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Vysoká škola báňská – Technická univerzita Ostrava
ENERGY MANAGEMENT
Didactic Text
René Pyszko
Ostrava 2014
Description: ENERGY MANAGEMENT
Author:
René Pyszko
Edition:
first, 2014
Pages:
54
Academic materials for the Metallurgy engineering study programme at the Faculty of
Metallurgy and Materials Engineering.
Proofreading: none.
Project designation:
Operation Programme of Education towards Competitive Strength
Description: ModIn - Modular innovation of bachelor and subsequent master programmes at
the Faculty of Metallurgy and Materials Engineering of VŠB - TU Ostrava
Ref. No.: CZ.1.07/2.2.00/28.0304
Realisation: VŠB – Technical University of Ostrava
© René Pyszko
© VŠB – Technical University of Ostrava
Energy Management
STUDY SUPPORT OVERVIEW
No.
1
2
3
4
5
Chapter summary
Study time
Concepts of energy, energetics. Types of energies, rest
mass energy, sources of energy. Energy consumption in
metallurgy.
Energy conversion. Basic concepts and laws of
thermodynamics, reversible changes, irreversible
changes.
Heat cycles. Reversible cycle. Carnot cycle. Irreversible
cycle.
Applicability of thermal energy. Entropy. Exergy, anergy.
Balancing energy facilities. Energy and exergy balance.
3h
8h
4h
6h
4h
Note: Completion of this study support without the use of compulsory literature is not sufficient
for the successful completion of the course.
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1.
CONCEPT OF ENERGY, SOURCES, CONSUMPTION AND
ENERGY CONVERSION
Study time: 3 hours
Objective: After reading this section, you will be able to:
 Define the basic concepts of energy and energetics.
 Explain the forms of energy, types of energy, energy resources
 Define the concept of secondary energy.
 Understand the forecast of trends in energy consumption
 Define the energy mix.
 Describe energy consumption in metallurgical processes.
PRESENTATION
Concept of energy, its forms, sources and consumption
Concepts of "energy", "energetics"
The word energy comes from the Greek ergon = act, energeia = activity.
Energy is the ability to perform mechanical work A = F·s (N m = J), where F is force and s is
distance), and it is measured in joules (J), the same as work.
Energy is associated with movement (or a possibility of movement).
Energy can take various forms and changes from one form to another.
Energy is the most important physical quantity.
Energetics is a science of energy forms and transformations of these forms.
Forms of energy
Energy can be divided into two basic forms - mechanical energy and internal energy.
Types of mechanical energy

Potential. In the gravitational field of the Earth Ep = G·h = m·g·h, where G is weight
(N), m is mass (kg) and h is height (m). Potential energy can exist in any field in
which there is acceleration, i.e. not only gravitational field.

Kinetic Classical Newtonian mechanics uses the relationship Ek = m·u2/2.
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Types of internal energy of substances

Thermal energy - is represented by the kinetic energy and potential energy of
molecules, represented by oscillating motion of atoms in crystals around the
equilibrium position or chaotic motion of atoms or molecules of liquids and gases (in
gases velocities at around 102 m·s-1 are involved).

Potential energy of molecules is determined by forces of attraction (long distances) or
repulsive forces (short distances).

Chemical energy is given by chemical coupling forces of electrical nature between the
atoms.

Nuclear energy is bound in the particles of matter and is associated with the rest mass
(rest mass energy E = m0c2)

Electrical (electromagnetic) energy caused by the movement in electrostatic and
magnetic fields. Radiation energy (electromagnetic waves) is also an electromagnetic
energy; depending on the wavelength, it ranges from radio waves to thermal radiation,
visible light, and finally X-ray and gamma radiation.
Rest mass energy
In his special theory of relativity, Albert Einstein proved the interdependence of
momentum and energy. Energy is indeed closely associated with momentum h = m u,
where m is the mass (kg) and (m s-1) is velocity, but momentum is not a measure of
energy. Momentum is a vector, while energy is a scalar. A. Einstein unified the laws of
conservation of mass, momentum and energy.
Kinetic energy is generally obtained by a force acting over a distance
dEk  F  ds
and after substituting from the 2nd Newton's law where force is equal to the change in
momentum in a time period, we get
F
dEk 
d(m  u )
dt
d(m  u )
ds
 ds , u 
dt
dt
where t (s) is the time and s (m) is the distance.
When substituting for the distance differential ds and expanding the differential,
dEk  d(m  u)  u
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dEk  (dm  u  m  du)  u
we will get the following relationship, which will be needed later
dEk  dm  u 2  m  u  du
(*)
From the special theory of relativity, we will use Einstein's mass transformation equation,
m
m0
1
u2
c2
where m0 is the rest mass of the body, m is the mass at a non-zero speed, c = 3 x 108 ms-1 is
the speed of light in vacuum.
Squaring this equation
m2  c 2  m2  u 2  m0  c 2
2
Differentiating the above equation (remember that m0 and c are constants)
2m  c2  dm  2m  u 2  dm  2m2  u  du
c2  dm  u 2  dm  m  u  du
The right hand side equals to the expression (*) for dEk, so
d Ek = c2·dm
and after integrating within the range from m0 to m
m
Ek  c   dm  c 2  (m  m0 )
2
m0
Ek = m·c2 – m0·c2
Ek = E – E0
We got a definite relationship, which shows that the total energy of the body E is the sum
of internal (rest) E0 energy and kinetic energy Ek
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E = m·c2 = E0 + Ek
Use of rest mass energy




chemical reaction (combustion) - about 10-10 of rest mass energy is utilised
nuclear fission reaction - 10-3 of the rest energy (about one per mille)
nuclear fusion - 10-2 of the rest energy (about one percent)
annihilation of particles - 100% conversion of mass into energy.
History of energy consumption
First in the history of production and transport, the energy of muscles was used (people,
animals), followed later by flowing water (water wheel, water turbine), wind (windmills,
sailboats), water steam (steam engine, steam turbine), chemical energy (combustion engine,
jet and rocket engine), electricity (dynamo, electric motor), nuclear energy (reactor).
Historical development of energy consumption per capita per year
The growth in energy consumption per capita per year is shown in Table 1.1.
Table 1.1 Historical development of energy consumption per capita per year
2 million years ago (homo habilis) human's own strength gained
from food
3 GJ
500 thousand years ago (homo
erectus)
use of fire
6 GJ
10 thousand years ago (farmers)
draught animals
20-30 GJ
turn of 18th and 19th centuries
(industrial revolution)
steam engine
100 GJ
today
coal, oil, nuclear power
20 GJ (Africa) to
350 GJ (USA)
Technical progress brings with it not only success and life improvements, but also
challenges. Former President of the Czech Academy of Sciences prof. Rudolf Zahradnik
spoke of the "law of conservation of joy and sorrow" which states that nothing in the world is
for free; in each discovery the humans gained and lost something at the same time
(environmental pollution, extinction of animal species, living in an artificial or virtual world,
etc.). It is a tax on progress.
Types of energy from the consumer's perspective

primary energy - crude form of energy (coal deposits, oil fields, uranium ore, down
slopes of watercourses, etc.). It also includes electricity generated by hydro,
geothermal and nuclear power plants and is measured in tonnes of black coal, called
tonne of coal equivalent (TCE), or in Anglo-Saxon countries in tonnes of oil
equivalent (toe) that would be consumed for its generation in a plant with an average
efficiency of 33%,
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
refined energy - primary energy (except electrical) must be refined for direct
consumption (coking plants, refineries, thermal power plants),

consumer energy - refined energy delivered to consumers. Transport of energy results
in losses by friction, local losses, and losses of pressure, volume, temperature,

effective energy - consumer energy transformed at consumers' facilities to heat,
motive power, etc.

