COURSE NAME: RENEWABLE ENERGY RESOURCES

COURSE CODE: 22OE139

COURSE COORDINATOR: Dr. Kumar Sourav (KS)
COURSE INSTRUCTOR: Dr. KS & Dr. Harinadha Gidituri
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The Boy Who Harnessed the Wind

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Course Summary
This course explores renewable energy resources, focusing on their
principles, potential, and applications. It covers solar radiation,
energy collection, and storage techniques; wind and ocean energy
systems; biomass conversion and geothermal energy utilization.
The course also introduces direct energy conversion methods,
including thermoelectric generators and fuel cells, emphasizing
efficiency, technologies, and practical applications.

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.

Course Summary
This course explores renewable energy resources, focusing on their
principles, potential, and applications. It covers solar radiation,
energy collection, and storage techniques; wind and ocean energy
systems; biomass conversion and geothermal energy utilization.
The course also introduces direct energy conversion methods,
including thermoelectric generators and fuel cells, emphasizing
efficiency, technologies, and practical applications.

Course Outcome
Understand principles and applications of renewable energy
sources.
Analyze energy collection, storage, and conversion
technologies.
Evaluate direct energy conversion systems and their efficiency.
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UNIT I - Principles of Solar Radiation

Role and potential of new and renewable source, the solar energy
option, Environmental impact of solar power, physics of the sun,
the solar constant, extraterrestrial and terrestrial solar radiation,
solar radiation on titled surface, instruments for measuring solar
radiation and sun shine, solar radiation data.

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UNIT I - Principles of Solar Radiation

Role and potential of new and renewable source, the solar energy
option, Environmental impact of solar power, physics of the sun,
the solar constant, extraterrestrial and terrestrial solar radiation,
solar radiation on titled surface, instruments for measuring solar
radiation and sun shine, solar radiation data.

UNIT II - Solar Energy Collection and Storage

Flat plate and concentrating collectors, classification of
concentrating collectors, orientation and thermal analysis,
advanced collectors.
Different methods, Sensible, latent heat and stratified storage,
solar ponds, Solar Applications.

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UNIT III - Wind Energy and Ocean Energy

Sources and potentials, horizontal and vertical axis windmills,
performance characteristics, Betz criteria
OTEC, Principles utilization, setting of OTEC plants,
thermodynamic cycles. Tidal and wave energy: Potential and
conversion techniques, minihydel power plants, and their
economics.

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UNIT III - Wind Energy and Ocean Energy

Sources and potentials, horizontal and vertical axis windmills,
performance characteristics, Betz criteria
OTEC, Principles utilization, setting of OTEC plants,
thermodynamic cycles. Tidal and wave energy: Potential and
conversion techniques, minihydel power plants, and their
economics.

UNIT VI - BIO-MASS and Geothermal Energy

Principles of Bio-Conversion, Anaerobic/aerobic digestion, types of
Bio-gas digesters, gas yield, combustion characteristics of bio-gas,
utilization for cooking.
Resources, types of wells, methods of harnessing the energy,
potential in India.

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UNIT V - Direct Energy Conversion

Need for DEC, Carnot cycle, limitations, principles of DEC.
Thermo-electric generators, Seebeck, Peltier and Joule Thomson
effects, Figure of merit, materials, applications, Fuel cells,
principles, Faraday’s laws, thermodynamic aspects, selection of
fuels and operating conditions.

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UNIT V - Direct Energy Conversion

Need for DEC, Carnot cycle, limitations, principles of DEC.
Thermo-electric generators, Seebeck, Peltier and Joule Thomson
effects, Figure of merit, materials, applications, Fuel cells,
principles, Faraday’s laws, thermodynamic aspects, selection of
fuels and operating conditions.

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Session 1: RENEWABLE ENERGY

Learning Outcome: The student will be able to identify renewable
energy sources and their potential applications in their locality.

Session 2: SOLAR CONSTANT

Learning Outcome: The student will be able to calculate the solar
constant and understand its importance.

Session 3: SOLAR RADIATION

Learning Outcome: The student will be able to differentiate
between extraterrestrial and terrestrial solar radiation.

Session 4: SOLAR RADIATION ON TITLED SURFACES

Learning Outcome: The student will be able to measure and
analyze solar radiation on inclined planes.
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Session 5: SOLAR RADIATION MEASURING INSTRUMENTS

Learning Outcome: The student will be able to operate basic solar
radiation measurement instruments.

Session 6: SOLAR RADIATION DATA

Learning Outcome: The student will be able to analyze solar
radiation data to identify optimal locations for solar power plants.

Session 7: QUIZ AND REVIEW SESSION

Learning Outcome: The student will be able to consolidate their
understanding of solar radiation concepts.

Session 8: FLAT PLATE AND CONCENTRATING COLLECTORS

Learning Outcome: The student will be able to consolidate their
understanding of solar radiation concepts.
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Session 9: ORIENTATION AND THERMAL ANALYSIS

Learning Outcome: The student will be able to evaluate the
thermal efficiency of collectors based on orientation.

Session 10: ADVANCED COLLECTORS

Learning Outcome: The student will be able to understand
advanced solar collection technologies and their applications.

Session 11: SOLAR ENERGY STORAGE METHODS

Learning Outcome: The student will be able to explain various
energy storage mechanisms.

Session 12: SOLAR PONDS AND APPLICATIONS

Learning Outcome: The student will be able to understand the
construction and applications of solar ponds

Session 13: Buffer Class 9/ 110

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Session 14: WIND ENERGY: SOURCES AND POTENTIALS

Learning Outcome: The student will be able to identify wind
energy sources and map potential wind farm locations in India.

Session 15: HORIZONTAL AND VERTICAL AXIS WINDMILLS

Learning Outcome: The student will be able to differentiate
between horizontal and vertical axis wind turbines and evaluate
their pros and cons.

Session 16: PERFORMANCE CHARACTERISTICS AND BETZ

LAW
Learning Outcome: The student will be able to calculate the
theoretical maximum efficiency of wind turbines using Betz’s law.

