Power plant engineering unit 1 by Varun Pratap Singh

Department of Mechanical Engineering, College of Engineering Roorkee (COER), Roorkee
Power Plant Engineering
Unit-1
NOTES
by
Varun Pratap Singh
Assistant Professor
Mechanical Engineering Department
College of Engineering Roorkee

Disclaimer
This document does not claim any originality and cannot be used as a substitute for prescribed
textbooks. The information presented here is merely a collection by the subject faculty members
for their respective teaching assignments. Various sources as mentioned at the end of the
document as well as freely available material from internet were consulted for preparing this
document. The ownership of the information lies with the respective authors or institutions.
Further, this document is not intended to be used for commercial purpose and the subject faculty
members are not accountable for any issues, legal or otherwise, arising out of use of this
document. The subject faculty members make no representations or warranties with respect to
the accuracy or completeness of the contents of this document and specifically disclaim any
implied warranties of merchantability or fitness for a particular purpose. The subject faculty
members shall be liable for any loss of profit or any other commercial damages, including but
not limited to special, incidental, consequential, or other damages.

SYLLABUS
Unit-I
Part-1 Introduction
Power and energy, sources of energy, review of thermodynamic cycles related to power plants,
fuels and combustion, calculations.
Part-2 Variable Load Problem
Industrial production and power generation compared, ideal and realised load curves, terms
and factors. Effect of variable load on power plan operation, methods of meeting the variable
load problem.
Part-3 Power plant economics and selection
Effect of plant type on costs, rates, fixed elements, energy elements, customer elements and
investor’s profit; depreciation and replacement, theory of rates. Economics of plant selection,
other considerations in plant selection.

Fundamental of Power Plant
Part-1 Introduction
Power and energy, sources of energy, review of thermodynamic cycles related to power
plants, fuels and combustion, calculations.
INTRODUCTION
The whole world is in the grip of energy crisis and the pollution manifesting itself in the
spiralling cost of energy and uncomforted due to increase in pollution as well as the depletion
of conventional energy resources and increasing curve of pollution elements. To meet these
challenges one way is to check growing energy demand but that would show down the
economic growth as first step and to develop non-polluting energy conversion system as second
step. It is commonly accepted that the standard of living increases with increasing energy
consumption per capita. Any consideration of energy requirement and supply has to take into
account the increase conservation measures. On the industrial font, emphasis must be placed
on the increased with constant effort to reduce energy consumption. Fundamental changes in
the process, production and services can affect considerable energy saving without affecting
the overall economy.
CONCEPT OF POWER PLANT
A power plant is assembly of systems or subsystems to generate electricity, i.e., power with
economy and requirements. The power plant itself must be useful economically and
environmental friendly to the society. The present subject is oriented to conventional as well
as non-conventional energy generation. While the stress is on energy efficient system regards
conventional power systems viz., to increase the system conversion efficiency the supreme
goal is to develop, design, and manufacturer the non-conventional power generating systems
in coming decades preferably after 2050 AD which are conducive to society as well as having
feasible energy conversion efficiency and non-friendly to pollution, keeping in view the
pollution act. The subject as a whole can be also stated as modern power plants for power viz
electricity generation in 21st century.

CLASSIFICATION OF POWER PLANTS
A power plant may be defined as a machine or assembly of equipment that generates and
delivers a flow of mechanical or electrical energy. The main equipment for the generation of
electric power is generator. When coupling it to a prime mover runs the generator, the
electricity is generated. The type of prime move determines, the type of power plants. The
major power plants, which are discussed in this study material, are,
1. Steam power plant
2. Diesel power plant
3. Gas turbine power plant
4. Nuclear power plant
5. Hydroelectric power plant
The Steam Power Plant, Diesel Power Plant, Gas Turbine Power Plant and Nuclear Power
Plants are called THERMAL POWER PLANT, because these convert heat into electric energy.
ENERGY
Energy is the capacity for doing work, generating heat, and emitting light. The equation for
work is the force, which is the mass time the gravity times the distance.
Heat is the ability to change the temperature of an object or phase of a substance. For
example, heat changes a solid into a liquid or a liquid into a vapour. Heat is part of the definition
of energy.
Another part of the definition of energy is radiation, which is the light and energy
emitted in the form of waves traveling at the speed of light.
Energy is measured in units of calorie, quad, and joule. A kilocalorie is the amount of
energy or heat required to raise the temperature of 1 kilogram of water from 14.5°C to 15.5°C.

The quad unit is used to measure energy needed for big countries. The final measurement of
energy is joules.
Energy is an essential input for economic development and improving quality of life.
India’s per capita consumption of Commercial Energy (viz., coal, petroleum and electricity) is
only one-eighth of the Global Average and will increase with growth in Gross Domestic
Production (GDP) and improvement in standard of living.
Commercial Energy accounts for a little over half of the total energy used in the
Country, the rest coming from non-commercial resources like cow-dung, fuel wood and
agricultural waste. Though the share of these non-commercial sources has been coming down,
consumption has increased almost double since 1953.
These renewable, non-commercial sources have been used extensively for hundreds of
years but in a primitive and ineffective way. Indiscriminate use of non-commercial energy
sources is leading to an energy crisis in the rural areas. Seventh Plan laid emphasis on the
development and accelerated utilisation of renewable energy sources in rural and urban areas.
A major Policy of the Government is directed towards increasing the use of coal in
household and of electricity in transport sector in order to reduce dependence on oil, which is
becoming scarce gradually.
The Government has formulated an energy policy with objectives of ensuring adequate
energy supply at minimum cost, achieving self-sufficiency in energy supplies and protecting
environment from adverse impact of utilising energy resources in an injudicious manner. Main
elements of the policy are:
1. Accelerated exploitation of domestic conventional energy resources-oil, coal, hydro and
nuclear power;
2. Intensification of exploration to increase indigenous production of oil and gas;
3. Management of demand for oil and other forms of energy;
4. Energy conservation and management;
5. Optimisation of utilisation of existing capacity in the country;
6. Development and exploitation of renewable sources of energy to meet energy requirements
of rural communities;
7. Intensification of research and development activities in new and renewable energy sources;
and
8. Organisation of training far personnel engaged at various levels in the energy sector.
TYPES OF ENERGY
There are various types of energy which, they include nuclear, electrical, thermal, chemical,
and radiant energy. In addition, gravitational potential energy and kinetic energy that combines
to produce mechanical energy.
Nuclear energy produces heat by fission on nuclei, which is generated by heat engines.
Nuclear energy is the world’s largest source of emission-free energy. There are two processes
in Nuclear energy fission and fusion. In fission, the nuclei of uranium or plutonium atoms are
split with the release of energy. In fusion, energy is released when small nuclei combine or
fuse. The fission process is used in all present nuclear power plants, because fusion cannot be
controlled. Nuclear energy is used to heat steam engines. A Nuclear power plant is a steam
engine using uranium as its fuel, and it suffers from low efficiency.
Electricity powers most factories and homes in our world. Some things like flashlights
and Game Boys use electricity that is stored in batteries as chemical energy. Other items use
electricity that comes from an electrical plug in a wall socket. Electricity is the conduction or
transfer of energy from one place to another. The electricity is the flow of energy. Atoms have

electrons circling then, some being loosely attached. When electrons move among the atoms
of matter, a current of electricity is created.
Thermal energy is kinetic and potential energy, but it is associated with the random
motion of atoms in an object. The kinetic and potential energy associated with this random
microscopic motion is called thermal energy. A great amount of thermal energy (heat) is stored
in the world’s oceans. Each day, the oceans absorb enough heat from the sun to equal the energy
contained in 250 billion barrels of oil (Ocean Thermal Energy Conversion Systems).
Chemical energy is a form of energy that comes from chemical reactions, in which the
chemical reaction is a process of oxidation. Potential energy is released when a chemical
reaction occurs, which is called chemical energy. A car battery is a good example, because the
chemical reaction produces voltage and current to start the car. When a plant goes through a
process of photosynthesis, what the plant is left with more chemical energy than the water and
carbon dioxide. Chemical energy is used in science labs to make medicine and to product power
from gas.
Radiant energy exists in a range of wavelengths that extends from radio waves that
many be thousands of meters long to gamma rays with wavelengths as short as a million-
millionth (10– 12) of a meter. Radiant energy is converted to chemical energy by the process
of photosynthesis.
The next two types of energy go hand and hand, gravitational potential energy and kinetic
energy. The term energy is motivated by the fact that potential energy and kinetic energy are
different aspects of the same thing, mechanical energy.
Potential energy exists whenever an object which has mass has a position within a
force field. The potential energy of an object in this case is given by the relation PE = mgh,
where PE is energy in joules, m is the mass of the object, g is the gravitational acceleration,
and h is the height of the object goes.
Kinetic energy is the energy of motion. An object in motion, whether it be vertical or
horizontal motion, has kinetic energy. There are different forms of kinetic energy vibrational,
which is the energy due to vibrational motion, rotational, which is the energy due to rotational
motion, and transnational, which is the energy due to motion from one location to the other.
The equation for kinetic energy is ½mv2, where m is the mass and v is the velocity. This
equation shows that the kinetic energy of an object is directly proportional to the square of its
speed.
POWER
Power is the rate doing work, which equals energy per time. Energy is thus required to produce
power. We need energy to run power plants to generate electricity. We need power to run our
appliances, and heat our homes. Without energy we would not have electricity.
The units of power are watts, joules per second, and horsepower,
where;
1 Watt = 1 joule per second
1 Kilowatt = 1,000 Watts
1 Megawatt = 1,000 kilowatts
= 1 horsepower
Electricity is the most convenient and versatile form of energy. Demand for it, therefore, has
been growing at a rate faster than other forms of energy. Power industry too has recorded a
phenomenal rate of growth both in terms of its volume and technological sophistication over
the last few decades. Electricity plays a crucial role in both industrial and agricultural sectors

