foundation of energy
TRANSCRIPT
Foundations of Energy
Course aimsThis module aims to give a very broad overview into the range of issues relevant
to energy as a commodity. Given the vast range of topics and disciplines this
covers, it will miss out many important aspects but the aim is to equip the reader
with enough background information the engage fully with the more detailed
modules which follow in the course, and to provide enough information and
pointers to be able to join the current debate on energy with some confidence.
The expected learning objectives are:
• You will gain a broad appreciation of the global and national patterns of energy use.
• You will acquire an overview over current energy resources and the technological extraction, conversion, and transmission.
• Issues of appropriate energy use, energy saving and energy efficiency will be addressed.
• The main current activities to formulate and implement global and national policies will be addressed, and
• The impacts of energy use on society and the environment will be introduced.
• With the insight gained, you will be able to assess critically, and
contribute constructively to, the ongoing debates and strategies for a
sustainable energy production.
While no prior knowledge is absolutely necessary, it is helpful to have (or revise)
the basic knowledge of thermodynamics gained in an engineering degree. More
specifically, familiarity with the first and second law of thermodynamics and an
understanding of standard thermodynamic cycles will be useful.
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Course materialThis course is based on basic material supplied by us, some of which is contained in this booklet and some of which is available from the web-based Virtual Learning Environment, http://webct.eps.hw.ac.uk. Take your time to familiarise yourself with the online material, as it contains material to be learned, for example in the folder ‘Basic Notes’, other material gathered by previous students, links to online resources, as well as places where you have to complete course work.
Other material is gathered by you during your studies for this module. It is expected that you use the following resources in your research as much as possible:
- this booklet
- the online notes
- standard text books on Energy, Thermodynamics, Fluid Mechanics, and other engineering topics relevant to Energy. A list of some basic text books is given below and in the ‘Library and Resources’ section in the online module.
- Professional and scientific journals, as available in your library or work place
- Online library resources provided to you from Heriot Watt Library through internet connection and Athens Authentication, in particular:
o The Encyclopedia of Energy. C. C. Cleveland, Elsevier, 2004
o ISI World of Knowledge, probably the most comprehensive citation index for engineering, science, and other research journal publications
o Electronic scientific journals
o The websites of organisations, e.g the International Energy Agency.
Links to these resources are all available from the ‘Library and Resources’ section in the online module.
Textbooks
The last few years have seen the publication or revision of a good number of textbooks on various aspects energy, and any recommendation will be based on personal preference. Books I have used include
G. Boyle, B. Everett and J. Ramage (Editors) (2003). Energy systems and sustainability. Open University.
J. Ramage. Energy: a guidebook. Oxford University Press, 1997, 2nd edition.
G. J. Aubrecht. Energy. Prentice-Hall, 1995, 2nd edition, and
E. S. Cassedy and P. Z. Grossman. Introduction to Energy – Resources, Technology, and Society. Cambridge University Press, 1998, 2nd edition.
James A McGovern: The essence of Engineering Thermodynamics, Prentice Hall
Y. A. Çengel and R. H. Turner (2001). Introduction to Thermal-Fluid Sciences, McGraw-Hill
B.S. Massey (now B.S. Massey and Ward-Smith). Mechanics of Fluids, Chapman&Hall or Stanley Thornes, now 7th or 8th edition.
Edward Hughes (2002). Hughes electrical and electronic technology. Prentice Hall, 8th edition
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Course assessmentThe course is assessed by a combination of coursework and examination:
1.) Online exercises, worth 20% of the module grade
In the online module, you will find a set of numeric exercises to test your
understanding of the basics covered in the ‘Basic Notes’ published in the online
module, and to test your ability to carry out basic calculations. These exercises are
mainly concerned with calculating energy consumption, CO2 emissions, heat transfer,
or turbine performance.
2.) A critical essay for Distance Learners or an oral presentation for On-campus students, worth 30%.
As a Distance Learner, you will have a high freedom to choose a topic of your
interest. Based on your research, drawing on the resources outlined above, you will
discuss a particular issue relevant to the theme of the module. You will find more
detailed information in the online module
3.) A 2-hour exam, usually timetabled for January, worth 50%
This exam will follow a similar format as that of the two coursework components.
Some of the questions will require you to carry out some basic calculations, while
others are essay-type questions, in which you will be given the opportunity to discuss
an issue from a list of topic choices.
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1 The energy challenge
Introduction
The aim of the chapter is to give a very brief, and broad-brush, overview of the role energy plays
in life and how energy is generated and consumed. While we will require increasing amounts of
energy to sustain the world's population, the demand must be met in a way which can be
sustained by the world, not only in economic but also social and environmental terms.
Read also the first section of the online ‘Basic Notes’.
Once you have read this section and the online notes, you are ready to complete the first set of
online exercises
1 Why bother with ‘Energy’?
A very succinct summary of why it is worthwhile to study energy can be found in the
first paragraph of Energy, a guidebook:
Energy matters. Many of us pay good money for it, and many more of us
walk for miles every day to find it. Some of us become very cross when we
can’t get it, and some of us even got to war over it. None of us could survive
without it..
J. Ramage. Energy, a guidebook, p. 3
There is not much more to say about the importance of energy. Just a few illustrations
from recent history may underline these statements:
• The fuel bill is the second-largest item on an average UK resident’s monthly bill
after their mortgage or rent.
• Farmers and lorry drivers in the UK managed to affect seriously daily life when
they protested in 2000 over what they felt were excessive fuel prices.
• Life and the economy in California were seriously affected when the energy
industry was unable to supply uninterrupted electricity to its customers in 2001.
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1. The energy challenge
• While countries tried to avoid military action in many parts of the world, swift
action was taken, especially by the USA and UK, when Kuwait was invaded by
Iraq, which obviously had serious implications on global oil supplies.
Serious interest in energy as global ‘problem’ resulted from the oil crisis in the 1970s,
and many textbooks were written in the late 70s or early 80s. At that time many
predictions of doom were published and discussed – fuel running out within the lifetime
of the current generation etc., but life did not stop and many returned to life as usual,
putting energy on the backburner. More recently, it has returned to the headlines, this
time in connection with its impact on our environment. Opinions to the severity (or
existence) of the problems or which actions should be taken are as divided as ever.
2 Energy use, now and projected
Before we go into the details of what energy is and how it is produced and consumed,
it is worth looking at the global perspective of energy in today’s and tomorrow’s world.
During the 20th century, humans have consumed ten times more energy than in the
preceding millennium. This trend is due to both the dramatic increase in the world's
populations and the increased per-capita energy consumption in the industrialised
world. Also the forms of energy used has been extended tremendously. Before the
invention and widespread use of powered transport and before industrialisation, most
energy was used in the form of heating, cooking, and lighting. Now, however, energy is
used in a much wider variety, from the traditional heat and light generation to
transport and many domestic and industrial applications of electricity. The per-capita
use varies accordingly between the different continents. The typical power
consumption per person is largest in the US with 10kW, while in Europe the
consumption is about 4kW per person. At the other end of the spectrum, in poor
developing countries the power consumption is as low as 0.1kW per person. The vast
difference in the energy demand could suggest that there might be a large potential for
using energy much more efficiently in highly industrial countries. A 'back of the
envelope' calculation of energy demand in terms of population and a measure of energy
efficiency is illustrated in Textbox 1
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1. The energy challenge
Textbox 1: Calculation of Energy Demand for population:
While the average energy demand per person varies, one can probably relate it roughly to the standard
of living of a person living in a specific country. Using such an estimate, together with the total
population will give an estimate for the total energy demand, either for a country or the whole world.
Symbols used: R Total energy required for population
N Number of people in population
E Per capita energy demand
S Standard of living
f coefficient to relate per capita energy demand to standard of living
The total annual energy requirement is linked to the population and the per-capita requirement by
R = E N.
The standard of living and the per-capita energy requirement are related by the empirical factor f,
S = f E.
The factor f might be interpreted as an efficiency of how much energy consumption contributes to the
standard of living. Combining these two equations gives
R = S N / f.
Currently the world’s population is roughly N0= 6 billion (6×109), and the energy consumption in 1997
was approximately 9500Mtoe. With this, the per capita consumption was E0= 1.58toe. This equals the
ratio S/f.
Current predictions are that the population grows by 2 to 3% per year, and that the standard of living
increases by 2 to 5%.
Let us normalise the measure of the standard of living, S, such that f=1toe–1 in 1997. Then, the standard
of living in 1997 was S0= 1.58. If we assume 2% for both quantities, then the population by 2020 will be
N = N0×1.0223= 9.5×109, and the standard of living will be S= 2.49. If there is no increase in the
efficiency of the energy used, then f will remain constant, and the total energy demand for 2020 will be
R= SN/f= 23,700Mtoe!
Exercise:
Assume a population increase of 2% per year, but an increase in the standard of living by 4%. Advances
in technology, however, have quadrupled the efficiency of energy conversion to f= 4toe–4. Calculate the
energy requirement for the year 2020. Could today’s energy production satisfy the demand?
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1. The energy challenge
The International Energy Agency (IEA) estimated the energy consumption in 1997 to
be 9521Mtoe (Mega tonnes oil equivalent), of which 26% could be allocated to the USA
and Canada, 18% to Europe, and 11% to Russia. A rough estimate of the projected
energy requirement by 2020 is estimated to be over 12000Mtoe compared to
6000Mtoe in 1973 and 9500Mtoe in 1997 (See Figure 1). Energy statistics for a few
selected countries are shown in Table 1.
05
101520253035404550
1980 1990 2000 2010 2020 2030Year
Ene
rgy
[EJ]
Petroleum Natural Gas Coal
Nuclear Hydropow er Other
Figure 1. Energy supply by fuel from 1971 to 2020 (Source: IEA, 2006)
Table 1. Energy statistics for the USA, UK, Norway, India, China, and Zambia for 1997.
Country Population
(106)
Total Energy
(Mtoe)
Energy
Per cap.
(toe/capita)
Coal
%
Oil
%
Gas
%
Nuclear
%
Renewable
%
Combustible
Renewable
%
USA 269 2182 8.1 23.6 38.9 22.8 8.5 1.8 3.5
UK 59 233 3.9 17.5 35.9 34.2 11.3 0.2 0.8
Norway 4.4 25.4 5.8 4.3 34 17.1 0 39.5 5
India 980 476 0.5 33.7 19 4.1 0.6 1.5 40.1
China 1239 1031 0.8 56.9 18.8 1.9 0.4 1.7 20.3
Zambia 9.7 6.1 0.6 1.6 9.1 0 0 10.8 78.5
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1. The energy challenge
3 What is ‘energy’ and ‘power’
The Oxford English Dictionary defines energy in the following way:
‘Energy: … 6. Physics. The power of ‘doing work’ possessed at any instant
by a body or system of bodies,…’ (Oxford English Dictionary)
3.a Energy units and conversions
The basic SI unit of energy is joule (J), although kilojoule (kJ= 1000J) is more common
among engineers. When considering national or global amounts of energy, the kilojoule
is far too small to be useful, and Megajoule (106J), Gigajoule (109J), Petajoule (1012J), and
even Petajoule (1015J) are used.
For electricity production which must balance the demand minute by minute, the
instantaneous load, or rate of energy consumption, is more relevant than the accumulated
energy supplied or used. The rate of change of energy is measured in joule/second =
watt (W), or kW. For accounting purposes, the accumulated energy is then calculated
in kWh (1kWh= 3600kJ= 3.6MJ).
Since the vast majority of our energy consumption is based on the burning of fossil
fuels, it is customary to base energy on the energy released by that fuel. A major
difficulty with this is that due to differences in the quality of fuels from different mines
or reservoirs varies substantially, it is strictly speaking not possible to give an exact
figure of the energy contained in a tonne of coal or oil.
The energy supplied in the form of fossil fuels is measured in tonnes of oil equivalent
(toe) or Mtoe. The corresponding energy content of electricity is given as
1TWh (= 109kWh)= 0.086Mtoe (Source: IEA).
3.b Types of energy
Energy comes in many forms, ranging from potential energy of an object sitting
somewhere high up and having the potential to fall down – converting the potential
energy to motion – ‘kinetic energy’, to electric and electromagnetic energy. This is
only a brief list to illustrate types of energy:
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1. The energy challenge
• Heat
Q= m Cp ∆T (or m CV ∆T ) = mass × specific heat × temperature difference
• Potential energy
Ep= m g h = mass × gravitational acceleration × height.
• Kinetic energy
Ek= ½ m v2 = half × mass × velocity squared.
• Work
W= F d = force × distance
• Pressure work
Wp= p V= pressure × Volume
This is the same as applying on an area, A, a pressure force, F= pA, and moving it by
a distance d. This will change the volume within the boundary by d×A:
W= F d= p A d = p V
• Electrical energy
Eel= V Q= volt × charge.
• Radiation
energy of a photon: h ν= h c / λ,
• with Planck's constant, h= 6.625× 10 − 34 J s, and the frequency, ν, or the speed of
light, c, and wavelength, λ.
• 'Chemical' energy
The heat released in a chemical reaction. This is specific to each reaction and is
usually given as energy unit mass (e.g. kJ/kg) or number of molecules (e.g. kJ/mol)
• Atomic energy
E= m c2 = mass × speed of light squared, with c= 3 × 108 m/s
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1. The energy challenge
3.c Quality of energy
While the first law of thermodynamics states that all forms of energy are equal (in
numeric terms), the second law states that some forms are 'more equal than others'. By
insisting that the entropy—a quantity related to temperature—cannot decrease during
processes in thermodynamic equilibrium, the second law implies that any energy
conversion leads to generation of waste heat. It is thus much harder to convert heat
to mechanical energy than the other way round.
To gain a measure of the difference between the different forms of energy, the concept
of quality is defined:
Quality is defined as the proportion of an energy source than can be converted
to mechanical work.
Energy supplies can be broadly divided into three divisions by quality:
1. Mechanical supplies are those energy sources, which provide directly work from
potential or kinetic energy for extraction. Examples of mechanical supplies are
hydro, wind, waves, and tides. The conversion of the kinetic or potential energy
into electricity is usually quite high, and so is the proportion of the energy
extracted from the available energy current. Typical proportions are for wind 30%,
hydro 60%, and wave and tidal about 75%.
2. Heat supplies are those which in the first instance provide heat—either for direct
consumption or further conversion into work or electricity. Examples are all the
fossil fuels and nuclear energy, together with biomass combustion and solar heating.
Even in ideal situations, the conversion of the heat to other forms is limited by the
second law of thermodynamics, and in practice the efficiency of conversion
processes is about half of the ideal maximum. For example, for thermal boiler heat
engines, a realistic maximum quality is about 0.35 or 35%.
3. Photon processes convert the solar radiation into other forms of energy. Examples
are photosynthesis and photochemistry used for the production of biofuels and
photovoltaic conversion to electricity. In theory, a photovoltaic cell matched to a
single photon frequency can convert the solar radiation by absorption into
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1. The energy challenge
electricity at very high efficiencies. Because the sunlight is not at a frequency,
however, it is difficult to optimise a PV cell for the solar frequency range, and
current typical conversion efficiencies are of the order of 15%.
Example: Based on the second law statement above we can expect that a particular
but common use of fossil fuels is a very inefficient use of the current energy flow
chain: Heating of water or buildings by electricity generated by burning of fossil fuels
or nuclear fission. These types of electricity generation convert heat to electricity in
turbines. Since the conversion of heat to mechanical energy requires substantial
waste heat, much heat is usually lost (a typical efficiency is of the order of 30%).
Therefore, most of the energy potentially available in heat has been lost already at the
first stage. An improvement of practice becomes immediately obvious: Combined
Heat and Power (CHP) generation.
The typical efficiency of illumination by incandescent ('normal') light bulbs is even
more dramatic: a generation efficiency of 30%, distribution efficiency of 90%, and
lighting by generating black body radiation due to Ohmic heating of the light bulb's
filament (5%) lead to an overall efficiency of 1.4%! Using more efficiently generated
electricity (80%, e.g. in a CHP plant) and using modern energy saving lighting (20%),
increases the efficiency tenfold to 14%.
3.d What is ‘power’
All these energy units represent the potential to do something. To actually do
something, one has to use energy of one form at a given rate and convert it to another
form. The rate at which energy is used and converted (or generated or lost) is the
definition of power:
Power is the rate of doing work.
The basic unit of power is the watt: 1 W= 1 J/s= 1 kg m2 s-3
This is usually used in 1 kW= 1000 W; 1 MW= 106 W and higher powers.
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1. The energy challenge
4 ‘Energy’ as a commodity
Nuclear Coal
Gas
Oil
Biomass
Hydro
Ocean Wind Nuclear
Res
ourc
es
etc
Fuel Electricity Energy type
Transport Manufacture
Building Service industry Domestic Use
rs
Par
amet
ers
Environment Society Policy Legislation
Thermal plant
Generator Refinery
Tra
snm
issi
on,
Con
vers
ion
He a t
Sun
Figure 2. Schematic diagram of the ‘energy world’ with the flow of energy from the resource to the consumer embedded in the wider issue of society and the environment.
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1. The energy challenge
4.a The flow of energy
A useful way of illustrating the flow from energy from the primary sources via
conversions to the end use is that of a 'spaghetti' diagram. A relatively simple flow is
found for Norway, as can be seen in Figure 3
Figure 3. Spaghetti diagram for the flow of energy from primary source to end use (taken from Twidell and Weir, 19).
4.b Efficient use of energy
The following strategies will help to meet projected energy targets by using the
supplied energy—from whatever source—more efficiently, rather than by increasing
the demand on the primary energy supply. In terms of the calculation in Textbox 1,
these strategies aim to increase the conversion factor, f, rather than just E or N.
Use of appropriate form of energy
From the arguments in section 3.a it becomes obvious that different energy sources are
most efficiently used for different purposes. Fossil fuels, nuclear, biomass, and solar are
best used for a final purpose of heat generation, while the mechanical sources are ideal
for electricity generation. A potential for more efficient use the energy products of
combustion in a thermal boiler heat engine, namely heat and work, is explored in the
development of CHP (combined heat and power) plant, currently promoted, for
example, by the EU and the UK government.
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1. The energy challenge
Minimisation of losses
It is obvious that energy savings could contribute to meeting the increasing demand. A
very obvious and practical approach is to insulate buildings adequately against heat loss
(in cold weather) or overheating (in hot weather). Examples of bad practice are
excessive use of heating in poorly built houses, or the excessive use of air conditioning.
A more problematic source of losses lies in the distribution of energy. Substantial
amounts of electrical energy are lost in the distribution network of national grid.
Solutions to this problems are far from easy, and might require new technology as well
as some radical changes in the way energy is distributed to customers.
Maximum conversion of energy
This strategy aims at improving the actual energy generating plant, such as using CGP
plant, or by improving current technology used in the energy extraction. Examples
might be more efficient turbine designs. At the consumer end, optimisation of
industrial processes, for example, or the development of new more efficient production
or reaction technology could yield substantial energy savings without affecting the
amount usefully consumed energy or the quality of the product. This strategy is called
Process intensification.
4.c Energy in the supply-and-demand chain
The planning of energy generation and distribution has to be carried out within an
integrated view of its eventual consumption. In an ideal world the generation exactly
balances consumption. Several factors, however, prevent this ideal situation. The
demand is highly fluctuating, where the fluctuation is of the same order of magnitude as
the mean demand. Many renewable sources, such as wind or solar, cannot be
scheduled, and even fossil fuel (which in principle could be burned to schedule) cannot
react to fluctuating demand at the time scale required. A typical fossil-fuel burning
power station may need up to a few hours to substantially alter its output, while the
demand may rise from average to peak within a few minutes.
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1. The energy challenge
Summary
In a short overview of current and projected energy use, it was stated that the global
annual energy consumption in 1997 was around 9,500Mtoe, compared to an expected
consumption of 12,000Mtoe by 2020. It became clear that the average energy per
person use for different countries varied over two orders of magnitude, with the US
citizen requiring about twice the amount as a European, and 100 times the amount of a
citizen of a poor developing country. The different forms of energy were discussed,
with the definition of quality of energy as the proportion of energy that can be converted
to mechanical energy.
