Exergetics - Exergy.se
EXERGETICS
by
Göran Wall
Bucaramanga 2009
EXERGETICS
Foreword
This is a guide book to introduce the concept of exergy and its
applications. The text assumes a basic skill in science and engineering.
The purpose is to give an understanding of the concept of exergy and
its potentials, enough for you to carry out exergy analysis of real systems
or processes.
The material is equivalent to about 200 hours study at graduate level.
All proposals to improve the text in order to understand the concept of
exergy are highly appreciated.
For those who want to put exergy into a social context I recommend
the book Time to turn by Bo Lundberg, 1996, ISBN 91-26-94842-7.
(http://www.exergy.se/ftp/timetoturn.pdf) Also an excellent book for those
not trained in science. From my homepage: http://exergy.se/, you can find
more publications on exergy.
The present situation for mankind is serious from an increasing lack of
natural resources and increasing environmental destruction. I recommend
a better understanding of our situation and possibilities by adopting the
concept of exergy and a focus on sustainable development.
Finally, I am deeply grateful for the kind support of Dr. Darwish M. K.
Al Gobaisi at the International Centre for Water and Energy Systems in
Abu Dhabi, who have made this work possible. I also want to thank Mei
Gong for preparing the manuscript and for all the comments to a previous
text I have received from readers on the web.
Bucaramanga in January, 2009
Göran Wall
Solhemsgatan 46
SE-431 44 Mölndal, Sweden
Tel./fax: +46-31-877579
[email protected]
http://exergy.se/
2
EXERGETICS
CONTENTS
NOMENCLATURE ......................................................................................................... 5 INTRODUCTION ............................................................................................................ 7 FUNDAMENTAL CONCEPTS ....................................................................................... 9 Energy and mass ....................................................................................................... 11 Energy power: ........................................................................................................... 14 Energy efficiency ...................................................................................................... 14 Exergy ....................................................................................................................... 17 Exergy power ............................................................................................................ 18 Exergy efficiency ...................................................................................................... 18 THERMODYNAMICS OR BETTER THERMOSTATICS .......................................... 20 Temperature, heat and internal energy ..................................................................... 20 The ideal gas model .................................................................................................. 22 Work, Mechanical Work and Pressure-Volume Work ............................................. 25 The 1st Law of Thermodynamics — Nothing disappears ........................................ 26 Work and Technical Work ........................................................................................ 28 Closed system ..................................................................................................... 28 Steady-State processes ........................................................................................ 28 State variables and State changes ............................................................................. 33 The 2nd Law of Thermodynamics — Everything disperse ...................................... 38 Additional examples and solutions: .......................................................................... 43 Exercises ................................................................................................................... 46 Answers..................................................................................................................... 47 Solutions ................................................................................................................... 47 EXERGY ........................................................................................................................ 51 Exergy of a general process ...................................................................................... 51 Exergy, work and entropy production ...................................................................... 53 Heat transfer .............................................................................................................. 54 Temperature exchange between two bodies ............................................................. 55 Friction ...................................................................................................................... 56 System in contact with a heat reservoir .................................................................... 57 System in contact with a heat and pressure reservoir ............................................... 57 Exergy of heat and cold ............................................................................................ 58 Exergy of black body radiation ................................................................................. 64 Exergy of materials ................................................................................................... 66 Exergy of nuclear fuel............................................................................................... 68 Exergy and information ............................................................................................ 70 Mixing entropy ................................................................................................... 72 Summary ................................................................................................................... 78 CYCLIC PROCESSES ................................................................................................... 79 Cyclic processes in general ....................................................................................... 79 The Carnot cycle ....................................................................................................... 80 Inverse Carnot cycle ........................................................................................... 82 Thermodynamics of steam ........................................................................................ 83 TS-diagram ......................................................................................................... 85 Mollier or hs-diagram ......................................................................................... 88 3
EXERGETICS
Steam power processes ............................................................................................. 89 Refrigerators and heat pumps ................................................................................... 99 HEAT TRANSFER ...................................................................................................... 103 Heat conduction ...................................................................................................... 103 Heat convection ...................................................................................................... 106 Free convection – free current .......................................................................... 106 Forced convection ............................................................................................. 107 Heat radiation .......................................................................................................... 109 Radiation between different bodies .................................................................. 110 Over all heat coefficient, U ..................................................................................... 113 -method ........................................................................................................... 115 U-method .......................................................................................................... 115 Heat exchangers ...................................................................................................... 117 COMBUSTION ............................................................................................................ 121 Enthalpy, Gibbs’ function and exergy of fuels ....................................................... 122 APPENDIX ................................................................................................................... 127 Appendix 1 Internal energy and entropy ................................................................ 127 Appendix 2 Exergy ................................................................................................. 128 Appendix 3 Heat exchange between many systems ............................................... 131 Appendix 4 Reference states................................................................................... 134 4
NOMENCLATURE
Symbol Quantity [Unit according to the SI-system]
A
a, A
COP
c
c, C
e, E
E/Q
F
g
g, G
h
h, H
H0
I
m
ni
ni0
P
P0
q, Q
R
S
S
S0
si
t
T
T0
U
u, U
U
v
v, V
W
xi
Z
µi
Area [m2]
Specific free energy (= Helmholtz’ function) [J/kg, Wh/kg] and free energy [J,
Wh], A = U  TS. Sometimes the symbol F is also used.
Coefficient of Performance, for heat pumps: produced heat (energy) divided by
used work (usually electricity) and for refrigerators: removed heat (energy)
divided by used work (usually electricity) [no unit]
Velocity of light in vacuum ≈ 2.997925×108 m/s
Specific heat capacity [J/kg K] or heat capacity [J/K], gives the amount of heat
(internal energy) which per unit mass is captured in a body as the temperature
increases by one degree Celsius, i. e. one Kelvin
Specific exergy [J/kg, Wh/kg] or exergy, availability or available work [J, Wh]
Exergy factor [no unit, %]
Power [N]
Constant of gravity ≈ 9.81 [m/s2]
Specific free enthalpy (= Gibbs’ function) [J/kg, Wh/kg] and free enthalpy [J,
Wh], G = H  TS.
Height [m]
Specific enthalpy [J/kg, Wh/kg] or enthalpy, H = U  PV [J, Wh]. In German
literature often signed by i and I.
Enthalpy for systems at environmental state, i.e. in equilibrium with the
environment
Electric current [A]
Mass [kg]
Number of mole of substance i [mol]
Number of mole of substance i for a system at equilibrium state [mol]
Pressure [Pa]
Environment pressure [Pa, bar] here 101.3 kPa = 1.013 bar
Specific heat [J/kg, Wh/kg] and heat [J, Wh]
Molar gas constant ≈ 8.314 [J/mol K], state equation for ideal gases: PV = nRT
Entropy, gives degree of disorder, the following relation holds (2nd Law):
dS ≥ Q/T [J/K]
Distance [m]
Entropy of a system at environmental state [J/K]
Specific entropy, entropy per unit mass, of substance i, [J/kg K]
Time [s, h], 1 year = 8760 h = 31,536,000 s
Temperature [K] (0 K = –273.15˚C)
Environment temperature [K] usually 20˚C = 293.15 K
Electric potential [V]
Specific internal energy [J/kg, Wh/kg] or internal energy [J, Wh]
Overall heat transfer coefficient, gives the heat transfer rate per unit area of a
substance, when the temperature difference is 1˚C, i.e. 1 K [W/m2 K].
Velocity [m/s]
Specific volume [m3/kg] and volume [m3]
Work [J, Wh], by definition equal to exergy
Molar fraction of substance i [no unit]
Height [m]
Chemical potential of substance i [J/mol], often equal to Gibbs function per mole
EXERGETICS
µi0
en
ex


Chemical potential of substance i in environmental state [J/mol]
Energy efficiency = Qout/Qin where Q states used energy [no unit or %]
Exergy efficiency = Eout/Ein where E states used exergy [no unit or %]
Density [kg/m3]
Temperature in degrees Celsius, see T above [˚C]
Index
i
0
Indicates a substance, i.e. element or summation index
Indicates environmental state
6
EXERGETICS
INTRODUCTION
This guide-book will introduce the exergy concept into engineering work. Especially
fundamental concepts of thermodynamics are treated to understand and be able to apply
the exergy concept. It is offered to engineers of different background and experience.
Thus, some parts might be well-known for some readers, but new for others. If you find
yourself well aware with what is treated — go ahead but make sure you understand the
content being treated.
In the text I use the following hints:
R:
L:
S:
Q:
Read carefully through the indicated text and solve treated examples or problems.
Look briefly through the text indicated to be aware of what is treated.
Solve indicated problems.
Question to answer, maybe a problem to calculate.
Indicated literature is written in short accordingly:
Exergy:
Exergy – a Useful Concept, Wall, G., Ph.D. Thesis (1986).
The following complementary paper is also recommended:
“Exergy, Ecology and Democracy — Concepts of a Vital Society”, Wall, G.,
Energy Systems and Ecology, Int. Conf. Krakow, Poland, July 5-9, 1993, pp. 111121.
These documents are found on the Internet: http://www.exergy.se.
There are also a number of excellent textbooks available, however none of them includes
the statistical treatment of entropy and the link to information theory, being carried out in
this material. This is a selection of most of them:
Ahern, J. E., The Exergy Method of Energy Systems Analysis, Wiley (1980)
Barclay, F. J., Combined Power and Process – an Exergy Approach, MEP (1995)
Barclay, F. J., Fuel Cells, Engines and Hydrogen: an Exergy Approach, Wiley
(1995)
Bejan, A., Tsatsaronis, G., Moran, M., Thermal Design and Optimization, Wiley
(1996)
Bejan, A. Advanced Engineering Thermodynamics (1988)
Brodyansky, V. M., Sorin, M. V., Le Goff, P., The Efficiency of Industrial
Processes: Exergy Analysis and Optimization, Elsevier (1994)
Dincer, I. & Rosen, M. A., Exergy, Energy, Environment and Sustainable
Development, Elsevier (2007)
Edgerton, R. H., Available Energy and Environmental Economics, Lexington
(1982)
Fratzscher, W., Brodjanskij, V. M., Michalek, K., Exergie – Theorie und
Anwendung, VEB, Springer (1986)
Kotas, T. J., The exergy Method of Thermal Plant Analysis, Butterwoods (1985)
Moran, M. J., Availability Analysis – A Guide to Efficient Energy Use, ASME
(1989)
7
EXERGETICS
Sato, N., Chemical Energy and Exergy - An introduction to Chemical
Thermodynamics for Engineers, Elsevier (2004)
Szargut, J., Morris, D. R., and Steward, F. R., Exergy Analysis of Thermal,
Chemical, and Metallurgical Processes, Springer (1988)
Yantovskii, E. I., Energy and Exergy Currents, Nova (1994)
Modern textbooks in thermodynamics also more often treat the exergy concept. Below
are some examples:
Van Wylen, G. J. and Sonntag, R. E., Fundamentals of Classical Thermodynamics, Wiley (1985)
Moran, M. J., Shapiro, H. N., Fundamentals of Engineering Thermodynamics,
Wiley (1995)
An excellent hypertext on physics including thermodynamics is found at:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html. This offers explanations
to most of the concepts and phenomena presented in this text.
Hints: Usually the theory becomes clearer after you have worked through some
exercises. Do not become frustrated if you do not understand everything
immediately. Also be prepared to repeat some sections more than one time. If you
get stuck, sometimes it might be better to read another section to come back later
with a new approach. If nothing else helps you should of course ask for help. You
are always welcome to ask me through my email address: [email protected]
But:
this self-instruction guide needs to be improved, e.g. poor language, misspelling
and misprinting. Please, let me know your corrections or comments to improve
the material!
When: you have worked yourself through this guide, and then you are prepared to study
done exergy analysis of real processes to later be able to carry out your own
exergy analysis.
Good Luck!
8
EXERGETICS
FUNDAMENTAL CONCEPTS
Energy vs. Exergy
Energy vs. Exergy Power
Energy vs. Exergy Efficiency
Energy: The word energy is derived from the Greek: en (in or internal) and ergon (force
or work). The concept was first formulated in the middle of the 19th century by lord
Kelvin and Joule, after many scientists (Benjamin, Thomson, Carnot, Mayer and others)
for decades had tried to find relations between mechanical work, power (horse power)
and heat.
Energy can be defined:
Energy = motion* or ability of motion
Motion might be a train running through the landscape or heat, i.e. moving molecules in a
body of temperature above 0 K. At 0 K or -273.15˚C all motion is assumed to stop.
We differ between different kinds of energy, such as:
Potential energy: A body of mass m [kg] at the height h [m] in a gravitational field
with gravitational constant g [m/s2], i.e. weight mg [N] has the potential energy
Wpotential = mgh.
Kinetic energy: A body of mass m and velocity v [m/s] has the kinetic energy
1
W kinetic  mv2 .
2
Pressure energy or external energy is energy stored as high pressure. A body with
pressure P and volume V represents the pressure energy (cf. work below)
Wpressure = PV.
Internal energy U (molecular kinetic energy) is energy stored in a body usually as
increased temperature. A body of mass m, specific heat c [J/K kg] and increased
temperature T is storing the internal energy
U = mcT.
Electrical energy: just as a weight has potential energy in a gravitational field a
charged particle has potential energy in an electrical field. An electric current of intensity
I [A] and electric potential U [V] during time t [s] is equivalent to the electric energy
Welectric = UIt.
*
Motion in a very general meaning, e.g. the molecular motion in a warm body.
9
EXERGETICS
Chemical energy is energy stored in substances as binding energy between its
components (atoms) and may be released as kinetic energy of the products at a chemical
reaction. Values of some substances are listed below
Substance
gchemical [MJ/kg]
Coal
Oil
Natural gas
Petrol
Dynamite
Firewood
Hydrogen
gchemical [kWh/kg]
32
42
50
44
4
14
124
9
12
14
12
1
4
34
Nuclear energy is in the same way stored energy as binding energy between its
internal components (nuclear particles) and may also be released as kinetic energy of the
final products at a nuclear reaction (fission or fusion). This energy is actually stored as
mass, i.e. we have the famous expression E = mc2 which will be discussed later. Values of
some substances are listed below
unuclear [MJ/kg]
Substance
Uranium ore
Uranium (Light Water Reactor)
Uranium (Breeder Reactor)
Deuterium (Fusion)
Mass (E = mc2)
30
1.9×105
1.3×107
3.5×108
9×1010
unuclear [kWh/kg]
8
5.3×104
3.6×106
9.7×107
2.5×1010
The following forms of energy are more than the other related to a process, that
something happens, i.e. an energy conversion.
Work – if a force F [N] acts over the distance S [m] this is equal to the work:
W = FS.
Electromagnetic radiation as light and heat radiation emitted from a body of area A
[m ] and temperature T [K] during time t [s] gives the energy (“Black body radiation”)
2
Qelectromagnetic = T4At
where  ≈ 5.67×10–8 [W/m2K4] is usually referred to as the Stefan-Boltzmann’s constant.
Heat – can be transferred to or from a body by changing the temperature, i.e. the
internal energy
Qheat = ∆U = U2 – U1 = mc(T2 – T1) = mc∆T.
10
EXERGETICS
Energy and mass
From the theory of relativity we know that energy and mass are equal (cf. nuclear energy
above)
E = mc2.
Einstein showed that the mass of a body depends on its speed according to
m
m0
1
v2
c2
where c is the speed of light in vacuum about 3×108 m/s. Thus, a body has a larger mass
when it is in motion than at rest. (We also note that the speed can never exceed the speed
of light.) The kinetic energy of a body with rest mass m0 and speed v becomes*
E kin  mc
2


 
v 2 
v 2  m v 2
2
2
2
m 0 c  mc 1  1 2   m 0 c 1 1  2  0
c 
2
  2c 

The energy forms we will mainly use are heat, light and mechanical work. Most of
the others may be regarded as subsidiary forms. The transformation between different
forms of energy is called conversion.
Everything that happens involves conversion of
energy
Whenever something happens energy is converted from one form to an other, i.e. an
energy conversion. Everything that can be described as a change in time involves an
energy conversion, from a supernova explosion to a thought of mind. The amount of
energy being converted may be large or small, but without exception energy is converted
in everything that happens.
Condensing power plants, see Fig. 1 below, are main energy converters in the society
of fossil or fissile fuels, i.e. chemical or nuclear fuels into heat and electricity. The
common principal in both cases is to boil water to get steam at high temperature and
pressure that through pipes is forced to move in one direction that can turn a turbine
connected to an electric generator in order to produce electricity.
From experience we know that there is something that always remains no matter what
kind of experiment we perform. This thing, which we do not know what it is, this we call
energy and energy is always conserved. We may also say that energy is everything and if
there was no energy there would not be anything. However, we do not understand what
energy actually is, this is beyond our intellectual capacity.
The SI unit of energy is J (Joule), Wh, kcal, eV. (1 eV = 0.16×10–18 J)
*((1-x)2
= 1-2x+x2≈1-2x when x<<1 and m≈ m0.
11
EXERGETICS
Some SI energy units:
2500 kcal = 2500×1.16×10–3 kWh = 2.9 kWh = 10.4 MJ;
1 kWh = 3.6×103 kJ = 3.6×106J; 1 TWh = 3.6 PJ
1 J = 1 Nm = 1 Ws
Condensing Power Plant
Exhaust gases
Steam
Electricity
Turbine
Boiler
Pump
Heat
Water
Condenser
Generator
Fuel
Cooling water
Waste heat
Fig. 1 The principle of condensing power plants is to boil water.
12
EXERGETICS
Prefix of multiple units
Factor
1024
1021
1018
1015
1012
109
106
103
102
101
10–1
10–2
10–3
10–6
10–9
10–12
10–15
10–18
10–21
10–24
Prefix
Notation
yotta
zetta
exa
peta
tera
giga
mega
kilo
hecto
deca
deci
centi
milli
micro
nano
pico
femto
atto
zepto
yocto
Y
Z
E
P
T
G
M
k
h
da
d
c
m
µ
n
p
f
a
z
y
Definition of 1 cal (between 1 – 100˚C approximately)
1 cal heats 1 g water 1K
1 kcal heats 1 kg water 1K
1.16 Wh heats 1 kg water 1K
4.2 kJ heats 1 kg water 1K
Exercise: How much energy is needed to make 1 kg of water at 5˚C to reach boiling
temperature? Answer in kWh.
Solution:*
Q = ∆U(5˚C100˚C) = mc∆T
where
Q =
U =
m =
c =
*
heat added
internal energy
mass
specific heat [J/kg K]
This will be treated further in the Chapter on thermodynamics below.
13
EXERGETICS
˚C  T + 273.15 K
∆˚C  ∆T K


here:
m
c
∆T
=
=
=
1 kg
4.2 kJ/kg K
∆100˚C – 5˚C = 95˚C “=” 95 K
 kg  kJ  K

 kJ  ≈
Q = ∆U = 1×4.2×95 

 kg  K
≈ 400 kJ = 400×103 J ≈ 400×103×0.28×10–6 kWh ≈ 0.11 kWh
Answer: 0.11 kWh (1 J = 0.2777…×10–3 Wh = 0.2777…×10–3×3600 Ws = 1 Ws)
Energy power*:
Energy power 
Energy
Time
Units for power are, e.g. hp, W, J/s, kWh/yr
1 hp
1W
≈
=
735.5 W
1 J/s
Definition of 1 hp: lifts 75 kg 1 m in 1 second, i.e. 75×9.80665×1/1≈735.5 W
[kg×m/s2×m/s = Nm/s = J/s = W]
Exercise: What energy power is at least needed to make 1 kg of water at 5˚C to boil in 5
minutes?
Solution: Energy power =
0.11 0.11  60
= 1.32 [kWh/h = kW]
5 
5
60
Q: How long time would it at least take with a 500 Watt heater?
Energy efficiency†
Energy efficiency 
Utilized energy
Used energy
N.B.: Efficiency must always refer to a system or a process!
Exercise: What is the energy efficiency if we must use 1.5 kW for 7 minutes in the
previous exercise?
* As we soon will see it is important to distinguish between energy and exergy, which then is valid also for
energy and exergy power.
† Analogously, as for power we also distinguish between energy and exergy efficiency.
14
EXERGETICS
Solution: The energy efficiency becomes =
Q:
0.11kWh 
0.11 60
≈ 0.63 = 63%.

7
1.5

7
1.5  kWh 
60
Try your own stove or heater. What energy efficiency do you get?
Units
Used units refer to the metric system or international system of units* (SI). Please, get
used to them!
Exercise: A. What is the energy power of a hot water-tap if the water is heated from 5˚C
to 70˚C? The flow rate is 0.2 kg/s.
B. What electric current does this corresponds to at a voltage of 220 V?
Solution: A.
Q
m
∆T
cWater
=
=
=
=
∆U = mc∆T
1 kg takes the time 1/0.2 s = 5 s
T1–T2 = (70–5)˚C
1 kcal/kg˚C ≈ 1.16 Wh/kg˚C ≈ 4.19 kJ/kg˚C
Q mc T

t
t
 kg  Wh/(kg  C)   C

11.16  (70  5)
 54288  54  103 
 W
P
5
hr


3600
P
or if you count in SI units mc in J/˚C
P
 kg  kJ   C

1 4.19  (70  5)
 54 
 kJ/s  kW 

5
 kg  C  s

Q: B. Try yourself by using the relation P = UI (electric power=voltage×current).
Q: What is your conclusion from this?
Exercise: Estimate the size of a water tank to support the heating needs of a single family
dwelling, about 20,000 kWh/yr? The water temperature is 80˚C?
Assume the temperature is changed from 80 to 40˚C?
Solution: Q = ∆U = mc∆T =20000×103 Wh
* The International System of Units (SI) ensures world-wide uniformity of measurements and their
traceability see http://www.bipm.org.
15
EXERGETICS
Thus: m 
20000  103
Q
 431034kg  431 ton  431m3

cT 1.16  (80  40)
Proposal of practical arrangement
Energy storage in a water tank
Solarpanel
House
Radiator
Water tank - 400m 3
80°C
The tank is charged during the summer with 20 000 kWh solar energy and reach the
temperature 80˚C by the winter season.
During the winter the tank is discharged 20 000 kWh and the temperature decrease to
40˚C by the summer. However, this proposal is far too simple to be economical. Better
options are to store the heat directly in the ground (clay or rock).
Q:
Consider the relation between heat content and heat losses. What does this imply?
Additional illustrations of energy conversions
Thus, different forms of energy are convertible into each other. When you lift a stone
you perform mechanical work, converting into increased potential energy of the stone. If
the stone is dropped this converts to kinetic energy in the motion, which later converts to
heat as it reaches the ground.
The combustion engine is an engineering application of energy conversions. The
stored chemical energy in the fuel is released through combustion and it is converted to
heat. The heat implies an increased pressure of the enclosed gas in the cylinder which
forces the piston to move, i.e. the volume expands. Through the connecting rod and the
crankshaft the linear motion is converted to a circular motion, i.e. rotation of the shaft.
We have a torque that may turn the wheels of a car thus generating kinetic energy again.
If the engine instead operates a pump the mechanical work may be converted to pressure
energy by decreasing the volume of an enclosed gas, to potential energy by lifting a
liquid to a higher level or to kinetic energy by increasing the flow rate.
16
EXERGETICS
In a hydro power plant the potential energy of water is converted to a torque of the
turbine shaft that is connected to an electric generator, where mechanical energy is
converted to electrical energy. In an electric motor the electrical energy is converted back
to a torque that might operate a fan which may generate potential, kinetic and pressure
energy in a gas.
As we see a certain form of energy may appear in many different parts of a process.
Potential energy is for instance the input in a hydro power plant but might be the output
in a pumping process. Electrical energy has the opposite position in these two cases.
But different forms of energy have different ability to convert into each other. Heat
and electricity are the most familiar forms of energy used in daily life. Of these electrical
energy may almost without exceptions be converted to any other form of energy without
substantially reducing the amount of energy. This makes electricity a high ranked form of
energy, with high availability. Heat or rather internal energy at high temperature can only
partly be converted to other forms of energy, e.g. mechanical energy. The rest of the
energy is converted to heat at lower temperature. If this lower temperature is close to
ambient, e.g. 30-40˚C it is hardly usable at all and must be low ranked. However, this low
ranked heat might be useful for space heating, but when the temperature reaches ambient
it is useless. The usefulness or availability of heat or internal energy is strongly
depending on the temperature of the heat and the environment.
In a furnace we convert the highly ranked chemical energy in the fuel into relatively
high ranked internal energy in the flame (high temperature), which is then converted into
low ranked internal energy in the radiators (30-50˚C). To understand these kinds of
processes exergy is a very useful concept.
Exergy
The word exergy is derived from the Greek ex (out or outer) and ergon (force or work),
(ex = outer, ergos = work, cf. energy: en = internal). The phenomenon behind this
concept was first noticed in 1824 by Carnot in the relation between heat and work.
Exergy can be defined as:
Exergy = work (ordered motion) or
ability to do work (ordered motion)
Exercise: Compare the following two energy conversions. 1 kJ converted to 1 kg of
water as:
(1) Internal energy, i.e. disordered motion of the water molecules.
(2) Kinetic energy, i.e. ordered motion of the water molecules.


1
kJ
U

 0.24  C 
  C  , i.e. a
mc 1 4.2
 kg  kJ/(kg  C)

hardly measurable change of the temperature.
Solution: 1. U  mcT  T 
mv 2
2E kin
2    
v 

 44.7m / s  161km / h , i.e. a
2
m
1
very extensive change of the speed.
2. E kin 
17
EXERGETICS
 J

Nm
(kg×m/s 2 )m
=
=
=m/s 
What about units?: 
kg
kg
 kg

Q:
What happens if we try to make the processes above in the opposite directions?
It is also important to distinguish between energy or exergy power and energy or
exergy efficiency.
Exergy power
Exergy power 
Exergy
Time
Exergy efficiency
Exergy efficiency 
Utilized exergy
Used exergy
Later we will see that everything that happens implies exergy consumption – the
consumption of exergy is actually the driving force of everything that happens.
Everything that happens involves consumption of
exergy
To remember
Energy = motion or ability of motion.
Everything that happens involves conversion of energy.
Energy power 
Energy efficiency 
18
Energy
Time
Utilized energy
Used energy
EXERGETICS
Exergy = work (ordered motion) or ability to do work (ordered motion)
Exergy power 
Exergy efficiency 
Exergy
Time
Utilized exergy
Used exergy
Everything that happens involves consumption of exergy
We will now introduce some thermodynamics and later come back to exergy.
19
EXERGETICS
THERMODYNAMICS OR BETTER THERMOSTATICS*
The history of thermodynamics might be summarized accordingly:
1769, James Watt built the first steam engine, thus starting the development for heat
engines. A theory of heat was needed to improve their performance.
1824, Sadi Carnot claimed that the efficiency of a heat engine relates to temperature,
what later became the so called Carnot factor.
1834, Clapeyron introduced the pressure-volume diagram to calculate work.
1842, Robert Meyer concluded the mechanical heat equivalence, which led to the
definition of the 1st law of thermodynamics.
1845, Joule developed the kinetic theory of gas, which explained heat as the result of
molecular motions.
1850 and -51, lord Kelvin and Clausius formulated the 2nd law of thermodynamics;
Kelvin also introduced the scale of temperature.
1865, Clausius introduced the concept of entropy, which was an important aid to the
theory of thermodynamics.
1872, Belpaire introduced the temperature-entropy diagram.
1873-78, Gibbs presented his phase rule, which increased the usability of
thermodynamics into new areas. Gibbs also established a base for the exergy concept.
1877, Boltzmann suggested that probability or order is linked to entropy.
1904, Mollier introduced the enthalpy-entropy diagram.
1906, Nernst formulates the 3rd law of thermodynamics.
1911, Planck generalized the 3rd law such that entropy and specific heat are zero at
absolute zero temperature for a chemically homogenous body of limited density.
1905 and -15, Einstein published his works on the relation between energy and mass
and the how the mass was related to its speed. Thus, thermodynamics was linked to the
modern physics, theory of atoms and quantum mechanics.
1948, Shannon verified the relation between entropy and probability, which linked
thermodynamics to information theory through the statistical mechanics.
1953, Rant proposed the word exergy.
Thermodynamics may be regarded as the theory of energy and its behavior.
A system in thermodynamic equilibrium may be described by state variables. In
thermodynamics the three state variables usually are pressure P, volume V and
temperature T.
Temperature, heat and internal energy
Temperature may be described as disordered motion of the substances smallest parts –
higher temperature implies faster motions. We distinguish between objects which we call
hot or cold. A hot object we say has a high temperature and a cold object has a low
temperature, but what we identify is based on that we touch the object, establish a
contact. Steel we experience as colder than tree even though they have the same
temperature. The reason is the difference in heat conductivity. Steel transfer heat more
efficient than tree. The temperature should be regarded as a quality of heat, the intensity
of the molecular motions, to be distinguished from the quantity of heat. This sometimes
makes it difficult in practice to differentiate between temperature and heat content, which
also historically was a problem.
*
The aspect of thermodynamics concerned with thermal equilibrium.
20
EXERGETICS
Q:
Why can you walk barefooted on live coal without burning?
Thermodynamics is built up of a number of laws, which also defines a number of
concepts. The zeroth law, which was defined after the first law, concerns the concept of
thermal equilibrium:
Two systems in thermal equilibrium with a third one are in thermal
equilibrium with each other
When two objects are in thermal equilibrium they are said to have the same temperature.
Thus temperature of a system has meaning only if the system is in thermal equilibrium.
Temperature is a concept for the whole system. Thus, it is meaningless to talk about the
temperature of a glass of lukewarm water where you just put a piece of ice.
The concepts of heat and heat content Q states the energy being transported between
systems. Thus, it is principally wrong to say that a system contains heat, instead it
contains energy. This energy, which is the sum of the containing particles’ kinetic and
potential energies, we call internal energy U, or sometimes it is called thermal energy.
One problem of understanding the concept of heat is that we can not experience heat,
instead what we experience when we feel “heat” is heat transfer.
mc(T)
Q
2
1
Q
T1
dT
T2
T
Assume we add the heat Q, to increase the temperature from to  of a
substance of mass m, and specific heat c, i.e. we change the internal energy from U till
U We have:
U2
Q
T2
 dU  m  cdT
U1
 mcˆ (T 2  T1 )  m cˆ T
T1
Q
U
m
c
=
=
=
=
C
T
cp
=
=
=
heat [J]
internal energy [J]
mass [kg]
specific heat capacity [J/kg K], the heat needed to raise the temperature
1 K of 1 kg of the substance. c often increases with temperature (but
not for water between 0 and 33.5˚C where it decreases!?). cˆ is the
average value of c in the temperature interval (T1, T2). c is also
depending on how the state changes occurs when the temperature
raises (see below for cp and cv).
mc = heat capacity [J/K]
temperature [K]
specific heat capacity when the state change occurs during constant
q 
dh
pressure (P = constant), i.e. c P 
. The concept

T P  constant dT
enthalpy, H or specific enthalpy, h we will define and examine below.
21
EXERGETICS
cv
=
specific heat capacity when the state change occurs during constant
q 
du
volume (V = constant), i.e. c V 
.

