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Review of TD
Thermodynamics is the science and technology that deals with certain laws those govern the transformation of energy from one form to another. The name thermodynamics is derived from Greek words thermo and dynam-ics meaning heat and mechanical power respectively.
The conceptions of thermodynamics are more or less idealistic. As an engineer we are always interested for bulk phenomena. Hence macro level conceptions are usual for an engineer.
Classification of Thermodynamics
Thermodynamics is applied to every wake of life. There is hardly any stream in science and technology which does not follow the laws of thermodynamics. A broad classification of the subject is shown in Table 1.1
Table 1.1 Classification of thermodynamics
Classical thermodynamics is an experimental science which deals with macroscopic or large scale properties of matter. General relations of parameters or quantities such as coeficients of expansion, compressibility, specific heat capacities, heats of transformation, magnetic and dielectric cofficients, etc can be established with the help of classical thermodynamics. Unfortunately, acutal magnitude of such quatities can not be estimated with the help of classical thermodynamics.
Kinetic theory of matter helps in estimating numerical value of the individual quantites. This class of thermodynamics deals with molecular model in which individual molecules follow the laws of mechanics.
http://nptel.ac.in/courses/112103016/module1/images/thermodynamics.jpgStatistical thermodynamics ignores the detailed consideration of molecules as individuals and applies statistical considerations to find the distribution of the very large number of molecules that make up a macroscopic piece of matter over the energy states of matter.
It is worth mentioning that these theories are applicable under equilibrium conditions.
Irreversible thermodynamics is another branch of thermodynamics dealing with non-equilibrium, irreversible processes.
Concepts of classical thermodynamics
Thermodynamic system
A thermodynamic system is defined as the quantity of matter or a region in space upon which attention is concentrated in the analysis of a problem. Here quantity of matter may be gas, solid or liquid, magnetic field, electric field or even photons. Minimum quantity of matter required to analyze a thermodynamic system is the order of .
Surroundings/Environment
Everything external to the system is called the surroundings or environment. The system is separated from the surroundings by the system boundary. The boundary may be fixed or flexible.
Fig 1.1 Thermodynamic System
System and surroundings together constitute the universe. Size of the universe depends on the size of the system and surroundings. A pin head can constitute a system and the bulb containing the pinhead may form the universe.
Some features of system approach
1. Thermodynamic system is arbitrary. Choice of system depends upon the type and nature of analysis
2. A system is always demarcated with a boundary
http://nptel.ac.in/courses/112103016/module1/images/avogadro.jpg3. It is hypothetical. Valid under the concept that system is continuum.
4. Energy and mass can enter or leave the system boundary
5. Fate of energy (heat and work) inside the system is neither known nor predictable
Classification of system
A system is classified as closed, open or isolated based on the interaction of mass and energy through the system boundary.
In a closed system no mass can transfer across the system boundary but energy can transfer through it.
Fig. 1.3 A Closed System
In an open system both the mass and energy transfer across the system boundary. Boudary for an open system is usually drawn with dotted or broken line
Fig. 1.4 An Open System
In case of an isolated system, no mass or energy can transfer across the boundary.
Fig. 1.5 An Isolated System
In all sort of thermodynamic problems it is essential to recognize the system for formulation. Once the system is identified, next step is to find the energy and mass interaction across the system boundary.
Fig 1.2 Thermodynamic System for a Pin Head
If we consider the earth (lithosphere) to be a system, the atmosphere may be termed as the surroundings. Hence earth and its atmosphere can be said to be the universe. Similarly, if we consider the solar system as our system, the milkyway may become the surroundings and universe may be infinite.
Behavoir of a thermodynamic system
The behavior of the system depends upon the interaction of energy with or without mass transfer across the boundary. Every system has certain characteristics by which its physical conditions may be described. Such behavior/characteristics of a system are called the properties of the system. There are 8 (eight) properties describing the behavior of a system. They are pressure, temperature, volume, entropy, internal energy, enthalpy, Gibbs function and Helmholtz functions. Pressure, temperature and volume are measurable properties and they are also known as physical properties (also known as macroscopic properties). Other properties are derived properties (they can not be measured directly). In sense of thermodynamics, pressure is always expressed in terms of absolute pressure and temperature is expressed in Kelvin.
When all the properties of a system have definite values, the system is said to exist in a definite state. State of a thermodynamic system can be indicated in a diagram with properties as coordinates.
Intensive and extensive properties
Thermodynamic properties can be divided into 2 (two) general classes such as intensive and extensive properties. An intensive property, is a physical property of a system that does not depend on the system size or the amount of material in the system. By contrast, an extensive property of a system does depend on the system size or the amount of material in the system. According to the definitions, density, pressure and temperature are intensive porperties and volume, internal energy are extensive properties. Symbols for representing properties: Extensive properties are symbolized by upper case (capital) letter such as V (volume), KE (kinetic energy), PE (potential energy), etc. Intensive properties are symbolized by lower case letters such as v (specific volume), ke (specific kinetic energy), e, u (specific internal energy), h (specific enthalpy), etc. Mole based properties are symbolized by lower
case letters with overbars. For example , , , and are molar specific voulme, molar specific energy, molar specific kinetic energy and molar specific potential energy respectively. Exceptions: Temperature (intensive), mass (extensive), and number of moles (extensive). The use of symbols for temperature, mass and moles are traditional.
Exceptions:
Table : 1.2 Symbols for Exceptional Properties
Parameter Property Traditional symbol
Temperature Intensive property T
Mass Extensive property m
Number of moles Extensive property n
Specific extensive properties, i.e., extensive properties per unit mass are intensive properties. For example: specific volume, specific energy, density, etc. It may be worth mentioning here that cyclic integral of the differential of any thermodynamic
property must be zero. For example,
However, change in volume between state points 1 and 2 is
Path and process
An operation in which one or more of the properties of a system changes is called a change of state. The sucession of states passed through during a change of state is called the path of the change of state.
The path of thermodynamic states that a system passes through as it goes from an initial state to a final state is known as the thermodynamic process.
Different thermodynamic processes are given in Table 1.3
Table : 1.3 Different Thermodynamic Processes
Sl No Name of the process Parameter Held
Constant
Remarks
1 Constant pressure(Isobaric) p = constant v = (mR/P)T
2 Constant volume(isochoric) process V = constant p = (mR/V)T
3 Constant temperature (isothermal)
process
T = constant pV = constant
4 Polytropic process n pVn = constant
5 Adiabatic process no heat flow across the
systemboundary
pv = constant
6 Isoenthapic process h = constant h = constant
Quiz 1
1. How does classical thermodynamics differ from kinetic theory and statistical thermodynamics?
2. Define thermodynamic path and process.
3. What are intensive and extensive properties? Give 3 examples for each.
4. Sketch a pressure cooker and show open, closed and isolated system around it.
5. Why pressure is an intensive property?
Review of TD I
Quasi-static process
Consider a system of gas contained in a cylinder-piston assembly (Fig 1.6). The system is initially in an equilibrium state represented by a set of properties (x1, y1). The weight of the piston just balances the upward force exerted by the gas. If the weight is removed there will be an unbalanced upward force between the system and the surroundings. The piston will move up till the stop because of the unbalance forces created by the gas. In this final position, the system will be again under equilibrium state with properties say, (x2, y2). Both the initial and final states can be located in a generalized 2 dimensional diagram. However, as the system was not in equilibrium between these two end states, they can not be presented in the same diagram. Such
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In this case also the gas is in initial equilibrium state 'i' with set of properties (xi, yi). Let us take out weight very slowly which is infinitesimally small. The system will come to equilibrium with new set of properties. Now remove the next weight slowly as before and this time also the piston will move up infinitesimally slow to another equilibrium state (x1, y1). State points and being in equilibrium can now be located in the coordinate. Experiment is repeated in similar manner till the last wieght is removed when we will attain the final state 'f' with properties at equilibrium condition (xf, yf).
From this experiment we find that each and every state points are under equilibrium state and they can be indicated in the coordinate. Now the initial, intermediate and final states, all being in equilibrium states, can be joined with a solid line. Such a process in which each of the state is in equilibrium state is known as quasi-static equilibrium state. A quasi-static process is thus a sucession of equilibrium states. The character of such a process is that it is infinitely slow process. Later on we will recognize the processes as reversible processes after learning the second law of thermodynamics. A reversible state is a quasi-static process without any dissipation. Till we learn about the reversible process, let us be content with the quasi static process.
