consumers
through transmission and distribution network to meet their electricity supply
requirements. Two important parameters that characterize the power system
are frequency and voltage.
The frequency of a power system is dependent entirely upon the speed at
which the generators are rotated by their prime-movers. All prime-movers,
whether they are steam or hydraulic turbines, are equipped with speed
governors to adjust the gate or control valve opening for constant speed. For
example, when a load is suddenly applied to the system, the individual
generators will meet this demand by the action of the prime-movers of
generators. The immediate effect of a sudden load is a reduction in speed of
the synchronous generators. However, the kinetic energy is normally sufficient
to maintain the energy balance until the reduction in speed is detected by the
drooping characteristic of the governor, which arrests the falling speed first
and then operates the gate or control valve opening to restore the output - input
balance by increasing the prime-mover torque. Thus the function of load –
frequency control on a power system becomes one of changing the gate or
control valve openings of generator prime-movers as a function of load
variations in order to hold system frequency constant.
The voltage of the power supply at the customer’s service terminals must be
held substantially constant. For example, motors operated at below normal
voltage draw abnormally high currents and may over heat, even when carrying
no more than the rated horsepower load. However, unlike frequency, voltage
can be controlled locally with suitable reactive power compensation. Service
voltage, usually specified by a nominal value, is maintained close to this
value, deviating not more than ± 5% of the nominal value.
1.1 Need for Voltage and Frequency regulation in Power System
In a Power System voltage and frequency need to be maintained for supplying
electricity with proper quality. The major reasons for this are:
1. Most types of AC motors run at speeds that are directly related to the
frequency.
2. The Generator Turbines, particularly Steam driven ones, are designed
to operate at specified speed with limited tolerance in variation for
maximum efficiency and less fatigue.
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3. The over all operation of a power system can be better controlled if
frequency error is kept within strict limits.
4. A large number of electrically operated clocks are used for power
system monitoring and control. They are all driven by synchronous
motors and the accuracy of these clocks is a function of the frequency
error.
5. All equipment and appliances are designed for a certain voltage level,
the rated or name plate voltage. If voltage V of the system should
deviate from that value, the performance of the device suffers and its
life expectancy drops. For example
V2
In Induction Motor,
Torque α
In a Lamp,
Light flux strongly varies with the
voltage, lower the voltage lower is the
flux.
There is, however, no need to regulate it within the same narrow
margins, as is the case with system frequency. Industry wide standards
exist, specifying tolerable voltage variations on a network.
6. The real line losses depend as much upon the reactive power as upon
the real line power flow. It is possible to minimize these losses by
selecting an optimum power flow, in terms of real and reactive powers.
A line flow depends greatly upon line end voltages, which thus become
a means of controlling the real losses.
Unusual deviations in frequency indicate that something is basically wrong
with the system. By reducing the normal frequency fluctuations to a faint
ripple, we will be able to detect the frequency disturbances following a fault at
an early stage. In modern energy systems the frequency constancy is normally
kept within ± 0.05 Hz.
1.2 System Load Variation
1.2.1 System Load Characteristics
Load is a device that taps energy from the network. The load ranges from few
watt night lamps to mega watt induction motor. The various load devices can
be classified into the following categories:
1.
2.
3.
4.
Motor devices;
Heating equipment;
Lighting equipment;
A diversity of electronic gear.
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From electrical point of view the multitude of devices are characterized by
vast difference in regard to
i.
ii.
iii.
iv.
Size
Symmetry (Single to 3 φ)
Load constancy (with reference to time, frequency and voltage)
Use cycle (regular or random use)
Characteristics of Typical System Load
1. Although individually of random type, the lumped or composite loads
as we encounter at substation levels are of highly predictable character.
2. Although the loads are time variant, the variations are relatively slow.
From minute to minute, we have an almost constant load. A minute is
long time period compared with the electrical time constants of the
power system and this permits us to consider the system operating in
steady state.
3. The typical load always consumes reactive power. Motors are always
inductive (with exception of over-excited synchronous machines).
4. The typical load is always symmetric. In case of large motors (> few
horse powers) this symmetry is automatic, since they are always
designed for balanced 3-phase operation. In case of 1-phase devices,
the symmetry comes about by statistical effects and intentional
distribution of loads between phases.
1.2.2 Loading pattern
The loading pattern can be expressed as either load curve or load duration
curve.
(b) Load Duration curve
15
Load in MW
5
10
15
10
5
Load in MW
20
20
(a) Load Curve
12
4
Midnight AM
8
12
Noon
4
8
PM
12
Midnight
0
4
8
12 16
20
24
Hours duration
Time of day
Fig.1.1 Load & Load Duration Curves
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Load Curve is the graph showing the variation in the demand for energy of
consumers on the supply system with respect to time. If the graph is plotted
for 24 hours it is called daily load curve; if the graph is plotted for one week,
one month or one year, we get weekly, monthly or annual load curves
respectively. The load curve is plotted chronologically.
The load curve helps to decide the operating schedule of the station i.e. how,
when and in what sequence various generating units should be started, run and
shut down.
Load duration curve gives the duration in hours for which the load has either
remained equal or exceeded the given value. The loads are arranged in the
order of descending magnitudes.
The area under the load duration curve is equal to that under load curve and
represents total energy consumed by load or delivered by generating station.