The second section of the chapter concentrated on environmental and health impacts
of combustion of fossil fuels, an aspect of energy policies and planning which has gained
significant importance in the recent years. The importance, and debate about its
relative importance to national economic interests were highlighted by an outline of a
few of the major international energy policy bodies and their progress.
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1. The energy challenge
Reading
Main textbook
J. Ramage. Energy: a guidebook. Oxford University Press, 1997, 2nd edition.
Other textbooks
G. J. Aubrecht. Energy. Prentice-Hall, 1995, 2nd edition.
E. S. Cassedy and P. Z. Grossman. Introduction to Energy – Resources, Technology, and Society. Cambridge
University Press, 1998, 2nd edition.
International organisations
International Energy Agency: http://www.iea.org
Monthly Oil Market Report; Biannual World Energy Outlook;
Global and national Statistics; Country Studies
Organisation for Economic Co-operation and Development (OECD): http://www.oecd.org
Commission of the European Union http://www.europa.eu.int/
e.g.: COM(95) 862: White Paper: An Energy Policy for the European Union
http://www.europa.eu.int/en/comm/dg17/whitepap.htm
UK organisations
Department of Trade and Industry.(DTI) : http://www.dti.gov.uk/energy/index.htm
Energy Digest 2000 ; Energy in Brief; Statistics; Policies
Royal Commission on Environmental Pollution: http://www.rcep.org.uk
Energy—The Changing Climate . Report and Summary, 2000
Department of the Environment, Transport and the Regions: http://www.detr.gov.uk
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2 Energy resources
Introduction
The aim of this section is to provide a definition of ‘energy resource’ and ‘fuel’ and to give a very
brief overview over the main fuels and some other energy resources.
Read also the second section of the online ‘Basic Notes.’
Once you have read this section and the online notes, you are ready to complete the first set of
online exercises
1 What are energy resources
Energy resources are either stores of energy which can be accessed for use when they
are needed, or they are currents of energy occurring naturally which can be converted
to more useful forms of energy.
Energy stores are fuels, such as wood, coal, oil and gas but also nuclear fuel.
Energy currents can be found in sun light, flowing rivers, the wind, or in the sea, such as
waves or tides. All these, apart from the tides are energy currents which are ultimately
driven by the sun; the sun drives the hydrological cycle by evaporating water to be
transported from sea levels to higher levels in the mountains; the sun drives the
atmospheric circulation, ie the winds; and the winds in turn generate ocean waves.
Another energy current is the flow of heat from the Earth’s interior to the Earth’s
surface. This ‘geothermal energy’ is a remnant of energy stored during the formation
of the Earth.
If you look at all these resources closely, you will see that the distinction between
stores and currents is a bit fuzzy: wood (and other biomass) is a store in that you can
harvest and store, but it is formed in a continuous way on an annual basis. Geothermal
energy is really a store of energy deep within the planet’s interior deposited during the
planet’s formation (from the gravitational collapse of galactic matter into a planet), but
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2. Energy resources
we can only access it through the current of heat from the hot interior to the surface
as the planet is gradually cooling down. As you can see, our distinction is made on the
way we can access the particular form of energy.
As energy stores can be seen as ‘lying around and waiting to be used’, they are always
quantified in terms of the energy content, i.e. in units of joule, million tonnes of oil
equivalent, quad, or other energy units.
As energy currents are naturally occurring fluxes of energy, they are quantified in
terms of the energy flux or power, i.e. in units of watt or similar.
2 Our ultimate energy sources
The energy potentially available for use comes ultimately from four sources,
1. Energy stored in the Earth crust during the formation of the Earth,
2. Energy released by the gradual cooling of the Earth’s interior,
3. Energy released from gravitational potential by the orbits of the Earth and
Moon,
4. Energy emitted by the sun and absorbed the Earth’s atmosphere and surface.
If we look at this list critically, we see that the most common sources for our daily
energy consumption, namely oil, gas, and coal do not seem to be included here. All
three resources are the remains of prehistoric vegetation which grew using primarily
the fourth energy category, sun light. During geological process, involving to a larger
or lesser degree all energy sources, this solar energy stored in the plants was then
concentrated in the deposits of coal, oil, and gas.
In a most pedantic definition, all energy sources are finite but the time scales associated
with the consumption of some of the energy sources is so much larger than even the
evolution of humans, that they can be regarded as effectively infinite. In the following
section a practical differentiation between finite and infinite energy resources will be
developed.
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2. Energy resources
3 Finite and Renewable energy sources
In this section, we are developing working definitions for the two contrasting types of
energy resources, which will be termed ‘finite’ and ‘renewable’, respectively. Before the
definitions, the ultimate origin of energy sources will be discussed, and the definitions
will be followed by an estimate of the renewable energy potentially available for
exploitation. The flow of energy available for renewable energy exploitation will finally
be divided into the forms in which they occur.
The resources, which we would intuitively classify as finite, are the fossil fuels and
nuclear fuels. The nuclear fuel reserves date back to the formation of the Earth and
belong to the first source, while the fossil fuse, namely coal, oil, and gas, derive from
the last category, as they are the remains of plants which grew using solar radiation.
These stores of energy are obviously of finite size, even if they ultimately originated
from the infinite source of solar radiation. However, the time scale it took to create
them is again very large, so that it will be impossible to renew these stores on a human
time scale.
Renewable resources, on the other hand, are usually associated with something that
happens anyway in the environment. The naturally occurring energy current is merely
accessed to extract a small portion of it for conversion to useful energy. Examples are
placing wind turbines in windy places, or covering roofs with photovoltaic cells.
3.a Definition of Finite Energy Resources
From the previous paragraphs it is becoming obvious that the common character of the
‘finite’ energy resources is that they are stores laid down sometime in the past, and
that they are accessed by some form of extraction. From this, we can now formulate a
definition for finite energy:
Finite Energy is energy obtained from stores of energy that remain bound
unless released by human activity.
In contrast to the definition of Twidell and Weir, I have changed ‘stores of static energy’
to ‘stores of energy’. The distinction lies in the difference between hydrocarbons and
radioactive stores. While hydrocarbon reserves are indeed static stores, the
radioactive elements do decay naturally while they remain in storage. The common
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2. Energy resources
factors are that they both a laid down in reserves in the Earth’s crust, and that they are
mined and transported to a convenient location before the conversion of the chemical
or nuclear energy to heat, work, or electricity is initiated.
A clear advantage of such stores is that the energy resource can be traded and
transported to arbitrary locations, such as power stations or petrol tanks in cars,
before its energy potential is realised.
3.b Definition of Renewable Energy Resources
From intuition, we know that renewable energy comes from sources such as sun light
(solar heating, photovoltaic cells), winds (wind turbines), or rivers and lakes (hydro).
This immediately leads to the picture of renewable energy an energy which flows
anyway, and which cannot be transported to a convenient location. Instead of bringing
the fuel to the power station or car, we have to take the power station (or car) to the
source of energy. This is indeed the picture given by Twidell and Weir, as
demonstrated in their comparison of renewable and finite energy in Figure 4. The
route of renewable energy flowing from A to C via B in the left diagram represents the
naturally occurring energy flow. The extraction, by human intervention, is realised by
placing a device in the flow, which extracts a (small) portion of the energy flow to the
path via D-E-F, where the energy is converted and delivered it to the end user, who
dumps any waste energy back into the systems (probably in a different form from the
initial energy). In contrast to this, the finite energy flow occurs by accessing the finite
source, or reservoir, as indicated by the switch D, followed by its conversion and use,
before dumping waste energy into the system, from where the original energy has not
originated (at least not on a human time scal
Figure 4. Comparison of Renewable Energy Resource use with Finite Energy Resource use (taken from Twidell and Weir).
17
2. Energy resources
There is one awkward energy resource, which does not seem to fit easily into either
category of energy, namely biomass. Biomass is a fuel, which can be stored, traded, and
transported to a convenient location, and thus does not really occur as a naturally
occurring flow of energy. On the other hand, the store of biomass is such that it is
created in a very short time scale, and is therefore not limited. Also, if one does a
energy and product balance over the time scale of the renewal of a biomass store, the
main energy derives from the sun, and the waste product, e.g. CO2, is released into the
atmosphere during use but then re-absorbed by the next season of biomass generation.
Because biomass is renewable on a human time scale, and because the effect on the
sink could theoretically be balanced in the source, biomass should clearly be classified
as a renewable. Another way of justifying this classification is to use the sun light as the
‘source’, the biomass creation as the ‘device’, and the combustion and further use as
the ‘use’ in Figure 4.
The definition of renewable energy proposed here is then
Renewable Energy is energy which is obtained from naturally occurring
currents of energy or from energy which can be regenerated by naturally
occurring energy currents on the time scale of a human life.
Apart from the case of biomass, all other current renewable energy resources are in
the form of naturally occurring energy currents. This obviously implies that one cannot
use, for example, the common way of producing electricity, which is the generation at
relatively few, very large power station, from where the electricity is distributed by a
grid. Electricity from renewable sources comes necessarily from a much larger number
of much smaller generators which are located where the energy is available.
Furthermore, as is immediately obvious from the nature of the winds, the energy
current itself is often not steady or even predictable. Two of the main problems faced
nowadays by renewable electricity generation are the integration of the many
generators into a stable grid, and the management of supply and demand when the
supply cannot be influenced (at least not to react to increased demand). The problem
of integration is a severe one, and a long-term solution has not been found yet. The
management of supply and demand requires additional devices, either to store excess
energy, or to actually control the supply not merely on a demand basis, but on a
prioritised demand basis where priority networks are guaranteed a steady supply but
18
2. Energy resources
where other networks are only supplied if the supply exceeds the demand of the
priority network.
4 A simple model of finite energy reserves
From Figure 1 in chapter 1 we saw that we continue to use more and more fossil fuels
to meet the energy demand. At the same time we know that there a finite amount of
those fuels was created during geological time scales. Several questions could be
useful here;
• How much was ever created? This would give us an absolute limit of energy
available from those sources, the ’ultimate resource’.
• How much of the total resource is actually recoverable?
The ‘ultimate recoverable resource’, Q∞.
• How much of the resource have we used by now?
The ‘cumulative production’, Qp.
• How much is left in the ground and recoverable?
The ‘remaining recoverable resource’, Q∞–Qp.
• How much do we know is left in the ground?
The ‘reserves’, Qr.
• How much do we know is left in the ground and is accessible
(both technologically and economically)?
While we may have a fairly good idea of the reserves we know about and how much it
would cost to extract them, we cannot know about the total resource or even the
remaining recoverable resource. We do, however, have to acknowledge that we
continue to find new reserves on a regular basis. To predict our energy balance based
on the reserves alone would therefore be unrealistically pessimistic. A simple model is
now widely used to estimate the ultimate recoverable resource for any finite energy
resource.
19
2. Energy resources
This model, which was initially proposed by Hubbert1 based on a simple statistical
curve, the normal curve or Gaussian curve, for the rate of production, and its cumulative
form, the logistic curve, for the accumulated production. The normal curve is a smooth
curve with a maximum production rate in the middle of the curve and starting with a
zero production rate at the beginning before the resource was seriously exploited. As
the resource becomes an established and reliable source of energy, the production will
increase rapidly while the reserves are abundant. As the resource becomes scarce, it
will become harder and more expensive to extract it and consequently the production
rate and consumption reduces again until virtually nothing is left. The accumulated
production is an S-shaped curve which also starts at zero before industrial use and
increases sharply if the production rate is large, becomes less steep if the production
rate falls, and eventually levels off at the ultimate recoverable resource, Qp, when the
production rate has dropped to virtually zero again.
1 M.K. Hubbert, U.S. Energy Resources, a Review as of 1972, U.S. Senate Committee on Interior and Insular Affairs report, Washington, D.C.: GPO, 1974.M.K. Hubbert, Am. J. Phys. 49, 1007 (1981).M.K. Hubbert, Chapter 8 in Resources and Man. Freeman, San Francisco, 1969
20
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.25
0.5
0.75
1
r (t-tpeak
)
P/(
rQ∞
)
-10 -8 -6 -4 -2 0 2 4 6 8 100
0.25
0.5
0.75
1
Qp/Q
∞
→
←
Figure 3. Production rate and cumulative production as derived from the logistic equation.
2. Energy resources
Textbox 1: Hubbert’s model of reserves:
The production rate, dtdQ
P p= , is the amount of resource extracted in a given time interval. The
assumptions that the curve describing this is the normal curve, or logistic equation, are
− If the past production has been successful, we expect that the demand is high and therefore the production rate; in a simple model, the production rate is proportional to the cumulative production, Qp.
− If the resource becomes scarce, it will be harder to extract the resource; the production rate is proportional to the remaining fraction of recoverable resource, (Q∞–Qp)/ Q∞ or (1 – Qp /Q∞).
This gives us
−==
rQdtdQ
P pp
p 1
The factor r measures the potential for increase.
Dividing the equation by the ultimate resource, ∞= QQq p / , we can rewrite this equation as a
relatively simple ordinary differential equation,
( )qqrdtdq −= 1 , or, after separation of variables, ( ) dtr
qqdq =−1
which can easily be integrated by referring to standard mathematical handbooks2 to give the fraction of the extracted resource, q, at time t, which depends on the factor r, and the fraction q0 at time t0:
( )00
0
11ln ttrqq
qq −=
−
−.
If we take our reference time, t0, to be the time of peak production, we recognise that the production is symmetric around this time with the result that we have recovered half of the ultimate recoverable resource, ie q0= 1/2. Rearranging the solution then gives
∞−
−
+= Q
eeQ tr
tr
p 1.
As a final step, we want to express the production curve as a function of time rather than of the cumulative production. So, we need to take our solution and insert it into the logistic equation to get
( ) ∞−
−
+= rQ
eeP
tr
tr
21.
These two curves are illustrated in Figure 3.
2 for example I.N. Bronstein, K.A.Semendjajew, G.Musiol, and H. Mühlig. Taschenbuch der Mathematik. Harri Deutsch, 2001 or M. Abramowitz and I. A. Stegun (eds). Handbook of mathematical functions with formulas, graphs and mathematical tables. National Bureau of Standards, 1964
21
2. Energy resources
5 Some energy currents and accumulated energy amounts
This section illustrates energy currents and how they accumulate into energy using two
common energy currents: sun light and wind.
The principle of accumulating the current into energy is analogous to electricity
consumption: your electrical appliances are rated in their consumption in watt but your
electricity bill calculates your electricity consumption over a period in kilowatt-hours,
kWh.
Example: You switch on a 100W light bulb: you are consuming electricity at a
rate of 100W. If you leave the light on for 100s, then you have used
100W×100s= 10,000Ws= 10,000J= 10kJ. If you leave the light on for 10 hours,
then you have used 100W×36000s= 3,600,000J= 3.6MJ. Since these are rather
large numbers, electricity is usually charged by the kWh: 100W×36000s=
100W×10h= 1,000Wh= 1kWh.
In short, if you consume or receive energy at a constant rate, then the accumulated
energy is equal to the product of the rate and the time. If the rate changes, then this
turns into the integral of the instantaneous power over the time interval of interest:
dtPE ∫=
5.a Sun light
The most significant supply of energy to the Earth is in the form of radiation from the
sun, which is largely radiation in the visible part of the spectrum. Overall, a 1m2 section
at the top of our atmosphere facing the sun receives 1367W. This number,
GS= 1367Wm–2, is known as the solar constant. Considering that about 30% is reflected
straight back to space, we can expect that 1m2 of ground on a clear day, with the sun
directly overhead receives about 1kW. When the sun is at an angle to the ground
(i.e. at higher latitudes and/or not at noon), that number is reduced by the cosine of
that angle.
Example: One square meter on the equator during equinox, the sun is directly
overhead at 12 noon, and the length of the day is 12 hours. At other times, the
surface make an angle between the sun of 90° or π at 6am, 45° at 9am, and so
22
2. Energy resources
on, giving a projected area (multiplied by the cosine) of cos[2π×(time–
12noon)/24h]:
( )
[ ]kWh
kWkW
dhkWdtkWdtPE
hh
pm
amhht
6.721sin1
cos1212cos1
2242/
2/224
2/
2/
6
62412
≈××=××=
××===
−
−
− ∫∫∫
ππ
ππ
π
π
θ
θθπ
This estimate compares well with the measured annual average of solar insolation
of about 6.2kWh m–2 day–1 for Al-Khartoum near the equator.
For a location at a higher latitude, this value is again reduced by the fact that
there is always a higher angle between the surface and the sun. We can estimate
the average clear day energy received per day on a latitude of 56° by multiplying
the equator value by the cosine of the latitude. This gives about 4.3kWh per
metre-square per day. As you can imagine, the sun does not always shine in
Scotland (not even in Edinburgh), and the actual value is about 2.3 kWh m–2 day–1.
5.b Wind
Wind is a very tricky energy resource as it is never constant. However, one can
measure an annual average wind speed. Taking this average wind speed, we know that
the energy flux through an area, A, facing the wind is given by
321 AUP ρ= ,
where ρ= 1.225kg m–3 is the air density.
One can then estimate the energy going through that area in a year by multiplying.
However, as the wind is fluctuating, this estimate has to be reduced by what is called
the capacity factor, which is about 30% (= 0.3) for wind power. Furthermore, no wind
turbine can extract 100% of the power flowing through the rotor, and the power going
through the rotor has to be multiplied by the efficiency or performance factor of the
turbine. Very good modern turbines reach a performance factor of about 50% at their
design point.
23
2. Energy resources
Example: For Bremen in northern Germany, the average wind speed is 5.3m s–1.
Using a capacity factor of CC=0.3, a performance factor of CP=0.5, and a year of
365.25days of 24hours= 8766hours per year, we can estimate a typical energy
flux of
2321 96 −== WmAUP ρ
and an energy yield per annum of
22321 1271270008766 kWhmWhhCCAUE PC ==×= −ρ .
For a location in Scotland with an average wind speed of 7 m s–1, we get
210W m–2 and 276 kWh m–2, respectively.
6 Fuels
This section summarises the most common type of energy resource, namely fuels. For
the purpose of this course, I use the word in the following meaning:
A fuel is a matter which releases heat by combustion (burning)
in a chemical reaction.
The only primary form of energy extracted from such a resource is heat which can
then be used either directly or converted to other forms, such as work in an engine or
electricity in a generator. This section includes both finite and renewable fuels in a
somewhat unorthodox order, starting with biomass, then moving on to coal before
arriving at the two giants, oil and gas.
Fuels are all energy stores. That means that they store energy as chemical energy
(except for nuclear fuel). This energy is then usually released in the form of heat by
combustion, and the fuel is characterised by how much heat is released in the quantity
called calorific value or heat content. Its base unit is J kg–1 but it is usually given in
MJ/kg or GJ/tonne (1000 J kg–1 = 1 MJ/kg = 1 GJ/tonne).
The resulting heat can then be used in all sorts of ways:
• Directly for heating; either to head a space, such as a building, or to provide
heat for a manufacturing process, such as making steel or paper.
• For transport, for example in internal combustion engines or jet engines
24
2. Energy resources
• For electricity generation
For all common fuels, such as wood, coal, oil, and gas, the combustion involves turning
a molecule which contains one or more carbon atoms into carbon dioxide (and usually
water) and energy. The very simplest fuel is coal.
6.a Biomass
As the name suggests, this is fuel which is derived from living organisms, and it includes
a vast variety of fuels, ranging from cow dung (used in Africa and India for heating and
cooking) and chicken litter (used in Scotland in a power plant3), wood, straw (used in
England in a power plant4) to biogas (mainly methane) being produced in farming and
landfill sites. This wide range brings with it obviously a wide range of energy contents
and techniques of burning the stuff.