and, therefore, consumption of electricity in the country is an indicator of productivity and
growth. In view of this, power development has been given high-priority in development
programme.
POWER DEVELOPMENT IN INDIA (For General Study Only)
The history of power development in India dates back to 1897 when a 200 kW hydro-station
was first commissioned at Darjeeling. The first steam station was set up in Calcutta in 1899.
By the end of 1920, the total capacity was 130mW, comprising. Hydro 74mW, thermal 50mW
and diesel 6mW. In 1940, the total capacity goes to 1208mW. There was very slow
development during 1935-1945 due to Second World War. The total generation capacity was
1710mW by the end of 1951. The development really started only after 1951 with the launching
of the first five-year plan to today.
National Power Grid will promote integrated operation and transfer of power from one system
to another with ultimate objective of ensuring optimum utilisation of resources in the Country.
India now has well integrated Regional Power Systems and exchange of power is taking place
regularly between a large numbers of state systems, which greatly facilitates better utilisation
of existing capacity.
National Power Grid will promote integrated operation and transfer of power from one system
to another with ultimate objective of ensuring optimum utilisation of resources in the Country.
India now has well integrated Regional Power Systems and exchange of power is taking place
regularly between a large numbers of state systems, which greatly facilitates better utilisation
of existing capacity.
RESOURCES FOR POWER GENERATION (For General Study Only)
The hydel power source plays a vital role in the generation of power, as it is a non-
conventional perennial source of energy. Unlike black oil, it is a non-conventional energy
source. A part of the endless cycle in which moisture is raised by the sun, formed into clouds
and then dropped back to earth to feed the rivers whose flow can be harnessed to produce
hydroelectric power. Water as a source of power is non-polluting which is a prime requirement
of power industry today.
The world’s total waterpower potential is estimated as 1500 million kW at mean flow. This
means that the energy generated at a load factor of 50% would be 6.5 million kW-hr, a quantity
equivalent to 3750 million tonnes of coal at 20% efficiency. The world hydel installed capacity
(as per 1963 estimate is only 65 million kW or 4.3% of the mean flow.
The next important source for power generation is fuel in the form of coal, oil or gas.
Unfortunately, the oil and gas resources are very much limited in India. Only few power plants
use oil or gas as a source of energy. India has to import most of the oil required and so it is not
desirable to use it for power generation. The known resources of coal in India are estimated to
be 121,000 million tonnes, which are localized in West Bengal, Bihar, Madhya Pradesh and
Andhra Pradesh. The present rate of annual production of coal is nearly 140 million tonnes of
which 40 million-tonnes are used for power generation. The coal used for power generation is
mainly low-grade coal with high ash content (20-40%).
The nuclear fuel which is commonly used for nuclear power plants is uranium. Deposits of
uranium has been located in Bihar and Rajasthan. It is estimated that the present reserves of
uranium available in country may be sufficient to sustain 10,000 mW power plants for its

thorium into nuclear Indian lifetime. Another possible nuclear power source is thorium, which
is abundant in this country, estimated at 500,000 tonnes. But the commercial use of this nuclear
fuel is tied up with development of fast breeder reactor which converts energy economy must
wait for the development of economic methods for using thorium which is expected to be
available before the end of twentieth century. The major hurdle in the development of nuclear
power in this country is lack of technical facility and foreign exchange required to purchase
the main component of nuclear power plant. Dr. Bhabha had envisaged 8000 mW of power
from nuclear reactors by 1980–81 which was subsequently scaled down to a more realistic
level of 2700 mW by Dr. Sarabhai out of this only 1040 MW has materialized which is less
than 1.5% of the country’s installed power capacity. Moreover, the performance of nuclear
plants has been satisfactory compared to thermal plants.
PRESENT POWER POSITION IN INDIA (For General Study Only)
The present power position in India is alarming as there are major power shortages in almost
all states of the country leading to crippling of industries and hundreds of thousands of people
losing jobs and a heavy loss of production.
The overall power scene in the country shows heavy shortages almost in all states. The
situation is going to be aggravated in coming years as the demand is increasing and the power
industry is not keeping pace with the increasing demand.
Many of the states in India depend to a large extent on hydro generation. The increase
in demand has far outstripped the installation of new plants. Also there is no central grid to
distribute excess energy from one region to another. The experience in the operation of thermal
plants is inadequate. All these have led to heavy shortages and severe hardship to people.
Very careful analysis of the problem and proper planning and execution is necessary to
solve the power crisis in our country.
Suitable hydrothermal mix, proper phasing of construction of new plants, training
personnel in maintenance of thermal plants.
POWER CORPORATIONS IN INDIA (For General Study Only)
1 NATIONAL THERMAL POWER CORPORATION
National Thermal Power Corporation (NTPC) was incorporated in November, 1975, as a public
sector undertaking with main objective of planning, promoting and organising integrated
development of Thermal Power in the Country.
2 NATIONAL HYDRO-ELECTRIC POWER CORPORATION
The National Hydroelectric Power Corporation (NHPC) was incorporated in November 1975,
with objectives to plan, promote and organize an integrated development of hydroelectric
Power in the Central Sector. NHPC is presently engaged in construction of Dulhasti, Uri and
Salal (Stage-II) Hydroelectric Projects (all in Jammu and Kashmir), Chamera Stage-1
(Himachal Pradesh), Tanakpur Project (UP) and Rangit Project (Sikkim). NHPC is also
responsible for operation and maintenance of Salal Project Stage-I (J & K), Baira Siul Project
(Himachal Pradesh) and Loktak Project (Manipur).
3. RURAL ELECTRIFICATION CORPORATION
The Rural Electrification Corporation (REC) was set up in July, 1969, with the primary
objective of promoting rural electrification by financing rural Electrification Schemes and
Rural Electric Cooperatives in the states.

4 DAMODAR VALLEY CORPORATION
Damodar Valley Corporation (DVC) was established in 1948 under an Act of Parliament for
unified development of Damodar Valley covering an area of 24,235 sq km in Bihar and West
Bengal.
Functions assigned to the corporation are: control of floods, irrigation, generation and
transmission of power besides activities like navigation, soil conservation and afforestation,
promotion of public health and agricultural, industrial and economic development of the valley.
The corporation has Three Thermal Power Stations at Bokaro, Chandrapura and Durgapur with
a total installed capacity (derated) of 1755 mw. It has four multi-purpose Dams at Tilaiya,
Maithon, Panchet and Konar. There are three Hydel Stations appended to Tilaiya, Maithon and
Panchet Dams with a capacity of 144 mW. DVC has also set up three Gas Turbine Units of 30
mW each at Maithon. The corporation is installing one more thermal units of 210 mW at
Bokaro ‘B’, Three Thermal Units of 210 mW each at Mejia and the fourth units of 210 Mw
each at the right bank of Maithon.
5. NORTH-EASTERN ELECTRIC POWER CORPORATION LIMITED
The North-Eastern Electric Power Corporation Ltd., was constituted in 1976 under the
Companies Act under the aim of developing the large electric power potential of the North-
Eastern Region. The corporation is responsible for operation and maintenance of the 150 mW
Kopili Hydro Electric Project which was commissioned in, June/July, 1988. The associated
220 kV and 132 kV transmission lines for supply power from this project to the constituent
states of the region, namely; Assam, Manipur, Mizoram and Tripura, have also been completed.
6 BHAKRA BEAS MANAGEMENT BOARD AND BEAS CONSTRUCTION BOARD
Under the Punjab Reorganization Act. 1966, Bhakra Management Board thereto managed
management of Bhakra Darn and reservoirs and works appurtenant. The construction of Beas
Project was undertaken by the Beas Construction Board. After completion of works of Beas
Project, management of the project was taken over by Bhakra Management Board redesignated
as Bhakra Beas Management Board (BBMB). BBMB now manages Hydro-electric Power
Stations of Bhakra-Beas Systems, namely. Bhakra Right Bank (660 mW), Bhakra Left Bank
(540 mW), Ganguwal (77mW), Kotla (77mW), Dehar Stage-I (660 mW), Debar Stage-II (330
mW). Pong Stage-I (240mW) and Pong Stage-I1 (120mW), all having a total installed capacity
of 2,704mW.
7 POWER ENGINEERS TRAINING SOCIETY (PETS)
The Power Engineers Training Society (PETS) was formed in 1980 as a autonomous body to
function as an Apex National Body for meeting the training requirements of Power Sector in
the Country. The society is responsible for coordinating training programmes of the various
State Electricity Boards, Power stations, etc. and supplementing these with its own training
activities. The society has Four Regional “Thermal Power Station Personnel Training Institutes
at Neyveli, Durgapur, Badarpur (New Delhi) and Nagpur. These Training Institutes conduct
regular induction courses, in-service refresher and short-term courses, on job and on plant
training programmes for Power Engineers, operators and technicians of Thermal Power
Stations/State Electricity Boards, etc. A Simulator installed at the Training Institute at Badarpur
(New Delhi) provides training to engineers and operators of 210mW Thermal Units.

8 CENTRAL POWER RESEARCH INSTITUTE (CPRI), BANGALORE
The Central Power Research Institute, which was set up in 1960 as a subordinate office under
the erstwhile Central Water and Power Commission (Power Wing), was reorganised and
registered as a Society under the Karnataka Societies Act, 1960, with effect from January,
1978. The CPRI functions as a National Laboratory for applied research in the field of Electric
Power Engineering. While the Central Power Research Institute has a Switchgear Testing and
Development Station at Bhopal, the main complex of its laboratories is at Bangalore. The
institute is an Apex Body for Research and Development in the Power Sector and conducts
tests of electrical apparatus in accordance with the National/International Standards so as to
meet fully the research and testing needs of electrical, transmission and distribution equipment.
The CPRI also serves as a National Testing and Certification Authority for transmission and
distribution equipment. The institute possesses Highly Sophisticated Laboratories comparable
to those in the Developed Countries.
9 NATHPA, JHAICRI POWER CORPORATION LIMITED
NJPC, a joint venture of the Centre and Government of Himachal Pradesh, was incorporated
on May 24, 1988, for execution of Nathpa Jhakri Power Project (6 × 250 mW) with equity
participation in the ratio of 3 : 1. The corporation has an authorized Share Capital of Rs 1,000
crore. It will also execute other Hydro-electric Power Projects in the region with consent of the
state government.

