CHAPTER 1 - FUNDAMENTALS
Instructor:
Prof. Dr. Uğur Atikol
Chapter 1
The Fundamentals
Outline
 Fundamentals
 Closed Systems
 Work Transfer
 Heat Transfer
 Energy Equation
 Open Systems
 Steady Flow Energy Equation
 Second Law of Thermodynamics
 Cycles
Conservation of Energy Principle
(First Law of Thermodynamics)
What is conservation of energy?
It simply states that during a prosess, energy can change from one state to another
but that the total amount of energy remains constant. (Energy can not be created
or destroyed)
Chemical energy to
kinetic energy and heat
Potenial energy to kinetic
energy and heat
Basic Dimensions and their Units
Basic
dimensions
Mass(m)
Length(L)
Time(t)
SI units
Kilogram (kg)
Meter (m)
Second (s)
English units
Pound-mass
(lbm)
Foot (ft)
Second (s or
sec)
1 lbm = 0.45359 kg
1 ft = 0.3048 m
Units Derived from Basic
Dimensions
Force  Mass  Acceleration
  In SI units the unit for force in Newtons (N)
F  m kg  a m 2
s
In English units, Pound - force (lbf)
Work  Force  Distance
W  F ( N )  d (m)
In SI units the unit of work is Joules (J)
In English units, British Thermal Units (Btu)
Energy has the same units as Work.
1 Btu  1.055 kJ
Forms of Energy
 thermal, mechanical, kinetic, potential, electric, magnetic,
chemical, nuclear
 Total energy in a process:
E  the sum of all forms of energy
 Thermodynamics deals with the change of energy instead
of its absolute value.
 It is appropriate to assign the total energy of a system a
value of zero (E=0) at some convenient reference point.
Forms of Energy
KINETIC ENERGY
m 
(kJ)
KE 
2
Rate of kinetic energy :
2
2



m
KE 
2
(kJ/s  kW)
INTERNAL ENERGY
U (kJ)
Is related to the degree of
molecular activity in a system.
U  m  Cv  T
Nomencature :
KE : kinetic energy (kJ)
PE : potential energy (kJ)
U : internal energy (kJ)
POTENTAL ENERGY
PE  m  g  z (kJ)
Rate of potential energy :
PE  m  g  z (kJ/s  kW)
m : mass (kg)
m : mass flow rate (kg/s)
 : velocity (m/s)
g : gravitatio nal acceleration (m/s2 )
z
: elevation from a referencepoint (m)
Cv : specific heat capacity at constant volume (kJ/kgK)
T : Temperatur e
Defining a System
 By defining a system we establish the surroundings of
the system.
 The surface that separates the system from the
surroundings is known as the boundary.
SYSTEM
BOUNDARY
SYSTEM
SURROUNDINGS
or
ENVIRONMENT
Total Energy of the System
 (m/s)
z
SYSTEM
(Total Energy E)
x
E  the sum of all forms of energy
m 2
E  U  KE  PE  U 
 mgz (kJ)
2
Pure Substances
 A Pure Substance is a substance that is chemically
homogenous and fixed in chemical composition.(e.g.
water, nitrogen, air & etc.)
 mixture of oil and water is not a pure substance.
 a mixture of two or more phases of a pure substance is
still a pure substance.
 The phase of a pure substance is the homogeneous,
chemical, and physical of aggregation of its molecules.
Three phases of a pure substance
The molecules in a solid are kept at their positions
by the large spring like inter-molecular forces.
At high temperatures, molecules overcome the
inter molecules forces and break away.
In the liquid phase the molecules are no longer at
fixed positions, and chunks of the molecules float
about each other.
In the gas phase the molecules are far apart
from each other, irregular and move about at
random colliding with each other. Molecules
are higher energy level than they are in liquid
or solid phases.
T-v diagram for the heating process
of water at constant pressure
First Law for Closed Systems
 Consider the following closed system:
E2 – E1
Q1-2
Q1-2 – W1-2
Heat
transfer
=
Work
transfer
W1-2
E2 – E1
Energy change
Energy is a thermodynamic property
Sign Convention
 Heat engine sign convention is used
 The purpose of a heat engine ( which is a closed
system) is to deliver work to its environment while
using up heat.
 Heat transfer into the system is (+) ve
 Work transfer out of the system is (+) ve
Q1-2
E2 – E1
Heat rejection
W1-2
Path dependence of Q1-2 and W1-2
• Energy interactions depend on the path followed but energy
change is path independent
dE
Q  W 
dt
• If the process in a closed system is a cycle then the energy change
will reduce to zero such that:
 Q  W  0
W1-2
W1-2
Q1-2
E2-E1
Path A
Q1-2
E2-E1
Path B
Q1-2 – W1-2 = E2 – E1
Energy change is
path-independent
Closed Systems
Surroundings
 Also known as control mass
 No mass can cross the
boundary
 Energy (in the form of heat
and work) can cross the
boundary
Net energy transfer
to (or from) the system
as heat and work
=
Q
System
W
Net increase (or decrease)
in the total energy
of the system
Q  W  E  U  KE  PE
m( 22  12 )
 (U 2  U1 ) 
 mg ( z2  z1 ) (kJ)
2
Example: constant-pressure process, initially saturated water vapor. Evaluate T2
Q – Wel =
=
-3.7 kJ – (-7.2kJ)= 0.025 kg (
= 2865.3 kJ/kg
- 2725.3 kJ/kg)
=300 kPa
= 2865.3
NOTE :