While biomass has been used as a fuel more many tens of thousands of years and is still
for many people the primary fuel supply (currently estimated to be between 10 and
20% of the world’s energy consumption), it is also becoming increasingly used as an
alternative to the fossil fuels introduced above in commercial applications. In the UK
alone, there are many power stations using a range of biofuels, such as straw or poultry
litter. Another by-product of human waste generation is the gas generated by
microbial activities in landfill sites. This biogas can be collected and used in the same
way as natural gas. Biodiesel as a renewable alternative to vehicle fuel from petroleum
oil is also increasingly used. Since biofuels cover such a vast range of origins it is
impossible to do anything other than a few sweeping statements about them as energy
resources.
The vast majority of biofuels are of the solid kind, and range from waste products of
human activity (eg the straw and poultry litter mentioned already) to energy crops which
are planted specifically for the purpose of energy production. Willow, for example is a
plant which grows fast enough to produce enough fuel at a reasonable rate. All biofuels
have the potential to be renewable resources. Since the generation of these fuels
3 http://www.william-siemens.co.uk/powgen_westfield.htm and http://www.eprl.co.uk/west1.htm 4 http://www.eprl.co.uk/ely1.htm
25
2. Energy resources
depends on human activity or intervention, the actual ‘renewability’ (if that is a word)
depends on proper planning of the production and consumption of these fuels.
All solid biofuels are based on plant material and thus come in a broad range of
chemical compositions. The main useful component of plants, the sugars, starch,
cellulose etc., are more complex molecules than the hydrocarbons. A good
representative of the fuel molecules are carbohydrates, which are long chains containing
carbon, hydrogen, and oxygen atoms with the generic chemical composition of
Cx(H2O)y. It is sufficient to just look at one unit of such a chain to appreciate what is
going on:
CH2O + O2 → CO2 + H2O + energy.
Because the actual composition for different biofuels is very different, the energy
released by burning 1kg of biofuel also varies greatly as can be seen in the few examples
given in Table 2.
Table 2. Selection of biofuels and their energy content.
Fuel MJ/kg Fuel – Wood MJ/kg
Straw 15 freshly cut 11
Corn (cob & stalk) 15 air dried 16
Dung 16 oven dried 20
Domestic refuse 9 charcoal 29
We can estimate the amount of carbon dioxide released into the atmosphere by
burning 1kg of a carbohydrate of the formula [C(H2O)]x but considering the reaction
outlined above and the molecular weight of the constituents (C: 12, H: 1, O: 16). The
carbohydrate unit on the left side of the reaction has a total molecular weight of 30
while the carbon dioxide molecule has a molecular weight of 44. Therefore, using up
30kg of carbohydrate releases 44kg of carbon dioxide, or 1.47kg of carbon dioxide per
kg of carbohydrate fuel. To appreciate why biofuels can be truly renewable and
carbon-friendly energy resources, it is useful to see, how plants make the
carbohydrates when they are growing in the process called photosynthesis,
CO2 + H2O + energy → CH2O + O2,
26
2. Energy resources
which is exactly the reverse reaction from the combustion. The energy supply for
photosynthesis is the sun’s light.
Therefore, each carbon dioxide molecule released by the burning is used exactly to
make the fuel again. This obviously does not take into account energy used in the
growing process, transporting it to the point of consumption, or any prior treatment
required.
6.b Coal
Coal is the most abundant fossil fuel, with estimates of reserves to last us over another
millennium, and the entire industrial revolution and industrialisation of the world was
largely built on coal as the primary energy source, and it is still a relatively cheap
resource. Yet it has become very unpopular for several reasons. It comes in solid bits
– much less easy to burn than a liquid or gaseous fuel, it is dirty – ‘a large coal-fired
power-station can produce enough ash in a year to cover an acre of ground to the
height of a six-storey building’ (Ramage, p.71), its combustion releases lots of
unpleasant and toxic substances as by-products, its extraction leads to unsightly open-
cast mines or deep mines which cause subsidence. In short, it is a very dirty energy
resource.
Coal is a dirty resource because it has a complex chemical composition — after all, it
has derived from decayed plants. No easy chemical formula can be given for coal, but it
is the carbon within coal which provides the energy. One of the major elements of
coal are rings of 6 carbon atom which also incorporate hydrogen, oxygen, and nitrogen,
with about equal numbers of carbon and hydrogen atoms. Up to ten percent of coal
may be other material which does not burn but remains as ash. The release of energy
occurs by oxidising the carbon to CO2. Therefore, whatever one does, using coal
generates CO2, whose presence in the atmosphere the world is trying to limit. The
actual combustion is not a simple of process of taking the carbon atoms burning them
to get carbon dioxide and energy but, because of the complex composition of coal,
involves several steps:
• First all moisture from the coal must be removed (up to 10%). This obviously
requires energy in the form of heat.
27
2. Energy resources
• As it heats up, several gases are released from the solid piece of coal, the so-
called volatile matter. This removes most of the hydrogen and oxygen, together
with carbon monoxide (CO) and some hydrocarbons. The hydrocarbons are
fuels and release energy in their combustion, up to half of the energy content of
the coal
• The remainder consists to the largest part of the so-called fixed carbon, which
does the direct chemical conversion of
C + O2 → CO2 + energy.
• Anything that cannot burn remains as ash.
Coal comes in a variety of qualities, depending on their age and environmental
conditions. The first stage in the coal formation is peat. A common type of low-grade
coal is lignite, also known as brown coal, with about equal proportions of fixed carbon,
volatile matter, and moisture. Sub-bituminous and bituminous coals, and some forms
of anthracite, are widely used for energy production. Some anthracite, however, is a
poor fuel despite its high carbon content due to the large amount of ash produced.
Natural graphite (with about 90% fixed carbon) is not used as an energy reserve. The
types of coal and their typical energy content are listed in Table 3.
Table 3. Comparison of energy content and composition of different types of coal. A representative sample of wood is included for comparison.
Type Energy content
(MJ/kg)
fixed carbon
(%)
volatile matter
(%)
Moisture
(%)
Ash
(%)
Wood ~20 ~10 ~40 ~45 ~5
Peat ~20 ~10 ~20 ~65 ~5
Lignite <20 ~30 ~30 ~35 ~5
Sub-bituminous 19–27 ~40 ~25 ~25 ~5
Bituminous 25–33 50–80 10–30 5–10 ~5
Anthracite ~30 >75 ~5 ~5 ~15
Graphite ~35 ~90 <5 <5 ~5
28
2. Energy resources
The coal with the highest calorific value contains about 90% of carbon and has a
calorific value of 34MJ/kg (or 34GJ/tonne); taking this into account, we have about
34/0.9= 38MJ/(kg of C), this then can be represented as
Mass and heat balance: 12kg + 32kg → 44kg + 38MJ/kg×12kg
1 + 2.67 → 3.67 + 38MJ/KG
We can see that a natural result of burning coal is the release of carbon dioxide to the
amount of about 0.1kg CO2 per MJ converted.
Considering that the efficiency of coal-fired power stations is about 33%, we produce
about 0.3kg CO2 per MJ electricity generated. Converting this to kWh (ie multiplying
by 3.6), gives us about 1kg of CO2 emitted per kWh generated – ie leaving your 100W
light bulb on for 10 hours adds 1kg of carbon dioxide to the air.
A large but typical coal-fired powerstation of generation capacity of 600MW therefore
emits 14400 tonnes of CO2 each day.
(600,000×24 kWh per day= 600,000×24 kg CO2= 600×24 tonnes of CO2)
6.c Hydrocarbons (Oil and Gas)
Oil and gas are very much easier and cleaner to burn and to use than coal. While
these two substances appear to be very different and are used for very different
applications, they are all part of the hydrocarbons and usually found together in
reservoirs. Hydrocarbons are a family of organic molecules which consist of chains of
carbon atoms to which hydrogen atoms are attached.
The simplest hydrocarbon is methane, CH4. Longer examples of simple chains can be
written as H3C–CH2–…–CH2–CH3 , or shortened to CnH2n+2.
The combustion and mass and heat balance of methane with a calorific value of 55MJ/kg
can be written as
CH4 + 2O2 → CO2 + 2H2O + heat
12+1×4 + 2×16×2 → 12+16×2 + 2×(1×2+16) + heat
16 + 64 → 44 + 36 + 55MJ/kg × 16
29
2. Energy resources
Because methane has a much higher heat content or calorific value than coal, we get
55MJ/kg heat and only 0.05kg per MJ converted.
Considering that gas-fired power stations have a
typical efficiency of about 50%, we can reduce the
carbon dioxide emissions from 1kg/kWh to about
0.3kg/kWh, reduction of 60-70%! Not surprisingly,
many nations have heavily invested in gas-fired power
stations as opposed to coal-fired power stations – and
this is only one advantage. Other advantages over
coal are that there is virtually no ash produced, and
that gas turbines can react to changing demand much
quicker.
The oil-based products, such as petrol and diesel are still the dominant fuel for
transport. The petrol can be reasonably well described by octane, C8H18, and diesel by
cetane, C16H34, both with a calorific value of about 48MJ/kg or about 35MJ per litre
(with a density of 0.7kg per litre for octane and 0.75kg per litre for cetane). The
chemical structure of methane and octane are shown in Figure 5.
We can extend the combustion equation for methane to the generic chain
CnH2n+2 + (1.5n+0.5) O2 → n CO2 + (n+1) H2O + heat
14n+2 + (1.5n+0.5)×32 → 44n + (n+1)×36 + 48MJ/kg × (14n+2)
This gives us 22n/(7n+1) kg of carbon dioxide per kg of fuel and 22n/([7n+1]×48) of
carbon dioxide per MJ of energy converted. If n is sufficiently large, the ‘+2’ in the fuel
can be ignored, and we approximate these numbers to
• 3.01kg CO2 per kg petrol and 3.12kg CO2 per kg diesel
• 2.5kg CO2 per litre petrol and 2.7kg CO2 per litre diesel
• ~ 0.064 kg CO2 per MJ of heat released
The actual energy released by the different hydrocarbons is listed in Table 4.
Figure 5. A methane and an octane molecule. The dark circles represent carbon atoms and the light circles hydrogen atoms1
30
2. Energy resources
Table 4. Energy content and carbon dioxide emission for different hydrocarbons.
Name Composition Molecularweight
Energycontent(MJ / kg)
Mass of CO2
released per kg fuel (kg)
Mass of CO2
released per energy (kg/MJ)
Methane CH4 16 55
Ethane C2H6 30 51
Propane C3H8 44 50
Butane C4H10 58 46
Pentane C5H12 72 48
Hexane C6H14 86 48
Heptane C7H16 100 48
Octane C8H18 114 48
Again, looking at the CO2 release of this process, we can see that each carbon atom
ends up in a CO2 molecule with a molecular weight of 44. With methane as the
example, the complete combustion of 16kg of methane releases 44kg of carbon
dioxide, or each kg of methane releases 44/16= 2.75kg of CO2.
Exercise:
Complete Table 4, and show in a graph how the carbon dioxide emission varies for the
different hydrocarbons in terms of amount of fuel (kg CO2 / kg fuel) and in terms of
the energy released by that fuel (kg CO2 / MJ). Compare the results with the
corresponding figures for burning coal.
31
2. Energy resources
6.d Nuclear fuel
Nuclear fuel does not release chemical energy by changing chemical bonds between
atoms but by changing the atoms themselves. There are two types of reaction possible,
nuclear fission, where an atom breaks into several parts, and nuclear fusion where two
or more atoms combine to make a new one. Obviously, the starting products for
nuclear fission will be large, heavy atoms, such as Uranium (with an atomic weight of
235), while the starting points for nuclear fusion will be small atoms, such as the three
varieties of hydrogen, standard hydrogen with a single proton, deuterium with a proton
and a neutron, and tritium with a proton and two neutrons.
Nuclear fission is a well-developed technology albeit, the disposal of its waste products
is not yet as well developed as most people would like to see….
The attraction of nuclear fission is that the reaction itself does not release carbon
dioxide, and that the fuel has a very high heat content: the energy release of uranium-
235, U235, is 82TJ/kg (= 82 million MJ/kg). This obviously has to be seen in the context
that natural uranium only contains about 0.72% of U235 and enriched uranium between
2% and 5%. Taking these numbers, we get an energy content of the base material,
natural uranium of 600,000 MJ/kg – still a factor of over 10,000 higher than the carbon-
based fuels –, and between 1.6 and 4 million MJ/kg of enriched fuel.
The actual output of a nuclear reactor can be about twice that amount, which is caused
by the fact that the products of the uranium fission themselves are radioactive and
release further energy as they undergo fission. This fact also result in the highly
radioactive waste a nuclear reactor produces…
Nuclear fusion, on the other hand, has been hailed as one of the ‘holy grails’ of energy
supply: its fuel, deuterium and tritium (the two heavy isotopes of hydrogen) can be
isolated from ordinary water, and its reaction products, hydrogen and helium in an
ideal life, are not only harmless but also useful – hydrogen is the other ‘holy grail’ if one
is allowed to have two...
Real life is not as simple: in reality there is usually some radioactive material produced
in fusion reactions due to the interaction of the released energy in the form of
neutrons and the material of which the reaction chamber is made, and nobody has as
yet achieved a sustained and effective fusion in a practical version.
32
3 Energy statistics:
production, consumption, and emissions
Introduction
The aim of the chapter is to introduce the way energy data are collected and presented in
national or global energy statistics. Some potential problems in their interpretations are
highlighted. The second part uses some energy statistics to discuss the link between energy
and living standard.
Read also section 3 of the Basic Notes available online
1 Energy statistics
Never trust a statistic unless you have fudged it yourself.5
Almost every single country on Earth collects and publishes data of their national
energy production and consumption. The data are usually published by (or on behalf
of) government departments. Digests of these national data are also compiled in
international databases, most notably by the International Energy Agency6 and by BP7, and
also by the UN and the OECD.
1.a Data and information reliability and use
A word of caution: while many organisations state that they are independent – and
may really try to stay independent – all collation, presentation, and interpretation of
data is necessarily biased by the authors opinions. For example, the IEA is an
5 While no direct source for this ‘quasi’-quote can be given, it is a caution frequently raised!6 http://www.iea.org7 BP Statistical Review of World Energy
33
3. Energy statistics
organisation closely founded by OECD countries, who represent the major oil-
consuming countries. As a result, one of their underlying tenets is to ensure a reliable
and affordable supply of oil and gas. Following the IEA World Energy Outlook report
in 2005 (which was in addition to its regular bi-annual reports during even years)
criticism was raised by several people that the IEA attempted to steer world policy
decisions to support oil and gas investment at the expense of investing in sustainable
renewable energy development8. You might ask yourself the question why the EIA felt
obliged to publish an World Energy Outlook report in 2005 when there was to be one
due in 2006.
Another example can be found in the wind debate. Some groups, either truly trying to
protect wildlife using wildlife and the country side in addition to a strong financial
involvement in the nuclear industry present all evidence that wind turbines are
unreliable, an eye-sore, and killing birds, whereas the respective national wind energy
association will present evidence that wind turbines have produced a substantial
amount of energy, that they attract tourists, and that bird fatalities are so much small
than other human activities that not installing wind power will kill more birds in the
resulting climate change….
1.b Data collation and presentation
To compile any statistic, the two fundamental questions are:
1. What do we want to know?
2. How should we organise the information?
The first question seems easy to answer:
• The energy which we put into the society/economy, the ‘primary energy’,
probably broken down into the different types of energy resources.
• Possibly some intermediate stages of the energy, for example when converting
it from, eg, coal to electricity.
8 EIA (2005). World Energy Outlook 2005 -- Middle East and North Africa Insights. EIA report summary: http://www.iea.org/textbase/npsum/WEO2005SUM.pdf Criticism raised by R. Rechsteiner (MP in Switzerland), Energy Bulletin, 26 Nov 2005: http://www.energybulletin.net/11701.html
34
3. Energy statistics
• The final amount of energy consumed,
probably broken down into type of energy consumed and/or type of consumer
While this sounds straightforward, there are some pitfalls:
• Is primary energy always well defined?
• How do we account for energy used in the energy generation, for example
getting the oil from the reservoir up, to shore, and refined into useful fuel?
• Where does the chain stop, or who is the final consumer? After all, the first
law of thermodynamics states quite firmly that energy cannot be ‘consumed’.
Before we go into the details of what exactly we put in the tables and figures, it is
useful to see a few of such figures. In Chapter 1, we have already seen some energy
graphs. It seems quite intuitive that the spaghetti diagram, Figure 4, is a clear graphical
description of the input and output of the energy, both by type and by end-user
category. The relative contribution of each energy resource, or energy form, is
represented by the thickness of each strand. Unfortunately, such a diagram only works
for very simple cases, such as Norway, which only used four types of resources and
used almost exclusively hydropower to generate electricity.
Figure 5 Spaghetti diagram of energy balance in Norway (from Twidell and Weir, Renewable Energy Resource, 1986, first edition)
For most countries, the primary input into electricity generation comes from a
multitude of sources, resulting in a true bowl of spaghetti where it is almost impossible
35
3. Energy statistics
to untangle individual strands. Often, a number of pie charts are used instead. Over
the next few sections, we will establish some statistics on the example of the
information about the energy balance in the UK for 1998. Unless stated otherwise, all
information are taken from DTI (1999) Energy in Brief9.
2 Primary energy
What is primary energy? Generally, it is the energy content of the original resource. In
the case of natural gas, it is the amount of gas extracted from the reservoir. Once it
has left the ground and thus entered the market, this primary energy resource could be
used for a variety of applications, ranging from cooking in the domestic kitchen to
electricity generation in a power station.
Simple!, but?
How do we count the water flowing down a river which may be used in a hydropower
scheme? After all, it cannot be used for anything other than electricity generation at
that particular location10?
In most statistics, we are given the energy content of the fuels and, in addition to that,
‘primary electricity’. Primary electricity is the form of electricity which derives from a
source which cannot be used for anything else. This includes, among others,
hydropower, wind power, and nuclear energy. This opens up a substantial difference
between the electricity produced by a nuclear power station and by a coal-fired power
station. For the nuclear power station, the output is used to fill in the energy
production while for the coal-fired station it is the input which goes in the statistics.
Problem: We expect less energy to come out of a coal-fired power station than we
put in (a typical proportion is that about 30% of the energy released by the burning of
the coal is converted to electricity). There is a fundamental unfairness in measuring the
input for one scheme but the output for another scheme. On the other hand, the
input into a primary electricity plant cannot be used for anything else, so the amount
we lose may be irrelevant, whereas coal could be used for something else where we
use the energy contained in the coal more efficiently.
9 http://www.dti.gov.uk/epa/10 The possibility of using a water mill to extract mechanical energy directly is ignored since this is not used to any noticeable extent anywhere in the world.
36
3. Energy statistics
Illustration 1: If a coal power station generates 1000kWh in a day, the primary coal
energy input on that day is 3000kWh and 2000kWh are lost as heat. If we were to
replace the coal power station by a hydro power station, we only need to generate the
1000kWh of energy and very little heat is generated (maybe about 100kWh). As a
result, the same contribution to the electricity will look very different in the energy
input side, depending on where we use it. In the first case, the entry is 3000kWh but
in the second case, it is only 1000kWh.
Illustration 2: If the same coal power station is replaced by a nuclear power plant,
the entry in the statistics will only be 1000kWh, but a nuclear power plant generates
the same amount of waste heat as a coal power station. While the entry in the primary
statistics is only 1000kWh, we still generate the waste heat of 2000kWh. If we were
able to use this waste heat constructively, we could use energy further down the line
which had never entered in the primary column.
Solutions: Three conventions to deal with this imbalance between primary fuels and
primary electricity can be found:
• acknowledge that they are very different sources of energy, and stick to the
simple output from the primary electricity generators as the input to the
statistics.
• multiply all output from primary electricity generators by a factor of 3 to show
the amount of energy which would be consumed if that electricity had been
generated by a traditional coal or gas power station.