Session 17: OCEAN ENERGY: OTEC PRINCIPLES

Learning Outcome: The student will be able to calculate the
theoretical maximum efficiency of wind turbines using Betz’s law. 10/ 110
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Session 18: TIDAL AND WAVE ENERGY

Learning Outcome: The student will be able to identify tidal and
wave energy conversion techniques and their potential applications
in India.

Session 19: MINI-HYDEL POWER PLANTS

Learning Outcome: The student will be able to design a
mini-hydropower plant for a specific location.

Session 20: REVIEW AND PROBLEM SOLVING

Learning Outcome: The student will be able to solve problems
related to wind and ocean energy systems.

Session 21: PRINCIPLES OF BIO-CONVERSION

Learning Outcome: The student will be able to explain the
processes of anaerobic and aerobic digestion.
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Session 22: BIO-GAS DIGESTERS

Learning Outcome: The student will be able to design a simple
biogas system.

Session 23: CHARACTERISTICS OF BIO-GAS

Learning Outcome: The student will be able to analyze the
combustion characteristics of bio- gas.

Session 24: GEOTHERMAL ENERGY RESOURCES

Learning Outcome: The student will be able to identify geothermal
resource regions in India.

Session 25: APPLICATIONS OF GEOTHERMAL ENERGY

Learning Outcome: The student will be able to explain practical
applications of geothermal energy.
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Session 26: REVIEW AND DISCUSSIONS

Learning Outcome: The student will be able to discuss the key
concepts of bio-mass and geothermal energy.

Session 27: QUIZ

Learning Outcome: The student will be able to assess their
knowledge of bio-mass and geothermal energy concepts.

Session 28: DEC AND CARNOT CYCLE

Learning Outcome: The student will be able to explain the need
for Direct Energy Conversion (DEC) and calculate efficiency using
Carnot cycle equations.

Session 29: THERMOELECTRIC GENERATORS

Learning Outcome: The student will be able to describe the
Seebeck, Peltier, and Thomson effects and their role in
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Session 30: MATERIALS FOR DEC

Learning Outcome: The student will be able to identify properties
and selection criteria for materials used in DEC systems.

Session 31: FUEL CELLS

Learning Outcome: The student will be able to explain the
principles of fuel cells and identify their components.

Session 32: SELECTION OF FUELS AND OPERATING

CONDITIONS
Learning Outcome: The student will be able to optimize the
operation of fuel cells under various conditions.

Session 33: APPLICATIONS OF DEC

Learning Outcome: The student will be able to evaluate real-world
applications of DEC technologies.
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Session 34: REVIEW AND DISCUSSIONS

Learning Outcome: The student will be able to consolidate their
understanding of DEC and apply it to a practical problem.

Session 35: COURSE REVIEW AND FINAL QUIZ

Learning Outcome: The student will be able to review the key
concepts of renewable energy and DEC and assess their
understanding through a final quiz.

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Session 34: REVIEW AND DISCUSSIONS

Learning Outcome: The student will be able to consolidate their
understanding of DEC and apply it to a practical problem.

Session 35: COURSE REVIEW AND FINAL QUIZ

Learning Outcome: The student will be able to review the key
concepts of renewable energy and DEC and assess their
understanding through a final quiz.

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Unit 1, Session 1: Renewable Energy

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Unit 1, Session 1: World Energy Consumption

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Unit 1, Session 1: Global energy use by end-use sectors

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Unit 1, Session 1: Total world primary energy supply by
fuel

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Unit 1, Session 1: Global electricity generation by fuel
type and source in 2015

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Unit 1, Session 1: Consequences of Fossil Fuel
Combustion

The switch from fossil fuels to renewable energy sources is

inevitable.
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Unit 1, Session 1: Sources of Renewable Energy

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 Unit 1, Session 1
Sources of Renewable Energy: Solar Energy

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 Unit 1, Session 1
Sources of Renewable Energy: Wind Energy

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 Unit 1, Session 1
Sources of Renewable Energy: Hydro Energy

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 Unit 1, Session 1
Sources of Renewable Energy: Geothermal Energy

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 Unit 1, Session 1
Sources of Renewable Energy: Biomass Energy

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Unit 1, Session 1: Activity

Visit https:
//iced.niti.gov.in/energy/electricity/generation
To do:
Make a table and pie-chart for electricity generation by fuel
type and source in 2024 for India.
Upload your document in CANVAS.

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Unit 1, Session 2: Solar Radiation

Any body above absolute zero temperature (-273K) emits

radiation.
A body that emits some radiation in the visible range is called
a light source. The sun is obviously our primary light source.
The electromagnetic radiation emitted by the sun is known as
solar radiation.
Almost half of solar radiation is light (i.e., it falls into the
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Unit 1, Session 2: Black-body Radiation

A blackbody is defined as a perfect emitter and absorber of

radiation. At a specified temperature and wavelength, no
surface can emit more energy than a blackbody. A blackbody
absorbs all incident radiation.
The radiation energy emitted by a blackbody per unit time
and per unit surface area was determined experimentally by
Joseph Stefan in 1879 is given as Eb (t) = σT 4 . Eb is the
blackbody emissive power.
where σ = 5.670 × 10−8 W /m2 K 4 is the Stefan-Boltzmann
constant and T is the absolute temperature of the surface in K 3030/ 110
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Unit 1, Session 2: Radiative Properties
Emissivity:
The emissivity of a surface is the ratio of the radiation
emitted by the surface at a given temperature to the radiation
emitted by a blackbody at the same temperature. The
emissivity of a surface is denoted by ϵ, and it varies between
0 < ϵ < 1. Emissivity is a measure of how closely areal surface
approximates a blackbody, for which ϵ = 1.
Irradiation:
The radiation flux incident on a surface from all directions is
called irradiation or incident radiation and is denoted by G . It
represents the rate at which radiation energy is incident on a
surface per unit area of the surface.
Absorptivity, Reflectivity, Transmissivity:
The fraction of irradiation absorbed by the surface is called
the absorptivity α, the fraction reflected by the surface is
called the reflectivity ρ, and the fraction transmitted is called 31/ 110
the transmissivity τ . 31 / 110
Unit 1, Session 2: Radiative Properties

Gabs + Gref + Gtr = G

α+γ+τ =1

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Problems: Problem 1
An opaque horizontal plate is well insulated on the edges and
the lower surface (below Figure). The irradiation on the plate
is 2500W /m2 , of which 800W /m2 is reflected. The plate has
a uniform temperature of 700K and has an emissive power of
9000W /m . Determine the total emissivity and absorptivity.

