MODULE I
INTRODUCTION TO ENERGY
RESOURCES
Lesson 1
Lesson 2
Lesson 3
Types of Power Plant as to its Sources
Intro to Renewable & Non-Renewable
Energy Resources
Power Plant Operation & Basic Problem
Solving
Module I
2
MODULE I
INTRODUCTION TO ENERGY RESOURCES

INTRODUCTION
This module presents Familiarization of the different Power plant as
to their flow of operations, and will discuss the power plants, as to its
purpose, as to its source, as to its use and reserves.
POWER PLANT ENGINEERING AND DESIGN- It is the art of selecting and
placing the necessary power-generating equipment so that the maximum of
return will result from the minimum expenditure over the working life of
the plant; and the operation of the completed plant in manner to provide a
cheap, reliable, and continuous service.
OBJECTIVES
After studying the module, you should be able to:
1. Evaluate the importance of the different power plants
2. Explain the different operations of a power plant.
3. Solve & analyze problems relative to the different power plant
discussed.
4. Evaluate & give synthesis the different existing power plants in
the Philippines.

DIRECTIONS/ MODULE ORGANIZER
There are three lessons in the module. Read each lesson carefully
then answer the exercises/activities to find out how much you have
benefited from it. Work on these exercises carefully and submit your output
to your tutor or to the COE office.
In case you encounter difficulty, discuss this with your tutor during
the face-to-face meeting. If not contact your tutor at the
COE office.
Good luck and happy reading!!! -mev
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Lesson 1

Types of Power Plants as to Sources
ELECTRICAL POWER GENERATING STATION (POWER PLANT)
A station in which are located prime movers, electric generators,
and auxiliary equipment for converting mechanical, chemical, and/or
nuclear energy into electrical energy.
TYPES OF POWER PLANTS AS TO SOURCE
1. THERMAL PLANT – an electric generating station using heat as a
source of energy.
1.1 Oil-fired Steam Plant – a thermal plant which makes use of heavy
fuel oil (HFO), Light fuel oil (LFO), or Bunker C Oil as fuel for the production
of electric energy. Example, Bataan 1 & 2 Thermal Plants.
1.2 Coal-fired Steam Plant – a thermal plant which makes use of
pulverized coal as fuel in producing electric energy. Example, Batangas
Coal-fired Thermal Power Plant.
1.3 Dendro-Thermal Power Plant – a thermal plant which makes use of
wood in producing electric energy.
1.4 Geothermal Steam Plant – a thermal plant which makes use of
generated heat from the earth’s magma as a fuel in producing electric
energy. Example, Mak-Ban Geothermal Plant
1.5 Nuclear Steam Plant – a thermal plant which makes use of the steam
generated in a reactor by heat from the fissioning of nuclear fuel as source
of producing electric energy. Example- Bataan Nuclear Power Plant (NonOperational, and was commissioned)
1.6 Diesel Plant – a thermal plant wherein the prime mover is an internal
combustion engine (ICE) and which makes use of diesel oil as the source of
electric energy.(1590 Energy Corp.-Bauang)
1.7 Gas-Turbine Plant – a thermal plant in wherein the prime mover is a
gas turbine engine and which makes use of combustible gas as fuel in
producing electric energy.
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1.8 Solar Thermal Plant - Solar thermal energy (STE) is a form of energy
and a technology for harnessing solar energy to generate thermal energy or
electrical energy for use in industry, and in the residential and commercial
sectors.
1.9 Biomass Power Plant- Biomass, as a renewable energy sources, is a
biological material from living, or recently living organisms. As an energy
source, biomass can either be used directly or converted into other energy
products such as biofuel. biomass is a plant matter used to generate
electricity with steam turbines and gasifies or produced heat usually by
direct combustion. Example (Phil-Bio Biomass Plant)
Thermal Plant Basic Block Diagram
Boiler – the pressure part of the steam generator wherein water is pumped
into it, vaporizes, and delivers to the steam turbine.
Steam Turbine – a versatile prime mover for the electric generator. Steam
strikes the blade of the turbine which makes the shaft to rotate hence
rotating the generator moving part to generate more power.
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In which, Combined heat Pressures from the turbine are categorized into
three (3) major parts a.) Low pressure turbine, b.) Intermediate pressure
turbine, c.) High pressure turbine.
Condenser – it is here where the steam is reduced to liquid and ready to be
pumped again in the broiler.
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Electric Generator – the mechanical power from the turbine converted into
electrical power through this equipment.
Exciter – provide excitation to the electric generator to generate emf and
hence electric power.
Transformer and Transmission Lines – power is distributed at higher
voltage by means of these equipment and devices.
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2.Hydro Electric Power Plant
HYDRO PLANT/HYDRO ELECTRIC POWER PLANT – an electric generating
station in which the prime mover is a turbine driven by falling water.
2.1 Run-of-River Plant – a hydroelectric generating station utilizing
limited pondage or the stream flow as it occurs.
2.2 Plant with Storage Capacity - a hydroelectric generating station
associated with a water storage reservoir.
2.3 Pumped-Storage Plant - a hydroelectric generating station wherein
electric energy during periods of relatively high system demand by utilizing
water which has been pumped into a storage reservoir usually during periods
of relatively low system demand.
Types of hydraulic turbines or waterwheels:
Impulse type – suitable for very high heads, for net heads 1,300 ft. & above
Reaction type/Francis type – suitable for medium heads,
Propeller type – suitable for very low heads, for net heads up to 70ft.
3.
Other types of Power Plant
3.1 Windmill Power Plant
•
- Wind power or wind energy is the use of wind to provide the
mechanical power through wind turbines to turn electric generators
and traditionally to do other work, like milling or pumping.
•
- Wind power is a sustainable and renewable energy, and has a
much smaller impact on the environment compared to burning fossil
fuels. Wind farms consist of many individual wind turbines, which are
connected to the electric power transmission network.
3.2 Sea Wave Power Plant
•
-Tidal power or tidal energy is the form of hydropower that
converts the energy obtained from tides into useful forms of power,
although not yet widely used, tidal energy has potential for future
electricity generation.
3.3 Solar Power
3.3.1 - Off-Grid Solar Power
- SMALL SCALE OR for MICRO GRID PURPOSES
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is a characteristic of facility and a lifestyle designed in an independent
manner without reliance on one or more public utilities and autonomous in
nature. The term "off-the-grid" traditionally refers to not being connected
to the “Grid” (ex. Luzon Grid).
Microgrids can serve hundreds to thousands of customers through a
community and support the penetration of local energy. In microgrid, some
houses may have some renewable sources that can supply their demand as
well as that of their neighbors within the same community. The community
microgrid may also be categorized as centralized source of energy & power.
3.3.2 HYBRID POWER
are hybrid power systems that combine solar power from a photovoltaic
system with another power generating energy source..(ex. Distribution
Utilities, solar-wind, solar-D.U.,solar-diesel. Etc.)
TYPE OF POWER PLANT AS TO USE:
1. Peaking Plant – a generating plant which is normally operated to
provide power only during peak load periods.
2. Regulating Plant – a generating plant capable of carrying load for
the time interval either during off-peak or peak periods, and
usually responds to changes in system frequency.
3. Base Load Plant – a generating plant that can carry a minimal
load over a given period of time and a plant that runs near its full
rating continuously, day and night, all year long.
TYPE OF PLANT RESERVE:
a) Cold Reserve- portion of the installed reserve kept in operable
condition and available for service but not for immediate loading.
b) Operating Reserve- refers to the capacity in actual service in excess
of peak load.
c) Hot Reserve- refers to units available, maintained at operating
temperature and ready for service although not in actual operation.
d) Spinning Reserve- generating capacity connected to the bus ready to
take load.
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DIFFERENT TYPES OF PRIME MOVERS:
1. STEAM
-Reciprocating Engine
-Turbines
2. HYDRAULIC
-Impulse wheel
-Reaction wheel
-Propeller wheel
3. INTERNAL COMBUSTION
-Oil, Diesel, Gas
THINK!
On your own point of view, what are the existing Power
plants available in our country today?
THINK!
Case:
On our existing power plant on our Generation Mix, which type
of power plant has the biggest part and why do you think that
this has the biggest share?
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Lesson 2

Intro to Renewable & Non-Renewable Energy
Our country today is in need and focused on the usage of renewable energy
as part of its sustainability to be partially independent from oil, diesel and
coal energy generation plants and to further lessen the greenhouse effect.
From Luzon Generation Mix, as of November 2013, 55.67% coal usage and
3.88% Diesel usage from generation plants, in which our country is highly
dependent from these plants.
In December 2008, our country enacted RA 9513, also known as the
renewable Energy Act of 2008, which conforms to internationally accepted
standards which enable our country to be part of a clean and green power
generation allies.
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Renewable energy
is derived from natural processes that are replenished constantly. In
its various forms, it derives directly from the sun, or from heat
generated deep within the earth. Included in the definition is
electricity and heat generated from solar, wind, ocean,
hydropower, biomass, geothermal resources, and biofuels and
hydrogen derived from renewable resources.
Non-Renewable Energy
A non-renewable resource (also called a finite resource) is a natural
resource that cannot be readily replaced by natural means at a quick
enough pace to keep up with consumption. An example is carbon-based
fossil fuel. The original organic matter, with the aid of heat and pressure,
becomes a fuel such as oil or gas. Earth minerals and metal ores, fossil
fuels (coal, petroleum, natural gas) and groundwater in certain aquifers
are all considered non-renewable resources, though individual elements are
always conserved (except in nuclear reactions).
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THINK!
Which is Cheaper in the Production of Energy?
THINK!
Write an essay why coal fired power plant dominates the
generation matrix in the production of Electrical Energy?
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Lesson 3

Power Plant Operation & Problem Solving
COAL FIRED POWER PLANT
Coal is the most abundant fossil fuel on the planet. It is the relatively
cheap fuel, with some of the largest deposits in region that stable
politically. Solid coal cannot directly replace natural gas or petroleum in
most applications, is mostly used for transportation and natural gas not used
for electricity generation is used for space water and industrial heating.
Coal can be converted to gas or liquid fuel, but the efficiencies and
economics of such processes can make them feasible. Vehicles or heaters
may require modification to use coal derived fuels. Coal can produce more
pollution than petroleum or natural gas.
A fossil fuel power plant station is a type of power station that burns
fossil fuels such as coal, natural gas or petroleum (oil) to produce
electricity. Central station fossil power plant is designed on a large scale for
continuous operation. In many centuries, such plants provide most of the
electrical energy used.
Fossil fuel power plant station have rotating machinery to convert
heat energy of combustion into mechanical, which then operates an
electrical generator. The prime mover may be steam turbine, a gas turbine
or in a small plant, reciprocating internal combustion engine. All plants used
the energy extracted from expanding gas-steam or combustion gasses. A
very few MHD generators have been built which directly convert the energy
of moving hot gas into electricity.
By products of thermal power plant operation must considered in
their design and operation. Waste heat energy, which remains due to the
finite efficiency of the Carnot, Rankine, or diesel power cycle is heat energy
released directly to the atmosphere, directly to the river or lake water or
directly to the atmosphere. Using a cooling tower with river or lake water as
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a cooling medium. The fuel gas from the combustion of the fossil fuel is
discharged to the air. The gas contains carbon dioxide, water vapor as well
as substances such as elemental nitrogen oxide, mercury traces of other
metals and for coal fired plants fly ashes. Solid waste ash from coal fired
boilers must also be removed. Some ash can be recycled for building
materials.
Fossil fueled power stations are major emitters of carbon dioxide a
greenhouse gas (GHG) which according to consensus opinion of scientific
organizations is a contributor of global warming as it has been observed over
the last years. Brown coal emits about three times as much as carbon
dioxide as natural gas, black coal emits about twice as much as carbon
dioxide per unit of electric energy. Carbon capture and storage of emission
is not expected to be available on a commercial economically viable basis
until government supported legislation enacted.
FUEL PROCESSING
Coal is prepared for use by crushing the rough coal to pieces less than
two inches or 5 centimeters in sizes. The coal is then transported from the
storage yard to plant storage silos by rubberized conveyor belts at rates u at
250 tons per hour under full, load to 4000 tons per hour.
In plants that burn pulverized coal, silos feed coal pulverizes (coal
mills) they take larger two inch or fifty-one millimeter pieces, grind them to
the consistency of face powder, sort them and mix them with primary
combustion air which transport the coal to the boiler furnace and preheats
the coal to the boiler in order to drive off the excess moisture content. A
500 MWE plant may have six such pulverizes, five of which can supply coal
to the furnace at 250 tons per hour under full load.
In plants that do not burn pulverized coal, the larger 2 inch or 51
millimeter pieces may be directly fed into the silos which then feed either
mechanically distributors that drop the coal on a travelling grate or the
cyclone burners a specific kind of combustor that can efficiently burn larger
pieces of fuel.
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Basic Conversion of Units, that is usable in solving in thermal power plants
especially, coal fired power plant.
Mass & weight (the weight of coal for every plant consumption)
1 Ton = 2,000 pounds (lbs.)
1Kg. = 2.2 lbs.
1 Long Ton = 2240 lbs.
1KgF = 9.80665 Newtons
1 Metric Ton = 1000 Kg. or 2,200 lbs.
1 Newton = 0.1019 Kg. = 0.0459 lbs.
1 KIP = 1,000 lbs.
1 Pound (lb.) = 0.4536 Kg. = 0.13825 N
1 slug = 14.59 Kg, = 32.17 lbs.
1 0z. = 28.35 grams
Work & Energy (thermal heat Efficiency, capacity)
1 BTU (British thermal Unit) – 778-ft-lb
- 252 cal.
- 1055 joules
- 778 ft-lb
- 0.2929 W-Hr
1Quad = 1x1015 BTU
1 Therm =1x105 BTU
1 Ton = 12,000 BTU/Hr.
1 Kcal (Kilo Calories) – 4.187 KJ or 1cal. = 4.187 joules
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1 ERG – 1 dyne-cm or 1x10-7 joule
1 Joule – 1 N-m
Power
1 Horsepower – 550 ft-lb/sec, or 33,000 ft-lb/min
- 2545 BTU/hr. or 42.4 BTU/min.
-0.746 KW or 746 watts
1KW – 3414 BTU/Hr. or 860 Kcal/Hr.
0r in Electrical Energy, thus:
1KWHR = 3414 BTU = 860 Kcal = 3,600KJ
1 Watt – 1 Joule/sec, 1 N-m/sec, or 1x10-7 ergs/sec.
1 Boiler HP = 33,480 BTU/Hr. or 35,322 KJ/Hr.
Overall Plant Thermal Efficiency
o 
Pout
HeatEnergyPower x PlantCapacity

Pin
Mass of coal x CalorificValue of coal to be used by the plant
As such,  o = Boiler Efficiency x Turbine Efficiency x Alt./Gen. Efficiency
Or,
 o = Thermal Efficiency x Electrical Efficiency
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Example 1:
1. To produce one KWHR, A coal fired power plant burns 0.9 lb. of coal
with a heating value of 13,000 BTU/lb. What is the heat rate of the
plant?
Sol. (By Unit analysis)
13,000 BTU/ lb. x 0.9lbs./1KWHR
= 11,700 BTU/KWHR /ans.
Example 2:
A certain coal fired power plant has a heat rate of 2.88x10 6 calories
per KWHR production, Coal cost’s P2,500 per ton. How much is the
fuel cost component of producing 1 KWHR?
Sol.
Kcal
x (1KWHR )
KWHR
P2500
x x
Fuel Cost=
BTU 860 Kcal 2.2lbs 100Kg
Ton
13,000
x
x
x
lb
3414BTU 1Kg
1Ton
2.88 x10 3
= P0.997 ≈ P1 per KWHR
Example 3:
A Coal fired power plant burns its imported fuel at a heating value of 14,000
BTU/lb. The plant has a turbine efficiency of 70% and an alternator
efficiency of 95%. Determine the fuel cost component in producing 1KWHR
in Philippine currency if coal cost 2 USD per pound. (Note: exchange rate for
the moment is 1USD = P52.45)
Sol.
 o = turbine x alternator = 0.70 x 0.90 = 0.63 x 100% = 63%
BTU
x ( KWHR )
KWHR
BTU 