W in energy equation means Wboundary  Wother
For constant pressure case : Q  Wother  H
=200ºC Table A-6
Specific Heats
The energy required to raise the temperature of a unit mass of a substance
by one degree.
Cv : specific heat at constant volume
Cp : specific heat at constant pressure
Helium gas:
V=constant
m=1kg
ΔT=1ºC
Cv=3.13 kJ/kgºC
(2)
(1)
P=const
m=1kg
ΔT=1ºC
Cp=5.2 kJ/kgºC
3.13kJ
5.2kJ
3-16
Cp > Cv
Because at constant
pressure, the energy
required for expansion work
must also be supplied to
system.
Three ways of calculating u and
h
h  h2  h1
from tables
2
h   c p (T ) dT
1
h  c p ,avg T
Similarly:
Work Transfer
V
P
Work interaction with the surroundings
is experienced if
a. A force is present on the boundary
and
b. Boundary must move
V  dV
P
W sh
 W   F  dr
System
ω
Work due to expansion of fluid
 Wrev  P  dV
Shaft work
 Wrev    d
d
d


Simple forms of energy storage and
corresponding work interactions
Systems can experience more than
one form of work interaction.
Therefore the previous equation
(Wrev = P dV) can be replaced by:
 Wrev   Yi dX i
i
where
Yi are generalized forces and
X i are generalized displacements
(Note: Negative of the pressure
used in the previous equation
due to work done on the system)
Source: Adrian Bejan, Advanced Engineering Thermodynamics 3rd Ed., John Wiley
Heat Transfer
TL
TH
 Heat transfer is the energy
interaction determined by the
temperature difference between
the system and its surroundings.
 «Heat transfer is the energy
interaction accompanied by
entropy transfer, whereas work
transfer is the energy interaction
that takes place in the absence of
entropy transfer.» Adrian Bejan
Q
Q
TL
Q
Sgen
Q
TH
Boundary
Temperature scales
T ( o C)  T ( K)  273.15
9
T (K )
5
5
T ( o C)  T ( o C)  32
9
T ( o F)  T ( R)  459.67
T ( R) 

Source: Adrian Bejan, Advanced Engineering Thermodynamics 3rd Ed., John Wiley

Energy Change
z
 (m/s)
x
First law for closed systems :
System
E
Q1 2  W1 2  E2  E1

W1-2
Surroundings
Energy change
The general expression for energy change :
1
1
E2  E1  U 2  U1  m 22  m12  mgz 2  mgz1





2 
2 

Energy change
Internal energy
Potential energy change
change
Q1-2
Kinetic energy change
Open Systems (or Control Volume)
Control volume can be thought of a region of space through which mass flows.
Control
Volume
(CV)
Mass
entering
Mass
leaving
Conservation of mass principle
Total mass  Total Mass   Net change in 




entering
CV
leaving
CV
mass
within
CV

 
 

m m  m
i
e
CV
Note: i-inlet , e-exit , CV-control volume
The rate form 
 m   m   m
i
e
CV
Flow Work in Open Systems
Mass
entering
Control
Volume
(CV)
Mass
leaving
(the fluid pushing
the fluid infront of it)
Flow energy or flow work
The energy required to push fluid into or out of a control volume.
The force applied on the fluid element :
F=PA
The work done in pushing the fluid element into the control volume:
i.e. the flow work : Wflow = F L = PAL = PV (kJ)
On a unit mass basis wflow  Pv
(kJ/kg)
Energy Equation for Open Systems
(also known as control volume)
Total energy
 Total energy of  Total energy of   Net change 

 
 
 

crossing boundary   mass entering the   mass leaving the   in Energy of 
as heat and work  control volume  control volume  control volume

 
 
 

Q  W   Ein   Eout  Ecv
1 2
[remember h  u  Pv therefore total energy e  uPv
  2   gz ]
h
In the rate form the energy equation in full :
dE  
1 2
1 2




 Q  W   m  h    g z    m  h    g z 
dt
2
2

 out 

in
Energy Equation for Steady Flow
Devices
Steady flow energy equation (SFEE)
Mass (and
energy) in
ECV  constant or ECV  0
Total energy
 Total energy of  Total energy of 

 
 

crossing boundary   mass entering the   mass leaving the   0
as heat and work  control volume  control volume 

 
 