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Problems: Problem 1
An opaque horizontal plate is well insulated on the edges and
the lower surface (below Figure). The irradiation on the plate
is 2500W /m2 , of which 800W /m2 is reflected. The plate has
a uniform temperature of 700K and has an emissive power of
9000W /m . Determine the total emissivity and absorptivity.

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Unit 1, Session 2: Solar Constant

Gs = Total Solar Irradiance =

1373 W /m2
Solar Constant (solar
irradiance) is the rate at which
solar energy is incident on a
surface normal to the sun’s
rays at the outer edge of the
atmosphere when the Earth is
at its mean distance from the
L = mean distance between the centers
Sun.
of sun and earth.
r = radius of the sun The value of the total solar
Tsun = Temperature of the sun surface irradiance can be used to
≈ 5800 K (= 5526.85 ◦ C) estimate the effective surface
temperature of the sun from
the requirement that
(4πL2 )Gs = (4πr 2 )σTsun
4
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Total Solar Radiation

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Solar Radiation Scattering

Mechanism that attenuates solar radiation as it passes

through the atmosphere is scattering or reflection by air
molecules and the many other kinds of particles such as dust,
smog, and water droplets suspended in the atmosphere.
Air molecules scatter blue light much more than they do red
light. At sunset, light travels through a thicker layer of
atmosphere, which removes much of the blue from the natural
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Problems:
Problem 2
Determine the solar constant by considering the effective
surface temperature of the sun as Tsun = 5778K , Diameter of
Sun D = 1.393 × 109 m and mean distance of
L = 1.496 × 1011 m from the EARTH.

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Problems:
Problem 2
Determine the solar constant by considering the effective
surface temperature of the sun as Tsun = 5778K , Diameter of
Sun D = 1.393 × 109 m and mean distance of
L = 1.496 × 1011 m from the EARTH.

Solution Hint
(4πL2 )Gs = (4πr 2 )σTsun
4

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Problems:
Problem 2
Determine the solar constant by considering the effective
surface temperature of the sun as Tsun = 5778K , Diameter of
Sun D = 1.393 × 109 m and mean distance of
L = 1.496 × 1011 m from the EARTH.

Solution Hint
(4πL2 )Gs = (4πr 2 )σTsun
4

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Problems: Problem 3

Problem 3
Determine the solar constant by considering the effective
surface temperature of the sun as Tsun = 5778K , Diameter of
Sun, D = 1.393 × 109 m and mean distance of
L = 5.91 × 1011 m from the PLUTO.

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Solar Radiation on a Tilted Surface

https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface
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Solar Radiation on a Tilted Surface

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Solar Radiation on a Tilted Surface

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Solar Radiation on a Tilted Surface

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Solar Radiation on a Tilted Surface

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Solar Radiation on a Tilted Surface

https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface
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Solar Radiation on a Tilted Surface

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https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface 42 / 110
Declination Angle

https://www.pveducation.org/pvcdrom/properties-of-sunlight/declination-angle

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Declination Angle

https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface

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Solar Radiation Measuring Instruments

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Solar Radiation Measuring Instruments

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Solar Radiation Measuring Instruments

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Solar Radiation Measuring Instruments

Pyranometer Pyrheliometer Sunshine Recorder

google images; https://www.youtube.com/watch?v=PmkbJx1jdV4

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Pyranometer
Pyranometer is an instrument used to measure the amount of
solar radiation (power) produced by the sun in a specific
location.
Measures direct and diffusive radiation

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google images
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Pyranometer construction and Working

Thermopile Pyranometer - widely used

Silicon Photocell Pyranometer - less expensive

google images 47/ 110

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Pyrheliometer
Measure only direct beam solar radiation
It is used in measuring climatic conditions
Outer structure looks like long tube telescope

google images 48/ 110

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Pyrheliometer Construction and working

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google images 49 / 110
Sunshine Recorder
It is a device that measure, how much sun light is received by
a region or area at any given time.
It is used to collect the data from same place over long time.

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Sunshine Recorder Construction and working

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google images 51 / 110
Sunshine Recorder Construction and working

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google images 52 / 110
Solar Radiation Data

TMY Data: Typical Meteorological Year Data

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Determining TMY Data
Various meteorological measurements are made at hourly
intervals over many years to build up a picture of local climate.
Typical meteorological year (TMY) is a collation of selected
weather data for a specific location, listing hourly values of
solar radiation and meteorological elements for a one-year.
The values are generated from a data bank much longer than
a year in duration, at least 12 years.
A simple average of the yearly data underestimates the
amount of variability, so the month that is most representative
of the location is selected.
For each month, the average radiation over the whole
measurement period is determined, together with the average
radiation in each month during the measurement period.
The data for the month that has the average radiation most
closely equal to the monthly average over the whole
measurement period is then chosen as the TMY data for that
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Sample Data

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CLASS TEST 1

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CLASS TEST 1

Determine the solar constant for Jupiter that is 777 millions

km away from the sun. Take the effective surface temperature
of the sun as Tsun = 55XX ◦ C and the diameter of the sun as
1.393 × 106 km. Take XX as last two digits of your roll
number.

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CLASS TEST 1

Determine the solar constant for Jupiter that is 777 millions

km away from the sun. Take the effective surface temperature
of the sun as Tsun = 55XX ◦ C and the diameter of the sun as
1.393 × 106 km. Take XX as last two digits of your roll
number.
Determine the declination angle at a location of latitude 23◦
15’ N and longitude 77◦ 30’ E on your date of birth. Also
determine the elevation angle.