(0.63) x 14,000

lb 

3414
m=
m = 0.387 lbs.
since, 1 lb. cost 2USD
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therefore,
2USD P52.45
x
1lb.
1USD
cost /KWHR = P 40.60
m = 0.387 lbs. x
Example 4:
A power plant consumes 21,542 Kg of coal per hour. The heating value of
coal is 6,652,800 calories per Kg and the overall plant capacity is 30%. What
is the MW output of the plant?
Sol.
1 KWHR
cal.
Kg .
x 21,542
x
Kg .
Hr .
860,000 cal.
Pin = 166,643.05 KW = 166.64 MW
Pin = 6,652,800
But,
o 
Pout
Pin
Therefore,
Pout = Pin  o
Pout = (166.64 MW) (0.30)
Pout = 50 MW
Example 5:
A coal fired thermal station has an overall efficiency of 15% and 0.75 Kg of
coal is burnt per KWHR by the station. Determine the calorific value of coal
in Kilocalories per Kilogram.
(in this portion of problem, I will show you how to solve in different
methods but will end up in the same answer...)
Solution 1.(Using 1KWHR=3.6x106 J)
1 KWHR 4.187 J
0.75Kg
Kg .
x
x
 8.723x10 7
6
KWHR 3.6 x10 J
1cal.
cal
1,146,405.541
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1Kcal
Kcal
x
1,146.405
Kg. 1,000cal.
Kg
Take the inverse of the value
20
1,146.405
Calorific Value =
Kcal
Kg
0.15
= 7,642.7
Kcal
/ans.
Kg
Sol.2 (Using 1KWHR = 860 KWHR)
0.75Kg 1 KWHR
Kg
x
 8.721 x10 4
KWHR 860Kcal
Kcal
= 1,146.667
Take the inverse
Kcal
Kg
1,146.667
Calorific Value =
Kcal
Kg
0.15
= 7,644.444
Kcal
/ans.
Kg
Sol.3 (Using thermal Efficiency method)
Given:
o 
Pout
HeatEnergyPower x PlantCapacity

Pin
Mass of coal x CalorificValue of coal to be used by the plant
 o = 15%
Using, 1KWHR= 860 Kcal
Kcal
KWHR
15% =
Kg
0.75
xq
KWHR
860
Therefore,q (calorific value) = 7,644.444 Kcal/Kg..../Ans.
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REMINDER!
In dealing with problems that involve heat-rate, amount of fuel
needed, cost of production of energy,use the mathematical
technique normally called “unit analysis method & cancellation of
units”..MEV
HYDROELECTRIC POWER PLANT
Hydroelectricity is a term referring to electricity generated by hydro power.
The production of electrical power through the use of the gravitational
force of failing or flowing or flowing water. It is the most widely used form
of renewable energy accounting for 16% of global electricity consumption
and 3427 terawatts-hour of electricity production in 2010 which continues
the rapid rate of increase experienced up to the present situation.
Hydro is also a flexible source of electricity since plants be ramped
up and down very quickly to adapt to changing energy demand. However,
damming interrupts, the flow of rivers and can harm local ecosystems and
building large. Dams and reservoirs often involves displacing people and
wildlife. Once a hydroelectric complex is constructed, the project produces
no direct waste and has a considerably lower output level of the greenhouse
gas carbon dioxide than fossil fuel powered energy plants.
CONVENTIONAL (DAMS)
Most hydroelectric power comes from the potential energy of
dammed water driving a water turbine and a generator. The power
extracted from the water depends on the volume and on the difference in
height between the sourced and water outflow. This height difference is
called head. The amount of potential energy in water is proportional to the
head a large pipe “PENSTOCKS” called delivers water to turbine.
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PUMPED STORAGE
Is a type of hydroelectric power generation used by some power
plants for load balancing, this method stores energy in the form of potential
energy of water, pumped from lower elevation reservoir to a higher
elevation? Low cost off peak electric power is used to run pumps. During
periods of high electrical demands, the stored water is stored/ water
released through the turbine to produce electric power. Although the losses
of the pumping process make the plant a net consumer of energy overall,
the system increases revenues by selling more electricity during periods of
peak demand, when electricity prices highest. Pumped storage schemes
currently provide the most commercially important means of large scale grid
energy storage and improve the daily capacity factor of the generation
system.
RUN-OF-THE-RIVER
Is a type of hydroelectric generation whereby little or no water
storage is provided? Run of the river plants may either have no storage is
provided. Run of the river plants may either have no storage at all, or a
limited amount or storage, in which case the storage reservoir is referred to
as a poundage. A plant without poundage has no storage and is, therefore,
subject to a seasonal river flows and serves as a peaking power while a plant
with poundage can regulate water flow and serve either as a peaking or
base load power plant.
ADVANTAGES OF HYDROELECTRIC POWER PLANT
FLEXIBILITY
Hydro is a flexible source of electricity since plants can be ramped up
and down very quickly to adapt to changing energy demands.
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LOW POWER COSTS
The major advantage of hydroelectric is elimination of the cost of
fuel. The cost of operating a hydroelectric is nearly immune to increases in
the cost of fossils such as oil, natural, gas or coal and no imports are
needed. The average cost of electricity from a hydro plant larger than 10
watts is 3 to 5% cents per kilowatt hour.
Hydroelectric plants still in after 50-100 years. Operating labor is also
usually low as plants are automated and have few personnel on site during
normal operations.
REDUCED CO2 EMISSIONS
Since hydroelectric dams do not fossils, they do not directly produced
carbon dioxide. While some carbon dioxide is produce during manufacturing
and construction of the products, this is a tiny fraction of the operating
emissions of equivalent fossil fuel electricity generations.
OTHER USES OF THE RESERVOIR
Reservoirs created by hydroelectric schemes often provide facilities
for water sports and become tourist attraction themselves some countries
aquaculture is common in reservoirs. Multi-use dams installed for irrigation
support agriculture with a relatively constant water supply. Large hydro
dams can control floods, which affect otherwise the people living
downstream of the project.
DISADVANTAGES
ECOSYSTEM DAMAGE AND LOSS AF LAND
Large reservoirs required for the operations of hydroelectric power
stations result submersion of extensive area upstream of the dams,
destroying biologically rich and productive lowland and river line, valley
forest, marshland and grasslands. The loss of land is often exacerbated by
habitat fragmentation of surrounding areas.
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SILTATION AND FLOW SHORTAGE
When water flows it has the ability to transport particles heavier than
itself downstream, this has a negative effect on dams and subsequently
their power stations particularly those on rivers or within catchment areas
with siltation. Siltation’s can fill a reservoir and reduce its capacity to
control floods along with causing additional horizontal pressure on the
upstream portion of dam. Eventually, some reservoirs can be full of
sediments and useless or over top during a flood and fail. Changes in the
amount of live storage in a reservoir therefore reducing the amount of
water that can be used for hydroelectricity.
Types & Process of Hydraulic Turbine Used In Hydro Electric Plant
a) Reaction Turbine – The water under the pressure is partly converted
into velocity before it enters the turbine runner.
Francis Type – The water enters the spiral case from the
penstock, passes through the stay ring guided by the stationary
stay ring vanes, then through the movable wicket gates
through the runner and into the draft tube through which it
flow into a tailrace or tail water reservoir.
Propeller Type – It is the same as Francis type but it has an
enshrouded * blades.
Axial Flow Turbine – is a propeller type runner with either fix
or adjustable blades.
b) Impulse Turbine – The water under pressure is partly converted into
velocity before it enters the turbine runner. It consists of one or more
free jets of water discharging into an aerated space and impinging on
a set of buckets attached around the periphery of a disk or wheel.
*a covering or enclosure with or if with a shroud
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How a Hydroelectric Power System Works?
Flowing water is directed at a turbine (remember turbines are just advanced
waterwheels). The flowing water causes the turbine to rotate, converting
the water’s kinetic energy into mechanical energy.
The mechanical energy produced by the turbine is converted into electric
energy using a turbine generator. Inside the generator, the shaft of the
turbine spins a magnet inside coils of copper wire. It is a fact of nature that
moving a magnet near a conductor causes an electric current.
Impulse Turbines
 Uses the velocity of the water to move the runner and discharges to
atmospheric pressure.
 The water stream hits each bucket on the runner.
 No suction downside, water flows out through turbine housing after
hitting.
 High head, low flow applications.
 Types : Pelton wheel, Cross Flow
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Pelton Wheels
 Nozzles direct forceful streams of water against a series of spoonshaped buckets mounted around the edge of a wheel.
 Each bucket reverses the flow of water and this impulse spins the
turbine.
 Suited for high head, low flow sites.
 The largest units can be up to 200 MW.
 Can operate with heads as small as 15 meters and as high as 1,800
meters.
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 drum-shaped
 elongated, rectangular-section nozzle directed against curved vanes
on a cylindrically shaped runner
 “squirrel cage” blower

water flows through the blades twice
 First pass : water flows from the outside of the blades to the inside
 Second pass : from the inside back out
 Larger water flows and lower heads than the Pelton.
Reaction Turbines
 Combined action of pressure and moving water.
 Runner placed directly in the water stream flowing over the blades
rather than striking each individually.
 lower head and higher flows than compared with the impulse
turbines.
Propeller Hydropower Turbine
 Runner with three to six blades.
 Water contacts all of the blades constantly.
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 Through the pipe, the pressure is constant
 Pitch of the blades - fixed or adjustable
 Scroll case, wicket gates, and a draft tube
 Types: Bulb turbine, Straflo, Tube turbine, Kaplan
Bulb Turbine
 The turbine and generator are a sealed unit placed directly in the
water stream.
Francis Turbines
 The inlet is spiral shaped.
 Guide vanes direct the water tangentially to the runner.
 This radial flow acts on the runner vanes, causing the runner to spin.
 The guide vanes (or wicket gate) may be adjustable to allow efficient
turbine operation for a range of water flow conditions.
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INSIDE A HYDROPOWER PLANT
Reservoir stores the water coming from the upper river or waterfalls.
Head Water is the water in the reservoir.
Spillway is a weir in the reservoir which discharges excess water so that the
head of the plant will be maintained.
Dam is the concrete structure that encloses the reservoir.
Silt Sluice is a chamber which collects the mud and through which the mud
is discharged.
Valve is a device that opens or closes the entrance of the water into the
penstock.
Trash rack is a screen which prevents the leaves branches and other water
contaminants to enter into the penstock.
Penstock is the channel that leads the water from the reservoir to the
turbine.
Surge Chamber is a standpipe connected to the atmosphere and attached to
the penstock so that the water will be at atmospheric pressure.
Tail Race is channel which leads the water from the turbine to the tail
water.
Tail Water is the water that is discharged from the turbine.
Draft Tube is a device that connects the turbine outlet to the tail water so
that the turbine can be set above the tail water level.
Intake - Gates on the dam open and gravity pulls the water through
the penstock, a pipeline that leads to the turbine. Water builds up pressure
as it flows through this pipe.
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Hydraulic Turbine- is a device that converts the energy of water into the
mechanical energy.
The water strikes and turns the large blades of a turbine, which is attached
to a generator above it by way of a shaft. The most common type of turbine
for hydropower plants is the Francis Turbine, which looks like a big disc with
curved blades. A turbine can weigh as much as 172 tons and turn at a rate of
90 revolutions per minute (rpm), according to the Foundation for Water &
Energy Education(FWEE).
Generator is a device that converts the mechanical energy of the turbine
into electrical energy.
Transformer - The transformer inside the powerhouse takes the AC and
converts it to higher-voltage current.
Power lines - Out of every power plant come four wires: the three phases of
power being produced simultaneously plus a neutral or ground common to
all three. (Read How Power Distribution Grids Work to learn more about
power line transmission.)
Outflow - Used water is carried through pipelines, called tailraces, and reenters the river downstream.
The water in the reservoir is considered stored energy. When the gates
open,
the
water
flowing
through
the
penstock
becomes kinetic
energy because it's in motion. The amount of electricity that is generated is
determined by several factors. Two of those factors are the volume of water
flow and the amount of hydraulic head. The head refers to the distance
between the water surface and the turbines. As the head and flow increase,
so does the electricity generated. The head is usually dependent upon the
amount of water in the reservoir.
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CONVENTIONAL (DAMS)
Pumped storage
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Run of River
AMBUKLAO HYDRO-POWER PLANT
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SAN ROQUE HYDRO-POWER PLANT
Theoretical Power Output of a Hydraulic Turbine
P = Qwh
where:
P = theoretical power of the turbine (watt)
Q = volume discharge of water (m3/sec)
w = specific weight of water (9810 N/m3)
h = head of water (m)
P
 o  out
Pin
P=
Qwh
k
where:
P = theoretical power of the turbine (horsepower)
K = constant for conversion factor
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Discharge of
water
ft3/sec
ft3/min
m3/sec
m3/min

Specific weight
(water)
62.4 lbs/ft3
62.4 lbs/ft3
9810 N/m3
9810 N/m3
Head or
Elevation
feet
feet
meter
meter
Constant for
conversion (k)
550
33000
746
44760
The Law of Proportionality of turbines of varying sizes (the variation of
power speed and discharge with runner size and head):
For constant runner
For variable diameter and
For constant head
diameter
head
P ∞ h3/2
P ∞ d2
P ∞ h3/2d2
N ∞ h1/2
N ∞ 1/d
N ∞ h1/2/d
Q ∞ h1/2
Q ∞ d2
Q ∞ h1/2/d2
where:
d = nominal diameter of the turbine runner
Basic Conversions:
Volume:
1L= 0.03531ft2 = 0.2642 gal.
= 0.22 British Gallon = 0.2642 US Gallon
1ft3 = 7.481 gal. = 28.3 liters
1 acre-ft. = 43,560 ft3
1 Gallon = 4 quarts = 3.79 liters
1 Quart = 2 pints = 32 fluid ounces
1 Quart = 0.95 liter
1 Pint = 0.47 liter
1 British gallon = 1.2 US gallons
1 Pound = 16 Ounces
1 cu. Meter = 1,000 liters = 35.5 cu.feet
Area:
1 ha = 104 m2 = 2.47 acres
1 acre = 4047 m2 = 43,560 ft2 = 0.405 ha
1 are = 100 sq.m
1 centare = 1 sq m.
Length (height for the head)
1 Km = 3281 ft. = 0.54 naut. mile = 0.6214 statute mile
1 naut.mile = 6080 ft.
1 statute mile = 5280 feet
1 mile = 5280 ft.
1 yd =3 ft.
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1m = 3.281 ft.
1cm = 0.3937 in
1 cable length = 720 ft.
1 fathom = 6 ft.
1 span = 9 inches
1 vara = 33 1/3 inches
1 furlong = 201.1684 meters
Velocity (water flow rate to be used) Q
1mph = 1.467 ft/s
= 0.8684 knot
1 knot = 1.688 ft/s
1km/h = 0.6214 mph