Work
Heat
Q  W   Ein   Eout  0
In the rate form (i  inlet, e  exit)
2
2
V
V
 Q  W   m e (he  e  gze )   m i (hi  i  gzi )
2
2
Mass (and
energy) out
Ex: Steam Turbine
For steady flow:
m 1  m 2  m
m = 600 kg/h
h1 = 3120 kJ/kg
V1 = 3 m/s
z1 = 3 m
V  V1
Q  W  m {h2  h1  2
 g ( z 2  z1 )}
2
Q W
V2
V2
  (h 
 gz )1  (h 
 gz ) 2  0
w
m m
2
2
600 kg
hr  0.167 kg
m 
Q = 630 kJ/h
sec
3600 sec
h2 = 2330 kJ/kg
hr
V2 = 130 m/s
kg
630
z2 = 0
hr  0.175 kW
Q 
3600 sec
hr
0.175 kJ
(100 m ) 2 9.81 m 2
(130 m ) 2
W
s
s
s
s  0)  0


 (3120 kJ 

)  (2330 kJ 
kg 2(1000 J ) 1000 J
kg 2(1000 J )
0.167 kg
0.167 kg
kJ
kJ
kJ
s
s
W
 1.05 kJ 
 (3120 kJ  5 kJ  0.029 kJ )  (2330  8.45  0)  0
kg
kg
kg
kg
0.167 kg
s
 W  131.183 kW
2
2
Second Law of Thermodynamics:
Is it required?
First Law says:
Potential energy lost = Kinetic energy gained
Can he fly backwards?
According to First Law:
Kinetic energy lost =
Potential energy gained
First Law Satisfied!?!?
We need a second law!
Processes can only
happen in one
direction and not in
the other.
Why 2nd Law?
A cup of hot coffee does not
get hotter in a cooler room.
Transferring heat to a paddle
wheel will not cause it to
rotate.
Transferring heat to a
wire will not generate
electricity.
Heat
It is clear from the above examples that processes take place in a certain direction
and not in the reverse direction. First law alone is not enough to determine if a
process will actually occur.
Another principle is needed: Second law of Thermodynamics
Clausius Statement of the 2nd Law
It is impossible for any device to
operate in such a manner that it
produces no effect other than the
transfer of heat from one body to
another body at a higher
temperature.
Warm environment
QH = 5kJ
Device
QL = 5kJ
Cold environment
Kelvin-Plank Statement of the 2nd
Law
It is impossible for any device to
operate in a cycle and produce
work while exchanging heat
only with a single reservoir (i.e.
no engine can have 100%
efficiency).
Thermal energy reservoir
Q H  100kW
Heat
engine
Wout  100kW
Definition of Performance for
Engineering Devices
Required Input
System
or
Device
Desired Output
Desired Output
Performance 
Required Input
or easy to remember version :
(for devices consuming fuel or electricity)
What I get
Performance 
What I pay for
Performance of Heat Engines:
Thermal Efficiency
High Temperature Reservoir at TH
Performance =
QH
Desired output
required input
=
What I Get
What I pay for
In heat engines:
Heat
Engine
the desired output = net work output = Wnet,out
Wnet,out = th x QH
QL
Low Temperature Reservoir at TL
the required input = heat supplied to system = Qin
Thermal efficiency is denoted with  th
th 
Wnet ,out
th 
Wnet ,out
Qin
QH
Qin  Qout
Qout

 1
Qin
Qin
 1
QL
QH
or
Performance of Cogeneration Units
TH
Heat rejection from a heat engine can be used
for heating processes as well
QH
H.E.
Wnet,out
Quseful
QL
TL
Energy Utilization Factor:

Wnet, out  Quseful
QH
Performance of Refrigerators:
Coefficient of Performance (COP)
 The efficiency of a refrigerator is expressed
in terms of the coefficient of performance
(COP).
 The objective of a refrigerator is to remove
heat (QL) from the refrigerated space.
For a refrigerator COPR =
Desired output
Required input
QL
Q L
COPR 
(or
)

Wnet , in
Wnet , in
Wnet,in  QH  QL
Notice that COPR, can be
greater than unity.
COPR 
( kJ )
QL
1

QH  QL QH
1
QL
Coefficient of Performance (COP)
of Heat Pumps
 another device that transfer heat from TL to TH.
 objective is different : maintain a heated space at
high temperature.
COPHP =
Desired output
Required input
COPHP 
The objective of a heat pump is to
supply heat QH into the warmer
space.
QH

Wnet,in
QH
1

QH  QL 1  QL
QH
Heat Pump Example
75,000 kJ/h
Q H
Q H
75,000

COPHP 
 Wnet,in 

W net,in
COPHP
2.5
Net power consumed by the heat pump :
W
 30,000 kJ h (or 8.33 kW )
QH  75,000 kJ
COPHP  2.5
net,in
Wnet ,in  ?
How much heat extracted from outdoors?
Q  Q  W
 (75,000  30,000)
L
H
Q L  ?
net ,in
Q L  45,000 kJ/h
0
h
COP of Absorption Refrigeration
Systems
Desired Output
Required Input
QL
Q