T V  constant dT
Thus, the following holds: (1) supplied heat at constant pressure is equal to the increase
in enthalpy and (2) supplied heat at constant volume is equal to the increase in internal
energy.
For liquids or solids cp ≈ cv since these usually can be regarded as incompressible.
Gases needs more heat to raise the temperature at constant pressure than at constant
volume, i.e. cp>cv. (The gas increases the volume to maintain constant pressure, i.e.
performs work by expansion on the environment in addition to the change of internal
energy given by the temperature change).
For gases the relation:
cP

cV
has a typical value for specific gases. At low pressures, i.e. more ideal gases,  becomes
almost constant accordingly.*
One atomic gases as inert gases:
 ≈ 1.66
Two atomic gases as N2, O2, H2 and air:
 ≈ 1.40
Three atomic gases as CO2 and steam H2O:
 ≈ 1.30
Before we look closer at the ideal gas model we need to define some further concepts:

 = density [kg/m3]
v = specific volume [m3/kg], v 
V 1

m 
M = molar weight of gas [kg/kmol], e.g. MH2 ≈ 2, MAir ≈ 29 and MO2 ≈ 32
S:
Calculate the molar weight of Carbon Dioxide and steam.
The ideal gas model
According to above the equilibrium state of a system is described by the state
variables P, V and T. These are not independent of each other – for all substances there
are state equations f(P,V,T) = 0. If the state equation is known all state variables may be
calculated if we know two of them.
I reality the state equation of a substance is often too complicated to derive
analytically, but from experiments we may get empirical relations. With modern
*
The heat capacity may be regarded as the capacity of a substance to store energy by the embodied
molecules. This can be done as kinetic energy from three dimensional motion and rotation. This energy
depends linearly with temperature, thus not adding any temperature dependens from the heat capacity. This
is valid for single atom gases as Ar, Ne and He. Contribution from vibrations and from electrons implies
that the heat capacity becomes temperature dependent, especially for two atoms gases as H2, O2 and air. For
many atoms gases as CO2 and H2O we have additional contributions from more possible states of
vibrations, which make an even stronger temperature dependance in the heat capacity.
22
EXERGETICS
computers, we may even describe liquids with fairly good correspondence between
theory and practice.
An ideal gas is a gas where the molecules only interact by collisions, and they do not
occupy any space, i.e. they are infinitely small.
For ideal gases there are three important experimental results:
Boyles-Mariotte’s law from 1660:
P
PV = constant, when T = constant
V
Charles’ law stated 1787:
P
P/T = constant, when V = constant
T
Gay-Lussac’s law from 1802:
V
V /T= constant, when P = constant
T
All these empirical relations can be summarized in a state equation:
PV = mRT
named the ideal gas equation, where:
P = pressure [N/m2 = Pa (Pascal)]
V = total volume [m3]
m = mass [kg]
8314.3
) [J/kg K] were M is the molar mass of
R = specific gas constant, R 
M
the substance, sometimes this symbol is also used for the universal gas
constant, i.e. R = 8314.3 [J/kmol K], which is experimentally measured.
23
EXERGETICS
T = temperature [K]
If we insert the specific volume
v
V
m
in the ideal gas equation we get:
Pv = RT
For the ideal gas we also have:
cP – cV = R
Q:
Derive this relation from the definitions of cP, cV , the enthalpy (h = u + Pv) and
the ideal gas equation.
Ex:
A specific amount of air has the temperature 290 K and pressure 0.2 MPa. What is
the pressure if the temperature increases to 310 K at constant volume? (For
instance car tires subject to heat by friction.)
S:
Assume ideal gas!
before: P1V = mRT1
after: P2V = mRT2
V = constant m = constant
P1 T1
PT
2  10 5  310
5

 P2  1 2 
 2.14  10 Pa = 0.214 MPa, i.e. 7%
P2 T2
T1
290
pressure increase. (We can also use Charles’ law.)
Ex:
If the volume in the previous exercise is 10 m3 how much heat is needed?
S:
Added heat becomes Q  U2  U1  mcˆ V (T2  T1 )
From the ideal gas relation we get the mass of the air:
m
P1 V
RT1
The state change appears during constant volume, but we only have data for cP.
c
cP
   1.4  c V  P
cV
1.4
cˆ P ≈ 1005 [J/kg K] (Average value between 290 and 310 K, however cˆ P ≈ 1.00
[kJ/kg K] is also of course acceptable).
24
EXERGETICS
PV
cP
2 105 10 103 1005
1
(310  290)  341.84  342  J 
Q

 (T2  T1 ) 

8314.3
1.4
RT1 1.4
 290
29
Nm 3 kg K J K

 Nm  J 
2

 m J K kgK

Work, Mechanical Work and Pressure-Volume Work
From Mechanics we have the following well-known relation: Work is equal to the Force
times the Distance,
W = FS
where: W = work [Nm = J]
F = force [N]
S = distance [m]
Assume a cylinder with a mobile piston (no friction) with the cross section area A. In the
cylinder we have a gas with pressure P.
State 1
P
S1
S2 – S1
S2
State 2
P
What work is done by the gas if the piston is allowed to move a short distance, so that it
does not influence the pressure?
If A is the area which is subject to the pressure P, the force becomes F = PA
The piston moves the distance S2 – S1
The work done by the piston, assuming constant pressure becomes:
W  F(S 2  S1 )  PA(S 2  S1 )  P(V2  V1 )
Since the volume in state 1 is V1 = AS1 and in state 2 is V2 = AS2.
If we allow the piston to move out a longer distance the pressure will drop according
to the ideal gas equation PV = mRT.
25
EXERGETICS
1
P
V1
2
dV
V2
V
The work done by the volume change dV is:
W = P(V)dV
The total work when the volume changes from V1 to V2 becomes:
V2
W   P(V )dV
V1
W is sometimes called pressure-volume work.
If we draw P as a function of V, i.e. P(V) in a PV-diagram, then W becomes the area
under the curve P(V) and between V1 and V2, i.e. the indicated area in the figure above.
Thus, the work performed by the gas depends on the shape of P(V), i.e. how the state
change occurs between 1 and 2.
The 1st Law of Thermodynamics — Nothing disappears
Energy can not be created nor destroyed.
A closed system: is a system where matter is not allowed to cross the system
boundary. The cylinder above is a closed system. The piston represents a moving
boundary allowing work to be extracted, i.e. energy to cross. In physics, a closed system
can exchange heat and work (energy), but not, with its surroundings. In contrast an
isolated system can exchange neither heat nor matter with the surroundings.
The total energy of an isolated system is constant. The energy before and after a
change of state is the same, i.e. we have an energy balance. Energy is defined as that
which is always conserved in every process, i.e. energy is always in balance.
Consequently, it is wrong to say energy production, instead use energy supply. By the
same reason energy consumption is impossible, what we actually mean is that an energy
form is consumed or that the energy is converted or maybe used but not consumed.
26
EXERGETICS
An open system is a system where mass may cross the system boundary. The mass of
the system is determined from in and out flows, mi and me (exit), see the figure below.
System
boundary
mi
me
Ex:
Water is falling 55 m at the rate of 500 m3/s. Estimate the temperature rise, if we
assume that all potential energy heats the water.
S:
From energy conservation the potential energy, Epot = mgh [J], completely
converts to increased internal energy, ∆U = mc∆T, i.e. the temperature increases.
E pot  U  mgh  mc T  gh  c T  T 
gh 9.81  55

 0.13K 
c 4.2  103


 Nm

Nmkg K NmK





K
Units?
kg J
J
 kg J

 kg K



Note that the flow rate is not needed.
Q:
If you try to verify this by experiment, you might find that the temperature drops.
Why, do you think? (Think about pictures of a water fall, e.g. the Niagara falls.)
Ex:
Estimate the energy power from the water fall above?
S:
 = 500 ton/s = 5×105 kg/s, i.e. in 1 second the energy Epot = mgh is
Mass flow: m
converted. The energy power P, which is energy per unit time then becomes:
P
E pot
t

mgh 5  10 5  9.81 55

 269,775,000 J / s  270 MW
t
1
 kg Nm Nm J


  W
Units? 
s
s
 skg

We will now introduce enthalpy, H, which assigns the total energy of a system, i.e.
the sum of the internal and external energies.
If the external energy is PV [J], where P = pressure [Pa = N/m2] and V = volume [m3] and
the internal energy is, U [J], then the enthalpy is
H = U + PV
27
EXERGETICS
Work and Technical Work
Closed system
Assume a system where no matter can cross the system boundary. We add heat, Q and
extract work, W and the internal energy changes from U1 to U2, but nothing else happens.
The 1st Law then becomes
Q = U2 – U1 + W = ∆U + W
Q = heat added [J]
U1 = internal energy [J] before heat is added
U2 = internal energy [J] after heat is added
W = extracted work [J].
Note that we regard input heat and output work as positive.
where
Q
W
SYSTEM
U1 U2
(Warning! By some authors input work is positive!)
Steady-State processes
In a steady state process the total mass and the total energy enclosed by the system
boundary is always conserved. This implies that the total amount of mass and energy
input must balance the total output. Such systems are pumps, compressors, fans, turbines,
power plants, engines, etc.
Assume a process with different inlet and outlet pressures.
mi
Pi
System boundary
me
m i = me = m
Pe
The mass input is mi, and from conservation the mass output me is the same as the
input, i.e. mi = me = m. Assume that the input specific internal energy is ui [J/kg]. The
surrounding air performs a specific external work Pivi [J/kg] on the input mass.
28
EXERGETICS
Analogously for the output specific internal energy ue and the mass performs an external
work Peve on the surrounding air.
We also add heat Q [J] to the system, which also performs work Wt [J], we call this
technical work.
m(Pi vi + u i)
Q
SYSTEM
Wt
m(Pe ve + ue )
The total energy of the system is constant according to the 1st Law, i.e. energy input is
equal to energy output.
Q  m(Piv i  u i )  m(Pev e  u e )  W t
By introducing the enthalpy concept h = Pv + u we get.
Q  mh i  mh e  W t  Q  H i  H e  W t Q  He  Hi  W t  H  W t
An important case is when no heat is added, Q = 0, i.e. an adiabatic process.
W t  Hi  H e = ∆H
Thus, the technical work is equal to the enthalpy change in an adiabatic process.
However, for a Steady-State process we also have that Q = ∆U + W = U2 – U1 + W,
where i=1 and e=2. U and W refers to the mass element m going through our system. We
have
Q = ∆U + W = ∆H + Wt
U2 – U1 + W = U2 + P2V2 – U1 – P1V1 + Wt
W = P2V2 – P1V1 + Wt
V2
W t  P1 V1  P2 V 2   PdV
V1
V2
since W   PdV
V1
29
EXERGETICS
P
1
P1
Wt
2
P2
V1
V2 V
If we indicate the pressure as a function of the volume, P(V), in a PV-diagram we see that
Wt is the indicated area in the diagrams above and below.
P
P1
P
1
V1
1
P1
2
P2
P
V2 V
P1
2
P2
V1
V2 V
V2
P1V1
P1V1   PdV
V1
The area can also be defined by the integral
P1
P2
P2
P1
Wt   VdP    VdP
30
1
2
P2
V1
V2 V
V2
P1V1   PdV P2 V 2
V1
EXERGETICS
Thus, we may write the energy balance for a closed system
V2
Q  U 2  U 1   PdV
V1
and for an open system
P2
Q  H 2  H1   VdP
P1
Assume the heat is added during constant volume, dV = 0, then we have
T2
Qv  m  c v (T )dT  mc v (T2  T1 )
T1
which in combination with the energy balance of a closed system gives
Qv  mc v (T2  T1 )  U2  U1  U
Thus, for a closed system we have that the heat transferred at constant volume is equal to
the change of the internal energy.
Instead, assume the heat is added during constant pressure, dP = 0, then we have
T2
Q P  m  cP (T )dT  mc P (T2  T1 )
T1
which in combination with the energy balance of an open system gives
QP  mc P (T2  T1 )  H2  H1  H
Thus, for an open system we have that the heat transferred at constant pressure is equal to
the change of the enthalpy.
This is valid for ∆U and ∆H at all change of state.
Ex/S: A closed isolated volume of 2 m3 with air of pressure 0.5 MPa and temperature
293 K, receives 2500 kJ of heat.
The internal energy?
We have a closed system, thus
Q = ∆U + W
No work is done  ∆U = Q = 2500 kJ
31
EXERGETICS
Final temperature of the gas?
V  const.  Q  mc v  T  T 
Q
mc v
The mass we can get from the ideal gas relation.
Air is approximately an ideal gas with  ≈ 1.40, and for an ideal gas we have
c P  c v  R
R
 c P    c v (   1)  R  c v 
 1
 c v
The temperature change:
T 

Q
Q
QT1 (  1)
 PV


R
mc v
P
1 1
V
1
1

RT1   1
2500   3  293(1.4  1)
 293K
5  105  2
The final temperature 2 = 1 + ∆ ≈ 586 K
The final pressure?
From the ideal gas relation we get
State 1:
P1V1 = mRT1
State 2:
P2V2 = mRT2
By dividing these relations we get
P1V1 T1
but since V1 = V2

P2 V 2 T 2
we get the final pressure
P2 
P1 T2 5  10 5  586
5

 10  10 Pa  1MPa
T1
293
Ex:
Air is compressed from 10 to 60 MPa, the temperature increases from 293 to 373
K and the flow rate is 3 kg/s. What is the work needed?
S:
Assume we can neglect the heat leakage from the compressor, i.e. an adiabatic
process.
32
EXERGETICS
mi
Pi T i
System boundary
me
m i = me = m
Pe T e
The process is an adiabatic steady state process, i.e. Q = 0. The work output
per second then becomes
Wt  H i  H e  m(hi  he )  mcˆP T  3 1009  (293  373)  242  kJ 
The minus sign indicates that work is needed, 242 kJ per second, i.e. the needed
exergy power is 242 kW.
State variables and State changes
Volume V, temperature T, enthalpy H and pressure P are called state variables. A
state variable has a specific value for a certain state independent from how the state was
attained. Thus, a state variable does not contain the history of the state. Work W,
technical work Wt and heat Q are depending on how the state of the substance changes,
i.e. they are “process related.”
Variables may also be extensive, i.e. they are depending on the size of the system, as
volume V and enthalpy H, or intensive, i.e. they are not depending of the size, as pressure
P and temperature T.
The four most common state changes or processes are defined
Isochoric process  V = constant, i.e. dV = 0
Isobaric process  P = constant, i.e. dP = 0
Isothermal process  T = constant, i.e. dT = 0
Adiabatic process Q = 0, no heat lost or added to the system. (This is also called
isocaloric. Note the “curly” differential sign since the change is not
unambiguous because it depends on the state change, as we just mentioned
above.)
In all four cases we assume that the state changes are reversible. This means that the
process is ideal, i.e. no losses occur when the system changes from 1  2 and back again
2  1. In reality there are no reversible processes. In real processes we have always
33
EXERGETICS
losses – more or less. A real process must be irreversible, i.e. not reversible, to have a
direction.
Let us now see what this means for the work W, technical work Wt and heat Q for an
ideal gas, i.e. PV = mRT.
Isochoric process 
P
 constant
T
p
1
Wt
2
V
V2
W   PdV  0
V1
P2
W t   Vdp  V ( P1  P2 )
P1
T2
Q  m  c V (T )dT  m ˆc V (T2  T1 )
T1
where cˆ V is the average value in the temperature region [T1,T2].
Isobaric process 
V
 constant
T
p
1
2
W
V
V2
W   PdV  P(V 2  V1 )
V1
34
EXERGETICS
P2
W t   VdP  0
P1
T2
Q  m  c P (T )dT  m ˆc P (T 2  T1 )
T1
where cˆ P is the average value in the temperature region [T1,T2].
Isothermal process  PV = constant
p
p
1
1
Wt
2
2
W
V
V2
W   PdV 
V1
V
V2

 
mRT
V 2 
  mRT ln  P1 
dV

mRT
ln
 V
 V1 
P2 
V1
P1
Wt   Vdp 
P2
Thus we have W = Wt.
P1
P
mRT
dP  mRT ln 1 
P
 P2 
P2

V2
Q  U 2  U 1   PdV  W  W t .
V1
T2
since U 2  U 1  m  c(T)dT  0 .
T1
Adiabatic process Q = 0. For a reversible adiabatic process we have Poisson’s
relation PV = constant, which also can be written TV-1 = constant or TP
p
p
1
1
Wt
2
2
W
Q:
V
Derive Poisson’s relation, i.e. show that PV = constant.
35
V

 1

 constant .
EXERGETICS
Hint: 1st Law and adiabatic process (Q = 0)  PdV + dU = 0, with dU = mcVdT
we get PdV + mcVdT = 0. By differentiating the ideal gas model we get: PdV +
R
VdP = mRdT. Eliminate dT and we have: PdV  VdP 
PdV or
cV
R
c
(1  ) PdV  VdP  0 . By using that cP = cV + R and   P we have:
cV
cV
dV dP
which after integration becomes – lnV = lnP + constant or PV =


V
P
constant. From the ideal gas model PV = mRT we also get the relations for T and
P and for T and V as above.
Work is
V2
W   PdV
V1
but from the 1st law we have
V2
Q  U 2  U 1   PdV  0 ,
V1
which gives
V2
W   PdV  U1  U 2  U .
V1
but
T2
 U  m  cV (T )dT
T1
which gives
T2
W  m  cV (T )dT .
T1
The technical work is
P2
W t   VdP
P1
but
P2
Q  H 2  H 1   VdP  0 ,
P1
and
T2
 H  m  cP (T )dT
T1
which gives
T2
T2
T1
T1
W t  m  c P (T )dT  m   c V (T )dT .
thus, we have Wt = W.
We will now study the polytropic process, which occurs during heat exchange with the
environment.
The process is characterized by
PVn = constant
36
EXERGETICS
n is the polytropic coefficient and can have any value larger than zero.
A polytropic process summarize all the
processes above, where the value of n
indicate the kind of process we have
Isochoric process
n=∞
Isobaric process
n=0
Isothermal process
n=1
Isentropic process
n = 

Q:
Show that when n  ∞ a polytropic process becomes an isochoric process.
1
Hint: PVn = constant we can write as V n   constant or
P
1
0
1 n
1
V   
 constant   
 constant  constant when n  ∞

P
P 
Ex:
A gas with cP = 1.0 kJ/kg K and  = 1.4 receives the heat 100 kJ/kg. How much
will the temperature rise if the heat is added at 1) constant pressure 2) constant
volume?
S:
1) Q  mcP (T2  T1 )  mcP T  T 
Q
100 103

 100 K .
mcP 11.0 103
Q  mcv (T2  T1 )  T 
Q 100 103 1.4



T


 140 K
cP
2)

3

mc
1
1.0
10


P

cV

Ex:
A gas with  = 1.4 and the initial values P1 = 0.4 MPa, V1 = 3 m3 and T1 = 473 K
is adiabatically expanding to V2 = 9 m3. Estimate final pressure, temperature and
the works W and Wt.
S:
For an adiabatic process we have PV = constant, where  

cP
, i.e.
cV

 V1 
5 3 


P1V1  P2 V 2  P2  P1
 4  10    0.86  10 5 Pa .
V 2 
9


 1
TV
 1
V 
 constant  T2  T1  1 
V2 
37
1.41
3
 473   
 305K
9 
EXERGETICS
V2
V2
V1
V1
W   PdV  constant 

1
1
 V  dV    1 (P V
1
1
 P2 V2 ) 
1
(0.4  10 6  3  0.86  10 5  9)  1.04  10 6 J  1.04  J
1.41  1
Wt = W ≈ 1.4×1.06≈ 1.4J
The 2nd Law of Thermodynamics — Everything disperse
By experience we know that heat spontaneous goes from a warm body to a cold body,
but never the opposite. We also know that the energy of heat can not completely be
transformed to work, even if we disregard losses as friction and heat losses. This implies
the second law, which has been formulated in many ways:
Clausius:
Heat can not “by itself” go from a low to a higher temperature.
Lord Kelvin:
Heat can not completely transfer to work.
A more poetic interpretation of the second law is offered in the poem of science fiction
written by the Swedish Nobel laureate Harry Martinson in 1956 Aniara, which is a story
of the space craft Aniara, that during a journey through space loses its course, and
subsequently aimlessly floats through space, without destination. As the space travel
continues, things become increasingly worn out, and many passengers are dying. The vast
halls of the golgonder are cold and empty as told in poem 99:
I paced the halls and it was very late,
paced Mima’s hall one night and I was cold.
Still colder, far from all things temperate
roared memory in my soul for Dorisworld.
Ever more mute and numb lay Aniara’s ship:
a proud goldonder once, now sarcophagus
which, having lost all power, through empty space was flung
in line with the loxodrome
to which in her fall she clung.
From the concept entropy S, which was introduced by Clausius the year 1865, we
may also make a formulation of the second law.
As we saw above work W and technical work Wt can be illustrated as areas in a
pressure/volume diagram PV-diagram. Similarly heat Q can be illustrated as an area in a
temperature/entropy diagram TS-diagram, see below.
38
EXERGETICS
Q
T
2
1
S1
S2
dS
S
If the area below the graph T(S) and between S1 and S2 is the heat content Q, then we have
Q = TdS
or
dS 
Q
T
,
with the unit J/K. When heat Q is transferred to a system it is the temperature T of the
heat that predicts the entropy dS that simultaneously is being transferred. By integration
we get
2
2
1
1
S   dS  S2  S1  
Q
T
The value of the integral is independent of the integration path, i.e. it is only depending
on the values of S1 and S2. Thus, the entropy is a state variable, i.e. a function only
depending on the state of the system. Therefore we may express the entropy as a function
of the state variables pressure, volume, and temperature, S(P,V,T). We will also notice
that the entropy is an extensive variable, i.e. depends on the size of the system as mass,
volume, and internal energy.
The 1st law says that energy is always conserved in any process. The 2nd law states
that every process mostly take place in such a way that the entropy is constant for a
reversible process or increases for an irreversible process, i.e.
∆S ≥ 0
Everything that happens implies an increase of the total entropy. Locally the entropy may
decrease but only if the entropy increases even more somewhere else so that the total
entropy increases.
Entropy may also be regarded as a value of order. (This we will treat in the section on
exergy of information.) Thus the 2nd law states that everything moves towards increased
disorder – increased disintegration. Locally we may still create increased order. In the
case of creating order on the earth, as in the living nature, this is established by the order
offered by the sun and the disorder offered by the space.
So far all the state changes we have studied have been reversible. A reversible process
is a process where you can always return to the initial state by running the process
backwards. Such a process has no losses and assumes all processes to take place during
thermodynamic equilibrium, i.e. no differences in temperature and pressure. This makes
39
EXERGETICS
such processes infinite in time and unrealistic. By the second law we may say that we
consider the irreversible processes which we have in reality.
Assume we have two containers with different gases, and . We know which gas we
have in each container, i.e. our system is ordered. If we open the valve connecting the
containers the gases will spontaneously diffuse into each other and we have lost some
order since we do not know where the -gas or -gas is, S increases. We may not predict
the exact distribution of and in the two containers.
Case 1: Order  S1 is small.
Case 2: Disorder  S2 > S1.
Also, we do not expect that the gases will spontaneously return to the initial state.
Let us see what this implies for different substances.
For a solid or liquid substance, i.e. an incompressible substance we have Q ≈ dU ≈
mcdT, which gives
T2
2
T 
mcdT
 S   dS  
 mc ln  2 ,
T1 
T
1
T1
where we have assumed the specific heat capacity to be constant.
Ex:
S:
1 kg water is heated from 20 to 90˚C. Estimate the entropy change when the
specific heat is constant and compare with steam tables.
T 
273.15  90 
 S  mc ln  2  1  4.184 ln 
 0.8959 kJ/K.
T1 
273.15  20 
The steam tables give ∆s ≈ 1.1925 – 0.2966 = 0.8959 kJ/kg K.
Thus, a very good resemblance.
For an ideal gas the relation becomes more complicated. From 1st law we have:
Q = dU + PdV
and for an ideal gas we have:

dU  mc V dT
P mR


T
V
Thus, the entropy change becomes
2
S  
1
V 
c
mc V
mR

dT  
dV  m  V dT  mR ln 2 
 V1 
T
T
T
V
1
1
1
Q
2
2
2
40
EXERGETICS
If the specific heat is constant, i.e. independent of the temperature we get
T 
V 
 S  mc V ln 2  mR ln  2 
T1 
V 1 
Analogously we have
Q = dH – VdP
and for an ideal gas we have

dH  mc P dT
V mR


T
P
Thus the entropy change becomes
2
S  
1
P 
c
mc P
mR
dT  
dP  m  P dT mR ln  2 
P1 
T
T
P
1
1
1
2
Q

T
2
2
If the specific heat is constant we get
T 
P 
 S  mc P ln 2   mR ln  2 
T1 
P1 
Ex:
In a container 2 kg of steam is condensed at temperature T = 100˚C, and a heat Q
= 4514 kJ is transferred to the surrounding air at temperature T0 = 20˚C. Calculate
the change in entropy in the container S, and the environment S0. Also calculate
the total entropy change Stot. The temperature of the environment is constant.
Q
T0
T
S:
If we assume that the heat is transferred at constant temperature we have for the
entropy
S
Container:
Q
T
4514
Q

 12.10 kJ/K, the minus sign indicates that
T 273  100
entropy and heat is transported out of the system.
S
Environment: S 0 
Q
4514

 15.41 kJ/K.
T0 273  20
41
EXERGETICS
The total entropy change becomes:
Stot = S +S0 ≥ –12.10+15.41 = 3.31 kJ/K ≥ 0, i.e. in accordance with the 2nd law.
We can summarize the following:
Laws
0th law defines the concept of temperature
1st law: Q = dU + W, defines energy as a conserved quantity and work: W = PdV
2nd law: dS ≥ 0, dSreversible = 0, defines the concept of entropy through heat: Q = TdS
3rd law defines the zero level for entropy
Concepts
Enthalpy: H = U + PV
Heat capacity: Q = CdT, Cv and CP
Model
Ideal gas: PV = n R T = mRT
Processes
n  0 isobaric
n  1 isothermal

Ideal gas: PVn = constant n  C P adiabatic

Cv

n


isochoric

Principally this is all you need to understand of classical thermodynamics to be able to
manage this part of science. All other relations can be derived from these relations.
Remember!
The energy of a closed system is constant (1st law).
Energy forms may be destroyed and created, consumed and produced.
Energy production should be called energy supply
Energy consumption should be called energy use
42
EXERGETICS
Additional examples and solutions:
We are now ready to study additional examples to repeat thermodynamics.
Try to solve the problem before looking at the solution. If you do not succeed, look at
the solution part by part and try to solve as much as possible yourself. Simultaneously,
repeat appropriate parts of the theory.
Ex:
A bottle of the volume 0.02 m3 contains hydrogen at 120 bar, 10˚C. What is the
weight of the gas?
S:
V = 0.02 m3
P =120×105 N/m2
8314 8314
R=
= 4157 J/kg K

M H2
2
T =273.15+ ≈ 273+10 = 283 K
PV 120  105  0.02

≈ 0.204 ≈ 0.20 kg
m=
RT
4157  283
Ex:
3 kg air of temperature 100˚C expands isothermally in a closed volume from 0.1
m3 to 0.3 m3. How much heat must be added?
m = 3 kg, T ≈ (273+100) K = 373 K, V1 = 0.1 m3 and V2 = 0.3 m3.
V 
For a isothermal process we have: Q  mRT1 ln  2 
V1 
8314 8314
The specific gas constant becomes R air 

 287 J/kg K.
M air
29
 0.3 
  352 kJ ≈ 98 Wh.
Thus Q  287  3  373  ln
 0.1 
In PV-diagram:
S:
W = Q since Q = (U2 – U1) + W where
U2 – U1 = mcv(T2 – T1) = 0.
Ex:
A closed container of 20 liter with air at 20˚C, 1 bar receives heat until the
pressure reaches 3 bar. How much heat is received?
43
EXERGETICS
S:
Heat is being transferred at constant volume, thus the process is an isochor.
R
Transferred heat is, Q = U2 – U1 = mcv(T2 – T1), cv 
, air = 1.4,
 1
8314
8314
R air 
 cv 
 716.7 J/kg K
29
29  0.4
The ideal gas model gives:
1 105  0.02
P1V1


RT1 8314  (273  20)
29
≈ 0.0238 [kg]
m
We get: T 2 
3
P2
T1   293  879 K
P1
1
And the added heat becomes: Q ≈ 0.0238××(879 – 293) ≈ 10 kJ
Ex:
Air (2 kg/s) at 6 bar, 400˚C expands in a turbine to the ambient pressure 1 bar.
What is the maximal electric power to be utilized?
P1 = 6 bar
T1 ≈ (273+400) = 673 K
P2 = 1 bar
dm
m 
 2 kg/s
dt
S:
The process is a steady state adiabatic process
Thus: Wt = W = H1 – H2 = mcP(T1 – T2)
 P (T1  T2 )
The power Wt is work Wt per time  Wt = Wt per second = mc
T P
We have: 1   1 
T2  P2 
 1

6
 
1
1.4 1
1.4
 60.286  1.67 , i.e. T 2 
The average value of cˆ P becomes: cˆ P 
44
673
 403 K.
1.67
c P (T1 )  c P (T2 )
 1039 J/kg K
2
EXERGETICS
(The specific heat is only depending on the temperature, since we assume an ideal
gas, thus independent of the pressure difference P1 – P2)
 kg  J  K

 J/s  W   561.1  560 kW
We get Wt  2 1039  (673  403) 
 s  kg  K

45
EXERGETICS
Exercises
Try to solve these problems by yourself, without checking the solutions.
1.
Which alternative use most energy? a) To remove the snow by truck that uses 5 liter
diesel to remove 5 tons of snow. b) To melt the snow by a diesel burner with 80%
efficiency. The enthalpy of melting is 334 kJ/kg. How much diesel will be used for 5
tons of snow?
2.
Snow equivalent to 20 mm water falls over a city. What is the energy difference per
m2 if this instead was as rain? The difference causes an extra cooling effect, but also
other effects. What other effects?
3.
How much fuel oil is needed to raise the temperature of the indoor air from 0˚C to
20˚C in a house of 125 m2 and height 2.4 m? The efficiency of the heater is 65%.
How is this effected if the volume is fixed or if the volume expands?
4.
10 kg hydrogen gas is heated 100K at constant volume. What is W and Wt? (Hint:
use the definition of technical work and the ideal gas model!)
5.
What is the volume of the gas in the previous problem if the initial state was +27˚C
and 1 bar? (Hint: use the ideal gas model.)
6.
148 kJ of heat is isothermally added to 1 kg air of volume 0.5 m3, so the final
volume becomes 2 m3. What was the initial temperature if Rair = 287 J/kg K? (Hint:
use 1st law and the ideal gas model for an isothermal process!)
7.
What was the initial pressure in the previous problem? (Hint: use the ideal gas
model.)
8.
A gas with  = 1.30 expands adiabatic from P1 = 6 bar to P2 = 2 bar. How many
percent does the volume increase? (Hint: adiabatic process.)
9.
The air of a combustion engine is compressed polytropically from P1 = 0.9 bar and
T1 = 40˚C. Determine the final pressure and temperature if the initial volume is 9
times the final volume and the polytropic coefficient is n = 1.35. (Hint: polytropic
process.)
10. A closed container of 5 liter contains air at 1 bar, 20˚C. The air is heated to 50˚C.
Calculate a) heat added b) final pressure (Hint: R = 285 J/kg K and the ideal gas
model. Calculate the average value of cP between 20-50˚C, and use the relation for
cP and cv.)
11. Air is adiabatically compressed from 1 bar, 20˚C to 6 bar. Calculate a) work needed
b) temperature after compression c) change of internal energy of the air. (Hint:
adiabatic process.)
12. Consider 0.6 m3 of air at 2 bar, 20˚C. Calculate a) internal energy, U b) enthalpy, H
if both of them are 0 at 1 bar, 0˚C. (Hint: isothermal and isobaric process.)
46
EXERGETICS
Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
To melt the snow “costs” about 60 liter diesel.
6.7 MJ/m3 or 2 kWh/m3
12 MJ or 3.3 kWh
W = 0, Wt = – 4200 (4157) kJ, i.e. we must add the technical work 4.2 MJ
The volume of the gas is 125 m3
The temperature of the air was 373 K ≈ 100˚C
The initial pressure was 2.1×105 Pa
The volume increase 130%
The final pressure is 17.5×105 Pa. The final temperature is 676 K ≈ 403˚C ≈ 400˚C
a) Added heat 130 (128) J, b) Final pressure 110 kPa
a) We add the work 200 (199) kJ/kg, b) Final temperature becomes 220 (216)˚C,
c) ∆U ≈ 140 kJ/kg
12. a) Internal energy at 20˚C is 20 kJ b) Enthalpy at 20˚C is 28 kJ.
Solutions
1.
To melt 1 kg snow you need 334 kJ  5 tons of snow need 5000×334×103 J ≈
1.7×109 J ≈ 1.7 GJ.
Heat of melting H melt
Efficiency is 80%, i.e.