REVIEW OF THERMODYNAMICS CYCLES RELATED TO
POWER PLANTS
Thermodynamics is the science of many processes involved in one form of energy
being changed into another. It is a set of book keeping principles that enable us to understand
and follow energy as it transformed from one form or state to the other.
The zeroth law of thermodynamics was enunciated after the first law. It states that if
two bodies are each in thermal equilibrium with a third, they must also be in thermal
equilibrium with each other. Equilibrium implies the existence of a situation in which the
system undergoes no net charge, and there is no net transfer of heat between the bodies.
The first law of thermodynamics says that energy can’t be destroyed or created. When
one energy form is converted into another, the total amount of energy remains constant. An
example of this law is a gasoline engine. The chemical energy in the fuel is converted into
various forms including kinetic energy of motion, potential energy, chemical energy in the
carbon dioxide, and water of the exhaust gas.
The second law of thermodynamics is the entropy law, which says that all physical
processes proceed in such a way that the availability of the energy involved decreases. This
means that no transformation of energy resource can ever be 100% efficient. The second law
declares that the material economy necessarily and unavoidably degrades the resources that
sustain it.
Entropy is a measure of disorder or chaos, when entropy increases disorder increases.
The third law of thermodynamics is the law of unattainability of absolute zero temperature,
which says that entropy of an ideal crystal at zero degrees Kelvin is zero. It’s unattainable
because it is the lowest temperature that can possibly exist and can only be approached but not
actually reached. This law is not needed for most thermodynamic work, but is a reminder that
like the efficiency of an ideal engine, there are absolute limits in physics.
The steam power plants work on modified Rankine Cycle in the case of steam
engines and isentropic cycle concerned in the case of impulse and reaction steam turbines. In
the case of I.C. Engines (Diesel Power Plant) it works on Otto Cycle, Diesel Cycle or Dual
Cycle and in the case of gas turbine it works on
Brayton Cycle, in the case of nuclear power plants it works on Einstein equation, as
well as on the basic principle of fission or fusion. However, in the case of non-conventional
energy generation it is complicated and depends upon the type of the system viz., thermos
electric or thermionic basic principles and theories et al.
CLASSIFICATION OF POWER PLANT CYCLE
Power plants cycle generally divided in to the following groups,
(1) Vapour Power Cycle
(Carnot cycle, Rankine cycle, Regenerative cycle, Reheat cycle, Binary vapour cycle)
(2) Gas Power Cycles
(Otto cycle, Diesel cycle, Dual combustion cycle, Gas turbine cycle.)

CARNOT CYCLE
This cycle is of great value to heat power theory although it has not been possible to construct
a practical plant on this cycle. It has high thermodynamics efficiency.
It is a standard of comparison for all other cycles. The thermal efficiency (η) of Carnot cycle
is as follows:
η = (T1 – T2)/T1
where, T1 = Temperature of heat source
T2 = Temperature of receiver
RANKINE CYCLE
Steam engine and steam turbines in which steam is used as working medium follow Rankine
cycle. This cycle can be carried out in four pieces of equipment joint by pipes for conveying working
medium as shown in below mentioned figure. The cycle is represented on Pressure Volume P-V and S-
T diagram as shown in below mentioned figures respectively.
Fig: Schematic Diagram of Rankine Cycle Fig: P-V Diagram
Fig: T-S Diagram
Efficiency of Rankine cycle = (H1 – H2)/ (H1 – Hw2)
where,
Hl = Total heat of steam at entry pressure
H2 = Total heat of steam at condenser pressure (exhaust pressure)
Hw2= Total heat of water at exhaust pressure

REHEAT CYCLE
In this cycle steam is extracted from a suitable point in the turbine and reheated generally to the original
temperature by flue gases. Reheating is generally used when the pressure is high say above 100 kg/cm2.
The various advantages of reheating are as follows:
(i) It increases dryness fraction of steam at exhaust so that blade erosion due to impact of water
particles is reduced.
(ii) It increases thermal efficiency.
(iii) It increases the work done per kg of steam and this results in reduced size of boiler.
The disadvantages of reheating are as follows:
(i) Cost of plant is increased due to the reheated and its long connections.
(ii) It increases condenser capacity due to increased dryness fraction.
Above mentioned figures shows flow diagram of reheat cycle. First turbine is high-pressure turbine
and second turbine is low pressure (L.P.) turbine. This cycle is shown on T-S (Temperature entropy)
diagram.
If,
H1 = Total heat of steam at 1
Hw4 = Total heat of water at 4
Efficiency = {(H1 – H2) + (H3 – H4)}/ {H1 + (H3 – H2) – Hw4}

REGENERATIVE CYCLE (FEED WATER HEATING)
The process of extracting steam from the turbine at certain points during its expansion and using
this steam for heating for feed water is known as Regeneration or Bleeding of steam. The arrangement
of bleeding the steam at two stages is shown in below mentioned figure.
Let,
m2 = Weight of bled steam at a per kg of feed water heated
m2 = Weight of bled steam at a per kg of feed water heated
H1 = Enthalpies of steam and water in boiler
Hw1 = Enthalpies of steam and water in boiler
H2, H3 = Enthalpies of steam at points a and b
t2, t3 = Temperatures of steam at points a and b
H4, Hw4 = Enthalpy of steam and water exhausted to hot well.
Work done in turbine per kg of feed water between entrance = H1 – H2
Work done between a and b = (1 – m2)(H2 – H3)
Work done between b and exhaust = (1 – m2 – m3)(H3 – H4)
Total heat supplied per kg of feed water = H1 – Hw2
Efficiency (η) = Total work done/Total heat supplied
= {(H1 – H2) + (1 – m2)(H2 – H3) + (1 – m2 – m3)(H3 – H4)}/(H1 – Hw2)
BINARY VAPOUR CYCLE
In this cycle two working fluids are used. Below mentioned figure shows Elements of Binary vapour
power plant. The mercury boiler heats the mercury into mercury vapours in a dry and saturated state.
These mercury vapours expand in the mercury turbine and then flow through heat exchanger where
they transfer the heat to the feed water, convert it into steam. The steam is passed through the steam
super heater where the steam is super-heated by the hot flue gases.
The steam then expands in the steam turbine.

REHEAT-REGENERATIVE CYCLE
In steam power plants using high steam pressure reheat regenerative cycle is used. The thermal
efficiency of this cycle is higher than only reheat or regenerative cycle. Fig. 1.8 shows the flow diagram
of reheat regenerative cycle. This cycle is commonly used to produce high pressure steam (90 kg/cm2)
to increase the cycle efficiency.
FORMULA SUMMARY
1. Rankine efficiency = (H1 – H2)/(H1 – Hw2)
2. Efficiency ratio or Relative efficiency = Indicated or Brake thermal
efficiency/Rankine efficiency
3. Thermal efficiency = 3600/m(H1 – Hw2), m = steam flow/kw hr
4. Carnot efficiency = (T1 – T2)/T1

FUELS AND COMBUSTION
The working substance of the energy conversion device viz., prime-mover (which convert the
natural resources of energy into power or electricity) is called fuel. The most common fuel is
fossil fuel viz., Coal, petrol, diesel or water gas in the case of steam power plants, I.C. Engines,
gas turbines, and hydro-electric power plants. Uranium 235(1U235) as fissionable and 1U238
as fertile fuel in the case of fission reactors of nuclear power plant and hydrogen as fuel in the
case of fusion nuclear reactor.
Combustion of the fuel is a must in any energy conversion device. It is defined as rapidly
proceeding chemical reaction with liberation of heat and light. This phenomenon incurved in
the case of thermal power plants especially in I.C. engines and gas turbines. But in the case of
fuel cell it is of the nature of chemical reaction i.e., transfer of ions, similarly in the case of
thermo-electric generator it is conduction of electron and holes, in the case of MHD power
plant it is drifting of positive and negative ion etc.
Fossil fuels originate from the earth as a result of the slow decomposition and chemical
conversion of organic material. They come in three basic forms: solid (coal). liquid (oil), and
natural gas. Coal represents the largest fossil-fuel energy resource in the world. In the United
States today (1983), it is responsible for about 50 percent of electric-power generation. Oil and
natural gas are responsible for another 30 percent. The remaining percentage is mostly due to
nuclear and hydraulic generation. Natural gas, however, is being phased out of the picture in
the United States because it must be conserved for essential industrial and domestic uses.
New combustible-fuel options include the so-called synthetic fuels, or synfuels, which are
liquids and gases derived largely from coal, oil shale, and tar sands. A tiny fraction of fuels
used today are industrial by-products, industrial and domestic wastes. and biomass.
This study material will cover the combustible fuels available to the utility industry, both
natural (fossil) and synthetic, and their preparation and firing systems. Nuclear fuels and d
renewable energy sources, and the environmental aspects of power generation in general will
be covered later in this text.
Fuel is a combustible substance, containing carbon as a main constituent, which on proper
burning gives large amount of heat, which can be used economically for domestic and
industrial purpose.
Example :
Wood, charcoal, coal, kerosene, petrol, diesel, producer gas, oil gas etc.
During the process of combustion, carbon, hydrogen, etc., combine with oxygen with a
liberation of heat.
The combustion reaction can be explained as
C + O2 CO2 + 94 kcals
2H2 + O2 2H2O + 68.5 kcals
The calorific value of a fuel depends mainly on the amount of Carbon and Hydrogen.

Requirements of a Good Fuel
A good fuel should have the following characteristics:
o High calorific value.
o Moderate ignition temperature.
o Low contents of non-combustible matters.
o Low moisture content.
o Free from objectionable and harmful gases like CO, SOx, H2S.
o Moderate velocity of combustion.
o Combustion should be controllable.
o Easy to transport and readily available at low cost.
Classification of Fuels
Fuels are classified into
(i) Primary or Natural fuels - These are found in nature.
(ii) Secondary or Artificial fuels - These are derived from primary fuels.
Primary and secondary fuels may also be divided into 3 classes namely solid, liquid and
gaseous fuels.
Fossil fuels
Fossil fuels are those, which have been derived from fossil remains of plant and animal life.
They are found in the earth’s crust.
All conventional fossil fuels whether solid, liquid or gaseous (coal, petroleum or Natural gas) contain
basically carbon and / or hydrogen. The fuels on combustion in presence of oxygen in the air release
heat energy.
This heat energy can be utilized for domestic and industrial purposes.
Advantages of Solid fuels
1. Handling and transportation of solid fuels are easy.
2. Solid fuels are cheap and easily available.
3. They have a moderate ignition temperature
4. This type of fuel can be stored conveniently without any risk.
Disadvantage of solid fuels:
1. During burning, solid fuels produce a large amount of ash and disposal of ash is a big problem.
2. The calorific value of solid fuel is comparatively low.
3. Since a lot of air is required for complete combustion, its thermal efficiency is not so high.
4. A large space is required for storage.
5. Combustion is a slow process and it cannot be easily controlled.