Energy interactions
In thermodynamics 2 (two) types of energy are considered. One is heat and the other is work. Both of these energies may interact at the boundary of the system. When work energy interacts at the boundary and converts to work only, we need not pay attention to its quality as final form of energy is work only. Heat to heat interactions always involve degradation. Work converted to heat is not useful, we call it as dissipation of energy. In thermodynamics basic interest is conversion of heat into work.
Laws of Thermodynamics
Interactions of energy at system boundary is governed by certain laws. There are 4 basic laws in classical thermodynamics. These are: Zeroth law, first law, second law and third law of thermodynamics. Apart from these laws, there is another law applied to irreversible thermodynamics developed by Onsagar in the year 1957. This law is termed as fourth law of thermodynamics.
Table 1.4 presents the laws of thermodynamics along with the scientists associated with their invention and year of invention.
Thermodynamic
Laws Scientists/Researchers Year
Zeroth Law Fowler and Guggenheim 1939
First Law Joule, Mayer, Thompson and Colding 1845
Second Law Carnot 1824
Third Law Nernst 1907
http://nptel.ac.in/courses/112103016/module1/animation/dotted%20line.swfhttp://nptel.ac.in/courses/112103016/module1/animation/slow.swfhttp://nptel.ac.in/courses/112103016/module1/animation/slow.swfhttp://nptel.ac.in/courses/112103016/module1/animation/slow.swfhttp://nptel.ac.in/courses/112103016/module1/images/joule.jpghttp://nptel.ac.in/courses/112103016/module1/images/meyer.jpghttp://nptel.ac.in/courses/112103016/module1/images/thompson.jpghttp://nptel.ac.in/courses/112103016/module1/images/colding.jpghttp://nptel.ac.in/courses/112103016/module1/images/carnot.jpghttp://nptel.ac.in/courses/112103016/module1/images/nernst.jpgFourth Law Onsagar 1968
Table 1.4: Different Thermodynamic Laws
The zeroth law and third law are more definitional in nature. The first and the second laws are more pragmatic and as an engineer we use both these laws for analysis.
Zeroth law gives the definition of temperature.
This law was framed by Ralph Fowler and Guggenheim. Fowler was a British Physicist and astronomer. He named zeroth law of thermodynamics in 1920. Fifteen Fellows of the Royal Society and three Nobel Laureates were supervised by Fowler between 1922 and 1939. He worked with Sir Arthur Eddington, Subrahmanyan Chandrasekhar, Paul Dirac, Sir William McCrea and Milne. Fowler introduced Paul Dirac to quantum theory in 1923. He supervised the doctoral studies of 64 students at Cambridge University, including John Lennard-Jones, Paul Dirac and Garrett Birkhoff.
Ralph Fowler Edward Armand Guggenheim
Fig: 1.8 Photo of Fowler and Guggenheim
Edward Armand Guggenheim (1901 1970) was an English thermodynamicist and professor of chemistry at the University of Reading in the year 1939, Guggenheim co-authored a volume entitled Statistical Thermodynamics with Ralph Fowler. The text book by Fowler and Guggenheim on Statistical Thermodynamics was published by Cambridge University Press in 1960.
Temperature is the only property which differentiates thermodymanics from other sciences. The first law is nothing but conservation of energy. Second law indicates the probability of happening of a cycle or process and indicates the direction of energy flow. Further, second law distinguishes the quality of energy. The third law gives the interpretation of a system while marching towards absolute zero temperature. Application of both the first and second law to a process give rise to some properties (internal energy and enthalpy from first law, entropy from the second law).
About the first law of thermodynamics
http://nptel.ac.in/courses/112103016/module1/images/onsagar.jpgSimple and most direct statement of the first law of thermodynamics is that energy is conserved. Energy can neither be created nor destroyed.
For a cyclic process involving heat and work First Law of Thermodynamics is given
by For a thermodynamic process with heat and work interaction the expression is given
by
where E, Q, and W are the energy stored in the system, heat and work interacting at the system boundary, respectively.
First Law gives a quantative measure of energy
It can not distinguish the different types of energy
It fails to indicate the direction of flow of energy
It can not indicate whether the process is cyclic or not.
First law applied to a process gives rise to a property (internal energy, U for a closed system and
enthalpy, h for a open system)
From the definition of first law, it is obvious that energy is always conserved. First law efficiency indicates conversion of energy in a cycle as given by Eq. (1.1)
1.1
The net output can be work or heat depending upon the application. In case of a heat engine, the net output is in the form of work whereas in case of a solar water heating system it can be the useful heat. The difference between the net output and input is the energy unused. It is always the need from the first law of the thermodynamics to reduce the energy waste from the system. First law efficiency can be improved to limited extent with heavy insulation to the system.
About the second law of thermodynamics
Second law of thermodynamics provides the criterion as to the probability of various processes. Sadi Carnot established this law in 1824. But it got importance from last two decades (1980 onwards) in view of conservation of energy. First law dictates that energy in a system is always conserved. There is no way to conserve energy by use of first law of thermodynamics. However, it is the exergy, which is a consequence of the second law that is never conserved. Unlike energy, exergy always decreases. So minimization of exergy loss is nothing but the principle of energy conservation. The net exergy output to the actual exergy input to the system is the second law efficiency.
1.2
where A = exergy
To improve a system, we always try to improve the second law efficiency. Exergy is a tool to identify the loss of energy. In complicated systems, exergy loss at different locations can give an estimation of such losses and measures can be taken up to reduce the same. Exergy analysis will be discussed in lectures 7-12 of Module 1.
Apart from the energy conservation, two important implications of second law of thermodynamics are:
Directional flow of energy
A spontaneous process occurs only in one direction. Heat always flows from a body at high temperature to a body at lower temperature, water always flows downhill, time always flows in forward direction. Reverse of these processes never happen spontaneously. The spontaneous process is due to a finite driving potential called FORCE or CAUSE. The outcome or result is called the FLUX, CURRENT or EFFECT.
Table: 1.5 Forces and fluxes
Sl No Force (Cause) Conjugate fluxes (Effect)
1 Temperature Gradient Heat Transfer
2 Concentration Gradient Mass Transfer
3 Electric Potential Gradient Flow of Electric Current
TRANSFER PROCESSES CAN NEVER SPONTANEOUSLY OCCUR FROM A LOWER
TO A HIGHER POTENTIAL. SECOND LAW OF THERMODYNAMICS PUTS
LIMITATION ON DIRECTION OF PROCESS OCCURANCE.
Qualitative measurement of energy
The second law distinguishes energy in two different forms (1) High grade energy and (2) Low grade energy. High grade energy is an orderly form of energy whereas low grade energy is in random form. Some high and low grade energy forms are given in Table 1.6.
Table: 1.6 Different form of energy
High Grade Energy Low Grade Energy
Electrical energy Thermal Energy from Fossil Fuel
Wind energy Nuclear Fission
Hydropower Nuclear Fusion
Kinetic energy of a water
jet
Waste Heat from Thermal Power Plant
Mechanical work Solar Thermal Energy
Tidal power Geothermal Energy
Quality of energy can be ascertained by applying the second law of thermodynamics to a process or a system.
Quiz 2
1. What is the force associated with heat transfer?
2. What is the major difference between a quasi-static and reversible process?
3. How do you distinguish the efficiency between the first and second law of thermodynamics?
4. Give examples of high and low grade energy.
Review of TD II
Thermodynamic Properties
Both the first and second law of thermodynamics while applied to processes give rise to some thermodynamic properties. First law of thermodynamics while applied to a process for a closed
system give rise to a property called internal energy U.
If Q is the amount of heat transferred to the system and W is the amount of work transferred
from the system during the process (Fig. 1.9(a)-(b)), the net energy transfer (Q - W) will be stored in the system.
Fig. 1.9 (a)-(b) Thermodynamic system and surrounding
As both the heat and work are energies in transit, they can not be stored. Hence, the quantity (Q -
W) is not either heat or work, but it is entirely a new form of energy, known as internal energy or simply the energy of the system.