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Unit 2 - Solar Energy Collection: Solar Water Collector

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Active Solar Water Heater

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

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Flat Plate Solar Collector

FR Solar heat removal factor

Kτ α function of the incident angle can be taken as 1.
U Overall heat transfer coefficient W /m2 ◦ C 60/ 110
Tc - Collector Temp.; Tw - Water Temp.; Ta - Air Temp. 60 / 110
Problems

Problem 1
The specifications of two flat-plate collectors are given as follows:
Single glazing: τ = 0.96 , α = 0.96, U = 9W /m2 ◦ C . Double
glazing: τ = 0.93 , α = 0.93, U = 6.5W /m2 ◦ C . The heat
removal factor for both collectors is 0.95, the solar irradiation is
550 W /m2 and the ambient air temperature is 23◦ C . For each
collector, determine
The collector efficiency when the water enters the collector at
45◦ C
The temperature of water at which the collector efficiency is
zero
The maximum collector efficiency. Take the incident angle
modifier to be 1.

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Solution

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Solution

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Solution

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

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Concentrating Solar Collector

Aa - Aperture area, Ar - Receiver area

CR - Concentration factor
Qr - Radiation supplied to the receiver
ηar - Optical Efficiency, ηc - Collector Efficiency
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Problems

Problem 2
Two concentrating collectors (collector A and collector B) have the
same concentration factor of CR = 9 and the optical efficiency of
ηar = 0.85. The collector temperature for both collectors is 350◦ F
and the ambient air temperature is 85◦ F. The heat loss coefficient
for collector A is 0.45 Btu/h.ft2 .◦ F and that for collector B is 0.63
Btu/h.ft2 .◦ F. The solar irradiation on collector A is 180 Btu/h.ft2 .
Determine the solar irradiation rate of collector B so that both
collectors have the same efficiency.
Use the following conversion factors: 1 Btu = 1055.06 J, 1ft =
5
0.3048 m, 1◦ F = K.
9

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Solution

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Solution

Using conversion factor, GB = 24700.606908365 J/h.m2 .

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Solar Concentrator Power Plant

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Solar Pond

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Solar Pond

Salt is added to the pond to stop natural convection by

making it denser (e.g., Dead Sea).
The cold water on the top of the pond acts as insulation.

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Problem

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Solution

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UNIT III: Wind Energy

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Wind Farms

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Horizontal axis turbines

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Vertical axis turbines

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Wind Farm simulation

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HAWT vs VAWT: Pros and Cons

Feature Horizontal Axis (HAWT) Vertical Axis (VAWT)

Orientation Horizontal Vertical
Efficiency Higher Lower
Maintenance Difficult (at high heights) Easier (near ground level)
Wind Direction Needs to face wind Omni-directional
Self-starting Yes Usually no
Application Large-scale wind farms Small-scale, urban environments

Table: Comparison of Horizontal and Vertical Axis Wind Turbines

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Wind Power Potential

A wind turbine converts the kinetic energy of the fluid into power.
If the wind is blowing at a location at a velocity of V, the available
wind power is expressed as
1
Ẇavailable = ṁV 2 (1)
2

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Wind Power Potential

A wind turbine converts the kinetic energy of the fluid into power.
If the wind is blowing at a location at a velocity of V, the available
wind power is expressed as
1
Ẇavailable = ṁV 2 (1)
2
This is the maximum power a wind turbine can generate for the
given wind velocity V. The mass flow rate is given by

ṁ = ρAV (2)
where ρ is the density and A is the disk area of a wind turbine (the
circular area swept out by the turbine blades as they rotate).
Substituting,

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Wind Power Potential

A wind turbine converts the kinetic energy of the fluid into power.
If the wind is blowing at a location at a velocity of V, the available
wind power is expressed as
1
Ẇavailable = ṁV 2 (1)
2
This is the maximum power a wind turbine can generate for the
given wind velocity V. The mass flow rate is given by

ṁ = ρAV (2)
where ρ is the density and A is the disk area of a wind turbine (the
circular area swept out by the turbine blades as they rotate).
Substituting,
1
Ẇavailable = ρAV 3 (3)
2
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The power potential of a wind turbine is proportional to the
density of air.

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The power potential of a wind turbine is proportional to the
density of air.
As a result, cold air has a higher wind power potential than warm
air.

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The power potential of a wind turbine is proportional to the
density of air.
As a result, cold air has a higher wind power potential than warm
air.
The density of air (treated as an ideal gas) can be determined
using P = ρRT .

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The power potential of a wind turbine is proportional to the
density of air.
As a result, cold air has a higher wind power potential than warm
air.
The density of air (treated as an ideal gas) can be determined
using P = ρRT .
πD 2
The disk area A =
4

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The power potential of a wind turbine is proportional to the
density of air.
As a result, cold air has a higher wind power potential than warm
air.
The density of air (treated as an ideal gas) can be determined
using P = ρRT .
πD 2
The disk area A =
4

Thus,

π PD 2 V 3
Ẇavailable = (4)
8 RT

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Problem: Wind power potential in a location

A wind turbine with a blade diameter of 2743.2 cm is to be

installed in a location where average wind velocity is 6.1 m/s. The
average temperature and pressure of ambient air in this location are
75◦ F and 1 bar, respectively. Determine the wind power potential.

