Automatic load frequency control

This topic belongs to Power System Operation and Control. It will help to all electrical engineering students as well as faculties.

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Chapter-05
Load Frequency Control, Control Area Concept
Introduction
Automatic Load frequency control (ALFC) in a power system regulates the power
flow between different areas while holding frequency constant. It divide the load
between the generators and control the tie line interchange schedules. The ALFC
loop will maintain control only during small and slow changes in load and
frequency. It will not provide adequate control during emergency situation when
large megawatt imbalance occurs. (Drawback of ALFC)
We shall first study as it applies to a single generator supplying power to a local
service area. The Real power control mechanism of a generator is shown in the fig.

The main parts are:
1) Speed changer
2) Speed governor
3) Hydraulic amplifier

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4) Control valve.
They are connected by linkage mechanism. Their incremental movements are in
vertical direction. In reality these movements are measured in millimeters; but in
our analysis we shall rather express them as power increments expressed in MW or
p.u. The movements are assumed positive in the directions of arrows.
When speed raises, linkage movements will be: “A” moves downwards; “C” moves
upwards; “D” moves upwards; “E” moves downwards. This allows more steam or
water flow into the turbine resulting incremental increase in generator output
power.
When the speed drops, linkage movements will be: “A” moves upwards ; “B”
moves upwards “C” moves downwards; “D” moves downwards; “E” moves upwards.
This allows less steam or water flow into the turbine resulting decrease in
generator output power.
Speed governing system:
The output command of speed governor is ΔP g which corresponds to movement
ΔXC. The speed governor has two inputs: (i) Change in the reference power setting,
ΔPref (ii) Change in the speed of frequency of the generator, Δf, as measured by ΔXB.
It is to be noted that a positive ΔPref will result in positive ΔPg.

If Δf= change in frequency
ΔP= Change in load
Then slope of the curve,

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The slope of the curve is also called as the regulation or drop.
Generally governors have a speed regulation of 5.6% from zero to full load.
Unit of R is Hz/MW. The output of speed governor mechanism ( ΔPg) is the
difference between the reference power (ΔPref) and change in load.









Taking Laplace transform of eq.(2) yields








Hydraulic valve actuator (or, Hydraulic amplifier)
The input position of the valve actuator(ΔXd) is the difference between governor
outputs (ΔPg) and the hydraulic actuator output (ΔPv)




Taking Laplace on both sides of eq(4)





For a small change in ΔPv, the oil flow into the hydraulic motor is proportional to
the position ΔXd of the pilot valve.

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Where K is a constant depends upon the orifice, cylinder geometries and the fluid
pressure.
Integrating both sides of above equation, we get















Taking Laplace on both sides of eq(5)











For transfer function, initial condition must be zero.







Using equation (5) in equation (8), we have

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Turbine generator
In normal steady state, the turbine power P T keeps zero acceleration and a
constant speed and frequency.
During transient state, let the change in turbine power be ΔP T and the
corresponding change in generator power be ΔPG
The accelerating power in turbine generator unit


Thus accelerating power



If ΔPT - ΔPG is negative, it will decelerate.
The turbine power increment ΔPT depends entirely upon the valve power increment
ΔPv and the characteristic of the turbine.

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Different type of turbines will have different characteristics.
Taking transfer function with single time constant for the turbine, we can write









The generator power increment ΔPG depends entirely upon the change in the load
ΔPD being fed from the generator.
The generator always adjusts its output so as to meet the demand changes ΔPD.
We can therefore set





accelerating power

















Static performance of speed governor:
The control loop shown in Fig. is open, No information can be obtained about the
static performance of the speed governor.

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The static performance can be obtained by letting s = 0.



From Fig.