• multiply only the output from nuclear power stations by a factor of 3, since they
also produce the same waste heat as other fuel-operated power stations. This
waste heat might be useful at some stage.
The UK statistics11 and the BP Statistical Review, for example, use the third policy in their
statistics, while the UN use the first option. Even though the choice of convention
affects the numbers, there is often no clear indication as to which convention is used in
many publications.
11 as explained in Appendix A in the DTI’s Energy Digest, see http://www.dti.gov.uk/epa/
37
3. Energy statistics
The primary energy production in the UK for 1998 is summarised in the table and
illustrated in a pie chart as
(Mtoe) %
Petroleum 145.1 51
Natural Gas 90.2 32
Coal 26.0 9
Primary electricity 24.1 8
Total 285.4 100
This is obviously only a snapshot for 1998. It is very instructive to compile data from
several years to see a trend in the energy production. The UK, for example, shifted
dramatically from a coal-based country to an oil-and-gas based country.
Exercises:
1. Compile comparative primary energy data for the UK from different years,
including the 1970s and the most up-to-date data available from the dti website.
2. Compile comparative primary energy data for different countries.
90.2
2624.1
145.1
Petroleum
Natural Gas
Coal
Primaryelectricity
38
3. Energy statistics
3 Energy consumption
Since most government statistics are interested in the energy market, it is universally
agreed that the ‘final consumer’ of energy is that person or company who pays for the
energy while it is in energy form. For example, the energy consumed in the process of
making a light bulb is counted against the manufacturer of the light bulb, the energy
consumed while transporting it to the customer is counted against the haulage
company filling up their lorries with diesel, but the energy consumed while the light
bulb is producing light in a house, is included in the energy bill of that domestic
customer.
Problem: Some countries, e.g. Switzerland and India, use to a noticeable degree
wood or other biomass to heat their building or do the cooking, where this biomass
has not been bought but collected free of charge by the individual (by felling a tree in a
communal forest or by picking up cow dung from the street). Because nobody pays for
this energy resource it cannot easily measured or included in the statistics.
Answer: Estimate....
Problem: Where do we count energy used by the energy industry: the energy
required to extract the fuel, the energy lost in the conversion to a useful energy, the
energy lost in transporting the energy...?
Answer: Not in the ‘Final use’
Problem: Fuels used in international transport. A national energy statistic will be
interested in the domestic use of fuels. Who includes the fuel used in open waters
(‘marine bunkers’ in UK statistics) or in the skies between two countries?
Answer: nobody really.
39
3. Energy statistics
The ‘final consumption’ for the UK is summarised by the DTI as
Mtoe %
Oil 66 42
Gas 56 36
Coal 6.5 4
Electricity 27 17
Total 155.5 99
Another way to look at the final
consumption, rather than by type of
energy, is by type of user:
Mtoe %
Industry 35.0 22
Service sector 21.7 14
Transport 53.6 34
Domestic 46.0 29
Total 156.3 99
We can attempt to compile the information as
010203040506070
OilGas Coa
l
Electri
city
Mtoe
Domestic
Transport
Services
Industry
0
10
2030
40
50
60
Industry
Servic
es
Transp
ort
Domestic
Mtoe
Electricity
Coal
Gas
Oil
53.6
21.7
3546 Industry
ServicesTransportDomestic
6.5
56
27
66OilGasCoalElectricity
40
3. Energy statistics
4 Production and consumption
Comparing the production and consumption can give some useful insights. Comparing
the primary energy production of 285.4Mtoe with the final consumption of 156.3Mtoe,
we see that there are some 129Mtoe missing. A closer look at the published data
shows
• Conversion losses: 51.7Mtoe, which is the energy lost as waste heat in the
electricity generation, and the energy lost in the refinement of crude oil to
ready-to-use fuels, ie petrol etc.
• Distribution losses and energy industry use: 20.8Mtoe.
• Import, export, stock changes
• Marine bunkers etc.
This figure shows the national production of the UK for 1998 in the left column, and
the national consumption in the right column. The middle column accounts for the
conversion losses and the energy used by the energy industry to provide the energy for
consumption. A few things can be noted by comparing all figures prepared so far.
• Much more oil is produced than consumed: the UK is an oil exporting country.
• More coal is used and lost in the conversion than produced: the UK is a coal
importer.
• The UK is a net energy exporting country
0
50
100
150
200
250
300
Production Conversion Consumption
Mtoe
ElectricityCoalGasOil
41
3. Energy statistics
• Gas is used in similar amounts as a final energy provider (heating, cooking etc.)
and as a fuel for electricity generation (visible in the conversion loss)
• While a large part of the primary electricity is ‘lost’ (this is due to incorporating
the waste heat into the nuclear electricity), more electricity is consumed.
Almost all of the electricity consumed has come with the penalty of about twice
the energy amount lost in the energy conversion.
• By far, the largest contributions to the energy consumption are in the fossil fuels
of oil and gas.
5 Change over time
It is useful to compare how the production and consumption have changed over time.
As a typical example, the 1970 figures are compared to the 1998 figures for the UK
• The UK moved from a coal producer/exporter and oil importer to a coal
importing and oil exporting country.
• The total consumption has increased very little, but the production has
increased dramatically
• Primary electricity has increased noticeable, mainly due to nuclear generation
1970 1998
0
50
100
150
200
250
300
Production Conversion Consumption Production Conversion Consumption
Mtoe
ElectricityCoalGasOil
42
3. Energy statistics
6 Energy data in relation to other statistics
The energy statistics presented so far have only given the bare numbers for energy
production and consumption. For these numbers to be meaningful they will have to be
placed in the context of other information. Energy consumption alone does not say
what has been done with the energy, or how efficiently is has been used. In this
section, three other types of information will be used to put the energy data into
context, namely the population, the ‘wealth creation’, and the ‘standard of living’ in a
country.
6.a Per caput consumption
It is obvious that a more populous
country consumes more energy
than a small country because it has
more people using energy.
The UK has a population of about 59 million. Converting the final consumption in 1998
to a per-capita consumption, the figures become those in the table to the right.
Exercise: Complete the table. The last column is the number of people which could
subsist on that energy if all they need is a daily food ration with an energy content of
2000Cal.
6.b Energy and GDP
A fairly common step to include the wealth generation
— which to a large degree depends on the energy
consumption — is to measure the per-capita energy
consumption against the national Gross Domestic
Produce (GDP). The GDP is the total inland production
annual daily ‘food’
GJ kWh # people
Industry 25
Service sector 15.4
Transport 38
Domestic 33
Subtotal 111.4
Losses 57
Total 168
But: the per-caput consumption does not measure the industrial output/’wealth generation’
MJ per £ GDP
Industry 1.7
Service sector 1.1
Transport 2.6
Domestic 2.3
Subtotal 7.7
Losses 3.9
Total 11.643
3. Energy statistics
of the national economic system, or the value of everything that the country has
produced in that year. For the year 1998, the UK National Office for statistics12 quotes
a GDP of £859,384 million, or £859×109. Dividing the annual total energy consumption
data by this figure, we obtain the amount of energy expended on generating one
pound of GDP.
6.c Energy and standard of living
The Gross Domestic Product is obviously only one indicator of a nation’s state. To
find a measure which is not just based on the economic performance of a country but
also on other factors which affect the standard of living is difficult. One such factor,
which has been defined by the United Nations, is the United Nation’s Human
Development Index (HDI)13. The HDI attempts to measure the human development of
a country by taking into account in equal parts indicators for the life expectancy, the
education (as measured by adult literacy and the enrolment into primary, secondary,
and tertiary education), and the GDP of a country. Each contribution is scaled such
that the value of each of the three indicators is between 0 (least development) and 1
(highest development). Taking a third of each, the life expectancy indicator, the
education index, and the GDP index, one can calculate the HDI, which again is between
zero and one. The UK has currently an HDI of 0.928, which identifies it as the 13th
most developed country (Top of the table is Norway with HDI= 0.942, and bottom of
the table at number 173 is Sierra Leone with HDI= 0.275). Figure 5 shows the Human
Development Index against the annual per-capita electricity consumption for most
countries included in the compilation of the HDI. A few things can be noted about the
figure. The horizontal axis is the electricity consumption rather than the total energy
consumption, so it may be somewhat skewed against highly developed countries which
might use a larger proportion of electricity rather than other fuels. Almost all
countries cluster around a well defined curve, which rises steeply from the minimum at
about 0.3 to a value between 0.7 and 0.8. Then the increase in the development index
slows down dramatically, and above an index of about 0.9, there is no clear gain in the
developmental index with increased energy consumption. It appears that one could
12 http://www.statistics.gov.uk13 United Nations Development Programme (UNDP), Human Development Report 1999, Oxford University Press, New York and Oxford, 1999. See also http://hdr.undp.org/
44
3. Energy statistics
argue that any electricity consumption below 2MWh per year per person is directly
used for the development of that country. Above 4MWh, however, there is not gain at
all by using more and more electricity. As an example, Spain and Italy have a
development index of about 0.9 at a per-capita consumption of about 4MWh, and the
USA and Canada have an index of about 0.93 but consume over three times that
amount of electricity. At first sight, there seems to be plenty of scope for increased
energy efficiency even when only using current technology.
Figure 6. Human Development index against annual per-capita electricity use (from Pasternak (2000)14
14 A.D. Pasternak (2000). Global Energy Futures and Human Development: a Framework for Analysis. US Department of Energy, Lawrence Livermore National Laboratory. Report UCRL-ID-140773; http://www.llnl.gov/tid/lof/documents/pdf/239193.pdf
45
3. Energy statistics
6.d Energy and Climate
Some basic facts:
The mean surface pressure is about 1000 mbar= 105Pa.
The mean Earth radius is 6371km → the surface area is 4πR2= 5.1×1014 m2
Using the pressure force and gravity in Newton’s second law, pA= F= mg, we can
estimate the mass of the atmosphere as m= pA/ g= 5.2×1018 kg.
Exercise: Refer back to the combustion products of the fossil fuels, coal, oil products,
and natural gas. These are mainly water and carbon dioxide. Consider the fact that
well over three quarters of our total energy consumption is based on these fuels,
estimate the anthropogenic emissions of carbon dioxide into the atmosphere. From
the total emission, calculate its contribution to the total atmosphere in ppm (parts per
million). Then, looking at the following graphs, think about the observation that the
carbon dioxide concentration in the atmosphere has increased in line with the
industrial revolution and energy consumption.
The following graphs were copied from reports of the Intergovernmental Panel on
Climate Change (IPCC):
46
3. Energy statistics
Figure 7. Measured mean surface temperature over the past 140years and 1000 years, respectively. (Source: IPCC,2001; Figure 2-3)15
Figure 8. Measured mean surface temperature over the past 140years and 1000 years, respectively. (Source: IPCC,2001; Figure 9-1a)
15 From IPCC Third Assessment Report: Climate Change 2001: Synthesis Report. Available from http://www.ipcc.ch/pub/reports.htm : pdf at http://www.ipcc.ch/pub/un/syreng/spm.pdf ;html at http://www.ipcc.ch/pub/syreng.htm
47
3. Energy statistics
Figure 9. Time series of atmospheric concentrations over the past 1000 years. (Source: IPCC,2001; Figure 2-1)
48
3. Energy statistics
Summary
Some ways of displaying energy statistics, both as data on their own and within the concept of energy use related to other national characteristics were presented. Some points which became apparent are
1. The current energy balance is very much dominated by oil and gas
2. A substantial fraction of the energy of gas and coal are lost in the conversion to electricity in power plants
3. Moderate electricity use contributes to the development of a country, but above a finite value of electricity consumption, no apparent gain in the country’s development can be detected.
4. In many countries, the energy consumption is split into three parts of similar proportions: transport, domestic, and industry (both manufacturing and service industry). In the UK, and many other countries, transport is the largest of the energy users. Transport is to over 90% based on oil.
5. The use of fossil fuels results necessarily in carbon dioxide emissions of the order of 3kg of CO2 for each kg of fossil fuel
Task:
Gather energy production and consumption data for your country of origin, or a
country of your choice. Put this into context of the development of that country (how
much has energy been linked to improvement in standard of living, how much is energy
a source of income or expenditure/ dependence on others). Find out what the national
policies and legislation are to guarantee a secure energy supply, yet comply with
international agreements (such as the Kyoto protocol), and how much those policies
are compatible with sustainable development. Think about your findings
Sources other than the main textbook
BP Statistical Review of World Energy
J. Ramage. Energy: a guide book
http://www.iea.org
http://www.energy.gov/
http://www.eia.doe.gov/emeu/international/contents.html
http://hdr.undp.org/
http://www.dti.gov.uk/epa/
http://www.statistics.gov.uk
other national statistics publications / web sites
49
4 Heat Transfer
Introduction
The aim of this section is to provide the basic concepts of heat transfer.
Read also the fourth section of the online ‘Basic Notes.’
Once you have read this section and the online notes, you are ready to complete the third set
of online exercises
1 What is heat transfer and why bother?
If we put a hot object in contact with a cold object, we know that heat will move from
the hot object to the cold object until the two objects are at the same temperature.
This fundamental observation that heat is transfer from a higher temperature to a
lower temperature region is a consequence of the second law of thermodynamics –
and we can’t do anything about this fact. We may be able to reduce or increase the
heat transfer rate but we cannot stop the process altogether.
The rate of heat transfer is measured in watt, i.e. it has the unit of power.
As the process of heat transfer always involves a rate at which this is happening, there
is always the time dimension involved. Thermodynamic cycles and processes are
analysed assuming thermodynamic equilibrium at any point in the process. This
removes any time dimension and a power cycle, for example, can be analysed by using
only energy or enthalpy values at a the beginning and end of each process. This cannot
be done in heat transfer. Heat transfer occurs if and only if the system is not in
thermodynamic equilibrium – i.e. if there is a temperature difference between different
parts of the system. The analysis of the heat transfer can tell us how long it will take to
achieve thermodynamic equilibrium, or at what rate energy is being transferred
between different parts of the system if each of these parts is kept away from
thermodynamic equilibrium.
50
4. Heat Transfer
An example for the former case is: How long does it take for your hot coffee to cool
down to a drinking temperature.
An example for the latter case is: What is the transfer rate through the outside wall of
a building between the room (heated to a comfortable temperature) and the cold
outside. This rate of heat transfer is equal to the power requirement on the heater in
that room to maintain the room temperature.
Space heating and transport are the two biggest energy consumers. Yet, looking at
heat losses and potential for energy savings is not a very popular subject to study.
Using very simple, and frequently cheap, insulation and heat saving measures, it is
possible to achieve substantial reductions in energy consumption. However, to
estimate the heating load for a room or building and the effectiveness of energy
efficiency and energy saving measures, one needs to have an understanding of heat loss
mechanisms. This section will briefly introduce the three main mechanisms for heat
transfer. In other applications, such as heat exchangers, you would obviously want to
maximise the heat transfer rate rather than minimise it..
The three basic forms of heat transfer are
• Conduction: heat diffuses through a solid (or stagnant fluid). This requires a
mass but no motion of the mass
• Convection: heat is physically moved by a moving fluid. The motion can either
be externally enforced, e.g., by a fan, or it can be a result of the temperature of
the fluid. Convection due to an externally imposed flow is called forced
convection, and that due to the temperature variations in the fluid is called either
free convection or natural convection.
• Radiation: Any object emits electromagnetic radiation which is a function of its
temperature. This radiation is called black-body radiation. This radiation
transport energy/heat through transparent objects as well as through vacuum.
Radiation does not require any matter or motion.
For the first and the last, we can write down a complete theory, and solve them in
principle, but convection involves fluid motion and is therefore very much harder to
describe completely.
51
4. Heat Transfer
2 Heat conduction
2.a Simple conduction and Fourier’s law
Heat conduction occurs through any material (but not in vacuum) and depends on the
temperature gradient (or temperature drop over a given distance) and the ability of the
material to conduct heat (i.e. is it a thermal conductor as most metals, or is it an
insulator such as fur or foam).
We can derive an expression which tells us how much heat we transfer given the
temperature drop, the type of material, its thickness, and the surface over which we
conduct the heat:
• The higher the temperature difference, the larger the heat loss
'If it is colder outside, we need more heating'
Q ∝ ∆T
• The larger the area is over which we can transfer heat, the larger the heat loss
'If the window is larger, we lose more heat through it'
Q ∝ A
• The thicker the material is, the less we lose heat,
'If the jumper is thicker, we stay warmer'
Q ∝ d–1
• If the material is better at conducting heat, the larger the heat loss,
'Aluminium is a good thermal conductor, wool a poor conductor'
This depends on the material and is a material property called thermal
conductivity. The symbol used varies but is usually k or λ, and sometimes α:
Q ∝ k
NB: Note that there are two conventions to denote the rate of heat transfer. One is
to use the symbol Q with a dot over it to denote that it is a rate of change or a rate of
transfer of heat. Since prehistory, however, people have also simply used the symbol Q
without the dot to denote the heat transfer rate – because it is so much to type in
52
4. Heat Transfer
text. Do not confuse this with the amount of heat. The amount of heat, also Q, has
units joule, the heat transfer rate, Q, has units joule per second or watt.
This leads us to Fourier's law of heat conduction as
Q = k A DT / d
If we reduce the thickness towards an infinitesimal distance and realise that the heat
goes in the direction against the temperature gradient (ie., the temperature gradient
point to the hot place and the heat goes to the cold place), this turns into
Q = – k A dT/dx
3 Convection
Real convection is a very complicated process which depends on the imposed or self-
excited fluid flow, the fluid properties which affect both the flow (such as viscosity) and
heat transfer (such as thermal conductivity or thermal diffusivity), but for many
practical purposes, one can express the heat transfer by convection in an equation
which is similar to the conduction equation.
Since the convection does not occur by conduction over a fixed distance, the ratio of
k/d in Fourier's law is simply replaced by an empirical factor, the heat transfer
coefficient. The symbol for that is usually h or α, and has units watt per metre-squared
per Kelvin, W m− 2 K −1:
Q = h A ∆T
Typical ranges of the heat transfer coefficient (from Çengel and Turner 2001).
Type of convection h W/m2/K
Free convection of gases 2 - 25
Free convection of liquids 10 - 1000
Forced convection of gases 25 - 250
Forced convection of liquids 50 - 20,000
Boiling and condensation 2,500 - 100,000
53
4. Heat Transfer
4 Radiation
Radiation occurs both in vacuum and in transparent media (glass, air, etc...)
Each object emits radiation according to its temperature and surface condition. The
most radiation emitted at any particular temperature is that emitted by a black body.
This radiation occurs over a wide range of frequencies of this radiation (see Planck's
distribution law, and the total amount emitted is given by the Stefan-Boltzmann law:
Q = σ A T4
where
• σ= 5.67 × 10− 8 W m− 2 K− 4 is the Stefan-Boltzmann constant,
• A the surface area of the black body, and
• T the surface temperature of the black body (in Kelvin)
Very few bodies are black, but most are 'grey', which means that the emission and
absorption are simply reduced by a factor, ε, called emissivity.
Sometimes, as in glass or air, the emissivity is very low at one range of frequencies (i.e.
the visible light), but very large (i.e. close to 1) at other ranges (e.g. the infrared)
The amount of heat transferred by radiation from one surface to the other does not
only depend on the surfaces and their temperatures, but also how much one surface
'can see' the other. This is expressed by the View factor:
The view factor from a surface i to a surface j, Fi→j is the fraction of the radiation leaving
surface i that strikes surface j directly.
Everything together, then results in a net heat transfer between surface i and surface j
as
Qi →j= Fi →j εσ A (Ti4 – Tj
4)
As with convection, it might be useful to reformulate the radiative heat transfer in a
form similar to Fourier’s law, using (a2–b2)= (a+)(a–):
Qi →j= Fi →j εσ A (Ti2+Tj
2)(Ti2– Tj
2) = Fi →j εσ A (Ti2+Tj
2)(Ti+Tj) ∆T
Giving a heat transfer coefficient of h= Fi →j εσ (Ti2+Tj
2)(Ti+Tj)
54
4. Heat Transfer
5 Combined heat transfer through several layers
If heat is conducted, for example, from the air in a room through the walls of the house
and the windows, the heat has to cross several hurdles.