Figure: A wind turbine

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Solution:

For Air, R = 287 J/kg·K

P 105
Therefore, ρ = = = 1.173kg /m3
RT 287 × 297.039

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Solution:

For Air, R = 287 J/kg·K

P 105
Therefore, ρ = = = 1.173kg /m3
RT 287 × 297.039
πD 2 π × 27.432
Area A = = = 590.94m2
4 4

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Solution:

For Air, R = 287 J/kg·K

P 105
Therefore, ρ = = = 1.173kg /m3
RT 287 × 297.039
πD 2 π × 27.432
Area A = = = 590.94m2
4 4
Mass flow rate ṁ = ρAV = 1.173 × 590.94 × 6.1 = 4228.35kg /s

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Solution:

For Air, R = 287 J/kg·K

P 105
Therefore, ρ = = = 1.173kg /m3
RT 287 × 297.039
πD 2 π × 27.432
Area A = = = 590.94m2
4 4
Mass flow rate ṁ = ρAV = 1.173 × 590.94 × 6.1 = 4228.35kg /s
Wind power potential
1 1
Ẇavailable = ṁV 2 = ×4228.35×6.12 = 78668.45W = 78.7kW .
2 2

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Solution:

For Air, R = 287 J/kg·K

P 105
Therefore, ρ = = = 1.173kg /m3
RT 287 × 297.039
πD 2 π × 27.432
Area A = = = 590.94m2
4 4
Mass flow rate ṁ = ρAV = 1.173 × 590.94 × 6.1 = 4228.35kg /s
Wind power potential
1 1
Ẇavailable = ṁV 2 = ×4228.35×6.12 = 78668.45W = 78.7kW .
2 2

Wind Power Density (WPD)

Ẇavailable 1
WPD = = ρV 3 (5)
A 2

Ẇ available 1 3
WPDavg = = ρV (6)
A 2
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Criteria for selecting Wind Turbine Location

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Problem

Consider two locations, location A and location B, with average

wind power densities of 250 W/m2 and 500 W/m2 , respectively.
Determine the average wind speed in each location. Take the
density of air to be 1.18 kg/m3 . If a turbine with a diameter of 40
m is to be installed in location A and a turbine with a diameter of
20 m is to be installed in location B, what is the ratio of wind
power potentials in location A to location B?

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Solution:

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Wind Turbine Efficiency

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Wind Turbine Efficiency

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Wind Turbine Efficiency

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Betz Limit for Wind Turbine Efficiency

ηwt,max = 0.5926

This value of ηwt,max represents the maximum possible

efficiency of any wind turbine and is known as the Betz limit.
All real wind turbines have a maximum achievable efficiency
less than this due to irreversible losses, which have been
ignored in this ideal analysis.

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Problem: Betz Limit

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Solution: Betz limit

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Ocean Thermal Energy Conversion (OTEC): Principle

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Ocean Thermal Energy Conversion (OTEC): Principle

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Carnot Heat Engine

η = Useful work/Heat Taken from the source

W QH − QC QC TC
η= = =1− =1− 97/ 110
QH QH QH TH 97 / 110
Problem

1) Calculate the efficiency of an OTEC plant with Ocean surface

and deep temperatures 28◦ C and 4◦ C respectively.

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Problem

1) Calculate the efficiency of an OTEC plant with Ocean surface

and deep temperatures 28◦ C and 4◦ C respectively.

2) Calculate the work output of an OTEC plant with input heat

flow to the plant is 2578 kW with plant efficiency 0.1 or 10 percent.

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Open-system OTEC Plant - Claude cycle

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Closed-cycle OTEC Plant - Anderson cycle

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OTEC Plant in Lakshadweep

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Wave Energy

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Tidal Energy
The tidal motion of ocean and seawater is due to gravitational
force of the moon and that of the sun.
These forces balance the centrifugal force on the water due to
rotation of the earth.

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Tidal Energy

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Hydel Plants

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Hydel Plants Advantages

Hydroelectric energy is a continuously renewable energy source

It is a much more concentrated energy resource than either
wind or solar power.
Hydroelectric energy is non-polluting—no heat or noxious
gases are released.
Hydroelectric energy has low operating and maintenance
costs, it is essentially inflation proof.
Small hydro plants can be tailored to the needs of the end use
market within the limits of water resources available.
There is no need of long transmission lines because the output
is consumed near the source.
High performing electrical equipment (alternator, control
circuit, battery storage, regulator, etc.) can be easily found in
the market.
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Hydel Plants Disadvantages
Hydro systems, unlike solar components for example do
require some maintenance.
The quality of small power station is not as good as that of
bigger one as these power plants are generally designed on the
basis of short term raw data. Thus the ground conditions of
operation are much different from those considered for the
design.
Majority of SHPs are located in remote places and not
connected with the grid. Therefore, transmission of surplus
power to other places is not possible.
revenue collection.
In the absence of adequate hydrological and geological data
there are always uncertainties about its potential as a resource.
Also, once commissioned there is no surety of a buyer at a
rate that is comparable to the outcome of the investment.
The rotation of turbines can kill fishes, especially young fishes
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Hydel Power
Available gross power, Pnet = ηo mgh
Problem: Estimate the power available from a proposed micro
hydro scheme at a site having a small stream with 100 litres
per second flow at a head of 30 m. Assume density of fresh
water as 996 kg/m3 and overall efficiency of the whole system
as 55%.

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Hydel Power
Available gross power, Pnet = ηo mgh
Problem: Estimate the power available from a proposed micro
hydro scheme at a site having a small stream with 100 litres
per second flow at a head of 30 m. Assume density of fresh
water as 996 kg/m3 and overall efficiency of the whole system
as 55%.

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RENEWABLE ENERGY RESOURCES

UNIT-IV BIOMASS ENERGY

Dr. Kumar Sourav

Assistant Professor
Mechanical Engineering Department
 Biomass Energy

● Biomass is an organic renewable energy. It is mostly produced from agriculture and

forest products and residues, energy crops, and algae.
● Biomass can be obtained from variety of resources called feedstocks.

● Biomass Resources: Dedicated Energy Crops, Agricultural Crops, Agriculture Crop

Residues, Forestry Residues, Aquatic Crops, Municipal Waste, Animal Waste
 CONVERSION OF BIOMASS TO BIOFUEL
1. Biochemical - based conversion processes
2. Thermo chemical-based conversion processes.

Biochemical - based conversion processes: In biochemical conversion

processes, enzymes and microorganisms are used as biocatalysts to convert
biomass or biomass-derived compounds into desirable products.