 0.8
Q input
Heat input
H
1.7  10 9
 Qinput  melt 
 2.1  109 .
0.8
0.8
1 liter diesel contains about 10 kWh or 36 MJ.
2.1  10 9
 58  60 liter, which is more then 10 times
Needed amount becomes:
3.6  107
what is needed for truck removal.
2.
Per square meter we get 20×10–3 m3 or 20 liter ice as snow, i.e. about 20 kg snow. To
melt 1 kg snow we need 334 kJ  20 kg snow need 20×334×103 ≈ 6.7×106 ≈ 6.7 MJ
≈ 2 kWh, i.e. the energy difference between snow and rain is about 6.7 MJ/m2 or
2 kWh/m2.
3.
Amount of air becomes 125×2.4 [m3] ×1.3[kg/m3] ≈ 390 kg.
Added heat is Q = cm∆T ≈ 1.0 [kJ/kg K] ×390[kg] ×20[K] = 7.8×106J.
Q 7.8  106
The need of fuel oil is Q fuel  
 12  106 J  3.3 kWh or about 1/3

0.65
liter.
4.
Work W   PdV  0 , because an isochor process, i.e. dV = 0.
Technical work W t   VdP (ideal gas model gives)  mR  dT  mR T 
8314.3[J/kmolK ]
 100[K ]  4.157  106 J  4.2 MJ, i.e. we must add the
2[kg/kmol]
technical work 4.2 MJ.
 10[kg] 
47
EXERGETICS
5.
The ideal gas model gives: V 
mRT

P
10[kg ] 
8314.3[J/kmolK ]
  [K ]
2[kg/kmol]

1 105[Pa ]
125 m3.
6.
1st law gives: Q  U  W   mcv dT  W  0  W , i.e. Q = W of an isothermal
process.
2
2
2
mRT
dV
V
Work W   PdV  
dV  mRT 
 mRT ln 2 of an ideal gas and
V
V
V1
1
1
1
isothermal process.
Q
148  103[J ]
T


 373 K or about 100˚C.
We get:
V2
2
mR ln
1[kg]  287[J/kgK ] ln
V1
0.5
7.
The ideal gas model gives:
mRT 1[kg ]  287[J/kgK ]  373[ K ]
 2.14  105 Pa.
P

3
0.5[m ]
V
8.
For an adiabatic process we have PV = constant, where  
1

cP
, i.e.
cv
1
P V 
V  P    6  105 1.3
  3  2.33 , i.e. the
P1V1  P2V2  1   2   2   1   
5 
P2  V1 
V1  P2 
 2  10 
volume increases 133%.
9.
For a polytropic process we have PVn = constant, where n is the polytropic
coefficient, i.e. the final pressure becomes P2:
n
1.35
V 
 9V 
PV  P V  P2  P1  1   0.9  105  2   0.9  105  91.35 
 V2 
 V2 
5
 17.5 10 Pa
From the ideal gas model (PV = mRT) we can rewrite this as:
n 1
(1.35  1)
 V1 
9V 2 

T 2  T1    (40  273.15)
 313  90.35  676 K ≈ 403˚C.
V 2 
 V 2 
n
1 1
n
2 2
10. a) To determine Q we must know cv in the temperature interval [T1,T2] and the mass
m. The mass we get from the ideal gas model (PV = mRT, where R is the special gas
R
constant, i.e. “per kg” according to R 
, where R is the general gas constant
M
and M is mol):
PV 1105  5 103
m 1 1 
 5.95 10-3 kg
RT1 8314  293.15
29
48
EXERGETICS
1.35
n
V 
 9V 
P1V1n  P2V2n  P2  P1  1   0.9  105  2  ≈ 0.9×105×91.35 ≈
 V2 
 V2 
≈17.5×105 Pa.
From the ideal gas model (PV = mRT) the relation above can be written as if we
c
have the following relations for cP and cv:   P and for air we have  ≈ 1.4. From
cv
tables we get the average value for cP in temperature range 20-50˚C to:
c (40 C)  cP (60 C)
cP (20 C)  P


c (20 C)  cP (50 C)
2
cˆP  cP (20 C,50 C)  P

2
2
1005  1009
1005 
cˆ
2

≈ 1006 J/kg˚C. Which gives cˆ v  P ≈ 719 J/kg˚C.
2

We can now determine the added heat from the definition of the heat capacity:
J
Q  mcˆv (T2  T1 )  mcˆv  21  5.95  103  719  30  128[kg
K  J]
kgK
(Always check the dimensions!)
b) To determine the final pressure we must know how the pressure changes during
the process, since it is an isochoric process we have (according to the ideal gas
model): P×constant = T or
P1 T1
T
323.15
  P2  P1 2  1 105
 1.10  105 Pa = 110 kPa.
P2 T2
T1
293.15
11. a & b) The process is adiabatic, i.e.:
 1
1.4 1
P  
 6  105  1.4

T2  T1  2   293.15
≈ 489 K, i.e. 2 ≈ 489–273 ≈ 216˚C.
5 
 1 10 
 P1 
We may now calculate the average value of the specific heat at constant pressure
accordingly:
c (20 C)  cP (216 C) 1005  1027
cˆP  cP (20 C,216 C)  P

≈ 1016 J/kg K.
2
2
Added work, i.e. technical work, when we have an open process, can be calculated:
1
w t  h1  h2   c PdT  cˆ P (T1  T2 ) ≈ 1016×(293.15–489.15) ≈ –199 kJ/kg. (The
2
minus sign indicate that work is being added to the process!)
c) The change of the internal energy can be calculated:
2
2
2
u2  u1   cV dT  
1
by, u21  
1
cP

 c dT
P
dT 
1


 wt

, i.e. the internal energy increases
(199)
≈ 140 kJ/kg.
1.4
12. The mass m we get from the ideal gas model (PV = mRT) accordingly:
PV
2  105  0.6
m 2 2 
≈ 1.4 kg.
RT2 287  293.15
49
EXERGETICS
2
2
a) For the internal energy: U 2  U1  m  cV dT  
1
1
cP

dT  m
cˆP

(T2  T1 ) 
cP (0 C)  cP (20 C)
c (0 C, 20 C)
2
m P
(T2  T1 )  m
(T2  T1 ) 




1005  1005
2
 1.4
(293.15  273.15) ≈ 20 kJ  U2 ≈ 20 +U1 ≈ 20 + 0 ≈ 20 kJ.
1.4
2
a) Analogously for the enthalpy: H 2  H1  m  cP dT  H1  mcˆ p (T2  T1 ) ≈ 0 +
1
1.4×1005× (293.15–273.15) ≈ 28 kJ.
50
EXERGETICS
EXERGY
We have now concluded that it is a fundamental law of nature that energy neither can
be created nor destroyed (the First Law of Thermodynamics). Energy is available in many
different forms and may be converted between these forms. However, a strict limitation is
always active. Different energies have different qualities, indicating to what extent they
are theoretically convertible to mechanical work. This limitation, a Law of Nature,
implies that the total energy quality always decreases in each conversion (the Second
Law of Thermodynamics).
The quality of energy is described by the concept of entropy. High entropy is equal to
low quality of energy. The Second Law states that conversions are possible only if the
total entropy increases. By introducing exergy, we may treat energy and entropy
simultaneously, i.e. “kill two birds with one stone”.
At this stage, introducing exergy is mere a matter of putting a name to something we
already know. By reading the more verbal treatment of the exergy concept you are well
prepared for the more analytic treatment.
Exergy of a general process
Assume a general (irreversible) process:
Q from the environment
at temperature T0
W
Hi
He
SYSTEM
U2 – U1
The processes involves a change of the internal energy of the system, U2 – U1, and a
change of the enthalpy of the flow through the system, He and Hi. Thus, the process is a
combination of a closed and open system.
Let us make the process reversible by adding a reversible heat engine as below.
Q 0 from the environment
at temperature T0
E
Wc
HEAT ENGINE
Wrev
Hi
Qrev
SYSTEM
He
U2 – U1
We can now describe the process in detail.
The work from the heat engine is Wc, and the total maximal work we may extract
from the system at a reversible process E (= exergy) is
51
EXERGETICS
E = Wrev + Wc
If we apply the first law on the modified system with a reversible process, we get
Qrev + Hi = U2 – U1 + Wrev + He
i.e.,
Wrev = Qrev + Hi – He – (U2 – U1)
Since, all processes are reversible the heat exchange with the environment must go
through the heat engine, as is shown in the Fig above. The first law applied to the heat
engine gives
Wc = Q0 – Qrev
and the second law gives
Q0
 S 0  S rev
T0
since work Wc is free of entropy (Sc =0) and were Srev is the entropy related to the heat
Qrev. The work from the heat engine becomes:
Wc = T0Srev – Qrev
the second law for the system gives:
Srev = S2 – S1 + Se – Si
Thus the work from the heat engine becomes
Wc = T0(S2 – S1 + Se – Si) – Qrev
By replacing Wrev and Wc in the Eq. for E, we get
E  Q rev  Hi  He  (U2  U1 )  T0 (S 2  S1  S e  S i )  Q rev
Which may be rewritten as
E  Hi  T0 S i  (He  T0S e )  (U1  T0 S1 )  (U 2  T0 S 2 )  E i  E e  E1  E 2
As we see Qrev disappears in the expression for the reversible work that can be extracted
from a general process, which involves heat at ambient temperature T0. This reversible
work, which is the maximum work that can be extracted, we call the exergy. This is the
part of the energy, which is completely convertible into any other forms of energy, i.e.
especially work. Other forms of energy, e.g. kinetic and potential mechanical energies
and electrical energy may just be directly added. Let us now look closer to special cases
of this general exergy expression.
R:
Appendix 2 in this material or Appendix A in G. Wall, “Exergy — A Useful
Concept within Resource Accounting”, 1977, pp. 40-43, that you find on the
Internet: http://exergy.se/. (This is an alternative derivation of the exergy concept
52
EXERGETICS
based on a closed system in an infinite environment. Thus the environmental
relations are stressed, but as we will see the environment only plays the role of
reference state. And, the exergy can be related to any reference state.)
Exergy, work and entropy production
We will now derive an expression for the total entropy production and the relation to
exergy.
Consider a defined thermodynamic system going through a process from state 1 to 2,
where work W is extracted. Further more heat Q is added from a reservoir, e.g. the
environment at temperature T0, see Fig. below.
The situation is completely general, the only restriction is that the system is defined.
W
12
Q
T0
The first law gives:
Q = U2 – U1 + W
(1)
The entropy change of the system is Ssys = S2 – S1. Further has the entropy of the
reservoir also changed? The reservoir has delivered the heat Q. The entropy change of the
reservoir then becomes
Q
S res 
(2)
T0
We use Stot to indicate the total entropy change of the system and the environment, which
becomes
S tot  Ssys  S res  S 2  S1 
Q
T0
(3)
By eliminating Q from Eq. 3 and Eq. 1 we get
W  U1  U 2  T0 ( S1  S 2 )  T0 S tot
(4)
The total entropy change Stot is always positive (Second Law)
S tot  0
so we will from now on call Stot the total entropy production.
53
(5)
EXERGETICS
Equations 4 and 5 give an upper limit of the work W which may be extracted from the
process. This upper limit is given from a reversible process, since the total entropy
production Stot of such a process is zero.
We now assume a reversible process, which utilizes the given heat reservoir and
converts the system from state 1 to 2. Any kind of extra apparatus is allowed. However,
these should be in the same state after the reversible process as they were before. The
maximal extractable work, i.e. the exergy then becomes according to Eq. 4:
E = U1 – U2 – T0(S1 – S2)
(6)
If we combine Eq. 4 and Eq. 6 we have an important relation
W  E  T0 S tot
(7)
The exergy obviously gives an upper limit to the work which is extractable from the
process, since the entropy production Stot is always positive. Thus the available work W is
limited by the exergy E.
At an irreversible process as above with the entropy production Stot, this corresponds
to a loss of available work or exergy. Thus, for an irreversible process we have:
Wloss  Eloss  T0 S tot
 tot
S  0
(8)
Thus, the relation between exergy and entropy is very fundamental. We may say that
exergy is the concept which links the universal concept of entropy with the conditions on
the Earth through the temperature. In the section on exergy and information we will
develop this fundamental relation further.
Exergy loss = Ambient temperature ×Entropy production
This relation may also be expressed as the following: the maximal available exergy, that
is equal to the maximal possible exergy loss, is equal to the ambient temperature times
the maximal entropy production.
The result of this section may be expressed in many ways as we will see in the
coming sections. This simple conclusion is still a very fundamental relation. We will now
calculate the entropy production of some simple cases.
Heat transfer
Assume that a certain amount of heat Q is transferred from a body with the
temperature T1 to a body with the temperature T2 and that the temperatures T1 and T2
remain constant. The entropy changes for the bodies are –Q/T1 and Q/T2. We may regard
this as if the heat flow is removing the entropy Q/T1 from “T1” and adding the entropy
Q/T2 to “T2”, and that the difference gives the entropy production. The heat exchange has
occurred through some intermediate medium. This medium is in the same state after the
process as before. The intermediate medium does not change its entropy.
54
EXERGETICS
Q
T1
T2
The total entropy change then becomes
S tot 
Q Q

T2 T1
(9)
If a heat Q is transferred from T1 to T2 and further to T3 then we get for each transfer
S12tot 
Q Q
Q Q
 and S 23tot  
T2 T1
T3 T2
(10)
and for the total transfer we get
S13tot 
Q Q

T3 T1
(11)
We could have stated Eq. 11 directly from Eq. 9.
It is important to note that the entropy production becomes less if the heat transfer
occurs at higher temperature. The entropy production at heat transfer from 610 K to
600 K is only one forth as big as if the same amount of heat is transferred from 310 K to
300 K.
Temperature exchange between two bodies
We have two bodies with the heat capacities C1, C2 and the temperatures T1 and T2. The
volumes are constant, e.g. two buckets of water with different temperatures.
C 1, T 1

C 2, T 2
C 1, T e
C 2, T e
Heat is exchanged to reach equilibrium at temperature Te. First law, i.e. energy
conservation gives:
Te 
C1T1  C2T2
C1  C2
The total entropy production becomes:
55
(12)
EXERGETICS
S
tot
Te
T
T
T
C dT e C dT
 1  2
 C1 ln e  C2 ln e
T2
T1
T
T
T1
T2
(13)
if we assume that the heat capacities are constant, i.e. do not depend on the temperature.
Ex:
Convince yourself that the total entropy production always is positive.
S:
To check this we set x = T2/T1, which gives:
S tot
C1
 C2
C1  xC2
 C1 ln
 C2 ln x
C1  C2
C1  C2
(14)
The entropy production is of course zero if x = 1. The derivative with respect to x
becomes:
dS tot
C1C2  1 

1  
dx
C1  C2 x  x 
(15)
The derivative is positive for x>l and negative for 0<x<l. Thus, the entropy
production is positive for 0<x<∞.
Friction
Assume wf to be the frictional work of a process, and assume this becomes heat, i.e. no
other effects. If the temperature of the frictional heat is T, the entropy production
becomes, according to Eq. 2 (Q = –wf):
s
wf
T
(16)
As we see the entropy production becomes less with increasing temperature, i.e. the
friction work is transferred to frictional heat of higher quality.
Q: Have you any ideas how this could be used? Do you know where?
If the frictional work Wf generates heat at temperature T1 and if the temperature then
decreases to T2 we get the entropy production from Eq. 9 and 16:

S
tot

 1
W
1  W
 f  W f     f
 T
T1
T1  T2
 2
(17)
Thus, we get the same entropy production as if the temperature of the frictional heat
originally was T2.
The total entropy production from friction is given by contributions defined by Eq.
16. The produced frictional heat also creates additional contributions to the total entropy
56
EXERGETICS
production according to Eq. 9, and finally the total entropy production becomes
according to Eq. 17.
Thus, if the total frictional work is Wf, and if all frictional heat is lost to the
environment at temperature T0, then the total entropy production from frictional work and
heat transfer to the environment becomes:
S tot 
Wf
T0
(18)
All frictional heat finally reaches the environment, however the entropy production will
appear at many sub processes.
System in contact with a heat reservoir
Our system is now in contact with a heat reservoir of temperature T0 during the process l
to 2. The initial and final temperatures must then be equal to T0:
T1  T2  T0
(19)
We also assume that the volume is constant:
V1 = V2
(20)
The process will then not generate any work on the environment.
These kinds of process usually occur for liquids and solids. The temperature might of
course change during the process.
The exergy, i.e. the available reversible work then becomes according to Eqs. 6, 19
and the definition of Helmholtz’ function or free energy, A = U – TS:
E = A1 – A2
(21)
W ≤ A1 – A2
(22)
From Eq. 7 we get:
Thus, the decrease of Helmholz free energy of the system gives the upper limit of the
available work of the process.
System in contact with a heat and pressure reservoir
In many processes the initial and final pressure is the same, i.e.:
P1 = P2 = P0
And often P0 is the ambient pressure, towards which the system performs a work
P0 (V 2  V 1 ) during the process. The available work then becomes:
57
(23)
EXERGETICS
W  W  P0 (V2  V1 )
(24)
The exergy then becomes according to Eq. 21, the definition of enthalpy H = U + PV and
Gibbs’ function or free enthalpy G = H – TS:
E = G1 – G2
(25)
W´≤ G1 – G2
(26)
For the available work we get:
Thus, the decrease of free enthalpy G for the system gives the upper limit of the available
work of the process, i.e. similarly as above for Helmholz’ function.
The Eqs. 22 and 26 explains why we use free energy for A and free enthalpy for G. It
is important to notice the conditions of these relations. As we have seen Helmholtz’ and
Gibbs’ functions are special cases of the exergy for special processes. In Appendix 2, you
will find a more detailed analysis between exergy and these and other concepts.
Exergy of heat and cold
We will now derive expressions for available work in some basic cases where heat
transfer is involved. The ambient temperature is T0.
Let us first assume that heat Q is transferred between two reservoirs at temperatures T
and T0, and T > T0 and find the exergy involved. A reversible heat engine working
between the temperatures T and T0 gives the maximal available work. By assuming that
the heat is completely lost we may calculate the maximal entropy production which gives
us the exergy being lost, i.e. the available exergy which is able to utilize.
T
Q
T0
From above we have:
Q Q
 T 
E  T0 S tot  T0     Q1  0 
 T 
 T0 T 
(T>T0)
(1)
If the temperature T <T0 , this only change the direction of the heat flow Q, which then
will go from T0 to T, the exergy now becomes:
Q Q 
T
E  T0     Q 0  1

T
T T0 
58
(T<T0)
(2)
EXERGETICS
The ratio between the exergy E and the energy of the transferred heat Q we call the
exergy factor E/Q. Thus, by multiplying the energy by the exergy factor we get the
exergy. The exergy factor of heat from a heat reservoir, i.e. at constant temperature T, that
differ from the ambient temperature T0 is
E T  T0

Q
T
(3)
This is a generalization of the Carnot relation, since it also covers temperature below
ambient. Let us exemplify this.
Ex:
What is the exergy factor of heat at 20˚C (≈ 293 K) in an environment at 5˚C (≈
278 K)? One example of this is exergy needed to maintain the inside temperature
at +20˚C when the outside temperature is 5˚C.
S:
From Eq. 3 we have:
E 293  278 15 15


 %  5%
Q
293
293 3
So, the exergy factor of heat at 20˚C in an environment at 5˚C is approximately 5
%. Thus, the exergy efficiency of an electric heat radiator (electric short circuit),
which maintain the indoor temperature at +20˚C when the outdoor temperature is
+5˚C is about 5 %. However, the energy efficiency is 100%.
Let us look closer to the relation between energy and exergy efficiencies and exergy
factor.
Ex:
Estimate the energy efficiency of an oil furnace when the exergy efficiency is 3%,
the exergy factors for in and out flows are 0.9 and 0.04 respectively?
S:
We have the following relations:
Exergy factor 
E
Exergy
, i.e.,
Energy
Q
Energy efficiency 
Energy output
Q
, i.e., en  out
Energy input
Q in
Exergy efficiency 
E
Exergy output
, i.e., ex  out
Ein
Exergy input
From these relations we may derive the following relation:
59
EXERGETICS
Eout
Q
Eout out
Q
Eout
Eout
en  out 

 ex
E
Q
Qin
Qin in
Ein in
Ein
Ein
Qout
Qout
Eout
 ex
Qin
Ein
Ein
Qin
Eout
Qout
Numerical values gives:
0 .9
 en  0.03
 0.675  70%
0.04
The energy efficiency becomes about 70 %.
Calculate the exergy factor for heat transferred between the following
temperatures in degrees Celsius and 0˚C:
S:




E
Q
–270 –200 –100 –20
…
…
…
0
… …
20 100 200 500 1000 5000
…
…
…
…
…
…
Check your values with the black line in the Fig. below.
The upper black curve, in Fig. below illustrates Eq. 3. (The lower gray curve will be
described below.) As we see cold has exergy, when energy is regarded as negative. We
also see that the exergy factor strongly increase for low temperatures, which explains the
difficulty to reach absolute zero temperature, i.e. 0K or 273.15˚C. At high temperatures
the exergy factor comes closer to 1 (the dotted line), i.e. exergy E and energy Q of the
heat becomes almost the same. As we know, high temperatures generate more work in a
heat engine.
E /Q
Liquid nitrogen
1.6
1.4
1.2
1
Furnace
0.8
Stove
0.6
Oven
0.4
Hot steam
0.2
0
T
T0
Boiling water
Waste heat
Freezer
0
Meltedsteel
Glowing steel
1
2
3
60
4
EXERGETICS
Equation 3 is useful to describe the exergy need to maintain the temperature in hot
spaces, e.g. furnaces, stoves, ovens, or houses, as well as in cold spaces, e.g. refrigerators
and freezers, or to keep a pleasant indoor climate by air-conditioning a hot summer day.
Exergy reflects better than energy that heat or cold is more “expensive” when we
need it the most. Thus, it is easier to generate cold in winter and heat in summer. By
using the energy concept it seems equal, since energy is not affected by the ambient
conditions. So, from an energy point of view it seems independent of whether it is
summer or winter. Thus, district heat should be more expensive in the winter than in the
summer. This is an example of something obvious which is not explained by the energy
concept, but is immediately explained by the exergy concept.
The exergy of a body with temperature T is slightly more complicated to calculate
since the temperature of the body as well as the delivered heat decreases when heat is
removed. Assume that the heat capacity of the body is C(T). Let us calculate the exergy
when the temperature decrease from T to T0. The heat content of a temperature change
dT´ is C(T´)dT´, then the exergy becomes an integral from T0 to T:
T
 T 
E   C (T )1  0 dT 
 T 
T0
(4)
It is not hard to see that this relation holds at both T>T0 and T<T0.
If the heat capacity is not depending on temperature, the exergy becomes


T 
T
T 
E  C  T  T0  T0 ln   C (T  T0 )1  0 ln 
T0 

 T  T0 T0 
(5)
Q  C (T  T0 )
(7)
E
T
T
 1  0 ln
Q
T  T0 T0
(8)
Since
the exergy factor E/Q becomes:
Where the absolute signs make sure that the relation also holds for T<T0. This relation is
indicated by the gray line in the Fig. above.
In the derivation of exergy for a body we have neglected change in volume, which is
usually acceptable for solids and for liquids.
Calculate the exergy factor for heat transferred between a body at the following
temperatures in degrees Celsius and 0˚C:
S:




˚C
E
Q
–270 –200 –100 –20
…
…
…
0
… …
20 100 200 500 1000 5000
…
…
…
…
…
…
Check your values with the gray line in the Fig. above.
(Please observe that at low temperatures the heat capacity C can not be regarded
as constant.)
61
EXERGETICS
In the Fig. above the exergies of some bodies at different temperatures are indicated,
e.g. liquid nitrogen, waste heat, boiling water, hot steam, glowing steel, and melted steel.
Let us compare the exergy factors between the two cases, reservoir and body, at
temperature T, see the exercises above. If T=100˚C then the exergy factor of the reservoir
is 0.27 and for the body 0.15, i.e. almost half.
As we would expect the exergy factor E/Q is lower for a body since the temperature
decreases with the heat released. At low temperatures, TT0, and for temperatures close to
ambient, T0, the exergy factor E/Q of a body is about half that of a reservoir. For high
temperatures the exergy factors for a reservoir and a body become closer. Thus, it is
important to make sure what kind of heat source is available, reservoir or body, for the
system to be studied.
We will now examine district heat more in detail. District heat is regarded as a body,
thus, the exergy factor becomes from Eq. 8:
T
E
T0
 1
ln supply
Q
Tsupply  T0
T0
(9)
where Tsupply is the temperature of the supplied heat(Tsupply>T0). Assume this to be 85˚C
when T0>2˚C and that it increases linearly with decreasing outdoor temperature to
120˚C when T0<20˚C. Then, we get the lower gray curve in Fig. below. We see that the
exergy factor is varying stepwise between about 0.10 and 0.22 when the temperature
decreases from 20 to –30˚C.
However, since a part of the supplied exergy is returned, we may calculate the
utilized exergy to:
T
E
T0
 1
ln supply
Q
Tsupply  Treturn Treturn
(10)
where Treturn is the temperature of the returned exergy, which we assume to 55˚C. Then,
we get the upper black curve in the Fig. below. As we expected the exergy factor
becomes higher, since the heat now is taken out at a higher average temperature. It now
varies stepwise between about 0.15 and 0.32.
62
EXERGETICS
E
Q
0.35
0.30
0.25
0.20
0.15
0.10
0.05
T0
0
-30
-20
-10
0
+10
+20
Ex:
Calculate the exergy factor of district heat when the outdoor temperature is 0˚C?
Also calculate the exergy factor of the utilized heat, see Eq. 10 above?
S:
First we must find the temperature on the supplied district heat Tsupply at the
prevailing outdoor temperature T0. When –20˚C <T0<+2˚C we have:
Tsupply  85  (120  85) 
2  T0
2  (20)
which gives Tsupply≈ 88.2˚C ≈ 361K. Numbers in Eq. 9 give:
273
361
273
E
ln
 1
 1
ln(1.3223)  0.13
361  273 273
88
Q
Thus, the exergy factor of the supplied heat is 0.13.
If we assume the return temperature to be 55˚C, i.e. 328K, then we get for the
utilized heat:
273
361
273
E
ln
 1
 1
ln(1.1006)  0.21
361  328 328
33
Q
Thus, the exergy factor of the utilized heat is 0.21.
S:
Verify the expression for the utilized district heat in Eq. 10 above.
We will return to exergy of heat and cold when we analyze the exergy conversions in
industrial processes.
63
EXERGETICS
Here will soon be add something about district cold also, this is interesting from the point
of putting a price on the energy. How do you price cold delivered to the consumer, when
the consumer is delivering the energy?
Exergy of black body radiation
For black body radiation* we have the following relations for energy and entropy
emission rates per unit area:
u  T 4
(1)
4
s  T 3
3
(2)
(Please, note that du  Tds .)
2 5 k 4
 5.67  108 W/K 4 m 2 is called the Stefan-Boltzmanns constant.
where  
3 2
15h c
Consider the reversible process, radiation from a black body at temperature T is
transmitted to another black body at temperature, T0:

Black body at
temperature T

. .
u, s
.
e
Radiation
. .
u0, s0
Black body at
temperature T0
Ambientat
temperature T0
.
q
The exergy flow e , which is associated with the heat radiation is derived by applying
the first and second law to an ideal reversible process. This process converts the radiation
to exergy and heat at ambient temperature. Assume that the converter can not receive
radiation without emit radiation, the principal of detailed balance.
The converter receives energy and entropy according to Eqs. 1 and 2. Simultaneously,
it emits energy, u0 and entropy, s0 according to Eqs. 1 and 2.
u0  T04
4
s0  T03
3
The first law gives:
u  u0  e  q
*
A black body by definition absorbs all incoming radiation.
64
(3)
EXERGETICS
Since the process is reversible, there is no entropy production, and the second law
gives.
q
s  s0 
(4)
T0
The work or exergy that can be extracted from the radiation becomes
e  u  u0  q  u  u0  T0 ( s  s0 ) 
 1  T0  4 4 T0 
4
 4
4
3
3 
4
  T  T0  T0 (T  T0 )  T 1    

3


 3  T  3 T 
(5)
Thus, the exergy factor of black body radiation is
4
e
1T  4 T
 1  0   0
u
3 T  3 T
(6)
Let us apply this result to sunlight*, where Tsun = 6000 K and Tearth =T0 = 300 K, the
exergy factor becomes from Eq. 6
4
1  300  4 300
 e 
 1 
 0.933
 
  
3  6000  3 6000
 u sunlight
Q:
What is the exergy factor for heat at heat transfer between these temperatures, 300
and 6000 K? What does the difference indicate?
Ex:
Compare the energy and exergy efficiencies of the following solar collectors, all
working in full sunlight, i.e. 1000 W/m2 and the ambient temperature is 15˚C:
1. A flat solar panel for heat generation of 1m2 which heats 3 dl of water from 10
to 40˚C per minute.
2. A concentration solar collector of 1 m2 which gives 1 g steam at ambient
pressure per minute from water at 10˚C.
3. Solar cell 95×65 mm which gives 0.45 V and 400 mA.
S:
1.
Energy power in Pen, in ≈ 1000×1 = 1000 W
Exergy power in Pex, in ≈ 1000×0.933×1 = 933 W
0 .3
 4200 ( 40  10)  630W
Energy power out Pen,out  C T 
60

T 
Exergy power out Pex,out  C  T  T0 ln out  
Tin 

0.3
313.15 

 4200 40  10  288.15  ln
  20.6W
60
283.15 

630
 63%
Energy efficiency  en 
1000

*
The sun and the earth can be regarded as almost black bodies.
65
EXERGETICS
Exergy efficiency  ex 
20.6
 2 .2 %
933
2. Energy power in Pen, in ≈ 1000×1 = 1000 W
Exergy power in Pex, in ≈ 1000×0.933×1 = 933 W
Energy power out Pen,ut  C T  H phase 


0.001
4200 (100  10)  2.3  10 6  44.6W
60

T 
T T
Exergy power out Pex,out  C  T  T0 ln ut   H phase  out 0 
Tin 
Tout

100  15 
0.001 
373.15 

6

4200100  10  288.15  ln

  2.3  10 

60 
283.15 
373.15 

≈ 0.7 + 8.7 = 9.4 W
44.6
 4%
Energy efficiency  en 
1000
9 .4
 1%
Exergy efficiency  ex 
933

3. Energy power in Pen, in ≈ 1000×95×65×10–6 ≈ 6.2 W
Exergy power in Pex, in ≈ 0.933×6.175 ≈ 5.8 W
Energy power in Pen, out = UI ≈ 0.45×0.4 = 0.18 W
Exergy power in Pex, out = UI ≈ 0.18 W
0.18
 2 .9 %
Energy efficiency  en 
6 .2
0.18
 3.1%
Exergy efficiency  ex 
5 .8
Q:
What are suitable applications of these different collectors?
Exergy of materials
A simplified introduction to the exergy of substances and materials is given below.
Assume the pressure P and the temperature T to be constant, i.e. P = P0 and T = T0,
then we have from the general expression of exergy, see App. 2:
~ 
~ )
E   ni (
i
i0
(1)
i
~
where i is the generalized chemical potential of substance i in its present state and
~ is the generalized chemical potential of substance i in its environmental state.

i0
Further, assume that the substance only depart from the environment in chemical
potential µ and concentration c. Also, assume that the general chemical potential may be
written
~   0  RT ln c

i
i
0
i
66
(2)
EXERGETICS
where  0i is the chemical potential for the material (substance) i in relation to its standard
state, given from chemical tables, usually the pure element. Then the exergy becomes
E   ni (i0  i00 )  RT0  ni ln
ci
ci 0
(3)
where i 0 is the chemical potential for the material in the environment in relation to its
standard state.
For only one material we have
0

c
E  n  0   00  RT0 ln 
c0 

(4)
Ex./S: Let us calculate the exergy of the Swedish iron ore production.
The Swedish iron ore has an average content of iron of about 60% (weight)
and usually consists of magnetite (Fe3O4). The molar weight of iron is 55.8 g,
which implies that 1 kg of iron ore contains 600/55.8 = 10.7 mol of iron = 3.58
mol magnetite = 0.83 kg magnetite.*
Assume that the reference (environmental) state of iron is hematite (Fe2O3) in
solid form and with molar concentration 2.7×10–4 and that oxygen O2 is in
gaseous form at partial pressure 20.40 kPa in the environment [Szargut (1980)].
The chemical potentials of iron in magnetite and hematite then become
[kJ/mol]:†
1
 0 ( Fe magnetite )  ( 1015.5  2  3.84)  335.9
3
1
 0 ( Fe hematite )  ( 742.2  1.5  3.84)  368.2
2
Where 3.84 kJ is the amount of exergy released when the partial pressure of 1
mol of oxygen gas (O2) decreases from 101.325 kPa to 20.40 kPa at 15˚C‡, since
this part is not included in the Gibbs free energy data.
c
 101.325 
RT0 ln   8.314  ( 273.15  15)  ln
  3840 J
 20.40 
 c0 
Thus, the specific exergy of iron ore and iron then becomes, [MJ/kg]:§

 0.83  0.43 
eiron ore  10.7  335.9  (368.2) 103  8.31 288  ln
4  ≈
 2  2.7  10 

≈ 0.51
*
Since we lack data on molar fraction of iron in iron ore, we use instead kg iron per kg iron ore.
Handbook of Chemistry and Physics, CRC, vol. 63, sida D-72: ∆Gf˚(Fe2O3) = –177.4 kcal/mol = –742.2
kJ/mol, ∆Gf˚(Fe3O4) = –242.7 kcal/mol = –1015.5 kJ/mol.
‡ ISA - International Standard Atmosphere is defined to 101.325 kPa, 15˚C and 0% humidity.
§ 3 of 7 atoms in the magnetite molecule is iron atoms, i.e. the iron concentration in magnetite is 3/7≈0.43.
†
67
EXERGETICS

1


 6.91
eiron  17.90  (368.2) 103  8.31 288  ln
4 
 2  2.7  10 

where 1 kg iron corresponds to 17.9 mol.
These values are in correspondence with those of others (McGannon, 1971,
Gyftopoulos et al., 1974).
The Swedish mining of iron ore amounted to 26.9 Mton in 1980, by assuming
this to be magnetite it corresponds to 14 PJ. The Swedish steel production was 3.5
Mton, i.e. about 24 PJ of exergy. To produce this 5.7 Mton of ore, i.e. about 3 PJ,
about 34 PJ of electricity and about 77 PJ of coal and other fuels where needed.
The exergy efficiency then becomes 21%.
Q:
Use the exergy calculator at http://www.exergoecology.com/excalc/ to see what
values you get and compare with the above results?
Ex./S: Let us also apply this to desalination and calculate the exergy of fresh water.
Assume that the reference (environmental) state of water is sea water. Sea water
has an average content of water, H2O of about 96.5% (weight) and the molar
weight is 18 g, which implies that 1 kg of sea water contains 965/18 ≈ 53.6 mol of
water. The rest is mainly sodium chloride (NaCl) with a molar weight of 58.5 g,
i.e. 1 kg of water contains 35/58.5 ≈ 0.598 mol of NaCl. However, when NaCl is
dissolved in water it becomes Na+ and Cl- ions, i.e. separate molecules. Thus 1 kg
of water contains 53.6 mol of H2O, 0.598 mol of Na+ and 0.598 mol of Cl-, i.e.
together about 54.8 mol. The mol concentration of water in sea water then
becomes about 97.8%. The exergy of pure water at ambient temperature 15˚C
then becomes
c
 1 
E  RT0 ln   8.314  ( 273.15  15)  ln
  54.9 J/mol
 0.978 
 c0 
where the chemical potentials vanishes since we assume no chemical effects. The
exergy of pure water then becomes about 3.05 kJ/kg or about 0.847 kWh/m3. If
we instead assume 4.5% of salt and the ambient temperature of 30˚C, then we get
72.0 J/mol, 4.0 kJ/kg or 1.1 kWh/m3. These exergies of fresh water corresponds to
the minimum amount of exergy needed to produce it from the assumed sea water,
respectively. The exergy of sodium chloride, on the other hand, becomes 184
kJ/kg and 182 kJ/kg respectively, i.e. less when the salt content in sea water
increases.
Exergy of nuclear fuel
When the nuclear particles, i.e. nucleons, regroup into more probable states, exergy is
released. In fission this occurs by splitting heavy atoms and in fusion by putting together
light atoms into heavy atoms.
68
EXERGETICS
8
7
Fusion
Bindning energy per nucleon [MeV]
9
6
5
4
3
2
Fission
1
0
0
20
40
60
80
100 120 140 160 180 200 220 240
Atomic weight
Energy per nucleon in an atom as function of atomic weight.
The difference in binding energies between the initial and the final state is released as
scattered motion in the produced particles, like neutrons and new elements. The energy
can be expressed as a change of mass, a mass shift, through the well known relation
E=mc2
where c is the speed of light in vacuum. However, the lost mass is very small compared
to the rest initial mass, e.g. in fission uranium-235 losses about 1‰ of its mass and in
fusion to helium deuterium (2H) losses about 64‰ of its mass. Thus, we write instead
U = ∆mc2
(1)
where we also use U for internal energy, in this case nuclear.
Since, a part of the released energy in most cases appear as neutrinos, which can be
regarded as invisible and non interactive, i.e. useless, we may write nuclear exergy as
E = ∆mc2 – Eneutrino
(2)
The energy of the neutrinos may sometime exceed up to 5% of the totally released
energy.
Q:
Estimate from the Fig. above the available exergy of 1 kg uranium? Compare this
with the extracted exergy (electricity), which in today’s Light Water Reactors is
about 1 TJ/kg uranium. What is this difference due to and what consequences
does it bring to the waste?
69
EXERGETICS
Exergy and information
Even though, thermodynamics treats the physics of systems which from a macroscopic
point of view may be very small (about 10–15 cm3) are they still from microscopic point of
view very big, therefor containing a large amount of particles (10–15 cm3 contains about
109 atoms). Thus, a detailed knowledge of the motion of every particle is impossible. In
statistical mechanics we consider the large amount of particles and use statistical methods
do explain the macroscopic phenomena as the result of microscopic rules. Statistical
mechanics, by this means the theoretical basis of understanding thermodynamics.
Statistical mechanics is also strongly linked to information theory, where we study
incomplete knowledge. Thus, many concepts and relations are valid in all these fields,
thermodynamics, statistical mechanics and information theory. Let us therefor look closer
into information theory.
Assume a system of N unique particles. The number of allowed states Ω of the system
is exponentially depending on N. Let the probability of the j:th state be Pj and the sum of
the probabilities of all states to be 1, i.e. the system is in at least one state

P
j 1
j
1
The entropy of the system is then defined from statistical mechanics as

S  k  Pj ln Pj
j 1
where k is the Boltzmann’s constant, 1.38054×10–23 J/K.
0
The probabilities at equilibrium Pj are such that they maximize the entropy S
independent of other restrictions on the system.

Seq  S max  k  Pj0 ln Pj0
j 1
The available negentropy, negative entropy, of the system then becomes


 
 S  Seq   Seq  S  k   Pj ln Pj   Pj0 ln Pj0 
j 1

 j 1
From information theory we have the information or the information capacity I in
binary units (bits)
I


1  
  Pj ln Pj   Pj0 ln Pj0 

ln 2  j 1
j 1

Let us exemplify by a system of N different particles with 2 possible states each, e.g.
0
0 or 1. Then we have Ω = 2N. If there are no other restrictions then all Pj must be 2–N, see
the Table below.
70
EXERGETICS
0
N
1
0,1
2
00,01,10,11
3
000,001,010,100,011,101,110,111
Ω = 2N Pj
1
21=2
2
1
22=4
4
1
23=8
8
etc.
Total information about the system (one of Pj equals 1 and all the others equal 0) gives I
= N,
2

1 
1  N 1

N
N
I
0

 2 N ( N ) ln 2  N
  2 ln 2  

ln 2 
2

j 1
 ln 2 
N
 
For every particle there is information corresponding to a “yes” or “no” to a specific
question. Every such answer corresponds to one binary unit (bit) of information.
Thus, negentropy and information are very closely linked
Seq  S  k I
where k = k ln2 ≈ 1.0×10–23 [J/K].
One bit of information, thus, is equivalent to 1.0×10–23 J/K of negentropy.
Exergy and negentropy is also related from a relation we already know:
E  T0 ( Seqtot  S tot )
”tot” means the total system, here we only use system. From these relations we now
get the following relation between exergy and information
E  k T0 I
where k T0  2.9×10–21 [J] is the amount of exergy related to 1 bit of information at
room temperature.
It is important to observe that information or information capacity, not necessary need
to be meaningful from a human point of view. Information is here used as a measure of
order or structure.
Within science information is just as fundamental as energy and matter. By small
amount of information, processes converting huge amount of energy and matter can be
controlled. As we have seen there is a fundamental relation between exergy and
information, defined from information theory. However, as we will see information
usually has a very small exergy value. Thus, it should be treated as a resource of one’s
own.
Let us apply the concepts above on a physical system. Consider a container of volume
V, with an ideal mixture of two different ideal gases 1 and 2, see the Fig. below. The total
71
EXERGETICS
number of molecules is N, divided into N1 and N2 respectively. The molar concentration
then becomes x1 = N1/N and x2 = N2/N = 1 – x1.
N1, P
N, P
The gases are mixed in the container
with pressure P.
N2, P
The gases are separated with the
same pressure P.
If we randomly pick one molecule from a specific place in the left container in the
Fig. we face a probability P = (x1, x2)* to get a specific molecule. The information we gain
when we discover which molecule it is, becomes according to above
I
1
 1 
1 
 x1 ln   x2 ln 
ln 2 
 x1 
 x2 
Let us now split the container in two parts with the volumes V1 = x1V and V2 = x2V
respectively, and arrange the molecules so that V1 only contains molecules 1 and V2 only
contains molecules 2, see the right part of the Fig. above. If we now randomly pick one
molecule from a specified part of the container, then we know which molecule we get,
and our knowledge and information increase by 0. By mixing the molecules information
is lost, i.e. I per molecule, which can be related to a loss of exergy as above. Let us look
closer to this.
Mixing entropy
Let us now calculate the maximal available work when we mix the two gases 1 and 2
as above. We may then get an expression of the amount of exergy, which should be
related to the information I, i.e. we will check the relation above.
Assume the container is in thermal equilibrium with the ambient at temperature T0.
Since the initial and final volumes are the same we may set the ambient pressure to 0.
First allow gas 1 and 2 independently expand isothermally, see the processes atob in the
Fig. below. The gases are then mixed reversible by ideal semi-permeable walls, see
process b to c. Please note there is no net force on the containers during the mixing
process b to c, and also there is not heat flow. So, the mixing is completely reversible, i.e.
the entropy is constant.
* Please note that we here are forced to use the same symbol for pressure and probability. This problem
usually occurs when you simultaneously treat different fields of science.
72
EXERGETICS
   
W1
Q1
T0


Q2
W2
a
b
c
A reversible mixing of the gases through an isothermal expansion (a to b), followed by an iso-exergetic
mixing (b to c).
When the gases are regarded as ideal the ideal gas model is valid, i.e. PV = nRT =
NkT and the work is obtainable as
V
V
Vi
Vi
Wi   Pi dV   
V 
1
N i kT0
dV   N i kT0 ln   NkT0 xi ln 
V
 Vi 
 xi 
which gives

1
 1 
W  W1  W2  NkT0  x1 ln   x2 ln   NkT0 ln 2 I
 x1 
 x2 

Thus, we have the following relation between the exergy per molecule E and
information per molecule I in the ordered system
E = kT0ln2I
(2)
This is analogous to what we had above.
Since the internal energy U for an ideal gas only is a function of the temperature, we
have for the isothermal process (a to b) that
Q1 + Q2 = W1 + W2
according to the first law. Thus, the entropy production S for the expansion (a to b)
becomes
S
Q
 Nk ln 2 I
T0
73
EXERGETICS
If we assume the molecules to be identical, 1 or 2, then I = 1, i.e. each molecule in the
ordered system is carrying 1 bit of information. The container may then be regarded as a
binary memory, and in principle one molecule is enough to store 1 bit of information. In
practice you need to make the reading easy and storing safe, and therefore repeat the
information by using several molecules, i.e. so-called redundancy. Therefore, it is
inadvisable to say that the ordered system contains N bits of information.
By information capacity we define the amount of information that a system can store
or transfer. This amount of information is limited by the available exergy, as above.
Since information needs such a small amount of exergy the demand of high
efficiency* has been weak. However, one area where increasing efficiency is an absolute
necessity is in highly packed integrated circuits, where the exergy losses are converted
into heat which may otherwise overheat the circuit. Increased efficiency also makes the
circuits faster and more powerful.
A limited efficiency is usually needed since it implies that the storage and transfer of
information becomes less sensitive for noise and other disturbances. Signals in electronic
systems are always subject to thermal noise, and the amount of exergy per bit must be
several magnitudes above the noise level kT. However, the information transfer in
biological systems when reproducing information and synthesizing proteins is far more
efficient and the exergy use is sometimes only 10-100 kT/bit. Even though, the
probability of errors, or mutations, is very low.† This is possible because of the specific
environment in which the information is transferred. Thus, the transfer of genetic
information occurs in a “shielded” environment. This is similar to integrated circuits
operating at super conductivity conditions.
Q:
Give examples of phenomena which seriously may damage the transfer of genetic
information.
Exergy is a measure of how much a system differs from equilibrium with the
environment. The more a system differs from the environment the more information is
needed to describe the system and the more information capacity can be carried by the
system. The relation between exergy and information (or information capacity) is therefor
as we have seen of fundamental significance.
Ex:
The net flow rate of information to the earth from the sun is about:
Pex
P  0.933
1.2  1017  0.933
 en

 4  1037  bits/s 
kT0 ln 2
kT0 ln 2
1.38  1023  300  0.693
Of this mankind make use of only a tiny fraction or about 5×1013 bits. The relation
between used and available information capacity becomes about 10–24. The same
relation for energy is about 3×10–5. Thus, mankind is a poor user of available
information or exergy. One reason to this will be mentioned below.
Information must be stored and transported safely. To reach this we must use
redundancy (over-explicitness) in codes and in copying, which implies access dissipation
of energy to make the process irreversible (safe) enough. The explicitness in the process
*
Since the relation between transferred or treated information and the used information capacity, both
measurable in exergy, is so small.
† It should also be noticed that errors in the information transfer of biological systems is a necessity. The
evolution is completely depending on mutations, i.e. errors in the reproduction of DNA.
74
EXERGETICS
increases on behalf of the information capacity. In biological systems there is a
continuous debugging or control of the transferred information. (As you might know this
becomes even more necessary in our increasingly computerized society, with the
belonging flourish computer viruses.)
In daily communication of information the exergy use is often far too high. Thus the
exergy per bit value becomes high, which implies that only a fraction of the available
information capacity is utilized, as mentioned before. In the living nature the solar exergy
is converted into highly ordered structures in the green plants. From an energy point of
view this is a small amount, but from an exergy or information point of view huge
amount is being utilized.
Let us compare the efficiency of information transfer, exergy per bit, for different
systems, see Table below. This has the dimension temperature, Ttransfer. The lower
temperature the more efficient is the information transfer, but if this temperature becomes
to low the thermal noise in the environment may ruin the information.
__________________________________________________________
Exergy per bit [J/bit]
Ttransfer [K]
__________________________________________________________
Electric typewriter
1
1023
Radio receiver
5×10-4
5×1019
Television
2×10-5
2×1018
Computer memory
10-12
1011
Human speech
10-16
107
Human ear
10-17
106
Human eye
5×10-18
5×105
Protein
biosynthesis in a cell
4.6×10-21
460
__________________________________________________________
The sensitivity of the retina is such that the human eye functions near the quantum
mechanical limit. It is in fact enough with only a few quanta to cause a reaction in the
eye. Storage of information in a computer memory has a characteristic temperature of
about 105 times the temperature for sight. But on the other hand the time resolution, and
thus the rate, is about 105 times higher than for the eye. The conclusion is that living
creatures and computers are each efficient in their use of exergy to receive and transfer
information. The biosynthesis in a cell is after all many times more efficient.
Electronic circuits, man’s ear and eye, and protein biosynthesis are drawn in a
logarithmic information rate (frequency) - power diagram in the Fig. below. The
minimum power requirements of integrated circuits or an elementary process in a circuit,
such as a measurement, a storage or a logic operation, requires an energy conversion
which is large in relation to kT ≈ 4×10-21 J (at room temperature). This is necessary to
avoid thermal fluctuations which are the cause of noise in electronic circuits, Brownian
movement, etc. The room temperature is marked with a straight line. A process must
ordinarily be far above this line to avoid serious disruption from thermal fluctuations. The
protein biosynthesis is surprisingly close to this critical line. The transfer from messenger
RNA to proteins, which uses half of the power requirements, is actually below this line.
This is possible because all of 4.3 bits of information are transferred at each
transformation. The ear and the eye occupy strikingly large areas in the diagram, covering
many orders of magnitude. Electronics is often regarded as the best technology we know
today. However, from an exergy efficient viewpoint, as the figure below shows, life itself
is far more efficient in its use of exergy to construct biological structures of matter.
75
EXERGETICS
Biological structures live by transforming energy from one form to another. The solar
exergy is used to build up complicated organic matter. This information is transferred
from generation to generation. The information which is stored in the genetic matter
(DNA molecule) directs the construction of matter when suitable conditions are given.
When biological material, e.g. wood or cellulose, is used as construction material, it is
these structures and this information that we benefit from.
Both exergy and information are measures of the deviation from a reference
environment. The exergy is the maximal amount of work that can be derived from such a
deviation, but the work is also needed to maintain and transfer information. Thus, the
relationship between exergy and information is very close.
Exergy Power
1W
10
Region of most
electronic circuits
-5
Micropower
circuits
10
10
-10
Human
ear
e
tur
a
r
pe
m
e
t
-15
t
ien
b
Am
Noise area
DNA
replication
10
Human
eye
-20
Information Rate
1
10
3
10
6
9
10 bits/s
Information rate - power diagram of information transfer
Q:
Consider the total information flow towards the earth during its lifetime. What
“happened”, from an information perspective after about 2-3 billion years, and
what was then the total amount of information that had reached the earth? What
“happened”, after another 2-3 billion years? The sun is now about middle age, so
if the evolution continues, what do you think will “happen” in the future, after
another 2-3 billion years and another 2-3 billion years? Why is this so hard to
understand?
These are some of my personal ideas about this. Let us consider the total information
flow towards the earth during its lifetime, i.e. about 10 billion years, see the Figure
below. Matter and the laws of nature were there from the start. After about 2-3 billion
76
EXERGETICS
years, when the earth had received about 1054 bits of information life was able to
develope on the earth into a myriad of life forms. Then after another 2-3 billion years the
ability to be aware of one self appeared, i.e. to be conscious, the mind or the soul. The
sun, which is “father”* of our solar system is now about middle age, so if the evolution
continues, what will appear in the future, after another 2-3 billion years and another 2-3
billion years? To answer this question is as difficult for us as it would have been for the
living organisms some billions of years ago, or a single bacterium of today to realize
Homo sapiens. Matter has organized itself into things as life, from attractive forces acting
on a microscopic level. In a similar way minds might organize itself into some form of
immaterial structure, where love might be one attractive force of importance. In this
immaterial universe, which is far beyond the so called Cyberspace, each mind would be
just as simple or stupid as molecules in the material world. However, this is just personal
speculations that are up to every one to fantasize about. The main reason for bringing up
this subject is to give a perspective of the human culture, it’s believes, myths and opinion
about the importance of itself, especially the European culture, i.e. the white man.
Finally let us compare the efficiency of a modern personal computer with other means
of information transfer, see the Table below.
Information system
Efficiency
Reading text by lamplight
10–20 - 10–21
Motorola Power PC or Intel Pentium processor
10–13
DNA-replication, protein synthesis
10–1 - 10–2
As we see a personal computer is far more efficient than reading by lamplight, but
compared with living nature a computer is still very poor. Today’s most sophisticated
computers are nothing but “steam engines” compared to future computers — please,
remember this!
*The natives of America use to say father sun and mother earth to describe the forces that creates our
world.
77
EXERGETICS
Summary
Let us close this Chapter with a schematic table of some forms of energy listed by
decreasing quality, from “extra superior” to “valueless”. The quality of the energy is
indicated by the exergy factor. The quality index ranges from 1 for potential energy,
kinetic energy and electricity (which are pure exergy and thus can be totally transformed
into all other forms of energy) to 0 for the exergy-lacking heat radiation from the earth.
The quality index of heat energy varies considerably from 0.6 for hot steam to zero for
heat radiation from the earth.
The
quality of different forms of energy.
_____________________________________________________
Form of energy
Quality index
(Exergy factor)
_____________________________________________________
Extra superior
Potential energy1
Kinetic energy2
Electrical energy
Superior
Nuclear energy3
Sunlight
Chemical energy4
Hot steam
District heating
Inferior
Waste heat
1
1
1
about 0.95
0.93
about 1
0.6
0.3
0.05
Valueless
Heat radiation from the earth
0
_____________________________________________________
1 e.g. highly situated water resources
2 e.g. waterfalls
3 e.g. the energy in nuclear fuel
4 e.g. oil, coal, gas or peat
78
EXERGETICS
CYCLIC PROCESSES
Cyclic processes in general
First we will consider cyclic processes in general.
A cyclic process is a process where the working medium, e.g. a gas goes through a
number of states in a cyclic pattern, i.e. each state is repeated in every cycle. We also
assume the process to be reversible.
The first law applied to the state change 1 to 2 along the upper curve in Fig. below
gives:
QH = U2 – U1 + WH
where QH is added heat and WH is the performed work, represented by the area between
the higher curve 1 to 2 and the V-axis, i.e.  PdV . When the medium is returned to the
initial state along the lower curve 2 to 1, we get as above:
–QL = U1 – U2 + (–WL)
where the heat –QL is added and the work –WL is extracted, now represented by the area
between the lower curve 2 to 1 and the V-axis.
P
QH
WH
1
W
QL
2
WL
V
For the complete cycle we have:
QH – QL = U2 – U1 + WH + U1 – U2 – WL
QH – QL = WH – WL = W
Thus, the net supply of heat QH – QL is equivalent to the net amount of performed work
W, which is indicated by the lined area in the Fig.
We can now define a thermal efficiency, t, of the cycle as the relation between the
work output and supplied heat.
t 
W QH  QL
Q
 1 L

QH
QH
QH
79
EXERGETICS
The Carnot cycle
We will now study the Carnot cycle, the cycle which has the highest theoretical
efficiency for thermal engines. According to the second law heat must go from high to
low temperature in order to produce work. The maximal work, to be extracted for two
heat reservoirs at temperatures TH and TL, is the rectangular area in a TS-diagram. This
cycle consists of two isothermal (T = constant) and two adiabatic or isentropic (S =
constant) processes, see Fig. below.
T
TH
P
1
QH
4
QH
2
2
Wnet
W
TL
1
W
4
QL
3
QL
S
S1
3
V
S2
A Carnot process consists of two reversible adiabatic and two reversible isothermal
processes.
1-2: isothermal “expansion”
2-3: adiabatic expansion
3-4: isothermal “compression”
4-1: adiabatic compression
From the Fig. we see: QH = TH(S2 – S1) and QL = TL(S2 – S1), which gives the work W
and the efficiency Carnot :
W  QH  QL  (TH  TL )( S 2  S1 )

Carnot 
W (TH  TL )( S 2  S1 )
T

 1 L


QH
TH  S 2  S1 
TH
Assume the following heat engine to realize the Carnot cycle. An ideal gas is
contained in a system of two heat exchangers, a turbine and a compressor, see Fig. below.
The turbine and the compressor are on the same axis, and the compressor is powered by
the turbine so the output work is
W = Wturbine – Wcompressor
80
EXERGETICS
QH
1
Wcompressor
2
Heat exchanger
Compressor
Turbine
Wturbine
Heat exchanger
4
3
QL
We must assume that process is ideal to be able to describe it, which means no heat
leakage or friction, i.e. completely reversible conditions. The high temperature in the
upper heat exchanger is TH and in the lower TL and TH>TL.
Let us follow one unit mass through the cycle.
1-2: isotherm, the heat QH at temperature TH is received in the upper heat exchanger.
2-3: reversible adiabatic, Q = 0, the gas is expanded in the turbine and work Wturbine is
extracted, the temperature decrease to TL.
3-4: isotherm, the heat QL at temperature TL is emitted in the lower heat exchanger.
4-1: reversible adiabatic, Q = 0, the gas is compressed, which uses work Wcompressor,
and the temperature increase to TH.
Let us now analyze the process step by step.
Isotherms:
Adiabatic:
V2
V1
1-2:
QH  RTH ln
3-4:
QL   RTL ln
2-3:
THV2 1  TLV3 1
4-1:
TLV4 1  THV1 1
V4
V
 RTL ln 3
V3
V4
which gives
 1
TH  V3 
 