secondary energy - is produced as waste in power plants, industry, agriculture (gas,
waste heat, pressure energy). It is counted in the balances when used in a plant or unit
other than the one in which it is produced.
The overall efficiency of an energy source is the product of efficiencies of extraction,
process of refining, storage, transport and the efficiency of the final equipment (a furnace, car,
turbine set, etc.)
c = extraction · refin · storage · transport · machine
Sources of energy
Source of energy is "stored work", which is released during the conversion of the carrier
into a different form.
Types of energy resources
A. Primary resources

non-renewable fuels (fossil) - solid, liquid and gaseous fuels containing carbon
(accumulated solar energy),

renewable fuels (wood, peat, biomass),

wind results from uneven warming of the atmosphere, leading to different densities and
therefore different pressures. Utilising wind is challenging - low concentration of energy
and power variations,

 vody  H  g  (W) , where p is the
watercourses: water turbine power P  Vvody  p   m
pressure gradient, Vvody is the volumetric flow rate (m3·s-1), m vody is the mass flow rate
(kg·s-1), H is the height of water level (m), g is the gravity of Earth, the efficiency is  
0.75,

controlled nuclear reactions: classic reactors - decay of U-235, fast reactors - use fast
neutrons to convert U-238 into Pt-239). The controlled nuclear fusion is not yet mastered,

tidal energy results from the pull of the Sun and Moon and the rotation of the Earth - Moon
system; differences in the levels of different seas and oceans are between 2 and 16 m,

energy of sea waves (about 35 times greater concentration of energy than the tidal power,
vibration of the water level is generated by the action of wind and celestial bodies),

solar energy: The solar constant is the radiant flux across the spectrum of electromagnetic
radiation in a place of the average distance of the Earth from the Sun, its value is
1,367·W·m-2. Due to the rotation of the Earth (alternation of day and night), the reflection
from the upper layers of the atmosphere and absorption in the atmosphere, the average
Energy Management
radiant flux falling onto a level surface in the Czech Republic is from 107 to 127 W·m-2.
When using energy transformation in photocells, the expected efficiency should be from
10 to 15% (theoretical efficiency limit of silicon cells is 31%, the best laboratory efficiency
achieved was 26 %). Figure 1.1 shows the average annual total solar radiation in the Czech
Republic. For example, the energy of 1,000 kWh that falls on the horizontal surface of
1 m2 in one year corresponds to the average power of 108 W.
Figure 1.1 Average annual total solar radiation in the Czech Republic.

geothermal energy The use of natural or artificial hot springs to obtain water steam.
The heat of the magma is formed by decay of uranium and thorium in the earth's core.
About 50 % of the primary energy is used, the rest is lost due to the low efficiency of
power equipment and machinery, transport and storage.
Approximate efficiency values of some power generating equipment:
Conventional thermal power plant 35 %, supercritical steam power plant 42 %, steam engine
15 %, industrial heating furnaces 30-50 %, fuel boiler 85 %.
Globally: The total amount of primary energy consumed by humans in 2000 was
10.4 Gtoe, which represents the average power of about 14 TW (7,000 Temelín nuclear power
plants). This energy represents less than 9 % of the solar energy that nature uses for
photosynthesis [14].
B. Secondary energy resources

waste fuel (converter gas and blast furnace gas),

waste heat (flue gas, enthalpy of scales, enthalpy of products, etc.)

waste pressure (blast furnace gas).
(Note: A resource is considered as secondary only if it is used in a process other than the
one where it is produced)
C. derived energy resources
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synthetic fuels (made from primary resources, such as fuel briquettes)
Prognosis energy consumption development
Depletion of resources is expected:

oil (with the current extraction rate by 2050),

natural gas (by 2175).
Other resources which will be used:

coal (at the current rate, will last 120 years),

tar sands and shale (mining by injection of hot water steam, by 2150).
In addition, electricity from the following primary resources will be used:

nuclear fission,

watercourses,

wind,

tide,

geothermal energy,

solar energy,

nuclear fusion (if the technology is developed and technically mastered),

new technologies, yet unknown.
Energy consumption in the Czech Republic:
black coal - when mining at 13 to 15106 t/year and with deposits of 2350106 t, it should last
until 2150. However, in recent years mining is in decline and the figure may change under
economic pressure and based on political decisions,
brown coal - if mining at 60106 t/year and with deposits of 3800106 t, it should last by 2060,
but in the prognosis for brown coal mining, the same applies as for black coal,
oil - domestic extraction is 0.1106 t/year, consumption of 12106 t/year, the difference is
covered by imports,
natural gas - extraction of 0.8109 m3/year, consumption 10109 m3/year, the difference is
covered by imports,
hydropower plants - 5 % of the total electricity generation,
nuclear energy - Dukovany 4 x 440 MW, Temelín 2 x 1,000 MW.
Realistic production in nuclear power plants (NPP) in the Czech Republic is 33 % of the
total electricity produced (the installed capacity of NPP is 21 %), the share of energy
produced in NPP in other countries: France 78 %, Lithuania 72 %, U.S. 19 %, EU 30 % ,
globally 16 %.
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History and forecast of mining and consumption of coal and the structure of electricity
production in the Czech Republic
As a general guidance, this section summarises in a graphical form some selected data on
the history and prognosis of black and brown coal. Besides, it shows the composition of the
energy mix in the Czech Republic in 2012, a map of the global installed capacity of nuclear
power plants, the structure and development of the primary energy resources consumption in
the Czech Republic, and finally the consumption of electricity in the Czech Republic.
source: http://www.csve.cz/
Figure 1.2 A chart of the energy mix in the Czech Republic (installed power)
http://www.tzb-info.cz
Figure 1.3 Global installed power of nuclear power plants
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Fuel consumption and energy in metallurgy
Despite the decline in steel production in recent years, metallurgical production is still the
most important industry in North Moravia. Therefore, it is important to understand the energy
demands and the types of energy sources used in the steel industry. Figures 1.4 and 1.5 show
the history of steel production, globally and in CR.
Figure 1.4 History of global steel production
Source: Jan Lasota, TŽ, a.s.
Figure 1.5 History of steel production in the CR
Energy Management
The cost of fuel and energy in the steelworks of the Czech Republic accounts for 20 % of
the cost of steel production. The share of energy consumption in steelworks in the overall
energy in the world is about 11 %, this figure the CR is 13-15%.
Energy consumption per tonne of steel decreased over the last few decades by 40 %.
Energy consumption for the production of one tonne of steel is about 20 to 25 GJ/t of steel.
Energy consumption has decreased due to the introduction of oxygen converters,
continuous casting and economic pressure associated with the rise in global energy prices. In
1965, energy consumption was about 38 GJ/t of steel.
Coal consumption per 1 tonne of steel produced from iron ore is about 630 kg. The share
of fuel and energy consumption per 1 ton of steel is shown in Table 1.4. Table 1.5 shows the
specific consumption of fuel and energy in the basic metallurgical plants.
Tab. 1.4 The share of energy and fuels consumption (expressed as specific energy
consumption) per 1 tonne of steel
coke
40 - 50 %
other solid fuels
10 - 20 %
liquid fuels
8 - 15 %
gaseous fuels
12 - 18 %
electric power (without production of
ferroalloys)
8 - 20 %
Table 1.5 Specific consumption of fuel and energy in the basic metallurgical plants.
Technological process
E (GJ / t of steel)
Share (%)
metallurgical production (agglomeration, blast
furnaces)
15
60
steel works
3
12
forming (rolling, forging, cold processing)
3
12
other consumption
4
16
Total
25
100
Reducing energy consumption as a result of modernisation of technological processes:

oxygen converters, secondary metallurgy, vacuum technologies, etc.