Specific Speed – is the speed of a hydraulic turbine to generate one
horsepower under a head of one foot.
N=
Nsh5/4
√BHP
where:
Ns = specific speed of the turbine (rpm)
N = actual speed of the turbine (rpm)
h = head of water available (foot)
BHP = brake horsepower output of the turbine
a. Gross Head, hg
Gross Head is the difference between head water and tail water elevation.
hg =hhw - htw
where:
hg= gross head
hhw = head water elevation
htw = tail water elevation
b. Friction Head Loss, hf
Friction Head Loss is the head lost by the flow in a stream or conduit due to
frictional disturbances set up by the moving fluid and its containing conduit
and by intermolecular friction.
a. Using Darcy’s Equation:
hf =
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b. Using Morse Equation
hf =
37
where:
hf = friction head (in m)
f = coefficient of friction
L = total length in m
g = 9.81 m/s2
D = inside diameter ( in m)
Note: Friction head loss is usually expressed as a percentage of the gross
head.
c. Net head or Effective head, h:
Net Head or Effective head is the difference between the gross head and
the friction head loss.
h = hg -hf
d. Penstock efficiency,ep:
Penstock efficiency is the ratio of the net head to the gross head.
ep =
e. Volume flow rate of water, Q:
The volume flow rate of water is the product of the velocity and the crosssectional area.
Q = AV
f. Water Power, Pw
Water power is the power generated from an elevated water supply by the
use of hydraulic turbines.
Pw= γQh
Where:
γ = specific weight of water = 9.81 kN/m3
g. Turbine efficiency, et:
Turbine efficiency is the ratio of the turbine power output to the water
power output.
et =
et =
or
Pt = γQhet
h. Electrical or Generator Efficiency, egen:
Electrical or Generator Efficiency is the ratio of the generator output to the
turbine power output.
eG =
egen =
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or
Pgen = Pt egen = γQhetegen
38
i.Generator Speed, N
N=
Where:
N = angular frequency, rpm
f = frequency (usually 60 hertz)
P = no. of poles (even number)
j. Hydraulic Efficiency, eh:
Hydraulic efficiency is the ratio of the utilized head to
the net head.
eh =
Where:
hw = utilized head
h = net head
k. Head of Impulse Turbine (Pelton)
Impulse Turbine is a power-generation prime mover in which fluid under
pressure enters a stationary nozzle where its pressure (potential) energy is
converted to velocity (kinetic) energy and absorbed by the rotor.
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l. Head of Reaction Turbine (Francis and Kaplan)
Reaction Turbine is a power – generation prime mover utilizing the steady
flow principle of fluid acceleration where nozzles are mounted on the
moving element.
m. Peripheral Coefficient, Φ
Peripheral coefficient is the ratio of the peripheral velocity (V p) to the
velocity of the jet (Vj).
Φ=
Φ=
Where:
D = diameter of runner
N = angular speed
H = net head
n. Specific Speed of Hydraulic Turbine, Ns
Specific speed is a number used to predict the performance of the hydraulic
turbines.
a. In English units:
where:
Ns =
N = angular speed, rpm
h = net head, ft
b. In SI units:
where:
Ns =
N = angular speed, rpm
h = net head, m
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40
o. Total efficiency, etotal:
etotal = ehemev
where:
eh = hydraulic efficiency
em = mechanical efficiency
ev = volumetric efficiency
p. Turbine type recommendation based on head
Net Head
Up tp 70 ft
70 ft to 110 ft
110 ft to 800 ft
800 ft to 1300 ft
1300 ft and above
Type of Turbine
Propeller Type
Propeller or Francis Type
Francis Type
Francis Type or Impulse
Type
Impulse Type
Example Problems:
Problem 1:
The volumetric flow of a river is 20 m 3/s. A hydraulic turbine is
proposed to be installed in the power plant site with an available head of
24 m. If the specific speed of the turbine is 68 rpm and with an
efficiency of 88%, determine the operating speed of the turbine.
Solution:
Qwh
(20)(9810)(24)
=
= 6312.06 hp
746
746
Pout = ŋPin = 0.88(6312.06) = 5554.6 hp
5/4
24 m 3.281
(68)
(
)
5/4
x
Nsh
1m
N=
=
√BHP
√5554.6
N = 214 rpm
Pin =
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Problem 2:
For a given head, the diameter of the runner of a hydraulic
turbine is 10 m while developing a BHP of 200. If the diameter of the
runner is increased to 15 m, what will be its new brake horsepower
rating?
Solution:
Using the law of proportionality:
P ∞ d2
for constant head
P1
d2
P2 = P1
= (
)
P2
d1
2
15
P2 = (200)
(
)
10
P2 = 450 BHP
(
d2
d1
)
2
Problem 3:
A hydroelectric plant generates 100 MW at an available head of
200 m and at overall efficiency of 75%, what quantity of water in cubic
meter per second is required?
Solution:
Pout
Pin
100
Pin =
0.75
Qwh
Pin =
Pin =
ŋ=
Pout
ŋ
= 133.33 MW
Q=
Pin
wh
133.33 x 106
(9810)(200)
Q = 67.956 m3 per sec
Q=
Problem 4:
A power plant gets water from dam from a height of 122.45 m at
the rate of 1000 cubic meters per minute. If the output of the plant is
15,000 kW, what is the plant efficiency?
Solution:
Pin =
Qwh
44760
Pout
=
Pin
ŋ = 74.92%
ŋ=
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(1000)(9810)(122.45)
= 26,837.23 hp
44760
15,000
X 100%
26837.23 (0.746)
=
42

Characteristics of cascaded units
ŋ overall =
Poutput(GEN)
Pinput(PM)
ŋoverall = (ŋPM)( ŋGEN)
where:
Poutput(GEN) = output power of the generator (watt)
Pinput(GEN) = input power of the generator (watt)
Poutput(PM) = output power of the prime mover (watt)
Pinput(GEN) = input power of the prime mover (watt)
ŋPM = prime mover efficiency
ŋGEN = generator efficiency
ŋoverall = total or overall efficiency of the cascaded units
Problem 5:
A waterfall is 60 meters high. It discharges at a constant rate of
1.0 cubic meter per second. A mini – hydroelectric plant is to be
constructed below the waterfalls. The turbine efficiency is 80% and the
generator efficiency is 95%. Calculate the kW output of the generator.
Solution:
Pin = Qwh = (1)(9810)(60) = 588.6 kW
Poutput = Pinput [(ŋturbine)(ŋgenerator)]
Poutput = 588.6(0.8)(0.95)
Poutput = 447.336 kW

Relationship between pulley diameter and speed
N1
N2
=
d2
d1
Note: The speed of a pulley is inversely proportional to the diameter
where:
N1 = speed of the driving pulley
N2 = speed of the driven pulley
d1 = diameter of the driving pulley
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d2 = diameter of the driven pulley
Problem 6:
A generator is rated 600 kVA, 240 V, 60 Hz, 3-phase, 6-poles and
wye connected. What will be the speed of the driving pulley if the driven
and driving pulleys are 1 ft and 2 ft in diameter respectively?
120f
N2
=
P
120(60)
N2
=
6
N2 = 1200 rpm
N1
N2
=
d2
d1
d2
)
d1
N1 = 600 rpm
N1 = N2