Advantages of Liquid fuels
1. Liquid fuels do not yield any ash after burning.
2. They require comparatively less storage space.
3. Calorific value of liquid fuel is higher than that of solid fuels.
4. Their combustion is uniform and easily controllable.
Disadvantages of liquid fuels:
1. Liquid fuels are comparatively costlier than the solid fuels.
2. They give unpleasant odour during incomplete combustion.
3. Some amount of liquid fuels will escape due to evaporation during storage.
4. Special type of burners are for effective combustion.
Advantages of gaseous fuels:
1. Gaseous fuels can be easily transported through the pipes.
2. They do not produce any ash or smoke during burning.
3. They have high calorific value than the solid fuels.
4. They have high thermal efficiency.
Disadvantages of gaseous fuels
1. They are highly inflammable and hence the chances for fire hazards are high.
2. Since gases occupy a large volume, they require large storage tanks.
Solid Fuels
1.Coal and its varieties (or) Ranking of Coal
Coal is an important primary solid fuel that has been formed as a result of alteration of
vegetable matter under some favourable conditions.
The process of conversion of lignite to anthracite is called coalification (or) metamorphism
of coal.
Coal is classified on the basis of its rank. The rank of coal denotes its degree of maturity.
Vegetable matter, under the action of pressure, heat and anaerobic conditions, gets converted
into different stages of coal namely,
Wood → Peat → lignite → sub-bituminous → bituminous coal → anthracite coal
With the progress of coal forming reaction, moisture content and oxygen content reduces and
% of carbon increases. Also calorific value increases from peat to bituminous. (Below
Mentioned Table)

Classification of coal
Coal is a general term that encompasses a large number of solid organic minerals with widely
differing compositions and properties, although all are essentially rich in amorous (without
regular structure) elemental carbon. It is found in stratified deposits at different and often great
depths, although sometimes near the surface.
It is estimated that in the United States there are 270,000 million tons of recoverable reserves
(those that can be mined economically within the foreseeable future) in 36 of the 50 states. this
accounts for about 30 percent of the world’s total.
There are many ways of classifying coal according to its chemical and physical properties
The most accepted system is the one used by the American Society for Testing and Materials
(ASTM), which classifies coals by grade or rank according to the degree of metamorphism
(change in form and structure under the influences of pressure, and water), ranging from the
lowest state, lignite, to the highest, Anthrasite (ASTM D 388). These classifications are briefly
described below in a deriding order.
a) Peat
Peat. Peat is not an ASTM rank of coal. It is, however, considered the first geological step in
coal’s formation. Peat is a heterogeneous material consisting of decomposed plant matter and
inorganic minerals. It contains up to 90 percent moisture. Although not attractive as a utility
fuel, it is abundant in many parts of the world. Several states in the United States have large
deposits. Because of its abundance, it is used in a few countries (Ireland, Finland, the USSR)
in some electric generating plants and in district heating.
1. Peat is the first stage in the formation of coal.
2. Its calorific value is about 4000-5400 k cal/kg.
3. It is an uneconomical fuel due to its high proportion of (80 -90%) moisture and lower
calorific value.
4. It is a brown fibrous mass.

b) Lignite
Lignite, The lowest grade of coal, lignite derives its name from the Latin lignum, which means
‘‘wood.’’ It is brown and laminar in structure, and remnants of wood fibre are often visible in
it. It originates mostly from resin-rich plants and is therefore high in both inherent moisture, as
high as 30 percent, and volatile matter. Its heating value ranges between less than 6300 to 8300
Btu/lb. (about 14,650 to 19,300 kJ/kg). Because of the high moisture content and low heating
value, lignite it is not economical to transport over long distances and it is usually burned by
utilities at the mine site. The lignite rank is subdivided into two groups: A and B.
1. Lignite is an intermediate stage in the process of coal formation.
2. Its calorific value is about 6500-7100 k cal/kg
3. Due to the presence of high volatile content, it burns with long smoky flame.
c) Bituminous coal
Bituminous coal is further sub-classified on the basis of its carbon content into three types as:
i) Sub- bituminous coal, ii) Bituminous coal and iii) semi-bituminous coal.
Bituminous coal. The largest group, bituminous coal is a broad class of coals containing 46 to
86 mass percent of fixed carbon and 20 to 40 percent of volatile matter of more complex content
than that found in anthracite. It derives its name from bitumen, an asphaltic residue obtained in
the distillation of some fuels. Bituminous coals range in heating value from 11,000 to more
than 14,000 Btu/lbm (about 25,600 to 32,600 kJ/kg). Bituminous coals usually burn easily,
especially in pulverized form.
The bituminous rank is subdivided into five groups: low-volatile, medium-volatile, and high
volatile A, B, and C. The lower the volatility, the higher the heating value. The low volatility
group is greyish black and granular in structure, while the high volatility groups are
homogeneous or laminar.
Subbituminous coal. This is a class of coal with generally lower heating values than
bituminous coal, between 8300 to 11,500 Btu/lbm (about 19,300 to 26,750 kJ/kg). It is
relatively high in inherent moisture content, as much as 15 to 30 percent, but often low in sulfur
content. It is brownish black or black and mostly homogeneous in structure. Subbituminous
coals are usually burned in pulverized form.
The subbituminous rank is divided into three groups: A, B, and C.
d) Anthracite
Anthracite. This is the highest grade of coal. It contains a high content, 86 to 98 mass percent
of fixed carbon (the carbon content in the elemental state) on a dry, mineral matter-free basis
and a low content of volatile matter, less than 2 to 14 mass percent chiefly methane, CHn.
Anthracite is a shiny black, dense, hard, brittle coal that — borders on graphite at the upper
end of fixed carbon. It is slow burning and has a heating value just below that of the highest
for bituminous coal. Its use in steam generators is largely confined to burning on stokers, and
rarely in pulverized form. In the United States it is mostly found in Pennsylvania.
The anthracite rank of coal is subdivided into three groups. In descending order of
fixed-carbon percent, they are meta-anthracite, greater than 98 percent anthracite, 92 to 98
percent; and semi-anthracite, 86 to 92 percent.
1. Anthracite is the superior grade of coal.
2. Its volatile, moisture and ash contents are very less.
3. Its calorific value is about 8650 k cal/kg.

Here peat is the most immature coal, hence it is lowest in rank whereas anthracite is the
most matured coal, and hence it is highest in rank
HEATING VALUE
The heating value, Btu/lbm or J/kg of fuel, may be determined on as-received, dry, or dry-ash-
free basis. It is the heat transferred when the products of complete American National
Standards
Institute/American Society for Testing and Materials.
combustion of a sample of coal or other fuel are cooled to the initial temperature of air and
fuel.
It is determined in a standard test in a bomb calorimeter given by ASTM Standards D 2015.
There are below mention table gives the proximate and ultimate analyses of some typical U.S.
coals.
Table: Proximate and ultimate analysis of some U.S. coals
Analysis
mass percent
Bituminous
medium velocity
Anthracite Subbituminous Lignite
Proximate
70.0
20.5
3.3
6.2
Fixed carbon
Volatile matter
Moisture
Ash
83.8
5.7
2.5
8.0
45.9
30.5
19.6
4.0
30.8
28.2
34.8
6.2
C
H2
S
O2
N2
H2O
Btu/lbm
83.9
2.9
0.7
0.7
1.3
2.5
13.320
80.7
4.5
1.8
2.4
1.1
3.3
14.310
58.8
3.8
0.3
12.2
1.3
19.6
10.130
42.4
2.8
0.7
12.4
0.7
34.8
7.210

PROXIMATE ANALYSIS
This is the easier of two types of coal analysis and the one which supplies readily meaningful
information for coal’s use in steam generators. The basic method for proximate analysis is
given by ANSI/ASTM Standards D 3172. It determines the mass percentages of fixed carbon,
volatile matter, moisture, and ash. Sulphur is obtained in a separate determination.
Moisture is determined by a standard procedure of drying in an oven. This does not account
for all the water present, which includes combined water and water of hydration. There are
several other terms for moisture in coal. One, inherent moisture, that existing in the natural
state of coal and considered to be part of the deposit, excluding surface water.
The volatile matter is that portion of coal, other than water vapour, which is driven off when
the sample is heated in the absence of oxygen in a standard test (up to 1750°F or 7 min). It
consists of hydrocarbon and other gases that result from distillation and decomposition.
Ash is the inorganic salts contained in coal. It is determined in practice as the non-combustible
residue after the combustion of dried coal in a standard test (at 1380°F). After the analysis of
volatile matter, the crucible with residual coal sample is heated without lid at 700 ± 50 °C for
30 minutes in a muffle furnace.
Fixed carbon is the elemental carbon that exists in coal. In proximate analysis, its
determination is approximated by assuming it to be the difference between the original sample
and the sum of volatile matter, moisture, and ash.
Sulphur is determined separately in a standard test, given by ANSUASTM Standards D 2492.
Being combustible, it contributes to the heating value of the coal. It forms oxides which
combine with water to form acids. These cause corrosion problems in the back end of steam
generators if the gases are cooled below the dew point, as well as environmental problems.

Significance (or) Importance of Proximate Analysis
Moisture
High moisture content in coal is undesirable because it,
i) Reduces Calorific Value of coal
ii) Increases the consumption of coal for heating purpose
iii) Lengthens the time of heating.
Hence, lesser the moisture content, better is the quality of coal.
Volatile Matter
During burning of coal, certain gases like CO, CO2, CH4, H2, N2, O2, hydrocarbons etc. that
come out are called volatile matter of the coal.
Fuels and Combustion 4.9
The coal with higher volatile content,
o Ignites easily (i.e : it has lower ignition temperature)
o Burns with long yellow smoky flame
o Has lower Calorific Value
o Will give more quantity of coal gas when it is heated in absence of air.
Ash
Ash is the combustion product of mineral matters present in the coal. It consists mainly of
SiO2, Al2O3 and Fe2O3 with varying amounts of other oxides such as Na2O, CaO, MgO etc.
Ash containing oxides of Na, Ca and Mg melt early. (Low melting ash). During coke
manufacture, the low melting ash forms a fused lumpy - expanded mass which block the
interspaces of the ‘grate’ and thereby obstructing the supply of air leading to irregular burning
of coal and loss of fuel.
High ash content in coal is undesirable because it (a) increases transporting, handling, storage
costs, (b) is harder and stronger, (c) has lower Calorific Value.
Fixed Carbon
It is the pure carbon present in coal. Higher the fixed carbon content of the coal, higher will
be its Calorific Value.