continuous casting of steel increases the yield by 11-13 %, which represents a saving of
1.2 to 1.8 GJ/t of steel. In addition to saving energy it eliminates deep furnaces and
blooming (savings of energy for heating ingots).
Reducing energy consumption due to the use of secondary energy resources:
Heat loss through the flue gas, cooling water, solid products and ambient loss amounts to
around 10-15 GJ/t of steel. Full use of the lost energy is not possible (low potential, irregular
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occurrence, etc.). If 20 % of losses were utilised, the savings would amount to at least 2 GJ/t
of steel.
Some technologies utilising secondary energy sources:

flue gas enthalpy in the waste heat boilers (steam production)

the use of pressure and temperature of flue gas in the gas expansion turbines,

steam production during evaporative cooling,

utilisation of the enthalpy of solid products (metal, coke, slag) to generate steam.
Energy resources in metallurgical plants
External resources (supplied to metallurgical facilities)






coal
oils
petrol
natural gas
technical gases
electric power
Internal resources (produced in the metallurgical facility)
- fuels (coal gas and formerly also producer gas),
- secondary energy resources
 waste fuels (blast furnace, converter, ferroalloy gases)
 waste heat (of flue gas, cooling water, hot metal, coke, slag)
 pressure energy (pressure of blast furnace gas).
Summary of chapter concepts






energy, energetics,
internal energy,
rest energy,
energy - primary, refined, consumer, effective, secondary
sources of energy
energy mix.
Questions on the topic of the chapter




List the types of energy from the extraction to the delivery to consumers.
What does the abbreviations toe and toc stand for?
Which energy resources do you know?
What affects the efficiency of converting solar energy into electricity?
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 Do you know any more methods of using solar energy that were not mentioned
here?
 What is the energy mix?
 Which phase of steel production has the highest energy demands?
Energy Management
2.
ENERGY TRANSFORMATION PROCESS
Study time: 8 hours
Objective: After reading this section, you will be able to:
 Define the first theorem of thermodynamics.
 Compare the volume (absolute) and the technical work
 Understand the implications of the second theorem of
thermodynamics.
 Explain the concept of enthalpy and calculate the change in
enthalpy.
 Explain the concept of entropy and calculate the change in
entropy.
 Compare the diagrams p – v and T – s, and know how to use
them.
 Explain the concepts of instantaneous and medium specific
heat capacity.
PRESENTATION
In terms of physics, all forms of energy are equivalent and have the same unit (J).
The various forms of energy can be converted to each other without limits (disregarding
losses) with one exception - the conversion of thermal energy to other forms.
A classic example of the mechanical energy conversion is a pendulum (cyclic conversion
of the potential to kinetic energy and vice versa).
Thermal energy is only convertible in a limited way.
Thermal energy which is in thermal equilibrium with its surroundings is completely
inconvertible.
First theorem of thermodynamics
The basic law for the conversion of energy forms is the Law of Conservation of Energy
(discovered by Mayer, Joule and Hemholtz). This law makes it impossible to construct the
first kind of perpetual motion machine in which e.g. the potential energy is cyclically
converted into the kinetic energy without losses. Such a machine was to move for an
indefinite period of time, or, according to some authors, it was to do work.
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An example is a wheel with tubes filled with mercury made by the Indian scholar Bhaskar
in the 12th century or a rotating ring shown in Figure 2.1. Some structures were very
sophisticated and kept moving for as much as several weeks, but eventually they all stopped
moving.
Source: [14]
Figure 2.1. Examples of the first kind of perpetual motion machine
During the transfer of thermal energy between the substance and the environment, a role is
played by the heat passed and the mechanical work done.
In thermo mechanics, the law of the conservation of energy is expressed in the first
theorem of thermodynamics, which is of quantitative character (the amount of energy is
involved, not the direction of conversion).
First formulation of the first theorem of thermodynamics
dq = du + da
dq = cv·dt + p·dv
Heat brought from the surroundings dq is utilised for a change in internal energy du and
external mechanical work da (also called volume, absolute, or one-off work) The quantity p is
the pressure (Pa), t (°C) is the temperature, v is the specific volume (m3·kg-1) and cv is the
specific heat capacity at constant volume (J·kg-1·K-1).
Second formulation of the first theorem of thermodynamics (appropriate for closed circulation
systems)
dq = di + dat
dq = cp·dt - v·dp
The supplied heat dq is changed into the enthalpy change di and technical work dat. The
quantity cp is the specific heat capacity at constant pressure (J·kg-1·K-1).
Energy Management
Properties of quantities in the equation of the first theorem of thermodynamics:
Internal energy
 is the sum of the kinetic energy and the potential energy of molecules,
 is a function of the state, independent of the course (integration path) of the change,
 is equal to the amount of heat applied at constant volume du = cv·dt
Enthalpy
 is a sum of the internal energy and pressure energy of a gas
i = u + p·v
di = du + d(p·v) = du + v·dp + p·dv
Note:
for p = const. is dp = 0, and therefore
di = du + p·dv
Comparing with the first formulation, it results in



di = dqp = cp·dt
di contains a complete differential d(p·v), in contrast with transferred heat dq which
contains a partial differential p·dv
is a function of the status, is independent of the way (integration path) of the change,
is equal to the amount of heat applied at constant pressure di = cp·dt
Volume work (or absolute, one-off work)
 is a work done with one change,
 is dependent on the manner (integration path) of the change,
 da = p·dv
Technical work
 is of importance for a closed cycle of a machine (e.g. piston machines)
 is equal to the sum of volume (absolute, one-off) work components for the whole cycle,
 is dependent on the manner (integration path) of the change during a cycle,
 in addition to the absolute work during the working change (working stroke), it also
includes the energy required to restore the state before the working stroke, i.e. restore
the ability of the machine to work,
 expansion 1→ 2 is expressed
v2
a1, 2   p  dv
v1
which is a volume work of the working stroke.
v2
Technical work of a closed cycle is at  p1  v1   p  dv  p2  v2
v1
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p2
or technical work can be expressed as at    v  dp
p1
Heat transfer
 is not a state variable
 is dependent on the manner (integration path) of the change,
Summary

Change in the internal energy and the enthalpy is not dependent on the manner of the
change, but only on the initial and final state.

Work and heat transfer is dependent on the relationship between pressure and volume
during the change, i.e. on the way of the change.
Second law of thermodynamics
Energy transfer is limited by the second sentence of thermodynamics.
Heat can only flow from a warmer body to a colder body.
According to the law of conservation of energy, the opposite direction is not excluded at
the level of elementary particles, but globally it is unlikely

some laws of energy transformation are probabilistic in nature.
Energy will be "degraded" when converting mechanical energy into thermal energy.
According to the second theorem of thermodynamics, heat energy is of an inferior form
compared to other forms of energy.

Heat cannot be fully transformed back to momentum energy to do work
According to the second theorem of thermodynamics, it is not possible to construct the
second kind of perpetual motion machine, i.e. a device which only takes heat from the
surroundings (e.g. sea) and converts it completely into kinetic energy.
Example:
Figure 2.2 shows a design of the second kind of perpetual motion machine. A double flask
connected with a tube is filled with a fluid with high thermal expansion. The fluid is heated on
one side and increases its volume, thereby shifting the center of gravity toward the axis of
rotation. The right flask is cooler, its center of gravity is therefore further away from the axis
of rotation. The whole system starts rotating by different moments of gravity forces.
However, the rotation stops because the machine designer did not realise that the fluid on the
right-hand side must cool again, and thus the gained heat must be transferred to the
surroundings.
Energy Management
Figure 2.2. Example of the second kind of perpetual motion machine
A realistic device can only work when removing heat from the warmer container T1,
which it partially converts into mechanical energy (work A) and partially passes heat to the
cooler container (cooler) T2 (Figure 2.3).
T1
A
T2
Figure 2.3 Diagram of a heat engine
The thermodynamic efficiency of converting thermal energy into mechanical energy is
therefore always less than 1
td ,1 
A1 Q1  Q2