The load duration curve gives guidance for merit order operation of generating
stations and helps in deciding about the base load, peak load and intermediate
load stations.
Fig. 1.1 shows both load and load duration curves.
1.2.3 Definitions of commonly used terms in load/ demand
Connected load:
Each electrical device has its rated capacity, which is normally given in W,
kW or MW. The sum of the continuous ratings of all the electrical devices
connected to the supply system is known as connected load.
Demand:
The demand of an installation or system is the load at the receiving terminals
averaged over a specified interval of time, say quarter of an hour or half an
hour or an hour.
Maximum demand:
The maximum demand of an installation or system is the greatest of all
demands, which have occurred during the specified period of time and is
called daily, weekly, monthly or annual maximum demand.
Coincident demand:
The coincident demand is the demand of a composite group of loads over a
specified period of time. The coincident maximum demand is the maximum
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sum of the simultaneous requirements of the individual demands occurring
over a specified period of time.
Demand factor (DF):
It is the ratio of the maximum demand of a system or group to the total
connected load of the system/ group.
Maximum demand
DF =
(1.1)
Total connected load
The demand factor is less than or equal to unity.
Diversity factor (FD):
The diversity factor (FD) is the ratio of the sum of the individual maximum
demands of the various groups of consumers to the coincident maximum
demand of the whole system.
FD =
Sum of individual maximum demands
Coincident maximum demand
n
∑
=
where Pi
Pc
=
=
Pi
i =1
(1.2)
Pc
Maximum demand of load group i
Coincident maximum demand of n load groups.
The diversity factor is greater than or equal to unity.
If CPi is the connected load group i and DFi its demand factor, then
n
Pi
=
∑
CPi * DFi
i =1
n
∑
∴
FD
=
CPi * DFi
i =1
Pc
(1.3)
If there is only one group, then FD = 1/DF.
Load Factor:
It is the ratio of the average load over a specified period of time to the peak
load occurring in that period.
Average load
LF =
(1.4)
Peak load
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It can also be defined as the ratio of the total energy consumed over a
specified period of time to the energy that would have been consumed had the
peak load occurred throughout the period. Plant load factor is defined in
similar manner except that the energy is produced instead of consumed and the
plant had operated throughout the period. The energy can be computed either
from load curve or from load duration curve.
Significance of load factor and diversity factor:
Load factor and diversity factor play important roles in the cost of supply of
electrical energy. Higher the values of load factor and diversity factor, lower
will be the overall cost per unit generated.
The capital cost of the power station depends upon the capacity of the power
station. Lower the maximum demand of the power station, the lower is the
capacity required and therefore lower is the capital cost of the plant. With a
given number of consumers the higher the diversity factor of their loads, the
lower will be the capacity of the plant required and consequently the fixed
charges due to capital investment will be much reduced.
Similarly higher load factor means more average load or more energy
generated for a given maximum demand and therefore overall cost per unit of
electrical energy generated is reduced due to distribution of fixed charges
which are proportional to maximum demand (and independent of number of
units generated).
Plant Utilization factor is the ratio of the maximum demand on the plant to
the rated capacity of the plant.
Plant Capacity factor or plant factor is the ratio of actual energy produced
to the maximum possible energy that could have been produced based on
installed plant capacity. This can also be defined as the ratio of average
demand to rated capacity and obtained directly or by multiplying plant load
factor with plant utilization factor.
Example 1.1: For the load curve shown in Fig. 1.1 find the load factor. If the
connected load is 25 MW, find the diversity factor.
Solution:
Energy consumed = 5 x 12 + 15 x 4 + 20 x 8 = 280 MWh
Average demand
=
Energy consumed
Time period
=
280
24
= 11.67 MW
Average demand
11.67 = 0.5835 = 58.35%
Load factor
=
=
20
Maximum demand
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Assuming demand factor as unity,
Diversity factor
Connected load
Coincident max. demand
=
25
20
=
= 1.25
Example 1.2: There are three consumers of electricity having different load
requirements at different times. Consumer 1 has a maximum demand of 5 kW
at 6 pm and a demand of 3 kW at 7 pm, and a daily load factor of 20%.
Consumer 2 has a maximum demand of 5 kW at 11 am, a load of 2 kW at 7
pm and an average load of 1.2 kW. Consumer 3 has an average load of 1 kW
and his maximum demand is 3 kW at 7 pm. determine (a) the diversity factor,
(b) the load factor and average load of each consumer and (c) the average load
and load factor of the combined load.
Solution:
a)
Consumer
Individual
MD (kW)
1
2
3
5
5
3
Load demand
At 11 am At 6 pm At 7 pm
kW
kW
kW
5
3
5
2
3
LF
%
Average
load (kW)
20
-
1.2
1
Maximum demand = 3 + 2 + 3 = 8 kW at 7 pm
Sum of individual maximum demands = 5 + 5 + 3 = 13 kW.
∴ Diversity factor = 13 / 8 = 1.625
(b) Consumer 1
Consumer 2
Consumer 3
Load factor = 20%
Average Load = 0.20 x 5
= 1 kW
Average Load = 1.2 kW
Load factor = 1.2 / 5
= 0.24 = 24%
Average Load = 1 kW
Load factor = 1 / 3 = 0.333 = 33.3%
(c) Combined average load
Combined load factor
= 1 + 1.2 + 1 = 3.2 kW
= 3.2 / 8 = 0.4 = 40%
Example 1.3:
A power station has to meet the following demand:
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Group A: 200 kW between 8 am and 6 pm
Group B: 100 kW between 6 am and 10 am
Group C: 50 kW between 6 am and 10 am
Group D: 100 kW between 10 am and next day 6 am.