Thermo chemical-based conversion processes: Heat energy and chemical

catalysts are used to break down biomass into intermediate compounds or
products.
 BIOMASS PRODUCTS

● A major product of biomass is biofuels, which are a replacement for

petroleum-based fuels. Biofuels can be liquid or gas.
1. Ethanol: Ethanol is made primarily from the starch in corn grain. Corn, sugar
beets, sugar cane, and even cellulose (wood and paper) are some of the
sources of ethanol.
C6 H12 O6 → 2 C2 H5 OH + 2 CO2
(Glucose) (Ethanol) (Carbon Dioxide)

Numerical 1: In the production of ethanol, the feedstock high in sugar content is first
converted to sugar, and the sugar (glucose) is fermented into ethanol. Consider 100 kg
of sugar beet roots whose sugar content represents 30 percent of total mass. How
much ethanol can be produced from these sugar beet roots?
Solution-
Given: Molar Mass of C= 12 kJ/kmol, H= 1 kJ/kmol, O=16 kJ/kmol
C6 H12 O6 → 2 C2 H5 OH + 2 CO2
C6 H12 O6: (12*6)+(1*12)+(16*6)= 180 kg/kmol
C2 H5 OH: (12*2)+(1*5)+(16*1)+(1*1) = 46 kg/kmol
Given total mass of Sugar Beet root: 100 Kg
And Sugar (Glucose) Content : 30 %
Therefore, total Sugar/Glucose (in Kg): 100 * 0.30 = 30 kg

Ratio of Molar mass of Ethanol/Glucose: 2 C2 H5 OH/C6 H12 O6 = (2*46) / 180 = 0.511

Hence If Glucose is 30 Kg then 0.51 times will be the methanol.
Therefore, methanol mass = 0.51 * 30 = 15 kg
2. Biodiesel: Biodiesel is ethyl or methyl ester that is produced through a process
that combines organically derived oils with ethanol or methanol in the presence of a
catalyst. Common sources of biodiesel include new and used vegetable oils, animal
fats, and recycled restaurant greases.

Vegetable oil Ethanol Ethyl Ester

This process is called transesterification, and using ethanol makes the biodiesel a
bit more renewable since ethanol can be produced from biomass like sugarcane or
corn.
3. Methanol: Methanol can be produced from fossil sources or biomass. Pure
methanol and its blend with gasoline have been extensively tested as an alternative
fuel to gasoline. Two common mixtures are M85 (85 percent methanol, 15 percent
gasoline) and M10 (10 percent methanol, 90 percent gasoline).

4. Pyrolysis Oil: Pyrolysis oil is produced when biomass is exposed to high

temperatures without the presence of air, causing it to decompose. A possible
reaction involves heating of cellulosic feedstock in grain form for a short period (less
than half a second) to a temperature of 400°C to 600°C and quenching it.

5. Biogas: Biogas can be produced from biological waste such as animal manure,
food waste, and agricultural waste. The process is called anaerobic digestion which
is the decomposition of organic waste into a gaseous fuel by bacteria action without
the presence of oxygen. Biogas, also called swamp gas, landfill gas, or digester gas.
Comparative chart of Heating value for the Bio fuel
S.No. Fuel HHV LHV

1 Gasoline 47,300 kJ/kg 44,000 kJ/kg

2 Ethanol/Ethyl Alcohol 29,670 kJ/kg 26,810 kJ/kg

3 BioDiesel 40,700 kJ/kg 37,200 kJ/kg

4 Methanol 22,660 kJ/kg 19,920 kJ/kg

5 Biogas with 50 % methane by 14,800 kJ/kg –

volume

6 Biogas with 80 % methane by 32,900 kJ/kg –

volume
Numerical 2: A certain biogas consists of 75 percent methane (CH 4) and 25 percent carbon
dioxide (CO2 ) by volume. If the higher heating value (HHV) of methane is 55,530 kJ/kg and the
lower heating value (LHV) is 50,050 kJ/kg, what are the HHV and LHV of this biogas?
Methane 75%
Solution: Given- Biogas
Carbon Dioxide 25 %

HHV for Methane: 55,530 KJ/Kg, LHV: 50,050 KJ/Kg

Molar Mass of Methane (CH4): 12+(1*4)= 16 Kg/Kmol

Molar Mass of Carbon Dioxide (CO2): 12+(16*2)= 44 Kg/Kmol

Total molar mass of Product of gas evolved from biogas: Methane + Carbon dioxide = [(0.75*16)+(0.25*44)] = 23 kg

Mass fraction of methane : mass of methane / Total mass = 12/23 = 0.522

Heating value of Carbon dioxide: 0

Since Methane (HHV)= 55,530 KJ/Kg therefore Biogas value will be : 0.522*55,530= 28986.66 KJ/Kg and

Methane (LHV)= 50,050 KJ/Kg therefore, Biogas value will be : 0.522*50,050= 28986.66 KJ/Kg
 ELECTRICITY AND HEAT PRODUCTION BY BIOMASS
● The production of electricity and heat from biomass is called biopower.
● There are three technologies used to convert biomass energy to heat and
electricity:

❖ Direct combustion
❖ Co-firing, and
❖ Anaerobic digestion
Direct combustion: Biomass consisting of waste wood products (i.e., wood pellet) can
be burned in conventional boilers to generate steam or hot water. This steam is run
through a turbine coupled with a generator to produce electricity.
Co-firing: It is replacing only a portion of fossil fuel in coal-fired boilers with biomass.
Anaerobic digestion: In this process, organic matter is decomposed by bacteria in the
absence of oxygen to produce natural gas consisting primarily of methane and other
by-products such as carbon dioxide.
 SOLID MUNICIPALITY WASTE
● Municipal Solid Waste (MSW) includes mostly organic materials such as paper, food scraps,
wood, and yard trimmings, but some fossil content such as plastic also exists.

● Most of MSW come from residences (55 to 65%), and 35 to 45% come from businesses,
schools, and hospitals. MSW does not include industrial, hazardous, or construction waste.
What is the best use of municipal solid waste? Is it better to burn or bury
waste when trying to recover energy and minimize emissions?