Q - 1.3
or,
1.4
If there are more such energy transfer quantities involved in the process, as shown in Fig.1.9(b), the first law gives,
Q2 + Q3 - Q1 2 + W3 - W1 - W4) 1.5
The following explanation can be given to justify internal energy as a thermodynamic property:
Consider a system, which changes from its state 1 to state 2 by following the path A, and returns
from state 2 to state1 by following the path B (Fig. 1.10). So the system undergoes a cycle.
Fig. 1.10 Thermodynamic processes
For path A:
QA A + WA 1.6
For path B:
QB B + WB 1.7
The processes A and B, together constitute a cycle for which
cycle cycle 1.8
or,
WA + WB = QA + QB 1.9
or,
QA - WA = -(QB - WB) 1.10
From Eqs. (1.6) and (1.7), it yields
A = - B 1.11
similarly, had the system returned from state 2 to state 1 by following the path C instead of path B,
A = - C 1.12
From Eqs. (1.11) and (1.12),
B C 1.13
Thus, it is seen that the change in energy between two states of a system is the same, whatever path the system follow in undergoing the change of state. Therefore, energy has a definite value for energy state of the system. Hence it is a point function and is a property of the system. The energy
is an extensive property. The specific energy (J/kg) is an intensive property. The cyclic
Different Forms of Stored Energy
Total energy stored in a system is represented by E. Basically there are two models in which energy may be stored in a system
1. Macroscopic energy mode
2. Microscopic energy mode
The macroscopic energy mode includes the macroscopic kinetic energy and potential energy of a
system. Let us consider a fluid element of mass m having the center of mass velocity (Fig. 1.11). The macroscopic kinetic energy EK of the fluid element by virtue of its motion is given by
1.14
Fig. 1.11 Macroscopic and Microscopic energy in a system
If the elevation energy of the fluid element from an arbitrary datum is z, then the macrosscopic potential energy Ep by virtue of its position is given by
Ep = mgz 1.25
The microscopic energy mode refers to the energy stored in the molecules and atomic structure of the system, which is called the molecular internal energy or simply internal energy U. Matter is composed of molecules. Molecules are in random thermal motion with an average velocity v, constantly colliding with one another and with the walls. Due to a collision, the molecules may subjected to rotation as well as vibration. They can have translational kinetic energy, rotational kinetic energy, vibrational energy, electronic energy, chemical energy and nuclear energy (Fig 1.12).
Fig. 1.12 Various components of internal energy stored in a molecule (a) translational kinetic energy, (b) rotational kinetic energy, (c) nuclear energy, (d) vibrational energy and (e) electron spin
If represents the energy of one molecule, then
1.16
If N is the number of molecules in the system, then the total internal energy is
1.17
For an ideal gas there are no intermolecular forces of attraction and repulsion, and the internal energy depends only on temperature. Thus, for an ideal gas
1.18
Other forms of energy which can also be possessed by a system are magnetic energy, electrical
energy and surface tension energy. In absence of these forms, the total energy E of a system is given by
E = (EK + EP) + U 1.19
where EK + EP Macro energy and U = Micro energy.
In absence of motion and gravity,
EK = 0, EP = 0 1.20
Hence,
E = U 1.21
Hence, Eq. (1.4) becomes
1.22
For open system, the enthalpy H = U + pV is a property. The specific enthalpy h = u + pv is an intensive property.
Relationship between the first and second law of thermodynamics
Consider a power plant converting a fraction of available energy A or W max to useful work W (Fig. 1.13)
Fig. 1.13 A thermodynamic cycle (a) sink at temperature T2 and (b) sink at ambient condition T0
For the desired output of W,
Amin = W 1.23
A = Wmax 1.24
I = Irreversibility = Wmax - W 1.25
1.26
Now,
1.27
where carnot is the Carnot efficiency of an engine operating between two fixed temperatures.
Efficiency of any thermodynamic cycle cannot be more than the Carnot cycle efficiency.
1.28
Hence,
1.29
Thermodynamics Processes
Second law of thermodynamics enables us to divide all the processes into two classes:
1. Reversible or ideal process
2. Irreversible or natural process
A process is reversible if, after the process has been completed, means can be found to restore the system and its elements of its and the surroundings to their respective initial states. Some of the reversible processes are:
Frictionless motion of solids
Slow frictionless adiabatic expansion/compression of a gas
Slow isothermal compression/expansion of a gas
Electrolysis of water
Flow of electric current through inductors and capacitors
Any process, which is not reversible, is irreversible. All natural (spontaneous) processes are irreversible. Some of the irreversible processes are:
Irreversibility due to finite temperature differences between combustion gas and working fluids in a
boiler
Internal irreversibilities in pipe fittings, bends etc- viscous dissipation
Unrestrained expansion-Joule Thomson effect
Mixing of two different fluid streams
Magnetization of materials exhibiting hysteresis
Joulean heat production due to flow of current through a electric conductor
Various processes can be expressed mathematically. The first as well as the second law can be combined together to analysis a thermodynamic system/process. Such mathematical relationships are elaborately discussed in Module 1 - Lecture 6.
Quiz 3
1. Under what condition the relationship dQ = dU + dW is valid?
2. How are the first and second law efficiencies correlated?
3. Give 3 examples of irreversible process
4. Give 5 examples of reversible process
Entropy
Second law of thermodynamics while applied to a process give rise to a property, called entropy.
Absolute entropy (S) or entropy is a measure of energy dispersion in a system.
Following are basic features of entropy:
Entropy transfer is associated with heat transfer (direction of flow of entropy is same as that of heat
flow)
Entropy is proportional to mass flow in an open system
Entropy does not transfer with work
Besides flow, entropy may be generated in a system (Table 1.7)
Table 1.7 Entropy linked with energy
Heat Work Entropy transfers
Work Work No entropy transfer
Work Heat No entropy transfers Entropy generates
When work is dissipated, energy dispersion decreases. As a result entropy increases. Consideration
of entropy affords a more convenient and straightforward means of establishing the reversibility and
irreversibility of processes.
Entropy in an isolated system
No interaction of the system with surrounding.
Fig. 1.14 An isolated system
http://nptel.ac.in/courses/112103016/module1/images/entropy.jpgFor a given system,
- 1.30
For isolated system
- 1.31
dE = 0 1.32
Again,
T dS = dQ = 0 1.33
1.34
Isolated system does not undergo any energy interaction with the surroundings (dQ = 0).
dSiso 1.35
For a reversible process,
dSiso = 0 1.36
or,
S = Constant 1.37
For irreversible or natural process
dSiso > 0 1.38
Hence, entropy of an isolated system can never decrease. It always increases and remains constant only when the process is reversible. This is known as principle of entropy increase or simplyentropy principle .
For isolated system,
dSuniv 1.39
or,
dSsys + dSsur 1.40
Fig. 1.15 Equilibrium state of an isolated system
Closed system
Consider a fixed mass of a substance in a closed system.
Entropy of any closed system can increase in two different ways
1. By heat interaction
2. By internal irreversibility or dissipative effects in which work or kinetic energy is dissipated into
internal energy increase.
Fig. 1.16 A Closed System
Let, be an infinitesimal amount of heat transfer rate from surroundings to the closed system at
the location on the boundary where temperature is T . Let, be the rate of work transfer due to this heat transfer to the system. Now, rate of entropy change of the surroundings is
1.41
If heat transfer does occur in many locations of the boundary we can write the entropy transfer due
to heat transfer as for the accomplishing the process within the system from state 1 to state 2. Entropy change of the system is
1.42
Taking the system and surroundings together, entropy change of the universe is
1.43
or,
1.44
For a reversible process
1.45
For an irreversible process
1.46
By definition
1.47
Thus for a closed system
1.48
or,
1.49
or,
1.50
where S2 - S1 = entropy change between two equilibrium states
= entropy transfer
Sgen = entropy production between the two equilibrium states
Sgen 1.51
Second Law of thermodynamics states that, in general any thermodynamics process is accompanied by entropy generation.