TL  V2 
 1
V 
  4 
 V1 

V2 V3

V1 V4
The thermal efficiency of the Carnot cycle becomes:
81
EXERGETICS
V3
W
Q
T
T
V4
t 
 1 L  1
 1 L 
 Carnot ,
V
QH
QH
TH TH
RTH ln 2
V1
RTL ln

which is what we should expect.
—Why is this efficiency not reachable in practice? At first, we have a temperature
difference in the heat exchangers and all mechanical parts suffer from friction.
Ex:
The surface water in a lake becomes warm in the summer about 20˚C, whereas the
bottom water is only 4˚C. What is the highest thermal efficiency we can achieve
with this temperature difference?
S:
The highest efficiency is the Carnot efficiency, thus:
Carnot  1 
TL
273  4
 1
 0.0546  5.5%
TH
273  20
Inverse Carnot cycle
Above we used the Carnot cycle as a heat engine to extract work from two heat
reservoirs. We will now reverse the cycle, which is possible because it is reversible. Then
instead work will be used to produced heat, QH according to the relation W = QH – QL.
P
1
QH
2
W
4
QL
3
V
One example of this is a heat pump or a refrigerator, which is actually the same thing.
In the heat pump we use the produced heat and in the refrigerator we use the
simultaneously produced cold, i.e. the removal of heat.
For the refrigerator we have a coefficient of performance (COP) for the produced
cold,
COPcold 
Removed heat
Used work
For a Carnot process we have:
82
EXERGETICS
COPcold, Carnot 
since
QL
QL
1
1



W QH  QL QH  1 TH  1
QL
TL
QH TH

for a Carnot cycle.
QL TL
For a heat pump we get analogously,
COPheat 
COPheat, Carnot 
Delivered heat
Used work
QH
QH
1
1
1




W QH  QL 1  QL 1  TL  Carnot
QH
TH
Thermodynamics of steam
Before we study steam cycles we will treat the thermodynamics of steam. Since, steam is
a gas close to condensation it can not be treated as an ideal gas. Steam cycles are the most
common cycles in power industry, e.g. fossil and nuclear fueled power plants. Let us first
look at the steam pressure curve, which shows the different phases of water in a PVdiagram, se the Fig. below.
P
Criticalpoint
Liquid
Gas
Steam
T
To the left of the curve we have only water as liquid, to the right only steam and gas. On
the curve we have liquid and steam simultaneously, and at the critical point the difference
between liquid and gas vanishes, i.e. a liquid is a compressed gas.
We will also study the transition between gas and liquid in a PV-diagram, see Fig.
below.
83
EXERGETICS
P
Tcritical
T4
T3
T2
liquid
T1
critical
point
2
gas –
superheated
steam
3
1
4
damped steam
(liquid + gas)
steam
low phase curve –
saturated liquid (f)
Vf
high phase curve –
saturated gas (g)
V
Vg
Let us “walk” along an isotherm, T = T1, and start at the left end in the PV-diagram.
Here we are in the liquid region, and if the pressure drops the volume increases only
slightly, the liquid is almost incompressible. At 1 we have saturated liquid that starts to
boil, and along 1 to 4 the liquid gradually boils to gas, the volume increases but the
pressure is constant, and at 4 all liquid has turned to saturated gas. if we continue to right
the pressure will decrease and the volume increase, as for an ideal gas.
By doing the same procedure for different temperatures we get a lot of data, for
saturated liquid and saturated gas. By connecting these we get the phase curves for
saturated liquid and saturated gas, and between them we have the damped steam, where
liquid and gas appear simultaneously.
The critical point we find at the top, where saturated liquid becomes saturated gas
without increasing volume. The critical point for water is:
Pcritical = 221.29×105Pa
Tcritical = 647.3 K  critical = 374.15˚C
vcritical = 3.1×10–3 m3/kg
At temperatures higher than Tcritical=374.15˚C water can only appear as gas. Along the
isotherm T = Tcritical to the critical point the liquid turns to gas with no change of volume.
From now we assign saturated gas by “g” and saturated liquid by “f” (fluid).
The specific gas content or quality of the steam, x, is the relative amount of gas in the
damped steam, i.e.
kg gas
x
kg damped steam
84
EXERGETICS
The specific liquid content then becomes 1–x, i.e.
1 x 
kg liquid
kg damped steam
The specific volume v [m3/kg] of the damped steam (liquid + gas) at a specific
temperature can be calculated from:
v = xvg + (1–x)vf
Ex:
A container contains a damped steam at 3 bar. The volume is 20 liter and the mass
of the mixture is 0.04 kg. What is the gas content?
S:
First calculate the specific volume of the damped steam!
20  103
v
 0.5 m3 /kg
0.04
From steam tables we find for P= 3 bar:


vf = 0.00107 m3/kg
vg = 0.60567 m3/kg
By rewriting the relation between x, v, vf and vg we get:
v  vf (1  x)  vg x  x 
v  vf
0.5  0.00107

 0.8.
vg  vf 0.60567  0.00107
TS-diagram
Since dS 
Q
, where S is entropy and Q is heat added of temperature T, then we
T
can calculate the added heat, Q, of a process 1 to 2, as:
2
Q12   TdS
1
In a TS-diagram the added heat is the area below the curve T(S) and between state 1 and
2, as shown in the Fig. below.
T
1
2
Q
S
85
EXERGETICS
In technical applications one usually set h 0 and s 0 for water as liquid at 0˚C and
the corresponding saturation pressure.
Let us make a Ts-diagram for steam, see Fig. below.
T
K
critical
point
T critical
P2
liquid
gas –
superheated steam
P1
damp steam
(liquid + gas)
273
saturated
liquid (f)
q
0
steam
saturated gas (g)
qs
r
s
0 sf
sg
ss
The isobars almost follow the curve of saturated liquid up to the boiling point.
When we add heat to the water at constant pressure we have q = h2 – h1 [J/kg], i.e. the
added heat is equal to the change of enthalpy.
Assume we have 1 kg 0 degree (273 K) water at p1 to be heated to boiling point. The
added heat q then becomes
q = hf – 0 = hf
where hf is the enthalpy at the boiling point and pressure P1, i.e. at the saturation curve.
Thus, q becomes the area in the Ts-diagram below the isobar P = P1.
The heat of evaporation, r, we get in the same way if we add heat during constant
pressure P1, i.e.
r = hg – hf
where hg is the enthalpy of the saturated gas and hf is the enthalpy of the saturated liquid.
r is also the surface in the Ts-diagram below the isobar P = P1, in the damped zone.
To determine the enthalpy of damped steam we use the specific gas content, x. In the
same way as we got the specific volume of the damped steam we can now determine the
specific enthalpy h [J/kg] of the damped steam
h = xhg + (1–x)hf
if we use that r = hg – hf we get
h = hf + xr.
86
EXERGETICS
Similarly, we also get the specific entropy s [J/kg K] of the damped steam
s = xsg + (1–x)sf
The heat content of the superheated steam then becomes
hs = q + r + qs = hf + r + qs = hg + qs
R:
Make your self acquainted with steam tables!
Ex:
Liquid water at P = 10 bar and  = 120˚C is heated to steam with x = 0.90 at
constant pressure. What is the specific heating need?
S:
From steam tables we get at P = 10 bar saturation state 179.88 ≈ 180˚C and hf =
762.63 kJ/kg.
If we assume that cP is constant in the interval 120-180˚C then the enthalpy of
water at 120˚C becomes
h180 = cP(180 – 0)
h120 = cP(120 – 0)
Which gives h120 
h180
762 .6
 120 
 120  509kJ/kg 
180
180
For the damped steam we get
h = xhg+(1–x)hf ≈ 0.90×2778+(1–0.9)×762.6 ≈ 2576 [kJ/kg]
The added heat then becomes
q = h – h120 ≈ 2576 – 509 = 2067 kJ/kg
Ex:
We have a closed 4 m3 container with 800 kg of water at pressure 0.1 bar. How
much heat is required to increase the pressure to 20 bar?
S:
The volume is constant, so we have
q = cv∆T = u2 – u1 = h2 – h1 – v(P2 – P1)
the specific volume is v 
4
 0.005 m3/kg
800
We assume only water in the container, which will be damped steam. From steam
tables we get:
State 1: P = 0.1 bar
vf = 1.01×10–3 m3/kg
vg = 14.67 m3/kg
v = 0.005 m3/kg
87
EXERGETICS
From v = vgx1+(1–x1)vf, we get x1 = 0.00027
When we know x1 we can calculate the enthalpy, i.e. the heat content of the
damped steam at P = 0.1 bar.
h1 = hgx1+(1–x1)hf ≈ 192.5 kJ/kg
State 2: P = 20 bar
In the same way we get x2 = 0.0388 and h2 = 981.9 kJ/kg.
The added heat then becomes
Q = m[h2 – h1 – v(P2 – P1)] ≈ 800[981.9×103 – 192.5×103 – 0.005(20×105 –
0.1×105)] ≈ 623 MJ
Mollier or hs-diagram
In the same way we have drawn PV-diagrams and Ts-diagrams for steam we can also
make hs-diagrams or Mollier-diagrams. In the Mollier-diagram we also have curves for
isobars and isotherms. See Fig. below or a steam table.
h
P2
P1
gas –
superheated steam
T2
T1
steam
liquid
critical
point
x1
damped steam
(liquid + gas)
saturated gas (g)
x2
saturated
liquid (f)
s
Ex:
10 kg steam per second at P = 20 bar,  = 400˚C expands adiabatically
(isentropically) to P= 2 bar. Determine the available power.
S:
The available power is work per unit time, and from the first law for an adiabatic
process, i.e. Q = 0, we have
Wt = H1 – H2 = m(h1 – h2)
88
EXERGETICS
The steam expands adiabatically, i.e. the entropy s is constant.
State 1: P1 = 20 bar, 1 = 400˚C, gives h1 ≈ 3.25 MJ/kg
State 2: P2 = 2 bar, s = constant. Find the point P1,1, then find P2 = 2 bar along
the line s = constant and get h2 = 2.70 MJ/kg.
wt = h1 – h2 ≈ 3.25-2.70 = 0.55 MJ/kg = 550 kJ/kg
So, when 10 kg steam per second expands the available power is Wt = 5500 kW.
Steam power processes
Steam is often used in industry for heating, drying and cleaning. The power industry uses
steam to transfer the power or exergy in a fuel (combustion or fission) trough a steam
generator in a furnace to motion in a turbine connected to a electric generator.
As we have seen the Carnot process is the most efficient process to transfer exergy of
heat to mechanical exergy (work). In the TS-diagram below the Carnot cycle is shown for
a steam process.
T
K
4
TH
PH
1
PL
TL
2
3
273
s
0
0
s3
s1
An example of such a heat power plant is shown below.
89
EXERGETICS
Heat from fuel
Saturated liquid
4
Wcompressor
QH
Heat exchanger Saturated gas
Evaporator
1
Compressor
Turbine
Wturbine
Steam =
Steam =
gas + liquid Heat exchanger gas + liquid
Condenser
3
2
QL
Heat to cooling water
In the evaporator (steam generator) water is converted from saturated liquid, 4 to
saturated gas, 1. The heat is usually extracted from the exhaust gases by heat exchangers,
and the evaporation occurs at constant pressure PH and constant temperature TH. In the
turbine the saturated gas expands adiabatically to 2 with a lower pressure PL. The steam is
this state of high quality, i.e. the gas content is high. According to the first law of a steady
state reversible adiabatic process, we have:
Wturbine = H1 – H2
The steam is condensed at constant pressure PL and constant temperature TL, to state 3.
The steam in this state is of low quality, i.e. the gas content is low and the liquid content
is high. By a heat exchanger the heat is delivered to cooling water. After that the steam is
compressed adiabatically to its initial state 4. In this compression the remaining gas will
turn to liquid, i.e. the compressor faces a very big change in volume implying an big and
expensive compressor. Thus, it is more economical to cool the steam to saturated liquid at
pressure PL and temperature TL, see Fig. below.
T
K
PH
TH
4
TL
273
1
PL
2
3
s
0
0 s3 = s4
s1
This process is called Clausius- Rankine process or just Rankine process. The pressure
increase from state 3 to 4 can now instead be obtained with a pump, se Fig. below. State 4
90
EXERGETICS
will now be in the liquid region since the pressure increase is reversible, s3 = s4. The
heating of the liquid to saturated liquid will now occur in the evaporator instead.
Heat from fuel
Liquid
4
Wpump
QH
Heat exchanger
evaporator
Pump
Saturated gas
1
Turbine
Wturbine
Saturated
liquid
3
Ex:
S:
Steam =
Heat exchanger gas + liquid
condenser
2
QL
Heat to cooling water
Determine how the energy and exergy conversions change when we choose a
Rankine process instead of a Carnot process in the case PH = 5 bar and PL = 1 bar.
Also estimate the thermal energy and exergy efficiencies (T0 = 293.15 K).
The processes are identical beside states 3 and 4.
In state 1 we have saturated gas at PH = 5 bar, i.e. from steam tables we get
h1 ≈ 2748.79 kJ/kg
s1 ≈ 6.8219 kJ/kg K
e1 = h1 – T0s1 ≈ 2748.79 – 293.15×6.8219 ≈ 748.95 kJ/kg
We now must determine the quality of steam in state 2, x2. The entropy in states 2
and 1 are the same, i.e.
s1 = s2 ≈ x27.3590 + (1 – x2)1.3028 ≈ 6.8219  x2 ≈ 0.91131.
From steam tables, we can now calculate the enthalpy and exergy in state 2:
h2 ≈ 417.550 + 0.91131×2257.71 ≈ 2475.03 kJ/kg
s2 ≈ 0.088691.3028 + 0.91131×7.3590 ≈ 6.8219 kJ/kg K
e2 = h2 – T0s2 ≈ 2475.03 – 293.15×6.8219 ≈ 475.19 kJ/kg
In state 4 we have for the Carnot process saturated liquid at PH = 5 bar, i.e. from
steam tables we get:
h4 ≈ 640.16 kJ/kg
s4 ≈ 1.8604 kJ/kg K
e4 ≈ 640.16 – 293.15×1.8604 ≈ 94.78 kJ/kg
For the Rankine process we have liquid at PH = 5 bar and s4 = s3. Let us therefor wait
to calculate this state. First we determine the exergy in state 3 for the Carnot process.
91
EXERGETICS
Then we need to know the quality of the steam, x3. The entropy in states 3 and 4 are
the same, i.e. from steam tables we get
s4 = s3 ≈ x37.3590 + (1 – x3)1.3028 ≈ 1.8604  x3 ≈ 0.09207.
h3 ≈ 417.550 + 0.09207×2257.71 ≈ 625.42 kJ/kg
s3 ≈ 0.90793×1.3028 + 0.09207×7.3590 ≈ 1.86039 kJ/kg K
e3 ≈ 625.42 – 293.15×1.86039 ≈ 80.05 kJ/kg
For the Rankine process it is easier since we have saturated liquid
h3 ≈ 417.550 kJ/kg
s3 ≈ 1.3028 kJ/kg K
e3 ≈ 417.550 – 293.15×1.3028 ≈ 35.63 kJ/kg
Thus we know, for state 4 that PH = 5 bar and s4 = s3 ≈ 1.3028 kJ/kg K, which
principally is enough to determine the state in detail, but it is better to first calculate
the pumping work. We know the specific volume of the liquid to be pumped and the
pressure levels. If we assume the liquid to be incompressible we get:
wpump = v(PH – PL) ≈ 0.00107(5 – 1)105/1000 ≈ 0.43 kJ/kg
Where we used the average value of the specific volume in the two states. As we see
the pump work is almost negligible. Thus, the state becomes
h4 = h3 + wpump ≈ 417.55 + 0.43 ≈ 417.98 kJ/kg
s4 = s3 ≈ 1.3028 kJ/kg K
e4 ≈ 417.98 – 293.15×1.3028 ≈ 36.06 kJ/kg, but also as above, i.e.:
e4 = e3 + wpump ≈ 35.63 + 0.43 ≈ 36.06 kJ/kg.
Thus, for the enthalpy we have:
h1 ≈ 2748.79, h2 ≈ 2475.03, h3 ≈ 625.42 (Carnot), h3 ≈ 417.55 (Rankine), h4 ≈ 640.16
kJ/kg (Carnot), and h4 ≈ 417.98 (Rankine) and we get the following values for the
energy conversion:
For the Carnot process the compressor work becomes:
wcompressor = h4 – h3 ≈ 640.16 – 625.42 ≈ 15 kJ/kg.
For the Rankine process the pump work becomes:
wpump = h4 – h3 ≈ 417.98 – 417.55 ≈ 0.43 kJ/kg, as we already know.
For the Carnot process the heat to cooling water becomes:
qL = h2 – h3 ≈ 2475.03 – 625.42 ≈ 1850 kJ/kg.
For the Rankine process the heat to cooling water becomes:
qL = h2 – h3 ≈ 2475.03 – 417.55 ≈ 2057 kJ/kg.
For the Carnot process the added heat becomes:
qH = h1 – h4 ≈ 2748.79 – 640.16 ≈ 2109 kJ/kg.
For Rankine process the added heat becomes:
qH = h1 – h4 ≈ 2748.79 – 417.98 ≈ 2331 kJ/kg.
The utilized energy in the turbine, which is the same in both cases becomes:
wturbine = h1 – h2 ≈ 2748.79 – 2475.03 ≈ 274 kJ/kg.
For the exergy we have: e1 ≈ 748.95, e2 ≈ 475.19, e3 ≈ 80.05 (Carnot), e3 ≈ 35.63
(Rankine), e4 ≈ 94.78 kJ/kg (Carnot), and e4 ≈ 36.06 (Rankine), which implies:
92
EXERGETICS
Carnot: wcompressor = e4 – e3 ≈ 94.78 – 80.05 ≈ 15 kJ/kg, as above.
Rankine: wpump = e4 – e3 ≈ 36.06 – 35.63 ≈ 0.43 kJ/kg, as above.
Carnot: qL = e2 – e3 ≈ 475.19 – 80.05 ≈ 395 kJ/kg.
Rankine: qL = e2 – e3 ≈ 475.19 – 35.63 ≈ 440 kJ/kg.
Carnot: qH = e1 – e4 ≈ 748.95 – 94.78 ≈ 654 kJ/kg.
Rankine: qH = e1 – e4 ≈ 748.95 – 36.06 ≈ 713 kJ/kg.
Carnot and Rankine: wturbine = e1 – e2 ≈ 748.95 – 475.19 ≈ 274 kJ/kg, as above.
The thermal energy efficiency, i.e.
en,Carnot 
wturbine  wcompressor
en,Clausius Rankine 

274  15
 12.3%
2109
qH
wturbine  wpump
qH

274
wturbine

 11.8%
qH
2331
The thermal exergy efficiency, i.e.
ex,Carnot 
wturbine  wcompressor
ex,ClausiusRankine 

Net work output
becomes in the two cases:
Added exergy as heat
274  15
 39.6%
654
qH
wturbine  wpump
qH
Net work output
becomes in the two cases:
Added energy as heat

wturbine 274

 38.4%
qH
713
As we expect the Rankine process has a lower efficiency than the Carnot process, which
is the most efficient process.
Q:
Check that we have both energy and exergy balances for the processes, i.e. that
both the first and second law is valid. What is the reason for the big difference
between the thermal energy and exergy efficiencies and why are they so poor?
Let us now also put the Rankine process in a Mollier diagram, i.e. a hs-diagram, see the
Fig. below.
93
EXERGETICS
h
PH
PL
1
h1
2
h2
x2
h4
h3
4
3
s
s3
s1
Let us see how to improve the efficiency of the Rankine process:
1.
Increase the pressure before the turbine, which means that the conversion 1-2 is
moved to left in the hs-diagram above. The difference h1 – h2 = e1 – e2 = wturbine
then increases, i.e. more work out.
2.
Lowering the pressure after the turbine implies that the cooling loss qL = e2 – e3
reduces and we utilize exergy which otherwise would be lost with the cooling
water.
3.
Superheating; in a real turbine we can not allow the quality of the steam to
become less than about 0.90, i.e. in state 2. The liquid will otherwise ruin the
turbine blades. Therefor the gas in state 1 must be superheated, which principally
also increases the efficiency – according to Carnot. However, material restrictions
in the turbine put an upper limit of about 600˚C in today's conventional
constructions.
4.
Feed water heating; an important reason of the poor efficiency is that a large
amount of the heating content of the steam is due to evaporation. By pre heating
of the water to the evaporator the efficiency could be improved. This is usually
done by draining the turbine of steam, which is mixed with the feed water to
increase its temperature.
In the treatment above we have assumed no losses in the components. In reality we
have a lot of losses, but with adequate combination of high pressures and superheating
before the turbine, low pressures after the turbine, and feed water heating we can still
reach efficiencies of about 40%.
We will now look closer at the losses. Before, we concluded that the available work
from a reversible adiabatic steady state process, i.e. Q = 0 and S = constant, is Wt = Ws =
H1 – H2 = E1 – E2.
In reality we have losses, i.e. Q ≠ 0 and S ≠ constant, so the output work is less and
we must add more work to the compressor or the pump. We can represent the internal
losses by the energy and exergy efficiency of the turbine, t,en and t,en, which are defined:
94
EXERGETICS
 t,en 
w
Actual work
 a   t,ex   t
Reversible work ws
Let us see what this implies for the turbine and the compressor in the Carnot process.
Turbine
PH
T
h
PH
1
PL
PL
1
2'
2'
2
2
s
s
The maximum work appear when ∆S = 0, but as we said before ∆S>0 for all real
processes. 1-2 gives maximum work, h1 – h2 and 1-2' gives the real work, h1 – h2', see Fig.
above. As we see h1 – h2 > h1 – h2'. The above is valid if no heat loss occur. The efficiency
of the turbine becomes:
h h
t  1 2'
h1  h2
which usually is about 0.80-0.85.
Compressor
The energy and exergy efficiencies of the compressor becomes accordingly:
c,en 
Reversible work ws

 c,ex  c
wa
Actual work
PH
T
PH
h
PL
4
PL
4'
3
4'
4
3
s
s
∆S>0 for the compressor since the process is irreversible, thus, we have
95
EXERGETICS
k 
h3  h4
h3  h4'
which is usually about 0.80-0.85.
Let us now return to the Rankine cycle and see how it usually looks, see Fig. below.
PH
T
h
PL
1
PL
4
3
PH
1
2
2
4
3
S
s
For a reversible Rankine cycle we have:
4-1 Liquid water converts to steam at pressure PH, we add heat QH.
1-2 The steam expands adiabatically, i.e. ∆S = 0, to pressure PL, just below saturation.
Work is transferred through the turbine to an electric generator in a power plant.
2-3 The steam is condensed at pressure PL, we remove heat QL.
3-4 Liquid water is pumped isentropically, i.e. ∆S = 0, to the boiler and the pressure
increases to PH. We feed the pump work Wp.
Net produced work becomes from first law:
Wnet = QH – QL – Wp
The pump work becomes, if we assume the liquid water to be incompressible and the
specific volume v = 0.001 m3/kg.
Wp≈0.001∆P = 0.001(PH – PL) [J/kg]
The theoretical thermal energy efficiency of the Rankine cycle then becomes:
 tt,en 
Output work
h h
 1 2
Input enthalpy h1  h4
And analogously for the theoretical thermal exergy efficiency:
 tt,ex 
Output work e1  e2

Input exergy e1  e4
96
EXERGETICS
As we already have concluded it is advantageously to superheat the steam to extract
more work from the turbine. In practice the pressure is below 200 bar and the temperature
below 550-600˚C at the admission state 1. As we see from the Fig. of the Rankine cycle
above, the actual work is depending of the pressure drop between the admission state 1
and the final state 2. The lower pressure (back pressure) the more work we get, however,
the content of liquid water in the steam must not exceed 10-12 weight-%, which gives the
lower limit of the pressure.
An other way to improve the steam power cycle is back pressure and cogeneration of
power and heat for district heating. If we have a simultaneous need of power and heat this
may be a good solution.
In the condensing power cycle we extract more power, but the low condensing
temperature makes the heat useless. In the cogeneration power cycle we extract less
power, but the condensing temperature is now high enough to make the heat useful. The
heat can be used for process heat or space heating. In this way the energy efficiency
increases, but the exergy efficiency is more or less the same.
There are a number of efficiencies to describe the different losses in a steam power
plant. Let us just mention some of them in energy units.
- Combustion losses in the boiler, the boiler energy efficiency:
b 
Energy in steam
  for big plants.
Energy in fuel
- The theoretical thermal energy efficiency depends on the design of the steam power
cycle:

tt ≈ 0.5 for a good design.
- The efficiency of the turbine (energy and exergy):

t ≈ 0.8-0.9
- Mechanical (friction) and electrical losses in the turbine and generator. The energy and
exergy efficiencies are usually:

m+el ≈ 0.9-0.95 for big plants.
97
EXERGETICS
The total energy efficiency from fuel to electricity then becomes about
tot = btttm+el ≈ 0.35-0.4 (NB, the pump is not included!)
Losses are also due to the use of power at the power plant, e.g. for pumps, fans and
internal transportation. These losses often add to between 2 and 5% of produced
electricity.
Ex:
A steam turbine has the admission state 60 bar, 500˚C, and the steam flow 10
ton/h. What is the power output if the back-pressure is 6 bar? (t = 0.80)
S:
From above we have:

t 
h  h2 '
Acual work
 1
Reversible work h1  h2
Reversible work: ws = h1 – h2
Actual work: wa = h1 – h2' = t(h1 – h2)
From steam tables we get:
h1 = 3425 kJ/kg and h2 = 2810 kJ/kg
(We find h2 by first finding state 1 in the Mollier chart and then follow an
isentropic line until it cross the pressure PL= 6 bar then we have state 2.
Pturbine  m  t (h1  h2 ) 
10  103
 0.8  (3425  2810)  1370kW
3600
In addition to this we also have mechanical and electrical losses, i.e. m+el ≈ 0.90.95, so Pel in reality becomes even less.
Ex:
a) An industrial cogeneration plant is working with the admission state 100 bar,
500˚C and back-pressure 5 bar. What is the maximum electricity production if the
heating need is 15 MW, t = 0.82 and el+m = 0.94.
b) The plant is used only for production of electricity, and the pressure after the
turbine is 0.08 bar. Determine the electric power and the specific amount of steam
at the real state after the turbine. c) Calculate the exergy output in the two cases.
(T0 = 20˚C)
S:
a) PH = 100 bar &  = 500˚C  h1 = 3.37 MJ/kg
PL2 = 5 bar & s = constant  h2 = 2.65 MJ/kg
h h
  t  1 2'  h2 '  h1   t (h1  h2 )  3.37  0.82  (3.37  2.65)  2.78
h1  h2
The condensing heat is used when the steam is condensed. The actual enthalpy
after the turbine is h2'. If h3 is the state when all the steam is condensed, i.e.
saturated liquid, then we get from steam tables:
PL2 = 5 bar and saturated liquid  h3 = 0.64 MJ/kg
The condensing heat is qL = h2' – h3 = 2.78 – 0.64 = 2.14 MJ/kg
98
EXERGETICS
We know that the heating need is 15 MW (15 MJ/s). From this we can determine
how much steam that must be condensed per second.
15  106
 7kg/s 
2.14  106
Thus, 7 kg steam is expanded in the turbine per second, which gives:
Pel  Wel  m (h1  h2 ) tmel  7  (3.37  2.65)      3.9MW .
b) The theoretical state after the turbine now instead becomes, from steam tables:
PL1 = 0.08 bar & s = constant  h2 = 2.06 MJ/kg
The real state is 2': h2' = h1 – t(h1 – h2) = 2.30 MJ/kg
Thus, we have at this state: h2' = 2.30 MJ/kg and P2 = 0.08 bar, and from the hsdiagram we get the gas content of the steam to: x ≈ 0.88
The electric output becomes:
Pel  Wel  m (h1  h2 ) tmel  7  (3.37  2.06)      7.1MW
c) The exergy in the produced heat in the cogeneration case becomes:
PL,ex  WL,ex  m (h3  h0 )  T0 ( s3  s0 ) 
 7  [(640.16  83.90)  (273.15  20)(1.8604  0.296)]  684kW   0.7MW 
The total exergy power then becomes: Pex = PL,ex + Pel ≈ 0.7 + 3.9 ≈ 4.6 [MW]
And in the case of pure condensation: Pex = Pel ≈ 7.1 [MW]
Let us summarize:
a) Cogeneration gives the energy power: 3.9 MW electricity and 15 MW heat.
b) The plant gives 7.1 MW electricity and the steam quality is 0.88.
c) Pex,a ≈ 4.6 MW (Pen,a ≈ 18.9 MW) and Pex,b ≈ 7.1 MW (≈ Pen,b)
Refrigerators and heat pumps
We have seen that heat can move from lower to higher temperature by adding work to
a Carnot cycles going backwards. This is what happens in refrigerators and heat pumps.
The most common systems are based on expansion, compression and absorption. We
will treat the compression process, since it is the most common.
The compression process is similar to a steam power cycle running backwards, see
the Fig. below. In short the boiling temperature of the refrigerant, i.e. the working fluid, is
moved by changing the pressure. A liquid boils (evaporates) at a low pressure, i.e. takes
up heat at low temperature, and then it condenses at high pressure, i.e. gives away heat at
high temperature.
99
EXERGETICS
Heat is released at TH
3
Saturated
liquid
QH
Heat exchanger
Condenser
Expansion valve
2
Gas
Compressor
W compressor
Saturated gas
4
Steam =
gas + liquid
1
Heat exchanger
Evaporator
QL
Heat is received atTL
For an ideal process we have:
1-2 The refrigerant is adiabatically, i.e. ∆s = 0, compressed to pressure PH.
2-3 The superheated refrigerant (gas) condenses at pressure PH and then releases heat QH
at temperature TH.
3-4 The refrigerant expands by an expansion valve to pressure PL. The expansion occurs
at constant enthalpy (H = constant) since neither heat nor work is involved (First
Law).
4-1 The refrigerant is in the damp region (gas + liquid). The liquid part evaporates and
takes up the heat QL at temperature TL.
In the Figs. below we see the process in a Ts and Ph-diagram.
In the ideal process the expansion occurs at constant enthalpy, which implies that
energy is neither added nor removed. However, a lot of exergy is lost in the expansion.
Let us apply the First Law:
Wcompressor = QH – QL
QL = H1 – H4 = H1 – H3, since H3 = H4
Wcompressor = H2 – H1
100
EXERGETICS
QH = H2 – H3
The efficiency as a refrigerator is defined by the Coefficient of Performance, COPcooling:
COPcooling 
Received heat
QL
H  H3

 1
Added work Wcompressor H 2  H1
The efficiency as a heat pump is similarly defined by COPheating:
COPheating 
Released heat
QH
H  H3

 2
Added work Wcompressor H 2  H1
The highest COP we get for a Carnot process.
COPcooling,Carnot 
COPheating,Carnot 
QL
TL