1. Waste to Energy (WTE)

2. Landfill-Gas-to-Energy (LFGTE)
 THANK YOU
FOR YOUR ATTENTION
RENEWABLE ENERGY RESOURCES

UNIT-IV
GEOTHERMAL ENERGY

Dr. Kumar Sourav

Assistant Professor
Mechanical Engineering Department
DEFINITION OF GEOTHERMAL
ENERGY
•Geothermal energy is the heat energy stored
beneath the Earth's surface, which can be
harnessed for electricity generation, heating, and
industrial applications. This energy originates from
the Earth’s core, where temperatures can exceed
6000°C.
 Classification of Geothermal Resources
Geothermal resources are classified based on their temperature, pressure, and composition, which
determine their potential for energy production.

The four main categories include hydrothermal, geopressurized, magma, and enhanced
geothermal systems (EGS).

1. Hydrothermal
Resources:
Characteristics
•These reservoirs are the most commonly used geothermal
sources.

•They are found near tectonic plate boundaries or regions with

high geothermal activity.

•Temperature varies but is typically above 150°C.

•Extraction involves drilling wells into the reservoir to bring the

hot fluid or steam to the surface.
 2. Geopressurized Resources
Definition
Geopressurized reservoirs contain hot liquid water trapped under extreme pressure (up to 600 bar) in deep
sedimentary formations. The temperature typically ranges from 150°C to 180°C.
Characteristics
•These reservoirs are found in deep, sediment-filled basins.
•The fluid contains high concentrations of dissolved solids, making it highly corrosive.
•Methane gas is often dissolved in the water, presenting an additional energy resource.
•Extraction is challenging due to the high pressure and corrosive nature of the fluids.

3. Magma Resources
Definition
Magma, or molten rock, is found beneath active volcanoes and contains extremely high thermal energy, with
temperatures above 650°C.
Characteristics
•These resources are the hottest known geothermal sources, but technology for direct energy extraction is
not yet fully developed.
•Magma is formed by the melting of rock deep inside the Earth due to high temperatures and pressures.
 4. Enhanced Geothermal Systems (EGS) (Hot Dry
Rock Systems)
Definition
Enhanced Geothermal Systems (EGS), also known as
hot dry rock systems, are artificial geothermal
reservoirs created by injecting water into naturally hot
rock formations deep underground.
Characteristics
•Unlike hydrothermal systems, these resources do not
have naturally occurring water.
•Water is pumped into the hot rock under high pressure,
creating fractures.
•The water absorbs heat from the rock, turns into steam,
and is extracted through a production well.
•Depths typically range from 3 to 5 km, with
temperatures reaching around 250°C.
 Comparison of Geothermal
Resource Types

Resource Type Temperature Range Depth Water Content Challenges

Limited to areas with

Hydrothermal 150°C+ 1-3 km Natural hot water
geothermal activity

High-pressure hot
Highly corrosive fluids,
Geopressurized 150°C - 180°C 3-6 km water with dissolved
high pressure
solids

No efficient technology
Magma 650°C+ 5-10 km Molten rock
for energy extraction

Artificial (water High cost, induced

Enhanced (EGS) ~250°C 3-5 km
injected) seismicity
 GEOTHERMAL APPLICATIONS
•Electricity Production – Used in geothermal power plants to generate electricity.

•Space Heating & Cooling – Provides heating and cooling for buildings.

•Cogeneration – Simultaneous production of electricity and heat.

•Geothermal Heat Pumps – Used for residential and commercial heating and cooling.

•Greenhouses – Supports plant and crop growth in controlled environments.

•Drying Processes – Used for drying lumber, fruits, and vegetables.

•Spas & Bathing – Hot springs used for relaxation and health benefits.

•Desalination – Helps in freshwater production by removing salts from seawater.

•Fish Farming – Maintains optimal water temperatures for aquaculture.

Geothermal Heat for Space Heating in
District Heating Systems
Geothermal energy is commonly used for space
heating in large-scale district heating systems. These
systems distribute heat to multiple buildings within a
district or community. However, direct circulation of
geothermal water is often avoided due to its
undesirable chemical composition and corrosive
properties. Instead, a heat exchanger system is used
to safely transfer heat from geothermal water to fresh
water, which is then circulated through the district.
Examples of Geothermal District Heating
•Reykjavik, Iceland – One of the world’s largest
geothermal district heating networks, supplying 90% of
the city’s heating needs.
•Paris Basin, France – Geothermal heating serves
thousands of homes using warm underground water
reservoirs.
•Boise, Idaho, USA – The oldest geothermal district
heating system in the U.S., operating since 1892.
 .

Suppose, The heater has some

efficiency,

Energy cost = Energy consumption * Unit price of energy

Numerical 1:
A residential district is currently heated using natural gas heaters with an average efficiency of 85%. The
cost of natural gas is $1.30 per therm, where 1 therm = 105.5 MJ.
It is proposed to heat the district using geothermal water. On an average winter day, geothermal water is
supplied at 93.33°C (200°F) at a mass flow rate of 24.95 kg/s (55 lbm/s) and returns at 54.44°C (130°F)
after transferring heat to the district.
If the heating system operates for 2800 hours per year under these average conditions and the
geothermal heat is sold at a 25% discount compared to natural gas, determine the total revenue that
can be generated annually.
 Energy Consumption Calculation Method
The Degree-Day Method is a simple and widely used technique for estimating the
annual energy consumption of heating and cooling systems in buildings. It is
based on the difference between the outdoor temperature and a reference indoor
temperature over time.
Degree-Days (DD): A measure of how much (and for how long) the outdoor temperature deviates from a
base temperature, typically 18°C (or 65°F) for buildings.
*24
 Numerical 2:
Consider a house with an overall heat loss coefficient of K overall = 0.5 kW/°C and a heating degree days of 2500°
C-days. Determine the annual heating energy consumptions for the following heating systems.
(a) Coal heater, heater efficiency = 0.75, Heating value of coal = 30,000 kJ/kg
(b) Natural gas heater, heater efficiency= 0.85
(c) Heat pump, COP = 2.5
(d) Resistance heater, heater efficiency = 1
(e) Geothermal, heater efficiency = 1

Solution:
(a)

(b) (c) (d) (e)

 GEOTHERMAL COOLING

Pressure: 3.5-5 bar,

Temperature: -10 to -15 ℃
GEOTHERMAL HEAT PUMP SYSTEMS
 Types of Geothermal Heat Pump Loops
Closed-Loop Systems

1.Horizontal Loop: Pipes are buried in shallow

trenches, suitable for residential properties with
enough land.