1. First it needs to get near the wall and then into the wall; this happens by
convection
2. Then it goes either through the plaster or the glass by conduction (and maybe
radiation)
3. Then maybe through an air gap or vacuum if we have double-glazing (by
conduction, convection, or radiation), or through bricks (by conduction)
4. and so on, until it gets to the outside surface.
5. There it leaves the wall/window and is carried away by convection and radiation
In the following, I will ignore radiation or assume that it has been incorporated in the
heat transfer coefficient, h.
We can treat the heat going through the wall section of the room separately from the
windows. They are obviously linked because they start with the same room and outside
temperatures
Let us take the easiest case of
1. the room with a heat transfer coefficient h1
2. a single wall with a thermal conductivity k1 and thickness d1, and
3. the outside with a heat transfer coefficient h2.
Even this has three stages. Since the amount of heat flow rate through each of the
stages is the same, and the surface areas are the same, we can work out what it is:
Q= (1/h1 + d1/k1 +1/h2 ) – 1 A ∆T
The expression for the several layers looks much easier, if we express Fourier's law
using the thermal resistance, R:
Q= A ∆T / R
where the units of R are K m− 2 W− 1.
55
4. Heat Transfer
For a single layer, R= d/k, and for a convective heat transfer, R=1/h
Looking back at the several layers, we now see, that the total thermal resistance of
several layers is the sum of the individual layers.
For our case of room - wall - outside, we have
R= 1/h1 + d1/k1 + 1/h2.
And we can easily add more and more layers....
Often, you might find something called the U-value. This is simply the inverse of the
thermal resistance, U= 1/R, so that
Q= A ∆T / R = U A ∆T
Reading
This section is best revised with a standard textbook on Thermodynamics and Heat
Transfer, such as the Eastop and McConkey, or the Çengel and Turner book
mentioned
56
5 Heat machines
Introduction
The aim of this section is to provide the very basic principles of the common types of engines
and turbines which operate by combusting fuel. To provide the required basic thermodynamic
concepts, the basic processes, such as isentropic expansion are briefly revised. These are then
applied to the range of engines and turbines from the internal combustion engine to gas
turbines.
Read also the fifth section of the online ‘Basic Notes.’
1 Basics for thermodynamic cycles
All machines turning heat into motion, for example to provide transport or drive an
electric generator, work on power cycles based on the first and second law of
thermodynamics:
1.) the energy in heat is equivalent to the energy in work but
2.) one cannot convert heat completely into work, i.e. there is always some waste heat
returned by the process.
All cycles consist of
1. the heat input (the fuel), Qin,
2. some work input (to get the fuel into the combustible state), Win,
3. the work output (what we want to get out), Wout, and
4. the waste heat or rejected heat, Qout.
1.a Efficiency
Because of the required work input and the waste heat, we never get as much energy
out as we put in. In other words, the efficiency, η, is always less than 1 (or 100%). The
basic definition is the ratio of the net useful work output over the required heat input:
57
5. Heat machines
η = ( Wout - Win ) / Qin
Since over a cycle Qin - Qout + Win - Wout= 0, we can write the definition of efficiency
also as:
η = ( Qin - Qout ) / Qin
1.b Open and closed cycles
The cycles to describe engines split into two major types, closed systems and open
systems:
Closed systems or non-flow systems
These systems work in a closed box. That means that no fluid enters the engine or
leaves it during the process. One can also say that all the stages take place in the same
container. The energy equation is the first law for the internal energy of the fluid:
∆u = Qin – Qout + Win – Wout = 0.
We have the zero on the right because we are looking at a cycle: the fluid returns at
the end of the process to its original state, and therefore must have the same internal
energy.
Examples are the Otto and Diesel engines, where the fluid throughput and the
combustion are two separate processes: they open some valves to take in the fuel but
then close the valves, do the compression, combustion and expansion while the
cylinder is closed.
Open systems or flow systems
These systems work have the working fluid flowing through them as it operates. This
also implies that the different stages take place in different equipment, and the change
from one part of the cycle to another takes place by the fluid moving from one part of
the engine to another. That means that fluid brings with it some energy and takes it
away during the process. The equation to be used describes the rate of change of
energy in terms of the enthalpy, h = u + p v, kinetic energy and potential energy of the
fluid passing through:
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5. Heat machines
Q' + W' = m' ∆( ½ v2 + gz + h ) = m' [ ½ (v22 – v1
2) + g(z2–z1) + h2–h1 ] ,
where the dash ' denotes the rate of change (usually denoted by a dot over the
symbol), v the fluid velocity, and z the vertical position of the fluid. Because they are
open, they don't really operate on a 'cycle' but one can close the loop if the outlet
conditions and inlet conditions are the same. Examples of flow system cycles are the
steam turbine (Rankine cycle) and the Gas turbine (Brayton cycle)
1.c The p-v diagram and the T-s diagram
Since work is defined as the integral ∫ p dv, it makes sense to draw the pressure of a
cycle against the volume (or specific volume if everything is considered per unit mass).
The integral of a function going round in a closed loop is the area enclosed by that
function. If we therefore draw the pressure against the volume for each point in a cycle,
we can work out what the work output is by measuring the area enclosed by the circle.
Since heat is defined as the integral ∫ T ds, the heat input is the area enclosed by the
graph of the temperature against the entropy.
1.d Isentropic expansion and compression of an ideal gas
The stages in a cycle which exchange work involve the expansion or compression of
the working gas. In many cases–and certainly in ideal engines–, it is possible to describe
the gas as an ideal gas, which follows the ideal gas equation of
p= ρ R T,
where p is the pressure, ρ the density, T the absolute temperature (in K) and R the
specific gas constant which is specific to each particular gas. In fact, the specific gas
constant can be expressed as the product of a universal gas constant,
Ru= 8.4133 KJ kmol– 1 K– 1, and its molar mass (in kg kmol– 1).
The change of the internal energy of an ideal gas is given by
∆U = m cV ( T2 – T1 )
or, if we divide by the mass involved:
∆u = cV ( T2 – T1 ) .
59
5. Heat machines
The change of the enthalpy of an ideal gas is given by
∆h = cp ( T2 – T1 ).
The factors cV and cp are the specific heat capacities of the gas at constant volume and
pressure, respectively. The ratio of those two values is usually denoted by the symbol γ
: γ= cp/cV
One can also see that R= cp – cV
If such an ideal gas expands without exchanging heat with the environment in an
isentropic way, the pressure, density, or temperature of the gas change as
p2 / p1 = ( V1 / V2 )γ
V2 / V1 = ( p1 / p2 )1 / γ
T2 / T1 = ( V1 / V2 ) γ - 1
T2 / T1 = ( p2 / p1 )( γ - 1 ) / γ
1.e The most ideal thermodynamic cycle: the Carnot cycle
The Carnot cycle is thermodynamically the simplest cycle of all. It consists of
• an adiabatic temperature increase (where work is done to get things going)
• an isothermal heat input (where energy is added in the form of heat)
• an adiabatic temperature decrease (where the cycle does the work)
• an isothermal heat rejection (where the waste heat is dumped to return the
system to the beginning of the cycle)
Since this description is described in
terms of heat and temperature, based on
adiabatic (no heat exchange) and
isothermal (constant temperature), it is
very easily drawn in a T-s diagram:
0 1 2 30
1
2
3The Carnot cycle
s
T 1
2
3
4
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5. Heat machines
The area enclosed by this cycle is the area under the upper line, T2 (s2–s1), minus the
area taken away again under the lower line, T1 (s2–s1)
Therefore the net heat added is (T2–T1) (s2–s1).
However, the original heat input was that added in stage 2, i.e. the upper line, and the
area under the upper line is T2 (s2–s1)
Therefore, we have an efficiency of the Carnot cycle of
η = Net heat / Heat input = (T2–T1) / T2 = 1 – T1/T2.
No real machine actually simulates the Carnot cycle but some get fairly close. In a
sense, this is the ultimate yard stick for heat-power cycles, whereas each real engine
has its own ideal cycle.
Example: In a spark-ignition internal combustion engine, the temperature of the inlet
gases may be around 30 degree or 303 K, and the temperature during combustion may
be 2500 K. The efficiency of a Carnot cycle operating between those temperatures will
be η = 1 - 300/2500= 0.88 = 88%. Compare this with the efficiency of the ideal Otto
cycle further down, and with a typical real efficiency of around 25%
2 Closed systems: internal combustion engines
The closed systems are identified by the fact that, at least during some of the power
cycle, the combustion chamber is physically closed. The igniting gas increases the
pressure in the chamber, and this pressure pushes a wall and increases the volume
available to the gas. The common closed system internal combustion engines are the
petrol engine and the diesel engine.
61
5. Heat machines
2.a The spark-ignition engine: the Otto cycle
This is a closed system and consists of
1. Isentropic compression as the piston moves forward (the work input):
T2 / T1 = ( v1 / v2 )( γ – 1 ) ,
2. Constant volume heat addition in the combustion chamber
qinput= u3 – u2= cv (T3 – T2).
3. Isentropic expansion as the piston moves back the cycle does the work:
T3 / T4 = ( v4 / v3 )( γ – 1 ) ,
4. Constant volume heat rejection. This is a bit of a fudge, since in the real cycle, the
exhaust valve is opened, and the piston moves forward again to expel the exhaust
gases, and then draws in the fresh air-fuel mix. To indicate this a horizontal line
denotes this piston movement. Since this does not enclose any area, we assume
that it does not cost any work. In real life, the pressure in the cylinder is slightly
above atmospheric when it expels the exhaust gas and slightly below atmospheric
when it draws the new fuel in. It then will enclose an area and the cycle goes round
the loop in a counter-clockwise direction whereas the other path form a clockwise
path. This means that this expulsion and drawing in of new fuel is in the other
direction of the main cycle - i.e. it is a work input reducing the net work output:
qwaste= cv (T4 – T1).
Since the volume is constant in stages 2 and 4, and the compression and expansion
cover the same range, we can introduce the volume ratio, r, or compression ratio. For
a typical petrol engine, the compression ratio is between 7 and 10.
As a result, we can write: T1/T2 = T4/T3 = 1 / r( γ – 1 ) ,
62
5. Heat machines
0 1 2 3 4 50
5
10
15
v
p
1
2
3
4
The efficiency of the Otto cycle is
η = 1 – 1/ r( γ – 1 ) .
A typical efficiency of a real petrol engine is around 25% to 30%
Example:
A petrol engine operating on an ideal Otto cycle has a maximum cylinder volume of
1.6 litres, a compression ratio of 10, and a peak cycle temperature of 2500 K. The
minimum cycle temperature and pressure are 300 K and 1 bar, respectively. The gas is
an ideal gas with γ = 1.4 and cv = 0.7 kJ/ (kg K) . Calculate
1. The cycle efficiency
2. The temperature at the end of the compression stage
3. The heat input
4. The temperature at the end of the expansion stage
5. The heat rejection
6. The work output
7. The net power output if the engine operates at 3000 cycles per minute
63
5. Heat machines
Solution:
1. The cycle efficiency is η = 1 – 1/ r( γ – 1 )= 60%
2. The mass in the cylinder is m= pV / RT= 0.00184 kg
The temperature at the end of the compression stage is
T2 = T1 r( γ – 1 ) = 300 * 100.4= 754 K.
3. The heat input is
Qinput= m cv (T3 – T2) = 0.00184 kg * 0.7 kJ/(kg K)* (2500 K – 754 K)= 2250 J.
4. The temperature at the end of the expansion stage is
T4 = T3 / r( γ – 1 ) = 2500 / 100.4= 995 K.
5. The heat rejection is
Qinput= m cv (T4 - T1) = 0.00184 kg * 0.7 kJ/(kg K)* (995 K - 300 K)= 897 J.
6. The work output is
woutput= m cv (T3 - T4) = 0.00184 kg * 0.7 kJ/(kg K)* (2500 K - 995 K)= 1939 J.
The work input is
winput= cv (T2 - T1) = 0.00184 kg * 0.7 kJ/(kg K)* (754 K - 300 K)= 585 J.
The net work output is wnet= 1939 - 585 J = 1354 J.
7. The power output is Poutput= 3000/60 * 1354= 68 kW.
2.b The compression-ignition engine: the Diesel cycle
This cycle is very similar to the Otto cycle. The only difference is that one assumes that
the combustion happens at a constant pressure rather than constant volume:
• Isentropic compression as the piston moves forward (The work input):
T2 / T1 = ( v1 / v2 )( γ – 1 ) ,
• Constant pressure heat addition in the combustion chamber
qinput= h3 – h2= cp (T3 – T2).
• Isentropic expansion as the piston moves back the cycle does the work:
T3 / T4 = ( v4 / v3 )( γ – 1 ) ,
• Constant volume heat rejection as with the Otto cycle:
qwaste= cv (T4 – T1).
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5. Heat machines
0 1 2 3 4 5 6 70
5
10
15
v
p
1
2
3
4
Because we have one constant pressure and one constant volume heat exchange
process, we need to define to ratios:
• The compression ratio, r= V2 / V1, which is typically between 12 and 23, and
• The cut-off ratio, rc= V3 / V2, which is the ratio of the volumes at the beginning
and end of the combustion stage. This is typically between 2 and 4.
The efficiency of the Diesel cycle is
η = 1 – 1/ r( γ – 1 ) [ (rcγ – 1 ) / (γ(r
c – 1)) ] .
A typical efficiency of a real Diesel engine is around 35% to 40%
65
5. Heat machines
3 Turbines: Open systems
In open systems, the heat releasing part of the cycle does not get closed off, but the
working fluid flows through that part of the turbine. Two examples are the steam
turbine and the gas turbine. These two also serve to illustrate the difference between
‘open-cycle system’, ‘open system’, and ‘internal/external’ combustion machines:
The steam turbine is a closed system because the water/steam fluid is recycled back
from the turbine outlet to the boiler by the feed pump, but it is modelled by an open
cycle because the steam never finds itself trapped in any part of the cycle, and it is an
external combustion machines because the heat is supplied through a heat exchanger in
the boiler by an external heat source.
The gas turbine is an open system because it draws in fresh air from the environment,
does its bit to the air/fuel mixture, and then spits it out at the end. It is also an open-
cycle system but it is an internal combustion turbine because the combustion occurs
within the working fluid in the (open) combustion chamber between the compressor
and the turbine.
3.a Steam turbines: the Rankine cycle
The steam turbine, which is modelled by the ideal Rankine cycle attempts to reproduce
the Carnot cycle. It exploits the property that the temperature does not change during
evaporation and condensation. The heat input is instead used for the phase change
from liquid to gaseous or back. If the fluid were always in a state where it consists of
both, liquid and gas, one could represent it reasonably well by the Carnot cycle. The
fact that the temperature of the fluid changes on heat input when it is either a pure
liquid or a pure gas means that the heat input stage in the boiler actually consists of
three sub-stages: first the heat input into the liquid before it starts to boil, then the
mixed-phase isothermal heat input, and finally the heating of the pure gas ("superheated
vapour"). Another point which makes the Rankine cycle look differently from the
Carnot cycle is the fact that liquids are virtually incompressible. As a result, the work
input (in the feed pump) only results in an absolutely minute pressure or temperature
increase. In fact, if the water were a completely ideal incompressible liquid, the
temperature increase would be zero.
66
5. Heat machines
The steam turbine consists of different parts which carry out the different parts of the
cycle of work input in the shape of a feed pump moving the fluid into the boiler, the
heat addition in the boiler, the work output in the turbine, and the heat rejection in a
condenser. As a result, we need to model the steam turbine as an open system (which
means that we need to look at enthalpy of the fluid, h, rather than the internal
energy, u.
This can now be summarised in the same way as the Carnot cycle (keeping the same
stage numbers) as
1. The work input: a feed pump moves the water from the condenser to the boiler:
wpump = ∆p v,
where pis the pressure difference across the pump and v the specific volume of the
liquid in the pump (i.e. 1/density).
The power requirement by the (ideal) pump can then be calculated from this and
the mass flow rate through the pump:
W'pump = m' wpump = m' ∆p vliquid
(this corresponds to the adiabatic temperature increase of the Carnot cycle)
2. The heat input into the fluid in the boiler (and superheater)
a. Heating of the water at constant pressure:
∆h = cp, liquid ( Tevaporation - Tfeed )
constant pressure/constant temperature evaporation
b. constant pressure isentropic heating of the steam in the superheater:
T2 / T1 = ( V1 / V2 )γ - 1
This is summarised by the enthalpies at the beginning and end of the total heating
stage: qinput= h3 - h2.
3. An adiabatic temperature decrease (by expansion in the turbine where the cycle
does the work): wout = h3 - h4,
The power output can then be calculated from this and the mass flow rate through
the turbine:
W'out = m' wout = m' ( h3 - h4 ).
67
5. Heat machines
4. Isothermal heat rejection by condensation (where the waste heat is dumped to
return the system to the beginning of the cycle): qwaste= h4 - h1.
0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
2
2.5
3
s
T
1
2a
2b
2c3
4
liquid
evaporation/condensation
Gas
, "s
uper
heat
ed s
team
"
(The length of the pumping process 1 is exaggerated to make it visible at all.)
The efficiency of the Rankine cycle is
η = [ (h3–h4) – (h2–h1) ] / [ h3–h2 ]
or
η = [ (h3–h4) – (p2–p1)vliquid ] / [ h3–h2 – (p2–p1)vliquid) ]
A typical efficiency of a real steam turbine is around 30% to 40%.
Since all these involve enthalpies, one has to resort to steam tables to find the
corresponding enthalpies of the working fluid at a particular pressure and temperature..
68
5. Heat machines
Example: (This is very abbreviated and should be followed by referring to a Thermodynamics book)
An ideal steam turbine operates with a boiler pressure of 5MPa and a condenser
pressure of 0.015 MPa. Steam leaves the superheater at a temperature of 400 degrees
C. The mass flow rate of the steam through the turbine is 0.5 kg/s. Calculate
1. The power output
2. The power input into the pump
3. the rate of heat rejection in the condenser
4. the cycle efficiency
The steam tables give us at 5 MPa and 400oC an enthalpy for the steam of h3= 3198
kJ/kg, and an entropy of s3= 6.65 kJ/ (kg K).
At 0.015 MPa we find an enthalpy for the steam-liquid mixture of h4= 2155 kJ/kg, and an
entropy of s4= s3.
At 0.015 MPa we find an enthalpy for the liquid of h1= 226 kJ/kg, and an entropy of s1=
0.755 kJ / (kg K).
1. The work output is
wout = h3 - h4= 3198 - 2154= 1044 kJ/kg.
With the mass rate, this results in a power output of
W'out = m' wout = 0.5 kg/s * 1044 kJ/kg= 522 kW.
2. The density of water is 1000kg/m3, i.e. the specific volume is 0.001 m3/kg, and the
power consumption by the pump is
W'pump = m' ∆p vliquid = 0.5 kg/s * ( 5,000,000 Pa - 15,000 Pa ) * 0.001 m3/kg= 2.5 kW
3. The rate of heat rejection is
Q'waste= m' (h4 - h1) = 0.5 kg/s * (2155 kJ/kg - 226 kJ/kg) = 0.96 MW
4. The efficiency is
η = [ (h3-h4) - (p2-p1)vliquid ] / [ h3-h2 - (p2-p1)vliquid) ]
= [ 3198 - 2155 - 5 ] / [ 3198 - 226 - 5 ] = 1038 / 2967 = 0.35
The efficiency of this ideal steam turbine is 35%
69
5. Heat machines
3.b Gas turbines: the Brayton cycle
Whereas a steam turbine usually operates in a closed cycle in which the steam in
condensed and pumped back into the boiler (but is modelled as an open system), the
gas turbine is usually a truly open system: fresh air enters the turbine and exhaust gases
leave the turbine. One still can close the thermodynamic cycle by regarding the
emission of the exhaust gases as the heat rejection stage. This ideal cycle for the gas
turbine is the Brayton cycle.