ULTIMATE ANALYSIS
A more scientific test than proximate analysis, ultimate analysis gives the mass percentages
of the chemical elements that constitute the coal. These include carbon, hydrogen, nitrogen,
oxygen, and sulphur. Ash is determined as a whole, sometimes in a separate analysis.
Ultimate analysis is given by ASTM Standards D 3176.
It means finding out the weight percentage of carbon, hydrogen, nitrogen, oxygen and
sulphur of the pure coal free from moisture and inorganic constituents.
This analysis gives the elementary constituents of coal.
It is useful to the designer of coal burning equipment’s and auxiliaries.
Determination of carbon and hydrogen in coal
A known amount of coal is burnt in presence of oxygen there by converting carbon and
hydrogen of coal into CO2 (C + O2 → CO2) and H2O (H2 + ½ O2 → H2O) respectively.
The products of combustion (CO2 and H2O) are made to pass over weighed tubes of
anhydrous CaCl2 and KOH, which absorb H2O and CO 2 respectively.
The increase in the weight of CaCl2 tube represents the weight of water formed while
increase in the weight of KOH tube represents the weight of CO2 formed. % of carbon and
hydrogen in coal can be calculated as follows.
Let X - the weight of coal sample taken
Y - the increase in the weight of KOH tube
Z - the increase in the weight of CaCl2 tube
a) Carbon
C + O2 → CO2
12 32 44
44 gms of CO2 contain 12 gms of carbon.
1 gm of CO2 contains 12/44 gms of carbon
Y gm of CO2 contains = 12/44 × Y gms of carbon
% of C in coal = 12/44 × (Y /X)× 100
b) Hydrogen
18 gms of water contains 2 gms of hydrogen.
1 gm of water contains 2 / 18 gms of hydrogen.
∴ Z gms of water contains =2/18× Z gms of Hydrogen.
% of hydrogen in coal = 2/18 × Z/X × 100
C) Determination of Nitrogen in coal
Nitrogen estimation is done by Kjeldahl’s method. A known amount of powdered coal is
heated with con. H2SO4 and K2SO4 in a long necked flask (called Kjeldahl’s flask), there by
converting nitrogen of coal to ammonium sulphate.
When the clear solution is obtained (i.e. the whole nitrogen is converted into ammonium
sulphate), it is heated with 50% NaOH solution.
(NH4)2 SO4 + 2NaOH Na2SO4 + 2NH3
The ammonia thus formed is distilled over and is absorbed in a known quantity of standard
0.1N HCl solution. The volume of unused 0.1 N HCl is then determined by titrating against

standard NaOH solution.
Thus, the amount of acid neutralized by liberated ammonia from coal is determined.
Let,
Volume of 0.1N HCl = A ml
Volume of unused HCl = B ml
Acid neutralised by ammonia = (A - B ) ml
We know that 1000 ml of 1 N HCl = 1 mole of HCl
= 1 mole of NH3
= 14 gms of N2
(A - B) ml of 0.1N HCl =[14*( A-B)*0.1]/1000*1N gms N2
d) Determination of Sulphur in coal
A known amount of coal is burnt completely in bomb calorimeter in presence of oxygen. Ash
thus obtained contains sulphur of coal as sulphate, which is extracted with dil HCl. The acid
extract is then treated with BaCl2 solution to precipitate sulphate as BaSO4.
The precipitate is filtered, washed, dried, and weighed, from which the sulphur in coal can be
computed as follows.
Let,
X = weight of coal sample taken
M = weight of BaSO4 precipitate formed.
S + 2O2 SO4 BaSO4
32 233
233 gms of BaSO4 contains 32 gms of sulphur
1 gm of BaSO4 contains 32 / 233 gms of sulphur
∴ M gms of BaSO4 contains (32 / 233) x M gms of sulphur

Significance (or) Importance of Ultimate Analysis
i) Carbon and Hydrogen
1. Higher the % of carbon and hydrogen, better the quality of coal and higher is its calorific
value.
2. The % of carbon is helpful in the classification of coal.
3. Higher the % of carbon in coal reduces the size of combustion chamber required.
ii) Nitrogen
1. Nitrogen does not have any calorific value, and its presence in coal is undesirable.
2. Good quality coal should have very little nitrogen content.
iii) Sulphur
Though sulphur increases the calorific value, its presence in coal is undesirable because
1. The combustion products of sulphur, i.e, SO2 and SO3 are harmful and have corrosion
effects on equipment’s.
2. The coal containing sulphur is not suitable for the preparation of metallurgical coke as it
affects the properties of the metal.
iv) Oxygen
1. Lower the % of oxygen higher is its calorific value.
2. As the oxygen content increases its moisture holding capacity increases and the calorific
value of the fuel is required.
Theoretical Calculation of Minimum Air required for
Combustion
In order of achieve efficient combustion of fuel, it is essential that the fuel is brought into
intimate contact with sufficient quantity of air to burn all the combustible matter under
appropriate conditions.
The correct conditions are
i). Intimate mixing of air with combustible matter and
ii). Sufficient time to allow the combustion process to be completed.
If these factors are inappropriate, inefficient combustion occurs.
The elements usually present in common fuels which enter into the process of combustion are
mainly C, H, S and O.
Nitrogen, ash and CO2 (if any) present in the fuel are incombustible matters and hence they
do not take any oxygen during combustion.
Air contains 21% oxygen by volume and 23% of oxygen by weight.
Hence from the amount of oxygen required by the fuel, the amount of air can be calculated.
From the combustion reaction equations, we can calculate the quantity of oxygen by weight
or volume and from this, the weight or volume of air required can be calculated.

i) Combustion of Carbon
12 parts by weight of carbon require 32 parts by weight of oxygen for complete combustion.
(or)
1 part by volume of carbon requires 1 part by volume of oxygen for complete combustion.
So, C parts by weight of carbon require = 32C/12 parts by weight of O2
ii) Combustion of Hydrogen
Oxygen when present in the fuel is always in combination with hydrogen. So, the quantity of
hydrogen in combination with oxygen, which is present in the fuel, will not take part in the
combustion reaction. Therefore, the quantity of hydrogen in combination with oxygen is
deduced from the total hydrogen in the fuel.
Now, the quantity of hydrogen available for combustion reaction will be H-O/8 where H is the
total quantity of hydrogen and O is the total quantity of oxygen in the fuel. (In water the
quantity of hydrogen in combination with oxygen is one-eighth of the weight of oxygen).
4 parts by weight of hydrogen require 32 parts by weight of oxygen for complete combustion.
(or)
2 parts by volume of hydrogen require 1 part by volume of oxygen for complete combustion.
∴ H parts by weight of hydrogen requires
iii) Combustion of Sulphur
32 parts by weight of sulphur requires 32 parts by weight ofoxygen for complete combustion.
(or)
1 part by volume of sulphur requires 1 part by volume of oxygen for complete combustion.
Consequently, theoretical amount of oxygen required for thecomplete combustion of 1kg of
solid or liquid fuel.

Amount of O2 required by the fuel will be given by subtracting the amount of O2 already
present in the fuel from the total or theoretical amount of O2 required by the fuel.

VARIABLE LOAD ON POWER STATIONS
Part-2 Variable Load Problem
Industrial production and power generation compared, ideal and realised load curves, terms
and factors. Effect of variable load on power plan operation, methods of meeting the variable
load problem.
Introduction
The function of a power station is to deliver power to a large number of consumers. However,
the power demands of different consumers vary in accordance with their activities. The result
of this variation in demand is that load on a power station is never constant, rather it varies
from time to time. Most of the complexities of modern power plant operation arise from the
inherent variability of the load demanded by the users. Unfortunately, electrical power cannot
be stored and, therefore, the power station must produce power as and when demanded to meet
the requirements of the consumers. On one hand, the power engineer would like that the
alternators in the power station should run at their rated capacity for maximum efficiency and
on the other hand, the demands of the consumers have wide variations. This makes the design
of a power station highly complex. In this chapter, we shall focus our attention on the problems
of variable load on power stations.
Variable Load on Power Station
The load on a power station varies from time to time due to uncertain demands of the
consumers and is known as variable load on the station.
Effects of variable load:
The variable load on a power station introduces many perplexities in its operation. Some of
the important effects of variable load on a power station are :
(i) Need of additional equipment.
(ii) Increase in production cost.

Types of Loads
A device which taps electrical energy from the electric power system is called a load on the
system. The load may be resistive (e.g., electric lamp), inductive (e.g., induction motor),
capacitive or some combination of them. The various types of loads on the power system are:
(i) Domestic load. Domestic load consists of lights, fans, refrigerators, heaters, television, small
motors for pumping water etc. Most of the residential load occurs only for some hours during
the day (i.e., 24 hours) e.g., lighting load occurs during night time and domestic appliance load
occurs for only a few hours. For this reason, the load factor is low (10% to 12%).
(ii) Commercial load. Commercial load consists of lighting for shops, fans and electric
appliances used in restaurants etc. This class of load occurs for more hours during the day as
compared to the domestic load. The commercial load has seasonal variations due to the
extensive use of air conditioners and space heaters.
(iii) Industrial load. Industrial load consists of load demand by industries. The magnitude of
industrial load depends upon the type of industry. Thus small scale industry requires load up
to 25 kW, medium scale industry between 25kW and 100 kW and large-scale industry requires
load above 500 kW. Industrial loads are generally not weather dependent.
(iv) Municipal load. Municipal load consists of street lighting, power required for water supply
and drainage purposes. Street lighting load is practically constant throughout the hours of the
night. For water supply, water is pumped to overhead tanks by pumps driven by electric motors.
Pumping is carried out during the off-peak period, usually occurring during the night. This
helps to improve the load factor of the power system.
(v) Irrigation load. This type of load is the electric power needed for pumps driven by motors
to supply water to fields. Generally, this type of load is supplied for 12 hours during night.
(vi) Traction load. This type of load includes tram cars, trolley buses, railways etc. This class
of load has wide variation. During the morning hour, it reaches peak value because people have
to go to their work place. After morning hours, the load starts decreasing and again rises during
evening since the people start coming to their homes.