Plot the daily load and load duration curves and determine (a) diversity factor,
(b) daily energy produced and (c) load factor.
Solution:
The given load cycle can be tabulated as follows:
Time (Hours)
Group A (kW)
Group B (kW)
Group C (kW)
Group D (kW)
Total load (kW)
0-6
100
100
6-8
100
50
150
8-10
200
100
50
350
10-18
200
100
300
18-24
100
100
From this table the load curve is plotted as shown in Fig. 1.2(a).
Load
in kW
Time (Hours)
0
24
Fig. 1.2(a) Load curve for Example 1.3
The table can be modified as follows to show the load and its duration:
Load (kW)
350
300
150
100
Duration (Hours)
2
8
2
12
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From this table the load duration curve is plotted as shown in Fig. 1.2(b).
Load
in kW
0
Time duration (Hours)
24
Fig. 1.2(b) Load duration curve for Example 1.3
The maximum demand on the system is 350 kW.
Sum of individual maximum demands of groups
= 200 + 100 + 50 + 100 = 450 kW
(a)
Diversity factor = 450/350 = 1.286
(b)
Daily energy produced = Area under load or load duration curve
= 350x2 + 300x8 + 150x2 + 100x12 = 4600 kWh
(c)
Average Demand = 4600/24 = 191.7 kW
Load factor = AD/MD = 191.7/350 = 0.548 = 54.8%
Example 1.4:
The annual load duration curve of a certain power station can be considered as
a straight line from 20 MW to 4 MW. To meet the load, three generators, two
rated at 10 MW each and one rated at 5 MW are installed. Determine (a)
installed capacity, (b) plant utilization factor, (c) units generated per annum,
(d) plant capacity factor and (e) load factor.
Solution:
Fig. 1.3 shows the annual load duration curve of the power station.
20
Load in MW
Fig. 1.3 Annual load
duration curve
4
0
8760
Hours of the year
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(a) Installed Capacity = 10 +10 + 5 = 25 MW
(b) From the load duration curve
Maximum Demand = 20 MW
Plant Utilization Factor = MD/ IC = 20/25 = 0.8 = 80%
(c)
Energy generated
= Area under load duration curve
= ½ (20 + 4) x 8760
= 105.12 x 103 MWhr = 105.12x106 kWhr or units
Average Demand
= Energy generated / Time
= 105.12 x 103/8760 = 12 MW
(d)
Plant Capacity Factor = AD/ IC = 12/25 = 0.48 = 48%
(e)
Plant Load Factor = AD/ MD = 12/20 = 0.6 = 60%
1.2.4 Load dependency on voltage and frequency
All loads are characterized by their dependency on voltage and
frequency. During fault and other abnormal situations, the voltage may vary
greatly resulting in major load fluctuations. Even minor changes in voltage
and frequency can cause load changes of practical significance.
Two important load types are
1. Impedance type loads;
2. Motor Loads.
Impedance Type Loads
Lighting, Heaters, Ovens, etc. are impedance type loads.
Motor Loads
Induction Motor load dominates this group. Its dependence upon
voltage and frequency is somewhat more complex to analyze.
Example 1.5
Consider an Inductive Load of impedance type which equals Z = R + jX. By
how many percent will the real load drop if the voltage is reduced by 1 per
cent?
Solution
S
= P + jQ = VI* = VV*Y*
= V2 Y*
= V2
= V2 R2+ jX2
R +X
1
R-jX
R
L
Fig.1.4 Impedance Type Load
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R
R +X2
∴
P
∴
X
R +X2
Real and Reactive loads both are proportional to V2.
Q
= V2
(1.5)
2
= V2
2
(1.6)
For a small voltage perturbation ∆V change in real power is calculated as
follows:
Taking logarithm on both sides of equation (1.5),
ln P
= 2 ln V + ln [R/( R2+X2)]
Differentiating partially w.r.t. V,
∆P
P
≈ 2
∆V
V
(1.7)
From equation (1.7) it may be noted that a small relative change in voltage
results in twice the relative change in mega watts.
∴
a 1% drop in voltage causes a 2 % drop in load.
Example 1.6
How would a 1 percent drop in frequency affect the real load in an inductive
load type having a power factor (cos φ) of 0.8?
Solution
Rewriting equation (1.5),
R
R +X2
P
= V2
∂P
∂f
= V2R -
4πXL
(R2+X2)2
[
2 X2
f(R2+X2)2
2
[
2
= -V R
∆P
∆P
P
≈
=-2
∂P
∂f
[
]
since X = ωL = 2πfL
]
2 X2
f(R2+X2)
=-P
[
]]
∆f
X2
R2+X2
]
∆f
f
(1.8)
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X2
R2+X2
∆P
P
∴
= sin2 φ = 1 – cos2 φ = 1-0.64 = 0.36
∆f
f
a 1% drop in frequency causes a 0.72 % increase in load.