This can now be summarised in the same way as the Carnot cycle (keeping the same
stage numbers) as
1. Isentropic compression (The work input): a compressor at the front moves the air
into the combustion chamber of the turbine:
T2 / T1 = ( p2 / p1 )( γ – 1 ) / γ ,
If the specific heat remains constant, the work input per kg of air is win= cp (T2 – T1 )
2. Constant pressure heat addition in the combustion chamber
qinput= h3 – h2= cp (T3 – T2).
3. Isentropic expansion in the turbine where the cycle does the work:
T3 / T4 = ( p3 / p4 )( γ – 1 ) / γ ,
If the specific heat remains constant, the work output per kg of air is
wout= cp (T3 – T4 )
4. Constant pressure heat rejection by condensation (where the waste heat is
dumped in the atmosphere to return the system to the beginning of the cycle):
qwaste= h4 – h1= cp (T4 – T1).
Since the pressure is constant in stages 2 and 4, the pressure in stage 3 reduces from p2
to p1 again. We only encounter one pressure ratio in the cycle:
rp= p2 / p1.
As a result, we can write:
T2/T1 = T3/T4 = rp( γ – 1 ) / γ ,
70
5. Heat machines
A T-s diagram and a p-v diagram of the ideal Brayton cycle.
The efficiency of the Brayton cycle is
η = 1 – 1/ rp( γ – 1 ) / γ .
A typical range of pressure ratios is between 5 and 20 and a typical efficiency of a real
Gas turbine is around 40%.
0 1 2 30
2
4
6
8
10
s
T
1
2
3
4
0 1 2 3 4 50
0.5
1
1.5
2
2.5
3
vp 1
2
3
4
71
5. Heat machines
Example: A power plant operating on an ideal Brayton cycle has a pressure ratio of 8.
The gas temperature at the compressor is 300 K and at the turbine inlet 1300 K. The
mass flow rate is 2 kg/s, and the gas is an ideal gas with γ = 1.4 and cp = 1.1 kJ/ (kg K) .
Calculate
1. the cycle efficiency
2. The gas temperature at the compressor exit
3. The gas temperature at the turbine exit
4. The heat input
5. The heat rejection
6. The power output
7. The power input into the compressor
Solution
1. the cycle efficiency is η = 1 - 1/ rp( γ - 1 )/γ = 45%
2. The gas temperature at the compressor exit is T2= T1 r( γ - 1 ) / γ = 300 * 80.4/1.4= 543 K
3. the gas temperature at the turbine exit T4= T3 / r( γ - 1 ) / γ = 1300 / 80.4/1.4= 718 K
4. The heat input is qinput= cp (T3 - T2) = 1.1 kJ/(kg K)* (1300 K - 543 K)= 833 kJ/kg.
With a flow rate of 2 kg/s, this is a rate of heat supply of 1.67 MW.
5. The heat rejection is qwaste= cp (T4 - T1) = 1.1 kJ/(kg K)* (718 K - 300 K)= 460 kJ/kg.
With a flow rate of 2 kg/s, this is a rate of heat supply of 920 kW.
6. The power output is woutput= cp (T3–T4) = 1.1 kJ/(kg K)* (1300 K - 718 K)= 640 kJ/kg.
With a flow rate of 2 kg/s, this is a rate of heat supply of 1.28 MW.
7. The power input into the compressor
winput= cp (T2 – T1) = 1.1 kJ/(kg K)* (543 K – 300 K)= 267 kJ/kg.
With a flow rate of 2 kg/s, this is a rate of heat supply of 534 kW.
8. The net power output is therefore 750 kW
9. The efficiency of this ideal gas turbine is 35%
72
5. Heat machines
Reading
This section is best revised with a standard textbook on Thermodynamics, such as the
Eastop and McConkey, or the Çengel and Turner book mentioned.
73
6 Fluid machines
Introduction
The aim of this section is to provide the very basic principles of the common types of engines
and turbines which exploit fluid motion or pressure. To provide the required basic fluid dynamic
concepts, the basic processes, such as Bernoulli’s equation are briefly revised. These are then
applied to the range of engines and turbines from reaction turbines to wind turbines.
Read also the sixth section of the online ‘Basic Notes.’
Once you have read this section and the online notes, you are ready to complete the fourth set
of online exercises, which is based on the material covered here and in the previous section.
1 Definition of fluid machines
This is an introduction to extracting work from a fluid flowing through a turbine. This
could be a reaction turbine in a hydropower station, driven by a pressure drop across
the turbine, or it could be a wind turbine extracting kinetic energy from the wind.
Turbines and engines are machines which extract energy or power
from a stream of fluid and convert it into mechanical energy
(and then possibly into other forms, such as electricity)
Pumps, fans, and compressors are machines which use mechanical
energy or power to increase the pressure or kinetic energy of a
fluid.
• Design parameters for pumps and fans:
1. Desired flow rate
2. Associated head loss
3. Power requirement
• Design parameters for compressors::
74
6. Fluid machines
1. Desired pressure increase
2. Associated flow rate
3. Power requirement
• Design parameters for turbines and engines:
Either
1. Desired power generation
2. Available head
3. required flow rate
Or, for wind turbines and similar,
1. Desired power generation
2. Available flow rate
Important quantities
1. Head or available head: hydrostatic pressure at machine inlet/outlet: p= ρgH .
When considering a turbine beneath a reservoir, the available head at the turbine
inlet is the height of the water level above the turbine minus the head loss in the
penstock feeding the water from the reservoir to the turbine (and possibly minus
residual head required at the turbine outlet for the water to clear the turbine).
2. Flow rate, Q.
3. Power:
this can be either the hydraulic power in cases where the static head is exploited:
p= ρgH,
or the power carried in an open stream, of velocity U, through the cross-section, A,
of the turbine facing the stream (in the case of a wind turbine, this is the circle
swept by the rotor blades of diameter D): P= ½ ρ AU3 .
75
6. Fluid machines
2 Basic Fluid Mechanic principles
This section quickly reviews the fundamental basics of fluid mechanics, based on the
conservation of mass, momentum, and energy.
2.a The control volume
If we want to be able to describe the forces and mass balances we need to define a
volume over which we do this. This volume of our choice is called the control volume,
abbreviated CV, and the surface which encloses this volume is called the control surface
(CS).
2.b Continuity equation
• The continuity equation states that mass is conserved. In an incompressible fluid, such as
water, this is equally to the statement that the volume or volume flow rate is conserved.
In other words: What comes in has to come out
ρin Qin = ρout Qout
ρ1 Q1 = ρ2 Q2
If fluid is incompressible (all liquids), ρ1 = ρ2 , and
Q1 = Q2
U1 A1 = U2 A2 ,
where Q is the volume flow rate, A the cross-sectional area through which the fluid
flows, and U the average velocity through that cross-sectional area.
2.c Momentum equation
Here, we have to remember that the best-known form of Newton’s second law,
F= m a, is a simplification of the proper definition as ‘The change of momentum is equal
to the applied forces',
F= d(mv)/dt. To apply this to a fluid, we have to rephrased it slightly if we look at a
position through which fluid is flowing, instead of looking at a solid object with mass m:
76
6. Fluid machines
The Change of momentum in our control volume is equal to the net gain of
momentum by momentum flowing into the control volume plus any forces
applied to the fluid within the control volume or at its surfaces.
By convention, one calculates the momentum flowing out of the control volume. This
gives a minus sign which disappears if we move that term to the left of the 'iis equal'.
Also, the forces can be divided into those which apply throughout the bulk of the
volume, 'body forces' (e.g. gravity), and those which contribute only at the outer surfaces
of the control volume (They are described by a stress tensor which includes forces due
to pressure and stress). With this, the momentum equation for a control volume can
be rephrased as
Rate of Change of Momentum + Net loss of Momentum
= Body Forces + Surface Forces .
You could also visualise the different meanings of the terms by realising that the
momentum carried by the fluid is carried by the mass flow rate = rate of change of
mass:
F= d(mv)/dt. = m dv/dt + v dm/dt = m a + v m’ = m a + ρ v Q= m a + ρ A U 2.
Remember that this is more of an illustration than a derivation. In particular, any force
applied is given by the difference between the value of ρ v Q at the inlet and the outlet.
If we are looking at a steady-state flow, then there will be no acceleration. If we
furthermore only consider a case where the only velocity component is that in the
direction of the mass flow rate, then the force becomes
F= ρ1 A1 U1 2 – ρ2 A2 U2
2.
The most important forces of all is the pressure force, F= pA.
77
6. Fluid machines
2.d Energy, energy per unit volume, and head
The most important forms of energy for fluid machines are
1. Potential energy: m g z
2. kinetic energy: ½ m U2
3. flow work or pressure energy: p V.
(all having the base unit: 1 joule= 1J= 1 kg m2 s −2)
In fluid mechanics one often uses energy per unit volume: 1 joule per m3= 1 kg m−1 s−2.
Note that this has the same units as pressure: 1 pascal= 1 Pa= 1 kg m−1 s−2.
1. Potential energy: ρ g z
2. kinetic energy: ½ ρ U2
3. flow work or pressure energy: p
Using the hydrostatic pressure equation, p= ρ g H, this pressure is equivalent to a
stationary column of fluid above the point you are looking at with a height of H.
As hydropower is usually associated with a reservoir a certain elevation above a
turbine house, and because pressure is most easily measured by the height of a fluid
column in a manometer, one often converts these into quantities with dimension (m)
by dividing each term by ρ g, and refers to them as head:
1. Elevation: z
2. dynamic head: 221 Uρ
3. static head: gpH
ρ=
2.e Bernoulli's equation
Bernoulli's equation is the statement of the first law of Thermodynamics that energy is
conserved. It considers the energy balance at two points along a path that a fluid takes,
a stream line.
In terms of energy per volume (unit pascal: 1 J/m3= kg m–1 s–2= Nm= Pa):
222
122
212
111 UgzpUgzp ρρρρ ++=++
78
6. Fluid machines
In terms of energy per mass (unit J/kg= m2 s–2)
222
12
2212
11
1 UgzpUgzp++=++
ρρ
In terms of head (unit m)
gUz
gp
gUz
gp
22
22
22
21
11 ++=++
ρρ
The last can be rewritten using the hydrostatic pressure due to a fluid column of height
h1: p= ρgh:
gUzh
gUzh
22
22
22
21
11 ++=++ ,
giving us Bernoulli’s equation using elevation, static head, and dynamic head.
Taking into account losses and work done on or by the fluid, the energy balance at
point 2 can be worked out in terms of the energy balance at point 1 and what has been
lost or gained in between. If energy is added to the fluid the work term is positive if
energy has been extracted, HW is negative:
PTL HHHg
Uzhg
Uzh −++++=++22
22
22
21
11 ,
where HL is the head loss (e.g., due to friction) between points 1 and 2, HT the head
extracted by a turbine, and HP the head added by a pump.
The head loss is usually due to a combination of friction and ‘minor losses’, quantified
by a friction factor, f, and minor loss coefficients, K, which result in a head loss of
( )g
UKfH DL
L 24
2Σ+=
If we only have a pump but no turbine in the system, we can re-arrange it to
( )g
UKfgUUzzhhH D
LP 2
42
221
22
1212 Σ++−+−+−= .
79
6. Fluid machines
As the left-hand side only quantifies the pump and the right only the two endpoints of
the fluid system, the right-hand side is called the system head, HS.. This equation is used
to calculate what flow rate and pressure one finds if one puts a given pump into a
pipeline system.
The same can be done for a turbine:
( )g
UKfgUUzzhhH D
LT 2
42
222
21
2121 Σ+−−
+−+−=
2.f Mechanical and Fluid power
Mechanical Fluid Hydraulic
Force F F Vm
Energy E F s
Torque T F d dVm t
Power P F U= F d ω = ω T QdVdVm tt ρωω = gHQρ
s is used for linear distance moved, while d is used for distance of selected point from
centre of rotation. ω is the angular velocity of the machine rotor.
Note that in the torque only the component of the force is counted which is at right
angles to the distance vector from the axis of rotation.
Euler’s turbomachine equation states that, in the absence of losses etc., the change in
hydraulic power across the turbine is fully transferred into shaft power by means of the
fluid applying a torque on the shaft in the form of the tangential velocity. In short, it
states that the three columns in the bottom row of the table above are all equal.
2.g Dimensional analysis
Dimensional analysis is tool used universally in all areas of the physical sciences albeit
frequently not overtly. It is a method to find parameters which describe the important
dynamics of a system in a way which is not restricted to that particular system.
Without it, it would be useful to build scale models of cars or aeroplanes and test them
in a wind tunnel. Dimensional analysis helps you to transfer the data from the wind
tunnel to the real thing.
80
6. Fluid machines
The procedure is to gather all relevant parameters, and then to reduce them to a
minimum set of nondimensional parameters using a set of base units. If we have
different situations where all nondimensional parameters are the same, then the
dynamics is the same, and we have similarity. Similarity is usually split into geometric
similarity, i.e. the shape is the same, and dynamic similarity, i.e. the other parameters are
the same. To have a truly similar situation, we need both geometric and dynamic
similarity.
It is probably best illustrated by an example. To read the theory, refer to a standard
Fluid Mechanics textbook
Example: Wind drag on a car.
The relevant parameters are the size and geometry of the car, the fluid properties of
the air through which the car is moving, the speed at which it is moving, and the drag
force: Size L, air density ρ, air viscosity µ, speed U, Force F. The base units are length,
m, mass, kg, and time, s.
The first condition is that we test the car using a scale model; trying to measure the
drag force on a Ferrari will be different to that on a minibus…
Secondly, we have 5 parameters and 3 base units. Since the procedure boils down to a
simultaneous set of equation for the five parameters, we know that we will end up with
5–3 = 2 nondimensional parameters.
One parameter to characterise the drag force will have the form Fa UbLcρdµe
The other one will have the form UfLgρhµ i
We can safely start with a=1. Putting this into the base units:
kg 1 m 1 s –2 × mb s –b × m c × kgd m –3d × kge m –e s –e = 0
and × mf s –2f × m g × kgh m –3h × kg i m –i s –i = 0
which can only be satisfied if each base unit drops out:
kg: 1 + d + e = 0
m: 1 + b + c – 3d –e = 0
s: – 2 – b – e = 0
81
6. Fluid machines
This reduces to :
b = – e – 2
c= – 2e – 2
d= – e – 1
We can now choose to set e=0, and get b= –2, c= –2, and d=–1:
The force parameter is F U –2L–2ρ –1 or, 22122 AU
FUL
FCD ρρ== , the drag coefficient.
Repeat the exercise for the second parameter (and making sure you don’t choose i=0!)
to get the Reynolds number, µρUL=Re .
As you saw, we had a choice to set one of the unknown exponents to zero. One can
end up with a different set of equally valid nondimensional parameters….
2.h Dimensional Analysis for fluid machines
Taking a reaction turbine as a representative, we recognise that the relevant
parameters are most likely to be:
• Shape of the turbine
1. Size of the turbine: usually the diameter of the rotor, D in [m]
2. Operating speed of the turbine, N in [rad/s]
3. Density of the fluid, ρ in [kg/m3]
4. Viscosity of the fluid, µ in [kg/m/s]
5. Available head of the fluid, H, or pressure, p in [kg/m/s2]
6. Flow rate through the turbine, Q in [m3/s]
7. Power output, P in [kg m2 / s3]
With 7 parameters and 3 base units, we expect 4 nondimensional parameters. If you go
through the analysis, trying to find one parameter proportional to H, and to Q, and one
to P, you are likely to end up with
82
6. Fluid machines
a Reynolds number, DQ
µρ=Re
a head coefficient, 22DNgH
H =Π
a flow coefficient, 3NDQ
Q =Π , and
a power coefficient, 53DNP
P ρ=Π .
This particular set is correct but has not been adopted as the most useful one. As there
is always some freedom in choice of nondimensional parameters, and because any
product of two nondimensional parameters is still a nondimensional parameters, we can
find one parameter which expresses the power in terms of head and flow rate, and
another parameter which in some way characterises the machine by its operating speed
without any reference to its size:
The new power coefficient becomes: gHQP
QH
P
ρ=
ΠΠΠ=Π . Recognising that the term
at the bottom is the hydraulic power, we see that we have derived an expression for the
efficiency.
2.i Specific Speed
The ‘speed’ coefficient, KN can be found by combining the power coefficient and the head
coefficient so that the size drops out, and that it is proportional to the operating speed:
( ) 4/5
4/1
552
244/1
55
1010
1062
24/1
5
2
gHPN
HgPN
HgDN
DNPK
H
PN ρρρ
=
=
=
ΠΠ=
A turbine with a high specific speed will provide a high power output for a low head,
where as a turbine with a low specific speed will provide power at a high head. Another
rule of thumb not immediately obvious from the equation is that of two turbines
providing the same power, the one with the higher specific speed will be more compact.
83
6. Fluid machines
Because the specific speed has been found to be very useful for many years, it has been
adopted in all sorts of units, from the standard SI units to Imperial units but also
‘practical units’ which then resulted in the specific speed not being a nondimensional unit
at all. Because all the turbines have almost exclusively been used for water (ρ
= 1000kg/m3), and on the Earth’s surface, (g= 9.81m/s), engineers have adopted the
specific speed using the short form 4/5−=H
PS NN (where I have used NS to distinguish it
from the nondimensional version derived above. Often, the speed is also taken in rev/s
rather than rad/s, which changes the specific speed value by a factor of 2π.
Exercise: Try to find the units of 4/5−=H
PS NN if you take N in rev/s (rather than
rad/s), P in MW, and H in m.
3 Types of turbines
3.a Reaction turbines
The action of driving the turbine shaft is by a gradual pressure drop over rotor.
machine through drop Pressurerotor through drop Pressure reaction of Degree' =
Depending on the size of the rotor and its orientation with the main flow direction,
they are classified into radial flow, axial flow, and mixed flow machines.
1. Radial flow turbine:
• Runner between outer inlet radius and inner outlet radius.
• specific speed range: 0.1−0.4 ;
• moderate head (up to 500m) and
moderate flow rates;
The most common example of a radial flow
reaction turbine is the Francis turbine. In fact, this turbine is in principle the same as
a centrifugal pump operating in reverse, and in cases where specifically designed
turbines are too expensive, centrifugal pumps have been used as turbines. This also
opens up the possibility of a pumped-storage hydropower station; when there low
84
6. Fluid machines
electricity demand, the power station operates the machines as pumps to fill up its
reservoir, and when electricity demand rises, the same machines are used as turbines
to generate electricity.
2. Axial flow turbine:
• Runner entirely within fluid, except for shaft.
• specific speed range: 0.3−1 ;
• low head (up to 150m for Kaplan, up to 25m for Bulb turbine)
and high flow rates.
Kaplan turbine='Propeller in housing' although usually with vertical axis.
Bulb turbine, usually with horizontal axis but with generator in bulb in line with
runner. Used at very low heads, such a tidal power stations.
85
6. Fluid machines
3.b Impulse turbines
Examples of impulse turbine are the Pelton Wheel
and the Turgo turbine:
All energy is converted to kinetic energy in a fluid jet
hitting the turbine runner. The runner is everywhere
at atmospheric pressure. As a result, the degree of
reaction is zero.
Because the size of the buckets at the rim of the runner limit the size of the jet, Pelton
Wheels operate best at fairly low flow rates. They are therefore best suited for high
heads. This is reflected in a specific speed range of 0.01 – 0.1.
3.c Turbine types according to their specific speed
• The diagrams only show the rotor, but not the stationary housing or guide
vanes.