Q1
Q1
The energy piped to the cooler container T2 cannot be transformed into useful work in the
same device. The energy piped to the cooler container can be utilised in a different machine or
device.
Machines can be cascaded in series (Figure 2.4). Due to a gradual decrease in temperature
of the reservoirs the usability of thermal energy brought from the preceding source is reduced.
In practice, therefore, a series of more than two to three devices does not make sense (e.g. flue
gas from a furnace is used in a fuel boiler for heating water, then a heat recovery unit can be
used at the output to heat air, e.g., heat pump, etc.).
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T1
A1
T2
A2
T3
An
Tn
Figure. 2.4 Cascading heat engines
Entropy
Entropy is a quantity characterizing the degradation of thermal energy. Entropy is the
amount of heat which has been supplied to the substance, relative to the temperature at which
the heat was supplied (entropy is sometimes inaccurately called a "reduced heat").
dS 
dQ
T
(J·K-1)
The lower the temperature of the reservoir, from which heat is pumped, the higher the
entropy.
Heat which is transferred at a lower potential (lower temperature) has lower utility value.
 the higher the entropy the worse the utility value of the heat.
Example:
Two reservoirs (Figure 2.5) contain the same amounts of heat Q, but have different
temperature of the heat transfer medium (different potentials), therefore, they have a different
utility value.
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QA = QB
SA < SB
QB = 1 GJ
TB = 50°C
QA = 1 GJ
TA =200°C
Figure. 2.5 Heat reservoirs with the same amount of heat and different temperatures.
Specific entropy is entropy related to 1 kg of a substance
s
S
m
(J.kg-1.K-1)
ds 
dq
T
(J.kg-1.K-1),
where q (J·kg-1) is heat related to 1 kg of a substance.
Entropy of an ideal gas
Calculation of the specific entropy according to the first formulation of the first law of
thermodynamics (entropy change in general)
dq = cv·dt + p·dv
ds 
(J.kg-1)
cv  dT p  dv

T
T
Substituting from the state equation of ideal gas p  v  r  T , or
ds 
cv  dT r  dv

T
v
after integrating, entropy between states 1 and 2 is
s2  s1  cv  ln
T2
v
 r  ln 2
T1
v1
p r
 we obtain
T v
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Calculation of the specific entropy according to the second formulation of the first law of
thermodynamics (analogous derivation)
ds 
c p  dT
T

r  dp
p
and after integrating
s2  s1  c p  ln
T2
p
 r  ln 2
T1
p1
Entropy is a state variable because it is dependent only on the state variables, its change is
therefore the difference between the initial and final state of the change.
Special cases (for theoretical changes)

Change in entropy of a system at a constant temperature T1 = T2 (isothermal change)
s2  s1  r  ln

Change in entropy of a system during heating from T1 to T2 at a constant pressure
(isobaric change) is equal to
s2  s1  c p  ln

p2
p1
T2
T1
for a polytropic change (the term polytropic changes is known from physics; in this
text it is explained later), the entropy change is usually determined from the
relationship
s2  s1  cn  ln
T2
T1
where cn (J·kg-1·K-1) is the specific heat capacity of a substance during a polytropic change.

For a reversible adiabatic change the change in entropy is zero
dq = 0  ds = 0
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Properties of entropy

the absolute value of entropy can not be determined, only the entropy change is
calculated,

zero entropy value is chosen in the tables either for 0 K or 273 K,

entropy is a state variable depending only on the state variables p, v, T, and on the
specific heat capacity of the substance,

in a spontaneous change of an isolated system, the entropy can only increase (e.g. due
to friction, irreversible changes), or remain constant (in the case of reversible
changes), which implies that the entropy of an isolated system can not decrease.
T - s diagrams
In addition to p - v diagrams, T - s diagrams are most frequently used in thermo mechanics.
The area under the curve which characterises the state change is proportional to the absolute
value of the supplied or removed amount of heat (Figure 2.6).
ds 
dq
T
dq  ds  T
2
q1, 2   T  ds
1
Figure. 2.6 T – s diagram
Basic reversible changes of states od ideal gases
The fundamental reversible changes include isobaric, isochoric, isothermal, adiabatic and
polytropic changes. Although in practice, the real changes only approach the theoretical
changes mentioned, these changes have a special meaning in energetics and thermo
mechanics. They are used for the purpose of theoretical derivation of thermo mechanical laws
and calculations. Using a sequence of the elementary reversible changes, any realistic change
can be modelled.
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The term "reversible change" means an ideal change without thermal, dissipative and
quantitative losses (i.e. no losses through substance leakage) which passes through
equilibrium states.
The polytropic change is of particular importance because all the basic reversible changes
can be seen as special cases of the polytropic change.
The equation describing the polytropic change is a relationship between pressure and
specific volume in the form of a power function. This function is an equation of the curve
called polytropic curve in a plane with coordinates p (pressure) and v (specific volume)
p  v n  konst
where n (1) is the polytropic exponent. In a polytropic change, its value is constant during
the change. In the case of actual changes n may not be constant.
The value of polytropic exponent determines what specific change is involved.
Table 2.1 lists the values of n for different types of basic reversible changes and their
definition equations expressing the relationship between pressure and specific volume.
Value k is the Poisson's ratio, also called "adiabatic exponent".
Its value is



1.67 for monatomic gases
1.4 for diatomic gases
1.33 for polyatomic gases.
Table. 2.1 Overview of the basic reversible changes
Value of the
Relationship between pressure and
polytropic
specific volume
exponent
p·v0 = const
n=0
p = const
p·v1 = const
n=1
T = const
Type of change
isobaric
isothermal
(ideal system cooling)
n=κ
p·vκ = const
dq = 0
adiabatic
(ideal heat insulation of the
system)
n=∞
p·v∞ = const; p1/∞·v = const
v = const
isochoric
Figure 2.7 shows the basic reversible changes in the diagrams p - v and T - s.
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Figure. 2.7 Comparison of the basic reversible changes in the diagrams p - v and T - s
In the p - v diagram the isotherm is a hyperbolic curve. The adiabat is a curve with state
points on it during an adiabatic change, i.e. during a change without heat transfer. The adiabat
is steeper than the isotherm.
The isobar in the T - s diagram is an exponential. The derivation can be done using the
equation from the first law of thermodynamics (second formulation)
dq  c p  dT  v  dp
which is for dp = 0 reduced to
dq  c p  dT
and heat can be expressed from the entropy
dq  T  ds
From both equations, it follows
ds  c p 
dT
T
and after integrating
s  c p  ln T  s0

T e
s  s0
cp
Analogously, one can derive an equation of the isochore (first formulation of the first law of
thermodynamics)
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T e
s  s0
cv
The isochoric curve in the T - s diagram is steeper than the isobar because cp > cv.
Specific thermal capacity
Specific heat capacity is the amount of heat required to heat a unit amount of a substance
by 1 K. The unit amount of the substance can be 1 kmol, 1 kg, or 1 m3 under normal physical
conditions. Specific heat capacity per unit mass is defined as
c
1 dQ
(J·kg-1·K-1)

m dt
Specific heat capacity per unit volume, which is mainly used for gases, is defined as
c
1 dQ

(J·m-3·K-1)
V dt
For the specific heat capacity of an ideal substance which does not depend on state
variables, per unit mass c (J.kg-1.K-1), the amount of heat required to heat m kg of a substance
is calculated from the relationship
Q = m · c · (t2 - t1) (J)
For gases, specific heat capacity is largely dependent on the type of change.
Comparing the amount of heat for heating by dt during an isobaric and isochoric change
For p = const, the first law of thermodynamics says that
dqp = cp.dt = du + da = cv·dt + da
For v = const
dqv = cv·dt = du
From the above it follows that for heating by dt during an isobaric change, it is necessary
to add more heat than for an isochoric change because heat is consumed not only to change
temperature, and therefore the internal energy, but also to produce work. Therefore cp > cv.
for specific heat capacities, Mayer equation applies
cp - cv
=
r
(J·kg-1·K-1)
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where r (J·kg-1·K-1) is the specific gas constant. Its value depends on the molar mass of the
gas, for example for air r =287 J·kg-1·K-1.
The ratio of the two specific heat capacities is equal to the Poisson's constant
cp
cv