synchronous speed
(
= 1200
(
1
)
2
Power developed (horsepower) on the
driven pulley of a generator
P =
2πNFr
33,000
where:
N = speed of the driven pulley (rpm)
r = radius of the driven pulley (foot)
F = force exerted by the driving belt (pound)
P = power developed on the driven pulley (horsepower)
In SI units,
P =
2πNFr
44,760
where:
N = speed of the driven pulley (rpm)
r = radius of the driven pulley (meter)
F = force exerted by the driving belt (newton)
P = power developed on the driven pulley (horsepower)
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Problem 7:
The pulley of an old gen-set has a diameter of 20 inches. The belt
exerts a pull of 353 pounds on the pulley. The gen-set runs at 900 rpm.
What is the approximate kW rating of the gen-set?
1 ft
2π(900)(353) (10 in x
)
2πNFr
12 ft
P
=
=
33,000
33,000
P = 50.41 hp
0.746 kW
P = 50.41 hp x
1 hp
P = 37.605 kW
REMINDER!
In dealing with hydro problems, take note on the consistency of
units from SI to English units…MEV
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DIESEL POWER PLANT
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46
Diesel Engine – is an excellent prime mover for electric power generation in
capacities of 101 hp to 5070 hp. These are also widely used in hotels, utility
companies, municipalities and private industries.
The design of diesel electric power plant includes the following elements;
the stationary diesel engine, fuel system, lubricating system, cooling
system, intake and exhaust system, starting system, and the governing
system.
Applications of Internal Combustion Engines (ICE)
1. Portable generating units in which may be moved from site
to site where electrical power is required temporarily.
2. Standby units, normally idle, which can be activated when
there is a failure of central station power where
interruption would mean financial losses or danger.
3. Engine-generator installed in power plants where they are
normal primary source of electrical power generated for
public, industrial, or institutional consumption.
Types of Internal Combustion Engines (ICE)
1. Gasoline Engine
2. Gas Engine
3. Hesselman Engine
4. Injection Gasoline Engine
5. Injection Gas Engine
6. Vaporizing Oil Engine
7. Diesel Engine
8. Gas-Diesel Engine
9. Dual-Fuel Engine
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Advantages of the diesel engines:
1.
2.
3.
4.
5.
6.
The cost of diesel fuel is cheaper than other fuels.
It needs no long warming up.
It has no standby losses.
It has uniformly high efficiency of all size.
It has a simple plant lay out.
It needs no large water supply.
PERFORMANCE OF DIESEL POWER PLANT
1. Heat supplied by fuel, Q,:
Q s = m fQ h
Where:
mf = mass flow rate of fuel
Qh = heating value of fuel
Sample problem: (Calculating the heat supplied by Fuel)
What is the heat that can be supplied by 39 kg/hr diesel fuel whose heating
value is 43,912 kJ/kg?
Solution:
Q s = m fQ h
Qs = (39/3600)43,912 kJ/s or kW
Answer:
Qs = 475.71 kW
2. Air – Fuel Ratio, A/F:
=
Where:
ma = mass of air
mf = mass of fuel
Sample Problem: ( Calculating the Air-Fuel Ratio)
The density of air entering the engine is 1.19 kg/m3 whose volume
flow rate is 0.15 m3/s. If the mass flow rate of fuel is 121.38 kg/hr, what is
the air-fuel ratio?
Solution:
=
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48
Where: mf = 121.38 kg/hr = 0.034 kg/s
Solving for the mass of air; ma:
ma = ρaVa = 1.19(0.51) = 0.6069 kg/s
then;
=
Answer:
= 18 kg air/kg fuel
3. Piston Displacement, VD:
Piston displacement is the volume displaced by the piston as it moves
from top dead center to bottom dead center.
VD = (
) LN ncnp
Where:
D = bore or diameter of the piston
L = length of stroke or stroke
N = engine speed
nc = no. of cylinders
np = no. of piston actions
NOTE: for 4-stroke engine
divide the engine speed “N”
by 2.
Sample problem: (Calculating the Piston displacement)
Determine the piston displacement of 35 cm x 45 cm, 4 –stroke, 1200
rpm, 8-cylinder diesel engine?
Solution:
Vd =
Vd =(
)LNncnp
)(0.45)[
](8)
Answer: Vd = 3.46 m3/s
4. Piston Speed:
Piston speed is the total distance a piston travels in a given time.
Piston speed = 2 LN
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49
Where:
2L = distance travelled by piston in one revolution
N = angular speed in rpm or rps
Sample problem: (calculating the piston speed of a Diesel Engine)
Solution:
The piston speed (V):
Piston speed = 2LN = 2(0.35)(1100/60)
Answer: Piston speed = 12.83 m/s
5. Indicated Power, Pind:
Indicated power is the power developed by the action of piston within
a cylinder, so named because it is measured by use of an indicator.
Pind = PmiVD
Where:
Pmi = indicated mean effective pressure
@ Calculating the indicated mean effective pressure using data provided
by the planimeter.
Planimeter measures the area of actual P – V diagram traced by engine
indicator
Pmi =
Where:
Ac = area of indicator card diagram
Sc = spring scale
Lc = length of indicator card diagram
@ If working cylinder and crankcase are to be considered:
Pmi = (
)wc – (
)cc
NOTE: Crankcase compression is
Used for scavenging.
Where:
wc = working cylinder
cc = crankcase
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Sample problem: (Calculating the indicated power of Diesel Engine)
The cylinder diameters of an eight – cylinder, single acting, four-stroke
diesel engine are 750 mm and the stroke is 1125 mm. The indicated mean
effective pressure in the cylinder is 586 kPa, when the engine is running at
110 rpm. Calculate the indicated power.
Soulution:
Pind = Pmi L A N nc np
= 586(1.25)[ (0.75)2][
Answer
](8)(1)
= 2,135.82 kW
6.
Brake Power, Pb
Brake power is the delivered power to a shaft. Brake power is always
less than the indicated power for a given engine, because some of the
work developed by the cylinders is used cylinders is used to overcome
the friction of running the engine. The often term of brake is shaft
power.
Pb =
and T=Fr
NOTE:Brake power is calculated
Where:
T = brake torque
N = engine rotative speed in rpm
F = brake force or brake load
R = brake arm or torque arm
using either prony brake or
dynamometer.
Sample problem: (Calculating the Brake power of a Diesel Engine)
The flywheel of a rope brake is 1.22 m diameter and the rope is 24mm
diameter. When the engine is running at 250rev/min the load on the brake
is 480 N on one end of the rope and 84 N on the other end. Calculate the
brake power.
Solution:
Pb =
The load (F) on the brake (F):
F = 480 – 84 = 395=6 N
The radius (r):
r = (1220 + 24) = 622 mm
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51
r = 0.622 m
The torque (T):
T = 396 (0.622) = 246.31 N-m
The Brake power (Pb):
Pb =
= 6,448.43 W
Answer Pb = 6.448kW
@ Brake power in terms of brake mean effective pressure and piston
displacement:
Pb = PmbVD
Where:
Pmb = brake mean effective pressure
Pb = brake power
VD = piston displacement
Sample problem: (Calculating the brake mean effective pressure)
A single acting, 8 cylinder, 4 stroke cycle diesel engine with a bore to stroke
of 142.1-mm x 210.45-mm operates at 1200 rpm. The load on the brake arm
at 1 m is 150 kg. What is the brake mean effective pressure in kPa?
Solution:
Pb = Pmb VD
The volume displacement (VD):
VD = [ (0.1412)2](0.21045)[
](8)(1)
= 0.267 m3/s
The brakepower (Pb):
Pb =
Where; the torque(T):
T = [150(0.00981)](1)
T = 1.4715 kN-m
The brakepower (Pb):
Pb =
Pb = 184.91kW
The mean effective pressure(Pmb):
184.91 = Pmb (0.267)
Answer Pmb = 692.55 kPa
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52
7. Friction Power, Pf:
Friction power is the power dissipated in an engine through friction.
Friction Power = Indicated Power – Brake Power
Pf
=
Pind
Pb
@ Calculating friction power using Morse test method applied to multicylinder engines:
Consider a six – cylinder engine
The indicated power if all six cylinders are firing
Pind(6) = Pb(6) – Pf(6)
If one cylinder is cut, or five cylinder are firing
Pind(5) = Pb(5) – Pf(5)
Derived from two equations above, equating the friction power:
Pind(6) – Pind(5) = Pind(1) = Pb(6) – Pb(5)
Thus, the total indicated power for six cylinder engine is,
Pind(6) = 6(Pb(6) – Pb(5))
Note: no matter how many cylinders are firing; the friction power is
constant.
(Pf(6) = Pf(5))
Sample problem: (Calculating the indicated power)
A six cylinder, four stroke diesel engine with 76 mm bore x 89 mm stroke
was run in the laboratory at 200rpm, when it was found that the engine
torque was 153.5 N-m with all cylinders firing but 123 N-m when one
cylinder was out. The engine consumed 12.2 kg of fuel per hour with a
heating value of 54,120 kJ/kg of air at 15.60C. Determine the indicated
power.
Solution:
Pind(6) = 6(Pb(6) –Pb(5))
Pind(6) =6[
-
]
Answer
Pind(6) = 38.83 kW
8. Mechanical Efficiency, em
Mechanical Efficiency is the ratio of the brake power to the indicated
power.
em =
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or
em =
53
where:
Pb = brake power
Pind = indicated power
Pmi = indicated mean effective pressure
Pmb = indicated mean brake power
Sample problem: (Calculating the indicated power)
What is the mechanical efficiency of a 0.5 MW diesel engine if the
friction power is 70 kW.
Solution:
The mechanical efficiency:
em =
The indicated power:
Pind = Pb + Pf
Pind = 500 + 70 kW
Pind = 570 kW
Substitute:
em =
Answer:
em = 0.8772 or 87.72 %
9. Electrical or Generator Efficiency, egen
Electrical or Generator Efficiency is the ratio of the generator power to
the brake power.
egen =
where:
Pgen = Generator Power
Pb = Brake Power
Sample problem: (Calculating the generator efficiency given the
brakepower)
A 16-cylinder V – type diesel engine is directly coupled to ¾ MW AC
generator. Calculate the generator efficiency of the engine if the brake
power is 833.33 kW.
Solution:
egen =
=
Answer:
egen = 0.9 or 90%
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54
10.
Thermal efficiencies,et
Thermal Efficiencies is the ratio of the work done by heat engine to the
heat supplied by the fuel.
a. Indicated thermal efficiency, eti
Indicated thermal efficiency is the ratio indicated power to the
heat supplied by the fuel.
eti =
Sample problem: (Calculating the indicated thermal efficiency)
A 3.5 MW diesel power plant uses 3500 gallons in 24 hours period. What
is the indicated thermal efficiency of the engine if the generator and
mechanical efficiencies are 90% and 92% respectively? Oil is 280 API.
Solution:
eti =
Solving for Qh:
Qh = 41,130 + 139.6 (28)
= 45038.80 kJ/kg
Solving for mfuel:
S.G.15.6 oC =
= 0.887
ρfuel = 0.887(1000) = 887 kg/m3
mfuel = ρfuel Vfuel
= 887[3500(
)] = 11750.53 kg
Mass of fuel per second:
mfuel =
= 0.136 kg/s
Then;
eti =
Answer: eti = 0.5714 04 57.14%
b. Brake thermal efficiency,eth
Brake thermal efficiency is the ratio of the brake power to the heat
supplied by the fuel.
eth =
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55
Sample problem:
(Calculating the brake thermal efficiency)
A supercharged six cylinder, four stroke cycle diesel engine of 10.48
cm bore and 12.7 cm stroke has a compression ratio of 15. When it
tested on a dynamometer with a 53.34 cm arm at 2500 rpm, the scale
reads 81.65 kg, 2.86 kg of fuel of 45822.20 kJ/kg heating value are
burned during a 6 min. test and air metered to the cylinders at the
rate of 0.182 kg/s. Find the brake thermal efficiency.
Solution:
etb =
Solving for Pb:
T = 81.65(0.00981)(0.5334)
T = 0.42725 N-m
Pb =
=
= 111. 854 kW
Solving for mf:
mf =
= 0.00794 kg/s
thus;
etb =
= 0.307
c. Combined or Over – all thermal efficiency, etc
Combined or over – all thermal efficiency is the ratio of the electrical
or generator power to heat supplied by the fuel.
etc =
Sample problem: (calculating the combined thermal efficiency)
A 16 – cylinder V –type diesel engine is directly coupled to 1 MW AC
generator. Calculate the combined thermal efficiency if the heat
supplied by the fuel is 2.5 MW.
Solution:
etc =
=
= 0.40 or 40%
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56
11.
Engine efficiencies, et
a. Indicated engine efficiency, eei
Indicated engine efficiency is the ratio of the indicated thermal
efficiency to the ideal thermal efficiency.
eei =
Sample problem: (Calculating the cycle efficiency)
What is the indicated engine efficiency of diesel engine if the
indicated thermal efficiency is 35% and the cycle efficiency is 45%.
Solution:
Answer:
eei =
eei = 77.78%
b. Brake engine efficiency: eeb
Brake engine efficiency is the ratio of the brake thermal efficiency to
the ideal thermal efficiency.
eeb =
Sample problem: (Calculating engine efficiency)
A 500 kW diesel has a rate of 12,000 kJ/kW-hr. The compression ratio
is 16:1 cut of ratio of 2.3. Assume k = 1.32. Calculate the engine
efficiency based on the output of 500kW.
Solution:
Let: eeb = brake engine efficiency
eeb =
The brake engine efficiency,etb:
etb =
= 0.30 or 30%
Solving for cycle efficiency, e
e = 1-
[
Answer
e = 0.5195 or 51.95%
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The engine efficiency, eeb:
eeb =
= .5777
eeb = 57.77%
c. Combined oe over – all engine efficiency, eec
Combined or over –all engine efficiency is the ratio of the combined or
over- all thermal efficiency to the ideal thermal efficiency.
eec =
Sample problem:
If the over-all thermal efficiency of a diesel engine is 33% and the
diesel cycle efficiency 49%, what is the combined engine efficiency?
Solution:
eec = 0.33/0.49
Answer:
eec = 67.35%
12.
Volumetric efficiency, ev
Volumetric efficiency is the ratio of the volume of air drawn into a
cylinder to the piston displacement.
ev =
where:
Va =
VD = piston displacement
Sample problem:
Determine the volumetric efficiency of 35 cm x 45 cm, 4 stroke 1200
rpm, 8 cylinder diesel engine if the air drawn in the engine is 3 m 3/s?
Solution:
The volume displacement, Vd:
Vd = (
=(
)LNn
)(0.45)[
Vd = 3.46 m3/s
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The volumetric efficiency, ev
ev =
Answer
ev =
= 0.87 or 87%
13.
Specific Fuel Consumption, m
Specific fuel consumption is the mass flow rate of fuel consumed per
unit power developed. It is also known as specific propellant
consumption.
a. Indicated specific fuel consumption, mi
mi =
Sample problem: (Calculating the indicated specific fuel
consumption)
What is the indicated specific fuel consumption of a six cylinder, four
stroke diesel engine with 76 mm bore x 89 mm stroke and indicated
power of 38.83 kW if the engine consumed 12.2 kg/hr of fuel?
Solution:
mi =
b. Brake Specific fuel consumption, mb
mb =
Sample problem: (Calculating the brake specific fuel consumption)
A four-stroke, 8-cylinder engine with bore and stroke of 9 in and 12 in
respectively and speed of 950 rpm has a brake mean effective
pressure of 164 psi. What is the brake specific fuel consumption in lb
per hp-hr if the engine consumed 468.55 lbs of fuel per hour?
Solution:
The brake mean effective pressure consumption, m b
mb =
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Solving for Pb:
Pb =
Where:
A=
( )2 = 0.44 ft2
Pb =
= 1,201.40 Hp
Then:
mb =
Answer: mb = 0.39 lb/ hp-hr
c. Combined or over – all Specific consumption ,mc
mc =
Sample problem: (Calculating the combined specific fuel consumption)
A 16-cylinder V - type diesel engine is directly coupled to 1 MW AC
generator. Calculate the combined specific fuel consumption if the
fuel consumed is 1 kg/s.
Solution:
mc =
Answer:
mc = 3.6 kg/kW-hr
14.
Heat Rate, HR
Heat rate is the rate of energy charge per unit of power. To calculate
the heat rate is to multiply the specific fuel consumption by the
heating value of the fuel.
A. Indicated Heat Rate, HRi:
HRi =
Sample problem: (Calculating the indicated heat rate)
A four cylinder, 4 stroke cycle, 20 cm x 25 cm x 55 rpm diesel engine
has a mean effective pressure of 150 psi. If the heat supplied by the
fuel is 50 kW, what is the indicated heat rate?
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Solution:
HRi =
Solving for Indicated power, Pind:
Pind = Pm L A N n
Pind = [150(
)](0.25)[
][
](4)
Pind = 148.875 kW
then;
HRi =
Answer:
HRi = 1209.07
B. Brake Heat Rate, HRb:
HRb =
Sample problem: (Calculating the brake heat rate)
A 305 mm 475 mm four stroke single acting diesel engine is rated at
150 kW at 2600 rpm. Fuel consumption at rated load is 0.26 kg/kW-hr
with a heating value of 43,912 kJ/kg. Calculate the brake heat rate.
Solution:
HRb =
HRb = mcQh
HRb = (0.26 kg/kW-hr)(43,912 kJ/kg)
Answer:
HRb = 11,417.12 kJ/kW-hr
C. Combined or over –all Heat Rate, HRc:
HRc =
Sample problem: (Calculating the combined heat rate)
The kilowatt output of a generator coupled to a diesel engine is 1.5
MW . If the mass of fuel with heating value 45,000 kJ/kg consumed by
the engine is 0.04 kg/s, what is the combined heat rate?
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Solution:
HRc =
Answer:
HRc = 4,320
15.
Generator Speed , N
N=
Where:
N = speed in rpm
f = frequency = 60 Hz(if not given)
P = no. of poles
Sample problem: (Calculating the generator speed)
An eight-cylinder , two cycle, single acting diesel engine rated at
1250 Hp at standard condition is to be directly coupled to a 24-pole
alternator, 3 phase, 60 cycles. What is the generator speed?
Solution:
N=
=
Answer:
N = 300 rpm
16.
Energy Stream
ITEM
1. Useful
Output
(brake output)
2. Cooling Loss
3. Exhaust Loss
4. Radiation,
Friction and
Unaccounted
Losses
Total
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HEATLOSS
(kJ/hr)
3600 Pb
HEAT INPUT
(kJ/kg)
mfQ f
mwCwΔTw
M fQ h
mgCρgΔTg
M fQ h
By difference
M fQ h
(%)
COOLING LOSS
100 – ∑ 1 to 3
100%
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17.
Diesel Engine with closed cooling system
By Energy Balance
Qwj = Qw
mwjCpwjΔTwj = mwCpwΔTw
mwjΔTwj = mwΔTw
18.
Waste Heat Recovery Boiler
By energy balance
Qsteam = Qgas
Ms(hs-hf) = mgCpgΔtg
Ms =
Boiler efficiency, eboiler:
eboiler =
19.
=
Engine operated at high altitudes
P = Ps (
P = Ps (
) SI units
)
English units
Where:
Ps = standard power or power at sea level
Pact = new pressure or actual barometric pressure in in.Hg
T = new temperature or actual absolute temperature in 0R
Note:
1. The decrease is pressure is approximately 1 in.Hg per
1000ft elevation.
Pact = 29/92 -
in.Hg
2. The decrease in temperature is approximately 3.60R per
1000ft elevation.
T = 520 –
0
R
Sample problem: (Calculating the maximum power delivered at a given
elevation)
What maximum power can be delivered by 1500 kW engine at 2800 ft
elevation considering the pressure effect alone.
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Solution:
P = Ps (
)
Solving for Pact
Pactual
= 29.92 -
= 29.92 -
= 27.12 in.Hg
T = 5200R (for ther’s no effect in temperature)
Answer: P = Ps (
)
= 1,359.63 kW.
GEOTHERMAL POWER PLANT
Geothermal Power Plant – is the facility in which the
electrical energy is produced from hot spring, steam vent or
geyser.
Geothermal Energy – is heat energy naturally occurring with
the earth. It comes from two words “geo” meaning earth and
“thermal” meaning concerning heat.
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Magma – a molten metal within the earth which basically
nickel-iron in composition whose stored energy heats the
surrounding water thereby producing steam or hot water. Its
temperature reaches as high as 12000C.
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Well-bore product the effluent coming out from the
geothermal well produced after drilling. This can be purely
steam or hot water, or a mixture of both.
Steam-dominated geothermal field refers to a geothermal
plant with its well producing all steam as the well-bore
product.
Liquid-dominated geothermal field refers to a geothermal
plant in which the well-bore product is practically all hot
water (Pressurized).
Fumarole is a crack in the earth through which geothermal
substance passes.
TYPES OF GEOTHERMAL PLANT
1. Dry or Superheated Geothermal Plant:
2. Separated Steam or “Single Flashed Geothermal Plant”:
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3. Separated Steam/Hot-Water-Flash or “Double Flash”
Geothermal Plant
4. Single Flashed Plant with Pumped Wells:
5. Binary Geothermal Plant
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Performance and Design of Geothermal Power Plant
1. Mass flow rate of steam, ms :
m s = x mg
where:
mg =mass of ground water
x = quality after throlling (Process ① →
②)
Solving for x:
h1 = h2 = hf2 + xhfg2
2. Turbine Output:
Wt = ms(h3 – h4) et
Where:
et = turbine efficiency
3. Generator Output:
Wgen = Turbine Output x Generation Eff.
Wgen = ms (h3 - h4) etegen
4. Heat Rejected in the Condenser:
QR = ms (h4 – h5)
5. Over-all Plant Efficiency:
ec =
ec =
SOURCES OF GEOTHERMAL ENERGY
1. Hydrothermal Fluids – basically made up of hot
water, steam and minerals. It is the only form of
energy currently being tapped for significant
commercial heat and electric energy supply.
2. Geopressurized brines – represent a special subset
of hydrothermal fluids typically found at depths
exceeding at 3km and is characterized as hot water
existing at pressure above the normal hydraulic
gradient and containing dissolved methane.
3. Hot dry rock – is a water free, impermeable rock at
high temperature and practically drilling depth to
extract energy, high pressure water maybe injected
through one or more wells to create new to enhance
existing natural fracture system with limited access
to ground water flow.
4. Magma – is characterized by or practically molten
rock with temperatures reaching as high as 12000C.
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THE GEOTHERMAL POWER PLANT IN THE PHILIPPINES
1. Tiwi Geothermal Power Plant, 330 MW (Albay)
2. Makiling – Banahaw (Mak-Ban) Geothermal Power
Plant, 330 MW (Los Banos , Laguna)
3. Tongonan Geothermal Power Plant, 112.5 MW (Leyte)
4. Palimpinon Dauin Geothermal Power Plant, 112.5 MW
(Negros Oriental)
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NUCLEAR POWER PLANT
NUCLEAR POWER PLANT is a power plant in which nuclear energy is
converted into heat to be used in producing steam for turbines, which in
turn drive generators that produce electric power.
TYPICAL NUCLEAR POWER PLANT
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Fuel Core is a radioactive material, U235 and U238 which is source of
energy.
Moderator is a device that slows down the neutrons to thermal
energy, made of Carbon and Beryllium.
Control Rods are boron coated steel rods used to control the reactor.
Reflector is a device made of lead or carbon which surrounds the
core to bounds back any leakage of neutrons.
Thermal Shield is a device that prevents escape of radiation from
reactor vessel.
Reactor Drum is a device that encloses the fuel core and
components.
Biological Shield is a concrete or lead which absorbs any leakage of
radiation and protects operators from exposure to radioactivity.
Control Cubicle is a device that contains the meters that show the
operating quantities in the reactor.
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Containment Vessel is a concrete the prevent the spread of radiation
in case of major explosion.
Coolant is a substance that absorbs the heat from the fuel core then
releases the heat to the water in the steam generator.
Coolant Pump is a device (pump) that circulates the coolant.
Turbine generator is a device that generates the electric power.
Condenser is a device that converts steam coming from the turbine
into liquid.
Feedwater Pump a device that delivers the feedwater to the steam
generator.
TYPES OF NUCLEAR POWER REACTORS
1. PRESSURIZED WATER NUCLEAR POWER REACTOR (PWR)
A pressurized water reactor uses water under pressure as
both the coolant and moderator in the reactor. The water
pressure must be sufficiently high so that we can have
water at 550 -6600F without boiling in the core. The fission
of the fuel produces heat that is carried away by circulating
water under pressure.
2. FAST-BREEDER NUCLEAR POWER REACTOR
A fast - breeder reactor depends upon the fission of fast
neutrons rather than thermal neutrons. Therefore, no
moderator or moderating material can be used. Further the
term “breeder” implies that the reactor produces more fuel
due to the absorption of neutrons than is burned up. This is
possible since on the average 2.5 neutrons are produced
due to its fission of a uranium atom and only one of this is
needed to keep the chain reaction going.
3. ADVANCE GAS – COOLED NUCLEAR POWER REACTOR
(AGCR)
Gas cooled nuclear power reactor uses carbon dioxide as
coolant.
Advantages of Gas Cooled Power Reactor
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1. High temperature is possible at low pressure since
boiling is not a problem.
2. Plant
thermal
efficiencies
increasing
cycle
temperatures Steam can be produced in heat
exchangers heated by gas.
3. Gases can be nonhazardous.
Disadvantages of Gas Cooled Power Reactor
1. The power required to pump the gas compared to
that for pumping liquid.
2. Large flow rates and high velocities are required.
3. Care must be taken not to use gases that are
corrosive to moderating materials at the high
temperatures employed.
4. LIQUID – COOLED NUCLEAR POWER REACTOR
There are two types of liquid – cooled reactor: the
water-cooled reactor and the liquid metal – cooled
reactor.
Advantages of water-cooled reactor
1. Low cost of coolant
2. Coolant is also a moderator
3. Pumping power is small and design of pumps is
relatively simple
4. Low viscosity
Disadvantages of water-cooled reactor
1. High-pressures are required to obtain high
temperatures.
2. Detrimental to large conversion ratios because
hydrogen in the water is a good absorber.
3. Problem of corrosion of materials in contact with the
water.
Advantages of liquid metal-cooled reactor
1. High thermal conductivity
2. Can operate atmospheric pressure
3. Good heat transfer characteristic
4. Stable at high temperatures
Disadvantages of liquid metal-cooled reactor
1. High Cost
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2. Danger of reaction with air or water makes for
difficulty in handling.
5. BOILING WATER NUCLEAR POWER REACTOR (BWR)
Boiling water nuclear power reactor is the simplest
nuclear reactor. The feed water from the power turbine
goes directly into the reactor and picks up the heat from
the fuel core, thus the feedwater also serves as the
coolant.
Advantages of boiling water reactor
1. Moderator and coolant are the same (water). This is
the same as in a pressurized-water reactor.
2. The steam formed goes directly to the turbine, thus
eliminating the extra heat exchange present in the
pressurized water plant.
3. The pumping power is much less than that for
pressurized water plants because of lower operating
pressures to produce the same temperature and
elimination of the extra loop.
4. The stresses in the pressure vessel are lower because
of lower operating pressures, thus decreasing costs.
Disadvantages of boiling water reactor
1. Presence of the steam in the reactor will result in
more neutron leakage and will require a more highly
enriched fuel.
(For example: enrichment of U235)
2. There will be some depositing radioactive elements in
the pump, turbine, and condensing. This can become a
hazard under normal maintenance procedures.
3. The leakage of the less dense steam is more apt to
occur than in a pressurized water reactor
Other Nuclear Reactors
6. Modulated Fuel Reactor (CANDU)
7. Heavy Water Reactor
8. Graphite Moderated Light Water
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Alternative Methods of disposing Nuclear Wastes