Substituting GH=1, we get



Again substituting GT =1 in equation (13),



Using eq.(18) in Eq(17) we have



Using Eq.(19) in eq.(2) we have







Using subscript 0 for static signas, Eq.(20) becomes







For study of static performance, three cases arise. These are
Case-A: The generator is synchronized to a network of very large size and the
frequency will be independent of any changes in the power output of this individual
generator(infinite network).
Since the frequency is independent of any changes of power output,


Substituting the value in eq.(21) becomes

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From equation (22), it is clear that for a generator operated at a constant speed ,
the turbine power is proportional to the reference power setting.
Case-B: Consider a finite network. (Frequency variable)
If the speed changer setting is constant, then


Substituting Eq.(24) in Eq.(21), we have












It is clear from equation (20), that for a constant setting of the speed changes, the
static increase in turbine power output is directly proportional to the static
frequency drop.
Case-C: In general case, changes may occur in both the speed changer (reference
power) setting and frequency. In frequency generation graph, equation (21) represents
a family of sloping lines as shown in fig. Each line corresponds to a specific reference
power setting. It is clear from fig that the generators working in parallel on the same
network should have the same regulation in order to share load changes in proportion
to size.

Example 1
A 100-MW, 50-Hz generator is connected to “infinite” network. How would you
increase its turbine power by 5 MW?

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Solution
Its turbine power can be increased by 5 MW by simply giving a “raise” signal of 5 MW
to the speed changer motor.

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1. A 300 MW turbo generator has a speed regulation of 0.045pu on its own rated
capacity as base. Determine the increase in power output when the frequency
drops from normal 50Hz to a steady state value of 49.95 Hz. (2015) [5]

Solution:

Given the generator power, = 100 MW
Speed regulation = 0.045 pu.
Drop in frequency, Δf = -0.05Hz
Increase in turbine power = ?
Regulation parameter,





Increase in turbine power at the frequency drop of -0.05Hz





Closing the ALFC Loop










The “old” load is a function of voltage magnitude and frequency. The old area load
depends on the frequency given by









For area power balance, the increase in turbine power is equal to the sum of old and
new load changes plus the rate of change of kinetic energy.
Therefore area power balance is given by
Turbine power = old load change + new load change + rate of change of kinetic energy







If Δf = change in frequency relative to f0
f0 = frequency at normal state
The new frequency, f = f0 + Δf
Substituting the value of f, we get

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Since Δf is very small as compared to f0, then

























































Dividing above equation by Pr , we get
























Taking Pr as base and writing above equation
















H is independent of Wkin
0. Typical value is 2-8 sec.















Taking Laplace transform on both sides of above equation, we get

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Equation (40) represents the missing link Δf(s) in the control loop of fig. the open loop
can be closed by adding a summing junction and a transfer block as shown in fig.

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Static response of Primary ALFC loop
The basic objective of the primary ALFC loop is to maintain constant frequency in
spite of changing loads. The primary ALFC loop as shown in the fig. has one output
and two inputs. ΔPref(s) and ΔPD(s)
From Fig., at the left hand summing point,








Again at right hand summing point, we get,




Substituting eq(29) in eq(30) we get,









But if reference input is constant, then ΔPref (s)=0
Therefore,

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For a step load change of constant magnitude,


Taking Laplace transform on both side of equation, we get





Substituting the value in above equation














Applying final value theorem to above equation
The static frequency drop is given by,
































But



















Substituting the values in above equation

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Where,














From the static response, frequency accuracy is obtained.
Dynamic Response of ALFC Loop
Dynamic response of ALFC loop gives information about tracking ability and stability
of the loop.
In eq(36), GH(s), GT(s) and GP(s) contains atleast one time constant each. Therefore,
the denominator will be of 3
rd order and the analysis will be complex. To simplify the
analysis, it is to be assumed that the action of speed governor plus the turbine
generator is instantaneous. Since the action of speed governor and turbine generator
is instantaneous the time constant for speed governor (TH) and the time constant for
turbine generator TT is equal to zero.



Substituting equation (52),(53) and (54) in equation (47), we get

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On solving












Using eq.(63) and (64) in eq.(62) we have,


















Taking inverse Laplace transform,













Fig shows the dynamic response of the primary ALFC loop to a step load increase.