Open system
In an open system, there are transfer of three quantities:
1. Mass of fluid
2. Energy (heat and wok)
3. Entropy
Fig. 1.17 An open system
For mass flow, continuity equation is
1.52
where = rate of mass accumulation in the system
The energy equation gives
1.53
where = rate of energy accumulation in the system
Now, entropy principle gives
1.54
where
= rate of entropy transfer into the system (control volume) due to mass mi
= rate of entropy transfer out of the system (control volume) due to mass me
= net rate of entropy transfer
= rate of change of entropy of control volume
Difference of entropy change of control volume and net rate of entropy transfer is the entropy generated within the control volume due to irreversibility.
1.55
By second law of thermodynamics
Sgen 1.56
From Eq. (1.54),
1.57
From Eq. (1.53),
1.58
1.59
1.60
or,
1.61
Eq. (1.61) is the expression for the rate of shaft work in the system.
At steady state,
1.62
Hence, energy equation becomes
1.63
Entropy equation is
1.64
Gouy-Stodola Theorem
Expression for the lost work can be obtained from the Gouy-Stodola theorem. According to this theorem rate of irreversibility or energy degradation is
1.65
1.66
1.67
1.68
On simplification,
=
1.69
Hence, lost work is proportional to the entropy generation rate.
Quiz 4
1. Write down the entropy balance equation for (a) closed system and (b) open system under steady
state condition.
2. State the principle of entropy increase
3. How does the entropy of a close system increase?
4. What is the meaning of lost work?
Thermodynamic Relations I
Mathematical Theorems
The equation of state provides the reationship between theromodynamic properties for ideal gas as well as for incompressible substances. However, simple equation of state is not enough to describe the thermodynamic properties for systems with real and complex materials. Hence, more mathematical equations are required to represent the realistic processes. Amongst the various thermodynamic properties, p, v, T are directly measurable whereas u, h, s are not directly measurable. This necessitates for developing thermodynamic relations apart from the equation of states.
The following three theorems are the base for developing the various thermodynamic relations.
Theorem-1 If a relation exists among the variables x, y, and z, then z may be expressed as a function of x and y .
or,
1.70
If,
1.71
and
1.72
Then,
dz = M dx + N dy 1.73
where M and N are functions of x and y. Differentiating M partially with respect to y and N with respect to x,
1.74
And,
1.75
Hence,
1.76
This is the condition of an exact (perfect) differential. If x and y are two independent thermodynamic properties, then Z (or z) will be also a thermodynamic property provided it satisfy Eq. (1.76). For example, let us consider the following relation,
dz = p dv + v2 dp 1.77
Here, M = p and N = v2
Now, and which does not satisfy Eq. (5.76). Hence Z is not a thermodynamic property.
Let us consider another relation,
z = p dv + v dp 1.78
With similar reasoning it can be easily established that Z is a thermodynamic property in this case.
With more than 3 variables, say x1, x2, x3 and Z,
Z=(x1, x2, x3) 1.79
1.80
For dz to be an exact differential,
1.81
For k number of variables
1.82
or,
1.83
Theorem 2 If a quantity f is a function of x, y, and z, and a relation exists amongst x, y and z, then fis a function of any two of x, y, and z.
1.84
or,
x = x(y, z) 1.85
1.86
1.87
1.88
1.89
From Eqs. (1.88) and (1.89), we have
1.90
or,
1.91
1.92
From Eqs. (1.91) and (1.92),
1.93
Eq. (1.93) can be rewritten as
1.94
Theorem 3 Among the variables x, y, and Z, any one variable may be considered as a function of the other two. Thus,
x = x(y, z) 1.95
1.96
Similarly,
z = z(x, y) 1.97
1.98
Hence,
1.99
or,
1.100
or,
1.101
Hence,
1.102
or,
1.103
Maxwell Relations
A pure substance existing in a single phase has only two independent variables. Out of the eight quantities p,V,T,S,U,H,F, and G any one may be expressed as a function of and other two quantities.
For a pure substance undergoing an infinitesimal reversible process
dU = TdS -pdV 1.104
dH = dU +pdV + Vdp = TdS + Vdp 1.105
dF = dU - TdS - SdT = -pdV - SdT 1.106
dG = dH - TdS - SdT = Vdp - SdT 1.107
Since, U, H, F and G are thermodynamic properties and exact differentials of the type dz = Mdx + Ndy (Eq. (1.73)), then from Eq. (1.76),
Applying this to the four equations (Eqs. (1.104-1.107))
1.108
1.109
1.110
1.111
Equations (1.108-1.111) are known as Maxwell relations.
The Maxwell relations need not be remembered. It can be easily found out from the thermodynamic mnemonic diagram. Construct a square with two diagonals as shown in Fig. 1.18.
Fig. 1.18 Thermodynamic mnemonic diagram
Mark the positions at the middle of the sides as well as at the corners. Write down the variables G, P, H, S, U, V, F and T starting from the middle of left hand side and in anti clockwise direction. The relations (Table 1.8) for du, dh, dF and dG can be easily memorized by using the phrase
G reat P hysicists H ave S tudied U nder V ery F amous T eachers
(G,P,H,S,U,V,F,T) by considering this thermodynamic mnemonic diagram. Diagonals are then drawn pointing away from the two bottom corners. Now, if an expression for dG is required (that is located at the middle of the line connecting the points T and p ), we first form the differentials of dT and dp, and then link the two with their conjugates as illustrated below:
dG = -S ( conjugate of T with the minus sign due to the diagonal pointing towards T ) x dT +V (that is conjugate of P with the plus sign due to the diagonal pointing away from P) x dp
Table 1.8 Maxwell relations
Differential Conjugate Maxwell Relation Remarks
u du = Tds - pdv T, -p
u = u(s, v)
h dh = Tds + pdv T, v
h = h(s,p)
f -sdT - pdv -s, -p
g dg = -sdT + vdp -s, v
g = g(T, p)
Quiz 5
1. Write down the Maxwell relations.
2. Can you interpret the need of Maxwell relations?
3. Check whether Z is a property in the following expressions:
a. dZ = pdv + vdp
b. dZ = p2dv + v2dp
c. dZ = p2dv - vdp
d.
Thermodynamic Relations II
FIRST T-dS RELATION
Consider,
S = S(T, V) 1.112
http://nptel.ac.in/courses/112103016/module1/images/74.50.png1.113
or,
1.114
Now, for a reversiable process.
dQ = dU + pdV = TdS 1.115
and,
1.116
where Cv is the heat capacity at control volume ( kJ/K).
Hence,
1.117
Therefore, applying the third Maxwell relations,
1.118
Equation (1.118) is known as First T - dS relation.
SECOND T - dS RELATION
Again, let us consider
S = S(T, p) 1.119
1.120
Now,
http://nptel.ac.in/courses/112103016/module1/images/81.50.png1.121
where, Cp is the heat capacity at constant pressure (kJ/K).
Therefore, applying Maxwell's fourth relation,
1.122
This is known as 2nd T - dS relation.
THIRD T - dS RELATION
Let,
S = S(V, p) 1.123
Hence,
1.124
Now,
1.125
1.126
or,
1.127
Again,
1.128
1.129
http://nptel.ac.in/courses/112103016/module1/images/43.30.pnghttp://nptel.ac.in/courses/112103016/module1/images/95.50.pngAgain,
1.130
1.131
or,
1.132
But,
1.133
Hence, Eq. (1.132) can be written as
1.134
Hence we obtain,
1.135
This is known as the third T - dS relation.
Difference in heat capacities
From the first and second T - dS relations we can write,
1.136
or,
1.137
or,
http://nptel.ac.in/courses/112103016/module1/images/109.50.png1.138
Again,
1.139
or,
1.140
Equating the coefficient of dV
1.141
1.142
Similarly,
1.143
From Theorem 3,
1.144
Hence,
1.145
Following interpretations can be made from Eq. (1.145)
Since is always positive and for any substance is always negative, the difference of
specific heats (CP - CV) is always positive.
As T 0, CP CV. In other words, CP = CV at absolute zero temperature.
When then CP = CV
For an ideal gas, pV = mRT
Hence, CP - CV = mR
or, CP - CV = R
Some definitions
In a single phase region, where pressure and temperature are independent, we can think of the volume as being a function of pressure and temperature.