LOAD CURVE:
It is the curve between load (MW) versus time. The curve showing the variation of load on the
power station with respect to (w.r.t) time is known as a load curve.
These load variations during the whole day (i.e., 24 hours) are recorded half-hourly or hourly
and are plotted against time on the graph. The curve thus obtained is known as daily load
curve.
The monthly load curve can be obtained from the daily load curves of that month. For this
purpose, average* values of power over a month at different times of the day are calculated
and then plotted on the graph. The monthly load curve is generally used to fix the rates of
energy.
The yearly load curve is obtained by considering the monthly load curves of that particular
year. The yearly load curve is generally used to determine the annual load factor.
Importance. The daily load curves have attained a great importance in generation as they
supply the following information readily :
(i) The daily load curve shows the variations of load on the power station during different
hours of the day.
(ii) The area under the daily load curve gives the number of units generated in the day.
Units generated/day = Area (in kWh) under daily load curve.
(iii) The highest point on the daily load curve represents the maximum demand on the station
on that day.
(iv) The area under the daily load curve divided by the total number of hours gives the
average load on the station in the day.
(v) The ratio of the area under the load curve to the total area of rectangle in which it is
contained gives the load factor.
(vi) The load curve helps in selecting* the size and number of generating units.
(vii) The load curve helps in preparing the operation schedule** of the station.
[* As you know that number and size of the generating units are selected to fit the load curve.
This helps in operating the generating units at or near the point of maximum efficiency.
** It is the sequence and time for which the various generating units (i.e., alternators) in the
plant will be put in operation.]

LOAD DURATION CURVE:
It is the rearrangement of all the load elements of a load curve in a descending order plotted as a
function of time. When the load elements of a load curve are arranged in the order of descending
magnitudes, the curve thus obtained is called a load duration curve.
The following points may be noted about load duration curve:
(i) The load duration curve gives the data in a more presentable form. In other words, it readily
shows the number of hours during which the given load has prevailed.
(ii) The area under the load duration curve is equal to that of the corresponding load curve.
Obviously, area under daily load duration curve (in kWh) will give the units generated on that
day.
(iii) The load duration curve can be extended to include any period of time. By laying out the
abscissa from 0 hour to 8760 hours, the variation and distribution of demand for an entire year
can be summarised in one curve. The curve thus obtained is called the annual load duration
curve.
Important Terms and Factors
The variable load problem has introduced the following terms and factors in power plant
engineering:
(i) Connected load.
It is the sum of continuous ratings of all the equipment’s connected to supply system.
A power station supplies load to thousands of consumers. Each consumer has certain
equipment installed in his premises. The sum of the continuous ratings of all the equipment’s
in the consumer’s premises is the “connected load” of the consumer. For instance, if a consumer
has connections of five 100-watt lamps and a power point of 500 watts, then connected load of
the consumer is 5 × 100 + 500 = 1000 watts. The sum of the connected loads of all the
consumers is the connected load to the power station.
(ii) Maximum demand:
It is the greatest demand of load on the power station during a given period.
The load on the power station varies from time to time. The maximum of all the demands that
have occurred during a given period (say a day) is the maximum demand. Thus referring back
to the load curve of Fig. 3.2, the maximum demand on the power station during the day is 6
MW and it occurs at 6 P.M. Maximum demand is generally less than the connected load
because all the consumers do not switch on their connected load to the system at a time. The
knowledge of maximum demand is very important as it helps in determining the installed
capacity of the station. The station must be capable of meeting the maximum demand.

(iii) Demand factor.
It is the ratio of maximum demand on the power station to its connected load i.e.,
Demand factor = Maximum demand/Connected load
The value of demand factor is usually less than 1. It is expected because maximum demand on
the power station is generally less than the connected load. If the maximum demand on the
power station is 80 MW and the connected load is 100 MW, then demand factor = 80/100 =
0·8. The knowledge of demand factor is vital in determining the capacity of the plant
equipment.
(iv) Average load. The average of loads occurring on the power station in a given period (day
or month or year) is known as average load or average demand.
(v) Load factor. The ratio of average load to the maximum demand during a given period is
known as load factor i.e.,
If the plant is in operation for T hours,
The load factor may be daily load factor, monthly load factor or annual load factor if the time
period considered is a day or month or year. Load factor is always less than 1 because average
load is smaller than the maximum demand. The load factor plays key role in determining the
overall cost per unit generated. Higher the load factor of the power station, lesser will be the
cost per unit generated.
(vi) Diversity factor. The ratio of the sum of individual maximum demands to the maximum
demand on power station is known as diversity factor i.e.,
Diversity factor = Sum of individual max. demands/ Max. demand on power station
A power station supplies load to various types of consumers whose maximum demands
generally do not occur at the same time. Therefore, the maximum demand on the power station
is always less than the sum of individual maximum demands of the consumers. Obviously,
diversity factor will always be greater than 1. The greater the diversity factor, the lesser is the
cost of generation of power.
Greater diversity factor means lesser maximum demand. This in turn means that lesser plant
capacity is required. Thus, the capital investment on the plant is reduced.
(vii) Plant capacity factor. It is the ratio of actual energy produced to the maximum possible
energy that could have been produced during a given period i.e.,

Thus if the considered period is one year,
The plant capacity factor is an indication of the reserve capacity of the plant. A power station
is so designed that it has some reserve capacity for meeting the increased load demand in future.
Therefore, the installed capacity of the plant is always somewhat greater than the maximum
demand on the plant.
Reserve capacity = Plant capacity − Max. demand
It is interesting to note that difference between load factor and plant capacity factor is an
indication of reserve capacity. If the maximum demand on the plant is equal to the plant
capacity, then load factor and plant capacity factor will have the same value. In such a case,
the plant will have no reserve capacity.
(viii) Plant use factor. It is ratio of kWh generated to the product of plant capacity and the
number of hours for which the plant was in operation i.e.
Suppose a plant having installed capacity of 20 MW produces annual output of 7·35 × 106
kWh and remains in operation for 2190 hours in a year. Then,
(viii) Plant Operating Factor
It is the ratio of the duration during which the plant is in actual service, to the total duration of the
period of time considered.
(ix) Utility Factor
It is the ratio of the units of electricity generated per year to the capacity of the plant installed in the
station. It can also be defined as the ratio of maximum demand of a plant to the rated capacity of the
plant. Supposing the rated capacity of a plant is 200 mW. The maximum load on the plant is 100 mW
at load factor of 80 per cent, then the utility will be = (100 × 0.8)/(200) = 40%.
(x) Dump Power This term is used in hydro plants and it shows the power in excess of the load
requirements and it is made available by surplus water.
(xi) Firm Power It is the power, which should always be available even under emergency conditions.
(xii) Prime Power It is power, may be mechanical, hydraulic or thermal that is always available for
conversion into electric power.

(xiii) Cold Reserve It is that reserve generating capacity which is not in operation but can be
made available for service.
(xiv) Hot Reserve It is that reserve generating capacity which is in operation but not in service
(xv) Spinning Reserve It is that reserve generating capacity which is connected to the bus and
is ready to take the load.
(xvi) Plant Use Factor This is a modification of Plant Capacity factor in that only the actual
number of hours that the plant was in operation is used. Thus Annual Plant Use factor is,
= (Annual kWh produced) / [Plant capacity (kW) × number of hours of plant operation]
Units Generated per Annum
It is often required to find the kWh generated per annum from maximum demand and load factor.
The procedure is as follows :
Typical Demand and Diversity Factors
The demand factor and diversity factor depend on the type of load and its magnitude.

Load Curves and Selection of Generating Units
The load on a power station is seldom constant; it varies from time to time. Obviously, a single
generating unit (i.e., alternator) will not be an economical proposition to meet this varying load.
It is because a single unit will have very poor* efficiency during the periods of light loads on
the power station. Therefore, in actual practice, a number of generating units of different sizes
are installed in a power station. The selection of the number and sizes of the units is decided
from the annual load curve of the station.
The number and size of the units are selected in such a way that they correctly fit the
station load curve.
Once this underlying principle is adhered to, it becomes possible to operat the generating units
at or near the point of maximum efficiency.
Illustration. The principle of selection of number and sizes of generating units with the help
of load curve is illustrated in below mentioned Fig. (i &ii ), the annual load curve of the station
is shown. It is clear from the curve that load on the station has wide variations; the minimum
load being somewhat near 50 kW and maximum load reaching the value of 500 kW. It hardly
needs any mention that use of a single unit to meet this varying load will be highly
uneconomical.
As discussed earlier, the total plant capacity is divided into several generating units of different
sizes to fit the load curve. This is illustrated in Fig. 3.11(ii) where the plant capacity is divided
into three units numbered as 1, 2 and 3. The cyan colour outline shows the unit’s capacity being
used.
The three units employed have different capacities and are used according to the demand on
the station. In this case, the operating schedule can be as under:
Time Units in operation
From 12 midnight to 7 A.M. Only unit no.1 is put in operation.
From 7 A.M. to 12.00 noon Unit no. 2 is also started, both units are in operation.
From 12.00 noon to 2 P.M. Unit no. 2 is stopped and only unit 1operates.
From 2 P.M. to 5 P.M. Unit no. 2 is started. Now both units are in operation.
From 5 P.M. to 10.30 P.M. Units 1, 2 and 3 are put in operation.
From 10. 30 P.M. to 12.00 midnight Units 1 and 2 are put in operation.
Thus by selecting the proper number and sizes of units, the generating units can be made to
operate near maximum efficiency. This results in the overall reduction in the cost of production
of electrical energy.