= - 0.72
Example 1.7
A 480 V 3 ϕ induction motor powers a compressor whose torque is assumed
speed-independent. The motor runs at an initial speed corresponding to a per
unit slip of s = 0.03. How will the motor load change if the voltage drops by
one per cent? Stator Resistance and reactance and rotor resistance and
reactance (referred to stator side) are 0.290 Ω / ϕ, 0.5 Ω / ϕ, 0.150 Ω / ϕ and
0.200 Ω / ϕ respectively.
R1
R2/s
X1+X2
+
I
V
-
Zm
Fig.1.5 Equivalent Circuit of 3 Ph Induction Motor
Solution
We will perform the analysis under the following simplifying assumptions:
1. We neglect the magnetizing impedance Zm.
2. We shall assume that we operate on the linear portion of the torque –
slip curve.
Motor Torque Tm
T
A’
A”
Load Torque
independent
of slip
O
s’
s”
0.5
1.0
Fig.1.6 Torque Vs Slip of induction motor
Pu slip
s
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Motor Torque Tm in this linear region
Tm α V2 s
(1.9)
where s - Pu slip
Initially, the motor operates at point A′ with slip s′ = 0.03. Following the
voltage drop the torque curve changes to the dashed one to maintain constant
torque and the motor will now operate at point A″ with a larger slip s″ derived
from equation (1.9) as
s″
|V’|2
=
s′
=
0.03 x (1.00 / 0.99)2 = 0.0306 Pu.
(1.10)
|V”|2
The real load P drained from the network equals
P
W/ϕ
= (R1 + R2/s) | I |2
(1.11)
Stator current
I
|V|
=
√ [(R1 + R2/s)
2
+ ( X1 + X2)
2
(1.12)
]
Using numerical values,
| I″|
=
1.0084 | I′|
The voltage drop of 1% causes a 0.84% current rise, a quite typical situation in
the case of induction motor load (Opposite situation in the case of an
impedance load).
P″
=
P″
=
(0.290 + 0.150/0.0306)
(0.290 + 0.150/0.03)
1.0084
1.0
2
P′
0.998 P′
This motor will reduce its power drain by only 0.2% for a voltage drop of 1%.
(Compare the 2% reduction in the case of an impedance load).
The above example illustrates that impedance loads give a considerably higher
power reduction than motor load under such conditions.
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1.3 Reserve Requirements
Reserve capacity:
The total capacity of a station should be more than the capacity required to
meet the maximum demand. This excess capacity as stand by is called
installed reserve capacity and is useful when exigencies arise due to tripping
of generating unit or reduction in generating capacity owing to failure of
certain auxiliary equipment or components in the generating plant. There are
different types of reserve depending upon the readiness of this capacity to
come into service.
Cold reserve is that portion of the installed reserve capacity kept in operable
condition and available for service, but not ready for immediate loadings. It
has to be brought to hot reserve state and then to spinning reserve before
loading.
Hot reserve is the reserve capacity ready for use with boiler in full steam; on
synchronization of the generator it becomes spinning reserve ready to take
load.
Spinning reserve is the generating capacity on line (running) in excess of
maximum demand and ready to take additional load.
Example 1.8
A generating station has a maximum demand of 25 MW, a load factor of 60%
and a plant capacity factor of 50%. Find (a) the daily energy produced, (b) the
reserve capacity of the plant and (c) the maximum energy that could be
produced daily if the plant were fully loaded.
Solution:
Load factor
= Average demand / Maximum demand.
Average demand
= 0.6 x 25 = 15 MW.
Daily energy produced=Average demand x hours=15 x 24= 360 MWhr
Plant capacity factor = Average demand / Installed capacity.
Installed capacity
= 15 / 0.5
= 30 MW.
Reserve capacity
= Installed capacity – Maximum demand.
= 30 – 25 = 5 MW.
Maximum energy corresponding to installed capacity.
= 30 x 24 = 720 MWhr
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1.4 Overview of system operation:
1.4.1 Load forecasting:
Load forecasting plays an important role in power system planning, operation
and control. Forecasting means estimating active load at various load buses
ahead of actual load occurrence.
There are two approaches for load forecasting, namely total load approach and
components approach (components may be domestic load, commercial load,
agricultural load, industrial load etc.). Total load approach has the merits that
it is much smoother and indicative of overall growth and is easy to apply. The
merit of the components approach is that abnormal conditions in growth trends
of any component can be detected, thus preventing misleading forecasts.
Further various components can have different growth rates contrary to same
growth rate used in the total load approach.
Load forecasting techniques may be different from application point of view long term forecasting required for planning and short term forecasting
needed for operation, dispatch and control. While sophisticated probabilistic
methods exist, the simple extrapolation technique is quite adequate giving
reasonable results for long term forecasting. Extrapolation technique involves
fitting trend curves to basic historical data adjusted to reflect the growth trend
itself. With the trend curve the forecast is obtained by evaluating the trend
curve function at the desired future point. Such a technique is deterministic,
since random errors in the data are not accounted for. Standard compounded
growth function is
Pm = Po (1 + g) m
where
(1.13)
Pm = Load at the end of mth year
Po = Initial load (load at the base year) and
g = Growth rate in p.u.
Future load of heavy industries may be obtained from survey of such
industries.
Some times correlation techniques of forecasting, which relate system loads to
various demographic and economic factors, are used. These factors are
population, GDP (Gross Domestic Product), employment etc.