MODULE SUMMARY
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Sending nuclear waste into outer space.
Burying them in polar ice caps
Dumping them deep into the ocean
Shooting them into the sun.
Land Burial.
75
Pressurized Water Reactor
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1 Gram of Uranium Pellet
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REMINDER!
In module I, you have learned about the introduction of the different
power plant as to their sources, application and how they are very much
available in our country. That is, Philippines is one of the countries who has
the most expensive energy and that’s for a fact that we as future Electrical
Engineers must be aware of the availability of the different plants that is
and will be discovered and may be able to be available in the near future to
be able to shift to plants that is more reliable and can lessen the cost of
energy in every household.
There are three lessons in module I. Lesson 1 consists of the different
types of power plants as to its sources, most particularly to thermal plant
that has most essential part in our energy grid, and most important is as to
its source, components and operation. A part of it also, is the hydro energy
that plays the major role in the production of energy and of its kind. The
characteristics of hydro plant makes it very unique in nature because of its
environmental part that its duality. Why duality? first it produces energy,
second, the excess water can also be used for farmer’s irrigation purposes
for their plants and harvests.
Lesson 2 deals with the discussion of renewable & non-renewable
power plant. Base load plants in our country, whether we like it or not are
more of a non-renewable type. That is, they carry big load of energy as its
big share, and mostly has a reciprocating effect in our environment, mostly
coal fired and diesel plant has its big share in the energy industry.
Lesson 3 deals now with the different power plant operation and how
it operates and its components from a raw material to its finished product
which is the energy output produced by this plants. The efficiencies of these
plants must be measured and through its high energy input, its energy
output must be compensated to achieve the highest possible energy
efficiencies.
Congratulations! You have just studied Module I. now you are ready to
evaluate how much you have benefited from your reading by answering the
summative test. Good Luck!!! -mev
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
SUMMATIVE TEST
Essay:
1. Enumerate and discuss the different thermal power plants and how
they are interlinked with each other.
2. From the different thermal plants discussed above, which of these
plants exists but do not/did not operate due to what probable
reasons.
3. From the hydro power energy, which of these are the best possible
design to further develop and had an advantage over the other types
of hydro power plants.
4. From the different hydro power plants discussed above, which of
these plants do not exist, and what could be the probable cause?
Problem Solving:
solve the requirement/s of the said unknown parameters: (show you
solution in an orderly manner).
1. A sixteen-cylinder, two cycle, single acting diesel engine rated at
1500 Hp at standard condition is to be directly coupled to a generator
running @ 3 phase for 13,200 volts output, what would be the
possible number of poles necessary to run the diesel engine at 150
rpm?
2. A coal fired thermal power plant has the following data:
Coal consumption per day: 68.54 slugs
Units of energy per day: 4,000 KWhr
Calorific value of fuel used: 12,000 Kcal/Kg
Alternator Efficiency: 95%
Turbine Efficiency: 93%
Solve a.) specific heat combustion of fuel
b.) overall efficiency
3.A hydroelectric station has a turbine efficiency of 90% & generator
efficiency of 89%. The effective head of water is 150m. Calculate the
amount of volume of water used in delivering a load of 50 MW for 8 hours
of operation.
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MODULE II
POWER PLANT ECONOMICS
Lesson 1
Lesson 2
Lesson 3
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Other Sources of Energy
Introduction to Power Plant Economics
Load Determination & Load Graphs
80
MODULE II
POWER PLANT ECONOMICS

INTRODUCTION
This module presents Familiarization of other Power plants sources as
to their flow of operations, and will discuss the power plants economics as
to its use, load determination, graphs, interpretations and reserves.
POWER PLANT ENGINEERING AND DESIGN- It is the art of selecting and
placing the necessary power-generating equipment so that the maximum of
return will result from the minimum expenditure over the working life of
the plant; and the operation of the completed plant in manner to provide a
cheap, reliable, and continuous service.
OBJECTIVES
After studying the module, you should be able to:
1.Evaluate the importance of the different power plants economics.
2.Explain the different operations of other sources of a power plant.
3.Solve & analyze problems relative to power plant economics &
other sources of power plants.
4.Give synthesis the different existing power plants in the
Philippines as to their economics.

DIRECTIONS/ MODULE ORGANIZER
There are three lessons in the module. Read each lesson carefully
then answer the exercises/activities to find out how much you have
benefited from it. Work on these exercises carefully and submit your output
to your tutor or to the COE office.
In case you encounter difficulty, discuss this with your tutor during
the face-to-face meeting. If not contact your tutor at the
COE office.
Good luck and happy reading!!! -mev
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Lesson 1

Other Sources of Energy
Wind power Plant (Windmill Power Plant)
Windmills are any various mechanisms, such as mill, pump or electric
generator, operated by the force of wind against vanes or sails radiating
about a horizontal shafts.
(old type design of a windmill typically used for water distribution on land
fields.
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PERFORMANCE OF WIND POWER PLANT
1. Pump Power, Wp
Wp =
Where:
γ = specific weight of water
γ = 9.81 kN/m3
Q = volume flow rate
H = net head
ep = pump efficiency
2. Kinetic Energy, KEair
KEair = maVa2
Where:
ma = mass flow rate of air
Va = average velocity of air
3. Volume of air, Va
Va =
If wind velocity is given:
Volume = Area x Velocity
Va =
x velocity
4. Aerodynamic Efficiency, ea:
ea =
Calculating the Power Output of a wind:
Power  k Cp1 / 2  AV 3
Where:
P = Power output, kilowatts
Cp = Maximum power coefficient, ranging from 0.25 to 0.45, dimension less
(theoretical maximum = 0.59)
ρ = Air density, lb/ft3
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A = Rotor swept area, ft2 or π D2/4 (D is the rotor diameter in ft, π = 3.1416)
V = Wind speed, mph
k = 0.000133 A constant to yield power in kilowatts. (Multiplying the above
kilowatt answer by 1.340 converts it to horse- power [i.e., 1 kW = 1.340
horsepower]).
The rotor swept area, A, is important because the rotor is the part of the
turbine that captures the wind energy. So, the larger the rotor, the more
energy it can capture.
Note:
The air density, ρ, changes slightly with air temperature and with elevation.
The ratings for wind turbines are based on standard conditions of 59° F (15°
C) at sea level. A density correction should be made for higher elevations as
shown in the Air Density Change with Elevation graph. A correction for
temperature is typically not needed for predicting the long-term
performance of a wind turbine.
Although the calculation of wind power illustrates important features about
wind turbines, the best measure of wind turbine performance is annual
energy output. The difference between power and energy is that power
(kilowatts [kW]) is the rate at which electricity is consumed, while energy
(kilowatt-hours [kWh]) is the quantity consumed. An estimate of the annual
energy output from your wind turbine, kWh/year, is the best way to
determine whether a particular wind turbine and tower will produce enough
electricity.
To get a preliminary estimate of the performance of a particular wind
turbine, use the formula below.
AEO  k D 2 V 3
Where:
k = 0.01328
AEO = Annual energy output, kWh/year
D = Rotor diameter, feet
V = Annual average wind speed, mph
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OTHER NON-CONVENTIONAL POWER SOURCES
① Tidal Power
② Thermoionic Converter
③ Fuel Cell
④ Low Thermal Head Plant
⑤ Magneto Hydrodynamic Plant
(Non-Conventional Power Plant) – Renewable Energy
SOLAR POWER PLANT
Solar Power Plant is the conversion of the energy of the sun’s
radiation to useful work.
TYPES OF SOLAR COLLECTORS
① Flat plate
② Concentrating
③ Focusing
SOLAR ENERGY RECEIVED AT THE EARTH’S SURFACE
Es = qS ( 1 – I ) A
Where:
Qs = solar energy without atmospheric interference
i = atmospheric interference
A = surface area of solar collector
Qsun = Qw + PE + Qloss
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Other types of Solar Plant
• 1.OFF-GRID POWER PLANT
• -SMALL SCALE OR MICRO GRID PURPOSES
• 2. HYBRID POWER
• 3. SOLAR THERMAL PLANT
(Solar Thermal Plant)
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(Hybrid Power Systems)
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OFF – GRID SOLAR POWER SYSTEM
(Small scale to Micro Grid system)
An off-grid solar system is designed for the power needs of mid- to
large-size homes upto communities of independent power. Unlike gridtied solar systems, off-grid systems have no connection to the utility grid,
and must make all the electricity necessary to power your home and
community. Off-grid solar systems operate from the stored energy in a
battery bank.
Advantages:
Disadvantages:
1. Solar Materials are High in cost beacause of importation.
2. Solar deep cycle batteries have short life especially on an under
designed off-grid power.But for computed and properly designed
off-grid power can last up to 3-6 years.
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The light from the Sun, made up of packets of energy called photons,
falls onto a solar panel and creates an electric current through a process
called the photovoltaic effect. Each panel produces a relatively small
amount of energy, but can be linked together with other panels to produce
higher amounts of energy as a solar array. The electricity produced from a
solar panel (or array) is in the form of direct current (DC).
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Charge Controllers
Pulse-Width Modulation (PWM)
•
The PWM charge controller is a good low cost solution for small
systems only, when solar cell temperature is moderate to high
(between 45°C and 75°C).
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Maximum Power Point Tracking (MPPT)
•
Maximum Power Point Tracking
The MPPT controller will harvest more power from the solar array.
The performance advantage is substantial (10% to 40%) when the solar
cell temperature is low (below 45°C), or very high (above 75°C), or
when irradiance is very low.
Irradiance
•
irradiance is the radiant flux (power) received by a surface per unit
area. The SI unit of irradiance is the watt per square meter (W.m⁻²).
The CGS unit erg per square centimeter per second (erg·cm⁻²·s⁻¹)
BASIC STEPS HOW TO DESIGN AN OFF-GRID SOLAR POWER
SYSTEM
1. Figure out the input power to the inverter when the load is on to decide
how many inverters are needed, especially the rating needed for the set-up
of the system.
- inverters must be “pure sine wave” in nature
There are inverters in the market which they claim to be pure sine
wave, but actually they are modified sine wave which will eventually
damage your load system especially your sensitive electronic appliances.
2. Figure out the amperage the charge controller provides to the inverter to
decide how many charge controllers needed for your design.
3. Calculate how much energy the load consumes per day basis.
- note: your energy is expressed in KWHR
For example: An electric Fan of 50 watts will operate at 12-hour basis = 0.6
KWHR
4. Calculate how many solar panels you will need in your specific location.
Note: solar panels are dependent on your specific location such as, too
cloudy, rainy areas, too hot season, areas where cold season like Baguio or
tagaytay.
Mono crystalline – used for cold, rainy seasons like in Baguio because of
slightly higher efficiency than poly.
Poly crystalline – are used for hot seasons and dependable to sustain
voltage regulation and wattage in your solar panel.
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Note: Do not mix any different kind of solar panel, check the characteristics
and specifications, because different brands different in its specifications.
-doing this, some installers connect a series of 12v to obtain a 24v solar
panel requirement, doing this may damage the solar charge controller and
may lessen its life and even voids the warranty.
5. Calculate how many batteries you will need for the system to work
properly and not stressed out, to further prolong its natural life span.
- over and under charging your batteries could further destroy and
shorten your batteries life span and can void the warranty.
-
Without sunlight for certain number of days.
Efficiency of the inverter:
ηinverter = output power/ input power
such that; Input power = output power/ ηinverter
Example: (An assumption of 50% efficiency of inverter at its full stressed
load.
Input power = output power/ ηinverter = 100 w/ 50% = 200 watts
Since, the input power to the inverter (200w) is less than its rated power
(450w), we only need one inverter (but still dependent on the inverter
model and capability.
Number of charge controller
-we know the power coming out of the charge controller to the inverter is
200w
Such that; P= V.I
And I= P/V , therefore 200/12 = 16.67Amps
-note (12v system is used for this example, it could either be 12, 24, 48
volts system where the charge controller could automatically detect.
-since the charge controller capacity is 10amps, and we are drawing
16.67Amps, therefore; we need 2 charge controllers, or 20 Amps rating of
charge controller @ 80% safety factor.
LOAD ENERGY USAGE
Example:
-we know the lightbulb consumes 100w for 2hours per day, we need to
convert that to KWHR per day:
100w x 1kw/1000w x 2hrs./day = 0.2 KWHR per day
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That means to say, that the energy coming from the charge controller is:
ηinverter =
output energy
input energy
input energy =
output energy
 inverter
=
0.2kwhr
=
50%
0.4kwhr
day
DE (direct sunlight energy) receives ; 2.88 Kwhr/m2 per day, this unit is
sometimes called sun-hours, also having units of hours/day of full sunlight.
Since you will use the solar panel that produces 20w full sunlight, you can
use it to calculate the kwh/day that the solar panel produces, likewise,
0.0576kwhr
day
1kw
2.88hours
20w
=
x
x
=
1 solar panel
day
1 solar panel 1,000w
-
You know you have to provide a 0.4 kwhr/day, so; to calculate
how many solar panels you need, that is:
0.4 kwhr
day
= 6.94 solar panels
0.0576 kwhr
day
1 solar panel
that is approximately = 7 solar panels@ 20w each
Month
January
Feb
March
April
May
June
July
Aug
Sep
Oct.
Nov
Dec
year
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RESULTS
Solar
AC Energy
Radiation(irradiance)
(KWH)
2
(KWH/m /day)
3.48
344
4.46
394
4.48
450
5.13
454
5.34
460
5.73
473
5.57
468
5.70
485
5.02
420
4.71
425
3.50
317
2.88
270
4.69
4960
Energy Value
30.27
34.67
39.60
39.95
40.48
41.62
41.18
42.68
36.96
37.40
27.90
23.76
436.48
95
An example how irradiance is used in an example given as it is
measured as its lowest in the month of December.
Note:
-this is just a basis approximation based on an experiment done
HOW MANY BATTERIES NEEDED?
-
-
To start, decide on number of day/s (hours if more specific) that
the batteries will need to operate w/o sunlight. For example, I
choose 3days in this exercise. You know that the load uses 0.4
kwhr/day and over three days, this is approximately equal to 1.2
kwhr, what would be the battery size in ampere-hours & how
many pieces?, if 12V-5AH is only available?
To convert 1.2 kwhr in amp-hours, you do the following:
1.2 kwhr x
1,000 ampere  hours
1
 0.1 kilo ampere  hours x
1 kilo ampere  hours
12 volts
= 100 ampere-hours
-
Since each battery is 5Amp-hrs only, you will now need 20
batteries to get 100 amp-hour
Note:
-
This Method is more likely applicable to light load only like light
bulbs, LED lightings.
But if with light inductive loads, specifically with motor, electric
funs, refrigerators, A/C..etc.. the formula for battery loading is
necessary:
Battery Bank (AH)