 L
Qgen  Wpump, in Qgen
COPabsorption 
Source: Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach,
McGraw-Hill (Book)
Reversible and Irreversible
Processes
In a reversible process the system changes in such a way that when the
process is reversed the system and the surroundings are restored to their
original states. Changes are infinitesimally small in a reversible process.
Surroundings
Surroundings
System
T + T
Q
T
System
T  T
Q
T
Vapor Cycles
Steady-flow energy equation for each component:
Brayton Cycles
qin
Pressure
ratio
Thermal efficiency
of the ideal
Brayton cycle as a
function of the
pressure ratio.
Refrigeration Cycles
Steady-flow
energy balance
The P-h diagram of an ideal vaporcompression refrigeration cycle.
CHAPTER 2 – THE SECOND LAW OF
THERMODYNAMICS
Instructor:
Prof. Dr. Uğur Atikol
Chapter 2
The Second Law of Thrmodynamics
Outline
 Closed Systems
 History
 One-thermal reservoir
 Reversibility and irreversible process
 Two-thermal reservoirs
 Any number of reservoirs
 Closed system entropy genaration
 Open Systems
2nd Law for closed systems
 History
 The early development of the steam engine took place
before 1800.
 The conversion efficiency of these devices at those times
were less than 5%.
 Sadi Carnot (A French Engineer) developed his first
theoretical analysis on heat engines in 1824.
 This was a first step in the development of the 2nd law.
 The most frequently cited statements of the 2nd law
came from Clausius and from Kelvin and Planck.
2nd Law for closed systems
 History
 It is observed that heat always flows by itself
(spontaneously) from a higher temperature to a region of
lower temperature. According to Clausius the reverse is
impossible without assistance (such as inputing work).
Heat flows from a cup of
hot coffee to cooler room.
Heat can flow from a refrigerated cold space
to warmer room with the aid of electric
power
Heat
Heat
Power
2nd Law for closed systems
 History
 The Kelvin-Plank concentrate on heat engines.
 The first law does not bring limitations to conversion of
heat to work or work to heat.
 Example of work conversion into heat would be frictional
process.
The first law does not bring
any restriction on the
conversion of heat into
work or work into heat.
2nd Law for closed systems
 Cycle in contact with one
temperature reservoir
Thermal energy reservoir
 Plank wrote in 1897: «It is impossible to
construct an engine which will work in a
complete cycle, and produce no effect
except the raising of a weight and cooling
of a heat-reservoir.»
Q H  100kW
Wout  100kW
Heat
engine
 A similar statement made in 1851 by William Thomson
(Lord Kelvin): «It is impossible, by means of inanimate
material agency, to derive mechanical effect from a
portion of matter by cooling it below the temperature of
the coldest of surrounding objects.
2nd Law for closed systems
 Cycle in contact with one
temperature reservoir
 A device which violates the above
statements are known as perpetual
motion machines of the second kind
(PMM2).
 PMM2 produces work solely by cooling
a body.
Thermal energy reservoir
Q H  100kW
Wout  100kW
Heat
engine
 If this was possible a self-acting machine would have
produced work by cooling the sea or the earth with no
other changes with the interacting system.
2nd Law for closed systems
 Cycle in contact with one temperature reservoir
 Kelvin-Planck statement implies that the net work produced in a heat
engine cycle can not be positive with only one reservoir:
If single reservoir, then first law implies:
𝑊𝑐𝑦𝑐𝑙𝑒,𝑛𝑒𝑡 = 𝑄𝑐𝑦𝑐𝑙𝑒,𝑛𝑒𝑡 (or
Single Thermal energy
reservoir
Q cycle, net
Wcycle, net
Heat
engine
𝛿𝑄 =
𝛿𝑊)
According to Kelvin−Planck, a system undergoing a
cycle with a single reservoir cannot deliver a net
amount of work to its surroundings. This
impossibility reduces to:
𝛿𝑊 ≤ 0
or in view of the first law for a cycle in a closed system:
𝛿𝑄 ≤ 0
2nd Law for closed systems
 Cycle in contact with one temperature reservoir
 For interpreting Kelvin-Planck statement, consider the
following example:
 Net work = Net heat transfer
 According to Kelvin-Planck net work of cycle can not be positive
 There may be Wnet to system
 Or Wnet may be zero
 Analytically this means
Thermal
Reservoir
𝛿𝑊 ≤ 0
 The less than sign implies irreversibilities
 The equal sign for reversible case
 Irreversibilities within the system
 Or internal irreversibilities is of interest
mass
2nd Law for closed systems
 Reversibility
 Two of the main objectives of the 2nd law analysis:
1. Determining the theorethical best performance of
devices
2. Quantifying the factors that diminish the performance
of devices
It is required to understand the concept of reversibility and
reversible process in general.
2nd Law for closed systems
 Reversible process
 All heat and work interactions which occur in the
initial process can be reversed; they are equal in
magnitude but opposite in direction.
 No net history is left in the surroundings when the
system gains its initial state.
Surroundings
Surroundings
W
System
T + T
W
Q
T
System
T  T
Q
T
2nd Law for closed systems
 Irreversible process
 The reversible process is an idealization. In real life
processes are irreversible.
 In real life we have friction, electric resistance,
inelasticity (in springs), heat losses (from insulated
pipes) etc., which are known as irreversibilities.
 Irreversibilities can be reduced but we can not
completely eliminate them.
 Any system which is returned to its initial state after
going through an irreversible process will leave a
history in the surroundings due to irreversibilities.
2nd Law for closed systems
 Cycles with Two Thermal Reservoirs
 Sadi Carnot introduced the reversible cycle in 1824.
 Carnot’s principle:
1. th,irrev is always less than th,rev operating between the
same reservoirs.
2. Thermal efficiencies of all reversible heat engines
operating between the same reservoirs are equal.
 An independent absolute temperature scale may be
defined which is independent of the nature of the
measuring substance (such as water).
2nd Law for closed systems
 Cycles with Two Thermal Reservoirs
 Mathematically the dependence of
(Q2/Q1) on T1 and T2 can be expressed by:
−𝑄2
= 𝑓(𝑇2 , 𝑇1 )
𝑄1 𝑟𝑒𝑣
 Note that f is an unknown function
 T1 and T2 are readings from the scale of
one thermometer
 Consider the following:
T1
Reservoir T1
Reservoir T1
Q1
Q1
Wa
Rev.
cycle
Q2
T2
Reservoir T2
Wb
Rev.
cycle
Reservoir T2
Q2
Wc
Rev.
cycle
T3
Q3
Q3
Reservoir T3
Reservoir T3
2nd Law for closed systems
In that case we can write the following equations:
−𝑄3
= 𝑓 𝑇3 , 𝑇1
𝑄1 𝑟𝑒𝑣
−𝑄3
= 𝑓(𝑇3 , 𝑇2 )
−𝑄2 𝑟𝑒𝑣
Dividing these two equations we obtain:
𝑓 𝑇3 , 𝑇1
𝑓(𝑇2 , 𝑇1 ) =
𝑓(𝑇3 , 𝑇2 )
−𝑄2
𝑓 𝑇3 , 𝑇1
=
𝑄1 𝑟𝑒𝑣 𝑓(𝑇3 , 𝑇2 )
𝑓(𝑇2 , 𝑇1 )
2nd Law for closed systems
𝑓 𝑇3 , 𝑇1
𝑓(𝑇2 , 𝑇1 ) =
𝑓(𝑇3 , 𝑇2 )
Does not depend on T3, therefore we can re-write as:
𝑓′ 𝑇1
𝑓(𝑇2 , 𝑇1 ) =
𝑓′(𝑇2 )
or letting 𝑓 ′′ = 1 𝑓′ :
−𝑄2
𝑓′′ 𝑇2
= 𝑓(𝑇2 , 𝑇1 ) =
𝑄1 𝑟𝑒𝑣
𝑓′′(𝑇1 )
At this point the choice of the mathematical form of 𝑓′′ 𝑇 is completely
arbitrary. Here Kelvin scale is defined by selecting 𝑓′′ 𝑇 =T. Therefore, for a
Carnot heat engine operating between T1 and T2:
𝑄2
𝑇2
=
𝑄1
𝑇1
2nd Law for closed systems
As a result, by introducing the concept of the thermodynamic
temperature scale (Kelvin temperature scale) for a reversible heat engine
we can write:
𝑄2
𝑇2
−𝑄2
𝑇2
=
or
=
𝑄1
𝑇1
𝑄1
𝑇1
The following second law statement can be derived:
𝑄1 𝑄2
+
≤0
𝑇1 𝑇2
This statement is a generalization of the Kelvin-Planck
statement and it is independent of the sign of W.
2nd Law for closed systems
 Cycles in contact with any Number of Thermal
Reservoirs
The Kelvin-Planck statement which was stated as