V = V (T, p) 1.146
or,
v = v(T, p) 1.147
Applying the chain rule of the calculus
1.148
The derivative represents the slope of a line of constant pressure on V - T a plane. A similar interpretation can be given for the second derivative. These derivatives are themselves intensive thermodynamic properties, since they have definite values at any fixed thermodynamic state. The first represents the sensitivity of the specific volume changes in temperature at constant pressure, and the second is a measure of the change in specific volume associated with a change in pressure at constant temperature. Isothermal compressibility, isentropic compressibility and volume expansivity are defined as
Isothermal compressibility:
Volume expansivity:
-efficient of linear
T T, Eq. (1.148) can be written as
- kT vdp 1.149
T are sometimes slowly varying functions of T and P. Another term in use is the isothermal compressibility defined as
1.150
1.151
Based on the definitions of isothermal compressibility and volume expansivity, the specific heat differences can be written as
Ratio of heat capacities
At constant entropy S , the two T - dS relations become
1.152
1.153
Dividing Eq. (1.152) by Eq. (1.153) and using Theorem 3
Energy Equation
For a system undergoing an infinitesimal reversible process between two equilibrium states, the change in internal energy is
http://nptel.ac.in/courses/112103016/module1/images/147.60.pngdU = Tds - pdV 1.155
Substituting the first T - dS relation,
1.156
1.157
If U = U(T, V), then
1.158
Comparing Eq. (1.157) with Eq. (1.158),
1.159
Eq. (1.159) is known as the energy equation.
Energy Equation
For a system undergoing an infinitesimal reversible process between two equilibrium states, the change in internal energy is
dU = Tds - pdV 1.155
Substituting the first T - dS relation,
1.156
1.157
http://nptel.ac.in/courses/112103016/module1/images/147.60.pngIf U = U(T, V), then
1.158
Comparing Eq. (1.157) with Eq. (1.158),
1.159
Eq. (1.159) is known as the energy equation.
Application of energy equation to an ideal gas
For an ideal gas in a closed system, equation of state is
1.160
1.161
1.162
U does not change with V at constant temperature. Further, applying Theorem 2,
1.163
Hence,
1.164
Since,
http://nptel.ac.in/courses/112103016/module1/images/148.60.pnghttp://nptel.ac.in/courses/112103016/module1/lec5/3.html1.165
From Eqs.(1.162) and Eq. (1.165) it is clear that U is neither a function of V nor a function of P at constant temperature. Only possibility is that internal energy is a function of temperature only.
For an open system
dH = Tds + Vdp 1.166
From second T - dS relation,
1.167
Now,
1.168
Let, H = H(T, p)
Hence,
1.169
Comparing Eq. (1.167) and Eq. (1.168),
1.170
Now,
1.171
For an ideal gas
1.172
http://nptel.ac.in/courses/112103016/module1/images/81.50.pngHence,
1.173
H does not change with p while temperature remaining unchanged.
From, we can infer that
H does not change with V while temperature remaining unchanged.
Application of energy equation for thermal radiation in equilibrium with the enclosure walls
Let, u = energy density
p = radiation pressure exerted by a black body in an enclosure ( from electro magnetic theory) =
Blackbody raditation is thus specified by the pressure, volume and temperature of radiation.
U = uV 1.174
1.175
1.176
1.177
Energy Equation is
1.178
Substituting in energy equation
1.179
1.180
On integration,
u = bT4 1.181
where b is a constant.
Eq.(1.181) is known as Stefan-Boltzmann equation.
Since,
U = uV = VbT4 1.182
1.183
And,
1.184
From first T - dS relation,
1.185
For a reversible isothermal change of volume, the heat is to be supplied reversibly to keep temperature constant [TdS = Q].
1.186
For a reversible adiabatic change of volume,
1.187
[Q = 0]
or,
1.188
or,
VT3 = Constant 1.189
If the temperature is one half of the original temperature, the volume of the blackbody radiation is to be increased adiabatically 8 times its original volume so that the radiation remains in equilibrium with the matter at that temperature.
Quiz 6
1. Write down the first and second T-dS relations.
2. Under what condition(s) specific heat at constant pressure and constant volume become same.
3. Why Cp > Cv ?
4. Express Cp - Cv in terms of volume expansivity and isothermal compressibility.
5. State the applications of energy equation.
Application of first law of thermodynamics for open systems
Throttling Devices
A throttling device is the generic name of any device or process that simply dissipates pressure energy by irreversibly converting it into thermal energy. Unlike nozzles and diffusers, throttling devices provide no form of useful energy recovery. They merely convert pressure energy into thermal energy through dissipative viscous flow processes. A throttle need not have same inlet and outlet flow velocities, and, therefore, it may have a significant specific kinetic energy changes across it. Throttles may or may not be insulated, but they are usually such small devices and have such high flow rates that the residence time of the fluid in them is too short for significant heat transport of energy to occur. Consequently, a throttling device is commonly taken to be adiabatic regardless of whether it is actually insulated or not.
Joule Kelvin (Joule-Thompson) Effect
The throttling phenomenon, also known as Joule Kelvin /Joule Thompson effect, may be demonstrated by considering a steady-state, steady flow (SSSF) process across a restriction, with a resulting drop in pressure. The restriction can be a porous plug, an orifice plate, a butterfly valve, any type of flow or pressure control valve. Even a geometry like sudden contraction or sudden expansion may give the desired effect [Fig. 1.19].
Fig: 1.19 (a) Orifice plate, (b) Porous plug, (Butterfly/throttle Valve ), (d) Any type of
flow or pressure control valve (e) sudden expansion, (f) Sudden contraction
The pressure and temperature of the fluid upstream and down stream of the restriction are measured with suitable manometers and thermometers [Fig. 1.20].
Fig. 1.20 Joule Thompson effect
Let pi and Ti be the arbitrarily chosen pressure and temperature before throttling and let us assume that they are constant. By operating the restriction (say, a valve) the fluid is throttled successively to different pressures and temperatures p , T ; p , T ; p , T and so on. These are then plotted on the T -p plane [Fig. 1.3].
Fig. 1.21 Isenthalpic states of a gas
All the points on this curve represent equilibrium states of some constant mass of the fluid. The curve passing through all these points is an isenthalpic curve or isenthalpe.
The initial pressure and temperature of the fluid (pi, Ti) are then set to new values and the fluid is throttled by controlling the restriction. Thus by throttling to different states we can have a family of isenthalpes for the fluid [Fig. 1.22]. The curve passing through the maxima of these isenthalpes is called the inversion curve.
Fig. 1.22 Inversion and saturation curves on T-s plot
The numerical value of the slope of an is enthalpe on the T - p diagram at any point is called the Joule Thompson coefficient j.
1.190
Locus of all points at which j is zero is the inversion curve. The region inside the inversion curve where j is negative is called the heating region and where j is positive is called the cooling region.
The difference in enthalpy between two neighboring equilibrium states is
dh = Tds + vdp 1.191
and the second T - dS relation (per unit mass)
1.192
1.193
The second term in the above equation stands only for a real gas, because for an ideal gas dh = CpdT
1.194
For an ideal gas:
pv = RT 1.195
1.196
There is no change in temperature when an ideal gas is made to undergo a Joule-Thompson expansion.
For an incompressible fluid:
1.197
Thus an incompressible fluid is heated while undergoing throttling process.
Application of Joule-Thompson effect:
1. Vapor compression refrigeration system
2. Production of dry ice or solid CO2 for low temperature refrigeration
Example:
Joule-Kelvin coefficient j is a measure of the temperature change during a throttling process. A similar measure of temperature change produced by an isentropic change of pressure is provided by the coefficient s where
Prove that:
Solution: The Joule Kelvin coefficient j is given by
Maxwell's Relations
Quiz 7
1. Give 2 (two) examples of Joule-Kelvin effect.
2. What is Joule-Kelvin coefficient?
3. What happens to the change of temperature for an ideal gas subjected to throttling process?
4. What happens to an incompressible fluid undergoing Joule-Kelvin expansion?
5. What is the numerical value of Joule-Kelvin coefficient at the point of inversion?
Exergy is the tool, which indicates how far the system departs from equilibrium state. The concept of exergy was put forward by Gibbs in 1878. It was further developed by Rant in 1957.