(1)
Both equations for an ideal gas give us
cp 
cv 

 1
r
1
r
 1
Dependence of specific heat capacity on the state variables
In actual substances, the specific heat capacity is largely dependent on the temperature and
pressure. For gases, the concept of imperfect gas has been introduced. It has the
characteristics of an ideal gas, but its specific heat capacity is dependent on the temperature.
Instantaneous and mean specific heat capacity
Instantaneous (or the "proper") specific heat capacity is defined as the heat required to warm
1 kg of the substance by dt
c
1 dQ

m dt
The mean specific heat capacity for the temperature interval <t1; t2>
If Q1,2 - the amount of heat required to heat m kg of a substance from the temperature t1 to t2,
then the mean heat capacity for the interval <t1; t2> is
[c] tt12 
Q1, 2
m  (t2  t1 )
The heat required for heating the substance from the temperature t1 to t2 is
Q1, 2  m  [c] tt12  (t2  t1 )
The mean specific heat capacity for the temperature interval <t1; t2> is calculated as an
integral average (Figure 2.8)
t
2
1
[c ] 
  c(t )  dt
t2  t1 t1
t2
t1
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c
c
0
t1
t2
t
Figure 2.8 Mean specific heat capacity
Mean specific heat capacity


geometrically, it is the height of a rectangle of the same area as the area under the
curve
mathematically, it is the median value of a continuous variable.
Calculating an integral from measured values is problematic. The literature therefore
provides calculated tabulated mean specific heat capacities for the interval 0 °C to t (°C)
t
1
c  [c] t0    c(t )  dt
t 0
Example:
Calculation of heat for the heating of m kg of water from the temperature t1 °C to t2 °C.
Let us compare the two procedures:
a) calculation by integrating the instantaneous specific heat capacities:
t2
Q1, 2  m   c(t )  dt
(J)
t1
It is necessary to integrate the empirical dependence of the instant heat capacity (usually
replaced by a polynomial function or numerical integration of tabular values via the
trapezoidal method).
b) calculation using a tabulated mean specific heat capacity for the interval <0, t>
The amount of heat is determined as a difference of heats required for the heating from 0
°C to the temperature t2 and from 0 °C to the temperature t1
Q1, 2  Q0, 2  Q0,1
Q1, 2  m  ([c] t02  t2  [c] t01  t1 )
(*)
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 1 t 2

 1 t1
 
Q1, 2  m    c  dt   t2    c  dt   t1 
0

 t1 0
 
 t2
or a reduced form
Q1,2  m  (c2  t2  c1  t1 )
In the calculation of the heat required for heating a substance by mean specific heat
capacities, it is not necessary to calculate the integral.
The equation (*) is multiplied by one
Q1, 2  m 
[c]t02  t 2  [c]t01  t1
(t 2  t1 )
t 2  t1
then the fraction in the previous equation corresponds to the mean specific heat capacity [c]tt12
in the interval <t1; t2>, and then the average specific heat capacity for the interval <t1; t2> is
determined by
[c]tt12 
[c]t02  t 2  [c]t01  t1
t2  t1
Note:
If the temperature interval <t1; t2> is very narrow, which is common in iterative numerical
t2
t1
t2
calculations, it is approximately [c]0  [c]0 , but it can vary greatly from [c] t1 , see
Figure 2.9.
c
c mean pro <0; t1 >
0
t1 t2
t
c instantaneous 
 c mean for <t1; t2 >
Figure. 2.9 Mean and instantaneous specific heat capacity for a narrow temperature interval
Therefore, in the equation
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Q1, 2  m  ([c]t02  t2  [c]t01  t1 )
it is not possible, due to the narrow interval <t1; t2>, to factor out before the parentheses the
t2
t1
mean specific heat capacity [c]0 , or [c]0 .
t2
Before the parentheses only the mean specific heat capacity can stand [c] t1 which due to
the narrow temperature interval <t1; t2> approaches the instant specific heat capacity
Q1, 2  m  [c] tt12  (t2  t1 )
Specific heat capacity of water
The instant heat capacity of water at 0 °C is 4.218 kJ.kg-1.K-1, for increasing temperature it
decreases, at 35 °C it reaches the minimum value of 4.178 and then rises again.
The mean c p begins at 0 °C at the same value 4.218, at around 60 °C it has the minimum
of 4.185, and then rises. A graph of the instantaneous (proper) specific heat capacity and the
mean specific heat capacity
of watertepelná
is shown inkapacita
Figure 2.10. vody
Měrná
4225
4220
4215
cp (J.kg-1.K-1)
4210
4205
4200
cp pravá
4195
cp střední
4190
4185
4180
4175
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
t (°C)
Figure. 2.10 Mean and instantaneous specific heat capacity for a narrow temperature interval
Example:
What is the absolute and relative error of the instantaneous specific heat capacity,
neglecting its change depending on the temperature, of a water heating system with a
temperature gradient of 90 / 35 °C?
Solution: The relative change in the instantaneous specific heat capacity between 35 and
90 °C is 0.72 %, the absolute change Δcp=30 J·kg-1·K-1.
Specific heat capacity of gas mixtures
Let us consider mixing two different gases of different temperatures and mass flows at a
constant pressure. The resulting mass flow is the sum of the mass flow rates of the two
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components. The amount of heat is an additive variable and is determined as a sum of the two
components (Figure 2.11).
p = const
m1, tvst
heater Q
m, t
m2, tvst
Figure. 2.11 Mixing gases with heating
Mass balance
m 1  m 2  m
Heat balance
Q  m 1  c p1  (t  tvst )  m 2  c p 2  (t  tvst )
Q  m  c p  (t  tvst )
cp 
m 1
m
 c p1  2  c p 2
m
m
n
1 n

c p    mi  ci   wi  ci
m 1
1
(J·kg-1·K-1)
where the mass component is marked with the letter w.
The following Table 2.2 summarises the basic formulas for the calculation of the heat
amount and the enthalpy of a gas mixture. The volume part is marked with an x.
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Table 2.2 Overview of basic relations for the calculation of the heat amount and the
enthalpy of a gas mixture
Specific heat capacity
Heat
Specific heat capacity of a gas
mixture
Specific enthalpy
Enthalpy:
Specific enthalpy of a gas
mixture
cv, cp
(J·kg-1·K-1)
cv, cp
(J·m-3·K-1)
Q = m · cp · (t2 - t1) (J)
Q = V · cp · (t2 - t1) (J)
c p   (c pi  wi ) (J  kg 1  K 1 )
c p   (c pi  xi ) (J  m3  K 1 )
i (J.kg-1)
i (J.m-3)
I = m · i (J)
I = V · i (J)
i   (ik  wk ) (J  kg 1 )
i   (ik  xk ) (J  m3 )
Note: In literature, the mean specific heat capacity is often referred to as cp, the same as the
instantaneous specific heat capacity, which may lead to confusion. In tables, values of cp
related to kg, m3, or kmol can be found, and marking is usually the same, so it is necessary to
pay attention to the units, making sure to read the comments.
Summary of chapter concepts
-
First theorem of thermodynamics,
Second theorem of thermodynamics,
perpetual motion machine of the second kind,
thermodynamic efficiency,
enthalpy,
entropy,
instantaneous and mean specific heat capacity
Questions on the topic of the chapter
-
Deduce the second formulation of the first law of thermodynamics from the first
formulation.
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-
Draw a general one-off change in the p - v diagram. Which area under the curve
characterises the absolute work and which characterises the technical work?
What is the perpetual motion machine of the first kind?
Why cannot the perpetual motion machine of the second kind work?
Draw the basic reversible changes in the diagrams p – v and T – s.
Calculate the specific heat capacity of air under normal conditions (assuming only a
mixture of oxygen and nitrogen)?
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3.
THERMAL CYCLES
Study time: 4 hours
Objective: After reading this section, you will be able to:

Define the thermal cycle.
 Determine the work of a cycle.
 Formulate the Carnot theorem.
 Compare reversible and irreversible cycle.
PRESENTATION
A thermal cycle is a number of changes in a series, after which the substance returns to its
original state (in a variety of ways). During each change, heat is supplied to or taken from the
substance and volume mechanical work is produced or supplied. The total work in the cycle is
the work of the cycle This is illustrated in Figure 3.1.
p
qa
a0
qb
v
Figure. 3.1 Thermal cycle
Work of a thermal cycle
The work of a closed cycle is the sum of single works (compression and expansion ones).
According to the first law of thermodynamics:
work = utilised heat
work = applied heat – removed heat
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a0 = qa – |qb|
Thermal efficiency of cycles

a0 qa  qb

qa
qa
the removed heat may be specified as negative qb <0, therefore, for clarity its absolute value is
given in the equation.
Carnot cycle
The reversible Carnot cycle is of theoretical significance (isothermal and adiabatic
expansion and compression, see Figure 3.2)
Figure 3.2 Carnot cycle [12]
Carnot's theorem:
The efficiency of all reversible Carnot cycles operating between temperatures Ta > Tb
(temperatures of the hot reservoir and cold reservoir) depends only on the two temperatures of
isothermal changes (it does not depend on the path of the change).

Ta  Tb
Ta
Values of the removed heat in the Carnot cycle are expressed as
qb Tb

qa Ta
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It is assumed that heat between the system and its surroundings is exchanged at a constant
temperature (isothermal expansion and compression).
Thermal efficiency of the reversible Carnot cycle would be nearly 100 %, if (which cannot
be implemented) the hot reservoir had a temperature Ta   or if the temperature of the
cooler was close to the absolute zero Tb  -273.15 °C.
Note: According to the third law of thermodynamics, the temperature of absolute zero (0 K)
cannot be reached.
The efficiency of the general reversible cycle is always lower than the Carnot cycle
efficiency operating between the same temperatures.
In practice, the Carnot cycle has no application. The first reason is that it is technically
difficult to implement.
The second reason is that the work of the Carnot cycle is small (closes a small area) in
comparison with other cycles.
If the Carnot cycle was to achieve the same extreme temperatures as a combustion engine,
the resulting pressures would be extremely high.
Irreversible changes
Every process in nature is directed towards equilibrium. Processes in nature are generally
irreversible (due to friction, leakage, ...). They can only approach a reversible process. The
degree of irreversibility may differ.
Example:
Impact of a lead bullet on a wall - a typically irreversible process.
Compression and expansion of gas in a low-speed compressor approaches a reversible
process.
The entropy of an isolated system in a reversible process is constant while in the case of
irreversible thermodynamic processes it grows.
 The change of entropy in irreversible processes is always positive (e.g. formation of
heat due to friction).
In nature there is a tendency to achieve greater disorder. Entropy grows, similarly to
disorder.
Note: Suppose we interconnect two vessels, each filled with a different gas. The system is
organized. After opening the valve, gases begin to mix, increasing disorder, i.e. the molecules
from one vessel are distributed throughout the system, and vice versa. Irreversible processes
have a "natural tendency". Irreversible processes are also represented by e.g. free expansion
and thermal conduction.
Adiabatic irreversible change of state
During an adiabatic change the system does not exchange heat with its surroundings
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dq = 0
The increase in entropy happens at the expense of mechanical energy due to friction.
ds > 0
ds 
dqtř
T
Note: A reversible adiabatic change is called isoentropic change because the entropy does
not change. In the case of an irreversible adiabatic change the entropy grows.
General irreversible change of state
Dual heat is supplied to a closed system (Figure 3.3)


heat from the surroundings dqvrat (as large as in a reversible change)
heat from the inside due to the irreversibility of the change (friction) dqtř
dqvrat
dqtř
Figure 3.3 Closed system
Entropy change due to the heat applied from the outside
ds1 
dqvrat
T
Change in entropy due to irreversibility
ds2 
dqtř
0
T
The total entropy change in an irreversible change is a sum of both entropy increases
dsnevrat  ds1  ds2 
dqvrat
T
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dsnevrat  T  dqvrat
This results in a general relationship referred to as a mathematical formulation of the
second law of thermodynamics
ds  T  dq
Verbal formulation: Product of the change in entropy of an irreversible change and the
temperature is greater than the heat supplied from the surroundings.
Irreversible cycles
A reversible engine is an ideal engine which can transfer heat in both directions (by
doing or, conversely, receiving the same amount of external work).
However, in the case of irreversible engines, the work done by the engine in the transfer of
heat from the warmer to the cooler reservoir is not enough to move the same amount of heat
back.
The work done by an irreversible engine is reduced by friction heat which is discharged
into the reservoir or the surrounding area (Figure 3.4).
Reversible engine
dqa
Irreversible engine
dqa
da0, nevr.=
= da0, vr - dqtř
da0, vr.
dqtř
dqb
dqb + dqtř
Figure 3.4 Reversible and irreversible heat engines
Summary of chapter concepts
-
thermal cycle,
cycle work,
Carnot cycle,
Carnot factor,
Questions on the topic of the chapter
-
What determines the shape of the cycle in the diagram p - v?
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-
-
Which cycle has the highest thermodynamic efficiency?
What are some of the irreversible changes in practice?
How much smaller is the work of an irreversible cycle compared to a reversible
cycle?
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4.
ABILITY TO UTILISE THERMAL ENERGY.
Study time: 6 hours
Objective: After reading this section, you will be able to:
 Define the concepts of exergy, anergy
 Determine the exergy of a substance and exergy of the
applied heat
 Establish utilisation of heat into mechanical work by means
of exergy efficiency and degree of perfection
PRESENTATION
Exergy is the maximum utilisable part of the energy which can be converted into
mechanical work.
Anergy is the part of energy which cannot be changed into mechanical work (useless).
Note: The anergy of mechanical or electrical energy is zero (can be fully utilised for
mechanical work, assuming no losses).
Example:
Illustration of the concepts of energy, exergy and anergy using the potential energy of
water. The total energy of a water flow is given by the difference of its level and the sea level.
The exergy, i.e. usable portion of the energy in a given location is lower. For example, it is
given by the difference in the levels above and below the dam. Anergy is the energy given by
the difference of levels below the dam and the sea (at a given point it is useless).
Exergy in thermo mechanics
Heat applied to a system is partly converted into mechanical work and the rest is
discharged into the cold reservoir. In addition, in an actual machine part of the applied heat
escapes in the form of losses and part of the work is wasted due to dissipative losses
(converted into frictional heat).
Maximum exergy is the part of the energy that can be utilised in a reversible Carnot cycle.
Exergy of an irreversible engine is lower by the heat generated by the irreversibility
(dissipative losses).
Irreversibility of the change (e.g. friction) leads to the formation of heat in the system
which reduces the work done (exergy) compared to the work done in a reversible change,
Figure 4.1.
The unused heat (anergy) may not always be irretrievably lost.
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The heat in the Carnot cycle of a heat engine goes into the cold reservoir; it is not lost
(losses), but only unused. Its utilisation by another machine is not ruled out.
In contrast, when throttling gas (typical irreversible change) no work takes place, no heat is
dissipated into the surroundings, but pressure energy of the gas is lost. No work was obtained
during the pressure drop. This wasted energy can no longer be used in any way.
Carnot engine
Q1=Ex+An
Irreversible engine
Q1
Anevrat=Ex
Amax= Exmax
Qnevrat
Q2=An
Q2, nevrat
Figure 4.1 Reversible and irreversible heat engine
We distinguish between the exergy linked to the substance and exergy of the applied heat.
Substance exergy (exergy bound to matter)
Exergy of a substance is equal to the maximum technical work which can be obtained from
the initial state of a substance p1, v1, T1 through a change to the state of the surroundings pa, va,
Ta .
In the deduction, the change is modelled as consisting of two elementary changes
(generally it is impossible to move from the initial state to any final state through a single
change only) Figure 4.2:
1. reversible adiabatic expansion to the pressure p2 and the ambient temperature Ta
2. reversible isothermal expansion to ambient pressure pa in which heat from the
surroundings is applied to the substance Ta.(sa – s1).
Figure 4.2 Reversible adiabatic expansion followed by a reversible isothermal expansion [14]
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Technical working ability of a substance (exergy) of a reversible process with heat input
from the surrounding environment can be expressed by the first sentence of thermodynamics:
dq = di + dat  dat = -di + dq
at,max = b1 = -[(i2 – i1) + (ia – i2)] + Ta. (sa – s1)
b1 = i1 – ia – Ta. (s1–sa)
(J.kg-1)
where
i1 is the enthalpy of the inputted substance (J·kg-1), ia is the enthalpy of the
outputted substance (in the state of the surroundings) (J.kg-1), Ta is thermodynamic ambient
temperature (K), (s1 - sa) is the gradient of the entropy of the substance due to the applied
external heat (J·kg-1·K-1).
The heat brought in from the surroundings increases exergy.
Technical working ability of a substance (exergy) in an irreversible process. Irreversibility
results in heat Ta. s at the expense of mechanical work.
b1 = i1 – ia – Ta. (s1–sa) - Ta. s (J.kg-1)
If the system is isolated and the processes are irreversible, then the increase in entropy s
due to irreversibility is positive and reduces exergy. It does not exchange heat with the
surroundings.
b1 = i1 – ia – Ta. s
(J·kg-1)
The product Ta·s is loss due to irreversibility (expressed as a change in entropy), called
lost exergy.
For a perfectly reversible process in an isolated system, the product of Ta s = 0 and the
exergy equals to the decrease in enthalpy
b1 = i1 – ia
(J·kg-1)
If the working medium and the surroundings form an isolated system (external heat is not
shared) and the processes occur irreversibly, than the maximum work can not be performed
because the heat generated by irreversibility is supplied at the expense of work
Ta s = Avrat - Anevrat
Definition of exergy:
Exergy is the work produced by isentropic (i.e. reversible adiabatic) expansion from the given
state to the state of the surroundings, less the heat applied by irreversibility at the ambient
temperature.
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Exergy of substances