𝛿𝑄 ≤ 0 for a single reservoir can be written as:
𝑄1
≤0
𝑇1
where T1 is the temperature of the single reservoir. For two
and n number of reservoirs:
𝑄1 𝑄2
+
≤0
𝑇1 𝑇2
and
𝑄1 𝑄2
𝑄𝑛
+
+⋯+
≤0
𝑇1 𝑇2
𝑇𝑛
2nd Law for closed systems
 Cycles in contact with any Number of Thermal
Reservoirs
Consider a continuous variation of system boundary temperature as
the cycle is performed in contact with an infinite order of reservoirs,
each supplying a heat interaction of 𝛿𝑄. The above equation can be
expressed as:
𝛿𝑄
≤0
𝑇
T is in Kelvin (or in Rankine) at the system boundary that is crossed
by 𝛿𝑄. For reversible cycles equal sign is used, such that, when the
cycles are reversed, the energy interactions are reversed but their
magnitudes are unchanged.
2nd Law for closed systems
 Cycles in contact with any Number of Thermal
Reservoirs
The concept of entropy evolves from the above equation, such that:
𝛿𝑄𝑟𝑒𝑣
=0
𝑇
In the above expression, 𝛿𝑄𝑟𝑒𝑣 𝑇 represents a property since the net
change during a reversible cycle is zero. This property is called entropy:
𝛿𝑄𝑟𝑒𝑣
𝑑𝑆 =
𝑇
The term «entropy» was coined by Clausius in 1865.
However, the same property was discovered before by Rankine. He
called it «thermodynamic function» and represented it with  instead of S
and regarded above equation as the general equation of thermodynamics.
2nd Law for closed systems
 Process in contact with any Number of Thermal
Reservoirs
Consider any arbitrary process (whose path and energy is not
specified) taking place between states 1 and 2 :
𝛿𝑄𝑟𝑒𝑣
𝑆2 − 𝑆1 =
𝑇
Using
𝛿𝑄
≤0
𝑇
2
1
𝛿𝑄
𝛿𝑄𝑟𝑒𝑣
⇒
+
≤0
𝑇
𝑇
1
2
2
⇒
A cycle composed of a
reversible and an
irreversible process.
𝛿𝑄
≤ 𝑆2 − 𝑆1
𝑇
1
Entropy
Entropy
transfer
change
(Nonproperty) (Property)
2nd Law for closed systems
 Closed-system entropy generation
For a process the second law requires that the entropy transfer never
exceeds the entropy change.
2
𝛿𝑄
⇒
≤ 𝑆2 − 𝑆1
𝑇
1
Entropy
Entropy
transfer
change
(Nonproperty) (Property)
This inequality can be transformed into an equality by the
introduction of the concept of entropy generation (or entropy
production).
2
⇒ 𝑆𝑔𝑒𝑛 = 𝑆2 − 𝑆1 −
𝛿𝑄
≥0
𝑇
1
2nd Law for open systems
At time t
Control
Volume
(CV)
Mass
entering
(Sin)
Mass
leaving
(Sout)
𝑊
At time t :
At time t +∆t :
𝑆𝑐𝑙𝑜𝑠𝑒𝑑,𝑡 = 𝑆𝑜𝑝𝑒𝑛,𝑡 + ∆𝑆𝑖𝑛
𝑆𝑐𝑙𝑜𝑠𝑒𝑑,𝑡+∆𝑡 = 𝑆𝑜𝑝𝑒𝑛,𝑡+∆𝑡 + ∆𝑆𝑜𝑢𝑡
𝑄
∆𝑆𝑖𝑛,𝑜𝑢𝑡 = (𝑠∆𝑚)𝑖𝑛,𝑜𝑢𝑡 = (𝑠𝑚)𝑖𝑛,𝑜𝑢𝑡 ∆𝑡
At time t +∆t
Mass
entering
(Sin)
The light blue area can be assumed to be fixed in
mass. Therefore it is said to be control mass or
closed system.
Control
Volume
(CV)
𝑄
If entropy generation from 𝑡 to t +∆t is ∆𝑆𝑔𝑒𝑛 :
Mass
leaving
(Sout)
𝑊
𝑄𝑖
∆𝑡 + 𝑚𝑠 𝑜𝑢𝑡 ∆𝑡
𝑇𝑖
− 𝑚𝑠 𝑖𝑛 ∆𝑡 ≥ 0
∆𝑆𝑔𝑒𝑛 = 𝑆𝑜𝑝𝑒𝑛,𝑡+∆𝑡 − 𝑆𝑜𝑝𝑒𝑛,𝑡 −
2 𝛿𝑄
obtained from 𝑆𝑔𝑒𝑛 = 𝑆2 − 𝑆1 − 1 𝑇 ≥ 0 and
considering any number of heat transfer sources i.
2nd Law for open systems
Consider the last equation obtained:
𝑄𝑖
∆𝑆𝑔𝑒𝑛 = 𝑆𝑜𝑝𝑒𝑛,𝑡+∆𝑡 − 𝑆𝑜𝑝𝑒𝑛,𝑡 − ∆𝑡 + 𝑚𝑠 𝑜𝑢𝑡 ∆𝑡 − 𝑚𝑠 𝑖𝑛 ∆𝑡 ≥ 0
𝑇𝑖
Dividing by ∆𝑡 or differentiating w.r.t. 𝑡:
𝑆𝑔𝑒𝑛
Entropy
generation
rate
=
𝑑𝑆
𝑑𝑡
−
Rate of entropy
accumulation
inside the
control
volume
𝑖
𝑄𝑖
𝑇𝑖
Entropy
transfer
rate due to
heat transfer
+
𝑚𝑠 −
𝑜𝑢𝑡
𝑚𝑠 ≥ 0
𝑖𝑛
Net entropy flow
rate out of the
control volume
due to mass flow
Note: Work transfer is not accompanied by entropy transfer.
CHAPTER 3 – INTRODUCTION TO ENTROPY
Instructor:
Prof. Dr. Uğur Atikol
Definition of Entropy
 Entropy is a measure of disorder of a system
 Entropy is created during a process
 Entropy can not be destroyed
The Clausius Inequality
𝛿𝑄
≤0
𝑇
Cyclic integral
This inequality is valid for all cycles, reversible or
irreversible
= for reversible
< for irreversible
The cyclic integral of δQ/T can be viewed as
the sum of all the differential amounts of heat
transfer divided by the temperature at the
boundary.
Reversible Cycles
High temp.
reservoir at TH
QH
Wnet
Rev.
H. E.
QL
High temp.
reservoir at TL
For reversible cycles:
QL TL
QL QH