Energy Analysis
Quality of energy
Quantative evaluation of energy in a cycle or in a process can be done using the first law of thermodynamics. The direction of flow of heat or work is known from the second law of thermodynamics. However, it is equally important to assign the quality to the energy. Energy can be broadly classified into high grade and low grade energy. High grade form of energy are highly organised in nature and conversion of such energy to some other high grade form (W W) is not
dictated by the second law of thermodynamics. Conversion of high grade energy to low grade energy is not desirable. However, there may be some conversion to low grade energy as work is converted into other useful form. This is because of dissipation of heat due to friction (example: mechanical work Electricity, some losses are there due to the friction in bearing of machineries).
Thus both the first and the second law of thermodynamics are to be considered for analysis.Low grade energy such as heat due to combustion, fission, fusion reactions as well as internal energies are highly random in nature. Conversion of such form of energy into high grade energy (Q W) is of
interest. This is due to the high quality of organised form of energy obtained from low quality energy. Second law of thermodynamics dictates that conversion of 100% heat into work is never possible. That part of low grade energy which is available for conversion is termed as available energy, availability or exergy. The part, which according to the second law of thermodynamics, must be rejected is known as unavailable energy. Exergy analysis helps in finding the following:
It can be used to determine the type, location and magnitude of energy losses in a system
It can be used to find means to reduce losses to make the energy system more efficient
At this point, it is worth mentioning that the environment plays an important role in evaluating the exergy (composite property).
Importance of exergy analysis
Let us take an example. In case of coal fired power plant, the first law indicates that the condenser greatly effects the power plant efficiency as large amount of heat is transferred to the cooling water without providing any clue on the real usefulness of this relatively low temperature fluid. Also, energy balances do not provide information about the internal losses such as throttling valve and heat exchanger. Second law or exergy balance, however indicates that there is hardly 1% exergy loss in the condenser with more than 60% in the boiler. The contribution in the boiler exergy loss accounts for irreversibilities associated with combustion and finite temperature differences. Hence, analysis of exergy plays a deterministic role in identification of processes and rectifying the components.
Available energy referred to a cycle
Consider a cyclic heat engine as shown in Fig. 1.23. The maximum work output obtainable from this engine is the available energy (AE). The minimum energy that has to be rejected to the sink (as per 2nd law of thermodynamics) is called the unavailable energy (UE).
Fig. 1.23 Available and unavailable energy in a cycle
From energy balance
Q1 = AE + UE 1.198
Again,
Wmax = AE = Q1 - UE 1.199
For the given source and sink temperatures T1 & T2 respectively
1.200
If T1 rev will increase with decrease of T2. The lowest practical temperature of heat rejection is the temperature of the surroundings, T0.
1.201
And,
1.202
Availability in a finite process
Let x-y be a finite process in which heat is supplied reversibly to a heat engine. Taking an elementary cycle, let dQ1 be the heat received by the engine reversibly at temperature T1. Then
1.203
For the whole process x-y,
Fig. 1.24 Unavailable energy by 2 nd law of thermodynamics
1.204
Hence, maximum work obtainable is
UE = Qx - y - Wmax 1.205
UE = T0(sy - sx) 1.206
Available energy from a finite energy source
Consider hot gas of mass mg at temperature T when the temperature of environment is T0 (Fig 1.25). Let the gas be cooled at constant pressure to T0 from state 1 and the heat given up by the gas Q1 is utilized in heating up reversibly a working fluid of mass mwf from state 3 to 1 along the same path so that the temperature difference between the gas and working fluid at any instant is zero and hence entropy increase of the universe is also zero.
The working fluid expands reversibly and adiabatically in an engine or turbine from state 1 to 2 doing work WE and then rejects heat Q2 reversibly and isothermally to return to the initial state 3 to complete the heat engine cycle.
Here,
Q1 = mgCpg(T - T0) = mwf Cpwf (T - T0) = Area 1-4-5-3-1 1.207
Fig. 1.25 Available and unavailable energy in finite processes
mg Cpg = mwf Cpwf 1.208
Now
( negative as T0 < T)
1.209
1.210
univ gas + wf = 0 1.211
1.212
Available energy
1.213
Hence, exergy of a gas of mass mg at temperature T is
1.214
Demonstration of quality of energy based on exergy
Consider a hot gas of mass m flowing through a pipeline (Fig. 1.26). Due to the heat loss to the surroundings, the temperature of the gas decreases continuously from the inlet state 'a' to the exit state 'b'. Thus, the process is irreversible. However, for sake of simplicity, let us consider the process to be reversible and isobaric between the inlet and outlet. For an infinitesimal reversible process at constant pressure,
Fig. 1.26 Energy quality degrades along the flow direction
1.215
or,
1.216
Thus the slope depends on temperature T. With increase in T, increases and vice versa.
Let, Q unit of heat lost to the surroundings as temperature of gas decreases from to , T1being average of the two temperatures.
Heat loss at section 1-1
1.217
Exergy lost with this heat loss at temperature T1 is
W1 = Q - T0 1 = T1 1 - T0 1 = (T1 - T0 1 1.218
At section 2-2, let heat loss be same as in section 1-1, Q(T2 < T1)
1.219
Exergy loss due to this heat loss at temperature T2 is
W2 = Q - T0 2 = (T2 - T0 2 1.220
Considering the exergy loss at temperatures T1 and T2
W1 = Q - T0 1 = T1 1 - T0 1 = (T1 - T0 1 1.221
W2 = Q - T0 2 = (T2 - T0 2 1.222
As T1 > T2 1 2
W1 > W2 1.223
Quiz 8
1. What do you mean by exergy?
2. Give the expression for the available energy for a thermodynamic cycle.
3. What is the availability in a finite energy source?
4. How does exergy represent the quality of energy?
5. Give 2 (two) practical applications of exergy analysis.
Example of Quality Degradation
Interpretations:
For the same heat loss at two different temperatures, exergy loss is more with higher temperature. For instance, exergy loss is more at 1000 K than that at 300 K for 1 kJ of heat loss.
The more the temperature, the more is the quality of energy. For example, quality of energy of a gas at 1000 K is superior to that at 400 K, since the gas at 1000 K has the capacity of doing more work than the gas at 400 K.
Some more examples of quality degradation
(a) Throttle Process
For frictionless, steady flow process of an ideal gas between sections (1)-(1) and (2)-(2) in a pipe (Fig. 1.27),
Fig. 1.27 Throttled flow through a pipe
h1 = h2 1.224
Again,
Tds = dh - vdp 1.225
and pv = RT
1.226
or,
1.227
or,
1.228
1.229
1.230
Equation (1.230) is the expression for the irreversibility or lost work in the system comprising of the sections (1)-(1) and (2)-(2). It is obvious from the expression that quality degrades logarithmically with pressure drop between the sections as well as with the mass flow rate for an initial pressure p1and surrounding temperature T0.
(B) Quality degradation for a flow with friction
Consider a steady and adiabatic flow of an ideal gas through the segment of a pipe. Applying the first law of thermodynamics between sections (1)-(1) and (2)-(2) (Fig. 1.28)
Fig. 1.28 Adiabatic steady flow of an ideal gas in a straight pipe
h1 = h2 1.231
Applying T-dS relationship
Tds = dh - vdp 1.232
or,
1.233
or,
1.234
1.235
where since and higher order terms are neglected. Hence, lost work or irreversibility is
1.236
Hence decrease in exergy is proportional to the pressure drop as well as mass flow rate.
(c) Mixing of two fluid streams
Two fluid streams 1 and 2 of an incompressible fluid or ideal gas mixing adiabatically at constant pressure as shown in Fig. 1.29.
Fig. 1.29 Mixing of two fluid streams
Mass balance:
m1 + m2 = m3 = m(say) 1.237
Let,
1.238
By first law of thermodynamics,
m1h1 + m2h2 = (m1 + m2)h3 1.239
or,
xh1 + (1 - x)h2 = h3 1.240
Since,
h = h(T) 1.241
From Eq. (1.240)
xT1 + (1-x)T2 = T3 1.242
1.243
where
1.244
By second law of thermodynamics,
1.245
1.246
1.247
1.248
or,
1.249
Let, Ns Entropy generation number =
Now, substituting from Eq. (1.243) in Eq. (1.249), we get
1.250
1.251
For, x = 1, Ns = 0 m2 = 0, the system is with single stream.
For, , Ns = 0, temperature of both the streams are same.