( Load in watts)(hrs. of operation)(DoD)(Batt. disc Loss Factor) ( BattAging factor)
( System voltage) (Comp..Efficiency) ( Inverter Efficiency)
Where: DoD = Depth of discharge @ 50% = 2
Batt.Discharge Factor = 1.3
Batt. Aging Factor = 1.25
Sys.Voltage = either 12,24,48 Volts system
Comp. Efficiency = 0.7, 70
Inverter Efficiency = 0.85 or 85%
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Load one 100w
AC light bulb
For 2 hours/day all year
long (computed req’ment)
Battery (ies)
5 amp-hr each
(needed 20 pcs)
100w
0.2 kwhr/day
Inverter
ηinverter = 50%
@450w maximum
(needed one)
200w
200w
0.4 kwhr/day
Note: in this design,
everything is 12 volts except
the output of the inverter
which is 220 volts
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Charge Controller
  100%
@ 10Amp max.
(Needed 2)
0.4 kwhr/day
Solar Photovoltaic
Module(s)
@ 20w each
(Needed 7)
97
Lesson 2

INTRODUCTION TO POWER PLANT ECONOMICS
GENERAL DEFINITIONS:
Before the economics of the power plant is studied in detail, the following
terms must be defined:
1. Load Factor: It is defined as the ratio of the average load to the peak
load during a certain prescribed period of time. The load factor of a
power plant should be high so that the total capacity of the plant is
utilized for the maximum period that will result in lower cost of the
electricity being generated. It is always less than unity.
2. 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
percent, then the utility will be
100 x 0.8
=
X 100 = 40 percent
200
3. 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.
4. Capacity Factor: It is the ratio of the average load on a machine or
equipment to the rating of the machine or equipment, for a certain
period of time considered.
5. Demand Factor: The actual maximum demand of a consumer to
always less than his connected load since all the appliances in his
resident will not be in operation at the same time or to their fullest
extent. This ratio of the maximum demand of a system to its
connected load is termed as demand factor. It is always less than
unity.
6. Diversity Factor: Supposing there is a group of consumers. It is
known from experience that the maximum demands of the individual
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consumers will not occur at one time. The ratio of the sum of the
individual maximum demands to the maximum demands of the total
group is known as diversity factor. It is always greater than unity.
7. Load Curve: It is curve showing the variation of power with time. It
shows the value of a specific load for each unit of the period covered.
The unit of time considered may be hour, days, weeks, months or
years.
8. Load Duration Curve: It is the curve for a plant showing the total
time within a specified period, during which the load equaled or
exceeded the values shown.
9. 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.
10. Firm Power: It is power which should always be available even under
emergency conditions.
11. Prime Power: It is power, may be mechanical, hydraulic or thermal
that is always available for conversion into electric power.
Recall on the first discussion as to plant as to its reserve:
12. Cold Reserve: It is that reserve generating capacity which is not in
operation but can be made available for service.
13. Hot Reserve: It is that reserve generating capacity which is in
operation but in service.
14. Spinning Reserve: It is that reserve generating capacity which is
connected to the bus and is ready to take the load.
COST ANALYSIS
The following cost analysis refers to the thermal stations. The cost analysis
for hydro-plants will be taken in the next article.
The annual cost of supplying electric energy from a thermal station is
split into the following items:
1. Fixed cost.
2. Operating or running cost.
3. Profit on the investment.
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Fixed Cost
The fixed cost depends upon the capital outlay or the total
investment of the plant. The total investment includes: cost of land, cost of
buildings and equipment, cost of installation, engineering fees, cost of
erection of the transmission lines and sub-stations. The annual fixed cost
consists of the interest on the total investment, taxes and insurance and the
annual contribution to the fund to pay off the capital expended after the
useful life of the plant. This remains fixed throughout the year or
throughout the useful life of the plant. It varies directly with the installed
capacity of the plant and it does not depend upon the factor whether the
plant supplies any electric power or not.
Cost of the land and building will depend upon the location of the
plant. If the plant is situated near the cities, the land will be costlier than
the case if it is located away from the cities or the residential areas. But it
is always better and economical to locate the plant near the load center to
reduce the transmission and distribution charges.
The cost of the equipment is carried over in the books from year to
year and there are two methods to do so; original cost, or replacement cost.
In the first method, the cost of the equipment is established at the time of
its erection and installation. This cost remains in the books as long as the
equipment is in operation with suitable allowances for the depreciation of
the equipment. In the second method, the cost of the equipment is
estimated periodically and the cost will depend upon the price level at the
time of the cost estimation. This is a very costly procedure.
Interest is included to represent the income which would have been
derived from the investment if it would have been used otherwise or loaned
in the market.
To avoid accidents from fire and explosions, the equipment and the
buildings must be insured and regularly inspected.
Depreciation is the most important item in the fixed costs. It
represents the decrease in the value of the property due to continuous wear
and tear and also due to obsolescence. The first factor can be reduced by
proper maintenance of the equipment and the buildings, but the second
factor is quite unpredictable. To account for the depreciation, a certain
fixed amount of money is set aside every year so that the total amount
accumulated at the end of the useful life of the equipment is the original
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cost of the equipment less the expected salvage value. There are two
methods to accumulate the amount: straight line method and the sinking
fund method. In the first method, the same fixed amount is set aside every
year throughout the useful life of the equipment. The interest earned on
this amount is not added to the depreciation reserve but is considered as
the income of the firm or the company. In the sinking fund method, the
interest earned is compounded periodically and it forms a part of the
depreciation reserve. The total amount accumulated by this method is the
sinking fund plus the interest earned and the sum should be equal to the
original cost of the equipment minus the salvage value.
Operating or Running Costs
These costs are based on the energy output as measured in kWhr and
these are directly proportional to the station load. These include:
(a) Cost of fuel including its handling and ash handling in the case of
steam plant.
(b) Cost of labor i.e. the salaries and wages.
(c) Cost of water for:
(1) Boiler feed, condensers, cooling and house service for steam
power plants.
(2) Jacket cooling water, and water to be used in cooling towers, for
i.e. engine power plants.
(3) Intercoolers and gas coolers in gas turbine.
(d) Repairs and maintenance.
(e) Oil, waste, stores and other supplies.
(f) Transmission and distribution costs.
(g) Customer’s expenses.
In the case of steam power plant, the cost of the coal is made up of
two parts: a small fixed component and a large running component. The
first component is that which is needed to keep the station in readiness to
meet the demand for power and it is independent of whether any energy is
sent out or not. The running component is the cost of coal equivalent of the
heat converted into electric power plus the heat rejected in the condensers.
It is directly proportional to the energy sent out in kWh. The amount of the
fuel required per kWh can be determined by dividing the heat rate by the
heating value of the coal. The amount of coal and its cost will depend upon
the plant capacity, its efficiency, load factors, transportation charges and
the cost of storage etc.
Labor is needed both for the operation and the maintenance of the
plant and its requirement is proportional to the capacity of the plant and
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the quality of the labor employed. Skilled labor will cost more than
unskilled labor but the plant will operate more economically and the
maintenance charges will be reduced.
The consumer’s expenses include the cost of meter reading, billing,
collecting of revenue, promotion of electrical appliances etc.
Investor’s Profit
If the power plant is the public property, as this case in India, then
the customers will be the taxpayers to share the burden of the government.
For this purpose, there is an item in the rates to cover taxes in place of the
investor’s profit. These taxes will be paid by the consumers in the form of
electric consumption bills. This amount is collected in twelve instalments
per year or six instalments per year.
COST OF HYDRO-PLANTS
The cost analysis of a hydro-plant is different from those of the other plants
in this respect that the fixed cost is the major item of the total cost and the
operating cost is relatively much smaller whereas in the steam and other
plants, the operating cost is a large part of the total cost.
The total annual cost of a hydro-electric scheme can be broken into
two components as usual:
1. Fixed Costs
(a) Interest on the capital.
(b) Amortization of the capital.
2. Running Costs
(a) Operating costs including salaries and wages.
(b) Maintenance and repairs.
(c) Rates and taxes.
(d) Stores, oil and other supplies.
It is a general practice to include the rates, taxes and the insurance
in the fixed charges. The fixed charges of a hydro-plant are about 60 to 70
percent of the total cost of power and these do not depend upon the station
output. The running charges depend upon the station output but not so
much as in the thermal power plant.
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The total capital outlay or the investment on a hydro-plant includes
the following items:
1. Preliminary surveys and investigations of the topography and
geology of the proposed plant site.
2. Purchase of water rights and the land needed to provide for
sufficient storage or poundage.
3. Compensation to the oustees.
4. Cost of preparation of detailed designs and specifications.
5. Costs of testing the materials of construction.
6. Costs of carrying out experimental work and model tests on
designs for hydraulic structures.
7. The actual costs of construction and the purchase and installation
of the equipment.
8. Interest on the capital during construction.
9. Working capital during the period of load development.
In addition to the above items, new roads, railway lines, residential
houses and even new towns will have to be constructed.
The total cost of construction of a hydro-plant is invariably higher
than that of a thermal plant of equal capacity. Therefore, the annual
charges for interest and depreciation are comparatively high. The cost per
kW for a hydro-plant is higher, the smaller the quantity of water stored. A
typical cost analysis of a hydro-plant is:
Reservoir, dam and water ways :
55%
Power plant and equipment
:
20%
Land
:
15%
Structures
:
10%
Another important item is called the transmission liability which
refers to the transmission charges for conveying the electricity from the
plant site to the load center.
Figure (17.1) shown a typical sub-division of the various items forming
the capital cost for the various plants. The areas of the circles represent the
comparative unit costs.
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1.
2.
3.
4.
5.
6.
7.
8.
9.
Turbo-generators and condensers
Land building and foundation
Misc.
Switch yard
Switching and wiring
Piping
Fuel handling
Turbine and generator
Cooling, fuel system and other auxiliaries.
ECONOMICS OF COMBINED-HYDRO AND STEAM POWER PLANTS
It has been already mentioned that if country or a region is neither rich in
fuel reserves nor in hydro resources then to generate the electricity at the
low economical cost, a combined operation of the hydro and steam power
plants will give the best results. The following benefits can be derived from
the combined operation:
1. Flexibility of Operation: The major drawback of steam power
plant is that appreciable time is required to start up and
synchronize the plant and then to increase the load. Therefore, it
is necessary that the capacity of the steam plant connected to the
system should be such as to cope with the rapid rate of load
increase and also it should possess adequate margin of safety
against breakdown. Moreover, to reduce the stresses and strains
on the boiler joints due to temperature variation, it is necessary
to keep sufficient boiler plant hot throughout the day and night to
meet the peak load requirements. This represents a continuous
loss of energy which cannot be avoided.
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On the other hand, a hydro-plant can be run up and
synchronized quickly and the load can be raised from no load to
full load very rapidly. This plant can also meet the sudden
variations of load demand easily and so it is particularly suitable
for peak load operation. Therefore, if the steam and hydro-plant
are combined in a system, its operation will become flexible. The
running of the steam plants will become easier as they can
operate within the flatter top portion of the daily load curve and
the violent fluctuations of load will be met by the hydro-plant.
2. Security of Supply: As far as the reliability of operation and
security of electric supply is concerned, there is not much to
choose between a steam plant and a hydro-plant. The reliability
of steam plant depends upon the coal supplies and that of the
hydro-plant on the steam flow. Due to greater diversity, a
combined steam and hydro-plant system would be more reliable
than the individual steam or hydro-steam.
3. Improved Utilization of Hydro-power: As already pointed out,
during the periods of heavy flow, the hydro-plant can generate
the base load and the steam plant may not generate as much
energy and this will result in a saving of fuel. During drought
periods, the steam plant can generate more energy and the water
can be held back in reservoirs to make the capacity of the hydroplant effective.
4. Spare Plant: In any electric system, a spare plant is needed to
provide for any unforeseen excess demand, for covering any
temporary loss due to breakdown and due to maintenance and
repairs of the main plant. A hydro-plant is simple and more
reliable than a steam plant and for this plant only few
maintenance and repairs are needed and if the combined system
adopted, the steam plant can be taken out of commission during
the periods of lower electric demand and a spare plant need not
be provided. But it is always judicious to carry a spare plant in a
combined system, it is possible to achieve an economy in the
overall amount of the spare plant.
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Economic Factors
The combined system of steam and hydro-plants can be grouped into three
categories:
1. Bulk of hydro power resources and fuel has to be imported.
2. Bulk of coal reserves with hydro resources deficient.
3. Considerable resources of both hydro and thermal power.
In the first case, there might be various plant sites and the
comparison will have to be made of the cost of producing equivalent
electric power and its transmission to the load center. Some sites will be
suitable for base load plants and others might be suitable for peak load
service. The cost of electric generating at various sites will have to be
studied for various load factors. The load factor to be adopted will be that
which will give the lowest overall cost of production.
There is one drawback of the above system. The maximum run-off
may not occur during the periods of maximum demand and for the hydroplant to supply the full demand throughout the year, a storage will have to
be built to cover the seasonal variations of run-off. This storage may be
either impracticable or very costly to provide. For this, a sufficient thermal
plant will have to be provided on the hydro system to make up its
deficiencies and also for supplying the peak loads. The load factor of the
thermal plant will be high only for a small period and its annual load factor
will be low.
In the second case it will be economical to develop the hydro-plants
for peak load service. The reason for this is that in hydro-plants, the cost of
storage is the major item in the total cost and this item does not depend
much on the annual load factor, therefore, the capital cost per kW will be
less for low load factor than for high load factor. It is found that the
incremental costs resulting from reduced load factor i.e. increased amount
of installed capacity is less than the capital cost of raising the installed
capacity of a steam plant. An advantage of increased capacity of the hydroplant is that the spoilage can be decreased at times of high run-off by
operating the plant for longer periods and thereby saving the cost of fuel in
the thermal stations.
For the third case, the hydro-plants must be planned to generate
enough electric power during adverse periods so that they can meet the
requirements of that portion of the load curve to which they have been
assigned.
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For any particular scheme, there must be some ratio of the hydroplant to the total demand which gives the lowest annual cost of electric
generation. In areas where both cheap fuel and favorable hydro sites are
abundant, this ratio will be much higher. But in areas where fuel is cheap
and the cost of development of hydro-plant is high, the ratio maybe quite
low.
For any combined system, the economic balance between the
capacity of the hydro station and thermal station will depend upon the
following factors:
1. Shape of load curve.
2. Cost of fuel.
3. Availability of condensing water.
4. Availability of suitable hydro sites and their distance from the
load center.
5. Run-off and its seasonal variation.
6. Cost of storage.
Combined Operation
The operation of all the stations in the integrated system should be
carefully coordinated to achieve the maximum overall economy. For this
purpose, there should be a centralized supervision station to supervise all
the operations including generation and switching. But a big centralized
control will be too unwieldy. So, the whole region is divided geographically
in smaller parts. Each part will be supervised from a group control center.
These control centers will have telephone connections with one another,
with the power stations under their respective control and also with the
system central control room.
PLANT SELECTION
After the capacity of the plant has been established, the question of
selection of the type of plant and its location arises. As already mentioned,
the major source of electric power in India are coal and water. The state
wise position was given in Chapter 1. The diesel engine or gas turbine or a
nuclear power plant can also be considered depending upon the availability
of oil fuel, natural gas and nuclear fuel. That type of plant is to be selected,
which will give the lowest cost of electric generation.
If the power plant is to be set up for an industrial complex, then the
problem of its location is automatically solved. If this site is near a lake,
river or sea then the condensing steam power plant may be economical. If
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water supply is scarce then the possibility of setting up either a steam plant
or a diesel plant using spray ponds or cooling towers should be investigated.
If the site is near a river with favorable hydro site, then the hydro-plant can
be set up. Anyhow before deciding the type of plant, the total annual cost
for each must be calculated.
If the power plant is to be set up for a regional system, then the
factors to be analyzed are more complex. There may be more than one
suitable site for the location of a steam plant or a hydro-plant. In studying
alternate plant sites, the final selection will depend upon the total annual
cost for each case. For a steam plant, the location will be governed by (1)
availability and cost of fuel and its transportation (2) water supply (3) cost
and nature of land (4) problems of ash handling, noise, smoke and heat if
the location near residential houses. For the hydro-plants, the factors to be
considered will be topography, geology and hydrology of the site. All the
above factors have been discussed in previous chapters.
Equipment Selection
Once the type of the power plant is selected, the size of the
equipment will depend upon many factors. With all the plants, higher the
load factor and larger the size of the plant, higher the working efficiency.
Moreover, the cost per unit installed is decreased if the plant size increases,
but a large plant will require bigger investment and it will work at low load
factor. The following considerations must be taken into account while
selecting the equipment for a steam power plant:
1. Operational consideration.
2. Economic considerations.
3. Thermodynamic considerations.
Operational Considerations: The equipment selected should be
reliable i.e. it should be of improved performance and design. The machine
should be simple in construction and operation so that unskilled labor can
be employed which will be cheaper. The machine should be easy to be
dismantled and reassembled and there should be accessibility for the
inspection purpose. Operation controls both for when the machine is
working and for emergency, should be provided with the machine.
Economic Considerations: Economic considerations are studied on
the basis of theoretical efficiency based on thermodynamic consideration
and depends upon kJ/kW; and practical efficiency based on cost
considerations i.e. cost of machine, depreciation, fixed charges and
interest, working efficiency at various loads; cost of coal including the cost
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of transportation etc. Cost of machine will depend upon the size of the
machine, condition of temperature and pressure and the year of
manufacture. The cost per kW, floor space per kW and volume of building
per kW decreases as the size of the machine is increased. Also the working
efficiency increases with the increase in size. However, there will be more
fixed charges with bigger size of the equipment. The capital cost per kW is
reduced and the thermal efficiency is increased with the increase in the unit
size.
If the plant is small, then there is no need to go in for higher
temperature and pressure, condition plant as it would prove uneconomical.
It is safe practice not to purchase the latest design of the machine. The
type of the steam turbine will depend upon its use. For industries, back
pressure or non-condensing and extraction or condensing machine may be
used but for power undertaking it is always better to use a condensing unit.
Thermodynamic Considerations: Following are the thermodynamic
considerations for the plant design:
1. Superheating of steam.
2. Use of high temperature and pressure.
3. Heat cycles.
Superheating is invariably practiced in modern power plants for the
following reasons: (a) more work can be obtained due to greater specific
volume and heat content of the steam (b) smaller plants (c) cost saving due
to decreased rate of steam. But the limitation of the superheating is the
decreased strength of steel at high temperature which makes it necessary to
use special alloy steels for super heater tubing when temperature of steam
exceeds about 425oC which makes the plant costlier. The decrease in the
rate of steam as a result of superheating does not mean only a saving in the
cost of fuel but it also means smaller size of the unit, saving in space, lesser
repairs and maintenance, less power consumption by the auxiliaries and
lesser breakdown.
High pressures and temperatures are desirable from the
thermodynamic considerations but as the pressure increases, the thickness
of metal increases with a resulting increase in cost. As the temperature
increases, better and the better materials are required resulting in the
increase in plant cost. Moreover, high temperatures introduce problems
such as thermal expansion, change in structural properties of materials.
The effect of heat cycles on the performance of a steam plant has
already been discussed in Chapter 6. The economy and the theoretical
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thermal efficiency increase by incorporating heaters and by the reheat
cycle.
In case of i.e., engines, the cycles become more efficient and the
efficiency increases as the comprehension ratio is raised. But higher
compression ratios increase the maximum cylinder pressure requiring
heavier cylinder walls, pistons, crankshafts etc. and the plant becomes
expensive. Regarding the choice between two stroke cycle and four stroke
cycle, the former is generally lighter in weight and less expensive and the
latter has higher thermal efficiency. Modern i.e. engines incorporate the
supercharger to increase the power per unit weight of the engine and at a
higher thermal efficiency. But the super charging will be justified only if
there is enough increase in power output which will balance the additional
cost of the supercharger. The type of fuel available will determine the type
of cycle to be adopted. Otto cycle is adopted if a plentiful and cheap supply
of gasoline is available and if cheap fuel oil is available then the diesel
engine will have favored. The latter case is usually adopted in practice since
fuel oil is cheaper than gasoline. A diesel engine usually operates at higher
thermal efficiencies as compared to a steam plant of the same capacity. But
the maximum unit size of the diesel engine is very much limited as
compared to a steam turbine and as the coal is cheaper than fuel oil, the
diesel engine plants have not found favor with large regional power systems.
They are usually suitable for municipal, commercial, institutional and
industrial purposes where steam is not required and water is scarce.
The thermal efficiency of a gas turbine increases as the simple cycle
is modified by incorporating reheaters, regeneration and intercooling but tis
additional expense will increase the cost of the plant. Due to limited unit
size, these plants are not suitable for base load power plants and are usually
adopted to supply peak loads.
COMBINATION OF WIND POWER WITH OTHER SOURCES OF POWER
No firm power can be obtained from wind unless there is some form of
storage. However, it may be used in combination with steam and hydropower.
Combination of Wind and Steam Power: For a power system
predominantly supplied by steam power plants, the value of wind power will
be only in the fuel saved by its operation in combination with steam plants.
For its economic justification the cost of wind energy may be compared with
the cost of generating additional units from the existing steam plants. Wind
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power would not be economical beyond the total incremental cost of
generating in the steam station. If the steam station is less efficient and
operates at low load factors, it might be economical to replace the energy
generated by the steam plant by the electric energy generated by wind
power.
Combination of Wind and Water Power: The firm capacity of an
electric system can be increased by operating the wind power in
combination with storage hydro-power stations, particularly if there is
diversity between the occurrence of rain and wind.
During windy periods, the wind energy can be fed into the electric
system. During this period, the output of the hydro-stations can be reduced.
This will result in the conservation of stored water which can be more
beneficially used afterwards.
SINKING FUND METHOD OF DEPRECIATION
In this method a fixed amount is set aside and put into the reserve fund
each year, but the interest earned on this amount collected in the sinking
fund plus the interest earned throughout the life of the equipment should
be equal to the original value of the equipment minus its salvage value. In
this method, the annual amount to be set aside will be less than that
calculated by straight line method.
Let
A = Annual deposit in the reserve fund, rupees
S = Sum to be provided at the end of useful life, rupees
n = expected useful life, years
I = Annual interest earned by the investment of A.
Accumulation of first year’s investment will be
= A (1 + I) n – 1
Accumulation of second year’s investment will be
= A (1 + I) n – 2
Similarly, the accumulation of (n – 2) year’s investment will be
= A (1 + I) 2
Accumulation of first year’s investment will be
= A (1 + I) n-1
Accumulation of the nth year’s investment will be
=A
Total accumulation
S = A [1 + (1 + I) + (1 + 1)2 + ….. + (1 + I)
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n–2
+ (1 + I)
n – 1]
111
This is geometrical progression with a common ratio of
r = (1 + I)
A [(1+I)n – 1]
= [(1+I)n – 1]
S=
A
1+I–1
I
I
A= S
(1+I)n -1
A i.e., the ratio of the annual depreciation to the
S capital invested is
termed as the depreciation rate.
The ratio
Example.1 If the expected life of an equipment is 15 years and the
interest earned on the capital invested is 5 percent, find the
depreciation rate.
Solution:
Depreciation rate,
n = 15 years
I = 5%
A
I
=
S
(1 + I)n – 1
0.05
=
(1 + 0.05)15 – 1
= 0.0463
Example.2 The original value of an equipment is Rs. 250,000 and its
salvage value at the end of its useful life of 20 years is Rs 25,000. Find
the value of the equipment at the end of 10 years of its use by the
following methods:
(a) Straight line depreciation.
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(b) Sinking fund depreciation, when it is compounded annually at
8%.
Solution:
(a) Straight line method
Original value = Rs. 250,000
Salvage value = Rs. 25,000
Total Depreciation = Rs. 250,000 – Rs. 25,000 = Rs. 225,000
225,000
Depreciation per year =
= Rs. 11,250
20
Depreciation at the end of 10 year
= Rs. 11,250 x 10 = Rs. 112,500
Value of the equipment at the end of years
= Rs. 250,000 – Rs. 112,500 = Rs. 137,500 Ans.
(b) Sinking Fund Method
Total amount to be provided in the sinking fund
end of 20 year = Rs. 250,000 – Rs. 25,000
S = Rs. 11,250
Now
I = 8%
I
0.08
Annual deposit, A = Rs. S [
] = Rs. 225,000 [
n
(1 + I) – 1
(1 + 1.08)20 – 1
Rs. 225,000 (
0.08
225,000 x 0.08
=
) = Rs.
4.66 – 1
3.66
Total amount collected at the end of 10 years will be
[(1 + I)n – 1]
[(1 + 1.08)10 – 1]
= Rs. A
= Rs. 4,920
I
0.08
Rs. 4,920 [
2.16 – 1
=
] = Rs. 71,250
0.08
Value of equipment at the end of 10 years
= Rs. 250,000 – Rs. 71,250 = Rs. 178,750
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at the
]
= Rs.
4920
113
Lesson 3