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It is clear that
The system can be made faster by reducing the value of R i.e. by increasing the value
of the static loop again.
Reduction of R reduces the static frequency error.
If the response of the turbine is not disregarded then the response will not be purely
exponential as shown in fig. above.
Physical Interpretation of Results
When the load is suddenly increased by 1% (20 MW), where did this power come from?
Certainly it must have come from somewhere as the load increase of 20 MW has been
met with instantaneously. In the milliseconds following the closure of the switch, the
frequency has not changed a measureable amount, speed governor would not have
acted and hence turbine power would not have increased. In those first instants the
total additional load demand of 20 MW is obtained from the stored kinetic energy,
which therefore will decrease at an initial rate of 20 MW. Release of KE will result in
speed and frequency reduction. As seen in eq.(66),













Initially frequency reduces. The frequency reduction causes the steam valve to open
and result in increased turbine power. In conclusion, the contribution to the load
increase of 20 MW is made up of three components:
• Rate of decrease of kinetic energy from the rotating system
• Increased turbine power
• “Released” old customer load
Initially the components 2 and 3 are zero. Finally, the frequency and hence the KE
settle at a lower value and the component 1 becomes zero. In between, component 1
keeps decreasing and components 2 and 3 keep increasing.
The Secondary (“Reset”) ALFC Loop:
It is seen from the previous discussion for a given speed changer setting, there is
considerable frequency drop for increased system load. So much change in frequency
cannot be tolerated. In fact, it is expected that the steady state frequency change must
be zero. In order to maintain the frequency at the scheduled value, the speed changer
setting must be adjusted automatically by monitoring the frequency changes.For this
purpose, INTEGRAL CONTROLLER is included. In this case, the speed changer is

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commanded by a signal obtained by first amplifying and then integrating the
frequency error.





Where K1 is integral control constant.
Negative sign in integral controller is choosen so as to cause a positive frequency error.
The gain constant k controls the rate of integration and the speed of response of the
loop.
Area Control Error(ACE): the signal fed into the integrator is referred to as area
control error(ACE) i.e. ACE=Δf.
When frequency error is reduced to zero, then the integrator output and the speed
changer position attains a constant value.
Taking Laplace transform,






The gain constant K1 controls the rate of integration and thus the speed of response of
the loop. For this signal ΔF(s) is fed to an integrator whose output controls the speed
changer position resulting in the block diagram configuration shown in fig.

As long as an error remains, the integrator output will increase, causing the speed
changer to move. When the frequency error has been reduced to zero, the integrator
output ceases and the speed changer position attains a constant value. Integral
controller will give rise to ZERO STEADY STATE FREQUENCY ERROR following a step
load change because of the reason stated above.

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Referring to the block diagram of single control area with integral controller shown in
fig. input to GHT is















Using this equation the block diagram in fig can be reduced as shown in fig.



Economic Dispatch Control
The primary ALFC loop makes the initial coarse readjustment of frequency. The
secondary ALFC loop takes over the fine adjustment of the frequency by resetting
through integral action, the frequency error to zero. Economic dispatch control can be
viewed as an additional tertiary control loop. As the control decisions in this loop are
based upon the solutions of the ODE’s it is necessary to incorporate a digital computer
as part of this control loop. This computer is placed in “energy control centre” and this
is linked to the various power plants via communication channel. Periodically every
five minutes the computer is provided with the megawatt setting in the power plants.

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Short Questions
1. What is meant by Load frequency control?
The frequency is closely related to the real power balance in the overall network.
The basic role of ALFC is to maintain desired megawatt output of a generator unit
and assist in controlling the frequency of the larger interconnection. The ALFC also
helps to keep the net interchange of power between pool members at
predetermined values.
Or
Load frequency control, as the name signifies, regulates the power flow
between different areas while holding the frequency constant.

2. What is the role of Hydraulic amplifier in load frequency control model?
Hydraulic amplifier consists of a main piston and pilot valve. Low power level pilot
valve movement is converted into high power level piston valve movement which is
necessary to open or close the steam valve against high pressure steam.
3. What is area control error?
Area control error (ACE) is the difference between scheduled and actual electrical
generation within a control area on the power grid, taking frequency bias into
account.
ACE = Pactual - Pscheduled
Whenever the ACE is greater than zero, it means that the area is over generating
and thus needs either to decrease generation or to sell more. Likewise, whenever
the ACE is less than zero, the area is under generating and thus needs either to
increase generation or to buy more. AGC works to keep the ACE close to zero.
4. What is the major control loops used in large generators?
(i) Automatic load frequency control loop: To maintain the frequency constant.
(ii) Automatic voltage regulator: To maintain the voltage constant.