QH  QL TH  TL
QH
TH

 1  COPcooling
QH  QL TH  TL
N.B.! T is the absolute temperature, i.e. in Kelvin.
In practice COPcooling = 0.4 to 0.6×COPcooling, Carnot and if all heat from the heat pump is
being used then COPheating = 1 + 0.4 to 0.6×COPcooling, Carnot.
Ex:
A house is kept at 20˚C by a compressor heat pump a cold winter day (–20˚C).
The heat source is the ground at 0˚C. The refrigerant is R-12, working between –
5˚C and 45˚C, which gives 40˚C in the radiators. a) Calculate the COPheating if the
system works ideally between –5˚C and 45˚C. b) What is then the COPheating,Carnot?
c) Estimate the COP in practice.
S:
a) From steam tables for R-12, we find:
State 1, L = –5˚C, saturated gas  PL ≈ 2.61 bar & h1 ≈ 349 kJ/kg.
State 3, H = 45˚C, saturated liquid  PH ≈ 10.84 bar & h3 ≈ 244 kJ/kg.
101
EXERGETICS
We now have to determine the enthalpy at state 2. Look for state 1 in the Phdiagram. Assume reversible process, i.e. s is constant. Follow the isentropic curve
to PH = 10.8 bar, and find h ≈ 376 kJ/ kg.
COPheating 
h2  h3 376  244

 4.9
h2  h1 376  349
b) COPheating,Carnot 
TH
45  273

 6.4
TH  TL (45  273)  (5  273)
c) In practice we have COPheating ≈ 1 + 0.4 to 0.6×COPcooling, Carnot ≈ 1 + 0.4 to
 5  273
≈ 3.1-4.2. Thus the result above seem to optimistic,
0.6×
(45  273)  (5  273)
after all we did assume no losses.
Ex:
A one stage refrigerator should have the cooling power 58 kW. The refrigerant is
ammonia, evaporating at –10˚C and condensing at 25˚C. Determine: a)
ammonium flow, b) COP, c) ammonium flow and COP when super cooling occur
at 15˚C in the condenser and d) the theoretically highest COP in b) and c).
S:
From the Ph-diagram for ammonia, we have:
At evaporation: L = –10˚C, PL= 2.9 bar and h1 = 1452 kJ/kg
At condensation: H = 25˚C, PH = 10.0 bar and h3 = 318 kJ/kg
At condensation and super cooling: H' = 15˚C, PH' = 10.0 bar and h3' = 270 kJ/kg
From the Ph-diagram we have h2 ≈ 1620 kJ/kg
Cooling power: Pcooling  m (h1  h3 )wherem is the refrigerant flow
h  h3
Coefficient of Performance: COPcooling  1
h2  h1
TL
The theoretically highest COP: COPCarnot 
TH  T L
Pcooling
58

 0.051 kg/s
a) m 
h1  h3 1452  318
h h
1452  318
 6.6
b) COPcooling  1 3 
h2  h1 1620  1452
Pcooling
58

 0.049 kg/s and
c) m 
h1  h3' 1452  270
h  h3' 1452  270
COPcooling  1

 6.8
h 2  h1 1620  1452
TL
273  10

 7.5
d) Case b: COPCarnot 
TH  TL 273  15  (273  10)
TL
273  10
Case c: COPCarnot 

 10.5
TH  T L 273  25  (273  10)
102
EXERGETICS
HEAT TRANSFER
Heat* is as we have mentioned before a consequence of that two bodies with different
temperature exchange internal energy. From the second law of thermodynamics heat
(internal energy) will spontaneously go from the warmer to the colder body. Heat should
therefor rather be regarded as a process like work, they are not true forms of energy like
mechanical and electrical energies. How heat is being transferred is a very complicated
science, i.e. involving material properties. This makes this area to a scientific field beside
thermodynamics. Heat can principally be transferred in three different ways:
1. Heat conduction
2. Heat convection, by self-convection or by external force
3. Heat radiation
Heat conduction
In a solid body heat is transferred from the warmer to the colder part by direct contact
between the micro particles in the body. The temperature drops linearly through a
homogenous body of rectangular shape.
T [K]

TH
Pen
Pen
Pex,H
Pex,L
TL
x [m]
d
The energy power Pen that is transferred through conduction is:
Pen,l 

d
A(TH  TL ) [W = J/s]
where TH and TL are the different temperatures [K], since it is a difference it is also valid
for ˚C,
A is the area which the heat flow is facing [m2],
 d is the distance the heat flow has to transfer [m] and
  is the heat conduction [W/m˚C or W/mK] for the body.
The exergy power Pex that is transferred through the body consists of firstly the exergy
flow in to the body at the temperature TH:
*
Heat is here reagarded both as energy and exergy.
103
EXERGETICS
E
T T
Pex,l,H    Pen,l  H 0 Pen,l
TH
 Q H
where T0 is the environmental temperature, and secondly of the exergy flow that leaves
the body at the temperature TL:
E
T T
Pex,l,L    Pen,l  L 0 Pen,l
TL
 Q L
The difference between these flows becomes:
Pex,l  Pex,l,H  Pex,l,L 

TH  T0
T T
Pen,l  L 0 Pen,l 
TH
TL
TH  T0 TL  TL  T0 TH P
en,l
THTL


d
A

T0 TH  TL 
Pen,l 
THTL
T0
TH  TL 2
THTL
This is the exergy needed to “fuel” the process, i.e. to maintain the temperature
difference.
The heat conductivity, , denotes how much heat passes per unit time through 1 m2 of
the body when this is 1 m thick and the temperature difference is 1 K (here equals 1˚C). 
is a material constant. The higher the  the more heat is conducted through the material.
If we want a good heat insulator, then we should look for a small . Metals are good heat
conductors (≈45 for steel), which is an advantage in e.g. heat radiators, whereas mineral
wool (≈0.036) and other porous materials are poor heat conductors, because of enclosed
stationary air. In liquids and gases we also have to consider convection, i.e. mass flows,
because of the temperature dependence of the density. This we will study below, but first
some numerical examples.
Ex:
A glass window has an area of 1m2, outside temperature is –5˚C and inside is
+17˚C. The thickness of the glass is 5 mm. What is the heat rate (energy power)
through the window, if we only consider heat conduction?
S:
A = 1 m2, H = 17˚C, L = –5˚C, d = 5 mm = 5×10–3 m,  = 0.9 W/m˚C at 20˚C,
from tables. The  value is depending on the temperature, however in most cases
we can neglect this dependence. From the equations above we get the energy
power:


0 .9
Pen, l  ATH  TL   A H   L  
 1  17  (5)  3960  4 kW 
d
d
5  10  3
Thus, a considerable amount of energy and energy loss. For typical two glass
windows the value for /d is about 2 W/m2˚C and is called overall heat transfer
coefficient and indicated by the symbol U.
Ex:
What is the exergy loss in the example above?
104
EXERGETICS
S:
The energy power leaking trough the window is Pen,l as above. Assume the
environmental temperature is equal to the outdoor temperature, i.e. T0 = TL then
the exergy will be zero for the heat when it reaches the outside temperature. The
exergy factor at the temperature TH becomes:
E
17  (5)
T T T T
  L
   H 0  H L  H

 0.075  7.5%.
273.15  17
TH
TH
TH
 Q H
The exergy loss then becomes:
E
Pex  Pex,l  Pen,l  0.075  3960  300[W ]
Q
The needed exergy power to maintain the temperature difference, thus, is far less
than the energy power. Let us compare this with a situation when we try to keep a
leaking bucket, which leaks 300 liter per hour full of water. Then adding more
than 300 liters per hour will only flow away, and to pour on more 4000 liter per
hour is ridiculous!
Since, the heat transfer is towards the environment the exergy flow only
becomes a loss, i.e. no exergy is being transferred.
Ex:
Let us compare dry and wet mineral wool. The area is 2 m2 and the temperatures
are 0˚C on outside and +20˚C on inside, the thickness is 10 cm. What are the
energy flows and the exergy losses in the two cases?
S:
We have: A = 2 m2, H = 20˚C, L = 0˚C, d = 10 cm = 0.1 m,  = 0.04 for dry wool
and for water we have  = 0.60 from tables. The energy flows in the two cases
become, if we neglect the wool in the wet case:


0.036
Pen,l,dry  A(TH  TL )  A( H   L ) 
 2  ( 20  0)  14.4 [W]
d
d
0 .1
0.60
Pen,l,wet 
 2  ( 20  0)  240 [W]
0 .1
Thus, insulation is very sensitive for water or moist. (Moist is also good if you
want to grow mold and fungi.)
The exergy factor becomes, in both cases, since T0 = TL:
200
 H  L


 0.068
273.15  20
Q
TH
Thus, the exergy losses becomes:
E
Pex,dry  Pex,l,dry  Pen,l,dry  0.068  14.4  1 [W ]
Q
Pex,wet  Pex,l,wet 
E
Pen,l,wet  0.068  240  16 [W ]
Q
The energy or exergy need to maintain the temperature difference increase by 16
times if the insulation is wet.
105
EXERGETICS
Heat conduction is of course also acting in gases and liquids, but heat transfer in these
phases mainly instead relates to mass transfer or convection, see below. This implies that
heat conduction is only of interest in gases and liquids if mass transfer is not aloud.
Heat convection
Convection is heat transfer by transfer of mass. The heat you can feel over a hot
radiator is due to that hot air rises when its density decreases, i.e. the air becomes lighter.
In the same way water on the stove is heated when the bottom becomes warm, the water
rises and replaces by colder heavier water, i.e. natural stirring. This does not occur in a
microwave oven, where the heat is being generated inside the substance, so we need to
manually “convect” the water, stir it. Thus, we differ between free and forced convection.
Free convection – free current
A wall of temperature TH is in contact with a gas (or liquid) of temperature TL, TH>TL.
From heat conduction, i.e. direct contact heat is being transferred to the medium in
contact with the wall. This medium then gets a higher temperature and lower density and
rises. Then the left space is being replaced by colder media, and so on. A free current of
the media has evolved.
T
Hot medium
P
Cold medium
TH

P
TL
Transferred energy power by convection then becomes:
Pen,c  A(TH  TL )
where  = heat transfer coefficient [W/m2K]
A is the area exposed to convection [m2]
The heat transfer coefficient , denotes how much heat [W] is transferred per unit time
and unit area at a temperature difference of 1 K [˚C] when we have convection.
As for heat conduction the exergy power becomes:
Pex,c,H 
T T
TH  T0
Pen,c  A H 0 (TH  TL )
TH
TH
106
EXERGETICS
Pex,c,L 
Pex,c 
T T
TL  T0
Pen,c  A L 0 (TH  TL )
TL
TL
T0 (TH  TL )
T
Pen,c  A 0 (TH  TL ) 2
THTL
THTL
Ex:
Assume hot water surrounding a pipe with the diameter 55 mm and length 4 m.
How much heat (energy and exergy) per unit time is transferred from the water to
the pipe if the water temperature is 50˚C and the pipe is 20˚C and if the heat
transfer coefficient is 5000 W/m2K? (0 = 0˚C)
S:
The area A is the area of the pipe, i.e. A = dL ≈ 3.14×55×10–3×4 ≈ 1 m2, H =
50˚C, L = 20˚C, d = 5 mm = 5×10–3 m. From the relations above we have:
Pen,c = A(TH–TL) = dL(H–L) ≈ 5000×3.14×55×10–3×4(50–20) ≈ 104 [kW]
The exergy power becomes:
Pex,c,H 
TH  T0
500
 
Pen,c  H 0 Pen,c 
 104  16 [kW]
TH
TH
273  50
Pex,c,L 
TL  T0
20  0
 
Pen,c  L 0 Pen,c 
 104  7 [kW]
TL
TL
273  20
Pex,c 
T0 (TH  TL )
273  (50  20)
Pen,c 
 104  9 [kW]
THTL
(273  50)  (273  20)
Thus more than half of the exergy being transferred is lost in the process.
Forced convection
Force convection appear when we increase the convection by external forces, e.g. by a
pump or a fan. As we are familiar with the heat transfer increases with the speed of the
fluid, e.g. we experience more heat losses when it is windy.
We have the following relations for the heat transfer coefficient at forced convection:
Air and steam
In pipes between +20˚C and +100˚C
 = c1(v) 0.79 [W/m2K]
where c1 is a constant depending on the shape of the pipe and the flowing medium,
the following approximate values are usually used:
107
EXERGETICS
Pipe diameter [mm]
c1 for air
c1 for steam
10
8.1
13.5
20
7.2
12.0
50
6.4
10.7
v is the speed of the flowing medium [m/s]
 is the density of the flowing medium [kg/m3]
For flat areas
 = 5.8+3.95v
v<5 m/s
 = 7.15v0.78
v>5 m/s
Water in pipes
 = c2v0.87 [W/m2]
At pipe diameters between 10 and 30 mm and temperatures between 10 to 50˚C c2 is
evaluated from:
c2 = 3370 + 55
( is the temperature in ˚C)
For condensing water we have  ≈ 10000 W/m2K.
Ex:
A hot pipe, length 10 m and diameter 10 cm, with water at the temperature 60˚C
passes a cold area at environmental temperature 0˚C. How much energy and
exergy is lost in this area? Assume that the area of the pipe can be regarded as a
flat surface, and assume that the surface of the pipe has the same temperature as
the flowing water.
S:
The area A becomes: A = dL ≈ 3.14×10×10–2×10 = 3.14 m2.
The heat transfer coefficient becomes from above (air surrounding a flat area): 
= 5.8 + 3.95v = 5.8 W/m2K, since v = 0 m/s.
The energy power becomes: Pen,c = A(H–L) = 5.8×3.14×(60 – 0) ≈ 1.1 [kW]
T (T  T L )
The exergy power, being lost, becomes:  Pex,c  0 H
Pen,c 
TH T L
T  TL
60  0
  L
 H
Pen,c  H
Pen,c 
 1100  200 [W]
TH
TH
273  60
A very efficient heat transfer occurs if the convection appears simultaneously as a
phase change, e.g. if a liquid is evaporation from a surface. This we experience when we
are wet or when we are sweating. At the evaporation heat is taken from the skin, which
then becomes extra cold.
108
EXERGETICS
Heat radiation
A body of temperature above the absolute zero (0K or –273.15˚C) emits
electromagnetic radiation, as light and heat radiation. High frequency radiation (blue
light, ultra violet, or x-ray) has more energy than low frequency radiation (red, infra-red
or heat radiation). The radiation is a way of transporting energy or exergy between body
which are not in physical contact. Thus, energy and exergy can be transported in vacuum
and long distances. The energy and exergy flow from the sun to the earth is one example
of this.
As we know the color has an important effect on the radiation. Dark clothes absorbs
more heat than light clothes. Á black body, by definition absorbs all incoming radiation
and it is at all temperatures the most emitting and absorbing body. A white body, by
definition reflects all incoming radiation. A gray body absorbs an equal amount of the
incoming radiation at all frequencies. A colored body radiates at certain frequencies
defined by the color.
A reflecting surface reflects the radiation at the same angle as the incoming radiation
and a rough surface reflects the incoming radiation in all directions, as scattered light.
The so called emission coefficient  denotes the amount of the radiation which is
emitted, i.e. leaves the surface. A surface with a low emission coefficient emits less of the
radiation to the environment and more of the radiation is reflected back into the body. A
mirror reflects most of the incoming radiation and only a small part is absorbed. A shiny
and white surface has a low emission coefficient, whereas a rough and black surface has a
higher emission coefficient and high absorption. A surface that absorbs all incoming
radiation or emits all outgoing radiation, has the emission coefficient  = 1, i.e. a black
body.
Q:
Perhaps you have seen a little wheel with wings on a vertical axis in a small glass
bulb. Usually four wings, like a water-wheel, one side black the other side white.
If you put it in the sunlight it starts to move, in which direction and why?
The emitted radiation per unit area and time from a black body is determined by
Stefan-Boltzmanns law:
Pen,r

4








T 

 AT  5.67  10 AT  5.67 A
100 
4
8
4





The energy power from a gray body instead becomes:




T 

 100 


Pen,r  AT 4   5.67  108 AT 4    5.67 A

4




where A is the emitting surface [m2]
  is the emission coefficient or absorption coefficient, which is found in tables.
T is the temperature in Kelvin.
As we see the radiation depends on the temperature raised to the 4th power, which
implies that when the temperature doubles the radiation increase by a factor of 16!
109
EXERGETICS
Q:
Assume that we concentrate sunlight by a big concave mirror. What is the
maximal temperature we could get in the focus and why? There are two different
explanations!
From above we have that the exergy of radiation is:
 1  T 4 4 T 
Pex,r  1   0   0  Pen,r
 3  T  3 T 
Radiation between different bodies
Since every body (over 0K) emits radiation we must consider the exchange of radiation
between different bodies to be able to calculate the net radiation from the warmer to the
colder body. Let us first consider two parallel surfaces of equal size, but with different
temperatures TH, TL and emission coefficients H, L, see below.
TH ,  H
T L,  L
PH
PL
TH  TL
We have the following radiation situation between the surfaces
 Pen,r,H   HATH4   H 5.67  108 ATH4

4
8
4
 Pen,r,L   LATL   L 5.67  10 ATL
The hot surface the receives the energy power Pen,r,L and the surface will absorb the
energy power HPen,r,L, while the rest (1-H)Pen,r,L is reflected back to the surface of lower
temperature, where the energy power L(1-H)Pen,r,L is absorbed, while the rest (1-L)(1H)Pen,r,L is reflected back, etceteras. The net energy flow, which of course always is
directed towards the surface of lower temperature, TL becomes:
Pen,r,HL 
1
H

1
1
L
1
ATH4  TL4 
which is valid for all colored surfaces (black or white) and liquid surfaces.
The net exergy flow becomes from above:
110
EXERGETICS
Pex,r,HL 
1
H

1
1
L
  1  T  4 4 T 
 1  T  4 4 T  
4
4
0
0
  TL 1   0   0  
ATH 1    
3
T
3
T
 3  TL  3 TL  
H
 H
 1  

For a hot surface completely covered by an other surface at lower temperature, see
Fig below, we get the net energy flow:
T H,  H
A
P HL
H
T L,  L
Pen,r,HL 
A
L

1
 TH4  TL4

1
1  1
  1

 H AH AL   L 

which is valid for all kinds of rooms where the hot body can be randomly placed. If the
net power P is negative then TH<TL. The exergy net power becomes as above:
Pex,r,HL
  1  T  4 4 T 
 1  T  4 4 T  
1
4
4
0
0
0
  TL 1   0  

 TH 1    

3  TL  3 TL  
   3  TH  3 TH 
1
1  1




  1  

 H AH AL   L 
Often the radiation conditions are more complicated and thus difficult to calculate,
but from handbooks and tables you usually find enough instructions. Other wise, you
have to simplify the problem to be able to solve it.
Ex:
How much heat (energy and exergy) does a naked person loose (surface
temperature 35˚C, H = 0.7 and area 1.5 m2) from radiation in a room with black
walls, L = 1, L = 20˚C and area 20 m2? (T0 = TL)
S:
We have a body contained in an other body, thus we have for the energy power,
see relation above, where TH ≈ 35+273 = 308 [K] and TL ≈ 20+273 = 293 [K]:
Pen,r,HL 
1
1
1 1 
   1
0.7  1.5 20  1 


5.67  10 8  1.5  3084  2934  145 [W]
The person will probably freeze, which we also might guess.
111
EXERGETICS
For the exergy we have:
Pex,r,HL 
1
1
1 1 
   1
0.7  1.5 20  1 
5.67  108 
 1  293  4 4 293 
 1.5  308 1   
  
  3.7 [W]
 3  308  3 308 
4
Thus, there are three ways for energy and exergy to be transferred as heat between
bodies: conduction, which was simple to describe, convection and radiation, which were
more complicated to describe. At temperatures close to ambient, i.e. ±20-30˚C we can
assume the following relation:
Pen,total  Pen,conduction  Pen,convection  Pen,radiation  UA(TH  TL )
Hereby, all phenomena are summarized in one expression and concept, the over all heat
coefficient U. Analogously, we have for the exergy flows:
Pex,total,H  Pex,conduction,H  Pex,convection,H  Pex,radiation,H 
TH  T0
Pen,total
TH
Pex,total,L  Pex,conduction,L  Pex,convection,L  Pex,radiation,L 
TL  T0
Pen,total
TL
The exergy loss then becomes:
Pex,total  Pex,total,H  Pex,total,L
T0  TH  TL 
T
2
Pen,total  UA 0  TH  TL 

THTL
THTL
As we see, the energy is always conserved, the first law, whereas exergy is always
lost, the second law. We also see that the exergy loss increases by the square of the
temperature difference TH–TL, which is an important conclusion.
Q:
Which form of heat has the highest exergy factor:
 Pex 
T  T0


conduction: 
T
 Pen conduction
 1  T0  4 4 T0 
 Pex 

 1    

or radiation: 
 Pen  radiation  3  T  3 T 
Thus, which form of heat is most ordered?
112
EXERGETICS
Over all heat coefficient, U
Regard the heat transport through a homogenous wall
1
T1
T1w
Pen
Pen
Pen

T2w
2
T2
d
Since the same energy flow passes all layers we can put up the following equations, we
neglect the radiation since the temperature differences (1 – T1w) and (T2w – T2) are
assumed to be small:
 Pen  1 A(T1  T1w )



 Pen  A(T1w  T2 w )
d

 Pen   2 A(T2 w  T2 )
By writing the expressions for the temperature differences we get:
 Pen
 A  T1  T1w
 1
 Pen d
 T1w  T2 w

 A
 Pen
 A  T2 w  T2
 2
By adding all these expressions we get:
Pen Pen d Pen


 T1  T1w  T1w  T2 w  T2 w  T2
1 A  A  2 A
or:
Pen  1 d 1 
     T1  T2
A  1   2 
which can be written:
Pen = UA(T1–T2)
113
EXERGETICS
1
1
1 d 1
   .
where the overall heat coefficient is defined as U  1 d 1 or
U 1   2
 
1   2
The unit of U is W/m2K, i.e. it gives the energy that per unit time and unit area passes
through the construction when the temperature difference is 1 K [˚C]. The U-coefficient
depends on the construction, d and  and the heat transfer coefficients,  1 and 2 . We
also see that at heat insulation, e.g. in houses the U-coefficient should be small, but at
heat conduction, e.g. in heat exchangers the U-coefficient should be large.
For a construction with several layers of materials, i and air, i.e. a sandwich
construction we write the U-coefficient accordingly:
1 1
1
d
1
 
 i 
U 1  2
U air
i i
where Uair combine conduction, convection and radiation for each air layer.
Ex:
A wall with the area 20 m2 has a mineral wool insulation with thickness 170 mm.
The indoor temperature is +20˚C and outdoor is –5˚C. How much heat (energy
and exergy) per unit time leaks trough the wall if the wind velocity is 1m/s? (0 =
–5˚C)
S:
We have the following data: d = 170 mm = 0.17 m, A = 20 m2, 1= +20˚C, 2= –
5˚C and  = 0.04 W/m˚C.
Pen = UA(1–2) where
1
1 d 1
   , since we only have one layer.
U 1   2
For flat surfaces we have from tables  = 5.8+3.95v if v<5 m/s
1 is the inside heat transfer coefficient
inside is v = 0  1 = 5.8 W/m2˚C
2 is the outside heat transfer coefficient
outside is v = 1 m/s  2 = 5.8+3.95 = 9.75 W/m2˚C
1
1 0.17
1



5
U 5.8 0.04 9.75
U
1
 0.2 [W/m2K]
5
Pen = UA(1–2) = 0.2×20(20–(–5)) = 0.2×20×25 = 100 W
TL  T0  Pex,L  0  Pex  Pex,H 
114
T0 (TH  TL )
Pen,total 
THTL
EXERGETICS
T0 (TH  T0 )
T T
20  (5)
 
Pen,total  H 0 Pen,total  H 0 Pen,total 
 100  8.5 [W]
THT0
TH
TH
273  20
Thus, 100 W energy is leaking out and 8.5 W exergy is lost.

There are two ways of determine the over all heat coefficient U: the -method and Umethod.
-method
In this method we consider each layer of the construction. The -value for a layer is the
average value of the -values of the different areas in proportion to its area perpendicular
to the direction of the heat flow, i.e. a, for a mixed layer is
a
 A

A
i
i
i

i
1 A1  2 A2  3 A3  ...
A1  A2  A3  ...
i
where i is the -value of layer i [W/mK]
Ai is the layer i’s area perpendicular to the heat flow [m2].
The overall heat coefficient U [W/m2K] then becomes:
U 
a
d
where d is the thickness [m].
 -method gives a slightly to high U-value compared to experiments.
Q: Why?
U-method
We now consider each cross-section of the construction along the direction of the heat
flow. The U-value for each section is calculated separately, and the total U-value is given
by the sum of the U-values for each cross-section in proportion to its area, i.e.
U A
U
A
i
i
i
i

U1 A1  U 2 A2  U 3 A3  ...
A1  A2  A3  ...
i
where Ui is the U-value of section i [W/m2K]
Ai is the area of section i perpendicular to the direction of the heat flow [m2]
The U-method give a slightly lower U-value compared to experiments.
Q: Why?
115
EXERGETICS
Ex:
Assume a wall, 20 m2, that from outside consists of 100 mm wood ( = 0.14
W/mK), 100 mm mineral wool ( = 0.036 W/mK) between wood bars (10% of the
area) and towards the inside a board, 40 mm ( = 0.05 W/mK). How much heat
leaks trough the construction per unit time if the outdoor temperature is –5˚C and
the indoor temperature is +20˚C? a) According the -method b) According the Umethod.
S:
a) -method: We calculate a for the mineral wool and the wood bars:
a 
1 A1  2 A2
A1  A2

0.036  0.9  0.14  0.1
 0.046 [W/m2 K]
0.9  0.1
The U-value for the construction becomes
1
0.1
0.1
0.04
d
 i 


 3.69  U wall  0.27 [W/m2K]
0.14 0.046 0.05
U
i i
The energy power becomes Pen = UA(1–2) ≈ 0.27×20×25 = 135 [W]
20  (5)
 
 135  11.5 [W]
The exergy power becomes Pex  1 2 Pen 
T1
273  20
b) U-method: the U-value where we have the wood bars becomes
1
U wood

i
di
i

0.1
0.1 0.04


 2.23  U wood  0.45 [W/m2 K ]
0.14 0.14 0.05
The U-value where we have the mineral wool becomes
1
U mineral wool

i
di
i

0.1
0.1
0.04


 4.29  U mineral wool  0.23 [W/m2 K ]
0.14 0.036 0.05
The U-value for the construction becomes
U
U1 A1  U 2 A2 0.45  0.1  0.23  0.9

 0.25 [W/m2 K]
0.1  0.9
A1  A2
The energy power becomes Pen  UA(1   2 )  0.25  20  25  125 [W]
20  (5)
 
 125  10.7 [W]
The exergy power becomes Pex  1 2 Pen 
T1
273  20
Thus, the-method gives that 135 W energy is leaking out and that 11.5 W
exergy is lost, but the U-method gives 125 W energy and 10.7 W exergy
respectively.
116
EXERGETICS
Ex:
Determine the U-value, the position where the temperature is 0˚C and where
condensation may occur in the following wall, from outside to inside: 100 mm
wood, 100 mm mineral wool and 40 mm board.
Inside
Outside
Air speed [m/s]
0
4
Temperature [˚C]
20
–10
Relative humidity [%]
80
50
A:
From tables we get
The U-value ≈ 0.23 W/m2K, the relative humidity exceeds 100% between the
wood and the mineral wool, where condensation may occur.
The temperature 0˚C passes in the mineral wool. The temperature increases
linearly in the wool. If we assume that 0˚C is x mm in the mineral wool from the
outside, then we have:
0 – (–4.6)
x
100 = 12.9 – (–4.6)  x ≈ 26 mm. (The problem can also be solved graphically.)
Heat exchangers
We will now look closer into heat exchangers, which are used to transfer heat
between two different media by heat transfer. We find heat exchangers as heat radiators,
hot water tanks, solar panels and car engine cooler. There are two main kinds of heat
exchangers, parallel and counter flow. Let us assume a simple model one tube in an other,
see Fig. below.
In a parallel heat exchanger the final temperature of the colder flow never can exceed
the final temperature of the hot flow, i.e. TH2>TC2. However, in a counter flow heat
exchanger we may have that TH2<TC2.
The energy power being transferred in a heat exchanger can be determined from
Pen = UA∆Tm [W]
where U is the overall heat coefficient [W/m2K]
A is the area of heat transfer [m2]
∆Tm is the average temperature difference, i.e.
Tm 
T1  T2
 T 
ln 1 
 T2 
where ∆T1 and ∆T2 is defined according to the Fig. below. (To use temperatures in
degrees Celsius is also allowed because the relation only contains temperature
differences.)
117
EXERGETICS
Parallel flow
Counter flow
TC1
TC2
TH1
TH1
TH2
TH2
TC2
TC1
TH1
TH1
² T1
TC2
TH2
² T2
TC2
² T1
TH2
² T2
TC1
TC1
Temperature profile
Temperature profile
Ex:
In a counter flow heat exchanger 2 kg per second of lubrication oil for a
ship engine is cooled from 70˚C to 40˚C by water. The water is then heated from
20˚C to 40˚C. The transfer surface is 6 m2. What is the U-value if the heat
capacity of the oil c is 1.9 kJ/kgK? What are the exergy losses if the ambient
temperature is 0 = 20˚C? Use: Pen  m c(1   2 )
S:
∆1 = 70–40 = 30˚C and ∆2 = 40–20 = 20˚C
The average temperature difference is:  m 
1   2 30  20

 24.7˚C
 1 
 30 
 ln 
ln
 20 


 2
The transferred energy power becomes: Pen = UA∆m
For the oil we have: Pen  m c( H1   H2 ) , where mÝ is the mass flow [kg/s], i.e.
Pen,oil = 2×1.9(70–40) = 114 [kW]
For the water we have: Pen,oil = Pen,water = UA∆m, i.e.
114×103 = U×6×24.7  U ≈ 770 [W/m2K]
For water we further have: Pen  m watercwater ( H1   H2 ) , i.e.
m water 
Pen
cwater ( H1   H2 )

114
 1.4 [kg/s]
4.18  (40  20)
118
EXERGETICS
For a heat content, which decreases its temperature with the heat revealed, we
have, according to the section above on exergy of heat and cold, if the heat
capacity is constant.
E
T
T
 1  0 ln
Q
T  T0 T0
which gives the following exergy flows for the hot oil:

T 
Pex,H1  m oilcoil  TH1  T0  T0 ln H1 
T0 


T 
Pex,H2  m oilcoil  TH2  T0  T0 ln H2 
T0 

The exergy power that the oil looses in the heat exchanger then becomes:

T 
Pex,H  Pex,H1  Pex,H2  m oil coil  TH 1  TH 2  T0 ln H 1  
TH 2 


T 
 m oil coil  H1   H2  T0 ln H1  
TH2 

273  70 

 2  1.9   70  40  (273  20) ln
  12.1 [kW]
273  40 

Similarly we have for the water on the cold side. The exergy power being
received by the water in the heat exchanger becomes:

T 
Pex,C  Pex,C2  Pex,C1  m water cwater  C 2   C1  T0 ln C 2  
TC1 

273  40 

 3.7 [kW]
 1.4  4.18  40  20  (273  20)ln

273  20 
Thus, the exergy loss in the heat exchanger becomes:
∆Pex = ∆Pex,H – ∆Pex,C ≈ 8.4 [kW]
The exergy efficiency of the heat exchanger becomes:
ex 
Pex,C 3.7

 31%
Pex,H 12.1
The energy efficiency is of course as we assumed 100%.
119
EXERGETICS
Ex:
A condenser heats water by steam at temperature 100˚C. The water temperature
increases from 30˚C to 75˚C and the condenser has a surface of 200 m2 and the Uvalue is 3000 W/m2K. How much hot water is produced per hour? What are the
exergy losses and the exergy efficiency when 0 = 20˚C?
S:
When steam is condensed the heat of evaporation is delivered to the colder flow.
The temperature differences at the in and out let becomes: ∆1 = 100–30 = 70˚C
and ∆2 = 100–75 = 25˚C. The average temperature then becomes:
 m 
1   2 70  25

 44˚C
 1 
 70 
 ln 
ln
 25 
  2 
The energy power which is being transferred to the water becomes: Pen = UA∆m
≈ 3000×200×44 = 26.4 [MW].
The water is being heated from 30 to 75˚C. Added energy power to the water
is Pen,w  m c(1   2 ) where c ≈ 4.18 [kJ/kgK], i.e. the specific heat capacity of
water from table. Thus we have
Pen,w  m c(1   2 )
which gives
m 
26.4  106
Pen

 140 [kg/s]
c(1   2 ) 4.18  103  (75  30)
From above we get the following exergy power being released from the steam
Pex,H 
TH  T0
100  20
 
Pen  H 0 Pen 
 26.4  5.7 [MW]
TH
TH
273  100
since the temperature is constant at the heat transfer.
In the same way as in the previous example we have for the water, which
receives heat, that the exergy power received becomes:

T 
Pex,C  Pex,C2  Pex,C1  m water cwater  C2   C1  T0 ln C2  
TC1 

273  75 

 140  4.18   75  30  (273  20) ln
  2.6 [MW]
273  30 

The exergy loss at the heat exchange then becomes:
∆Pex = ∆Pex,H – ∆Pex,C ≈ 3.1 [MW]
The exergy efficiency becomes:
ex 
Pex,C 2.6

 46%
Pex,H 5.7
Please, remember that the energy efficiency usually is assumed to be 100%!
120
EXERGETICS
COMBUSTION
First some common concepts:
nm3 – normal cubic meter. The gas volume is usually measured in nm3, which is the
amount of gas at the temperature 0˚C, i.e. 273.15 K and the pressure 1 bar, i.e. 0.1
MPa. Normal cubic meter, nm3.
mol – one mol is the number of molecules or atoms that are in 0.012 kg of the element
carbon, C 12. Thus, one mol is a number, actually 6.0231027 pieces. This means
that 1 mol of a substance has the same weight in gram as is given by its atom
weight, and 1 kmol corresponds to kg.
Hs – Calorimetric heat value or high heat value – is measured by the amount of enthalpy
being released when a certain mass is combusted in a bomb calorimeter,
principally a completely isolated container. The combustion products are cooled
so that the moist in the fuel and the produced steam is condensed. This means that
the enthalpy of the generated steam is included in Hs.
Hi – Effective heat value or low heat value – is the enthalpy released at normal
combustion, i.e. the generated steam is excluded. Thus H i  H s .
The combustion reaction 2H2  O2  2H2O means that if 1 kmol H2, i.e. 2 kg, burns
you get 1 kmol H2O, i.e. 18 kg. In weight, you get 9 times as much water as hydrogen.
Thus, if we know the amount of hydrogen and moist in the fuel we may approximately
calculate Hi from Hs. The enthalpy of steam is about 2.5 MJ/kg at 293 K (≈20˚C).
For solid and liquid fuels we have accordingly:
Hi = Hs – 2.5(9H + F) [MJ/kg]
where H is the weight percentage of hydrogen
F is the weight percentage of moist (water)
For dry fire wood, independent of kind we have Hi ≈ 19.2 MJ/kg and for damped:
Hi = 19.2 – 21.7F [MJ/kg]
Ex:
At fuel oil combustion, Hi = 42 MJ/kg the air surplus is 40%. Calculate the
amount of exhaust gas and the combustion temperature?
S:
Hi = 42 MJ/kg  theoretical amount of exhaust gas go = 11.8 nm3/kg, from tables
and theoretical amount of air (dry air) lo = 11.1 nm3/kg
The air surplus is 40%, i.e.
l  lo
l
 0.4  airfactor: m   1.4
lo
lo
Real amount of exhaust gas g = go + lo(m – l) = 11.8+11.1(1.4-1) = 16.24 nm3/kg
The heat content of the exhaust gas h max 
and from tables we get ≈1920 K.
121
Hi
42
3

 2600kJ / nm
g 16.24
EXERGETICS
Ex:
A boiler combusts 2 kg oil/hr, Hi = 42 MJ/kg, with an air surplus of 20%. The
temperature of the exhaust gases are 620 K. Estimate the heat power, losses and
efficiency of the boiler.
Enthalpy, Gibbs’ function and exergy of fuels
When we have chemical reactions it is necessary to define a common reference state.
This is usually defined at 25˚C (298.15 K) and 101.3 kPa (T0,P0) and that gases are
treated as ideal gases. Regard the following stationary combustion process, solid carbon
is combusted with oxygen (ideal gas), where both are supplied at the reference state,
T0,P0. The produced carbon dioxide (ideal gas) is leaving the combustion chamber at the
reference state, T0,P0. If we could measure the heat transfer to the process (the reaction
chamber) it would be –393,522 kJ/kmol CO2 (the minus sign because heat is released, an
exothermic reaction).
Q - –393522 kJ
1 kmol C
T0 ,P0
1 kmol CO2
T0 ,P0 = 298.15 K, 101.3 kPa
1 kmol O2
T0 ,P0
The chemical reaction is:
C + O2  CO2
The first law gives:
Q  HR  HP
where the indices R and P refer to reactants and products. First law is usually written
Q   n i hi   ne h e
R
P
where the sum involves all reactants and products respectively, n the number of mol and
h indicates the enthalpy per mol of the substance. The enthalpy of formation is tabulated
for most substances in standard states at T0,P0. Usually this is put to zero for the substance
in its standard state, pure and normal form. For carbon, C this is pure carbon in solid state
0
0
and for oxygen as O2 in gas state, i.e. (hf ) C,s,graphite  0 and (hf ) O2 ,g  0 . Thus the
0
enthalpy of formation for carbon dioxide becomes, (hf ) CO2 ,g  –393522 kJ/kmol. If we
instead assume the enthalpy of formation of O2 and of CO2 as gases to zero, i.e.
(h0f ) O2 ,g  0 and (h0f ) CO2 ,g  0 , then we get the enthalpy of formation of C as solid
(h0f ) C,s,graphite , or shorter h C = 393522 kJ/kmol. This is usually called the heat content of
the fuel at constant pressure (and temperature), in the same way the internal energy of
formation UC the heat content at constant volume (and temperature). Sometimes you also
say the “high” and “low” heat content as above. The high value refers to the case when
122
EXERGETICS
water in the exhaust gases is as liquid, and the low value when it is as gas. Note, that this
sometimes gives confusing values of the energy efficiency.
The enthalpy at an arbitrary temperature is:
h T ,P  (h 0f )  h T0 ,P0  T, P
where h T0 ,P0 T , P represents the difference in enthalpy between a given state at T,P and
at T0,P0.
If we apply the second law to this combustion process and change the heat extracted
Q by the maximal extractable (reversible process) exergy E we get
E  ER  EP.
As above this can be written
E   ni e i   n e e e
R
P
where the sum includes all reactants and products respectively and e is the exergy per
mol of the substance in relation to its environmental state, which is called exergy of
formation and is tabulated for the most common substance, see App. 4.
When the reactants and the products are in pressure and temperature equilibrium with
the environment we can write
E   ni gi   n e g e
R
P
where g is the Gibbs’ function per mol.
Let us calculate the change in Gibbs’ function when forming CO2 from the reaction
C + O2  CO2
where C, O2 and CO2 are all separated and at T0,P0.
The change in Gibbs’ function can be calculated accordingly
G P  G R  HP  HR  T0 (S P  S R )
i.e.
 n (g
e
P


0
0
)  ni ( g 0f )i   ne (h f0 ) e  ni (h f0 )i T0  ne ( s298
) e  ni ( s298
)i 
R
P
R
R
P

0
f e
From tables, e.g. Table A.13, Van Wylen, G. J. and Sonntag, R. E., Fundamentals of
Classical Thermodynamics, Wiley (1985) we get
0
0
0
( g 0f ) P  ( g 0f ) R  (h f0 )CO2  0  298.15[(s298
)CO2  ( s298
)C  ( s298
)O2 ]
≈ –393522 – 298.15(213.795 – 5.686 – 205.142) ≈ – 394407 kJ/kmol.
If the Gibbs’ function for the reactants are set to zero, i.e. ( g0f ) R  0 , the Gibbs’
function for forming CO2 becomes
( g 0f ) P  ( g 0f ) CO 2 ≈ – 394407 kJ/kmol.
123
EXERGETICS
If we instead assume Gibbs’ function of O2 and CO2 to zero, as we did above for the
enthalpy, we have that gC  (g 0f )C ≈ 394407 kJ/kmol. Thus, we see that hC < gC, i.e. the
enthalpy is less than the Gibbs’ function for C, because the entropy S of the products are
much larger than for the reactants – “heat” Q = T0S is captured by the products.
Let us now calculate the exergy change for this reaction. The reactants and the
products are now instead related to their individual environmental states. The difference
between the enthalpy and Gibbs’ function is that in Gibbs’ function we also include the
change of entropy of the reactants and the products in relation to their individual standard
states. The difference between Gibbs’ function and the exergy is that in exergy we also
include the environment state, i.e. in what regard the reactants and products differ from
environmental state, e.g. in concentration or chemically. From tables of exergy for
different substances, App. 4 we have
(e f0 )C,s,graphite ≈ 410260 kJ/kmol
(e f0 )O2 ,g ≈ 3970 kJ/kmol
(e f0 )CO2 ,g ≈ 19870 kJ/kmol,
which gives the exergy change
eC   ni ei   neee ≈ 410260 + 3970 – 19870 = 394360.
R
P
Let us summarize, for C we have
Gibbs’ function
hC ≈ 393522 kJ/kmol
gC ≈ 394407 kJ/kmol
Exergy
eC ≈ 394360 kJ/kmol
Enthalpy
The relation between the exergy and enthalpy for C is
eC 394360

≈ 1.002 or 100.2%
hC 393522
Q:
Why is enthalpy less than exergy?
We have seen that from exergy tables of different substances we may calculate the
exergy change of any chemical reaction.
In same way as for enthalpy we may also calculate exergy at any temperature
eT , P  (e f0 )T0 , P0  eT0 , P0 T , P
the term eT0 , P0 T , P represents the difference in exergy between a given state at T,P and
the exergy at T0,P0. This exergy sometimes is called the thermal exergy to differ from the
chemical exergy.
Finally, we will just see what happens with the exergy in a real process, e.g. an
combustion engine, se the figure below.
124
EXERGETICS
Exergy
Compression Combustion
Work
Exergy
of Fuel
Expansion Work, etc.,
i.e. Exergy to the
Environment
Exergy
Losses
from
Irreversibilities
Combustion
Losses
and
Pollutants
to the
Environment
Heat transfer
Mechanical Work, i.e.
the Exergy Output
Utility
Turning Point of the Piston
As we see most of the exergy is used to get the exhaust gases into the environment, i.e. a
pressure-volume work, P0∆V, performed on the surrounding air.
Q:
Reflect on what this means for a car with a combustion engine.
Example: 1000 ton of CO2 or 1000 ton of cooling water
Let us compare an emission of 1000 ton of CO2 at ambient pressure and temperature with
1000 ton of cooling water. From exergy tables we have for the different chemical
compounds of C and H:
Substance
C
C
CCl4
CN
C2N2
CO
CO2
…
H2O
H2O
H3PO4
H2S
H2SO4
…
State
s, graphite
s, diamond
l
g, cyano
g, cyanogen
g
g
g
l
s
g
l
Mol mass
m
Enthalpy of formation
( h0f )
Exergy
( ef0 )
[kg/kmol]
12.01115
12.01115
153.823
26.01785
52.0357
28.0105
44.0095
[kJ/mol]
393.509
395.406
578.95
858.00
1096.14
282.984
0
[kJ/mol]
410.26
413.16
473.1
845.0
1118.9
275.10
19.87
18.01534
18.01534
98.0013
34.080
98.077
0
–44.012
–76.26
946.61
153.25
9.5
0.9
104.0
812.0
163.4
125
EXERGETICS
Thus we have for 1000 ton of CO2
ECO2


19.87  1000  1000  kJ  kg 

 4.5  105 MJ   0.45TJ 


kg
44.0095
 mol 

kmol 

For water the chemical exergy becomes:
EH 2O,chemical 
0.9  1000  1000
 5  10 4 MJ   0.05TJ 
18.01534
The thermal exergy becomes:
T
 T 
Ethermal   C (T )1  0 dT 
 T 
T0
If the heat capacity C is not depending on temperature, the exergy becomes

T
Ethermal  C  T  T0  T0 ln 
T0 

Assume C = 4.2 kJ/kg and To=298.15K, then Ethermal=0.4 TJ correspond to a temperature
increase for 1000 ton of water of about 400K, i.e. some of the water will be steam. The
exergy of the water at boiling temperature, i.e. at 373.15 K is
373.15 

10
Ethermal  4.2  103  1000  103  373.15  298.15  298.15 ln
  3.4  10 J 
298
.
15


Thus 0.4–0.034=0.366 TJ = 366 GJ is left for steam production at ambient pressure. The
enthalpy of steam is about 2300 kJ/kg so the exergy of steam becomes
 kJ 
 T 
 298.15 
esteam  hsteam 1  0   23001 
  462 
 T
 373.15 
 kg 
The mass converted to steam them becomes
msteam 
Esteam 366  GJ 

 0.792  106 kg   792ton 


462 462  kJ/kg 
Thus, from an exergy point of view, an emission of 1000 ton of CO2 is equivalent to 792
ton of steam and 208 ton of water at 373.15 K, if the environment is at ambient pressure
and temperature, i.e. 101.3 kPa and 298.15 K.
126
EXERGETICS
APPENDIX
Appendix 1 Internal energy and entropy
To be written.
127
EXERGETICS
Appendix 2 Exergy
A derivation of the exergy concept can be made from simple thermodynamic relations of
state changes which are related to the concept of work.
 P,T ,  i
A 
 U,V ,S,n i
 P , T ,
A   0 0 i0
U 0 ,V 0 ,S0 ,n i 0
Fig. 1 The system A in an environment A0
Assume we have a system A in a very large homogenous environment (reservoir) A0
˜ i0 (pressure, temperature and
which is defined by the intensive parameters P0, T0 and 
˜ i . The
generalized chemical potential*). The intensive parameters of A are P, T and 
corresponding extensive parameters of A and A0 are U, V, S and ni (internal energy,
volume, entropy and number of mol of different substances) respectively U0, V0, S0 and ni0
for A0, Fig. l. Also assume that A and A0 each are in internal equilibrium.
Assume that all extensive parameters of A are much smaller than those of A0 and
further that the total system A  A 0 is isolated from the surrounding except for the
extraction of work W which is extracted from the total system, then we have the
following relations
U  U 0

 V  V0
n  n
i0
 i
(1)
U  U 0  W  constant

V  V0  constant
n  n  constant
 i i0
(2)
dU  dU 0  dW  0

dV  dV0  0
dn  dn  0
i0
 i
(3)
Interaction between A and A0 can take place in a controlled way through the interface
of A. Since A is small, Eq. 1, this does not change the intensive parameters of A0,
*
The generelized chemical potential includes all potentials realted to the substance, e.g. chemical,
electrical, magnetical, mechanical, and gravitational potentials.
128
EXERGETICS
 dT0  0

 dP0  0
d~  0
 i0
(4)
The entropy differential of the environment A0 is
dS 0 
1
˜ i0 dn i 0 )
(dU 0  P0 dV0   
T0
i
(5)
which can be written by using Eq. 3
dS0  
1
~ dn   dW
 dU  P0 dV   
i0
i
T0 
i
 T0
(6)
The total entropy differential of the system and the environment is
dS tot  dS  dS0  
1
~ dn   dW
 dU  P0 dV  T0 dS   
i0
i
T0 
i
 T0
(7)
This may be written as
dS tot  
1
(dE  dW )
T0
(8)
where we have introduced exergy E,
˜ i0 ni
E  U  P0 V  T0 S   
(9)
~n
U  TS  PV   
i i
(10)
~ 
~ )
E  S (T  T0 )  V ( P  P0 )   ni (
i
i0
(11)
i
If we use the Gibbs relation
i
in (9) we get
i
which shows that E vanishes at equilibrium, i.e.
P  P0 

T  T0   E  0
~ 
~ 

i
i0 
129
(12)
EXERGETICS
Assume now that A evolves towards equilibrium with its environment A0 and the
work ∆W is performed during this process. The exergy is then changed by -E from E to 0
tot
tot
and the total entropy S is changed by S . By integrating (8) we then get
S tot  
1
 E  W 
T0
(13)
and thus
Since according to the second law
W  E  T0 S tot
(14)
S  0
tot
(15)
∆W ≤ E
(16)
we get
where equality only holds when ∆Stot = 0, i.e. reversible processes.
Thus, the exergy E is the maximum work that can be extracted
from a system through interaction with its environment.
We may subtract from (A.7) the corresponding equation at equilibrium, i.e.
~ n
Eeq  U eq  P0Veq  T0 Seq   
i 0 i ,eq
(17)
i
Since E vanishes at equilibrium, i.e. E eq  0 we then find
~ n  n 
E  U  U eq  P0 V  Veq   T0 S  Seq    
i0 i
i ,eq
(18)
i
which is a useful relation for practical determinations of exergy.
In special cases exergy differences are related to differences of other, better known,
thermodynamic potentials, e.g. Gibbs’ function, Helmholtz’ function and enthalpy as
described in the Table below. By differentiating the definition of exergy, Eq. 9, we can
easily find the following relations to these potentials
Table:
Relations between differences in exergy and in other thermodynamic potentials.
_______________________________________________________________
Definition of X
Usually named
Case
∆E=∆X
_______________________________________________________________
∆E=∆G0
G0=UP0V–T0S
∆ni=0
∆E=∆F0
F0=U–T0S
∆ni=0, ∆V=0
∆E=∆H0
H0=UP0V
∆ni=0, ∆S=0
∆E=∆G
G=UPV–TS
Gibbs’ function
∆ni=0, P=P0, T=T0
∆E=∆F
F=U–TS
Helmholz’ function
∆ni=0, ∆V=0, T=T0
∆E=∆H
H = UPV
Enthalpy
∆ni=0, ∆S=0, P=P0
_______________________________________________________________
130
EXERGETICS
Appendix 3 Heat exchange between many systems
Let us derive an expression for the exergy when heat is exchanged between many
systems.
T1
C1
T2
C2
•
•
•
Ti
Ci
Q1
Q2
dE
Q i
•
•
•
We have:
Energy conservation:
Q
i
 dE  0
(1)
i
 dS
Entropy conservation:
i
0
(reversible process)
(2)
i
dS i 
We have for the entropy:
and for the heat:
Eqs. l and 4:
Q i
Ti
(3)
 Qi  Ci (Ti )dTi
(4)
dE   Ci (Ti )dTi
(5)
C i (Ti )dTi
0
Ti
(6)
i
Eqs. 2, 3, and 4:

i
Tif
Integration gives:
E    Ci (Ti )dTi
i
(7)
Ti0
Tif
C i (Ti )dTi
0
Ti
Ti0

i
(8)
where T i0 is the initial temperatures and T if is the final temperature.
Thus, we have two equations which completely describes the available exergy of the
system. If T if = T f for all i then Eq. 8 gives T f which in Eq. 7 gives the exergy.
131
EXERGETICS
Assume that the heat capacities are constants and that the final temperature is the
same for all subsystems, i.e. Ci(Ti) = Ci and T if = T f :
E   Ci Ti  T
0
f
i
C
(9)
i
i
1

C i
T f  (Ti0 )Ci i
 i

(10)
E   Ci (T  T )
(11)
Equation 9 may be written:
0
f
i
where
T0
C T

C
0
i i
i
(12)
i
i
0
f
T is a weighted arithmetic average value and T is a weighted geometric average value.
We can also find that the work obtainable from the system becomes:
S
tot
  C i (ln T – ln T )
0
f
(13)
i
And the exergy becomes:
*
E  T S
tot
(14)
where
T0 Tf
T 
ln T 0  ln T f
*
(15)
Thus, we are able to calculate the exergy of a system of arbitrary numbers of bodies
with arbitrary heat capacity and temperature and with a reference environment. Let us
look at some simple special cases, which we already are familiar with.
Ex:
Assume only two systems, one limited and the other unlimited (ambient). We get:


T
T 
T 
 1  ln   C  T  T0  T0 ln 

T
T0 
T0 

 0


E  CT0 
where T and C are temperature and heat capacity of the limited system and T0 is
ambient temperature. This we recognize from before.
Ex:
Assume all subsystems are identical, but with different temperatures, i.e.:
132
EXERGETICS
Ci = C for all i then we get:
1


0
n


T

E  nC  i    Ti 0  
 
 i n  i

where n is number of subsystems. For n = 2 we get:

E  C T10  T20
133

2
EXERGETICS
Appendix 4 Reference states†
Table A.4.1
Enthalpy and exergy of formation for inorganic substances at reference
state (T0 = 298.15 K, P0 = 101.325 kPa)
Substance
Ag
Ag2CO3
AgCl
AgF
AgNO3
Ag2O
Ag2O2
Ag2S
Ag2SO4
Al
Al4C3
AlCl3
Al2O3
Al2O3·H2O
Al2O3·3H2O
Al2S3
Al2(SO4)3
Al2SiO5
Al2SiO5
Al2SiO5
Al2Si2O5(OH)4
3AI2O3·2SiO2
Ar
As
As2O5
Au
AuCl
AuCl3
AuF3
Au2O3
B
B2O3
Ba
BaCO3
BaCl2
BaF2
BaO
BaO2
Ba(OH)2
BaS
State
Molar mass
[kg/kmol]
107.870
275.749
143.323
126.868
169.875
231.739
247.739
247.804
311.802
26.9815
143.959
133.3405
101.9612
119.9765
156.0072
150.155
342.148
162.046
162.046
162.046
258.1615
426.0536
39.948
74.9216
229.8402
196.967
232.42
303.326
253.962
441.932
10.811
69.6202
137.34
197.35
208.25
175.34
153.34
169.34
171.36
169.40
s
s
s
s
s
s
s
s, 
s
s
s
s
s, , corundum
s, boermite
s, gibbsite
s
s
s, andalusite
s, kyanite
s, sillimanite
s, kaolinite
s, mullite
g
s
s
s
s
s
s
s
s
s
s, II
s, II
s
s
s
s
s
s
†
Enthalpy
[kJ/mol]
47.48
–17.38
0
47.60
–76.91
63.91
70.69
787.79
104.50
930.69
4694.51
467.18
185.69
128.35
24.13
3313.81
596.80
28.03
25.94
0
68.25
630.11
0
462.44
0
0
45.49
123.09
246.46
–80.81
636.39
0
747.77
–75.18
48.69
–53.24
194.15
113.38
45.93
1012.88
Exergy
[kJ/mol]
70.2
115.0
22.2
118.5
43. I
57.6
172.1
709.5
139.6
888.4
4588.2
444.9
200.4
195.3
209.5
2890.7
529.7
43.9
45.1
15.4
197.8
618.8
11.69
494.6
216.9
15.4
62.2
155.5
437.3
114.7
628.5
69.4
747.7
26.3
61.3
57.2
224.6
169.3
132.9
901.9
Szargut, J., Morris, D. R., and Steward, F. R., Exergy Analysis of Thermal, Chemical, and Metallurgical
Processes, Springer (1988)
134
EXERGETICS
Substance
BaSO4
Bi
Bi2O3
Bi2S3
Br2
C
C
CCl4
CN
C2N2
CO
CO2
CS2
Ca
CaC2
CaCO3
CaCO3·MgCO3
CaCl2
CaF2
CaFe2O4
Ca2Fe2O4
Ca2Mg5Si8O22(OH)2
Ca(NO3)2
CaO
CaO·Al2O3
CaO·2Al2O3
3CaO·Al2O3
12CaO·7Al2O3
CaO·Al2O3·2SiO2
Ca(OH)2
Ca3(PO4)2
CaS
CaSO4
CaSO4·1/2H2O
CaSO4·2H2O
CaSiO3
Ca2SiO4
Ca3SiO5
Cd
Cd
CdCO3
CdCl2
CdO
Cd(OH)2
CdS
CdSO4
CdSO4·H2O
Cl2
State
Molar mass
[kg/kmol]
233.40
208.980
465.958
514.152
159.812
12.01115
12.01115
153.823
26.01785
52.0357
28.0105
44.0095
76.139
40.08
64.10
100.09
184.411
110.99
78.077
215.77
271.85
812.41
164.0898
56.08
158.04
260.00
270.20
1386.68
222.038
74.09
310.18
72.14
136.14
145.15
172.17
116.16
172.24
282.32
112.40
112.40
172.41
183.31
128.40
146.41
144.46
208.46
226.48
70.906
s, barite
s
s
s
l
s, graphite
s, diamond
l
g, cyano
g, cyanogen
g
g
l
s, II
s
s, aragonite
s, dolomite
s
s
s
s
s, tremolite
s
s
s
s
s
s
s, anortite
s
s, 
s
s, anhydrite
s, 
s, gypsum
s, volastonite
s, 
s
s, 
s, 
s
s
s
s
s
s
s
g
135
Enthalpy
[kJ/mol]
0
286.94
0
2607.05
—
393.509
395.406
578.95
858.00
1096.14
282.984
0
1934.09
813.57
1541.18
0
0
178.21
0
161.07
321.00
425.49
–124.90
178.44
351.66
541.71
716.72
3415.71
273.92
69.04
0
1056.57
104.88
83.16
0
90.24
232.28
424.94
357.10
356.51
0
126.04
98.95
38.26
920.60
149.24
84.79
160.44
Exergy
[kJ/mol]
3.4
274.5
61.4
2237.3
101.2
410.26
413.16
473.1
845.0
1118.9
275.10
19.87
1694.7
712.4
1468.3
1.0
15.1
87.9
11.4
104.0
194.7
81.6
–18.1
110.2
275.4
460.4
500.6
2526.8
218.3
53.7
19.4
844.6
8.2
12.1
8.6
23.6
95.7
219.8
293.8
293.2
40.6
73.4
67.3
59.5
746.9
88.6
80.6
123.6
EXERGETICS
Substance
Cl
Co
CoCO3
CoCl2
CoO
Co3O4
Co(OH)2
CoS
CoSO4
Cr
Cr3C2
Cr7Cl
CrCl2
CrCl3
Cr2O3
Cs
CsCl
CsNO3
Cs2O
Cs2SO4
Cu
CuC03
CuCl
CuCl2
CuFe2O4
CuO
Cu2O
Cu(OH)2
CuS
Cu2S
CuS04
Cu2SO4
D2
D2O
D2O
F2
Fe
Fe3C
FeCO3
FeCl2
FeCl3
FeCr2O4
Fe0.947O
FeO
Fe2O3
Fe3O4
Fe(OH)3
FeS
State
Molar mass
[kg/kmol]
35.453
58.9332
118.9426
129.839
74.9326
240.7872
92.9479
90.997
154.995
51.996
180.010
400.005
122.902
158.355
151.990
132.905
168.358
194.910
281.809
361.872
63.54
123.55
98.99
134.45
239.23
79.54
143.08
97.55
95.60
159.14
159.60
223.14
4.02946
20.02886
20.02886
37.9968
55.847
179.552
115.856
126.753
162.206
223.837
68.8865
71.846
159.692
231.539
106.869
87.911
g
s, , hexagonal
s
s
s
s
s, pink
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
g
g
l
g
s, 
, cementite
s, siderite
s
s
s
s, wustite
s
s, hematite
s, magnetite
s
s, 
136
Enthalpy
[kJ/mol]
201.90
297.06
–22.38
144.96
59.12
0
–0.86
942.27
134.22
569.86
2415.85
5007.63
361.91
281.05
0
362.68
0
–80.22
407.75
30.53
201.59
0
144.57
141.95
60.62
44.27
234.56
–6.37
873.87
1049.10
155.65
377.15
249.199
0
–45.401
406.07
412.12
1654.97
65.06
230.77
253.29
107.10
124.01
140.16
0
117.98
–48.14
1037.54
Exergy
[kJ/mol]
87.1
265.0
45.8
118.8
52.8
38.2
50.7
792.2
99.8
544.3
2372.0
4874.2
311.9
261.6
36.5
404.4
51.5
18.2
521.8
127.0
134.2
31.5
76.2
82.1
36.1
6.5
124.4
15.3
690.3
791.8
89.8
253.6
263.8
31.2
22.3
466.3
376.4
1560.2
125.9
197.6
230.2
129.1
113.3
127.0
16.5
121.6
39.6
885.6
EXERGETICS
Substance
FeS2
FeSO4
FeSi
FeSiO3
Fe2SiO4
FeTiO3
H2
H
HCl
HDO
HDO
HF
HNO3
H2O
H2O
H3PO4
H2S
H2SO4
He
Hg
Hg2CO3
HgCl2
Hg2Cl2
HgO
HgS
HgSO4
Hg2SO4
I2
K
KAlSi3O4
K2CO3
KCl
KClO4
KF
KNO3
K2O
KOH
K2S
K2SO3
K2SO4
K2SiO3
Kr
Li
Li2CO3
LiCl
Li2O
LiOH
Li2SO4
State
s, pyrite
s
s
s
s, fayalite
s
g
g
g
g
l
g
l
g
l
s
g
l
g
l
s
s
s
s, red, orthorhombic
s, red
s
s
s
s
s, adularia
s
s
s
s
s
s
s
s
s
s
s
g
s
s
s
s
s
s
137
Molar mass
[kg/kmol]
119.975
151.909
83.933
131.931
203.778
151.75
2.01594
1.00797
36.461
19.0213
19.0213
20.0064
63.0129
18.01534
18.01534
98.0013
34.080
98.077
4.0026
200.59
461.189
271.50
472.09
216.59
232.65
296.65
497.24
253.8088
39.102
278.337
138.213
75.555
138.553
58.100
101.1069
94.203
56.109
110.268
158.266
174.266
154.288
83.80
6.939
73.887
42.392
29.877
23.946
109.940
Enthalpy
[kJ/mol]
1684.72
209.11
1249.42
118.07
255.30
118.90
241.818
338.874
108.82
0.21
–44.38
52.82
–53.19
0
–44.012
–76.26
946.61
153.25
0
63.82
–12.39
0
22.86
–27.01
731.08
81.73
110.36
—
356.63
66.26
–43.58
0
6.67
–7.77
–135.90
350.04
52.72
1024.40
300.47
4.62
75.9
0
328.10
–166.33
0
57.38
–35.73
–52.86
Exergy
[kJ/mol]
1428.7
173.0
1157.3
161.7
236.2
131.4
236.1
331.3
84.5
18.8
10.0
80.0
43.5
9.5
0.9
104.0
812.0
163.4
30.37
115.9
179.8
60 8
144.5
57.3
674.8
146.0
223.4
174.7
366.6
99.9
85.1
19.6
136.0
62.2
–19.4
413.1
107.6
943.0
302.6
35.0
137.9
34.36
393.0
70.1
70.7
225.7
74.1
204.3
EXERGETICS
Substance
Mg
MgAl2O4
MgCO3
MgCl2
MgFe2O4
MgO
Mg(OH)2
Mg(NO3)2
Mg3(PO4)2
MgS
MgSO4
MgSiO3
Mg2SiO4
Mg3Si2O5(OH)4
Mg3Si4O10(OH)2
Mg2TiO4
Mn
Mn3C
MnCO3
MnCl2
MnFe2O4
MnO
MnO2
Mn2O3
Mn3O4
Mn(OH)2
MnS
MnSO4
MnSiO3
Mo
Mo2C
MoO2
MoO3
Mo2S3
MoS2
N2
N2, atmospheric
NH3
NH4CL
NH4NO3
(NH4)2SO4
NO
NO2
N2O
N2O4
N2O5
Na
NaAlO2
State
Molar mass
[kg/kmol]
24.312
142.273
84.321
95.218
200.004
40.311
58.327
148.3218
262.879
56.376
120.374
100.396
140.708
277.134
379.289
160.52
54.9381
176.82545
114.9475
125.844
230.630
70.9375
86.9369
157.8744
228.8119
88.9528
87.002
151.000
131.022
95.94
203.89
127.94
143.94
288.07
160.068
28.0134
28.1541
17.0305
53.491
80.04348
132.138
30.0061
46.0055
44.0128
92.0110
108.0104
22.9898
81.9701
s
s, spinel
s
s
s
s
s
s
s
s
s
s
s
s, chrysolite
s, talc
s
s, 
s
s
s
s
s
s
s
s
s, amorphous
s, green
s
s
s
s
s
s
g
g
g
s
s
s
g
g
g
g
g
s
s
138
Enthalpy
[kJ/mol]
725.71
274.17
23.43
244.65
121.53
124.38
42.73
–64.34
76.59
1105.11
166.22
87.73
188.35
117.06
140.26
231.48
520.03
1958.20
19.42
199.18
118.36
134.81
0
81.09
172.26
66.47
1031.23
180.20
110.08
745.09
1838.88
156.15
0
3302.43
1960.78
0
0
316.62
249.43
118.08
511.84
90.25
33.18
82.05
9.163
11.30
330.90
128.40
Exergy
[kJ/mol]
633.8
230.3
37.9
165.9
77.9
66.8
40.9
57.4
130.0
901.6
80.7
22.0
74.9
61.3
36.5
134.3
482.3
1862.3
81.8
165.4
121.4
119.4
21.1
89.4
171.6
107.3
873.5
142.4
102.3
730.3
1824.6
201.2
68.2
2891.2
1723.1
0.72
0.69
337.9
331.3
294.8
660.6
88.9
55.6
106.9
106.5
125.7
336.6
151.7
EXERGETICS
Substance
NaAlSi2O6·H2O
NaAlSi3O8
Na2CO3
NaCl
NaHCO3
NaI
NaNO3
Na2O
NaOH
Na2S
Na2SO3
Na2SO4
Na2SiO3
Na2Si2O5
Na4SiO4
Ne
Ni
Ni3C
NiCO3
NiCl2
NiO
Ni(OH)2
NiS
Ni3S2
NiSO4
NiSO4·6H20
O2
O
O3
P
P
P4O10
Pb
PbCO3
PbCl2
PbO
PbO
PbO2
Pb3O4
Pb(OH)2
PbS
PbSO4
PbSiO3
Pb2SiO4
Rb
Rb2CO3
RbCl
Rb2O
State
s, analcime
s, low albite
s
s
s
s
s
s
s
s
s
s
s
s
s
g
s
s
s
s
s
s
s
s
s
s, , tetragonal, green
g
g
g
s, , white
s, red, triclinic
s, hexagonal
s
s
s
s, yellow
s, red
s
s
s
s
s
s
s
s
s
s
s
139
Molar mass
[kg/kmol]
220.055
262.2245
105.9891
58.443
84.0071
149.8942
84.9947
61.9790
39.9972
78.044
126.042
142.041
122.064
182.149
184.043
20.183
58.71
188.14
11 8.72
129.62
74.71
92.72
90.77
240.26
154.77
262.86
31.9988
15.9994
47.9982
30.9738
30.9738
283.8892
207.19
267.20
278.10
223.19
223.19
239.19
685.57
241.20
239.25
303.25
283.27
506.46
85.47
230.95
120.92
186.94
Enthalpy
[kJ/mol]
35.41
72.75
75.62
0
–101.94
–135.62
243.82
23.79
1014.84
297.63
0
11.31
13.28
151.45
0
239.74
1180.09
–49.93
94.85
0
–48.13
883.15
1967.14
92.25
–266.75
0
249.17
142.67
840.06
822.49
376.21
305.64
0
106.67
88.32
86.65
28.24
198.53
32.48
930.64
111.12
70.88
159.07
350.38
–33.90
0
370.60
Exergy
[kJ/mol]
104.2
105.5
41.5
14.3
21.6
136.1
–22.7
296.2
74.9
921.4
287.5
21.4
66.1
67.6
256.6
27.19
232.7
1142.9
36.4
97.2
23.0
25.5
762.8
1720.2
90.4
53.6
3.97
233.7
169.1
875.8
863.6
825.3
232.8
23.5
42.3
46.9
45.9
19.4
105.2
20.6
743.7
37.2
31.2
75.5
388.6
152.4
48.6
491.3
EXERGETICS
Substance
S
SO2
SO3
Sb
Sb2O3
Sb2O4
Sb2O5
Se
Si
SiC
SiCl4
SiO2
SiO2
SiO2
SiS2
Sn
Sn
SnCl2
SnO
SnO2
SnS
SnS2
Sr
SrCO3
SrCl2
SrO
SrO2
SrS
SrSO4
Ti
TiC
TiO
TiO2
Ti2O3
Ti3Os
TiS2
U
UCl3
UCl4
UCl5
UO2
UO3
U3O8
V
VC
VO
VO2
V2O3
State
Molar mass
[kg/kmol]
32.064
64.0628
80.0622
121.75
291.50
307.50
323.60
78.96
28.086
40.097
169.898
60.085
60.085
60.085
92.214
118.69
118.69
189.60
134.69
150.69
150.75
182.82
87.62
143.63
158.53
103.62
119.62
119.68
183.68
47.90
59.91
63.90
79.90
143.80
223.70
112.03
238.03
344.39
379.84
415.30
270.03
286.03
842.085
50.942
62.953
66.941
82.940
149.882
s, rhombic
g
g
s, III
s
s
s
s, black
s
s, , hexagonal
l
s, , quartz
s, , cristobalite
s, amorphous
s
s, I, white
s, II, gray
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s, rutile
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
140
Enthalpy
[kJ/mol]
725.42
428.59
329.70
485.97
263.07
62.63
0
–
910.94
1241.69
544.81
0
1.46
7.45
2149.23
580.74
578.65
416.08
294.97
0
1205.74
1863.8
826.34
0
157.93
234.30
191.96
1098.99
98.66
944.75
1154.16
425.14
0
368.66
375.10
2060.45
1230.10
577.35
499.39
536.93
145.19
0
115.49
775.30
1067.96
343.51
57.74
322.60
Exergy
[kJ/mol]
609.6
313.4
249.1
435.8
251.2
83.7
52.3
346.5
854.6
1204.6
481.9
1.9
2.8
7.9
1866.3
544.8
544.9
386.4
289.9
29.1
1056.1
1604.6
730.2
6.2
72.6
170.2
140.4
891.8
7.1
906.9
1136.7
418.5
21.4
385.5
413.2
1875.9
1190.7
550.1
475.2
513.6
162.9
43.9
218.5
721.1
1032.6
318.9
61.9
299.7
EXERGETICS
Substance
V2O5
W
WC
WO2
WO3
WS2
Xe
Zn
ZnCO3
ZnCl2
ZnFe2O4
ZnO
Zn(OH)2
ZnS
ZnSO4
Zn2SiO4
State
Molar mass
[kg/kmol]
181.881
183.85
195.86
215.85
231.85
249.98
131.30
65.37
125.38
136.28
241.06
81.37
99.38
97.43
161.43
222.82
s
s
s
s
s
s
g
s
s
s
s
s
s, 
s, sphalerite
s
s
141
Enthalpy
[kJ/mol]
0
842.87
1195.84
253.18
0
2084.51
0
419.27
0
583.93
74.08
70.99
19.18
938.71
161.87
112.74
Exergy
[kJ/mol]
32.5
827.5
1199.5
297.5
69.3
1796.6
40.33
339.2
23.9
93.4
36.4
22.9
25.7
747.6
82.3
17.8
EXERGETICS
Table A.4.2
Enthalpy and exergy of formation for organic substances at reference state
(T0 = 298.15 K, P0 = 101.325 kPa)
Chemical
formula
Namne
CH2
CH3
Methylene
Methyl
State
g
g
Molar
mass
kg/kmol
Enthalpy
Exergy
of formation
[kJ/mol] [kJ/mol]
14.02709
15.03506
1032.9
889.9
1030.5
900.5
802.33
1427.79
2045.4
2658.4
3274.3
3247.2
3889.3
3857.6
4464.7
5074.4
5684.2
6294.0
6908.6
7518.8
8129.0
8739.2
9349.4
9959.6
831.65
1495.84
2154.0
2805.8
3463.3
3461.8
4118.5
4114.5
4761.7
5413.1
6064.9
6716.8
7376.9
8029.4
8682.0
9334.5
9984.8
10639.7
1959.2
2388.7
3691.4
3658.3
3673.5
4295.5
4914.3
4873.8
5483.3
6094.3
2043.2
2516.2
3914.3
3909.2
3910.8
4556.9
5228.5
5205.9
5857.7
6511.5
1323.1
1927.7
2542.9
1361.1
2003.9
2659.7
Enthalpy
Exergy
Aliphatic saturated hydrocarbons CnH2n+2
CH4
C2H6
C3H8
C4H10
C5H12
Methane
Ethane
Propane
n-Butane
n-Pentane
C6H14
n-Hexane
C7H16
C8H18
C9H20
C10H22
C11H24
C12H26
C13H28
C14H30
C15H32
C16H34
n-Heptane
n-Octane
n-Nonane
n-Decane
n-Undecane
n-Dodecane
n-Tridecane
n-Tetradecane
n-Pentadecane
n-Hexadecane
g
g
g
g
g
l
g
l
l
l
l
l
l
l
l
l
l
l
16.04303
30.07012
44.09721
58.1243
72.15139
86.17848
100.20557
114.23266
128.25975
142.28684
156.31393
170.34102
184.36811
198.3952
212.42229
226.44938
Cycloparaffins CnH2n
C3H6
C4H8
C6H12
Cyclopropane
Cyclobutane
Cyclohexane
C6H12
C7H14
C8H16
Methylcyclopentane
Methylcyclohexane
Ethylcyclohexane
C9H18
C10H20
n-Propylcyclohexane
n-Butylcyclohexane
g
g
g
l
l
g
g
l
l
l
42.08127
56.10836
84.16254
84.16254
98.18963
112.21672
126.24381
140.2709
Olefins (ethylenic hydrocarbons) CnH2n
C2H4
C3H6
C4H8
Ethylene
Propylene
1-Butylene
Chemical
Namne
g
g
g
State
142
28.05418
42.08127
56.10836
Molar
EXERGETICS
formula
C6H12
1-Hexylene
C7H14
1-Heptylene
g
l
g
l
mass
kg/kmol
of formation
[kJ/mol] [kJ/mol]
84.16254
3772.9
3742.2
4388.0
4338.5
3970.5
3967.9
4625.5
4604.6
98.18963
Acetylene hydrocarbons CnH2n-2
C2H2
C3H4
C4H6
C6H10
C7H12
C8H14
C9H16
Acetylene
Propyne
1-Butyne
1-Hexyne
1-Heptyne
1-Octyne
1-Nonyne
g
g
g
g
g
g
g
26.03824
40.06533
54.09242
82.1466
96.17369
110.20078
124.22787
1255.6
1850.9
2465.6
3696.3
4311.4
4923.2
5537.9
1265.8
1899.5
2552.3
3865.1
4520.5
5170.3
5825.1
g
g
40.06533
68.11951
1472.1
2789.2
1523.8
2914.8
Diene hydrocarbons CnH2n-2
C3H4
C5H8
Propadiene
Pentadiene
Aromatic hydrocarbons (benzene derivatives)
C6H6
Benzene
C7H8
Toluene
C8H10
Ethylbenzene
C8H10
C9H12
C10H14
C16H26
o-Xylene
n-Propylbenzene
n-Butylbenzene
n-Decylbenzene
g
l
g
l
g
l
l
l
l
l
78.11472
106.1689
120.19599
134.22308
218.38562
3171.6
3137.7
3774.4
3736.4
4390.0
4347.7
4332.8
4957.5
5567.7
9198.3
3303.6
3298.5
3943.4
3931.0
4598.8
4587.9
4573.1
5249.1
5892.0
9700.8
s
s
s
s
s
s
s
s
s
s
s
128.17526 4984.2
134.22308 5533.0
142.20235 5574.9
148.25017 6131.6
162.27726 6739.1
178.2358
6850.9
178.2358
6835.9
182.26768 7250.9
254.50356 11116.7
244.33937 9579.7
306.41106 11850.1
5255.0
5880.0
5881.4
6516.0
7171.0
7218.1
7201.8
7665.9
11937.4
10109.2
12490.3
92.14181
106.1689
Solid hydrocarbons
C10H8
C10H14
C10H10
C11H16
C12H18
C14H10
C14H10
C14H14
C18H38
C19H16
C24H18
Naphthalene
1,2,4,5-Tetramethylbenzene
2-Methylnaphthalene
Pentamethylbenzene
Hexamethylbenzene
Anthracene
Phenanthrene
1,1-Diphenylethan
n-Octadecane
Triphenylmethane
1,3,5-Triphenylbenzene
Chemical
Namne
State
143
Molar
Enthalpy
Exergy
EXERGETICS
formula
C25H20
mass
kg/kmol
Tetraphenylmethane
s
of formation
[kJ/mol] [kJ/mol]
320.43815 12544.1
13231.6
Organic compounds containing oxygen
CH2O
CH2O3
Formaldehyde
Formic acid
CH4O
C2H6O
Methanol
Ethyl alcohol
C2H6O
C2H4O
C2H4O
C2H6O2
C2H4O2
Dimethyl ether
Acetic aldehyde
Ethyleneoxyde
Ethylene glycol
Acetic acid
C3H8O
C3H6O
Propylalcohol-2
Acetone
C4H8O
C4H8O
C4H40
Butylaldehyde-1
Butylketone-2
Furane
C4H8O2
C4H8O2
C5H12O
C5H12O
C5H10O
C5H6O2
C6H14O
C6H12O
C7H16O
C7H8O
C4H10O4
C4H6O4
C4H4O4
C4H4O4
C6H6O
C6H14O6
C6H14O6
C6H12O6
C6H12O6
C7H6O2
C7H6O3
C8H4O3
C8H6O4
Butyric acid
Ethyl acetate
Amyl alcohol
2-Methylbutanol-2
Cyclopentanol
Furfuryl alcohol
Hexyl alcohol-l
Cyclohexanol
Heptyl alcohol-l
Benzyl alcohol
Erythrite
Succinic acid
Malonic acid
Fumaric acid
Phenol
Dulcite
Mannit
-D-Galactose
L-Sorbose
Benzoic acid
Hydroxybenzoic acid
Phthalic acid anhydride
Phthalic acid
Chemical
Namne
g
g
l
l
g
l
g
g
g
l
g
l
l
g
l
l
l
g
l
l
l
l
l
l
l
l
l
l
l
s
s
s
s
s
s
s
s
s
s
s
s
s
State
144
30.02649
46.02589
32.04243
46.06952
46.06952
44.05358
44.05358
62.06892
60.05298
60.09661
58.08067
72.10776
72.10776
68.07588
88.10716
88.10716
88.15079
88.15079
86.13485
98.10237
102.17788
100.16194
116.20497
108.14121
122.1219
118.09002
116.07408
116.07408
94.11412
182.17488
182.17488
180.15894
180.15894
122.12467
138.12407
148.11928
166.13462
Molar
519.4
259.1
213.0
638.4
1278.2
1235.9
1328.1
1105.5
1220.5
1058.6
834.1
786.6
1830.6
1690.9
1659.6
2296.5
2264.1
2024.4
1996.7
2018.8
2073.6
3060.7
3017.2
2878.7
2418.6
3668.9
3465.4
4285.6
3563.4
1874.7
1356.9
1271.3
1249.1
2925.9
2729.6
2739.6
2529.6
2544.6
3097.2
2888.1
3173.8
3094.3
538.4
301.3
291.7
718.0
1363.9
1357.7
1419.5
1163.3
1284.4
1207.3
919.0
908.0
1998.6
1791.5
1788.5
2463.3
2432.6
2118.8
2118.2
2215.8
2269.6
3311.7
3275.7
3109.7
2687.7
3961.1
3750.8
4619.2
3795.8
2193.0
1609.4
1495.7
1471.5
3128.5
3196.3
3204.8
2928.8
2939.0
3343.5
3151.2
3434.8
3412.6
Enthalpy
Exergy
EXERGETICS
formula
C12H10O
C12H22O11
C12H22O11
C12H24O12
C12H24O12
C16H34O
C16H32O2
C2H2O4
mass
kg/kmol
Diphenyl ether
-Lactose
Saccharose
-Lactose monohydrate
-Maltose monohydrate
Cetyl alcohol
Palmitin acid
Oxalic acid
s
s
s
s
s
s
s
s
170.2129
342.30254
342.30254
360.31788
360.31788
242.44878
256.43224
90.03584
of formation
[kJ/mol] [kJ/mol]
5903.1
5154.2
5166.2
5152.2
5173.2
9731.3
9290.3
202.7
6282.4
5988.1
6007.8
6043.3
6063.4
10493.9
10052.3
368.7
Organic compounds containing nitrogen and oxygen
C2H4N4
C3H6N6
C5H5N5
C6H4N2
C12H11N
Dicyanodiamide
Melamine
Adenine
2-Cyanopyridine
Diphenylamine
s
s
s
s
s
84.08098
126.12147
135.1291
104.11218
169.22817
1296.5
1835.6
2664.9
3106.1
6188.1
1477.4
2120.5
2941.0
3246.9
6540.7
CH4ON2
CH6O2N2
C2H5O2N
C3H7O2N
C4H7O4N
C4H8O3N2
C4H2O4N2
C4H7ON3
C4H9O2N3
C4H6O3N4
C5H9O4N
C5H4ON4
C5H4O2N4
C5H5O3N4
C5H5ON5
C9H9O3N
C12H5O12N7
Urea
Ammonium urethane
Aminoacetic acid
D,L-Alanine
L-Aspartic acid
L-Asparagine
Alloxan
Creatinine
Creatine
Allantoin
D-Glutamic acid
Hypoxanthine
Xanthine
Uric acid
Guanine
Hypuric acid
1,3,5-Hexanitrodiphenylamine
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
60.05583
78.06037
75.06765
89.09474
133.10469
132.11996
142.07154
113.11989
131.13523
158.11742
147.13178
136.11383
152.11323
169.1206
151.1285
179.17698
439.21335
544.7
474.8
867.6
1462.4
1445.2
1749.9
813.1
2179.7
2123.1
1580.0
2047.9
2337.3
2068.8
1950.6
2385.2
4014.3
5397.2
689.0
666.7
1049.5
1689.4
1743.8
2061.3
1053.8
2440.6
2442.8
1909.9
2393.2
2602.4
2361.7
2289.1
2691.2
4388.1
6167.8
62.134
62.134
76.161
76.161
90.188
90.188
90.188
90.188
2164.9
2173.2
2784.3
2783.6
3386.2
3378.9
3392.9
3386.2
2134.0
2145.4
2794.7
2795.0
3438.7
3434.3
3446.3
3442.6
Enthalpy
Exergy
Organic compounds containing sulfur
C2H6S
C2H6S
C3H8S
C3H8S
C4H10S
C4H10S
C4H10S
C4H10S
Ethyl mercaptan
Dimethyl sulfide
Propyl mercaptan
Methylethyl sulfide
Butyl mercaptan
2-Methylpropyl mercaptan
Diethyl sulfide
Methylpropyl sulfide
Chemical
Namne
l
l
l
l
l
l
l
l
State
145
Molar
EXERGETICS
formula
C4H4S
C4H10S2
C5H12S
C5H6S
C5H6S
C6H6S
C6H14S2
C6H10O4S2
C3H7OSN
C6H12O4S2N2
mass
kg/kmol
Thiophene
Ethylbutyl disulfide
Amyl mercaptan
2-Methylthiophene
3-Methylthiophene
Thiophenol
Dipropyl disulfide
Milk acid ,'-dithiane
1-Cysteine
1-Cystine
l
l
l
l
l
l
l
s
s
s
146
84.140
122.252
104.215
98.168
98.168
110.179
150.306
210.272
105.159
240.302
of formation
[kJ/mol] [kJ/mol]
2865.9
4117.1
3994.8
3374.5
3376.1
3876.6
5336.5
4053.1
2219.3
4212.8
2847.0
4055.4
4091.3
3396.5
3398.2
3916.1
5358.4
4168.6
2285.8
4415.5
EXERGETICS
Table A.4.3
Enthalpy and exergy of formation for organic substances by groups
Gases
Enthalpy
Exergy
of formation
[kJ/mol]
[kJ/mol]
Group
Liquids
Enthalpy
Exergy
of formation
[kJ/mol] [kJ/mol]
C
398.57
462.77
403.54
462.64
CH
509.77
557.40
485.75
545.27
CH2
CH3
614.91
713.47
654.51
747.97
607.38
715.35
651.46
752.03
C
440.53
513.35
443.16
473.02
CH
CH2
C
C
CH
551.86
660.26
543.04
510.20
625.37
576.31
678.74
554.23
519.58
630.28
535.08
680.26
539.28
494.34
623.86
569.95
675.68
559.21
515.27
634.34
(ring)
413.34
461.01
379.03
425.11
CH (ring)
522.78
561.37
468.76
543.05
CH2 (ring)
629.05
662.29
614.16
653.63
C
442.71
466.41
559.18
576.65
542.95
568.28
414.22
436.03
416.14
435.03
415.47
440.00
426.22
436.45
528.26
549.91
519.93
547.15
C
(ring)
CH (ring)

C




C

HC
147
—
—
EXERGETICS
Gases
Group
(ring)

O
O
O

O
Liquids
Enthalpy
Exergy
of formation
[kJ/mol]
[kJ/mol]
Enthalpy
Exergy
of formation
[kJ/mol] [kJ/mol]
–111.59
–246.86
–117.42
–89.11
–245.09
–97.12
–131.17
–91.46
–126.27
–106.64
–89.96
–83.59
–84.43
–73.13
–137.54
–80.08
–86.52
(
C
)
–68.42
–165.48
OH to
(
CH
)
63.90
–66.78
OH (to
OH (to
CH2 )
CH3 )
–56.66
–77.00
–42.89
–25.52
–84.82
–76.87
–51.34
–33.97
OH (to
CH ring)
-65.16
–46.78
–70.47
–58.16
OH (attached to
aromatic)
–66.12
–52.01
81.64
–47.57
C
O
262.38
293.87
231.58
281.36
O
388.64
412.68
356.72
400.21
O
65.69
108.30
35.90
101.15
296.94
382.66
244.81
362.70
OH to
–85 11
–52
H
C
O
C
O
C
O
O
C
O
C
OH
O
O
C
—
H
168.04
207.01
250.09
—
155.11
183.96
O
C
(ring)
—
305.66
148
—
277.76
EXERGETICS
Gases
Liquids
Enthalpy
Exergy
of formation
[kJ/mol]
[kJ/mol]
Group
Enthalpy
Exergy
of formation
[kJ/mol] [kJ/mol]
H
C
O (attached to
aromatic)
382.87
415.07
379.60
410.21
N
97.03
142.05
64.60
131.09
NH
181.49
213.38
137.18
195.56
NH2
258.43
290.20
235.43
284.39
N
56.82
72.98
–103.43
N
24.18
23.06
0
29.97
NH (ring)
186.19
209.24
151.20
199.37
NH2 (attached to
aromatic)
237.80
240.16
216.84
269.24
NH (attached to
aromatic)
153.58
196.27
77.07
134.06
69.08
81.68
72.34
83.33
–42.30
1.45
–58.32
12.16
19.66
–89.91
585.26
516.93
761.07
862.06
692.04
18.89
–89.77
592.73
527.50
636.88
724.36
553.78
—
–121.71
551.97
510.53
741.74
848.67
696.47
—
–23.88
584.03
522.14
642.32
732.26
566.88
660.53
562.95
686.39
(attached to
aromatic)

N

N
NO2
N
O
O
N
C
S
SH
SO
SO
NO
O2
C
149
EXERGETICS
SO 2
439.87
Enthalpy
Exergy
of formation
[kJ/mol]
[kJ/mol]
Group
(ring)
414.68
Liquids
Enthalpy
Exergy
of formation
[kJ/mol] [kJ/mol]
764.08
633.45
755.42
687.25
S
781.15
653.56
762.45
654.41
F
Cl
F (attached to aromatic)
Cl (attached to aromatic)
Ortho (1,2)
Meta (1,3)
Para (1,4)
1,2,3 Position
1,2,4 Position
1,3,5 Position
1,2,3,5 Position
1,2,3,5 Position
1,2,4,5 Position
1,2,3,4,5 Position
1,2,3,4,5,6 Position
3 Atom saturated ring
4 Atom saturated ring
5 Atom saturated ring
6 Atom saturated ring
7 Atom saturated ring
8 Atom saturated ring
Pentene ring
Hexene ring
–15.00
46.08
15.08
46.08
4.35
1.76
1.26
12.89
9.05
6.28
14.06
12.80
12.38
17.99
19.66
62.30
50.84
–50.38
–83.05
–73.81
–72.59
–50.38
–83.05
16.25
26.24
45.16
26.24
7.66
3.47
4.60
20.08
13.60
14.27
26.94
23.93
24.27
35.36
62.34
49.04
43.76
–45.52
–61.25
–46.32
–33.47
–45.52
–61.25
–8.85
42.89
–8.85
42.89
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
83.68
87.82
0.00
–28.79
34.70
32.01
34.70
32.01
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
83.68
82.30
0.00
0.00
0.00
–28.79
0.00
0.00

S
373.78
Gases

150
EXERGETICS
Table A.4.4
Exergy of substances in ideal water solution (molarity 1 mol/kg H2O), (T0
= 298.15 K, P0 = 101.325 kPa)
Exergy
Solved
Chemical
formula
Unsolved
[kJ/mol]
Form
[kJ/mol]
AgCl
CaCl2
59.4
—
Ag2+, Cl–
Ca2+, 2Cl–
77.9
19.7
CuCl
CuCl2
—
—
Cu+, Cl
Cu2+, 2Cl–
114.7
61.2
FeCl2
—
Fe2+, 2Cl–
158.7
29.57
2H+, CO3,2 –
H+, HCO3,–
H+, Cl–
124.6
H2CO3
HCl
HF 56.6
HNO3
H3PO4
H2S
H2SO4
48.9
H+, F–
74.6
H+, NO3
20.9
3H+, PO4,3 –
2H+, HPO4,2 –
80.6
817.9
—
65.6
48.5
13.0
204.4
—
134.1
—
2H+, SO4,2 –
108.6
K+, Cl–
14.6
–20.7
KCl
KNO3
15.0
–19.8
KOH
K2SO4
46.9
—
LiCl
MgCl2
NH3
NH4OH
58.1
—
327.4
328.8
Na2CO3
—
2Na+, CO3,2 –
37.7
5.5
21.4
Na+, Cl–
5.1
21.4
NaCl
NaHCO3
NaNO3
–29.4
NaOH
Na2SO4
O2
37.7
—
20.3
PbCl2
ZnCl2
ZnSO4
K+, NO3,–
K+, OH
2K+, SO4,2 –
Li+, Cl
Mg2+, 2Cl
—
NH4, OH
Na+, HCO3,–
Na+, NO3,–
Na+, OH–
46.9
41.3
29.7
40.4
—
322.1
–30.2
—
37.7
22.2
—
—
Pb2+, 2Cl–
69.5
—
Zn2+, 2Cl–
53.2
51.8
Zn2+, SO4,2 –
80.2
2Na+, SO4,2 –
151
Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Related manuals

Download PDF

advertisement