Load Determination & Load Graphs
The load demand on a power system is governed by the consumers
and for a system supplying industrial and domestic consumers, it varies
within wide limits. This variation of load can be considered as daily, weekly,
monthly or yearly. Typical load curves for a large power system are shown in
Fig. 17.3 these curves are for a day and for a year and these show the load
demanded by the consumers at any particular time. Such load curves are
termed as “Chronological load Curves”. If the ordinates of the chronological
load curves are arranged in the descending order of magnitude with the
highest ordinates on left, a new type of load known as “load duration curve”
is obtained. Fig. 17.4 shows such a curve. If any point is taken on this curve
then the abscissa of this point will show the number of hours per year during
which the load exceeds the value denoted by its ordinate. Another type of
curve is known as “energy load curve” or the “integrated duration curve”.
This curve is plotted between the load in kW or MW and the total energy
generated in kWh. If any point is taken on this curve, abscissa of this point
show the total energy in kWh generated at or below the load given by the
ordinate of this point. Such a curve is shown in Fig. 17.5. In Fig. 17.4, the
lower part of the curve consisting of the loads which are to be supplied for
almost the whole number of hours in a year, represents the “Base Load”,
while the upper part, comprising loads which are required for relatively few
hours per year, represents “Peak Load”.
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Ideal and Realized Load Curves: From the stand-point of equipment
needed and operating routine, the ideal load on a power plant would be one
of constant magnitude and steady duration. However, the shape of the
actual load curve (more frequently realized) departs far from this ideal, Fig.
17.3. The cost to produce one unit of electric power in the former case
would be from ½ to ¾ of that for the latter case, when the load does not
remain constant or steady but varies with time. This is because of the lower
first cost of the equipment due to simplified control and the elimination of
various auxiliaries and regulating devices. Also, the ideal load curve will
result in the improved operating conditions with the various plant machines
(for example turbine and generators etc.) operating at their best efficiency.
The reason behind the shape of the actual realized load curve is that the
various users of electric power (industrial, domestic etc.) impose highly
variable demands upon the capacity of the plant.
Effect of variable Load on Power Plant Design: The characteristics and
method of use of a power plant equipment is largely influenced by the
extent of variable load on the plant. Supposing the load on the plant
increases. This will reduce the rotational speed of the turbo-generator. The
governor will come into action operating a steam valve and admitting more
steam and increasing the turbine speed to its normal value. This increased
amount of steam will have to be supplied by the seam generation. The
governor response from load to turbine is quite prompt, but after this point,
the governing response will be quite slower.
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The reason is explained as given below:
In most automatic combustion control systems, steam pressure
variation is the primary signal used. The steam generates with imbalance
between heat transfer and steam demand long enough to suffer a slight but
definite decrease in steam pressure. The automatic combustion controller
must then increase fuel, air and water flow in the proper amount. This will
affect the operation of practically every component of auxiliary equipment
in the plant. Thus, there is a certain time lag element present in
combustion control. Due to this, the combustion control components should
be of most efficient design so that they are quick to cope with the variable
load demand.
Variable load results in fluctuating steam demand. Due to this it
becomes very difficult to secure good combustion since efficient combustion
requires the coordination of so many various services. Efficient combustion
is readily attained under steady steaming conditions. In diesel and hydropower plants, the total governing response is prompt since control is needed
only for the prime mover.
The variable load requirements also modify the operating
characteristics built into equipment. Due to non-steady load on the plant,
the equipment cannot operate at the designed load points. Hence for the
equipment, a flat topped load efficiency curve is more desirable than the
peaked one.
Regarding the plant units, if their number and sizes have been
selected to fit a known or correctly predicted load curve, then, it may be
possible to operate them at or near the pint of maximum efficiency.
However, to follow the variable load curve very closely, the total plant
capacity has usually to be sub divided into several power units of different
sizes. Sometimes, the total plant capacity would be more nearly coincide
with the variable load curve, if more units of smaller unit size are employed
than a few units of bigger unit size. Also, it will be possible to load the
smaller units somewhere near their most efficient operating points.
However, it must be kept in mind that as the unit size decreases, the initial
cost per kW of capacity increases.
Again, duplicate units may not fit the load curve as closely as units of
unequal capacities. However, if identical units are installed, there is a
saving in the first cost because of the duplication of sizes, dimensions of
pipes, foundations, wires insulations etc. and also because spare parts
requires are less.
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Effect of Variable Load on Power Plant Operation: In addition to the
effect of variable load on power plant design, the variable load conditions
impose operation problems also, when the power plant is commissioned.
Even though the availability for service of the modern central power plant is
very high, usually more than 95%, the public utility plants commonly remain
on the “readiness-to-service” bases. Due to this, they must keep certain of
their reserve capacity in “readiness-to-service”. This capacity is called
“spinning reserve” and represents the equipment standby at normal
operating conditions of pressure, speed etc. normally, the spinning reserve
should be at least equal to the least unit actively carrying load. This will
increase the cost of electric generation per unit (kWh).
In a steam power plant, the variable load on electric generation
ultimately gets reflected on the variable steam demand on the steam
generator and on various other equipment. The operation characteristics of
such equipment are not linear with load, so, their operation becomes quite
complicated.
As the load on electrical supply systems grow, a number of power
plants are interconnected to meet the load. The load is divided among
various power plants to achieve the utmost economy in the whole system.
When the system consist of one base load plant and one or more peak load
plants, the load in excess of base load plant capacity is dispatched to the
best peak system, all of which are nearly equally efficient, the best load
distribution needs thorough study and full knowledge of the system.
Coordinate Base Load and Peak Load Power Plants: If the load
represented by Fig. 17.4 is to be supplied from one power plant only, then
the installed capacity of the plant should be equal to the peak load. Such a
plant will be uneconomical since the peak load occurs only for a short
period in a year and therefore the capacity equal to the difference of peak
load and base load will remain idle for the major part of the year. Hence
such a power would not be supplied from a single power station. There
would be some station supplying the base load and others, possible of
different type, supplying the peak load.
One method of meeting this varying load demand is to coordinate the
operation of hydro and steam stations. The steam plant capacity and the
available water power energy are fitted into load curve. Peak load demand
can be conveniently met by hydro-stations, the base load being supplied by
the steam power plants. A hydro-station can be started up quickly at any
time to meet a sudden emergency. Also the load on hydro-station can be
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reduced more quickly than is possible with steam plant. There are two
methods for utilizing the hydro-electric power for supplying the peak load.
1. By storing the natural run-off from a catchment area during hours of
light load and employing the water to operate the station at full
capacity during periods of peak demand.
2. Pumped storage system. In this water is pumped into a high level
reservoir at off-peak periods, and is utilized to drive the turbines and
generators at the time of peak demand. The system has already been
discussed in Chapter 16.
Peak load can also be supplied by diesel engine power plant and gas
turbine power plants. Base load stations operate almost continuously i.e. at
a load factor of about 80%. They are shut down only for a small periods for
maintenance and overhaul. The load factor of peak load plants is very low,
e.g. 5 to 15%, since they operate only for a small period in a day, week,
month or a year. As already discussed the cost supplying the electric energy
may be divided into two parts:
(a) Fixed cost, which mainly consists of the interest on the capital cost
and depreciation. It is independent of the amount of electric energy
actually supplied. It is however, approximately proportional to the
capacity of plant installed i.e. proportional to kW.
(b) Running cost, which depends upon the actual energy generated, i.e.
it is proportional to the kWh.
Since the electric power plants are very expensive to install,
therefore, it is desirable and even essential to generate as much energy as
possible in order to spread the fixed cost over the highest possible number
of units (kWh) supplied. Therefore, the plants should run at a high load
factor and then the cost per unit will be minimum. If the plant is idle for
most of the period, it will generate only a small number of units and hence
the fixed charges will be shared by small number of units resulting in high
cost per unit supplied. Therefore, the load factor has a very important
effect on the cost of the electric energy supplied from a power plant.
Hence, the base load are cheaper to be supplied whereas the peak load
units are expensive to produce. A base load power station should have the
highest possible efficiency. For peak load plants, since the units to be
generated are small, efficiency is not of much importance of course, the
capital cost should be minimum since it is to be distributed over the small
number of units supplied or generated. Since a peak load plant may have to
be started once or perhaps twice in a day and possibly for some unexpected
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emergency condition, it should be capable of quick starting and quick load
pick up.
Significance of Various Factors
(a) Load Factor: High load factor is a desirable quantity. Higher load
factor means greater average load, resulting in greater number of
power units generated for a given maximum demand. Thus, fixed
cost, which is proportional to the maximum demand, can be
distributed over a greater number of units (kWh) supplied. This will
lower the overall cost of the supply of electric energy.
(b) Diversity Factor: High diversity factor (which is always greater than
unity) is also a desirable quantity. With a given number of consumers,
higher the value of diversity factor, lower will be the maximum
demand on the plant, since,
Diversity Factor =
Sum of the individual maximum demands
Maximum demand of the total group
So, the capacity of the plant will be smaller, resulting in lower fixed
charges.
(c) Plant Capacity Factor: Since the load and diversity factors are not
involved with ‘reserve capacity’ of the power plant, a factor is
needed which will measure the reserve, likewise the degree of
utilization of the installed equipment. For this, the factor “Plant
factor, capacity factor of Plant capacity factor” is defined as,
Plant Capacity Factor =
Actual Energy Produced
Maximum possible energy that might have
been produced during the same period
Thus, the annual plant capacity factor will be,
=
Annual kWh produced
Plant capacity (kW) x hours of the year
The difference between load and capacity factors is an indication of
reserve capacity.
(d) 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
are used. Thus, annual plant use factor is,
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=
Annual kWh produced
Plant capacity (kW) x number of hours of plant operation
Example.3. Explain a method of constructing a load duration curve using
a load curve. The following data were collected from the daily load
curves of a power system during a year.
Load (kW)
15,000
12,000 and over
10,000 and over
8,000 and over
6,000 and over
4,000 and over
2,000 and over
Duration
(Hrs)
87
876
1752
2628
4380
7000
8760
Construct annual load duration curve and find load factor of
system (Assume a smooth curve).
Solution: The load duration curve is shown in Fig. 17.6
To get average demand find energy consumed or area under curve
first. To find energy used:
Energy consumed = Area under the curve
= Area 1 + Area 2 + … + Area 12
= 500 x 114.8 MW hrs
Energy consumed in one year
Average demand =
No. of hrs. in a year
500 x 114.8
=
= 6.45 MW
365 x 24
Max. demand = 15 MW
6.54
Load factor of system =
X 100 = 4.36%
15
To draw Load Duration Curve Using Load Curve: This can be best
explained by considering an example. Now take the load curve shown in Fig.
17.7 and 17.8.
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At various ordinates draw horizontal lines (show dotted) and note the
length over which the curve cuts these lines. Then at corresponding
ordinates on load duration side for the abscissa equal in length to the total
no. of lines cut by load curve. 3 cases have been illustrated.
After obtaining a no. of points like this draw a smooth curve through
them. This gives the load duration curve.
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Example.4. Draw the chronological daily load curve and the load duration
curve from the following observation. If the capacity of the plant serving
this load is 100 MW, find the load factor and the plant utilization factor.
The changes in load are linear.
Time
12 P.M.
2 A.M.
6
8
12 M
Load, MW
20
10
10
50
50
Time
12:30 P.M.
1
5
6
12
Load, MW
40
50
50
70
20
Solution: The chronological load curve and the load duration curve are
shown in Fig. 17.9.
From the load duration curve, the average load can be estimated.
Lave = Average load for the period = E/h
where E = total energy in load curve for period
h = total number of hours in period
Lave comes out to be about 42.8 MW
Lave
Load factor =
Lmax
Now Lmax = 70 MW
42.8
Load factor =
= 61.14%
70
L
Utilization factor = max
Cap
where
Cap = Rated capacity of the plant
Utilization 70
= 70%
factor = 100
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Example.5. The yearly duration curve of a certain plant can be
considered as a straight line from 20 MW to 3 MW. To meet this load,
three turbine-generator units, two rated at 10 MW each and one at 5
MW, are installed.
Determine:
(a) Installed capacity
(b) Plant factor
(c) Maximum demand
(d) Load factor
(e) Utilization factor.
Solution: The load duration curve is shown below in Fig. 17.10.
(a) Installed capacity
= 2 x 10 + 5
= 25 MW
(b) Plant factor, capacity factor or use factor
Lave
=
Cap
8760 x 3 + ½ x 8760
E
x 17
Now
Lave =
=
= 11.5 MW
h
8760
Plant 11.5
= 46%
factor = 25
(c) Maximum demand = 20 MW
(d)
11.5
Load factor Lave
=
= 57.5%
= Lmax
20
(e) Utilization factor Lmax
20
=
= 80%
= Cap
25
Example.6. A residential consumer has 10 lamps of 40 W each connected
his residence. His demand is
Mid-night to 5 AM ……………….. 40 W
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5 AM to 6 PM
6 PM to 7 PM
7 PM to 9 PM
9 PM to 12 mid-night
………………..
………………..
………………..
………………..
no load
320 W
360 W
160 W
(a) Plot the load curve
(b) Find Average load
(c) Max. load
(d) Load Factor
(e) Energy consumption during day.
Solution:
(a) The load curve: It is drawn in Fig. 17.11.
(b)
Average load
=
=
Energy consumed in 24
hrs.
=
24 hrs.
40 x 5 + 320 x 1 + 2 x 360 + 3 x
160
24
Area of load
curve
24
=
(c) Max. load = 360 W
Average demand
71.7
x 100
Load factor =
= 360
Maximum
=
demand
(d) E
Energy consumed during one day
= Average load x 24 hrs
= 71.7 x 24 watt hrs
= 71.7 x 24 x 10-3 kWh = 1.72 kWh
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1720
= 71.7 W
24
717
360 = 19.9%
124
Example 7. A power system has the following loads
1. Residential lighting load
Maximum demand = 1000 kW
L.F. = 20%
Diversity between = 1.3
consumers
2. Commercial load
Maximum demand = 2000 kW
L.F. = 30%
Diversity between consumers = 1.1
3. Industrial Load
Maximum demand = 5000 kW
L.F. = 0%
Diversity between consumers = 1.2
Overall diversity factor may be taken as 1.4.
Find:
(a) Max. demand on system.
(b) Daily energy consumption of each type of load and total energy
consumption.
(c) Overall load factor.
(d) Connected loads of each type assuming that demand factor for
each is 100%.
Solution:
(a) Group diversity factor
Sum of individual max. demands
=
Actual max. demand of the group
Maximum demand of the system
1000 + 2000+ 5000
8000
=
=
= 5628 kW
1.4
1.4
(b) Average demand
Average demand
= Maximum demand x L.F.
= 1000 x 0.2 + 2000 x 0.3 + 5000 x 0.8
= 4800 kW
Daily energy consumption = 4800 x 24 = 115,200 kWH
(c)
Overall
L.F. =
(d)
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Average
Demand
Maximum
Demand
4800
=
5628
= 85.3%
125
Maximum demand
Demand factor
Now maximum demand
= 1000 x 1.3 + 2000 x 1.1 + 5000 x 1.2
= 1300 + 2200 +600 = 9500 kW
Connected load = 9500 x 1 = 9500 kW
Connected load =
Example.8. A consumer has following connected load: 10 lamps of 60 W
each 2 heaters of 1000 W each
Max. demand = 1500 W. on the average he uses 8 lamps, for 5 hrs
a day and each heater for 3 hrs a day. Find his average demand, load
factor and monthly energy consumption.
Solution:
(a) Average demand:
Daily energy consumption = (60 x 8 x 5 + 2 x 1000 x 3) 10-3 x 1.2
= (2400 + 6000) 10-3 kW = 8.4 kW
This is for 24 hrs.
8400
Average Demand =
= 350 watts
24
(b)
Load factor =
Ave. Demand
Max Demand
(c) Monthly energy consumption
350 x 100
1500
=
= 23.3%
= Daily energy consumption x 30
= 8.4 x 30 = 252 kWh
Example 9. The following data is given for a steam power plant:
Maximum demand
25,000 kW
Load factor
40%
Coal consumption
0.86 kg per kWh
Boiler efficiency
85%
Turbine efficiency
90%
Price of coal
Rs. 55 per tonne
Determine:
1. Thermal efficiency of the station.
2. Coal bill of the station for one year.
Solution:
Thermal efficiency of the plant = Boiler efficiency x Turbine efficiency
= 0.85 x 0.90 = 76.6%
Now average demand on the station
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= Maximum demand x load factor
126
= 25000 x 0.4 = 10000 kW
Energy generated per year = 1000 x 8760 kWh
Coal consumption = 1000 x 8760 x 0.86 N per year
Coal bill per year =
10000 x 8760 x 0.86 x 55
1000
= Rs. 4,143,480
Example.10. The load shown by the load duration curve, Fig. 17.12, is to
be carried by a base-load station having a capacity of 18000 kW and a
standby station having a capacity of 20000 kW. Determine the load
factors, use factors and capacity factors of the two station.
Solution: The load shown by the shaded area is to be met by the standby
station. The remaining load is to be met by the base-load station.
From the curve,
Annual standby output = 7,350,000 kWh
Annual base load station output = 101,350,000 kWh
Peak load on standby station = 30,000 – 18,000 = 12,000 kW
Time during which standby station was in use = 2190 hours
(from the figure)
1. Standby Station:
Average load
7350000
Load factor =
=
= 0.2797
Maximum load
12000 x 2190
Units of electricity generated per year
Use factor =
Capacity of the plant (units generated during actual use)
7350000
=
= 0.167
20000 x 2190
Average load or units generated per year
Capacity factor =
Capacity of plant (units) per year
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127
=
7350000
20000 x 8760
= 0.043
2. Base Load Station:
Load factor =
101350000
18000 x 8760
= 0.642
Since the plant works throughout the year, the use factor and the capacity
factor will also be 0.642.
Example 11. A 10,000 kW steam plant is erected at a cost of Rs. 2000 per
kW. Assume that the bonds in the amount of the total cost were sold.
They are to mature in 15 years which is also the estimated life of the
plant. Salvage value is estimated at 5% of the first cost. Interest on bonds
is 4%, on sinking fund deposit 3 ½ %. Determine the amount of annual
payment on the investment and also the sinking fund accumulation after
5 years.
Solution: In equation (17.1), the ratio A/S is frequently referred to as the
“sinking fund factor” which will be
I
=
(1 + I)n – 1
I = 3.5%, n = 15 years
0.035
= 0.052
(1.035)15 – 1
Sinking fund annual payment, A = S x 0.052
Sinking fund factor =
S = 10000 x 2000 x 0.95
A = 10000 x 2000 x 0.95 x 0.052 = Rs. 988000
Annual interest payment = 0.4 x 10000 x 2000 = Rs. 800000
Total = Rs. 1788000
0.035
(1.035)15 – 1
A
Accumulated amount after 5 years, S =
0.186
988000
=
0.186
Now five year accumulation factor
=
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= 0.186
= Rs. 5311828
128
In module II, you have learned about the introduction to power
plant’s load economics and analysis through graphs that shows emphasis on
the relavance on its usage.
There are three lessons in module II. Lesson 1 deals with the
additional resources of energy that could be very useful to society and our
environment, just like solar energy that is fastly growing in today’s
scenarios of renewable energy. While Lesson 2 deals with the introduction
to power plant economics that clearly illustrates the different power plants
usage as to its, dependability, fuel consumption, number of hours of
operation that will show its power plant demand, utilization factors, etc.,
Lesson 3 deals with load determination and load graphs, that will illustrate
the learner the different usage of power plants as to its efficiency, demand
and as to how it operates without any shutdown or interruptions that can
cause its losses and dependability.
Congratulations! You have just finished the course. now you are ready
to evaluate how much you have benefited from your reading by answering
the summative test. Good Luck!!! -mev