In short-term load forecasting, hour-by-hour predictions are made for the
particular day under consideration. Weather forecast plays an important role.
Seasonal and cyclic loads are also given due importance in the forecast.
Random factors such as unexpected storms, strikes, sudden telecast of
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interesting TV programmes can upset the predictions. Regression analysis is
often used for obtaining the short-term load forecasting required for
economical load dispatch.
1.4.2 Unit commitment:
The system load varies over a day or week and hence it is not economical to
keep all the units on line for the entire duration. A proper schedule for starting
up or shutting down the units can save costs significantly and this is called unit
commitment problem.
The unit commitment (UC) problem usually covers a time range from 24
hours (1 day) to 168 hours (1 week) ahead and is handled by the operator in
the pre-dispatch stage. In this problem the operator needs to take decisions on
how to commit (keep running) or de-commit (shut down) the available units
over the next day or next week. The input to the operator is the demand
forecast for the next day or next week, as the case may be, aggregated for the
whole system.
Similar to the formulation of economic load dispatch (ELD) problem, the
operator seeks to minimize the system costs over the planning horizon in an
UC problem, while meeting the forecast demand to decide upon the unit
up/down status for every hour.
UC problems are much more complex to solve compared to the simple ELD
problem due to the presence of binary decision variables on unit status (on /
off) and cost involved with starting up and shut down.
1.4.3 Load dispatching:
The objective of the system operator is to schedule the generation dispatch so
as satisfy the system load in the best possible way, that is, in the most reliable,
secure and economic manner. The activities of the system operator can be
divided over three distinct time periods:
(i)
Pre-dispatch planning activities:
The pre-dispatch stage comprises a period of a week ahead of
actual operation to a day ahead. A short term forecast of the hourly
aggregate system load is usually available to the operator based on
which the schedule for unit operation is drawn up for the plan
period. This can also include scheduling of available hydro
resources in the system depending upon their reservoir levels,
irrigation commitments and other factors. Also plan for power
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exchange with other regions can be drawn up and included while
formulating the unit operation schedules.
(ii)
Dispatch - short-term scheduling:
The dispatch stage begins about 30 minutes ahead of the actual
operation and in this stage the operator carries out short-term
planning activities such as system power flow and economic
scheduling. The operator has a fair idea of the load expected on the
system during this stage, including power exchanges scheduled,
and accordingly decides upon the dispatch and if necessary load
curtailment. He is also responsible for maintaining adequate
reserves in the system at suitable locations, which can be called
upon at a short notice, in case of emergencies.
(iii)
Instantaneous dispatch – activities in real time:
This stage can range from 5 minutes ahead of real time to actual
operation. In this stage the operator implements the plans chalked
out at the pre-dispatch stage and fine-tuned at the dispatch stage.
His activities include continuous improvement of pre-dispatch and
dispatch decisions based on real time data, initiating secondary
frequency control actions if required and updating the generation
schedule based on participation factors of the units and information
on load.
A number of programs and simulation software are available to the operator to
aid his decision making process and make power system operations more
economic and reliable. Depending on the relative accuracy and computational
burden, these facilities are used in the dispatch or pre-dispatch stages.
1.5 Overview of System Control
In real time control of power system as the demand deviates from its normal
value the state of the system will change. The automatic control system must
detect these changes and initiate in “real time” a set of counter control action,
which will eliminate the state deviations as quickly and effectively as possible.
Computer control of modern power systems is proposed to improve economy,
to maintain quality of power supply such as to keep the system voltage and
frequency as constant and for better security. Such a control is feasible,
primarily if all meter readings and other information pertaining to the
operating state of the system are processed in real time into a more useful form
so that control decisions are made using them.
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Power System Operation and Control
Fig 1.7 shows the system being used in modern power generators.
Secondary ALFC Loop
Primary ALFC Loop
Tie Line
Powers
Hydraulic
amplifier
Integrator
∆Pref
Speed
Governor
Voltage error
f
Area Control
Error (ACE)
|V|ref
Comparator
∆PC
Amplifier
Exciter
Speed n
Signal Mixer
Speed
changer
Steam/Hydro
valve
Hyd. Pr
Ex. Pr
∆PT
Field ckt
+
Speed
sensor
Turbine
|Vi|
Rectifier
& Filter
-
Genr.
PT
Transformer
∆PG
Freq. sensor
∆PL Local
Load
Bus i
Fig. 1.7 System of modern generator
While the plant level control is at micro level, system level control is at macro
level. Any decision taken at system level is passed on to plant level for
implementation.
System level control
System level control ensures that the over all power system parameters such as
frequency, tie-line power flow and voltages at important grid points are
maintained. This activity is performed by energy management centre or load
dispatch centre. The Supervisory Control and Data Acquisition System
installed in the centre receives data over signal transmission network from
Remote Terminal Units (RTUs), generating stations, sub-load dispatch
centres, if any, etc. Any mismatch between generation and demand is
translated into a control signal and the control order is transmitted for
compliance.
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∆PI
To N/W
Power System Operation and Control
Plant level control
Control of the plant during normal and abnormal conditions is at the plant
level. This ensures that the plant is operating within its capability curve –
stator current limit, rotor current limit, turbine limit, stability limit to name a
few. The dispatch order received from the load dispatch centre at generating
station level is further subdivided to plant level depending on the number of
plants available for control. The boiler (if thermal based plant), turbine and
generator operate corresponding to this dispatch value.