• The values in the various specific speeds are only approximate ranges. The
usual conversion is Kn= 2π Kn (rev) = 0.0052 NS (metric)= 0.023 NS (British)
• The specific speed for the Pelton Wheel applies to a single jet. For multiple jet,
the power output is proportional to the number of jets.
86
6. Fluid machines
Name Image of the runner typical
head (m)
Kn Kn
(rev)
NS
(metri
c)
NS
(Britis
h)
Impulse turbines
Pelton Wheel > 300 <0.2 <0.03 < 30 < 10
Turgo turbine
(cross flow)
steam turbine
(axial flow)
87
6. Fluid machines
Reaction turbines
Radial flow reaction turbines
Francis 500 – 30 0.25 –
1.3
0.04 –
0.2
50 –
250
10 –
60
Mixed flow reaction turbines
100 – 15 0.6 – 2 0.1 –
0.3
120 –
360
30 –
100
1.3 – 2.5 0.2 –
0.4
250 –
500
60 –
120
2 – 3 0.3 –
0.5
360 –
600
100 –
150
88
6. Fluid machines
Axial flow reaction turbines
Kaplan 50 – 4 2 – 6 0.3 –
1
360 –
1200
100 –
300
Bulb or Pit < 20 > 3 > 0.5 > 600 > 150
Wells
3.d Controlling the output
As the turbine will slow down if energy is drawn off at a higher rate, the response of a
turbine to an increase of the electricity demand is a change in speed. Frequencies can
be measured extremely accurately, and this is a way to monitor whether a turbine is
matching the electricity demand. If the frequency drops, the flow rate through the
turbine has to increase. This control of the flow rate is achieve by different means.
The flow rate of the water jets in a Pelton Wheel is usually controlled by a spear valve,
which can change the volume flow rate but does not affect the jet velocity. In reaction
turbine, wicket gates and guide vanes are used to both, control the flow rate and
condition the velocities to enter the turbine runner smoothly.
89
6. Fluid machines
Example: A large hydropower station
A new 150MW hydropower station is to be built at a site where the available head is
estimated at 350m. The generators require the turbine to rotate at 10Hz.
1. Find the volume flow rate to generate 150MW of hydraulic power.
2. Find the specific speed (KN (rev)) to generate 150MW.
3. If the only available turbines are Francis turbines with KN= 0.08 rev, find the power
output from one turbine, and the number of turbines required to generate 150MW.
Solution
1. 137433509810000000150 −=
×== sm
gHP
Q h .,,ρ
2.( ) ( )
147034341000
00000015010 251
45 .
,,. ===
gH
PNKnρ
3. ( ) ( )MWW
NgHK
P n 441024410
343410000806
2
522
2
522=×=
××== .
. ..ρ
So, we would need 4 turbines
4
90
6. Fluid machines
5 Euler's Turbomachine equation
In an ideal world, all power transferred to or from the machine rotor is transferred
from or to the fluid power, and all the power transferred to or from the fluid in the
rotor is transferred to or from the hydraulic power of the fluid. As the power
transmitted by a rotating shaft is due to a torque, the component at of the fluid velocity
which acts perpendicular to the direction to the axis of rotation, the momentum flow
rate associated with the tangential velocity is the crucial quantity in the power transfer
mechanism. The deficit in angular momentum of the fluid found between the fluid
entering the turbine and leaving it must have been transferred elsewhere, ideally to the
turbine shaft. This is quantified in Euler’s turbomachine equation:
( ) gHQVRVRmTP tt ρωω =−== 1122
where
ω: angular velocity of the machine rotor
T: torque on the rotor shaft
R1: inlet radius of the rotor
R2: outlet radius of the rotor
Vt1: tangential fluid velocity at the rotor inlet
Vt2: tangential fluid velocity at the rotor outlet
m : mass flow rate through the machine
Q: volume flow rate through the machine
H: head difference between inlet and outlet of the machine
The problem is to know or find the tangential fluid velocities, Vt,
• To find the tangential velocities, we need to know the radial fluid velocities and
the rotor blade velocities
• The rotor blade velocity, U, is found from the rotation rate: ωrU =
91
6. Fluid machines
• The radial fluid velocity, Vr, is found from the mass flow rate: brmVr πρ2
=
where b is the 'height' of the blade, i.e. 2πrb is the cross-section of the
cylindrical opening through which the fluid flows
To be able to find relationships between all velocities, we assume that the fluid is
always flowing parallel to the blades.
The control volume is the space between two blades, which is moving with the rotating
runner . In this moving frame of reference, the fluid whirl/tangential velocity is
apparently reduced by blade velocity, ωRU =
The usual notation for the different velocity components is
U: blade velocity,
Vr: radial component of fluid velocity
Vt or Vw: tangential component of fluid velocity, aka whirl velocity
R or W: Fluid velocity relative to blade
This is the effective velocity in the control volume!
5.a Principle of a Francis Turbine
Euler’s turbomachine equation, ( )1122 tt VRVRmTP −== ωω , is illustrated using the
Francis turbine as an example. The rotation rate is fixed by the grid frequency, the
inlet and outlet radius of the runner are fixed by design, but the mass flow rate and
inlet whirl velocities are variable. The mass flow rate is adjusted by opening or closing
a valve.
By the mass flow rate, the radial component of the fluid velocity is specified, by
Am
rV ρ= , where A is the cross-sectional area of the inlet or outlet, respectively.
An inlet whirl velocity can be added by deflecting the water before it enters the
turbine. This is done by guide vanes surrounding the runner
92
6. Fluid machines
The outlet whirl velocity is determined by the outlet radial velocity, the blade speed,
and the orientation of the blade. (‘blade angle’ refers to the angle between the blade
tangent and the blade motion, U)
Figure 10.Control volume for the Francis turbine: the space between two turbine blades
Using velocity vectors, we can locally fit a Cartesian co-ordinate system to a fluid
particle near a blade with the x-axis in the tangential direction and the y-axis in the
radial direction.
The fluid velocity is then
−=
r
t
V
VV and the blade velocity is
=
0
UU , but we also
know that the fluid is moving along the blade, which has an angle of β with the
tangential direction. Therefore, the fluid velocity relative to the blade is
W
−=
β
β
sin
cosW . The combination of the fluid moving along the blade while the blade
is moving, happens at a velocity of
−=+
β
β
sin
cos
W
WUUW , which describes the fluid
velocity. Therefore, we have two expressions for the fluid velocity, which obviously
β1
β2
β1
W
U
Vr
Vt
U
W β
To centre
Blade, moving at velocity U
Fluid flowing along blade, W
V
Vt
Vr
93
6. Fluid machines
have to be equal: VUW =
−=
−=+
r
t
V
V
W
WU
β
β
sin
cos. The second component gives
us βsin
rVW = , which we can use in the first component to get:
βββ cot
sincos
rrt VUVUV −=−=−
or, in terms of the rotation rate and flow rate, and using only the magnitude of the
velocities:
βω cotAQ
t RV −=
Ideal operation
It is usually the best to have the water leave the turbine with relatively little tangential
velocity. However, it is not possible to have zero outlet whirl for all flow rates.
The ideal power output is given by
( ) ( ) ( )[ ]( ) ( )[ ]
( ) ( )( ) 2
1122
21
22
2
212
21
22
2
1212
22
1112221122
112
11
22
11
22
12
Qbb
QRR
QQRR
RRQ
RRRRQVRVRQP
AR
AR
AQR
AQR
AQ
AQ
tt
−−−=
−−−=
−−−=
−−−=−=
ββπρωρω
ββρωρω
βωβωρω
βωβωρωρω
tantan
cotcot
cotcot
cotcot
Ideal outlet conditions
Once you have decided on the flow rate at which the fluid leaving the turbine should
have no whirl, the outlet blade can be designed by setting 222 20 βω cotA
Qt RV −==
or22
2 ARQω
β =tan
94
6. Fluid machines
Inlet blade angle
Having decided on the flow rate and chosen the appropriate outlet blade angle, we can
fix the inlet blade angle to give the ideal power at that flow rate.
( )11111 1βωρωρω cotA
Qt RRQVRQP −−=−= or Q
ARQR
PA 112
1
11
ωρω
β −=cot
Ideal inlet conditions
Given all the parameters in the power output equation, it is desirable to give the fluid
the right inlet flow conditions before it encounters the turbine blade. This is done by
the guide vanes. They give the fluid the correct tangential velocity component. By
convention, this angle, α, is often (but not always) measured with respect to the radial
direction (in contrast to the blade angle which often (but not always) is measured with
respect to the tangential direction!!!)
( )1
211
111111
1
1 21 βωπβωβωα cotcot
cottan −=−=
−==
QRb
QRA
Q
RA
VV A
Q
r
t
If α is defined with respect to the tangential, then tan α= Vr / Vt …
Real performance
With good design and control, the
real efficiency of a Francis turbine
can be in excess of 90% over a
relatively large range around the
best efficiency point, from about
50% to 110% of the rated power
of the turbine.
0
100
200
300
400
500
600
700
800
900
0 10 20 30 40 50 60 70 80 90
Q (m3/ s)
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Figure 12. Power input (blue diamonds), Power output (red squares), and efficiency (triangles) for a typical Francis turbine
95
6. Fluid machines
5.b Example
A small Francis turbine, turning at 11rev/s, has a runner diameter of 0.5m, a constant
blade height of 60mm, and an outlet diameter of 0.3m. At maximum efficiency it
delivers 200kW and has a specific speed of 0.09rev.
1. Calculate the optimum blade outlet angle.
2. Calculate the inlet angle, assuming that the best efficiency is 95%.
3. Calculate the angle of the guide vanes at the best-efficiency point
4. Calculate the guide angle and power output at 75% and 50% of the flow rate,
assuming an efficiency of 90% and 75%, respectively.
Solution
1. For the optimum blade angle, we want that the fluid leaves the turbine without any
whirl, ie Vt is zero: tan β2= Q/(ωR2A2)
We know
ω= 11 rev/s= 69rad/s
R2= 0.15m
A2= 2πbR2= 2π×0.06×0.15=0.0566m2
To solve this, we still need the flow rate.
From the specific speed, we can get the head at which the turbine delivers 200kW:
H= (N/KN)4/5 (P/ρ)2/5 / g= (11/0.09)4/5 (200)2/5 / 9.81= 39.7m
If the turbine has an efficiency of 95% at this point, we know that the (hydraulic) power
going in is 200kW/0.95= 210.5kW.
The hydraulic power, Ph= ρgHQ, gives us then the flow rate:
Q= Ph / (ρgH)= 210500/ (1000×9.81×39.7)= 0.541m3/s.
Finally, we can use the blade angle equation
tan β2= Q/(ωR2A2)= 0.541 / (69×0.15×0.0566)= 0.9228
which gives us β2= 42.7°.
96
6. Fluid machines
2. To calculate the inlet angle, we use Euler's equation which relates the difference of
the fluid's angular momentum at the inlet and outlet. Since we know from 1. that we
have designed the outlet so that the fluid has no angular momentum when it leaves the
turbine, the entire angular momentum at the inlet is available for power generation, and
we can use the equation for the inlet blade angle,
cot β1= A1P/ (ρωR1Q2) − ωR1A1/Q.
We know
ω= 69rad/s
R1= 0.25m
A1= 2πbR1= 2π×0.06×0.25=0.0942m2
P= 200kW
Q= 0.541m3/s,
and we get
cot β1= 0.0953×200000 / (1000×69×0.25×0.5412) − 69×0.25×0.0942/ 0.541= 0.7174
or tan β1= 1/ cot β1= 1.3939 which gives β1= 54.3°.
3. For the angle of the guide vanes, we need to work out the radial fluid velocity and
the tangential fluid velocity.
The radial velocity is given from the inlet area, A1, and the flow rate, Q:
Vr,1= Q/ A1= 0.541m3/s / 0.0942m2= 5.74m/s.
The tangential velocity can be found from re-arranging Euler’s equation, P= ωρQ R1Vt,1,
to Vt,1= P/ (ωρQ R1)= 200,000/ (69.1× 1000× 0.541× 0.25)= 21.4m/s.
If we take the angle with respect to the radial direction,
we get tan α= Vt,1/ Vr,1= 21.4/5.74= 3.728, or α= 75°.
If we take the angle with respect to the tangential direction (ie the same as the blade
angle), we get tan α= Vr,1/ Vt,1= 5.74/21.4= 0.268, or α= 15°.
97
6. Fluid machines
4. For the power output, we use the hydraulic power at the reduced flow rate and the
efficiency.
For the guide vane, we use the reduced radial velocity and the tangential velocity to get
the ideal (hydraulic) power out.
Q= 75% Q= 50%
Q 0.541*0.75= 0.406 0.541*0.5= 0.270 m3/s
efficienc
y
η 90% 75%
Pin =ρgHQ 9810*39.7*0.406= 158 9810*39.7*0.270= 105.3 kW
Pout = η Pin 0.90*158= 142 0.75*105.3= 78.9 kW
Vr,1 = Q/A 0.406/0.0942= 4.30 0.270/0.0942= 2.87 m/s
Vt,1 = P /
(ωρQ R1)
158,000/ (69.×1000×
0.406×0.25)
22.5 105,300/ (69.×1000×
0.270×0.25)
22.5 m/s
tan α = Vt,1/
Vr,1
22.5/ 4.3 5.232 22.5/2.87 7.848
α =atan α 79 83 °
6 Propellers and wind turbines
To get a good idea what one might get out at best, one can simplify the problem
greatly. Instead of looking at the detailed flow through the turbine, we can treat the
turbine itself as a black box and only look at its effect on the nearby fluid stream. One
simplification is to regard the black box as a very thin disk just enclosing the rotor, the
actuator disk. We can then define a control volume which encloses some fluid
upstream of the rotor, the actuator disk, and some downstream fluid. Because the only
object within that control volume is the actuator disk, we can use Bernoulli's equation
everywhere, except across the disk. But the disk is the only thing which can affect the
flow, so it is the only thing where a force can be exerted.
• Propellers generate thrust by accelerating the fluid through the rotor.
• Wind (and tidal stream) turbines generate power by converting kinetic energy
of the fluid flow into rotation of the rotor.
98
6. Fluid machines
• They have no casing to guide the fluid through the machine
• They cannot maintain a pressure drop between upstream and downstream of
the machine.
• Simple wind mill designs may have flat rotor blades which act like deflector
plates. The torque on the rotor is given by how much the air stream is
deflected by the blades. The real efficiency is much reduced because one cannot
have ideal flow conditions across a moving deflector blades if it is to do work
(use the velocity triangles: because the entry angle and the exit angle of the
blade are the same, we cannot do the work gradually along the blade.
• Rotor blades are generally shaped
like areofoils - they generate a lift
force perpendicular to the fluid flow
along the blade (remember that the
blade is moving at the same time).
Propellers want to generate forward
thrust while wind turbines want to
generate torque.
6.a Actuator disk theory
To get a good idea what one might get out at best, one can simplify the problem
greatly. Instead of looking at the detailed flow through the turbine, we can treat the
turbine itself as a black box and only look at its effect on the nearby fluid stream. One
simplification is to regard the black box as a very thin disk just enclosing the rotor, the
actuator disk. We can then define a control volume which encloses some fluid
upstream of the rotor, the actuator disk, and some downstream fluid. Because the
only object within that control volume is the actuator disk, we can use Bernoulli’s
equation everywhere, except across the disk. But the disk is the only thing which can
affect the flow, so it is the only thing where a force can be exerted.
L i f t
P r o p e l l e r
T h r u s t
W i n d t u r b i n e
T o r q u eL i f t
99
6. Fluid machines
Actuator disk Control Volume
The control volume wants to enclose
the actuator disk completely but not
look at the fluid flowing past it. We
also want to use streamlines as the
side boundaries so that we know that
there is no fluid leaving the control
volume through sides and that we can
use Bernoulli’s equation along the side.
The streamline which just touches the
edge of the disk is called the slipstream. Also, we need to extend the control volume to
far enough away from the disk so that we look at simple, unperturbed flow:
• Pressure far upstream and downstream is unaffected: 014 == pp
• Mass flow through control volume: 44332211 uAuAuAuAm ρρρρ ====
• The disk is very thin: AAA == 23 , where A is the swept area of the rotor.
• By continuity, 32 AuAu = : 32 uu =
• The force on the disk by the flow is the pressure difference across the disk:
( ) AppF 32 −=
• Bernoulli before disk: ( )22
212
12 uup −= ρ
• Bernoulli after disk: ( )22
242
13 uup −= ρ
• Inserting pressures gives force: ( )24
212
1 uuAF −= ρ
• Force on disk is also the net change in the momentum flow rate:
( )41 uumF −=
• with Momentum flow rate into C.V.: 1umJ in =
• Momentum flow rate out of C.V.: 4umJout =
u 1 u 2 u 3 u 4
A 1A 2
A 3
A 4
p 1 p 2 p 3 p 4
100
6. Fluid machines
• Equating both forces, using ( ) ( ) ( )414124
21 uuuuuu +−=− ,
and re-arranging gives ( )4121
2 uuu +=
• Power transmitted by disk: ( )( )4124
214
12 uuuuAFuP +−== ρ
• Using 1uU = and 1
4uu
=χ :
( )( ) ( )3234123
41
2 111 χχχρχχρ −−+=+−== AUAUFuP
6.b Wind turbine
The flow of kinetic energy by the wind
through an area A is
321 AUPair ρ=
The efficiency of a stationary actuator disk is
therefore:
( )3221 1 χχχη −−+==
airPP
6.c Propeller
If we consider a propeller, the power
conversion between actuator disk and fluid is
given by the force and the fluid velocity
through the disk, 2u , but the useful power is
that which is related to the actual speed of
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7 Wind turbine efficiency
Eff
icie
ncy η
χ =u 4 /U
1 2 3 4 5 6 7 8 9 10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1 Propeller efficiency
χ =u 4 /U
Eff
icie
ncy
η
101
6. Fluid machines
the aircraft, which is U, but the force is obviously still the same. So the useful output is
( )2321 1 χρ −== AUFUPout .
The efficiency is χη
+==
12
turbine
outP
P.
Example: Design of an aircraft propeller.
An aircraft is powered by to turboprop engines. Assume that the actual propellers
work at the ideal limit according to actuator disk theory where the wind speed behind
the propeller is 50% higher than the travelling speed. Each engine must provide a thrust
of 50kN to achieve a speed of 300mph at an altitude where the density of air is
0.8kg/m3.
1. Calculate the efficiency of the propellers
2. Calculate the power requirement for the engines.
3. Calculate the diameter of the propellers.
4. Calculate the pressure changes across the propeller.
Solution
U= 300mph= 185m/s.
χ= 1.5=3/2
1. %202.0512
12
25
====+
=χ
η
2. F= 50,000N
u2= ½(U+u4)= 1.25U= 231m/s.
P= Fu2= 11.55MW
3. ( )221 1 χρ −= AUF --> A= 2.92m2 and D= 1.93m.
4. mbarbarkPaAFp 17017.017 ====δ
102
6. Fluid machines
Design of a helicopter rotor
A helicopter design requires from the rotor that it can carry a mass of 5,000kg. The
diameter of the rotor is not to exceed 8m. Assume that actuator disk theory gives a
useful indication of the situation. Near ground, the air density is 1.2kg/m3.
1. Sketch a diagram of the helicopter rotor as an actuator disk, and outline the
slipstream boundary and the velocities at crucial points.
2. Determine the air velocity through the disk required to produce the force to
balance the weight of the helicopter.
3. Calculate the power requirement for the motor powering the rotor.
4. Estimate the torque on the rotor shaft if the rotor rotates at 300rpm
Solution
The swept area of the rotor is 24 27.50
2mA D == π
The upstream velocity, u1, is zero, and the velocity through the disk, u2, is therefore half
of the downstream velocity, u4.
The force generated by the change in velocity is ( ) 2214 2 AuuumF ρ=−= .