is a state variable (like enthalpy), but it also depends on the ambient temperature,
it is the usable part of enthalpy
some authors call it a technical working capacity of a substance.
Calculation of change in exergy of a substance
Expresses how much (J·kg-1) the working ability of a substance changes after the change is
performed (states 1 and 2 are now two general states of the substance).
For a reversible process
b = b1 – b2 = i1 – ia – Ta. (s1–sa) – [i2 – ia – Ta. (s2–sa)]
b = i1 – i2 – Ta .(s1 – s2)
Special cases
In an isothermal change the enthalpy change is equal to zero, therefore the change in
exergy is equal to outputted (inputted) heat to (from) the surroundings
b = – Ta·(s1 – s2) = Ta·r·ln(p1/p2)
as the entropy change in the isothermal change according to the second formulation of the
first law of thermodynamics (where di = cp·dT = 0) is (see above)
p
1 2
s2  s1     v  dp 
T p2
p2
1
p
 dp   r  ln 2
p
p1
p2
 r  
Question: Why is a minus sign in the equation? Answer: In an isothermal expansion it is
necessary to supply heat to the substance, therefore, at the end the entropy is higher than at
the beginning, s2 - s1 > 0. After the expansion it is p2 < p1, logarithm is negative, therefore the
minus sign is necessary.
In a reversible isentropic expansion, dq = 0, and therefore the entropy change is zero,
then the change in exergy is
b = i1 – i2
(J.kg-1).
In an isobaric change the exergy change is calculated using the relationship
b = i1 – i2 – Ta. (s1–s2) = cp·[T1 – T2 – Ta·ln(T1/T2)]
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because the entropy change in an isobaric change is, according to the second formulation
of the first theorem of thermodynamics, where v·dp = 0 (see above)
s = s2 – s1 = cp·ln(T2/T1)
Exergy of heat introduced into circulation
The maximum of usable heat energy convertible into mechanical work (called heat exergy)
is obtained in a Carnot cycle (i.e. a reversible cycle) working with the temperature of a cold
reservoir equal to the ambient temperature. Exergy is the product of the heat and the
efficiency of the Carnot cycle, called Carnot factor
bq ,1  q1 
T1  Ta
T1
For an irreversible cycle, it is necessary to deduct the irreversible heat produced at the
expense of mechanical work
bq ,1  q1 
T1  Ta
 Ta  s
T1
Exergy heat depends on the temperature at which heat is applied, and the ambient
temperature.
If the temperature of the substance is equal to the ambient temperature, heat is unusable.
The ambient temperature does not necessarily mean the temperature of the surroundings,
but in fact it is the temperature of the cold reservoir into which the heat is removed. In
different parts of the actual thermal cycle there may therefore be different ambient
temperatures.
Example:
In a steam turbine, the ambient temperature is the condenser temperature.
In a condenser the ambient temperature is the cooling water temperature.
In a cooling tower it is the temperature of the surrounding air.
For a heating furnace the ambient temperature is the surrounding environment temperature.
Exergetic efficiency
It is used to assess processes in terms of losses caused by their irreversibility. It is a ratio of
actually obtained work through an irreversible process to the maximum work obtained by a
reversible process
ηe 
at,nevrat
(1)
at,vrat
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The work done in an irreversible process of heat exchange is equal to the change in the
substance exergy and the difference of exergy of the heat exchanged with the surroundings
at ,1, 2  b1  b2  q1 
T1  Ta
T T
 q2  2 a  Ta  s
T1
T2
where b1 is the exergy of the working substance at the beginning, b2 is the exergy of the
substance at the end, q1 is the heat supplied to the working cycle, q2 is the heat removed from
the working cycle.
The work produced in a reversible process with heat exchange
at ,1, 2  b1  b2  q1 
T1  Ta
T T
 q2  2 a
T1
T2
Exergy efficiency of a perfectly reversible process is equal to e = 1.
Total efficiency (degree of perfection)
The overall efficiency is defined as the ratio of the sum of the output energies, including
the work, to the sum of the input energies. The overall efficiency expressed in terms of
energies can be written down using the fact that the work done is the difference between the
input and output energies
ν
where
E1 is the input energy,
the energy lost by irreversibility
E2 E1  EZ

E1
E1
E2 is the output energy including the work, EZ is
If we are working with exergies, then a balance equation applies
sum of the input exergy =
= Sum of the output exergy + work + loss by irreversibility
The degree of perfection is then equal to