1.5.1 Cross Coupling between P – f and Q –V control channels
The dependency of f and V on P and Q can be explained as follows:
1. A surplus of mega watts tends to increase the frequency of system. The
frequency is a system-wide variable, uniform throughout the system.
Surplus of megavars tends to increase voltage level of a system. The
changes are not uniform but will be greatest at the buses where the
system strength is the weakest.
2. As we change the mega watt output of one or several generating units
in order to maintain frequency constant, no resulting measurable
changes occur in voltage levels.
On the contrary, as we change the Q input at a certain bus thereby
affecting its voltage level (V) we do immediately also change the real
voltage-dependent load of the bus. This mega watt change will have an
effect on frequency.
1.5.2 Governor control:
When the electrical load on a generator is suddenly increased, the electrical
power output needed exceeds the mechanical power input. This deficiency in
power is made good by releasing the kinetic energy stored in the rotating
system. The reduction in kinetic energy causes the fall in turbine speed and
consequently the frequency. The change in speed is sensed by the turbine
governor, which acts to adjust the opening of the turbine input value so as to
change the mechanical power and bring the speed back to normal steady state
value.
1.5.2.1 Load Frequency Mechanism
The frequency is closely related to real power balance in the over all network.
The electric energy production rate must equal the consumption rate at each
moment of time. Should power balance not exist then the difference would
enter into or exit from Kinetic energy storage. As the kinetic energy depends
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Chapter-1 Introduction
19
Power System Operation and Control
upon generator speed, a power imbalance will thus translate into deviation in
speed and frequency.
For example, suppose it is desired to keep the velocity of the train controlled
at constant value in spite of the fact that the trains will experience fluctuating
gradients along its course. If Total Pengine > Total Ppull exerted by the freight
cars, the speed will increase as the difference is transformed into Kinetic
energy storage. Similarly the shortage of engine power results in a slow down.
The rate of speed change depends upon the magnitude of the power imbalance
and train inertia. In electric system, surplus of generator power causes increase
in the total speed (frequency) and the rate of speed or frequency increase
depends on surplus power and total moment of inertia of the rotating masses.
Example 1.9
All the generators and motors in a power system represent a total kinetic
energy of 1250 MJ or MWs as measured at rated frequency 50 Hz. The system
experiences a momentary power surplus of 5 MW. At what rate will the
frequency increase?
Solution:
Wkin
½Iω2
=
(1.14)
2
Wkin
=
ω
ωo
=
f
fo
Wkin
=
Wokin
I
=
Inertia Kg/m2
ω
=
Speed rad / sec.
Wkin
=
Kinetic energy
-----Wokin
2
where
f
fo
2
J or Ws
(1.15)
Corresponding power imbalance ∆P is given by
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Chapter-1 Introduction
20
Power System Operation and Control
∆P
=
d/dt (Wkin)
=
=
2 fWokin/ fo2
df
dt
d/dt
Wokin
df
dt
since the change in frequency is very small and f ≈ fo
∆P
=
2 Wokin/ fo
f
fo
2
(1.16)
Given
∆P
=
Wokin =
=
fo
+ 5 MW
1250 MJ
50 Hz
Substituting,
df
dt
= 0.1 Hz / s.
1.5.3 Load Frequency Control (LFC)
As the system load changes, it becomes necessary to adjust the generation so
that the power imbalance is continuously zeroed. Load Frequency Control
(LFC) is in fact a basic control function in a power system. The control signal
is computed in the central Energy Control Centre and the signal corresponding
to each distributed control centre is transmitted to them. The distributed
control centres, in turn, compute and transmit the change in dispatch to each
generating plant that participates in the LFC scheme.
The load frequency control system has two feed back loops, primary and
secondary to achieve real power balance or load tracking in the system, that is,
the power balance is maintained by appropriate adjustment of the turbine
torque. By means of the primary loop, a relatively fast (in seconds) course
frequency control is achieved. The secondary loop works in a slow (in
minutes) reset mode to eliminate the frequency error. This loop also controls
the power exchange between the pool members. The control signal referred as
area control error contains both frequency error and error in tie line power
flow (power exchange) and is used in this loop.
1.5.4 Economic Dispatch Control (EDC)
Similar to LFC, economic dispatch control also controls the generator output
based on economic considerations rather than frequency control. The solution
of optimal dispatch equations optimizing the total cost of energy production
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Power System Operation and Control
forms the basis for the economic dispatch control decisions. The computer
installed in energy control centre, which is linked to the various power plants
via communication channels, microwave, fibre-optic, power line carrier
communication, telephone etc. is used for this purpose. Periodically, say every
five minutes, the computer is provided with the megawatt settings in the
power plants. These settings are compared with the optimal settings derived
from the solution of optimal dispatch equations. If the actual settings are off
from the optimal values the computer sends back instructions to the plants to
readjust the megawatts outputs accordingly.
1.5.5 System Voltage Control
The voltage is closely related to reactive power balance at every bus in the
network. There are various methods of controlling voltage in a power system.
These methods are outlined below.
1. Excitation control of generators: This is the foremost means of control
and it focuses on maintaining good voltage control at the generator
buses.
2. Switched shunt capacitors or reactors: These provide the capability of
controlled reactive power injection into or drain from a bus.