SUMMATIVE TEST
Solve for the following unknown parameters
1. The annual peak load on a 15,000-kW power plant is 10,500 kW. Two
substations are supplied
by this plant. Annual energy dispatched through
substation A is 27,500,000 kW-hr with a peak
load at 8,900 kW,
16,500,000 are sent through substation B with a peak load at 6,650 kW.
Neglect line losses. Find the diversity factor between substations and
capacity factor of the power plant.
a.
b.
1.48, 0.446
1.48, 0.335
c. 1.75, 0.335
d. 1.75, 0.446
2. What is the annual capacity factor of the plant if the annual energy
produced in a 150 MW
power plant is 500,000,000 kW-hrs?
a. 38.05 %
c. 56.785 %
b. 44.04 %
d. 34.44 %
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3. Calculate the use factor of a power plant if the capacity factor is 35% and it
operates 8000 hrs .during the year?
a.
b.
38.325 %
33.825 %
c. 35.823 %
d. 32.538 %
4. A 75 MW power plant has an average load of 35,000 kW and a load factor of
65%. Find the reserve over peak.
a.
b.
21.15 MW
23.41 MW
c. 25.38 MW
d. 18.75 MW
II. Define the following according to its function and uses.
1.Reserve Capacity
2. Use Factor
3. Capacity Factor
4. Demand Factor
5. Annual Load Factor?
EE158 module -mev