The force required to counteract the weight of 5,000kg is F= mg= 49kN.
The air velocity through the rotor is then smA
Fu /2022 ==
ρ.
The power carried by the air is
kWD
FA
FA
FFFuumP 9801222
33
22
21
4======
πρρρ
The torque is kNmPT 3152
000,980 =×
==πω
.
103
6. Fluid machines
6.d Some remarks on wind turbines
One of the design and operating conditions not used in the actuator disk theory is the
rotational speed of the turbine rotor. It is obvious, however, that wind of a given
speed, U0, can turn the rotor only at a certain speed. The parts of the rotor moving
fastest are the tips of the rotor blades. A rotor with a radius, R, and turning with an
angular velocity, ω, has blade tips moving at Utip= R ω.
In fact, two other constraints limit the useful speed of such a turbine or propeller:
1. If any part of the rotor is moving at speeds approaching the speed of sound, the
compressibility of air will affect the performance
2. If the rotor is moving too fast (or if the blades are too close together), a blade will
follow in the wake of the previous blade rather than receive 'fresh' air.
While the first is more relevant to aircraft propellers, the second is important for wind
turbines. As a result, smaller turbines may turn faster and/or have more blades, while
larger ones tend to have fewer blades and turn slower. Equally, if you want to reduce
the number of blades for a given rotor radius or diameter, you must increase the
rotation rate to achieve the same output.
Most current large wind turbines have three blades, and operate at a tip speed ratio of
about 8:1. This means that the tips move about 8 times faster than the mean wind.
Reading
This section is best revised with a standard textbook on basic Fluid Mechanics, such as
that by Massey.
104
7 Electricity generation and transmission
Introduction
The aim of this section is to provide the bare bones of the basic principles of electricity
generation and transmission.
7 Generation
7.a Faraday’s law
Faraday’s induction law relates the creation of an electric voltage in a wire moving
through a magnetic field or, conversely a magnetic field moving past a wire.
The voltage (electromotive force) induced in a coil is proportional to the rate of change
of the magnetic flux. Since this voltage is set up across a single coil, we can add up the
voltage if a wire makes several coils:
dtdNV φ=
where V is the voltage induced, N the number of windings, φ the magnetic flux.
Generation in a generator or alternator follows this principle where the motion of the
coils is rotational, and where the magnetic field is not generated by a permanent
magnet but by an electromagnet. So, a generator has rotor coils which provide the
magnetic flux, and stator coils around the rotating shaft which generate the voltage.
The voltage induced in each stator coil alternates in polarity as the magnetic poles of
the rotor passing through change polarity. If they are designed not too badly, the
induced voltage can be described by a sine function – giving the an AC voltage. If that
voltage is linked to an electric circuit, it results in an AC current.
105
7. Electricity generation and transmission
The power output from a generator coil is P= VI, where V is the voltage (in volt) and I
the current (in ampere). Each coil will also result in a sinusoidal power output. To
provide more control over the output, a generator consists of many stator coils which
are wired such that there are three sets of coils, all the coils in one set generating in
phase but the three sets at different phases. The average power at any time provided
by those three phases is almost constant:
Figure 11. Power output from three phases individually (blue thin solid line, red dash-dotted line, and green dotted line), and the average power output from all three phases together.
7.b Power transformers
The very same principle of power generation applies to transformation. If a coil wound
around an iron core carries a time-varying current, it sets up a time-varying magnetic
flux in the iron core proportional to the number of windings of that coil, according to
Faraday’s law. The magnetic flux in the core then induces a voltage in another coil
proportional to the number of windings of that coil around the core. We therefore
get an equation for the induced voltage, V2, in coil 2 as
11
22 V
NNV = ,
where V1 is the voltage in coil 1, and N1 and N2 are the number of windings in coils 1
and 2, respectively.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time
P
106
7. Electricity generation and transmission
If the transformer works without any losses, then the power, P= VI, will not change,
and consequently, the current will change in the inverse relation as the voltage.
A transformer with 10 windings in coil 1 and 100 windings in coil 2, receiving a current
of 1A at a voltage of 240V in coil 1, will turn this into a current of 0.1A at 2400V.
7.c Losses in transmission
The answer to a question you might have asked in the previous section: Why are we
interested in transforming power, and why do we have high voltage power lines?
Losses in an electric conductor are given by Ohms law, which states that the voltage
drop (or loss), VR, due to its resistance along a conductor is proportional to the
current:
RIVR =
where R is the Ohmic resistance of the conductor (in ohm).
This can be used to express the power dissipated, PR, due to resistance in that
conductor is
2RIIVP RR == .
We can see that the power loss is proportional to the square of the current;
decreasing the current by a factor 2 will decrease the losses by a factor 4. If we want
to minimise losses, then we need to minimise the current.
In the example of the previous section, let us assume we had lost 10% of the power
had we used the low voltage to transmit the power from the generator to the
consumer. With the 1:10 transformer, we have reduced the losses by a factor of
1:102= 1:100. As a result, we would only lose 0.1% of the power.
Reading
This section is best revised with a standard textbook, such as Hughes electrical and
electronic technology.
107
8 Outlook
Introduction
This section highlights some of the very many points not even touched upon in the previous
sections. If I were to discuss, or even provide answers, to some of the questions raised in this
chapter, I would fill many more hundreds of pages… I have therefore decided to pose
questions and leave it up to you to find out more using newspapers, professional journals, and
scientific articles address, and, above all, your critical mind.
1 Energy and sustainable development
Currently, over 80% of our energy supply come from fossil sources, coal, oil, gas, and
nuclear. We are consuming fossil fuels at a rate that they will certainly be virtually used
up within a few hundred years whereas it took some million years to create them.
Apart from nuclear, they all were originally produced as biomass. If we make a rough
estimate that one can get about 50t/ha/year as a theoretical maximum (Wheat
productivity in the UK is a little under 50t/ha/year), using a typical net calorific value of
10GJ/t, we can estimate proportion of annual biomass production we are currently
using up. The total land surface is about 13 billion hectares = 1.3×1014 m2 of which
about 1/3 is arable, say 4×109 ha. This makes a global maximum energy supply of
TPES= 10×109J/t×50t/ha/yr×4×109 = 8×1021J. A more realistic value will be closer to
1% than 10% of that figure. Our current annual energy consumption is around 5×1020 J.
Are our current consumption patters compatible with the principle of sustainable
development?
Do we have the moral duty to our offspring and the planet to
a) say we are working towards sustainable development
b) make some effort towards sustainable development
108
8. Outlook
c) change our fundamental way of life so that we can achieve sustainable
development?
2 Energy saving and energy efficiency
Energy saving is if we use less energy, energy efficiency is if we get out more from the
same energy used. While they seem to be saying the same thing in terms of numbers,
there is a fundamental difference, linked to our human nature: if we achieve energy
efficiency, we feel justified to increase what we get out of the energy – so that we may
get more at the end but do not use less energy.
True energy saving is if we actually use less energy, regardless of what we get for it.
Should we priorities economic development using highest energy efficiency technology,
or should we prioritise energy reduction? The latter is obviously completely anathema
to the modern philosophy of achieving economic development and growth. This can
always be justified by saying that economic development will result in wealth, health,
and new technologies.
3 Technological, behavioural, and attitudinal solutions
‘Technological fixes’ as opposed to behavioural and attitudinal fixes are currently the
only seriously acceptable solutions to tackle the energy challenge, as they both seem to
be good for the environment but also bolster technological and economic
development. Furthermore, it is the only way of remaining electable as the population
does not like being told they cannot use their car, fridge, cooker, or computer
whenever they like.
However, technological fixes tend to have their own problems; look at recent
technological fixes, e.g. drugs to address one illness usually come with side effects
which may result in decades of scarred people (example thalidomide); nuclear power is
a wonderful technological fix to supply vast quantities of electricity without vast
amounts of carbon dioxide emissions; just a pity that no real solution to the waste
problem has been found.
109
8. Outlook
4 Is Renewable Energy always good?
Obviously, one form of technological fix proposed to address both the finite nature of
fossil fuels and their environmental impact, is the development of renewable energy
technologies.
On the face, they are a good way of reducing emissions and reducing reliance on fossil
fuels. However, they also have some problems attached. One of the main problems is
that their energy density tends to be much lower than that of the fuels. One needs so
many installations and requires so much land that they do more damage the landscape
than a compact power station (be it in the form of nuclear, or coal/gas fired power
station which captures the CO2 for secure storage; only problem is how to store it
practically and securely at the scale required?)
Wind energy, for example, is controversial among nature lovers: they argue that wind
turbines kill birds – but then, how many birds are going to be killed by climate change?
It is argued that wind cannot really supply enough energy: once there is more than
10% or 20% of the electricity supplied by wind, the fluctuating nature of the wind
resource makes the entire electricity network unstable; required backup generation in
the form of fossil generators must be on stand-by and may cause as much emissions as
if they were the energy supplier without wind power.
Biomass has another problem associated with it: sustainable production of it. While
the use of waste and residue seems reasonable, what about biomass crop. Since the
UK’s potential for biofuel (ethanol as a petrol substitute and biodiesel) currently seems
to be able to cover may be up to 20% of the current fuel consumption, developing
countries are now proposing to chop down their tropical rain forest to produce biofuel
for the West while destroying habitat for endangered species as well as tinkering with
an important part of the climate system.
5 Energy and climate
There is no doubt that the vast majority of the increased carbon dioxide in the
atmosphere is a direct result of our energy consumption in the form of burning the
carbon-based fossil fuels.
110
8. Outlook
Even assuming we could find more and more oil and gas if we just look hard enough
and invest enough (as the EIA implicitly assumed in its 2005 World Energy Outlook), we
are still facing the problem that we are rapidly heading to a climate situation with CO2
levels never seen before.
Climate change is not just an academic debate. Results of measured and predicted
climate change at present include: increased droughts in dry areas, increased likelihood
of flooding and other areas, increased occurrence of sever storms, general rise of sea
level. All this is expected to lead to a substantial and rapid shift in habitats for our
fauna and in growing patterns and yield of crops. This in turn will lead to global shift
and uncertainty of national economies and security.
A small UK example is the rise in house insurance premiums as buildings are more
likely to suffer from flooding of nearby rivers.
6 Costing energy
Currently, the price of energy contains the investment required to extract it from the
ground, process it ready for end-use, transport it to the consumer and support this
entire chain. In addition, it is more or less heavily taxed to provide the government
with income, to invest in supporting this infrastructure, e.g. by building roads, or invest
in the nation’s welfare, e.g. health, education, defence, or to invest in development of
technological solutions to provide energy security. With the carbon tax in the UK, an
attempt has been made to include an environmental cost.
How do we define the total cost of energy?
How do we calculate the total cost of energy?
Task
Read the newspapers critically. Follow up controversial stories by doing background
research. Read the material you find in your research critically – there is no such thing
as a truly objective or unbiased account. Try to be objective by looking at both sides of
the argument.
111
Appendix A: Energy units and conversions
A.1 Basic definitions 1 joule = 1J = 1Nm: The energy expended to move against a force of 1N by a distance
of 1m.
1 kilowatt-hour = 1kWh: At a constant rate of energy consumption of 1kW, the energy consumed during the time interval of 1 hour.
1 calorie = 1cal= 4.19J: the heat energy needed to warm 1g of water by 1°C,.(1kcal= 1Cal= 1000cal: the heat to warm 1kg of water by 1K)
1 British thermal unit = 1BTU= 1055J: The heat energy needed to warm one pound of water by one degree Fahrenheit. (1therm= 100,000BTU= 105 BTU)
1 quad = 1quadrillion British thermal units = 1.055EJ
1 tonne of oil equivalent: the heat released from burning 1t of oil. The exact amount of heat released depends on the quality of the oil. A generally accepted ‘standard’ tonne of oil gives 1toe=42GJ
A.2 Basic unitsIn the following tables: To convert from the unit in the left column to the unit in the top row, multiply by the factor in the intersecting cell.
Example: The electricity bill states that you have used 869kWh during the preceding 90 days. How much is that in MJ?
The entry in the cell with kWh to the left and MJ to the top is 3.6.
⇒ 869kWh = 3.6 × 869 = 3128MJ.
J cal BTU
J 1 0.239 9.48×10–4
cal 4.19 1 3.97×10–3
BTU 1055 252 1
112
Appendix A: Energy units and conversions
A.3 Small unitsJ kJ MJ kWh
J 1
kJ 1
MJ 1 0.28
kWh 3.6 1
A.4 Medium unitsJ kWh GJ toe
J 1
kWh 1
GJ 1
toe 1
A.5 Large unitsJ PJ TWh Mtoe
J 1
PJ 1
TWh 1
Mtoe 1
A.6 Orders of magnitudeSymbol prefix Power of ten Name Examples:
E exa- 1018 quintillion 1EJ
P peta- 1015 quadrillion 1PJ
T tera- 1012 trillion 1TWh
G giga- 109 billion 1GW
M mega- 106 million 1MJ
k kilo- 103 thousand 1kWh
113
Appendix B: Example exam
Q.1 [10 points]
A large coal-fired power station has a generating capacity of 600MW. Its turbines
operate with an efficiency of 35%.
It has operated during one month at a capacity factor of 0.8 using coal with a heat
content of 29 GJ/t, a carbon content of 88% of carbon and an ash content of 7%.
Calculate
a) The overall electricity generated
b) The amount of fuel consumed
c) The amount of ash produced
d) The amount of carbon dioxide produced
Q.2 [20 points]
A room has an outside wall which consists of a brick wall of area 20m2, thickness
200mm, and a thermal conductivity 0.60Wm –1 K –1. It also has windows of area
8m2, with panes of thickness 3mm and thermal conductivity 0.78W m – 1 K – 1. The
heat transfer coefficient from the room air to the wall is 20W m – 2 K – 1. The heat
transfer coefficient between the wall and the air is 150 W m – 2 K – 1.
a) Calculate the heating load to maintain the room at 20°C if the outside
temperature is 5°C.
b) Calculate the heat loss for a room temperature of 18°C.
c) Calculate heat loss if an insulating layer of thickness 3mm and thermal
conductivity 0.026Wm –1 K –1 is added to the walls.
d) Calculate the heat loss if a second pane of glass of 3mm thickness is added
to the windows, leaving an air gap of 10mm. Assume the only form of heat
transfer through the air gap is conduction. Thermal conductivity of air is
0.026Wm –1 K –1.
114
Appendix A: Energy units and conversions
e) Calculate the temperature at each glass surface for the double-paned
window.
f) Estimate the net radiative heat transfer between the two glass panes, if both
surfaces have an emissivity of 0.8, and comment on the validity of ignoring
radiation in part (d)
Stefan-Boltzmann constant: σ= 5.67×10–8 Wm –2 K –4.
g) Comment on the respective savings made in parts (b), (c), and (d).
Q.3 [20 points]
An ideal gas turbine operates by burning natural gas (72% methane and 19% ethane
by volume) such that it combusts with 27% of the oxygen present in the air (assume
this is 5.4% of the air). The air is drawn in from the environment at atmospheric
pressure and a temperature of 296K. The combustion chamber can withhold
temperatures of up to 1827 K.
Assume the following constant properties for the air and air-gas mixtures
• cp= 0.97 kJ/kg and
• heat content of methane= 56 MJ/kg,
• heat content of ethane= 53 MJ/kg.
a) Calculate the pressure as the gas enters the turbine
b) Calculate the temperature of the exhaust gas
c) Calculate the efficiency of the entire process
d) Calculate the heat content of the air-fuel mix
e) Calculate the energy output per cycle
f) Calculate the mass flow rate of the air-fuel mix required to generate a net
power output of 100MW.
g) Calculate the rate of consumption of natural gas to generate a net power
output of 100MW.
115
Appendix A: Energy units and conversions
Q.4 [20 points]
Describe the situation of energy production and consumption for a country of your
choice. Discuss this in view of its role in the economic development of that country
over recent history. Mention steps this country has undertaken or planned to meet
forthcoming challenges such as oil prices, security of supply, or compliance with
international treaties, such as the Kyoto Protocol.
Q.5 [30 points]
Discuss a topical issue for the field of Energy. Briefly describe the issue and explain
what is topical or controversial about it. Discuss some alternative proposals or
solution to address this issue, making reference to literature you have read.
116
2. Energy resources
The coal with the highest calorific value contains about 90% of carbon and has a
calorific value of 34MJ/kg (or 34GJ/tonne); taking this into account, we have about
34/0.9= 38MJ/(kg of C), this then can be represented as
Mass and heat balance: 12kg + 32kg → 44kg + 38MJ/kg×12kg
1 + 2.67 → 3.67 + 38MJ/KG
We can see that a natural result of burning coal is the release of carbon dioxide to the
amount of about 0.1kg CO2 per MJ converted.
Considering that the efficiency of coal-fired power stations is about 33%, we produce
about 0.3kg CO2 per MJ electricity generated. Converting this to kWh (ie multiplying
by 3.6), gives us about 1kg of CO2 emitted per kWh generated – ie leaving your 100W
light bulb on for 10 hours adds 1kg of carbon dioxide to the air.
A large but typical coal-fired powerstation of generation capacity of 600MW therefore
emits 14400 tonnes of CO2 each day.
(600,000×24 kWh per day= 600,000×24 kg CO2= 600×24 tonnes of CO2)
6.c Hydrocarbons (Oil and Gas)
Oil and gas are very much easier and cleaner to burn and to use than coal. While
these two substances appear to be very different and are used for very different
applications, they are all part of the hydrocarbons and usually found together in
reservoirs. Hydrocarbons are a family of organic molecules which consist of chains of
carbon atoms to which hydrogen atoms are attached.
The simplest hydrocarbon is methane, CH4. Longer examples of simple chains can be
written as H3C–CH2–…–CH2–CH3 , or shortened to CnH2n+2.
The combustion and mass and heat balance of methane with a calorific value of 55MJ/kg
can be written as
CH4 + 2O2 → CO2 + 2H2O + heat
12+1×4 + 2×16×2 → 12+16×2 + 2×(1×2+16) + heat
16 + 64 → 44 + 36 + 55MJ/kg × 16
29
2. Energy resources
Because methane has a much higher heat content or calorific value than coal, we get
55MJ/kg heat and only 0.05kg per MJ converted.
Considering that gas-fired power stations have a
typical efficiency of about 50%, we can reduce the
carbon dioxide emissions from 1kg/kWh to about
0.3kg/kWh, reduction of 60-70%! Not surprisingly,
many nations have heavily invested in gas-fired power
stations as opposed to coal-fired power stations – and
this is only one advantage. Other advantages over
coal are that there is virtually no ash produced, and
that gas turbines can react to changing demand much
quicker.
The oil-based products, such as petrol and diesel are still the dominant fuel for
transport. The petrol can be reasonably well described by octane, C8H18, and diesel by
cetane, C16H34, both with a calorific value of about 48MJ/kg or about 35MJ per litre
(with a density of 0.7kg per litre for octane and 0.75kg per litre for cetane). The
chemical structure of methane and octane are shown in Figure 5.
We can extend the combustion equation for methane to the generic chain
CnH2n+2 + (1.5n+0.5) O2 → n CO2 + (n+1) H2O + heat
14n+2 + (1.5n+0.5)×32 → 44n + (n+1)×36 + 48MJ/kg × (14n+2)
This gives us 22n/(7n+1) kg of carbon dioxide per kg of fuel and 22n/([7n+1]×48) of
carbon dioxide per MJ of energy converted. If n is sufficiently large, the ‘+2’ in the fuel
can be ignored, and we approximate these numbers to
• 3.01kg CO2 per kg petrol and 3.12kg CO2 per kg diesel
• 2.5kg CO2 per litre petrol and 2.7kg CO2 per litre diesel
• ~ 0.064 kg CO2 per MJ of heat released
The actual energy released by the different hydrocarbons is listed in Table 4.
Figure 5. A methane and an octane molecule. The dark circles represent carbon atoms and the light circles hydrogen atoms1
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