Rate of exergy loss due to mixing is,
Wlost = I = T0Sgen 1.252
1.253
Quiz 9
1. Derive the expression for quality degradation of energy in a throttle process.
2. How does the exergy decrease in a flow through a pipe with friction differ from that of a throttle
process?
3. What is entropy generation number? What is its physical significance?
4. How does quality degradation of energy take place in a steam generator?
5. Which component of a thermal power is responsible for maximum quality degration?
Energy Balance Equation and Tools
Exergy balance for closed and open system
Exergy balance for the closed and open systems can be used to determine the locations, types and magnitude of losses of potential energy resources (fuels) and ways can be found to reduce such losses for making the energy system more efficient.
Closed system
For a closed system (Fig. 1.30), exergy or availability transfer occurs through heat and work interactions. No mass is transferred across the system boundary.
Fig. 1.30 A closed system
Ist Law of thermodynamics:
1.254
2nd Law of thermodynamics:
1.255
or,
1.256
Subtracting Eq. (1.256) from Eq. (1.254), we get,
1.257
Now, Defining availability function as
A = E + p0V- T0S 1.258
1.259
1.260
In the form of rate equation,
1.261
where, Tj instantaneous temperature at the boundary
rate of change of system volume.
For an isolated system,
A2 - A1 -I 1.262
Since, I > 0, the only processes allowed by the second law of thermodynamics are those for which the exergy of the isolated system decreases. In other words,
The exergy of an isolated system can never increase.
(Counterpart of entropy principle, which states that entropy of an isolated system can never decrease.)
Open system: Exergy balance for a steady flow system
Fig. 1.31 An open system
Consider an open system (Fig. 1.31). Applying the first law of thermodynamics:
1.263
Second law of thermodynamics:
1.264
or,
1.265
From Eqs. (1.263) and (1.265),
1.266
or,
1.267
In the form of rate equation at steady state,
1.268
where,
1.269
For a single stream entering and leaving, the exergy balance is given by
1.270
Tools for exergy analysis
The following are the tools for exergy analysis
1. Value/Grassman diagram
2. Pinch point technology
In general, conversion of energy or production of chemical substances in a plant requires determination of all the exergy flows that are transferred between distinguished apparatuses or unit operations of the plant. The resulting exergy losses can provide useful information with regard to the overall performance of the plant. However, it is generally difficult to judge the thermodynamic losses without any reference. Evaluation of plant performance will usually require a comparison of the thermodynamic performance of specific apparatuses or unit operations with available data from previously built plants. Thus, exergy efficiencies have appeared to be more useful, for large plants or integrated plants. It bears no significance to components. Exergy efficiencies combined with exergy flow diagrams can provide the thermodynamic performance of even for a complex system.
Q - T Diagram
Consider the heat transfer between two fluids in a recuperative type of heat exchanger. Temperature versus heat flow(T - Q) between the two fluids has been presented in Fig. 1.32.
Fig. 1.32 Heat transfer process represented in a Q-T diagram
This diagram may be more informative if modified as follows:
Abscissa: indicates the heat transfer between the two streams
Ordinate: temperature T is replaced by the term . At , T = T0 the ordinate is zero and at T
exergy loss of heat transfer :
Fig. 1.33 Heat transfer process represented in a value diagram
If it is assumed that an infinitesimal small amount of heat dQ is transferred from the secondary flow, the flow that is cooled down in the heat exchanger, the resulting decrease in temperature dTs may be neglected. For the energy of this amount of heat can be written as
1.271
In the value diagram (Fig. 1.33) the area 1-3-4-6-1 equals the amount of heat dQ, whereas the area
1-2'-5'-6-1 equals the exergy of this amount of heat. The term indicates which part of the considered heat can be converted into work and can in principle be converted into work and can be seen as the exergy fraction of this amount of heat. The total exergy from the secondary flow van be determined by integrating Eq.(1.271) from the inlet temperature Ts,i to the outlet temperature Ts,out
1.272
This amount of exergy equals the whole area below the temperature curve in the value diagram. Within the heat exchanger the heat dQ is transferred to the primary flow. The exergy of the heat supplied to this flow is
1.273
In the value diagram, this exergy is represented by the area 1-2-5-6-1. The area 2-2'-5'-5-2 indicates the exergy lost due to temperature difference necessary to transfer heat from the secondary to the primary flow. The total exergy absorbed by the primary flow is
1.274
where subscript p indicates the primary flow. The absorbed exergy by the primary flow equals the area below the temperature curve for this flow in the value diagram. The total exergy loss due to heat transfer, Exs - Exp, is represented by the area between the two temperature curves.
Thus,the amount of exergy as well as the exergy loss can be easily presented with the help of value diagram.
Quiz 10
1. Write down the rate equation for exergy balance in a closed system.
2. Why does exergy always decreases*?
3. Derive the exergy balance equation for an open system under steady flow condition.
4. What are the tools for exergy analysis?
Pinch Point Technology
A method of using the overall thermal balance of a process
Theoretically predicts minimum energy consumption
Predicts the construction costs of the plant using a heat recovery system
First put into practical use in 1984, at Linhoff March Co.
Pinch point technology, or process integration, is the name given for a technique developed by Prof. Linnhof and co-workers (1978) at Leeds University, UK to optimize the heat recovery in large complex plants with several hot and cold streams of fluids. To illustrate the basic principle take a case of a plant with two hot and two cold streams, as shown in Table 1.9.
Table 1.9 Data for 4 (four) fluid streams
The hot streams can be combined into an equivalent composite stream as follows: From Table 1.9, it is clear that both stream 1 and 2 are having common temperature drop between 170C to 70C. For the common processes, we go for process integration (heat recovery) by considering composite thermal capacity.
The hot composite curve will consists of the following
1. Stream1 from 200C to170C with heat capacity 1.98 kW/K,
2. Stream (1+2) between temperature 170C to 70C, a combined stream of thermal capacity rate
(2.2+3.9) = 6.1 kW/K
3. Stream2 between temperature 70C to 50C, stream 1 with heat capacity rate 1.98 kW/K.
To plot the composite heating curve the calculations can be estimated as shown in Table 1.10.
Table 1.10 Composite Heating Curve
Similarly, the cold streams can be combined. The cold composite curve will consists of
1. Stream3 from 40C to 100C with heat capacity 2.8 kW/K,
2. Stream (3+4) between temperature 100C to 150C, a combined stream of thermal capacity rate
(2.8+5.12) = 7.92 kW/K
3. Stream4 between temperatures 70C to 50C, stream 1 with heat capacity rate 1.98 kW/K.
To plot the composite cooling curve the calculations can be estimated as shown in Table 1.11.
Table 1.11 Composite Cooling Curve
The two composite streams are then plotted on a temperature heat load graph. The temperature and rate of change of enthalpy for cold stream and hot stream are estimated in Table 1.12.
= 2.2 50 = 110 Kw Similarly, at temp 40C th = 2.8 40
= 112 Kw Similar calculations are done for the selected data.
Table 1.12 Estimation of hot and cold steam
The two composite streams are then plotted on a temperature heat load graph as shown in Fig. 1.34.
Fig 1.34 Temperature verses rate of change enthalpy change for composite hot and cold streams
The pinch point is defined as the point where temperature difference between the two composite curves is minimum.
The temperature difference at the pinch point depends on the design of the heat exchanger. Smaller the temperature difference, the more expensive is the heat exchanger. A high value of pinch point indicates high thermal losses due to external irreversibility.
Example 1: Pinch point temperature of 7C
Pinch point temperature of 7C, then the cold stream (composite) can be moved from left to right on the diagram horizontally, keeping the hot composite curve fixed, until the temperature difference at pinch point is 7C. It is then seen from Fig 1.35 that the external heating load of 30 kW and external cooling load of 102kW are required for the system, all other energy changes can be achieved by the heat exchangers between the various streams.
Fig 1.35 Temperature verses rate of change enthalpy change for composite hot and cold streams for 7C Pinch
Mathematically to obtain the pinch point at 7C the values of cold steams are increased by 100KW, keeping the values of hot streams constant. Required estimation of hot and cold streams for 7C pinch point is given in Table. 1.13.
Table 1.13 Estimation of hot and cold streams for 7C pinch point
The mathematically the cooling and heating above and below 7C Pinch point is mentioned in Table 1.14.