Solved Examples
Example 1. The maximum demand on a power station is 100 MW. If the annual load factor
is 40% , calculate the total energy generated in a year.
Example.2. A generating station has a connected load of 43MW and a maximum demand of
20 MW; the units generated being 61·5 × 106
er annum.
Calculate:
(i) the demand factor and
(ii) load factor.
Example 3. A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours and is
shut down for the rest of each day. It is also shut down for maintenance for 45 days each year. Calculate
its annual load factor.
Example 4. A generating station has a maximum demand of 25MW, a load factor of 60%, a
plant capacity factor of 50% and a plant use factor of 72%.
Find
(i) the reserve capacity of the plant
(ii) the daily energy produced and
(iii) maximum energy that could be produced daily if the plant while running as per schedule, were fully
loaded.

Example 5. A diesel station supplies the following loads to various consumers: Industrial
consumer = 1500 kW; Commercial establishment = 750 kW Domestic power = 100 kW; Domestic light
= 450 kW If the maximum demand on the station is 2500 kW and the number of kWh generated per year
is 45 × 105, determine (i) the diversity factor and (ii) annual load factor.
Example 6. A power station has a maximum demand of 15000 kW. The annual load factor is
50% and plant capacity factor is 40%. Determine the reserve capacity of the plant.

Example 7. A power supply is having the following loads:
Example 8. At the end of a power distribution system, a certain feeder supplies three distribution
transformers, each one supplying a group of customers whose connected loads are as under:
Solution. Below mentioned figure shows a feeder supplying three distribution transformers. Sum
of max. demands of customers on Transformer 1 = connected load × demand factor = 10 × 0·65 = 6·5
kW

Example 9. It has been desired to install a diesel power station to supply power in a suburban
area having the following particulars :
(i) 1000 houses with average connected load of 1·5 kW in each house. The demand factor and diversity
factor being 0·4 and 2·5 respectively.
(ii) 10 factories having overall maximum demand of 90 kW.
(iii) 7 tube wells of 7 kW each and operating together in the morning.
The diversity factor among above three types of consumers is 1·2. What should be the minimum capacity
of power station?
Example 10. A generating station has the following daily load cycle:
Time (Hours) 0 —6 6 —10 10 —12 12 —16 16 —20 20 —24
Load (M W) 40 50 60 50 70 40
Draw the load curve and find (i) maximum demand (ii) units generated per day (iii) average load and
(iv) load factor.

Example 11. 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 (i) diversity factor (ii) units generated per day (iii) load
factor.
From this table, it is clear that total load on power station is 100 kW for 0—6 hours, 150 kW
for 6—8 hours, 350 kW for 8—10 hours, 300 kW for 10— 18 hours and 100 kW for 18—24
hours. Plotting the load on power station versus time, we get the daily load curve as shown in
Fig. 3.7. It is clear from the curve that maximum demand on the station is 350 kW and occurs
from 8 A.M. to 10 A. M. i.e.,
Maximum demand = 350 kW
Sum of individual maximum demands of groups = 200 + 100 + 50 + 100
= 450 kW

Example 12. The daily demands of three consumers are given below:

Example 13. A power station has the following daily load cycle:
Solution. Below mentioned figure (i) shows the daily load curve, whereas Fig. (ii) shows the daily
load duration curve. It can be readily seen that area under the two load curves is the same. Note that
load duration curve is drawn by arranging the loads in the order of descending magnitudes.
Example 14. The annual load duration curve of a certain power station can be considered as a
straight line from 20 MW to 4 MW. To meet this load, three turbine-generator units, two rated at 10
MW each and one rated at 5 MW are installed. Determine (i) installed capacity (ii) plant factor (iii)
units generated per annum (iv) load factor and (v) utilisation factor.

Example 15. A power station has a daily load cycle as under:
260 MW for 6 hours; 200 MW for 8 hours: 160 MW for 4 hours, 100 MW for 6 hours. If the power
station is equipped with 4 sets of 75 MW each, calculate (i) daily load factor (ii) plant capacity factor
and (iii) daily requirement if the calorific value of oil used were 10,000 kcal/kg and the average heat
rate of station were 2860 kcal/kWh.

Important Points in the Selection of Units
While making the selection of number and sizes of the generating units, the following points
should be kept in view:
(i) The number and sizes of the units should be so selected that they approximately fit the
annual load curve of the station.
(ii) The units should be preferably of different capacities to meet the load requirements.
Although use of identical units (i.e., having same capacity) ensures saving in cost, they often
do not meet the load requirement.
(iii) The capacity of the plant should be made 15% to 20% more than the maximum demand to
meet the future load requirements.
(iv) There should be a spare generating unit so that repairs and overhauling of the working
units can be carried out.
(v) The tendency to select a large number of units of smaller capacity in order to fit the load
curve very accurately should be avoided. It is because the investment cost per kW of capacity
increases as the size of the unit’s decreases.
Example 16. A proposed station has the following daily load cycle:
Solution. The load curve of the power station can be drawn to some suitable scale as shown in
Fig.
Units generated per day = Area (in kWh) under the load curve
= 103
[20 × 8 + 40 × 3 + 50 × 5 + 35 × 3 + 70 × 3 + 40 × 2]
= 103
[160 + 120 + 250 + 105 + 210 + 80] kWh
= 925 × 103
kWh
Selection of number and sizes of units: Assuming power
factor of the machines to be 0·8, the output of the generating units
available will be 8, 16, 20 and 24 MW. There can be
several possibilities. However, while selecting the size and number of units, it has to be borne in mind
that (i) one set of highest capacity should be kept as standby unit (ii) the units should meet the
maximum demand (70 MW in this case) on the station (iii) there should be overall economy.
Keeping in view the above facts, 4 sets of 24 MW each maybe chosen. Three sets will meet the
maximum demand of 70 MW and one unit will serve as a standby unit.
Operational schedule. Referring to the load curve shown in Fig., the operational schedule will be
as under :
(i) Set No. 1 will run for 24 hours.
(ii) Set No. 2 will run from 8.00 hours to midnight.
(iii) Set No. 3 will run from 11.00 hours to 16 hours and again from 19 hours to 22 hours.

Example 17. A generating station is to supply four regions of load whose peak loads are
10 MW, 5 MW, 8 MW and 7 MW. The diversity factor at the station is 1·5 and the average annual
load factor is 60%. Calculate:
(i) the maximum demand on the station. (ii) annual energy supplied by the station.
(iii) Suggest the installed capacity and the number of units.
Base Load and Peak Load on Power Station
The changing load on the power station makes its load curve of variable nature. Fig. shows the
typical load curve of a power station. It is clear that load on the power station varies from time
to time. However, a close look at the load curve reveals
that load on the power station can be considered in two
parts, namely;
(i) Base load (ii) Peak load
(i) Base load. The unvarying load which occurs almost
the whole day on the station is known as base load.
Referring to the load curve of Fig., it is clear that 20 MW
of load has to be supplied by the station at all times of day
and night i.e. throughout 24 hours. Therefore, 20 MW is
the base load of the station. As base load on the station is
almost of constant nature, therefore, it can be suitably supplied (as discussed in the next Article)
without facing the problems of variable load.
(ii) Peak load. The various peak demands of load over and above the base load of the
station is known as peak load.
Referring to the load curve of Fig., it is clear that there are peak demands of load excluding
base load. These peak demands of the station generally form a small part of the total load and
may occur throughout the day.

Method of Meeting the Load
The total load on a power station consists of two parts viz., base load and peak load. In order
to achieve overall economy, the best method to meet load is to interconnect two different
power stations. The more efficient plant is used to supply the base load and is known as base
load power station. The less efficient plant is used to supply the peak loads and is known as
peak load power station. There is no hard and fast rule for selection of base load and peak
load stations as it would depend upon the particular situation. For example, both hydro-electric
and steam power stations are quite efficient and can be used as base load as well as peak load
station to meet a particular load requirement.
Illustration. The interconnection of steam and hydro plants is a beautiful illustration to meet
the load. When water is available in sufficient quantity as in summer and rainy season, the
hydroelectric plant is used to carry the base load and the steam plant supplies the peak load as
shown in Fig (i).
However, when the water is not available in sufficient quantity as in winter, the steam plant
carries the base load, whereas the hydro-electric plant carries the peak load as shown in Fig.
(ii)

Power Plant Economics and Selection
Part-3 Power plant economics and selection
Effect of plant type on costs, rates, fixed elements, energy elements, customer elements and
investor’s profit; depreciation and replacement, theory of rates. Economics of plant selection,
other considerations in plant selection.
Choice of size and number of Generating Units
1. Economy: large size units’ have
a. low capital cost/kW,
b. need less land area
c. requires less operating labour
d. have better efficiency
e. system strength
2. Power plant capacity: neither small nor large number of units
3. Transmission facility
4. Reserve requirements
5. Status of technology
6. For hydro stations unit size depends on nature of flow, availability of head and water to
generate maximum power possible.
Factor Effecting Power Plant Design
Following are the factor effecting while designing a power plant.
(1) Location of power plant
(2) Availability of water in power plant
(3) Availability of labour nearer to power plant
(4) Land cost of power plant
(5) Low operating cost
(6) Low maintenance cost
(7) Low cost of energy generation
(8) Low capital cost
EFFECT OF POWER PLANT TYPE ON COSTS
The cost of a power plant depends upon, when a new power plant is to set up or an existing
plant is to be replaced or plant to be extended. The cost analysis includes
1. Fixed Cost
It includes Initial cost of the plant, Rate of interest, Depreciation cost, Taxes, and Insurance.
2. Operational Cost
It includes Fuel cost, operating labour cost, Maintenance cost, Supplies, Supervision, Operating
taxes.
1.(a) Initial Cost
The initial cost of a power station includes the following:
1. Land cost
2. Building cost
3. Equipment cost
4. Installation cost

5. Overhead charges, which will include the transportation cost, stores and storekeeping charges,
interest during construction etc.
To reduce the cost of building, it is desirable to eliminate the superstructure over the boiler
house and as far as possible on turbine house also.
Adopting unit system where one boiler is used for one turbo-generator can reduce the cost on
equipment. Also by simplifying the piping system and elimination of duplicate system such
as steam headers and boiler feed headers. Eliminating duplicate or stand-by auxiliaries can
further reduce the cost.
When the power plant is not situated in the proximity to the load served, the cost of a primary
distribution system will be a part of the initial investment.
1. (b)Rate of Interest
All enterprises need investment of money and this money may be obtained as loan, through
bonds and shares or from owners of personal funds. Interest is the difference between money
borrowed and money returned. It may be charged at a simple rate expressed as % per annum
or may be compounded, in which case the interest is reinvested and adds to the principal,
thereby earning more interest in subsequent years. Even if the owner invests his own capital
the charge of interest is necessary to cover the income that he would have derived from it
through an alternative investment or fixed deposit with a bank. Amortization in the periodic
repayment of the principal as a uniform annual expense.
1. (c) Depreciation
Depreciation accounts for the deterioration of the equipment and decrease in its value due to
corrosion, weathering and wear and tear with use. It also covers the decrease in value of
equipment due to obsolescence. With rapid improvements in design and construction of plants,
obsolescence factor is of enormous importance. Availability of better models with lesser
overall cost of generation makes it imperative to replace the old equipment earlier than its
useful life is spent. The actual life span of the plant has, therefore, to be taken as shorter than
what would be normally expected out of it.
The following methods are used to calculate the depreciation cost:
(1) Straight line method
(2) Percentage method
(3) Sinking fund method
(4) Unit method.
(1) Straight Line Method. It is the simplest and commonly used method. The life of the
equipment
or the enterprise is first assessed as also the residual or salvage value of the same after the
estimated life span. This salvage value is deducted from the initial capital cost and the balance
is divided by the life as assessed in years. Thus, the annual value of decrease in cost of
equipment is found and is set aside as depreciation annually from the income. Thus, the rate of
depreciation is uniform throughout the life of the equipment. By the time the equipment has
lived out its useful life, an amount equivalent to its net cost is accumulated which can be utilized
for replacement of the plant.