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5. What is the function of speed changer in a turbine speed governing system
of power system?
The speed changer provides a steady state power output setting for the turbine.
The downward movement of the speed changer opens the upper pilot valve so that
more steam is admitted to the turbine under steady condition. The reverse
happens when the speed changer moves upward.
6. what is the function of load frequency control?
The basic role of ALFC is to maintain desired megawatt output of a generator unit
and assist in controlling the frequency of the larger interconnection. The ALFC also
helps to keep the net interchange of power betwe en pool members at
predetermined values. Control should be applied in such a fashion that highly
differing response characteristics of units of various type (hydro, nuclear, fossil,
etc.) are recognized. Also, unnecessary power output changes should be kept to a
minimum in order to reduce wear of control values.

7. Explain the need of fly ball speed governor.
It is purely mechanical speed sensitive device coupled directly to the hydraulic
amplifier which adjusts the control valve opening via the linkage mechanism.
8. What do you understand by Tie line bias control in a power system?
Tie line bias control in a power system is the control of power flow between two
areas.
9. What is tie line in a power system?
The transmission lines that connect an area to its neighboring area are called tie-lines.
Power sharing between two areas occurs through these tie-lines.
10. What are the components used for the automatic load frequency control of a
single area system.
11. What are typical conditions needed to be taken care of while distributing
loads among the plants of a system?
12. Why frequency should remain constant? Explain in brief.
Reasons for the need of maintaining constant frequency:
(i) The speed of a.c. motors are directly related to the frequency.
(ii) If the normal operating frequency is 50 Hz and the turbines run at speeds
corresponding to frequencies less than 47.5 Hz or above 52.5 Hz, then the
blades of the turbines may get damaged.
(iii) The operation of a transformer below the rated frequency is not desirable.
When frequency goes below rated frequency at constant system voltage then the
flux in the core increases and then the transformer core goes into the
saturation region.

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(iv) With reduced frequency the blast by ID fans and FD fans decrease, and so
the generation decreases and thus it becomes a multiplying effect and may
result in shut down of the plant.

13. What is meant by p-f control of power system?
p-f control deals with the control of loading of the generating units for the system at
normal frequency. The load in a power system is never constant and the system
frequency remain at its nominal value only when there is a match between the active
power generation and the active power demand.
14. What are the typical conditions needed to be taken care of while distributing
loads within the units of a plant and the same among plants?
15. Draw the static response characteristics of the primary ALFC loop.
16. Differentiate static and dynamic response of an ALFC loop.
(i) Static response of an ALFC loop will inform about frequency accuracy.
(ii) The dynamic response of an ALFC loop will inform about the stability of the
loop.
17. State the functions of AVR.
The function of AVR excitation control is to regulate generator voltage and
relative power output. As the terminal voltage varies the excitation control, it
maintains the terminal voltage to the required standard and the demand of the
reactive power is also met by the excitation control unit.
18. List the various components in AVR loop.
Exciter, comparator, amplifier, rectifier, synchronous generator.
19. What is the need for compensator in the AVR loop?
Stability compensation improves the dynamic response characteristics without
affecting the static loop gain.
20. Define (i) Load factor (ii) Diversity factor.
Load factor:
The ratio of average load to the maximum demand during a given period is
known as load factor.
Load factor = (average load) / (maximum demand)
Diversity factor:
The ratio of the sum of individual maximum demand on power station is known
as diversity factor.
Diversity factor = (sum of individual maximum demand) / (maximum demand).
21. What is meant by free governor operation?
Only governor control is called as free governor action. It can be obtained by
deactivating the integral controller.
22. Compare the functions of speed governor and speed changer of a turbine -
generator set.

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23. How the ALFC loop is affected by AVR loop?
AVR affect the magnitude of generated emf Eg, This generated emf affects the
generated real power. Therefore changes in AVR loop affect ALFC loop.