3. Synchronous condensers: These are active devices that permit both
continuous and sign sensitive control of reactive power.
4. Tap changing of Transformers: This is mostly used in distribution
transformers.
5. Series capacitors: These are used in long transmission lines.
6. Static VAR compensators: These can smoothly control reactive power
and voltage.
7. Flexible AC Transmission System (FACTS): These are multi-purpose
devices that can control various system parameters including voltage.
1.5.5.1 Automatic Voltage Regulator (AVR)
The generator terminal voltage is maintained constant by the automatic
voltage regulator. The bus voltage is measured utilizing a potential
transformer and is compared to a reference value after being rectified and
filtered. The resulting error voltage, after amplification, serves as input to the
excitation control system whose output directly feeds the generator field. A
drop in the terminal voltage causes a boost in the field current. This increases
the reactive power output of the machine, thus tending to offset the initial
voltage drop.
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Power System Operation and Control
1.5.6 Security Control:
Under normal operating conditions a power system may face a contingency
condition such as outage (complete or partial) of a generating unit or a
transformer or a transmission line or a sudden increase or decrease in power
demand on the system. The system operator has to analyze the effect of highly
probable contingencies so that the operator can quickly take corrective action
in the event of their occurrence. This helps in enhancing the system security.
The security assessment and its control form an important part of planning and
operation of power system.
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Power System Operation and Control
Short answer questions
1.1 What are the two basic parameters that need to be maintained in a power
system?
The parameters are frequency and voltage magnitude.
1.2 What is the objective of power system control?
The objective of power system control is to maintain frequency and
voltage for supplying electricity with proper quality.
1.3 What is the need for voltage regulation in Power System?
In a Power System voltage needs to be maintained for supplying electricity
with proper quality. The major reasons for this are:
i.
All equipment and appliances are designed for a certain voltage
level, the rated or name plate voltage. If voltage V of the system
should deviate from that value, the performance of the device
suffers and its life expectancy drops.
ii.
The real line losses depend upon the line flow which in turn
depends greatly upon line end voltages.
1.4 What is the need for frequency regulation in Power System?
In a Power System frequency needs to be maintained for supplying
electricity with proper quality. The major reasons for this are:
i.
Most types of AC motors run at speeds that are directly related to
the frequency.
ii.
The Generator Turbines, particularly Steam driven ones, are
designed to operate at specified speed with limited tolerance in
variation for maximum efficiency and less fatigue and wear and
tear.
iii.
The over all operation of a power system can be better controlled if
frequency error is kept within strict limits.
iv.
A large number of electrically operated clocks are used for power
system monitoring and control. They are all driven by synchronous
motors and the accuracy of these clocks is a function of the
frequency error.
1.5 State whether changes in AVR loop will be reflected in ALFC loop.
Yes, but marginally. Any change in voltage caused by AVR loop changes the
voltage-dependent load and hence the frequency, which will get reflected in
the ALFC loop.
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Power System Operation and Control
1.6 State about real time control of power systems.
The system operator continuously monitors the system parameters like
frequency, voltage, power dispatch, load and power flow over the lines and
transformers and take suitable action to maintain the normal condition of the
power system in real time using SCADA.
1.7 What happens to frequency if the load on the generator increases?
Increase in load is met from the kinetic energy of the rotating masses, because
of which the speed decreases and consequently the frequency’
1.8 State the purpose of system generation control. Or
1.9 What is meant by load frequency control?
Any change in load causes frequency change and generation is controlled to
bring the frequency back to the normal value. Sometimes the generation
control is effected to maintain the tie-lie flow. This is called ALFC.
1.10 Define spinning reserve.
Spinning reserve is the generating capacity on line (running) in excess of
maximum demand and ready to take additional load.
1.11 What are the different types of load?
The various load devices can be classified into the following categories:
i.
Motor devices;
ii.
Heating equipment;
iii. Lighting equipment;
iv.
A diversity of electronic gear.
1.12 How system loads are classified?
System loads are classified as
1. Constant power
2. Constant current
3. Constant impedance
1.13 Define load curve.
Load Curve is the graph showing the variation in the demand for energy of
consumers on the supply system with respect to time. If the graph is plotted
for 24 hours it is called daily load curve; if the graph is plotted for one week,
one month or one year, we get weekly monthly or annual load curves
respectively. The load curve is plotted chronologically.
1.14 Define load duration curve.
Load duration curve gives the duration in hours for which the load has either
remained equal or exceeded the given value. The load elements of a load curve
are arranged in the order of descending magnitudes.
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Power System Operation and Control
1.15 What is connected load?
Each electrical device has its rated capacity, which is normally given in W,
kW or MW. The sum of the continuous ratings of all the electrical devices
connected to the supply system is known as connected load.
1.16 What is maximum demand?
The maximum demand of an installation or system is the greatest of all
demands, which have occurred during the specified period of time and is
called daily, weekly, monthly or annual maximum demand.
1.17 Define demand factor.
The demand factor is the ratio of the maximum demand of a system or group
to the total connected load of the system/ group.
DF =
Maximum demand
Total connected load
The demand factor is less than or equal to unity.
1.18 Define diversity factor.
The diversity factor (FD) is the ratio of the sum of the individual maximum
demands of the various groups of consumers to the maximum demand of the
whole system.