(2) Percentage Method. In this method the deterioration in value of equipment from year to
year is taken into account and the amount of depreciation calculated upon actual residual value
for each year. It thus, reduces for successive years.
(3) Sinking Fund Method. This method is based on the conception that the annual uniform
deduction from income for depreciation will accumulate to the capital value of the plant at the
end of life of the plant or equipment. In this method, the amount set aside per year consists of
annual instalments and the interest earned on all the instalments.
Let,
A = Amount set aside at the end of each year for n years.
n = Life of plant in years.
S = Salvage value at the end of plant life.
i = Annual rate of compound interest on the invested capital.
P = Initial investment to install the plant.
Then, amount set aside at the end of first year = A
Amount at the end of second year
= A + interest on A = A + Ai = A(1 + i)
Amount at the end of third year
= A(1 + i) + interest on A(1 + i)
= A(1 + i) +A(1 + i)i
= A(1 + i)2
Amount at the end of nth year = A(1 + i)n
– 1
Total amount accumulated in n years (say x)
= sum of the amounts accumulated in n years
i.e., x = A + A(1 + i) + A(1 + i)2
+ ...... + A(1 + i)n – 1
= A[1 + (1 + i) + (1 + i)2
+...... + (1 + i)n – 1
] ...(1)
Multiplying the above equation by (1 + i), we get
x(1 + i) = A [(1 + i) + (1 + i)2
+ (1 + i)3
+ ...... + (1 + i)n
] ...(2)
Subtracting equation (1) from (2), we get
x.i = [(1 + i)n
– 1] A
x = [{(1 + i)n – 1}/i]A, where x = (P – S)
P – S = [{(1 + i)n
– 1}/i]A
A = (P – S)[i/{(1 + i)n
– 1}]A
(4) Unit Method. In this method some factor is taken as a standard one and, depreciation is
measured by that standard. In place of years’ equipment will last, the number of hours that
equipment will last is calculated. This total number of hours is then divided by the capital value
of the equipment. This constant is then multiplied by the number of actual working hours each
year to get the value of depreciation for that year. In place of number of hours, the number of
units of production is taken as the measuring standard.

2. OPERATIONAL COSTS
The elements that make up the operating expenditure of a power plant include the following
(1) Cost of fuels.
(2) Labour cost.
(3) Cost of maintenance and repairs.
(4) Cost of stores (other than fuel).
(5) Supervision.
(6) Taxes.
(1) COST OF FUELS
In a thermal station fuel is the heaviest item of operating cost. The selection of the fuel and the
maximum economy in its use are, therefore, very important considerations in thermal plant
design. It is desirable to achieve the highest thermal efficiency for the plant so that fuel charges
are reduced. The cost of fuel includes not only its price at the site of purchase but its
transportation and handling costs also. In the hydro plants the absence of fuel factor in cost is
responsible for lowering the operating cost.
Plant heat rate can be improved by the use of better quality of fuel or by employing better
thermodynamic conditions in the plant design.
The cost of fuel varies with the following:
(1) Unit price of the fuel.
(2) Amount of energy produced.
(3) Efficiency of the plant.
(2) LABOUR COST
For plant operation labour cost is another item of operating cost. Maximum labour is needed
in a thermal power plant using. Coal as a fuel. A hydraulic power plant or a diesel power plant
of equal capacity requires a lesser number of persons. In case of automatic power station, the
cost of labour is reduced to a great extent. However, labour cost cannot be completely
eliminated even with fully automatic station, as they will still require some manpower for
periodic inspection etc.
(3) COST OF MAINTENANCE AND REPAIRS
In order to avoid plant breakdowns maintenance is necessary. Maintenance includes periodic
cleaning, greasing, adjustments and overhauling of equipment. The material used for
maintenance is also charged under this head. Sometimes an arbitrary percentage is assumed as
maintenance cost. A good plan of maintenance would keep the sets in dependable condition
and avoid the necessity of too many stand-by plants.
Repairs are necessitated when the plant breaks down or stops due to faults developing in the
mechanism. The repairs may be minor, major or periodic overhauls and are charged to the
depreciation fund of the equipment. This item of cost is higher for thermal plants than for
hydro-plants due to complex nature of principal equipment and auxiliaries in the former.
(4) COST OF STORES
The items of consumable stores other than fuel include such articles as lubricating oil and
greases, cotton waste, small tools, chemicals, paints and such other things. The incidence of
this cost is also higher thermal stations than in hydro-electric power stations.

(5) SUPERVISION
In this head the salary of supervising staff is included. A good supervision is reflected in lesser
breakdowns and extended plant life. The supervising staff includes the station superintendent,
chief engineer, chemist, engineers, supervisors, stores in charges, purchase officer and other
establishment. Again, thermal stations, particularly coal fed, have a greater incidence of this
cost than the hydro-electric power stations.
(6) TAXES
The taxes under operating head includes the following:
(i) Income tax
(ii) Sales tax
(iii) Social security and employee’s security etc.
EFFECT OF PLANT TYPE ON RATES (TARIFFS OR ENERGY ELEMENT)
Rates are the different methods of charging the consumers for the consumption of electricity.
It is desirable to charge the consumer according to his maximum demand (kW) and the energy
consumed (kWh). The tariff chosen should recover the fixed cost, operating cost and profit etc.
incurred in generating the electrical energy.
REQUIREMENTS OF A TARIFF
Tariff should satisfy the following requirements:
(1) It should be easier to understand.
(2) It should provide low rates for high consumption.
(3) It should encourage the consumers having high load factors.
(4) It should take into account maximum demand charges and energy charges.
(5) It should provide less charges for power connections than for lighting.
(6) It should avoid the complication of separate wiring and metering connections.
TYPES OF TARIFFS
The various types of tariffs are as follows,
(1) Flat demand rate
(2) Straight line meter rate
(3) Step meter rate
(4) Block rate tariff
(5) Two-part tariff
(6) Three-part tariff.
The various types of tariffs can be derived from the following general equation:
Y = DX + EZ + C
where
Y = Total amount of bill for the period considered.
D = Rate per kW of maximum demand.
X = Maximum demand in kW.
E = Energy rate per kW.
Z = Energy consumed in kWh during the given period.
C = Constant amount to be charged from the consumer during each billing period.
Various type of tariffs are as follows:

(1) Flat Demand Rate. It is based on the number of lamps installed and a fixed number of hours of
use per month or per year. The rate is expressed as a certain price per lamp or per unit of demand (kW)
of the consumer. This energy rate eliminates the use of metering equipment. It is expressed by the
expression.
(2) Straight Line Meter Rate. According to this energy rate the amount to be charged from
the consumer depends upon the energy consumed in kWh which is recorded by a means of a
kilowatt hour meter. It is expressed in the form
Y = EZ
This rate suffers from a drawback that a consumer using no energy will not pay any amount
although he has incurred some expense to the power station due to its readiness to serve him.
Secondly since the rate per kWh is fixed, this tariff does not encourage the consumer to use
more power.
(3) Step Meter Rate. According to this tariff the charge for energy consumption goes down
as the energy consumption becomes more. This tariff is expressed as follows.
Y = EZ If 0 ≤ Z ≤ A
Y = E1Z1 If A ≤ Z1 ≤ B
Y = E2Z2 If B ≤ Z2 ≤ C
And so on. Where E, E1, E2 are the energy rate per kWh and A, B and C, are the limits of
energy consumption.
(4) Block Rate Tariff.
According to this tariff a certain price per units (kWh) is charged for all or any part of block of
each unit and for succeeding blocks of energy the corresponding unit charges decrease.
It is expressed by the expression
Y = E1Z1 + E2Z2 + E3Z3 + E4Z4 +.....
where E1, E2, E3.... are unit energy charges for energy blocks of magnitude Z1, Zz, Zg,
respectively.

(5) Two Part Tariff (Hopkinson Demand Rate). In this tariff the total charges are based on
the maximum demand and energy consumed. It is expressed as
Y = D . X + EZ
A separate meter is required to record the maximum demand. This tariff is used for industrial
loads.
(6) Three-Part Tariff (Doherty Rate). According to this tariff the customer pays some fixed
amount in addition to the charges for maximum demand and energy consumed. The fixed
amount to be charged depends upon the occasional increase in fuel price, rise in wages of
labour etc. It is expressed by the expression
Y = DX + EZ + C.
EFFECT OF PLANT TYPE ON FIXED ELEMENTS
Various types of fixed element are:
(1) Land
(2) Building
(3) Equipment
(4) Installation of Machine
(5) Design and planning
The fixed element means which are not movable, and for any types of power plant, the fixed
elements play a major role. Since each cost is added to the final cost of our product
(electricity in case of Power plant). So when a power plant is established, the first selection is
fixed element. Effect of plant on land is as cost of land.

Power plant engineering unit 1 by Varun Pratap Singh

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