2. Vertical Loop: Deep boreholes (100–500 feet)

are drilled, making this ideal for small properties.

3. Pond/Lake Loop: Pipes are submerged in a

body of water, providing cost-effective installation.

Open-Loop Systems
•Use groundwater directly from a well or pond,
pumping it through the heat pump and discharging
it back to the source.
Horizontal Loop Type
Vertical Loop Type
 Compare performances of an
Air-source heat pump and a ground-source heat
pump

The maximum COP of the ground-source heat pump is 67

percent greater than that of the air-source heat pump.
THANK YOU FOR YOUR ATTENTION
 UNIT-V
DIRECT ENERGY CONVERSION

Dr. Kumar Sourav

Assistant Professor
Department of Mechanical Engineering
 Direct Energy Conversion (DEC)
Direct Energy Conversion (DEC) refers to the process of converting one form of energy directly
into another without intermediate mechanical processes. It is widely used in various
applications where conventional mechanical energy conversion is inefficient or impractical.
Why is DEC Needed?

1.Higher Efficiency – Traditional energy conversion methods (e.g., thermal power plants) involve
multiple stages, leading to energy losses. DEC eliminates these intermediate steps, improving
efficiency.
2.Compact and Lightweight Systems – DEC devices, such as thermoelectric generators, fuel cells,
and photovoltaic cells, are more compact and suitable for space applications, portable electronics, and
biomedical implants.
3.Reliability and Durability – Since DEC systems have fewer moving parts, they are more reliable,
require less maintenance, and have longer operational life spans.
4.Sustainability – DEC enables the direct conversion of renewable energy sources (e.g., solar,
geothermal) into usable electrical energy, reducing dependence on fossil fuels.
5.Silent Operation – Unlike traditional mechanical engines, DEC systems operate silently, making them
ideal for medical, space, and military applications.
6.Usability in Extreme Conditions – DEC technologies function effectively in space, deep-sea, and
 Principles of DEC (Direct Energy Conversion)

Direct Energy Conversion (DEC) involves converting energy from one form to another without
involving mechanical work. The following principles guide DEC systems:

1. Thermoelectric Principle
•Utilizes the See-beck effect, where a temperature gradient in a semiconductor or metal generates an
electric voltage.

See-beck effect
Application: Thermocouple as temperature sensor, thermo electric coolers
 2. Electromagnetic Induction
•Based on Faraday’s Law, where changing magnetic fields induce an electric current.
•Applied in MHD (Magnetohydrodynamic) generators and inductive power transfer.

3. Photoelectric Effect
•Discovered by Albert Einstein, it explains how photons can release electrons from a material,
generating electricity.
•Used in solar photovoltaic (PV) cells for direct sunlight-to-electricity conversion.
4. Electrochemical Reactions
•Involves redox reactions where chemical energy is directly converted to electricity.
•Used in fuel cells and batteries for efficient energy storage and conversion.

5. Nuclear Energy Conversion

•Direct conversion of nuclear radiation into electricity using radioisotope thermoelectric
generators (RTGs).
•Used in space probes, deep-sea research, and military applications.
6. Piezoelectric Effect
•Converts mechanical stress into electrical charge using piezoelectric materials.
•Used in sensors, energy harvesting from vibrations, and biomedical applications.
 CARNOT
CYCLE
•The Carnot Cycle is an idealized thermodynamic cycle designed to describe the most efficient heat engine
possible.

It operates between two thermal reservoirs:

•A hot reservoir at temperature TH
•A cold reservoir at temperature TL

It consists of four reversible processes:

•Isothermal Expansion (1 → 2): Heat QH absorbed at TH

•Adiabatic Expansion (2 → 3): No heat exchange, temperature drops to TL

•Isothermal Compression (3 → 4): Heat QL rejected at TL

•Adiabatic Compression (4 → 1): No heat exchange, temperature rises back to TH

 Assumptions of the Carnot Cycle:

1. Working fluid is an ideal gas that obeys PV=n*R*T

2. All processes are reversible (quasi-static and without friction)

3. No heat loss to surroundings—perfect insulation during adiabatic

steps

4. Perfect thermal contact with reservoirs during isothermal steps

5. No friction or mechanical losses (100% ideal components)

6. Cycle is closed—the system returns to its initial state

7. No entropy generation within the system

8. Temperature of reservoirs remains constant

1

2
4
Efficiency of Heat
Engine
Numerical
1:
Numerical
2:
 HYDROGEN: AN ENERGY CARRIER
• Hydrogen is a fuel with a higher heating value of 141,800 kJ/kg and a lower
heating value of 120,000 kJ/kg.
• One of the greatest advantages of hydrogen is that the exhaust of a hydrogen
engine does not contain carbon monoxide, sulfur, hydrocarbon, or carbon di-ox
ide emissions.

• The steam reforming is a much cheaper method of producing hydrogen

compared with the water electrolysis.
• However, using a fossil fuel source (natural gas) for producing hydrogen is not
a sustainable path.
• The efficiency of a typical electrolyzer is about 80 percent.
Why Hydrogen Storage Is Challenging
 Metal Hydride Storage
• Some metal alloys can absorb hydrogen atoms into their structure, forming
metal hydrides.
• Hydrogen is stored in a solid form, safely "trapped" inside the material. When
needed, a small amount of heat releases the hydrogen as a gas.
Summary: