EE6603 - Power System Operation & Control

Vel Tech High Tech Dr.Rangarajan Dr.Sakunthala Engg. College EE6603 – Power System Operation & Control
Prepared by Prabaakaran K AP/EEE Page1
UNIT – 1 INTRODUCTION
PART – A
1. State the need of voltage regulation in power system. M-J 2013
Voltage regulation helps in maintaining the voltage at the load terminals within prescribed limits
under conditions, by employing suitable voltage control equipment.
2. Explain is the effect of load factor in the cost of generation. M-J 2013
Load factor is inversely proportional to the cost of generation.
3. Define Plant capacity factor. N-D 2013
This is the ratio of actual energy produced to the maximum possible energy that could have
been produced during a given period.
Capacity factor =
Actual energy produced
Maximum energy that have been produced
4. List out the various needs for frequency regulation in power system. N-D 2016, 2013
The speed of the system is directly proportional to the frequency of the system.
N α f
Whenever there is change in the speed of the system, frequency of the system also varies. In order to
maintain the frequency of the system in a constant level we have to undergo frequency regulation.
1. In any power system, if the frequency changes there won‟t be required receiving end voltage. If we
connect two systems in parallel, it will spoil the system.
2. The generator turbines, particularly steam driven ones are designed to operate at a very precise
speed.
3. Most of AC motors runs at speeds that are directly related to the frequency
When two systems working at different frequencies are to be tied together to make same frequency,
frequency converting stations or links are required.
5. State maximum demand. M-J 2014
Maximum demand is the greatest demand of load on the power station during a given period. The
greatest of all “short time interval averaged” during a given period on the power station is called the
maximum demand.
6. Define Plant use factor. M-J 2014
Plant use factories defined as the ratio of the actual energy generated during a given period to the
product of plant capacity and the number of hours for which the plant was in operation during the period of
time.
Plant use factor =
𝐓𝐨𝐭𝐚𝐥𝐤𝐖𝐡𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐞𝐝
𝐑𝐚𝐭𝐞𝐝𝐜𝐚𝐩𝐚𝐜𝐢𝐭𝐲𝐨𝐟𝐭𝐡𝐞𝐩𝐥𝐚𝐧𝐭𝐱𝐍𝐮𝐦𝐛𝐞𝐫𝐨𝐟𝐨𝐩𝐞𝐫𝐚𝐭𝐢𝐧𝐠𝐡𝐨𝐮𝐫𝐬
7. Distinguish between load curve and load duration curve. N-D 2014
Load factor Diversity factor
 Load factor is defined as the ratio of
average load to the maximum demand
during a certain period of time.
 The average load is less than the maximum
demand; load factor is therefore, always
less than unity.
Load factor =
AVEARGE LOAD
MAXIMUM DEMAND
*100
 Maximum demand on the installation is,
therefore always less than the sum of
individual maximum demands of all
consumers connected to it.
 The value of diversity factor will be
always greater than unity.
Diversity factor =
𝐒𝐮𝐦𝐨𝐟𝐢𝐧𝐝𝐢𝐯𝐢𝐝𝐮𝐚𝐥𝐦𝐚𝐱𝐢𝐦𝐮𝐦𝐝𝐞𝐦𝐚𝐧𝐝
𝐌𝐚𝐱𝐢𝐦𝐮𝐦𝐝𝐞𝐦𝐚𝐧𝐝
8. Define the term Diversity factor. N-D 2014
The maximum demand of all the consumers supplied from an installation do not occur
usually at the same time, maximum demand on the installation is, therefore always less than the sum of
individual maximum demands of all consumers connected to it.

The ratio of the sum of individual maximum demand of all the consumers supplied by it to the
maximum demand of the power station is known as diversity factor.
Diversity factor =
𝐒𝐮𝐦𝐨𝐟𝐢𝐧𝐝𝐢𝐯𝐢𝐝𝐮𝐚𝐥𝐦𝐚𝐱𝐢𝐦𝐮𝐦𝐝𝐞𝐦𝐚𝐧𝐝
The value of diversity factor will be always greater than unity.
9. List the significance of load factor. N-D 2015
Significance of Load Factor are listed as follows
 Load factor is always greater than unity, because average load is smaller than maximum demand.
 It is used to determine the overall cost per unit generated.
 If the load factor is high, cost per unit generated is low.
10. Define demand factor. N-D 2015
Demand factor is the ratio of actual maximum demand on the system to the total rated load
connected to the system. It is always less than the unity; the demand factor decides the maximum constraints
on power system operation based on load.
Demand factor =
𝐂𝐨𝐧𝐧𝐞𝐜𝐭𝐞𝐝𝐥𝐨𝐚𝐝
Clearly, the idea of demand factor was introduced due to the fact that all the equipment connected to the
system will not be worked at a time in practice and the kW or kVA maximum demand of a group of
electricity consuming devices will always be less than the capacities of the devices.
11. Classify the system load. M-J 2016
The load on a power system varies from time to time due to uncertain demands of the consumers,
known as variable load on the station.
The system load in an area depends on residential commercial, industrial, agricultural, municipal and
traction loads. System load variation occurs due to special events on religious and social occasions, etc.
Types of load are listed as follows:
 Lighting load
 Heating loads.
 Induction motors.
 Electronic devices.
12. Define spinning reserve.N-D 2018M-J 2016
Spinning reserve is that generating capacity which is connected to the bus and is ready to take load.
13. Define Load duration Curve. N-D 2016
This type of curve which indicates the variation of load, but with the loads arranged in
descending order of magnitude, i.e., the greatest load on the left and lesser loads towards right.
From this curve, the load factor of the station can also be determined.
14. State the need of load forecasting.N-D 2018,A-M 2017, 2015

 To meet out the future demand.
 Long term forecasting is required for preparing maintenance schedule of the generating units,
planning future expansion of the system.
 For day-to-day operation, short term load forecasting is needed in order to commit enough
generating capacity for the forecasting demand nd for maintaining the required spinning reserve.
 Very short term load forecasting is used for generation and distribution. That is, economic generation
scheduling and load dispatching.
 Medium term load forecasting is needed for predicted monsoon acting and hydro availability and
allocating.
15. Define Blackout &brownouts. A-M 2017
A brownout is an intentional or unintentional drop in voltage in an electrical power supply system.
Intentional brownouts are used for load reduction in an emergency. The reduction lasts for minutes or hours,
as opposed to short-term voltage sag (or dip). The term brownout comes from the dimming experienced by
incandescent lighting when the voltage sags. A voltage reduction may be an effect of disruption of an
electrical grid, or may occasionally be imposed in an effort to reduce load and prevent a power outage,
known as a blackout.
16. Define the term load curve and load duration curve. N-D 2017
The curve drawn between the variations of load on the power station with reference to time is known
as load curve.
This type of curve which indicates the variation of load, but with the loads arranged in
descending order of magnitude, i.e., the greatest load on the left and lesser loads towards right.
17. Define load forecasting in power system. N-D 2017
The load on their system should be estimated in advance. This estimation in advance is known as
load forecasting. Load forecasting based on the previous experience without any historical data.
18. State the types and advantages of computer based control in power system. A-M 2015
Types : SCADA , EMS, ECC
Advantages : Computerized control systems provide benefits that can be categorized into four distinct
groups:
 Time
 Money
 Information management
 Improved work conditions.
19.What are the requirements of planning the operation of a power system?
Planning the operation of a power system requires load studies, fault calculations, the design of
means for protecting the system against lightning and switching surges and against short circuits
and studies of the stability of the system.
Steps to be followed.
 Planning of power system.
 Implementation of the plans.
 Monitoring of the system.
Compare with results
20.What is meant by Residential load and Commercial load?
Residential load: It consists of domestic lights, fans, and other appliances such as heaters,
refrigerators, television, radio, air-conditioning, etc., has a high peak during evening.
Commercial load: It includes Lightings for shops, advertising hoardings, hospitals, hotels, shopping
complex, theatres, etc., It has two peaks, morning and evening
Demand factor = 0.9 to 1.0;
Diversity factor = 1.1 to 1.2;
Load factor = 0.25 to 0.3

21. State the factors affecting the load forecasting. A/M 2018
 To meet out future demand
 Long term forecasting – Preparing maintenance schedule of generating units, future expansion.
 Medium term – Prediction of monsoon action and hydro availability
 Short term – Spinning reserve & Unit commitment
22. Write the implications of high diversity factor and list any two methods employed to increase the
diversity factor. A/M 2018
 Giving incentives to some consumers to use electricity in the night or light load periods.
 Using day – light saving
 Staggering the office timings
 Having two part tariff in which consumer has to pay an amount dependent on the maximum demand
of consumer uses.
UNIT – 2 REAL – FREQUENCY
1. State the types of ALFC for interconnected power system. N-D 2017
 Flat frequency Control
 Flat tie line control
 Frequency bias tie- line control
2. Illustrate the conditions for proper synchronizing of alternators. N-D 2017
The conditions for operating two synchronous machines in parallel are:
 Terminal voltage must be same.
 The Speed & system frequency must be same.
 Phase sequence must be same.
3. Explain the need for integral controller in ALFC. A-M 2017
Thereis a considerable droop in speed on frequency of the turbine for a given speed changer setting.
Such a large deviation (± 0.5 Hz) cannot be tolerated and we must develop some suitable control strategy to
achieve much better frequency constancy. For this purpose, a signal from Δf is fed through an integrator to
the speed changer. The integral controller actuates the load reference point until the frequency deviation
becomes zero. Integral controller gives zero steady state error.
4. Define control area. A-M 2017 N-D 2016, 2015, 2014
A control area is defined as a system to which a common generation control scheme is applied. The
electrical interconnection within each control area is very strong as compared to the ties with the
neighbouring areas. All the generators in control area swing in coherently or it is characterized by a single
frequency. It is necessary to be considered as many control areas as number of coherent group. Each control
area of a power system should help to maintain the frequency and voltage profile of the overall system.
5. Specify the use of static & dynamic response of the ALFC. N-D 2016
Static response of an ALFC loop will inform about frequency accuracy, whereas, the dynamic
response of ALFC loop will inform about the stability of the loop.
6. State the objective of tie –line bias control. M-J 2016
The objective of tie-lines is to trade power with the systems or areas in the neighbourhood whose
costs for operation create such transactions cost-effective. Moreover, even though no power is being
transmitted through the tie-lines to the neighbourhood systems/areas and it so happens that suddenly there is
a loss of a generating unit in one of the systems. During such type of situations all the units in the
interconnection experience an alteration in frequency and because of which the desired frequency is
regained. Let there be two control areas and power is to be exchanged from area 1 to area 2.

7. Define area control error. N-D 2018, M-J 2016
ACE is the change in area frequency which when used in integral control loop
focused the steady state frequency error to zero.
ACE =∆Ptie + b ∆f p.u MW (for multi area system)
ACE =∆f (for single area system)
Where, b = Area frequency bias.
∆Ptieis the change in tie line power,
∆f is the change in frequency.
8. State the advantage of AVR loop over ALFC. N-D 2015
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.
9. Brief the application of secondary ALFC loop in power system networks. N-D2018,A-M 2015
The application of secondary ALFC loop in power system network is to maintain desired megawatt
output of a generator unit and controlling the frequency of the larger interconnection. The frequency is
controlled directly using ALFC so that the real power of power system is to be controlled. The frequency in
interconnected system is always subject to change due change in load and power capability of tie-line.
10. State the control objective of two area load frequency control. N-D 2014
 Under normal operating condition, each control area should have the capacity to meet its own load
from its own spinning generator, plus the scheduled interchange between the neighboring areas.
 Under emergency conditions, the energy can be drawn from the spinning reserves of all the
neighboring areas immediately due to the sudden loss of generating unit
11. State any two necessities to put alternators in parallel. M-J 2014, N-D 2013
 Local or regional power use may exceed the power of a single available generation.
 Parallel alternators allow one or more units to be shut down for scheduled or emergency maintenance
while the load is being supplied with power.
12. State the assumptions made in dynamic response of uncontrolled case. M-J 2014
 Neglect the turbine dynamics
 The speed changer action is instantaneous.
13. Define the function of load frequency control. M-J, N-D 2013
The function of load frequency control on a power system is to change the valve settings or gate
openings of a prime mover as a function of load variations in order to hold system frequency at a constant
level. The prime mover is controlled hence in order to that the input of turbine can be controlled easily. This
is the main function of Load Frequency Control in power system. The speed governor is the main and
important primary tool used in the Load Frequency Control (LFC).
14. Define Speed droop. M-J 2013
In electrical power generation, droop speed control is a speed control mode of a prime mover driving
a synchronous generator connected to an electrical grid. This mode allows synchronous generators to run in
parallel, so that loads are shared among generators in proportion to their power rating.
15.Why are governors employed in Power System?
Governors are employed in power systems for sensing the bias in frequency which is the result of the
modification in load and eliminate it by changing the turbine inputs such as the characteristic for speed
regulation (R) and the governor time constant (Tg). Governor‟s aims to limit the deviation in frequency in
the presence of changing active power load. Consequently, the load reference set point can be utilized for
adjusting the valve/gate positions so as to cancel all the variations in load by controlling the generation of
power rather than ensuing deviation in frequency through Governors employed in power system.

16.Define regulation.
Regulation is defined as percentage rise in full load at the specified power factor is witched
off, the excitation being adjusted initially to give normal voltage.
Regulation =
E0−V
V
Where, E0 = No load voltage
V = On load voltage
The regulation of the real power output of the generator and its frequency (speed) is dealt by the Automatic
Load Frequency Control (ALFC)
17.What are the factors to be considered for increasing response in AVR? (N/D 14)
To make excitation faster, the following may be considered.
 Provision of separate excitation for the self-excited.
 Reduction of number of field turns to reduce the time constant.
 Provision of new armature wining for higher voltage operation.
 Increasing the ceiling voltage by decreasing the permanent external field circuit resistance or by
increasing the excitation voltage E.
 Increase of both excitation voltage and field resistance in the same ratio , so that ceiling voltage is
unaltered but time is constant.
18.List the various components in AVR loop. (A/M’08)
The various components in AVR loop are,
 Exciter
 Comparator
 Amplifier
 Rectifier
 Synchronous generator.
19.What is the inter relation between AVR loop and ALFC loop?
Control actions in the AVR loop affects the magnitude of the generator e.m.f “E”.
As the internal e.m.f determines the magnitude of the ideal power, changes in the AVR loop must be felt in
the ALFC loop.
PG =
𝑉 𝐸
𝑋 𝑑
sin𝛿 +
𝑉 2
2
[
1
𝑋 𝑞
−
1
𝑋 𝑑
] sin 2𝛿 MW/ph.
Hence the changes in Automatic Voltage Regulator (AVR) will be reflected in Automatic Load Frequency
Control (ALFC) loop.
20. List the components of speed governing mechanism. A-M 2018
 Fly ball speed governor
 Speed changer
 Hydraulic Amplifier
 Linkage Mechanism
21. Distinguish between primary and secondary feedback loops in LFC. A-M 2018
Primary loop Secondary loop
The speed change from synchronous speed initiates
the governor control action resulting in all the
participation generator – turbine units taking up the
change in load and stabilizes the system frequency. It
control the steam valve leading to the turbine.
It adjust the load reference set points of selected
turbine – generator units so as to give nominal value
of frequency.
UNIT – 3 REACTIVE POWER CONTROL
1. Show that the shunt compensation improves critical voltage as well as the power factor. A-M 2015
Voltage stability of a system is affected by reactive power limit of the system. FACTs devices
improve the reactive power flow in system thereby improving voltage stability and these are used for
controlling transmission voltage, Power flow, dynamic response and reducing reactive losses in transmission
lines. This paper explores the effect of SVC on static voltage stability and presents the effect of Static VAR

compensator (SVC) on Voltage Profile & Reactive Power for variable load conditions is investigated and
presents static methods like Modal Analysis, Two Bus Thevenin Equivalent and Continuation Power Flow
methods to predict the voltage collapse of the bus in the power system. FACTs controllers help to increase
the load ability margin of the power network.
2. Distinguish between rotor angle stability and short term voltage stability. A-M 2015
Rotor Angle Stability Short-term Voltage Stability
It is the ability of the system to remain in
synchronism when subjected to a disturbance.
It is the ability of the system to maintain steady
state voltages at all the system buses when
subjected to a disturbance.
The rotor angle of a generator depends on the
balance between the electromagnetic torque due
to the generator electrical power output and
mechanical torque due to the input mechanical
power through a prime mover.
Unlike angle stability, voltage stability can also be
a long term phenomenon.
Remaining in synchronism means that all the
generators electromagnetic torque is exactly
balanced by the mechanical torque.
On the other hand if voltage variations are due to
slow change in load, over loading of lines,
generators hitting reactive power limits, tap
changing transformers etc. then time frame for
voltage stability can stretch from 1 minute to
several minutes.
The main difference between voltage stability and angle stability is that voltage stability depends on
the balance of reactive power demand and generation in the system where as the angle stability mainly
depends on the balance between real power generation and demand
3. State the function of load frequency control. N-D 2017
Loading of units are allocated to serve the objective of minimum fuel cost is known as automatic
load scheduling.
4. State the advantages and disadvantages of synchronous compensators. N-D 2017, 2013
Advantages:
 Flexibility of operation for all load conditions.
 As the losses are considerable compared with static capacitors and the power factor is not zero.
Disadvantages:
 The cost of installation is high
 Losses of synchronous condensers are much higher compared to those of capacitors.
5. Comment on the use of series capacitors in transmission lines. A-M 2017
 The reactive power produced by the series capacitor increases with the increase in power transfer.
 Series capacitors are connected in series to compensate the inductive reactance of line. This reduces
the transfer reactance between the buses to which the line is connected.
 It increases maximum power that can be transmitted and reduces reactive power loss.
6. Define exciter ceiling voltage. A-M 2017
 It is the maximum voltage that may be attained by an exciter under specific conditions.
 The voltage of the main exciter should be controlled from zero to ceiling voltage, the maximum
voltage that may be attained by the exciter under specified conditions, to obtain rapid correction of
exciter voltage after disturbance or fault.
7. State the various functions of an excitation system. N-D 2016, 2013
The basic function of an excitation system is to
 Provide necessary direct current to the field winding of the synchronous generator.
 The excitation system must be able to automatically adjust the field current to maintain the required
terminal voltage. The DC field current is obtained from a separate source called an exciter.The
excitation systems have taken many forms over the years of their evolution

8. Mention the purpose of series compensation. N-D 2016
Purpose of series compensation:
 Increase in Power Transfer Capability
 Improvement in System Stability
 Load Division among Parallel Line
 Control of Voltage
9. Define SVC. M-J 2016
SVC comprises of switched or fixed capacitor bank and switched reactor bank in parallel. These
compensators draw reactive power from the line thereby regulating voltage, improve stability (steady state
and dynamic), control overvoltage and reduce voltage and current unbalances. In HVDC application these
compensators provide the required reactive power and damp out sub harmonic oscillations. Static VAR
compensators use switching for VAR control. These are also called static VAR switches or systems.
10. Explain how voltage and reactive power interrelated. M-J 2016, 2013
 Voltage control in an electrical power system is important for proper operation for electrical power
equipment to prevent damage such as overheating of generators and motors, to reduce transmission
losses and to maintain the ability of the system to withstand and prevent voltage collapse.
 In general terms, decreasing reactive power causing voltage to fall while increasing it causing
voltage to rise. A voltage collapse occurs when the system try to serve much more load than the
voltage can support.
 When reactive power supply lower voltage, as voltage drops current must increase to maintain power
supplied, causing system to consume more reactive power and the voltage drops further.
 If the current increase too much, transmission lines go off line, overloading other lines and
potentially causing cascading failures.
11. State the role of exciter. N-D 2015
The exciter is the main component in AVR loop. It delivers the DC power to the generator field. It must
have adequate power capacity and sufficient speed of response (rise time less than 0.1 sec). The exciter
control is the major important concern in the reactive power control of power system. By controlling the
voltage can control the reactive power. The main function of the exciter is to provide Direct Current to the
synchronous motor. The control functions which are provided for exciter are control of voltage and reactive
power flow and the enhancement of system stability.
12. Define stability compensation. N-D 2015
Compensation consists of injecting reactive power to improve power system operation to keep voltages
close to nominal values. It reduces the line currents and hence network losses then improve the stability
enhancement. It Compares the Generator terminal voltage with a preset reference voltage. If the Generator
terminal voltage is less than the reference voltage, the AVR increases D.C. voltage across the Generator
field, maintaining the constant voltage as per the setting. Stability compensation improves the dynamic
response characteristics without affecting the static loop gain.
13. State the relation between voltage, power and reactive power at a node. N-D 2014
The voltage drop in the transmission line is directly proportional to the reactive power flow (Q-flow)
in the transmission line.Most of the electric load is inductive in nature. In a day, during the peak hours, Q-
flow will be heavy, resulting more voltage drop. However, during off-peak hours, the load will be very small
and the distributed shunt capacitances throughout the transmission line become predominant making the
receiving-end voltage greater than the sending-end voltage (Ferranti effect). Thus during off-peak hours
there may be voltage rise in the transmission line from sending-end to receiving-end.
𝜕𝑄
𝜕𝑉
=
𝐸−2𝑉
𝑋
If X is small,
𝜕𝑄
𝜕𝑉
is large.
14. State the advantages of Switched capacitors in voltage control. N-D 2018, 2014
The SC behaves like a resistor whose value depends on capacitance CS and switching frequency f. The SC

resistor is used as a replacement for simple resistors in integrated circuits because it is easier to fabricate
reliably with a wide range of values. It also has the benefit that its value can be adjusted by changing the
switching frequency (i.e., it is a programmable resistance).
 Power factor improvement.
 Efficiency of transmission and distribution of power is high.
 Reactive power compensation.
 Less costly.
 Improve feeder voltage control.
15. Where are synchronous condensers installed? M-J 2014
In Industries in order for Power factor correction and need to maintain Power factor within limits.
16. State the different types of static VAR compensators.N-D 2018, M-J 2014
 Saturated Reactors [SR]
 Thyristor Controlled Reactors [TCR]
 Thyristor Switched Capacitors [TSC]
 Thyristor Controlled Transformers [TCT]
 Thyristor Switched Capacitors [TSC]
17. State the functions of AVR. M-J 2013
The function of the Automatic Voltage Regulator (AVR) is to maintain constant voltage and
power line conditioning to the load under a wide variety of conditions, even when the utility input
voltage, frequency or system load vary widely.
The Automatic Voltage Regulator (AVR) is mainly responsible for regulating system voltage
without violating the limit which indirectly controls the reactive power of a system.
18. Draw the typical block diagram of DC excitation system.
DC excitation system utilizes DC generators as sources of excitation power and provide current to the rotor
of the synchronous machine through slip rings.
19.Write the different methods of voltage control.
The different methods of voltage control are,
 Excitation control,
 Static Shunt capacitors
 Static Shunt reactor,
 Synchronous compensator
 Tap changing transformer,
 Booster Transformer,
 Regulating Transformer and

Prepared by Prabaakaran K AP/EEE
Page10
 Static VAR compensator.
20. Write the objectives of FACTS controllers in the power system network and the main areas of
application of FACTS devices.
Objectives of FACTS controllers
• Better the control of power flow (Real and Reactive) in transmission lines.
• Limits SC current
• Increase the load ability of the system
• Increase dynamic and transient stability of power system
• Load compensation
• Power quality improvement
FACTS mainly find application in following areas,
• Power transmission
• Power Quality
• Railway Grid Connection
• Wind power grid Connection
• Cable Systems
21. State the main objective of Reactive power and voltage control in power system. A-M 2018
For efficient and reliable operation of power system should have the following.
 All machines are designed to operate at a certain voltage – Above / below rated voltage cause
damages to them
 System stability is increased due to maximum utilization of transmission line.
 Reactive power flow is minimized so as to reduce I2
R and I2
X losses and to operate the transmission
system efficiently.
22. Outline the role of Synchronous generators adopted for generation and absorption of reactive
power. A-M 2018
Over excited – Generate Reactive power
Under excited – Absorption of reactive power
UNIT – 4 – UNIT COMMITMENT & ECONOMICS DISPATCH
1. Find the incremental transmission losses for a two area power system, where the bus voltages are
kept fixed and the line power flow is a function of line angle. Power loss is a function of generation of
area B only. A-M 2015
It is evident that,
Pl =f (PgB)
This also suggests that the increment transmission loss for grid A will be zero and the incremental
transmission loss of the line will be governed by the grid B only.
Thus,
(ITL)A = 0

Page11
(ITL)B=
𝜕𝑃 𝑙
𝜕𝑃 𝑔𝐵
=
𝜕𝑃 𝑙
𝜕𝛿
x
𝜕𝛿
𝜕𝑃 𝑔𝐵
Economic operation being dictated by the criterion,
λ =
(IFC )A
1−(ITL )A
=
(IFC )B
1−(ITL )B
In this case for economic operation,
λ =
(IFC )A
1− 0
=
(IFC )B
1−(ITL )B
2. Define spinning reserve. A-M 2015, M-J 2014, 2015
Spinning reserve is the total amount of generation availability from all units synchronized on the system
minus the present load and losses being supplied.
Spinning Reserve = [Total amount of generation] – [Present Load + Losses]
Spinning reserve must be established, so that the loss of one or more units does not cause drop in system
frequency. (i.e. if one unit is lot, the spinning reserve unit has to make up for the loss in a specified time
period.
Spinning reserve is the reserve generating capacity running at zero load or no load.
3. Define FLAPC. N-D 2017, 2013
FLAPC - Full Load Average Production Cost
Full load average production cost = {net heat rate at full load} *{ fuel cost }
FLAPC =
𝐶𝑖(𝑃 𝐺𝑖 )
𝑃 𝐺𝑖
=
𝐾.𝐻𝑖(𝑃 𝐺𝑖 )
𝑃 𝐺𝑖
4. State the conditions for the optimal power dispatch in a lossless system. N-D 2017, 2013
Thenecessary condition for the existence of a minimum cost operating condition is that the incremental cost
rates of all the units be equal to some undetermined value (λ) called Lagrangian multiplier.
λi =
𝑑 𝐹𝑖
𝑑𝑃 𝑖
5. State the constraints in unit commitment. A-M 2017, N-D 2014
 Spinning reserve constraints
 Thermal unit constraints are,
• Minimum up time
• Minimum down time &
• Crew constraints.
 Other Constraints
 Hydro Constraints
 Must Run Constraint
 Fuel Constraint
For a single value turbine the governing is done by throttling of steam and for such units, the input-output
curve is substantially a straight line within its operating range.
6. Define incremental cost in power dispatch. A-M 2017
Incremental Cost =
𝑆𝑚𝑎𝑙𝑙𝐶 ℎ𝑎𝑛𝑔𝑒𝑖𝑛𝑖𝑛𝑝𝑢𝑡
𝑆𝑚𝑎𝑙𝑙𝐶 ℎ𝑎𝑛𝑔𝑒𝑖𝑛𝑜𝑢𝑡𝑝𝑢𝑡
=
𝑑𝐶 𝑖(𝑃𝐶 𝑖)
𝑑𝑃 𝐺𝑖
7. Write the coordination equation taking the effect of transmission losses. N-D 2016
In this case for economic operation,
λ =
(IFC )A
1− 0
=
(IFC )B
1−(ITL )B
8. Write the relationship between fuel energy input and input fuel cost, give the quadratic expression
of fuel cost.

Page12
Relationship between fuel energy input and input fuel cost
Input fuel cost Fi (Pi)=K.Fi(Pi) Rs/hr
Where, K = Cost of the fuel inRs/MKcal.
Fi (Pi) = Fuel energy input in MKcal/hr.
Quadratic expression of fuel cost
Fi (Pi)= = αi (Pi) + βi (Pi) + γiRs/hr
Where αi, βi, γi are constants.
(Pi)=Power generation.
9. Define priority list method. M-J 2016
Priority list method is the simplest unit commitment solution method which consists of creation a
priority list of units.
The priority list can be obtained by noting the full-load average production cost of each unit.
Full load average production cost = {net heat rate at full load} *{ fuel cost }
FLAPC =
𝐶𝑖(𝑃 𝐺𝑖 )
𝑃 𝐺𝑖
=
𝐾.𝐻𝑖(𝑃 𝐺𝑖 )
𝑃 𝐺𝑖
Assumptions
 No load costs are zero
 Unit input –output characteristics are linear between zero output and full load.
 Start-up costs are a fixed amount.
Ignore minimum up time and minimum down time.
10. Define incremental transmission loss. M-J 2016
𝜕𝑃𝐿 𝐿
𝜕𝑃𝐺𝑖
= 𝐼𝑇𝐿
11. Draw incremental fuel cost curve. N-D 2015, M-J 2015, 2016
The slope of fuel cost curve i.e.,
𝑑𝐶𝑖
𝑑𝑃 𝐺𝑖
Rs/ MWh can be obtained by differentiating equation with respect to
PGi.
12. Define crew constraints. N-D 2018, 2015
If a plant consist of two (or) more units, all the units cannot be turned on at the same time since there
are not enough crew members to attend both units while starting up. In a plant with more than one unit there
may not be enough personnel to attend both the units if both are turned on or off at the same time and hence
at the same time both cannot be turned on or off. A certain amount of energy is expended to bring the unit
on-line. This is not generated and is included in the unit commitment problem as a start-up cost.
13. Draw the incremental cost curve of thermal unit. N-D 2014
Incremental cost curve of thermal unit is plotted with the derivative of the fuel cost curve versus real power

Page13
14. Write the equality and inequality constraints considered in the economic dispatch problems
(N/D’11)
The power system has to satisfy several constraints while in operation. These may be broadly divided into
two types,
Equality constraints:
The first of these arises out of the necessity for the system to satisfy load balance and are called equality
constraints.
Inequality constraints:
A number of other constrains due to physical and operational limitations of the units and components will
arise in economic scheduling. These are in the form of inequality constraints.
Each generator in operation will have a minimum and maximum permissible output and the production must
be constrained to ensure that
P1min< P1<P1max
15. Distinguish between economic dispatch & unit commitment. N-D 2018, A-M 2018, M-J 2014
Economic Dispatch Unit Commitment
Optimum allocation of generation to each
station. (At each generating station at various
station load levels)
Optimum allocation of number of units to be
operated (to determine the units of a plant
that should operate for a given load is the
problem of unit commitment)
The problem assumes that there are „n‟ units
already connected to the system.
There are number of subsets of the complete
set of „n‟ units that would satisfy the
expected demand.
Purpose of economic dispatch problems is to
find the optimum operating policy for these
„n‟ units.
Purpose of unit commitment is to find the
optimal subset among the subsets which
provide the minimum operating cost.
16. Write the condition for minimum up time.(A/M’11)
Once the unit is running, it should not be turned off immediately. This is the condition for minimum
uptime. Generally more generators are started up at around the peak load, and few units are started up at
light loads based on full load average production cost. To satisfy minimum up time constraint the units are
set continuously in ON condition.
17.List the techniques for the solution of the unit commitment problem.
The various techniques involved in unit commitment problem are listed as follows.
They are,
 Priority list method
 Dynamic programming
 Brute force technique.
A recent literature review identified nine of these methodologies: priority list method, dynamic
programming, Lagrangian relaxation, genetic algorithms, simulated annealing, particle swarm optimization,
fuzzy logic algorithm, and evolutionary programming.

Page14
18.List out the reasons for choosing the LR algorithm. (A/M’13)
 Speciﬁed for the Unit Commitment Program.
 Flexible in dealing with various types of constraints.
 Flexible to incorporate additional coupling constraints that have not been considered so far.
 Flexible because no priority ordering is imposed
Computationally much more attractive for large system since the amount of computation varies with
the number of units.
19.What are loss coefficients?
Transmission loss can be calculated for a two unit system as
PL = 𝑷 𝑮𝟏 𝑷 𝑮𝟐
𝑩 𝟏𝟏 𝑩 𝟏𝟐
𝑩 𝟐𝟏 𝑩 𝟐𝟐
𝑷 𝑮𝟏
𝑷 𝑮𝟐
In vector form, loss coefficient is nothing but [R + jX] is a drop of transmission line.
Transmission line drops are constraint in matrix form is called loss coefficient or B matrix or Bmn
coefficient.
B =
𝑩 𝟏𝟏 𝑩 𝟏𝟐
𝑩 𝟐𝟏 𝑩 𝟐𝟐
20. What is the need of economic dispatch control?
a. Economic dispatch determines the best way to minimize the current generator operating costs.
b. The λ-iteration method is a good approach for solving the economic dispatch problem:
i. Generator limits are easily handled.
ii. Penalty factors are used to consider the impact of losses.
c. Economic dispatch is not concerned with determining which units to turn on/off (this is the unit
commitment problem).
Basic form of economic dispatch ignores the transmission system limitations.
21. Relate the necessary condition for the existence of minimum cost operation for the thermal power
system. A-M 2018.
 Optimal Loading
 Regular Maintenance
 Adopt energy efficient equipment and process.
UNIT – 5 SCADA
1. State the role of SCADA plays in electrical power systems. A-M 2015
Modern SCADA systems are already contributing and playing a key role at many utilities towards
achieving:
• New levels in electric grid reliability – increased revenue.
• Proactive problem detection and resolution – higher reliability.
• Meeting the mandated power quality requirements – increased customer satisfaction.
Real time strategic decision making – cost reductions and increased revenue
2. Define the responsibilities of regional load dispatch centre. A-M 2015
The main responsibilities of RLDCs are:
♦ System parameters and security.
♦ To ensure the integrated operation of the power system grid in the respective region.
♦ System studies, planning and contingency analysis.
♦ Daily scheduling and operational planning.
♦ Facilitating bilateral and inter-regional exchanges.
♦ Computation of energy dispatch and drawn values using SEMs.
♦ Augmentation of telemetry, computing and communication facilities.
3. Define SCADA. N-D 2017

Page15
SCADA stands for supervisory control and data acquisition system. It allows a few operators to
monitor the generation and high voltage transmission systems and to take action to correct overloads. The
supervisory system may be combined with a data acquisition system by adding the use of coded signals over
communication channels to acquire information about the status of the remote equipment for display or for
recording functions. It is a type of industrial control system (ICS).
4. Define State estimation. N-D 2017, M-J 2016, N-D 2013, M-J 2013
State estimation is defined as the process of assigning a value to an unknown system state variable based
on the measurements from that system according to some criteria.
1. Real-time data primarily come from SCADA
2. SE supplements SCADA data: filter, fill, smooth.
3. To provide a consistent representation for power system security analysis
5. List out the conditions for normal operation of a power system. A-M 2017
 Controller needs to operate the system as economic.
 Voltage & Frequency are needs to keep close to normal.
6. Define energy control centre. A-M 2017
Whenthe power system increases in size, their operation and interaction become more complex. So, it
becomes essential to monitor this information simultaneously for the total system which is called Energy
control Centre.
7. State the functions of SCADA. N-D 2016, M-J 2016, N-D 2013, M-J 2013
The data acquisition and control system performs the following general functions.
a. Data acquisition and validation,
b. Real time variable computations,
c. Alarm monitoring and display,
d. Performance and deviation calculation,
e. Trends, events, reports and logs,
f. Sequential control,
g. Modulating control
h. Other related functions.
8. State the major functions that are carried out in an operational control centre.N-D 2016
 System Load Forecasting-Hourly energy, 1 to 7 days.
 Unit commitment-1 to 7days.
 Economic dispatch
 Hydro-thermal scheduling- up to 7 days.
 MW interchange evaluation- with neighbouring system
 Transmission loss minimization
 Security constrained dispatch
 Maintenance scheduling, Production cost calculation
9. Explain the states of power system. N-D 2015
 Normal state
 Alert state
 Emergency state
 Extremis state
 Restorative state
10. State the functions of control centre. N-D 2015
Functions of Control Centre.
 System Monitoring

Page16
 Data Acquisition and Control
11. Define power system security. N-D 2014
It is defined as the degree of risk in the ability to survive imminent disturbances (contingencies) without
interruption of customer service.
It depends upon the reserve capacity available in a given situation and the contingency probability of a
disturbance.
12. List the application of SCADA. N-D 2014
As the power system deals with power generation, transmission and distribution sectors, monitoring is the
main aspect in all these areas. Thus the SCADA implementation of power system improves
The overall efficiency of the system for optimizing
Supervising and controlling the generation and transmission systems,
Greater system reliability and
Stability for integrated grid operation.
13. State the objectives of AGC. M-J 2014
The objectives of AGC are
 To hold frequency at or very close to a specified normal value.
 To maintain the current value of interchange power between control areas.
 To maintain each units generation at the most economic value.
14. Define restorative state. M-J 2014
To bring the extremis state back to normal through the restorative state this is a slower process.
From this state, the system may be brought back either to alert state or secure state.The latter is a slow
process. Hence, in certain cases, first the system is brought back to alert state and then to the secure state.
This is done using restorative control action.
15. Define Network Topology in a Power System.
In order to run the state estimation, we must know how the transmission line are connected to the load and
generation buses. This information is called network topology.
16. List out the functions of Energy management system. A-M 2018
 MW interchange evaluation- with neighboring system
 Maintenance scheduling
 Production cost calculation.
17. Point out the objectives of Load Frequency Control.
Load frequency control (LFC) has to achieve three primary objectives, which are stated below in priority
order:
 To maintain frequency at the scheduled value.
 To maintain net power interchanges with neighbouring control areas at the scheduled values.
 To maintain power allocation among units at economically desired values.
18. Write about RTU and Master Station in SCADA.
Remote Terminal Units (RTU): RTUs are microprocessor controlled electronic devices deployed in field at
specific sites and locations. They collect necessary data and transmit them to SCADA for processing.

Page17
Master Unit: It is a large computer system which serves as a central processor.
Communication Links Fibre optic / satellite / microwave communications are employed to link RTUs with
SCADA.
19. Draw the general layout of Master station and Remote control station.
20. List out the National Regional Electricity Boards.
The national regional electricity boards are,
 Northern Regional Electricity Board
 Western Regional Electricity Board
 Southern Regional Electricity Board
 Eastern Regional Electricity Board
 North-east Regional Electricity Board
21. State the Weighted Least Square Criterion. A-M 2018, N-D 2018
The objective is to minimize the sum of the squares of the weighted deviations of the estimated
measurements [F(X)] from the actual measurements [Z].
22. What are the priorities for operation of modern power system? N-D 2018
 Economic Operation
 Frequency Regulation
 Voltage Regulation
 Power factor maintenance
 Load forecasting and Scheduling
 Ability to Restructure.

EE6603 – Power System Operation & Control (R2013)
Prepared by: Prabaakaran K AP/EEE Page No:1
UNIT – 1 INTRODUCTION
PART – B
1. A Power station has to meet the following demand :
Group A: 200 KW between 8 a.m and 6 p.m
Group B: 100 KW between 6 a.m and 10 A.m
Group C: 50 KW between 6 a.m and 10 A.m
Group D: 100 KW between 10 a.m and 6 p.m and then between 6 P.M and 6 A.M.
Plot the daily load curve and determine:
(a) Diversity Factor
(b) Units generated per day
(c) Load factor (16)(N/D’18,14)
SOLUTION:
The given load cycle can be tabulated as follows:
TIME(HOURS) & GROUP 0-6 6-8 8-10 10-18 18-24
A 0 0 200 200 0
B 0 100 100 0 0
C 0 50 50 0 0
D 100 0 0 100 100
TOTAL LOAD ON POWER
STATION
100 150 350 300 100
Load curve:
(4)
(a) Maximum demand = 350 KW
Diversity factor =
𝑆𝑈𝑀 𝑂𝐹 𝐴𝐿𝐿 𝐼𝑁𝐷𝐼𝑉𝐼𝐷𝑈𝐴𝐿 𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷
𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷 𝑂𝑁 𝑆𝑇𝐴𝑇𝐼𝑂𝑁
=
200+100+50+100
350
= 1.286 (4)
(b) Units generated per day = Area under the load curve
= 100*6 + 150*2 + 350*2 + 300*8 + 100*6
= 4600 KW/hr (4)
(c) Load factor =
AVERAGE LOAD
MAXIMUM DEMAND
𝐴𝑉𝐸𝑅𝐴𝐺𝐸 𝐿𝑂𝐴𝐷 =
𝑈𝑁𝐼𝑇𝑆 𝐺𝐸𝑁𝐸𝑅𝐴𝑇𝐸𝐷 𝑃𝐸𝑅 𝐷𝐴𝑌
𝐻𝑂𝑈𝑅𝑆 𝐼𝑁 𝐴 𝐷𝐴𝑌
= 4600/24 = 191.666 Kw
LOAD FACTOR = 191.666/350 = 54.76%(4)
2. A Power supply is having the following loads :
Type of load
Maximum
Demand (kW)
Diversity factor of
Group
Demand factor
Domestic 10,000 1.2 0.8
Commercial 30,000 1.3 0.9
Industrial 50,000 1.35 0.95
If the overall system diversity factor is 1.5, determine (a) the maximum demand, (b) connected load of each
type. (12)

Vel Tech High Tech Dr.Rangarajan Dr.Sakunthla Engineering College
Solution:
(a) Total maximum demand of loads=10000+30000+50000=90000 kw
System diversity factor=1.5
Maximum demand=Total demand/system diversity factor
=90000/1.5
= 60000 kW (4)
(b) Connected load of each type Domestic load:
Diversity factor of domestic load=
Maximum domestic demand
Maximum domestic load demand
= 12000/0.8
= 15000 kw (4)
(c) Commercial load:
Connected commercial demand=
Maximum demand
Demand factor for commercial load
= 39000/0.9
= 43333.33 kw (4)
(d) Industrial load:
Connected industrial load=
Maximum demand
demand factor of industrial load
= 67500/0.95 = 71052.63 kw
3. A generating station has a maximum demand of 400 MW. The annual load factor is 60% and
capacity factor is 50%.Find the reserve capacity of the plant. (4)
Energy generated per annum = Maximum demand x load factor x hours in a year = 400 x 0.6 x 8760
= 2102.4 x 103
MWhr
Capacity Factor =
Units generated per annum
Plant capacity x hours in a year
Plant capacity = Unit generated per annum/capacity factor hours in a year =
(2102.4 x 103)
0.5 x 8760
= 480 MW
Reserve capacity = Plant capacity – Maximum Demand
= 480 - 400 = 80 MW
4.A generating station has the following daily load cycle: (ND’15) (8)
TIME
(HOURS)
0-6 6-10 10-12 12-16 16-20 20-24
LOAD
(MW)
20 25 30 25 35 20
Draw the load curve and find:
(1) Maximum demand, (2) Units generated per day, (3) Average load, (4) Load factor.
Solution:
(4)
(1) Maximum Demand = 35 MW = 35 x 103
KW (3)
(2) Units generated per day = Area under the load curve in Kw/hr
= (6*20+4*25+2*30+4*25+4*35+4*20)*103
= 600*103
Kwhr (3)
(3) Average Load =
UNITS GENERATED PER DAY
HOURS IN A DAY
= 66*103
/24 = 25000kw (3)
(4) Load factor =
AVERAGE LOAD
𝑀𝐴𝑋𝐼𝑀𝑈𝑀𝐷𝐸𝑀𝐴𝑁𝐷
= 25000/35*103
= 71.43% (3)
0
20
40
0-6 6-10 10-12 12-16 16-20 20-24
LOAD (MW)
LOAD (MW)

5. A power station has to meet the following load demands:
Load A : 50 KW between 10AM and 6PM
Load B : 30 KW between 6PM and 10PM
Load C : 20 KW between 4PM and 10AM
Plot the daily load curve and determine i) diversity factor, ii) units generated per day, iii) load factor
(M/J’ 16) (8)
SOLUTION:
The given load cycle can be tabulated as follows:
TIME(HOURS) & GROUP 10-6 6-10 4-10
A 50 0 0
B 0 30 0
C 0 0 20
TOTAL LOAD ON POWER
STATION
50 30 20
Load curve:
(4)
(d) Maximum demand = 350 KW
Diversity factor =
𝑆𝑈𝑀𝑂𝐹𝐴𝐿𝐿𝐼𝑁𝐷𝐼𝑉𝐼𝐷𝑈𝐴𝐿𝑀𝐴𝑋𝐼𝑀𝑈𝑀𝐷𝐸𝑀𝐴𝑁𝐷
𝑀𝐴𝑋𝐼𝑀𝑈𝑀𝐷𝐸𝑀𝐴𝑁𝐷𝑂𝑁𝑆𝑇𝐴𝑇𝐼𝑂𝑁
=
50+30+20
100
= 1 (4)
(e) Units generated per day = Area under the load curve
= 50*8 + 30*4 + 20*6
= 640KWhr (4)
(f) Load factor =
AVERAGE LOAD
MAXIMUM DEMAND
𝐴𝑉𝐸𝑅𝐴𝐺𝐸 𝐿𝑂𝐴𝐷 =
𝑈𝑁𝐼𝑇𝑆 𝐺𝐸𝑁𝐸𝑅𝐴𝑇𝐸𝐷 𝑃𝐸𝑅 𝐷𝐴𝑌
𝐻𝑂𝑈𝑅𝑆 𝐼𝑁 𝐴 𝐷𝐴𝑌
= 640/24 = 26.66 Kw
LOAD FACTOR = 26.666/100 = 26.00%(4)
6. A generating station has the following daily loads:
TIME
(HOURS)
0-6
6-8
(Hrs)
8-12
(Hrs)
12-14
(Hrs)
14-18
(Hrs)
18-20
(Hrs)
20-24
(Hrs)
LOAD
(KW)
4500 3500 7500 2000 8000 2500 2500
Sketch the load duration curve and determine the load factor and plant capacity factor if
the capacity of the plant is 12 MW. (16)(A/M’11)
Solution:
(4)
0
20
40
60
0 2 4 6 8 10 12 14 16 18 20 22 24
DemandMW
Time Hrs
LOAD CURVE
0
2000
4000
6000
8000
10000
0-6 6-8
(Hrs)
8-12
(Hrs)
12-14
(Hrs)
14-18
(Hrs)
18-20
(Hrs)
20-24
(Hrs)
LOAD (KW)
LOAD (KW)

Daily Load Curve
TIME (HOURS) 4 8 12 18 20 22 24
LOAD (KW) 8000 7500 5000 4500 3500 2500 2000
(4)
LOAD DURATION CURVE
Capacity of the plant = 12.0 MW
Max. demand of the generating station = 8000 KW
Units generated in 24 hrs = 8000*4 + 7500*4 + 5000*4 + 4500*6 + 3500*2 + 2500*2 + 2000*2
= 1,25,000Kwhr (2)
Average Load =
UNITS GENERATED PER DAY
HOURS IN A DAY
= 1,25,000/24 = 5208 KW(2)
Load factor =
AVERAGE LOAD
MAXIMUM DEMAND
*100 = 5208/8000*100 = 65.1% (2)
Plant capacity factor =
Average demand
Rated capacity of power plant
* 100
= 5208/12000 *100 (2)
Plant capacity factor = 43.4%
7. Consider an inductive load of type Z = R + jX. By how much percentage the real load drop, if the voltage
is reduced by 5%?
How would 2% drop in frequency affect the real load, if the load power factor is 0.8.Derive the relations used?
(M/J’ 16) (16)
Solution:
P + j Q = |𝑉2| ∗ 𝑌∗
= |𝑉2| ∗
1
𝑅−𝑗 𝑋
(4)
Multiplying by complex conjugate of denominator and numerator, we get
𝑉2
𝑅 − 𝑗𝑋
∗
𝑅 + 𝑗𝑋
𝑅 + 𝑗𝑋
= |𝑉2| ∗
𝑅 + 𝑗𝑋
𝑅 + 𝑋2
Equating real and imaginary parts, we get
𝑃 = |𝑉2|
𝑅
𝑅2 + 𝑋2
𝑄 = |𝑉2|
𝑋
𝑅2 + 𝑋2
A small change in voltage results in twice the relative change in MW. In this case 5% drop in voltage causes a 10% drop in
load.
𝑃 = |𝑉2|
𝑅
𝑅2+ 𝑋2 (4)
Differentiate p w.r to f,
∆𝑃
𝑃
=
− 2 𝑋2
𝑅2+ 𝑋2
∆𝑓
𝑓
(4)
For cos∅ = 0.8
Sin ∅ =
𝑋
𝑍
=
𝑋
𝑅2+ 𝑋2
∆𝑃
𝑃
= −0.72
∆𝑓
𝑓
(4)
2 % Frequency Drop results in a 1.44% Load Increase
0
2000
4000
6000
8000
10000
1 2 3 4 5 6 7
LOAD (KW)
TIME (HOURS)

8. Explain in detail about plant level & system level control. (N/D 18)

9. Types of Load Forecasting Methods in load prediction techniques. (A/M ’18)

UNIT – 2 REAL POWERS AND FREQUENCY CONTROL
1. What are the components of speed governor system of an alternator? Derive its transfer function with
an aid of a block diagram. (A/M 18)(A/M’,13) (A/M’ 15) (N/D’ 14) (N/D’ 17 (M/J’ 16)
Answer:
The real power control mechanism of a generator is shown in bel ow Fig. The main parts are:
1)Speed changer
2)Speed governor
3)Hydraulic amplifier
4)Control valve
SPEEDCHANGER (4)
It is connected by linkage mechanism. The 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. MW as the case may be. The movements are assumed positive in the directions of
arrows.
Corresponding to “raise” command, 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 point “B” moves upwards and again generator output power will
increase.
SPEED GOVERNOR (4)
The output commend of speed governor is ΔPg which corresponds to movement ΔxC. The speed governor
has two inputs:
1) Change in the reference power setting, ΔPref
2) Change in the speed of the generator, Δf, as measured by ΔxB.
HYDRAULIC AMPLIFIER & CONTROL VALVE (8)
The generator is synchronized to a network of very large size, solar in fact, that its frequency will be
essentially independent of any changes in the power output of this individual generator (“infinite” network).
Since Δf0=0. Thus for a generator operating at constant speed, (or frequency) there exists a direct
proportionality between turbine power and reference power setting.

2. Draw the block diagram of controlled and uncontrolled single area load frequency system and explain
the salient features under static conditions.
Static analysis or steady state response of Uncontrolled Case: (8)
Consider the speed changer has a fixed setting. Under this condition ∆PC = 0 and the load demand changes. This
is known as free governor operation. The block diagram is shown in figure drawn from substituting ∆PC = 0.
Using Block reduction technique, the block diagram is shown below,
∆𝒇 𝒔𝒕𝒂𝒕=
−∆𝐏 𝐃
(𝐁+
𝟏
𝐑 𝟏
+
𝟏
𝐑 𝟐
+⋯……….+
𝟏
𝐑 𝐧
)
Static analysis or steady state response of Controlled Case: (8)
In this case, there is a step change ∆PC force for speed changer setting and the load demand remains fixed i.e.,
∆PD=0.
The block diagram is shown in figure,

∆𝒇 𝒔𝒕𝒂𝒕
∆𝑷 𝑪
=
𝟏
𝐁+
𝟏
𝐑
3. With the help of block diagram of controlled and uncontrolled single area load frequency system
explain the dynamic response of it. (N/D’10)(A/M’11)
Answer:
A static response of AFLC loop will inform about frequency accuracy, whereas the dynamic response of ALFC
loop will inform about the stability of the loop.
To obtain the dynamic response representing the change in frequency as a function of time for a step change in
load. The block diagram reduces as shown in figure, (2)
Put ∆ PC (s) = 0
The block diagram reduces as shown in Fig.
(4)
(6)
Dynamic Analysis of Controlled Case:
∆𝑷 𝑫(𝒔) = 𝟎
(4)

4. Develop the state variable model of system and state the advantages of the model.
(M/J’2014)

5. Two synchronous generators operating in parallel. Their capacities are 300 MW and 400 MW. The
droop characteristics of their governor are 4% and 5% from no load to full load. Assuming that the
generators are operating at 50 Hz at no load, how would be a load of 600 MW shared between them.
What will be the system frequency at this load? Assume free governor action. N-D 2015, 2014, A-M 2014

6. A two area system connected by a tie line has the following parameters with base MVA for each area.
Area 1 2
Turbine output power 2000 MVA 1000 MVA
Nominal frequency 50 Hz 50 Hz
Speed regulation 3 % 5 %
Power system gain (kP) 50 Hz/ p.u. MW 40
Governor time constant 0.3 0.2
Turbine time Constant 0.6 0.4

The synchronous power coefficient is computed from the initial operating condition T12 = 2.0 p.u. A load
change of 400 MW occurs in area 1. Determine the steady state frequency and the change in the tie line
flow, comment on the results. A/M 2017
Sol :

UNIT – 3 REACTIVE POWER – VOLTAGE CONTROL
1. Develop the block diagram of AVR and obtain its transfer function and explain the static & dynamic response.
A/M’ 18, (M/J’ 16) N/D’ 18

Synchronous Generator

2. Explain in detail about Tap changing transformer for voltage improvement. (N/D 18)
Off load Tap Changing :
Its shows the disconnection of transformer while in tap changing is carried
out.
It is operated frequently during load change.
On load Tap Changing:

3. Explain in detail about Boost transformer for voltage control. (N/D 18)

4. Derive the expression for relation between voltage, Power and reactive power at a node ‘n’ in power system.
(16)[ND’12, MJ’ 16]
The phase voltage ‘V’ at a node is a function of real and reactive power at that node.(4)
i.e., V = f(P,Q)
Differentiating 𝑑𝑉 =
𝜕𝑉
𝜕𝑃
. 𝜕𝑃 +
𝜕𝑉
𝜕𝑄
. 𝑑𝑄
𝑑𝑉 =
𝑑𝑃
𝜕𝑃
𝜕𝑉
+
𝑑𝑄
𝜕𝑄
𝜕𝑉
The change in voltage at a node is defined by
𝜕𝑃
𝜕𝑉
and
𝜕𝑄
𝜕𝑉
.
Consider a short transmission line with series impedance R + jX, as shown below.
E = V + IZ = V + I (R + jX)
𝑉⃗ = 𝑉∠0o
(reference)
𝐼⃗⃗ =
𝑆∗
𝑉∗
=
𝑃 − 𝑗𝑄
𝑉
Change in voltage ∆𝑉 = E – V = I (R + jX) =
𝑃−𝑗𝑄
𝑉
(𝑅 + 𝑗𝑋)
=
𝑃𝑅 + 𝑄𝑋
𝑉
+ 𝑗
𝑃𝑋 − 𝑅𝑄
𝑉
E – V =
𝑃𝑅+𝑄𝑋
𝑉
Calculate real power:(4)
(E - V) V = PR + QX
PR = (E - V) V – QX
Real power P =
(𝐸−𝑉)𝑉−𝑄𝑋
𝑅
=
𝐸
𝑅
−
𝑉2
𝑅
−
𝑄𝑋
𝑅
Partially differentiating P with respecting to V, we get
𝜕𝑃
𝜕𝑉
=
𝐸 − 2𝑉
𝑅
Calculate reactive power:
(E - V) V = PR + QX
QX = (E - V) V – PR
Reactive Power Q =
(𝐸−𝑉)𝑉−𝑃𝑅
𝑋
=
𝐸𝑉
𝑋
−
𝑉2
𝑋
−
𝑃𝑅
𝑋
Partially differentiating with respecting to V, we get
𝜕𝑄
𝜕𝑉
=
𝐸
𝑋
−
2𝑉
𝑋
=
𝐸 − 2𝑉
𝑋
We know dV = =
𝑑𝑃
𝜕𝑃
𝜕𝑉
+
𝑑𝑄
𝜕𝑄
𝜕𝑉
=
𝑑𝑃
(
𝐸−2𝑉
𝑅
)
+
𝑑𝑄
(
𝐸−2𝑉
𝑋
)
=
𝑅 𝑑𝑃
𝐸 − 2𝑉
+
𝑋 𝑑𝑄
𝐸 − 2𝑉
For constant voltage V and ∆𝑉,
R dP + X dQ = 0
X dQ = - R dP
𝑑𝑄 =
−𝑅 𝑑𝑃
𝑋
(4)
The quantity
𝜕𝑄
𝜕𝑉
can be determined by using a network analyzer by the injection of a known quantity of VARs at
the node and measuring the difference in voltage produced at that node.
Partially differentiating Q with respecting to V, we get
𝜕𝑄
𝜕𝑉
=
𝐸 − 2𝑉
𝑋

If X is small,
𝜕𝑄
𝜕𝑉
is large.
∆𝑉is small, when the number of lines meeting at a node is more, therefore X is small. So
𝜕𝑄
𝜕𝑉
is large and has the
value of 10 – 15 MVAR/KV.
𝝏𝑸
𝝏𝑽
and short circuit at a node :(4)
If the three phases at the receiving end are short circuited, E = V.
Substituting in
𝜕𝑄
𝜕𝑉
,we get
𝜕𝑄
𝜕𝑉
=
𝐸−2𝑉
𝑋
=
𝐸−2𝐸
𝑋
Amp =
−𝐸
𝑋
Amp
5. Explain in detail about Static Var Compensation with Phasor Diagram. A/M 2018

6. The load at receiving end of a 3 phase overhead transmission line is 25 MW, 0.8 p.f lags at the line
voltage of 33 KV. A Synchronous compensator is situated at receiving end and voltage at both ends of
the lines is maintained at 33 KV. Calculate the MAVR of the compensator. The line has resistance and
reactance of 5 Ω/ph, 20 Ω/ph. (16) [MJ’14]

UNIT – 4 – ECONOMIC DISPATCH & UNIT COMMITMENT
1. What is priority list method of unit commitment? Explain with an example.(A/M ‘15)
Unit commitment: (8)
The life style of a modern man follows regular habits and hence the present society also follows
regularly repeated cycles or pattern in daily life. Therefore, the consumption of electrical energy also follows a
predictable daily, weekly and seasonal pattern.
There are periods of high power consumption as well as low power consumption. It is therefore possible
to commit the generating units from the available capacity into service to meet the demand.
The computational aspects for allocating load to a plant in the most economical manner. For a given
combination of plants the determination of optimal combination of plants for operation at any one time is also
desired for carrying out the aforesaid task.
The plant commitment and unit ordering schedules extend the period of optimization from a few
minutes to several hours. From daily schedules weekly patterns can be developed. Likewise, monthly, seasonal
and annual schedules can be prepared taking into consideration the repetitive nature of the load demand and
seasonal variations.
Unit commitment schedules are thus required for economically committing the units in plants to service
with the time at which individual units should be taken out from or returned to service.
Constraints in Unit Commitment (8)
Many constraints can be placed on the unit commitment problem. The list presented here is by no means
exhaustive. Each individual power system, power pool, reliability council, and so forth, may impose different
rules on the scheduling of units, depending on the generation makeup, load-curve characteristics.
The simplest unit commitment solution method consists of creating a priority list of units. A simple
shut-down rule or priority-list scheme could be obtained after an exhaustive enumeration of all unit
combinations at each load level.
The priority list of could be obtained in a much simpler manner by noting the full-load average
production cost of each unit, where the full-load average production cost is simply the net heat rate at full load
multiplied by the fuel cost.

Priority List Method:
Priority list method is the simplest unit commitment solution which consists of creating a priority list of
units.
Full load average production cost= Net heat rate at full load X Fuel cost Assumptions:
1. No load cost is zero
2. Unit input-output characteristics are linear between zero output and full load
3. Start up costs are a fixed amount
4. Ignore minimum up time and minimum down time
Steps to be followed
1. Determine the full load average production cost for each units
2. Form priority order based on average production cost
3. Commit number of units corresponding to the priority order
4. Calculate PG1, PG2 ………….PGN from economic dispatch problem for the feasible combinations only
5. For the load curve shown
Assume load is dropping or decreasing, determine whether dropping the next unit will supply generation &
spinning reserve.
If not, continue as it is
If yes, go to the next step
6. Determine the number of hours H, before the unit will be needed again.
7. Check H< minimum shut down time.
If not, go to the last step
If yes, go to the next step
8. Calculate two costs
1. Sum of hourly production for the next H hours with the unit up
2. Recalculate the same for the unit down + startup cost for either cooling or banking
9. Repeat the procedure until the priority list
Merits:
1. No need to go for N combinations
2. Take only one constraint
3. Ignore the minimum up time & down time
4. Complication reduced
Demerits:
1. Startup cost are fixed amount
2. No load costs are not considered.
2. Describe the forward dynamic programming approach for solving unit commitment problem.
[M/J ’18, 16] [N/D 18]
Dynamic-Programming Solution (8)
Dynamic programming has many advantages over the enumeration scheme, the chief advantage being a
reduction in the dimensionality of the problem. Suppose we have found units in a system and any combination
of them could serve the (single) load. There would be a maximum of 24 - 1 = 15 combinations to test.
However, if a strict priority order is imposed, there are only four combinations to try:
Priority 1 unit
Priority 1 unit + Priority 2 unit
Priority 1 unit + Priority 2 unit + Priority 3 unit
Priority 1 unit + Priority 2 unit + Priority 3 unit + Priority 4 unit
The imposition of a priority list arranged in order of the full-load average cost rate would result in a
theoretically correct dispatch and commitment only if:

1. No load costs are zero.
2. Unit input-output characteristics are linear between zero output and full load.
3. There are no other restrictions.
4. Start-up costs are a fixed amount.
In the dynamic-programming approach that follows, we assume that:
1. A state consists of an array of units with specified units operating and
2. The start-up cost of a unit is independent of the time it has been off-line
3. There are no costs for shutting down a unit.
4. There is a strict priority order, and in each interval a specified minimum the rest off-line. (i.e., it is a
fixed amount).amount of capacity must be operating.
A feasible state is one in which the committed units can supply the required load and that meets the
minimum amount of capacity each period.
Forward DP Approach
One could set up a dynamic-programming algorithm to run backward in time starting from the final
hour to be studied, back to the initial hour. Conversely, one could set up the algorithm to run forward in time
from the initial hour to the final hour.
The forward approach has distinct advantages in solving generator unit commitment. For example, if the
start-up cost of a unit is a function of the time it has been off-line (i.e., its temperature), then a forward
dynamic-program approach is more suitable since the previous history of theunit can be computed at each
stage.
There are other practical reasons for going forward. The initial conditions are easily specified and the
computations can go forward in time as long as required. A forward dynamic-programming algorithm is shown
by the flowchart.
Algorithm and flowchart: (8)
The recursive algorithm to compute the minimum cost in hour K with combination
Fcost(K,I)= min[Pcost(K,I)+Scost(K-1,L:K,I)+Fcost(K-1,L)] ---------------------------------- (1)
Where
Fcost(K, I ) = least total cost to arrive at state ( K , I )
Pcost(KI, ) = production cost for state ( K ,I )
Scost(K - 1, L: K , I)= transition cost from state (K - 1, L) to state ( K , I )
State (K, 1) is the Zth combination in hour K. For the forward dynamic programming approach, we
define a strategy as the transition, or path, from one state at a given hour to a state at the next hour.
Note that two new variables, X and N, have been introduced in Figure 2.11 X =
number of states to search each period
N = number of strategies, or paths, to save at each step.
These variables allow control of the computational effort (see below Figure).For complete enumeration, the
maximum number of the value of X or N is 2n
– 1
Compute the minimum cost
Figure: Forward DP Approach

3. Derive the coordination equations for economic dispatch with and without loss. (16) (N/D ‘14)
Coordination equation without loss.
The optimization problem can now be stated as
Minimize C = ∑ 𝐶𝑖(𝑃𝐺𝑖)𝑁
𝑖=1
Where, C = Operating cost
i = Number of generators on the bus.
Subject to H (PG1, PG2,….PGN) = PD -∑ 𝐶 𝐺𝑖
𝑁
𝑖=1
Using Lagrange multiplier 𝜆,
Lagrange function C* = C + 𝜆𝐻
𝐿 = 𝐶∗
= ∑ 𝐶𝑖(𝑃𝐺𝑖)𝑁
𝑖=1 + 𝜆(𝑃𝐷- ∑ 𝑃𝐺𝑖
𝑁
𝑖=1 )
For maximum objective function, differentiate equation
𝜕𝑐∗
𝜕𝑃 𝐺𝑖
= 0
𝜕𝑐∗
𝜕𝑃 𝐺𝑖
=
𝜕𝐿
𝜕𝑃 𝐺𝑖
=
𝜕𝐶 𝑖
𝜕𝑃 𝐺𝑖
+0– 𝜆 =0
𝜆 =
𝜕𝐶 𝑖
𝜕𝑃 𝐺𝑖
i= 1,2,……..N
For N units,
𝜕𝐶1
𝜕𝑃 𝐺1
=
𝜕𝐶2
𝜕𝑃 𝐺2
= ⋯ =
𝜕𝐶 𝑁
𝜕𝑃 𝐺𝑁
= 𝜆……………………….(1)
Where,
𝜕𝐶 𝑖
𝜕𝑃 𝐺𝑖
= Incremental cost of the unit i.
The optimal loading of generator corresponds to the equal incremental cost point of all the generators.
Equation (1) is called as coordination equation neglecting losses. (8)
Coordination equation with loss.
Minimize C = ∑ 𝐶𝑖(𝑃𝐺𝑖)𝑁
𝑖=1 I=1,2,…n
Subject to constraint
∑ 𝑃𝐷𝑖+ 𝑃𝑙
𝑁
𝑖=1 = ∑ 𝑃𝐺𝑖
𝑁
𝑖=1

∑ 𝑃𝐷𝑖+ 𝑃𝑙
𝑁
𝑖=1 - ∑ 𝑃𝐺𝑖
𝑁
𝑖=1 = 0
Apply Lagrange multiplier 𝜆
𝐶∗
= ∑ 𝐶𝑖(𝑃𝐷𝑖)𝑁
𝑖=1 + 𝜆(∑ 𝑃𝐷𝑖+ 𝑃𝑙
𝑁
𝑖=1 - ∑ 𝑃𝐺𝑖
𝑁
𝑖=1 )
For maximum objective function, differentiate equation
𝜕𝑐∗
𝜕𝑃 𝐺𝑖
= 0
𝜕𝑐∗
𝜕𝑃 𝐺𝑖
=
𝜕𝐶 𝑖
𝜕𝑃 𝐺𝑖
+
𝜆𝜕𝑃 𝐿
𝜕𝑃 𝐺𝑖
– 𝜆 =0
On deriving we get
𝜆 =
(𝐼𝐶) 𝑖
1−(𝐼𝑇𝐿)𝑖
= 𝐿𝑖(𝐼𝐶)𝑖
Where Li is the penalty factor of plants
So 𝜆i =
1
1−(𝐼𝑇𝐿)𝑖
(8)
Above equation is the exact coordination equation when transmission losses are considered.
4. The fuel inputs per hour of plants 1 and 2 are given as
F1 = 0.2 𝐏𝟏
𝟐
+ 40 𝐏𝟏 + 120 Rs/hr
F2 = 0.25 𝐏𝟐
𝟐
+ 30 𝐏𝟐 + 150 Rs/hr
Determine the economic operating schedule and the corresponding cost of generation if the
maximum and minimum loading of each unit is 100 MW and 25 MW. Assume the
transmission losses are ignored and the total demand is 180 MW. Also determine the saving
obtained if the load is equally shared by both the units.(16) A/M 2018
Solution:
λ1 =
𝑑𝐹1
𝑑𝑃1
= 2 x 0.2 𝑃1+ 40 = 0.4 P1 + 40
λ2 =
𝑑𝐹2
𝑑𝑃2
= 2 x 0.25 P2 + 30 = 0.5 P2 + 30
PD = P1 + P2 = 180 MW
λ =
PD+
b1
2 a1
+
b2
2 a2
1
2 a1
+
1
2 a2
=
180+
40
0.4
+
30
0.5
1
0.4
+
1
0.5
= 75.5556 Rs/MW hr
PG1 =
𝜆− 𝑏1
2 𝑎1
=
75.5556−40
0.4
= 88.889 MW
PG2 =
𝜆− 𝑏2
2 𝑎2
=
75.5556−30
0.5
= 91.1111 MW
Cost of generation for unit 1, F1 = 0.2 x (88.8889)2
+ (40 x 88.8889) + 120
F1 = 5255.8 Rs/hr
Cost of generation for unit 2, F2 = 0.25 x (91.1111)2
+ (30 x 91.1111) + 150
F2 = 4958.64 Rs/hr (8)
If each unit supplied
180
2
= 90 MW
Increase in cost for unit 1 is C1 = ∫ (
𝑑𝐹1
𝑑𝑃1
)
90
88.8889
x dP1
= ∫ 0.4 𝑃1 + 40
90
88.8889
dP1
= [
0.4
2
𝑃1
2
+ 40 𝑃1]
90
88.889
= 84.197 Rs/hr
Increase in cost for unit 2 is C2 = ∫ (
𝑑𝐹2
𝑑𝑃2
)
90
91.111
x dP2
= ∫ 0.5 𝑃2 + 30
90
91.111
dP2
= [
0.5
2
𝑃2
2
+ 30 𝑃2]
90
91.111
= -83.6411 Rs/hr
Net saving = 84.197 + (-83.6411) = 0.5559 Rs/hr (8)

5. What is unit commitment problem? Discuss the constraints that are to be accounted in unit
commitment problem. (6) (M/J ‘13)
Constraints in Unit Commitment
Many constraints can be placed on the unit commitment problem.
Spinning 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 the present load and losses being supplied.
Spinning reserve must be carried so that the loss of one or more units does not cause too far a drop in system
frequency.
Reserve capacity
Capacity bin excess o0f that required to carry peak load
Reserve generating capacity (2)
The amount of power that can be produced at a given point in time by generating units that are kept
available in case of special need.
Reserve margin
The percentage of installed capacity exceeding the excepted peak demand during specified period
Thermal unit constraints.
The thermal unit constraints are minimum uptime, minimum downtime and crew constraints.
Minimum up time.
Once the unit is running, it should not be turned off immediately, this is the condition for minimum
uptime. Generally more generators are started up at around the peak load, and few units are started up at light
loads based on full load average production cost. To satisfy minimum up time constraint the units are set
continuously ON condition.
Minimum down time. (2)
The unit is recommitted; there is a minimum time before it can be recommended this is the condition for
minimum down time. The units with small minimum up and down times have more changes in their status,
while the units with large minimum up and down times will require less change. A simple way to achieve this is
by the categorization of units as base load, sub-base load, peaking units, must run, and can run.
Crew constraints. (2)
If a plant consist of two (or) more units, all the units cannot be turned on at the same time since there are
not enough crew members to attend both units while starting up. In a plant with more than one unit
there may not be enough personnel to attend both the units if both are turned on or off at the
same time and hence at the same time both cannot be turned on or off. A certain amount of energy is
expended to bring the unit on-line. This is not generated and is included in the unit commitment problem as a
start-up cost.
6. Obtain the priority list of unit commitment using full load average production cost for the
given data: (A/M’ 18)
Heat rate of unit 1, H1 = 510 + 7.2 PG1 + 0.00142 PG1
2 MW/hr
2 MW/hr
2 MW/hr
Unit Minimum (MW) Maximum (MW) Fuel Cost (K)
1 150 600 1.1
2 100 400 1.0
3 50 200 1.2
PD = 500 MW. (10) (M/J ‘13)
Solution:
Step 1: Full load average production cost (FLAPC)
FLAPC1 =
K1 x H1 (PG1 max)
PG1 max
=
1.1 𝑥 [510+(7.2 𝑋 600)+(0.00142 𝑋 6002)]
600
= 9.79
FLAPC2 =
K2 x H2 (PG2 max)
PG2 max
=
1.0 𝑥 [310+(7.85 𝑋 400)+(0.00194 𝑋 4002)]
400
= 9.4

FLAPC3 =
K3 x H3 (PG3 max)
PG3 max
=
1.2 𝑋 [78+(7.97 𝑋 200)+(0.00482 𝑋 2002)]
200
= 11.188 (5)
Step 2: Priority order (Arrange FLAPC in ascending order)
Unit FLAPC Min(MW) Max(MW)
2 9.4 100 400
1 9.79 150 600
3 11.188 50 200
Step 3: Unit Commitment
Combination Minimum MW from
Combination
Maximum MW from
combination
2 + 1 +3 300 1200
2+1 250 1000
2 100 400
All the three units would be held on until load reached 1000 MW.
Unit 2 and 1 would be held on until the load reached 400 MW, then unit 1 would be dropped.
For demand of 500 MW, Units 2 and 1 would be operated. (5)
7. Describe the direct method and X iteration method for solving economic dispatch problem.
[M/J ‘ 16]
Case 1. If operating limits for power generation are not specified.
Step 1: Assign initial trail value of 𝜆.
Step 2 : compute PGi corresponding to 𝜆.
Step 3 : compute Power generation in each unit.
Step 4 : check power balance equation.
Step 5 : calculate demand in each stage.(8)
Case 2. If operating limits for power generation and production cost are specified.
Step 1:compute initial trail value of 𝜆.
Step 2 : compute PGi using equal incremental cost basis
Step 3 : check the minimum and maximum demand in station.
Step 4 : check PGi limit violations.
Step 5 : redistribute remaining load in the system
Step 6 : compute new value of 𝜆.
Step 7 : check optimality condition.(8)

8. Draw the flow chart for obtaining the optimum dispatch strategy of N-bus system neglecting the
system transmission loss. (8)[N/D ‘15]
9. The fuel cost of two units are given by
F1 = F1 (PG1)= 1.5 + 20 PG1 + 0.1 PG1
2Rs/hr
F2 = F2 (PG2)= 1.9 + 30 PG2 + 0.1 PG2
2Rs/hr
If the total demand on the generator is 200 MW. Find the economic load scheduling of the two
units. (4)(M/J ‘13)
Solution:
λ1 =
𝒅𝑭 𝟏
𝒅𝑷 𝑮𝟏
= 20 + (2 x 0.1)PG1

λ2 =
𝒅𝑭 𝟐
𝒅𝑷 𝑮𝟐
= 30 + (2 x 0.1)PG2
λ =
PD+ ∑
bi
2ai
N
i=1
∑
1
2ai
N
i=1
=
𝟐𝟎𝟎+ [
𝟐𝟎
𝟎.𝟐
+
𝟑𝟎
𝟎.𝟐
]
𝟏
𝟎.𝟐
+
𝟏
𝟎.𝟐
λ =
𝟒𝟓𝟎
𝟏𝟎
= 45
PG1 =
𝝀 − 𝒃 𝟏
𝟐𝒂 𝟏
=
𝟒𝟓 − 𝟐𝟎
𝟐(𝟎.𝟏)
= 125 MW
PG2 =
𝝀 − 𝒃 𝟐
𝟐𝒂 𝟐
=
𝟒𝟓 −𝟑𝟎
𝟐 (𝟎.𝟏)
= 75 MW
10. Create a unit commitment using the priority list method for the following three units.
The fuel cost equations are as follows:
Unit 1: F1(P1) = 561 + 7.92 P1 + 0.001562 P1
2 150≤P1≤600
Unit 2: F2(P2)= 310 + 7.85 P2 + 0.00194 P2
2 100≤P2≤400
Unit 3: F3(P3) = 93.6 + 9.56 P3 + 0.005784P3 50≤P3≤200 (8) ((N/D ‘14)
Sol : -
FLAPC =
𝐹𝑖(𝑃 𝐺𝑖)
𝑃 𝐺𝑖
Step1:
FLAPC1 =
561+(7.92 𝑥 600)+(0.001562 𝑥 6002)
600
FLAPC1 = 9.79
FLAPC2 =
310+(7.85 𝑥 400)+(0.00194 𝑥 4002)
400
FLAPC2 = 9.40
FLAPC3 =
93.6+(95.6 𝑋 200)+(0.00578 𝑋 2002)
200
FLAPC3 = 11.184
Unit FLAPC Min Max
2 9.40 100 400
1 9.79 150 600
3 11.184 50 200
Combinations Min MW from Combination Max MW from Combination
2 + 1 + 3 300 1200
2 + 1 250 1000
2 100 400
All three units would be held on until load reached 1000 MW.
Units 2 and 1 would be held on until load reached 400 MW, then Unit 2 would be dropped.
11. A heat rate of 100 MW fuel fired generator is
10 Mkcal / Mwhr at 25% rating;
9 Mkcal / Mwhr at 40% rating;
8 Mkcal / MWhr at 100 % rating;
Cost of the fuel is Rs.2 /Mkcal.
Find (i) Fuel cost at 25, 40, 100% loading;
(ii) Increment cost;
(iii) Fuel input rate, using quadratic approximation to deliver 101MW.

Solution:
Energy input Fi(PGi) = PGi. HI (PGi) = Ci’ + bi’ PGi + ai’ PGi
2
Input fuel cost Ci(PGi) = K Fi (PGi) = Ci + bi PGi + ai PGi
2
Heat Rate Hi (PGi) =
𝐶𝑖
′
𝑃 𝐺𝑖
+ 𝑏𝑖
′
+ 𝑎𝑖
′
PGi
H1(PG1) = H1(25) =
𝐶𝑖
′
25
+ 𝑏𝑖
′
+ 𝑎𝑖
′
x 25 = 10 ………………………………………(1)
H2(PG2) = H2(40) =
𝐶𝑖
′
40
+ 𝑏𝑖
′
+ 𝑎𝑖
′
x 40 = 9 ………………………………………...(2)
H3(PG3) = H3(100) =
𝐶𝑖
′
100
+ 𝑏𝑖
′
+ 𝑎𝑖
′
x 100 = 8 ……………………………………..(3)
Solving equations 1, 2, 3 find the values of 𝑎1
′
, 𝑏1
′
, 𝑐1
′
[
0.04 1 25
0.025 1 40
0.01 1 100
] [
𝑐𝑖
′
𝑏𝑖
′
𝑎𝑖
′
] = [
10
9
8
]
∆ = 0.675
∆1 = |
10 1 25
9 1 40
8 1 100
| = 45
𝑐𝑖
′
=
∆1
∆
=
45
0.675
= 66.66
∆2 = |
0.04 10 25
0.025 9 40
0.01 8 100
| = 495
𝑏𝑖
′
=
∆2
∆
= 7.33
∆3 = |
0.04 1 10
0.025 1 9
0.01 1 8
| = 0
𝑎𝑖
′
= 0
Fi (PGi) = Hi (PGi) x PGi = Ci’ + bi’ PGi + ai’ PGi
2
= 66.66 +7.33 PGi(8)
a) At 25% rating
F (25) = 66.66 +7.33 x 25 = 250 MKcal/hr
C (25) = K x F (25) = 2 x 250 = 500 Rs/hr
b) At 40% rating
F (40) = 66.66 + 7.33 x 40 = 360 MKcal/hr
C (40) = 2 x F (40) = 720 Rs/hr
c) At 100% rating
F (100) = 66.66 + 7.33 x100 = 800 MKcal/hr (4)
C (100) = 2 x F (100) = 1600 Rs/hr
(ii) Incremental cost:
IC =
dC(PGi)
dPGi
=
K x dF(PGi)
dPGi
= 2 x
dF(PGi)
dPGi
= 2 x 7.33 = 14.66 Rs/MWhr
(iii) Cost using quadratic approximation:
C(PGi) = 2 F(PGi) = [2 x 66.66]+ [2x 7.33 PGi]
C(101) = [2 x 66.66] + [2x 7.33 x 101] = 1614 Rs/hr (4)

UNIT – 5 – COMPUTER CONTROL OF POWER SYSTEMS
1. Explain the Hardware configuration and function of SCADA. [N/D’18’ 15]
SCADA: (8)
There are two parts to the term SCADA Supervisory control indicates that the operator, residing in the
energy control center (ECC), has the ability to control remote equipment. Data acquisition indicates that
information is gathered characterizing the state of the remote equipment and sent to the ECC for monitoring
purposes.
The monitoring equipment is normally located in the substations and is consolidated in what is known
as the remote terminal unit (RTU). Generally, the RTUs are equipped with microprocessors having memory
and logic capability. Older RTUs are equipped with modems to provide the communication link back to the
ECC, whereas newer RTUs generally have intranet or internet capability.
Relays located within the RTU, on command from the ECC, open or close selected control circuits to
perform a supervisory action. Such actions may include, for example, opening or closing of a circuit breaker
or switch, modifying a transformer tap setting, raising lowering generator MW output or terminal voltage,
switching in or out a shunt capacitor or inductor, and the starting or stopping of a synchronous condenser.
Information gathered by the RTU and communicated to the ECC includes both analog information and
status indicators. Analog information includes, for example, frequency, voltages, currents, and real and
reactive power flows. Status indicators include alarm signals (over-temperature, low relay battery voltage,
illegal entry) and whether switches and circuit breakers are open or closed. Such information is provided to the
ECC through a periodic scan of all RTUs. A 2 second scan cycle is typical.
SCADA Functional Block
Functions of SCADA Systems (8)
1. Data acquisition
2. Information display.
3. Supervisory Control(CBs:ON/OFF, Generator: stop/start, RAISE/LOWER command)
4. Information storage and result display.
5. Sequence of events acquisition.
6. Remote terminal unit processing.
7. General maintenance.
8. Runtime status verification.
9. Economic modeling.
10. Remote start/stop.
11. Load matching based on economics.
12. Load shedding.

Control functions
The following are some of the main functions of SCADA.
i) Data Acquisition: To provide data, measurements and status information to the operator.
ii) Automatic generation control: To control the generations at the power plants.
iii) Load Shedding: To make automatic load shedding in emergent conditions to avoid system collapsing.
iv) Load Restoration: To restore the loads in steps to bring the system to normal state.
v) Supervisory Control: To operate the circuit breaker remotely.
vi) Logging: To log all data and information in a systematic manner.
vii) Alarms: To send alarm signals in case of undesirable operating conditions
To fulfill the above operations SCADA has the following components.
Sensors and Control Relays: Analog and digital sensors along with control relays which can interface with the
system.
2. Briefly explain the various functions of SCADA with a neat diagram. [ND’10,’11,’12, MJ’12]
Remote Terminal Units (RTU): (2)
RTUs are microprocessor controlled electronic devices deployed in field at specific sites and locations.
They collect necessary data and transmit them to SCADA for processing.
Master Unit: (2)
Large computer system which serves as a central processor.
Communication Links Fiber optic / satellite / microwave communications are employed to link RTUs with
SCADA.
Necessary Software’s: (5)
To execute different operational problems
The form of communication required for SCADA is telemetry. Telemetry is the measurement of a
quantity in such a way so as to allow interpretation of that measurement at a distance from the primary
detector. The distinctive feature of telemetry is the nature of the translating means, which includes provision
for converting the measure into a representative quantity of another kind that can be transmitted conveniently
for measurement at a distance. The actual distance is irrelevant.
Telemetry may be analog or digital. In analog telemetry, a voltage, current, or frequency proportional
to the quantity being measured is developed and transmitted on a communication channel to the receiving
location, where the received signal is applied to a meter calibrated to indicate the quantity being measured, or
it is applied directly to a control device such as a ECC computer.
Forms of analog telemetry include variable current, pulse-amplitude, pulse-length, and pulse-rate, with
the latter two being the most common. In digital telemetry, the quantity being measured is converted to a code
in which the sequence of pulses transmitted indicates the quantity. One of the advantages to digital
telemetering is the fact that accuracy of data is not lost in transmitting the data from one location to another.
Digital telemetry requires analog to digital (A/D) and possible digital to analog (D/A) converters.
The earliest form of signal circuit used for SCADA telemetry consisted of twisted pair wires; although simple
and economic for short distances, it suffers from reliability problems due to breakage, water ingress, and
ground potential risk during faults. (3)
Block Diagram of Telemetering System
Improvements over twisted pair wires came in the form of what is now the most common, traditional
type of telemetry mediums based on leased-wire, power-line carrier, or microwave. These are voice grade
forms of telemetry, meaning they represent communication channels suitable for the transmission of speech,
either digital or analog, generally with a frequency range of about 300 to 3000 Hz.

SCADA Functional Block (4)
Leased-wire means use of a standard telephone circuit; this is a convenient and straightforward means
of telemetry when it is available, although it can be unreliable, and it requires a continual outlay of leasing
expenditures. In addition, it is not under user control and requires careful coordination between the user and
the telephone company. Power-line carrier (PLC) offers an inexpensive and typically more reliable alternative
to leased-wire. Here, the transmission circuit itself is used to modulate a communication signal at a frequency
much greater than the 60 Hz power frequency.
Most PLC occurs at frequencies in the range of 30-500 kHz. The security of PLC is very high since the
communication equipment is located inside the substations. One disadvantage of PLC is that the
communication cannot be made through open disconnects, i.e., when the transmission line is outraged. Often,
this is precisely the time when the communication signal is needed most. In addition, PLC is susceptible to
line noise and requires careful signal-to-noise ratio analysis. Most PLC is strictly analog although digital PLC
has become available from a few suppliers during the last few years.
Microwave radio refers to ultra-high-frequency (UHF) radio systems operating above 1 GHz. The
earliest microwave telemetry was strictly analog, but digital microwave communication is now quite common
for EMS/SCADA applications. This form of communication has obvious advantages over PLC and leased
wire since it requires no physical conducting medium and therefore no right-of-way. However, line of sight
clearance is required in order to ensure reliable communication, and therefore it is not applicable in some
cases.
A more recent development has concerned the use of fiber optic cable, a technology capable of
extremely fast communication speeds. Although cost was originally prohibitive, it has now decreased to the
point where it is viable. Fiber optics may be either run inside underground power cables or they may be
fastened to overhead transmission line towers just below the lines. They may also be run within the shield wire
suspended above the transmission lines.
3. Describe the security analysis and controls of power system. [M/J’ 16]
The operation of an electric power system is characterized with a number of control functions. Some of them
are automatic and others require operator initiation. Consider, for example, a single unit power plant. One can
recognize a number of control loops:
(1) The Voltage Control Loop. (4)
The objective of this control loop is to regulate the voltage at the terminals of the generator. It consists
of the voltage regulator and exciter system. Inputs to this control loop are the reference voltage Vref, which
may be selected by the system dispatcher or automatically by computers (VAR dispatch), and the actual
voltage at the terminals of the generator Vg.
(2) The Power System Stabilizer (PPS) Loop. (4)

The objective of this control loop is to slow down the oscillations of the generator following a
disturbance. It consists of a feedback system which injects a stabilizing signal into the exciter system.
Feedback quantities may be: frequency, f, real power, Pg, etc.
(3) The Secondary Automatic Generation Control Loop. (4)
The objective of the secondary automatic generation control loop is
to regulate the net interchange, unit real power output, and speed (frequency). It consists of a feedback system
which injects a signal into the speed regulator (governor). The signal, referred to as the Unit Control Error
(UCE), is constructed from measurements of frequency, interchange schedule, unit real power output, etc.
Reference quantities for this control loop are: (a) Scheduled interchange of real power, Psched, (b) Scheduled
frequency, fsched, and (c) Scheduled unit real power output, Pdes. This control loop uses integral feedback of
frequency and therefore regulates the system real time (integral of frequency).
4. Draw the state transition diagram of a power system. Explain the state transition that may occur due
to system disturbance and also different control actions that can be taken to improve the security level
of the system. [ND’18’09’11’12’13, MJ’11’13, 18]
Operating states (6)
1. Normal state
2. Alert state
3. Emergency state
4. Extremis state
5. Restorative state
Normal state: (2)
A system is said to be in normal if both load and operating constraints are satisfied. It is one in which the total
demand on the system is met by satisfying all the operating constraints.
Alert state: (2)
A normal state of the system said to be in alert state if one or more of the postulated contingency states,
consists of the constraint limits violated. When the system security level falls below a certain level or the
probability of disturbance increases, the system may be in alert state. All equalities and inequalities are
satisfied, but on the event of a disturbance, the system may not have all the inequality constraints satisfied. If
severe disturbance occurs, the system will push into emergency state. To bring back the system to secure state,
preventive control action is carried out.
Emergency state: (2)
The system is said to be in emergency state if one or more operating constraints are violated, but the
load constraint is satisfied. In this state, the equality constraints are unchanged. The system will return to the
normal or alert state by means of corrective actions, disconnection
of faulted section or load sharing.
Extremis state: (2)
When the system is in emergency, if no proper corrective action is taken in time, then it goes to either
emergency state or extremis state. In this regard neither the load or nor the operating constraint is satisfied,

this result is islanding. Also the generating units are strained beyond their capacity. So emergency control
action is done to bring back the system state either to the emergency state or normal state.
Restorative state: (2)
From this state, the system may be brought back either to alert state or secure state. The latter is a slow
process. Hence, in certain cases, first the system is brought back to alert state and then to the secure state.
This is done using restorative control action.
5. What is EMS? What are its major functions in power system operation and control?
Functionality Power EMS:
 MW interchange evaluation- with neighboring system
 Maintenance scheduling
 Production cost calculation
 Data acquisition: Provides telemeter measurements and status information to operator.
 Supervisory control: Allows operator to remotely control devices, e.g., open and close circuit
breakers. A “select before operate” procedure is used for greater safety.
 Tagging: Identifies a device as subject to specific operating restrictions and prevents unauthorized
operation.
 Alarms: Inform operator of unplanned events and undesirable operating conditions. Alarms are sorted
by criticality, area of responsibility, and chronology. Acknowledgment may be required
 Logging: Logs all operator entry, all alarms, and selected information.
 Load shed: Provides both automatic and operator-initiated tripping of load in response to system
emergencies.
6. Discuss the various functions of Energy Control Centre. [MJ’ 18’ 08’12, ND’18’12’13, 14]
The various functions of an energy control center can be enumerated as under:
 Load forecasting
 Automatic generation control
 Planning studies

EE6603 - Power System Operation & Control

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