FD =
Sum of individual maximum demands
Coincident maximum demand
n
∑
=
where Pi
Pc
=
=
Pi
i =1
Pc
Maximum demand of load group i
Coincident maximum demand of n load groups.
The diversity factor is greater than or equal to unity.
1.19 Define load factor.
The load factor is the ratio of the average load over a specified period of time
to the peak load occurring in that period.
Average load
LF =
Peak load
It can also be defined as the ratio of the total energy consumed over a
specified period of time to the energy that would have been consumed had the
peak load occurred throughout the period.
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Power System Operation and Control
1.20 Define plant load factor.
Plant load factor is defined as the ratio of the total energy produced over a
specified period of time to the energy that would have been produced had the
plant operated throughout the period. The energy can be computed either from
load curve or from load duration curve.
1.21 What is the significance of load factor and diversity factor?
Load factor and diversity factor play important roles in the cost of supply of
electrical energy. Higher the values of load factor and diversity factor, lower
will be the overall cost per unit generated.
Higher the diversity factor of loads, the smaller will be the capacity of the
plant required and consequently the fixed charges due to capital investment
will be much reduced.
Similarly higher load factor means more average load or more energy
generated for a given maximum demand and therefore overall cost per unit of
electrical energy generated is reduced due to distribution of standing charges
which are proportional to maximum demand (and independent of number of
units generated).
1.22 Define plant utilization factor.
Plant Utilization factor is the ratio of the maximum demand on the plant to the
rated capacity of the plant.
1.23 Define plant factor.
Plant Capacity factor or plant factor is the ratio of actual energy produced to
the maximum possible energy that could have been produced based on
installed plant capacity. This can also be defined as the ratio of average
demand to rated capacity and obtained by multiplying plant load factor with
plant utilization factor.
1.24 State the differences between P-f and Q-V controls.
The dependency of f and V on P and Q can be explained as follows:
1. A surplus of mega watts tends to increase the frequency of system. The
frequency is a system-wide variable, uniform throughout the system.
Surplus of megavars tends to increase voltage level of a system. The
changes are not uniform but will be greatest at the buses where the
system strength is the weakest.
2. As we change the mega watt output of one or several generating units
in order to maintain frequency constant, no resulting measurable
changes occur in voltage levels.
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Power System Operation and Control
On the contrary, as we change the Q input at a certain bus thereby
affecting its voltage level (V) we do immediately also change the real
voltage-dependent load of the bus. This mega watt change will have an
effect on frequency.
1.25 Give the two major control loops of large generators.
They are ALFC (Automatic Load Frequency Control) and AVR (Automatic
Voltage Regulator) loops.
1.26 What is base load?
Base load is the minimum load on a system that is fed throughout the period of
system operation.
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Power System Operation and Control
Exercises
Problem 1.1:
A generating station has a maximum demand of 80 MW and a connected load
of 150 MW. If the energy generated in an year is 400x103 MWhr, calculate (a)
load factor and (b) demand factor.
(Ans.: 57%, 53.3%)
Problem 1.2:
Load duration data of a system are given below.
Load (MW)
2
4
6
8
10
12
15
Duration
(Hours)
8760
7000
4000
3000
2000
1000
100
Plot the load duration curve and determine the load factor.
(Ans. 39%)
Problem 1.3:
Peak demand of a generating station is 90 MW. The load factor and the plant
capacity factor are 0.6 and 0.5 respectively. Determine (a) daily energy
produced, (b) installed capacity, (c) reserve capacity and (d) utilization factor.
(Ans.: 1296 MWhr, 108 MW, 18 MW, 83.3%)
Problem 1.4:
The daily demands of three consumers connected to a substation are given
below.
Time
12 midnight to 8 A.M.
8 A.M. to 2 P.M.
2 P.M. to 4 P.M.
4 P.M. to 10 P.M.
10 P.M. to midnight
Consumer 1
No load
600 W
200 W
800 W
No load
Consumer 2
200 W
No load
1000 W
No load
200 W
Consumer 3
No load
200 W
1200 W
No load
200 W
Plot the load curve and determine (i) maximum demand and load factor of
individual consumer and (ii) diversity factor and load factor of the station.
(Ans.800 W, 1000 W, 1200 W, 45.8%, 16.7%, 13.8%, 1.25, 29.1%)
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Power System Operation and Control
Problem 1.5:
A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours
and is shut down for the rest of each day. It is also shut down for maintenance
for 45 days each year. Calculate its annual load factor.
(Ans. 18.26%)
Problem 1.6:
A diesel station supplies the following loads to various consumers:
Industries
= 1500 kW
Commercial establishment = 750 kW
Domestic power
= 100 kW
Domestic lighting
= 450 kW
If the maximum demand on the station is 2500 kW and the units generated per
year is 45x105, determine (i) the diversity factor and (ii) annual load factor.
(Ans. 1.12, 20.5%)
Problem 1.7:
A generating station has the following daily loads:
0 – 6 hrs
4500 kW
6 – 8 hrs
3500 kW
8 – 12 hrs
7500 kW
12 – 14 hrs
2000 kW
14 – 18 hrs
8000 kW
18 – 20 hrs
2500 kW
20 – 24 hrs
5000 kW
Sketch the load duration curve and determine the load factor and plant
capacity factor, if the capacity of the plant is 12 MW.
(Ans. 65.1%, 43.4%)
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