Table 1.14 Cooling and heating above and below 7C pinch point
From Table 1.14, and Fig 1.35 we can infer that for a pinch point of 7C, we require external heating load of 29.7kW and cooling load of 101.7kW above and below the pinch point respectively. The process integration is now as follows:
Above pinch point, stream 1 and 3 exchanges 204.6 kW heat, stream 2 and stream 4 exchanges 245.7 kW. Below pinch point, stream 2 and stream 3 exchange 168 kW heat. Stream 3 and 4 are to be externally heated with (-224+204.6 =-19.4 kW) and (-256+245.7 =-10.3 kW) respectively to meet the deficit /demand of 29.7 kW. Similarly, stream 1 has to be cooled externally with 125.4 kW and 3 has to be heated externally with (144.3-168 =-23.7 kW) heat exchangers respectively to meet the demand of 101.7 kW.
Possible processes are shown in Fig. 1.36.
Fig 1.36 Possible plant to heat and cool four fluid streams for a minimum 7C temperature difference.
Example 2: Pinch point temperature of 23.5C
Similar to example1 the cold stream (composite) is moved from left to right on the diagram (Fig 1.37) horizontally, keeping the hot composite curve fixed, until the temperature difference at pinch point is 23.5C. It is then seen that the external heating load of 130 kW and external cooling load of
202kW are required for the system, all other energy changes can be achieved by the heat exchangers between the various streams.
Fig 1.37 Temperature verses rate of change enthalpy change for composite hot and cold streams for Pinch point temperature of 23.5C
Mathematically to obtain the pinch point at 23.5C the values of cold steams are increased by 200KW, keeping the values of hot streams constant. Required estimation of hot and cold streams for 23.5C pinch point is given in Table. 1.15.
Table 1.15 Estimation of hot and cold streams for 23.5C pinch point
Different heat loads are shown in Table 1.16
Table 1.16 Cooling and heating above and below 23.5C pinch point.
From Table 1.16, mathematically with reference to the Fig 1.37 we can infer that for a pinch point of 23.5C, we require external heating load of 130.35kW and cooling load of 202.35kW above and below the pinch point respectively.
The process integration is now as follows:
Above pinch point, stream 1 and 3 exchanges 168.3 kW heat, stream 2 and stream 4 exchanges 181.35kw. Below pinch point, stream 2 and stream 3 exchange 168 kW heat. Stream 3 and 4 are to be externally heated with (224-168.3 = 55.7 kW) and (256-181.35 = 74.65 kW) respectively to meet the deficit /demand of 130.35 kW. Similarly, stream 1 and 3 are to be cooled externally with 161.7 kW and (208.65-168 = 40.65 kW) heat exchangers respectively to meet the demand of 202.35 kW.
Possible processes are shown in Fig. 1.38.
Fig 1.38 Possible plant to heat and cool four fluid streams for a minimum 23.5C temperature difference.
The following rules should be followed in process integration
1. Do not transfer heat from one fluid to another across the pinch point
2. No external heating below pinch point
3. No external cooling above the pinch point
4. A heat exchanger should operate on one side of the pinch, either taking a heat supply from below
the pinch, or rejecting heat to a fluid above the pinch
5. A heat pump should operate across the pinch from a cold stream below the pinch to a hot stream
above the pinch.
Summary:
1. Exergy is the maximum work potential of a system
2. Exergy transfer with heat, work and mass
3. For an isolated system exergy always decreases
4. Exergy remains constant in a reversible process
Anything that generate entropy is responsible for decrease of exergy
Quiz 11
1. What is a pinch point?
2. What happens if the pinch point is small?
3. How do you draw a composite curve?
4. What are the rules applied for process integration?
Third law of thermodynamics
Marching towards absolute zero temperature.
Motivation
Application of first law and second law of thermodynamics to reactive systems become difficult due to the non availability of a standard reference entropy value of various substances. There is a need to have a reference entropy value for all substances for evaluating the efficiency of a reactive system. Third law of thermodynamics provides a base value for the entropy. The third law of thermodynamics was formulated during the early part of twentieth century. The initial work was done primarily by W. H. Nernst [1864-1941] and Max Planck [1858-1947].
Attaining low temperature
1. Below 5 K is possible by Joule Kelvin expansion, by producing liquid helium.
2. Still lower temperature can be attained by adiabatic demagnetization of a paramagnetic salt.
3. Temperature as low as 0.001 K has been achieved by magnetic cooling.
Magnetic Property
1. Diamagnetic: substance is repelled by magnet
2. Paramagnetic: substance attracted by magnet, such as Iron, Gadolinium sulphate
Adiabatic Demagnetization of a paramagnetic salt:
Gadolinium sulphate is used here.
Salt is hung by a fine nylon thread inside the salt tube such that it does not touch the sides.
Fig: 1.39 Adiabatic demagnetization of a paramagnetic salt
The salt is first cooled slightly below 1 K by surrounding it with liquid helium boiling under reduced
pressure.
Next, the salt is exposed to a strong magnetic field of about 25000 Gauss. Heat produced due to
magnetization of the salt is transferred to the liquid helium without causing an increase in salt
temperature.
With the magnetic field still present, the inner chamber containing the salt is evacuated of gaseous
helium.
Finally, the magnetic field is removed. The molecules disalign themselves, which require energy.
This energy is obtained by the salt getting cooler in the process.
Repetition of the process lowers down the temperature of the salt. Temperature as low as 0.001 K
have been achieved so far.
Measurement of temperature
law gives the most convenient method for measurement of at low temperature (approximately).
1.275
where is the magnetic susceptibility of the salt T is the absolute temperature
The fundamental features of all cooling process are that the lower the temperature achieved, the harder it is to go still lower.
Interpretations of adiabatic demagnetization:
Final temperature Tf i (initial temperature)
First demagnetization produces a temperature half that at start
Second demagnetization produces a temperature
Third demagnetization produces a temperature
rth demagnetization produces a temperature
Infinite number of adiabatic demagnetizations will be required to attain absolute zero temperature.
FOWLER-GUGGENHEIM STATEMENT OF THIRD LAW:
It is impossible by any procedure, no matter how idealized, to reduce any condensed system to the
absolute zero of temperature in a finite number of operations.
Any isothermal magnetization from 0 to H (magnetic intensity) such as k - i1, f1 - i2 etc. is associated with the release of heat, i.e., a decrease in entropy (Fig. 1.40).
Fig. 1.40 T-H and T-S diagrams of a paramagnetic substance to show the equivalence of three statements of the third law
The processes i1 - f1, i2 - f2, i3 - f3 etc. represent reversible adiabatic demagnetizations with temperatures getting lower and lower. Repeated cycles of isothermal magnetization and adiabatic demagnetization would bring about a very low temperature. It is seen that [S(T, H) - S(T,0)] decrease III II I. Thus entropy change associated with an isothermal reversible process of a condensed system approaches zero. It is called Nernst-Simon statement of third law of thermodynamics.
NERNST-SIMON STATEMENT:
The entropy change associated with any isothermal reversible process of a condensed system
approaches zero.
Another statement of third law can be given like this:
It is impossible by any procedure, no matter how idealized, to reduce the entropy of a system to
zero point value in a finite number of operations.
Physical and chemical facts which substantiate the third law:
1. For any phase change that takes place at low temperature, Clausius-Clayperon equation hold good,
1.276
2. From third law of thermodynamics, , since vf - vi is not zero. It shows that
1.277
3. This is substantiated by all known sublimation curves.
4.
Fig. 1.41 The temperature dependence of the change in the Gibbs function and in the
enthalpy for an isobaric process
5.
- 1.278
6. is very small.
1.279
7. Which confirms that .
8. From the Gibbs-Helmholtz equation
1.280
9. (Fig. 1.41)
1.281
10. Similarly, using Gibbs Helmholtz equation containing U and F
1.282
11.
Cp = Cv = 0 as T 0 1.283
Quiz 12
1. State the Nernst-Simon statement of third law of thermodynamics.
2.
3. What do you mean by paramagnetic and diamagnetic substance?
4. Give 3 (three) examples of paramagnetic salts
5. What is adiabatic demagnetization?
6. Explain with the help of neat sketch the working principle of adiabatic demagnetization of a
paramagnetic salt.
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