T2  Ta
 at ,1, 2
T2
T T
b1  q1  1 a
T1
b2  q2 
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Overall efficiency of the Carnot cycle
In the case of a closed cycle, the exergies of the input substance and output substance are
equal b1 = b2. Maximum of usable heat energy in the Carnot cycle is equal to the exergy of
the applied heat
bq ,1  q1 
T1  Ta
T1
If the temperature of the cold reservoir is higher than the ambient temperature, then the
exergy of the heat removed from the Carnot cycle equals
bq , 2  q2 
T2  Ta
T2
The work done using the Carnot cycle is equal to the difference of exergies of the supplied
and removed heat
at = bq,1 – bq,2
Technical work done by the Carnot cycle is equal to
at  q1 
T1  T2
T1
Substituting for exergy of the supplied and removed heat and after an adjustment, the total
efficiency of the Carnot cycle is equal to v = 1 (reversible cycle).
Limiting factors of energy transformation




limiting energy flux density (given by dimensions, material resistance),
economic constraints (prices for equipment, return on investment)
psychological (full shading by solar panels, etc.)
political (related to the finiteness of the earth's surface area, states, limiting natural
resources
Summary of chapter concepts
-
exergy, anergy.
Exergetic efficiency
total efficiency, degree of perfection
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Questions on the topic of the chapter
-
Explain the concepts of exergy and anergy.
How was exergy of a substance derived?
How do you define the exergy of a substance and exergy of heat?
In which cycle is thermal energy best utilised?
Which parameters allow evaluation of processes in terms of losses caused by their
irreversibility?
What is the total efficiency of the Carnot cycle?
Energy Management
5.
BALANCING ENERGY FACILITIES
Study time: 4 hours
Objective: After reading this section, you will be able to:
 Understand the principle and purpose of balancing fuels and
energies.
 Decide on the decomposition of the system into subsystems
 Prepare an energy and exergy balance of a heating furnace
PRESENTATION
Energy balances
They express the quantitative aspects of energy conversion in a device (process, company,
...). Energy balances of a facility involve all types of energies used in the business.
Balancing is based on the first law of thermodynamics.
Performed activities:
 research into the fuel switching possibility,
 determining the optimum calorific value of fuel gases for each appliance
(low calorific value for blast heaters and wind of coke oven batteries, high calorific
value for heating furnaces)
 the needs for purchasing energies (the difference between consumption and own
production)
 analysis of input and output of heat involving material balances and heat flows.
Sankey diagrams are used to provide graphical representation of energy flows. They may
express items of the first formulation of the first theorem of thermodynamics (internal energy
and volume work) or, more frequently, the second formulation (enthalpy and technical work).
Example: Sankey diagrams of reversible changes. Figure 5.1 shows a Sankey diagram of
energy balance of polytropic expansion (work will consume part of the internal energy and
the supplied heat)
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Figure 5.1 Sankey’s diagram of polytropic expansion [14]
Figure 5.2 shows a diagram of energy balance of isobaric change (the applied heat results
in an increase in the internal energy and its part is consumed for work)
Figure 5.2 Sankey’s diagram of isobaric expansion [14]
Heat balances of heating furnaces
Energy consumption in furnaces has a major impact on specific consumption of
metallurgical plants. The aim of balancing is to assess the thermal work of furnaces and find
the possibility of reducing consumption.
Balances refer to


unit of time (for continuously working furnaces - for example a shove furnace)
technological cycle or its part (for furnaces working periodically, the determination is
more complex as it is a non-stationary process).
Balance is focused on


the entire device (e.g. furnace + recuperator + waste heat recovery unit)
part of the device (e.g. only for the furnace chamber, etc.)
Balance from the system perspective:


balance of the entire system,
balance of subsystems
Example: Heating furnace
Balance of the entire system of a heating furnace
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Pch
Puž
S
Pz,c
Figure 5.3 System of heating furnace
The balance equation is
Pch  Puž  Pz, c
where Pch (W) is the flow of the chemical heat of fuel (neglecting the perceptible heat of
fuel and air without preheating), Puž is useful heat flux of the charge (W), Pz,c is the total loss
(W).
System efficiency
celk 
Puž
Pch
A more detailed analysis is done by decomposing the system into subsystems of the
heating system and the furnace chamber
heating
working area
Pch
Ppp
Puž
S1
S2
Pz,sp
Pz,pp
chimney
loss
S
losses through
walls etc.
Figure 5.4 Decomposition of the heating furnace system into two subsystems
The efficiencies of the subsystems can be expressed
Overall system efficiency
Pch  Pz,sp
1 
Ppp
2 
Puž
Puž

Ppp Pch  Pz,sp
Pch

Pch
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celk 
Puž
 1 2
Pch
If the subsystems are connected in series, the efficiency is multiplied.
Subsystem S1 (heating) can be further decomposed into subsystems of burners H and
recuperator R. Using the diagram in Figure 5.5, the balance equation and efficiency can be
expressed
Pch  PR  Ppp
celk 
Ppp
Pch  PR
The subsystem of heating must, in the case with or without a recuperator, supply the
required power Ppp to the working space of the furnace.
If a recuperator is used, the heat input of the chemical heat Pch is lower, than in the
operation without a recuperator. It is reduced by the heat gained in the recuperator PR by
removing part of the heat of the flue gas PzH leaving from the burners.
heating
PR
Pch
Ppp
H
PzH
S1
R
PzR =Pz,sp
loss to
chimney
Figure 5.5 Decomposition of the furnace heating subsystem into burners and recuperator
Sankey’s diagram of a furnace
Figure 5.6 shows a Sankey’s diagram of a heating furnace with a recuperator A detailed
description of the diagram and each item can be studied in the literature [15].
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Key: Qch chemical heat of the fuel, Qr recuperated heat, Qspp heat of flue gas in the flue after
the furnace Qsrp heat of flue gas in the flue after the recuperator [15]
Figure 5.6 A Sankey’s diagram of a heating furnace with a recuperator.
After removing the recuperator from the process, all heat in the flue gas leaves to the
smoke stack without being used (Figure 5.7). Part of the input energy which was supplied by
the recuperator must be supplied in the form of the chemical heat of fuel, thus increasing fuel
consumption.
Source: adapted from [15]
Figure 5.7 Modifying the Sankey diagram of a heating furnace after removing the recuperator
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Exergy balance
Unlike energy balances, exergy balances take account of the qualitative aspects of energy,
i.e. ability to utilise the energy
Exergy of the supplied heat
B1  Q1 
where
(K).
T1  Ta
T1
(J)
Q1 is the applied heat at the temperature T1 (J), Ta is the ambient temperature
Energy Management
REFERENCES
Reading
[1] KREITH., F., BLACK, W. Z. Basic heat transfer. New York: Harper and Row, 1980.
[2] LIENHARD IV, J. H., LIENHARD V, J. H. A Heat Transfer Textbook. 4th ed.
Cambridge: Phlogiston Press, 2012. http://web.mit.edu/lienhard/www/ahtt.html
[3] BARRIE JENKINS, PETER MULLINGER. Industrial and Process Furnaces: Principles,
Design and Operation. Butterworth-Heinemann, 2011. ISBN 0080558062,
9780080558066. 544 p.
[4] BEJAN, A., KRAUS, A. D. Heat Transfer Handbook. John Wiley & Sons, 2003.
ISBN 978-0-471-39015-2.
[5] DAVID J.C. MACKAY. Sustainable Energy – without the hot air. UIT Cambridge,
2008. ISBN 978-0-9544529-3-3. Available free online from www.withouthotair.com.
[6] CHANSON, H. APPLIED HYDRODYNAMICS. An Introduction to Ideal and Real
Fluid Flows. Leiden: CRC Press, Taylor & Francis Group, 2009, ISBN 978-0-415-49271-3.
[7] KHAVKIN, Y. I. Combustion System Design. tulsa: PennWell Books, 1996. ISBN 087814-462-5.
[8] KAZANTSEV, E. I. Industrial Furnaces. Moscow: Mir Publishers, 1977.
[9] SIENIUTYCZ, S., JEŻOWSKI, J. Energy Optimization in Process Systems. Oxford:
Elsevier, 2009. ISBN 978-0-08-045141-1.
[10] Energy strategy for Europe. [online]. 2013. http://ec.europa.eu/energy/index_en.htm
[11] EU Energy in Figures. 1. vyd. Luxembourg: Publications Office of the European Union,
2012. ISBN 978-92-79-22556-7.
Other
[12] Enenkl, V. – Hloušek, J. – Janotková, E.: Termomechanika. (Skriptum). Brno, VUT,
1983. 290 s.
[13] http://www.transformacni-technologie.cz
[14] KALČÍK, J. Technická termodynamika. Praha: ČSAV, 1963.
[15] RÉDR, M. Tepelné hospodářství hutí, I, II. Skriptum VŠB. Ostrava, 1983. (selected
chapters)