24. What is the use of secondary loop?
A slower secondary loop maintains the fine adjustment of the frequency, and also by
‘reset’ action maintains proper megawatt interchange with other pool members .
This loop is insensitive to rapid load and frequency changes but focuses instead on
drift like changes which take place over periods of minutes.

25. What is the advantage of AVR loop over ALFC loop? AVR loop is much faster
than the ALFD loop and therefore there is a tendency, for the AVR dynamics to
settle down before they can make themselves felt in the slower load –frequency
control channel.

26. What is the function of AVR?
The basic role of the AVR is to provide constancy of the generator terminal voltage
during normal, small and slow changes in the load.

27. Specify the disadvantages of ALFC loop?
The ALFC loop will maintain control only during normal (small and slow) changes in
load and frequency. It is typically unable to provide adequate control during
emergency situations, when large megawatt imbalances occur. Then more drastic
‘emergency’.
28. What is the control area?
Most power systems normally control their generators in unison. The individual
control loops have the same regulation parameters. The individual generator
turbines tend to have the same response characteristics then it is possible to let
the control loop in the whole system which then would be referred to as a control
area.
29. Specify the use of static and dynamic response of the ALFC loop.

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The static response of the ALFC loop yielded important information about frequency
accuracy. The dynamic response of the loop will inform about ‘tracking’ ability and
stability of the loop.
30. What is the basic principle in pool operation?
A basic guiding principle in pool operation must be that each area, in normal steady
state, absorbs its own load.
31. What are the sources of reactive power? How it is controlled?
The sources of reactive power are generators, capacitors, and reactors. These are
controlled by field excitation.
32. Write about static VAR compensator (SVC).
These comprise capacitor bank fixed or switched or fixed capacitor bank and switched
reactor bank in parallel. These compensators draw reactive (loading or lagging)
power from the line thereby regulating voltage, improves stability (steady-state and
dynamic), control over voltage and reduce voltage and current unbalances. In
HVDC application these compensators provide the required reactive power and
damp out sub harmonic oscillation.
33. List the advantages of series compensation.
The advantages of series compensation are, a. Series capacitors are inherently self-
regulating and a control system is not required. b. For voltage stability, series
capacitors lower the critical c. Series capacitors possess adequate time-overload
capability. d. For the same performance, series capacitors are often less costly than
SVCs and losses are very low.
34. What are the parts of speed governing system?
The parts of speed governing system are, a. Speed governor, b. Linkage mechanism c.
Hydraulic amplifier, d. Speed changer

Numerical:
Example-1
Two generators rated 200MW and 400MW are operating in parallel. The drop
characteristics of their governors are 4% and 5% respectively from no load to
full load. Assuming that the governors are operating at 50Hz at no load, how
would a load of 600MW be sha red between them? What will be the system
frequency at this load? Assume free governor operation.
Solution:
Since the generators are operating in parallel, they will operate at the same
frequency at steady load.
Let load on generator-1 = x MW
Load on generator-2 = 600 – x MW.
Reduction in frequency = Δf

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From equation(1),




From equation(2),




Comparing equation (2) and (3), we have







Therefore load on generator-1 is 231 MW
Load on generator-2 is 600-x = 600-231 = 369MW
System frequency =




Example-2:
Two generators rated with 221MW and 429MW are operating in Parallel. The
drop characteristics of their governors are 4.15% and 5.35% respectively
from no-load to full load. The speed changers are so set that the generators
operate at 50 Hz sharing the full load of 650MW in the ratio of their ratings.
If the load reduces to 550 MW, what will be t he load shared by each
generator? Also find out the system frequency under this condition. 2016 [5]

Solution:
Since the generators are operating in parallel, they will operate at the same
frequency at steady load.
Let load on generator-1(221MW) = x MW
Load on generator-2(429MW) = 550 – x MW.
Reduction in frequency = Δf

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From equation(1),




From equation(2),




Comparing equation (2) and (3), we have







Therefore load on generator-1 is 219.493 MW
Load on generator-2 is 600-x = 550-219.493 = 330.506 MW
System frequency =




Example-03

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Example-04



Example-05

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Example-06

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Example-07

Example-08

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Example-09