QH TH
TL
TH
Q
 QH
 QL
(
)


 T rev  TH  TL
1
1

 QH    QL

TH
TL
Q
Q
 H  L 0
TH
TL
Q
note :  (
) rev  0
T
For irreversible cycles:
High temp. reservoir at TH
QH
QH
Wnet
Rev.
H. E.
QL ,irrev  QL
IRREV.
Wnet, irrev
H. E.
QDiff
QL
QL, irrev
High temp. reservoir at TL
Note:
Q
or
QL ,irrev  QL  QDiff
QH QL ,irrev


T
TH
TL
QDiff
QH QL QDiff


0
TH TL
TL
TL
Q
(
 T )irrev  0
(
)irrev 
For all cycles, the two results are combined:
𝛿𝑄
> 0 violates the 2nd law of thermodynamics
𝑇
𝛿𝑄
has to be always negative.
𝑇
𝛿𝑄
≤0
𝑇
Entropy is a property
Internally reversible
 dV  0
The net change in volume (a property) during a cycle is
always zero.
Any property change during a cycle is zero.
Q 
Q 
 0,   must represent a

Since  
 T int rev
 T int rev
property in the differential form.
2
1
Q
Q
Q
(
)

(
)

(
 T int rev 1 T A 2 T ) B  0
2
Q
Q
  ( ) A   ( )B
T
T
1
1
2
The value of the integral depends on the
end states only and not the path followed
Q
dS  ( )int rev
T
This represents the change of a property
(kJ )
K
This property is called entropy, S .
Entropy is an extensive property.
Q
)int rev (kJ )
K
T
1
2
The entropy change of a system:
 S  S 2  S1   (
Example: air temperature is raised from T1 to T2
0
Thermal
insulation
 Q   W  dU
 Q  d U  mCv dT
2
Q
Air
2
mCv dT
T2
S   (
)int rev  
 mCv ln
T
T
T1
1
1
Q
Special Case:
Internally Reversible Isothermal heat transfer processes:
Particularly useful for determining the entropy changes of thermal energy
reservoirs that can absorb or supply heat indefinitely at constant temperature.
Q
Q
1
 S   ( )int rev   ( )int rev   ( Q)int rev
T
TC
TC 1
1
1
2
Liq+vap mixture
T = 300K = const.
Q
T
 2.5 kJ/K
 S sys 
Q = 750 kJ
S 
Q
TC
2
kJ K 
Constant absolute
temperature
2
Increase of Entropy Principle:
Entropy Generation
 The inequality dS  Q/T implies that for irreversible cases
dS is greater than Q/T
 Therefore dS  Q/T > 0 and this quantity is known as
entropy generation
 For any closed system:
T1
T2
Tn
dQ
 S gen
T
or if there are several heat transfer
d S sys 
positions on the boundary :
n
d Qi
d S sys  
 S gen
i 1 Ti
Q1
Q2
Closed
system
Qn
Increase of Entropy Principle:
Entropy Generation
 Consider equation
 For an isolated system SQ=0  0
 The entropy of an isolated system always increases (due to
irreversibilities) or if reversible, remains constant.
Entropy Balance
 The property entropy is a measure of
molecular disorder or randomness of a
system.
 Enropy can be created but it cannot be
destroyed
 Total   Total   Total   Change in the 

 
 
 

 entropy    entropy    entropy    total entropy 
 entering   leaving   generated   of the system 

 
 
 

or
S in  S out  S gen   S system
Entropy Change of a System Ssys
Entropy change of a system  Entropy at final state  Entropy at initial state
 S sys  S final  S initial
Note : S sys  0 during steady state operation.
When the properties of the system are not uniform, the entropy
of the system can be determined by :
S sys   s  m   s dV
volume
density
Entropy Change of Pure Substances
 TdS relations:
ℎ = 𝑢 + 𝑃𝑣 → 𝑑ℎ = 𝑑𝑢 + 𝑃𝑑𝑣 + 𝑣𝑑𝑃
𝑇𝑑𝑠 = 𝑑ℎ − 𝑣𝑑𝑃
𝑇𝑑𝑆 = 𝛿𝑄 → 𝑇𝑑𝑠 = 𝑑𝑢 + 𝑃𝑑𝑣
 Hence useful relations can be obtained for ds:

 𝑑𝑠 =
𝑑𝑢
𝑃𝑑𝑣
+
𝑇
𝑇
and 𝑑𝑠 =
𝑑ℎ
𝑣𝑑𝑃
−
𝑇
𝑇
 We must know the relationship between du or dh and T
 𝑑𝑢 = 𝑐𝑣 𝑑𝑇 or for ideal gases 𝑑ℎ = 𝑐𝑝 𝑑𝑇
 For ideal gases: 𝑃𝑣 = 𝑅𝑇 and hence:
 𝑑𝑠 = 𝑐𝑣
𝑑𝑇
𝑑𝑣
+𝑅
𝑇
𝑣
𝑜𝑟 𝑑𝑠 = 𝑐𝑝
𝑑𝑇
𝑑𝑃
−𝑅
𝑇
𝑃
 For liquids and solids assume incompressible hence 𝑑𝑣 ≅ 0
𝑑𝑢
𝑐𝑑𝑇
 Hence for liquids and solids: 𝑑𝑠 =
=
since 𝑐𝑝 = 𝑐𝑣 = 𝑐
𝑇
𝑇
and 𝑑𝑢 = 𝑐 𝑑𝑇
Entropy Change of Ideal Gases
 Specific heats vary with temperature
 Therefore
2
𝑑𝑇
𝑑𝑃
𝑑𝑇
𝑑𝑃
𝑑𝑠 = 𝑐𝑝 − 𝑅 → 𝑠2 − 𝑠1 = 1 𝑐𝑝 (𝑇) − 𝑅
𝑇
𝑃
𝑇
𝑃
 Choose absolute zero as reference T and define:
𝑠° =
𝑇
𝑑𝑇
𝑐 (𝑇)
0 𝑝
𝑇
 Table A17 in Çengel* tabulate 𝑠°
2
𝑑𝑇
 Therefore 1 𝑐𝑝 𝑇
= 𝑠°2 − 𝑠°1
𝑇
𝑑𝑃
 Hence 𝑠2 − 𝑠1 = 𝑠°2 − 𝑠°1 − 𝑅
𝑃
Mechanisms of entropy transfer,
Sin and Sout
 Entropy can be transferred by the following two
mechanisms:
 Heat transfer
Heat is a chaotic form of
energy and some chaos
(entropy) flows with heat
 Mass flow
Mass contains entropy and
entropy is carried with it.
Entropy increases with mass
 No entropy is transferred by work
Entropy transfer by heat transfer
Entropy transfer by work:
When temperature is
not constant or
different throughout
the boundary
Heat transfer is always accompanied by
entropy transfer in the amount of Q/T,
where T is the boundary temperature.
No entropy accompanies work as it crosses the
system boundary. But entropy may be
generated within the system as work is
dissipated into a less useful form of energy.
Entropy transfer by mass flow
Entropy transfer by mass:
When the properties of the mass
change during the process
Mass contains entropy as well as
energy, and thus mass flow into or
out of system is always
accompanied by energy and
entropy transfer.
Entropy generation, Sgen
Sin with mass
Sin with heat
System
Ssys
Sgen≥0
Sout with mass
Sout with heat
Entropy generation, Sgen
 The term Sgen represents the entropy within the
system boundary only
 External irreversibilities are not accounted for in the
term Sgen.
Entropy generation
outside system
boundaries can be
accounted for by
writing an entropy
balance on an
extended system that
includes the system
and its immediate
surroundings.
Entropy balance of control masses
(closed systems)
T0
 S sys



Entropy change
of a closed system

Qk
T
k

Sum of net entropy
transfer through the
system boundary by
heat transfer
 S gen

Entropy
generated
Q 0
(kJ/K)
System
Q1
Q 2
Q 3
T1
T2
T3
W
Example:
Entropy generation in a wall
Determine the rate of entropy generation in a wall of 5-m x 7-m and thickness 30 cm.
The rate of heat transfer through the wall is 1035 W.
0 (steady heat flow)
Sin  Sout 



Rate of net entropy
transfer by heat
and mass
S gen


Rate of entropy
generation
d S sys
dt

Rate of
change in
entropy
 Q   Q 
      S gen  0
 T in  T out
1035 W 1035 W 

 S gen  0
293 K
278 K
 therefore S gen , wall  0.191 W/K
The total rate of entropy generation (including the indoors and outdoors)
can be found by taking into account the indoors and outdoors temperatures
(extended system):
1035 W 1035 W 

 S gen  0
300 K
273 K
 therefore S gen, total  0.341 W/K
Entropy balance of control volumes
(open systems)
T0
The entropy of a control
volume changes as a result of
mass flow as well as heat
transfer.
m in
Qk
 T   mi si   me se  S gen  (S
2  S1 ) CV

k
(kJ/K)
or in the rate form :
Q k
m i si  m e se  S gen 
T 


k
Entropy

Net entropy flow rate
(kW/K)
 S CV
Entropy transfer
rate by heat
transfer
out of the control volume
via mass flow
generation
rate
dS CV
dt

Rate of entropy
accumulation in
the control volume
Q 0
System
Q1
Q 2
Q 3
T1
T2
T3
m out
W
Entropy balance of control volumes
(open systems)
Q k
dS
 T   m i si   m e se  S gen  dtCV
k
The entropy of a
substance always
increases (or remains
constant in the case of a
reversible process) as it
flows through a singlestream, adiabatic,
steady-flow device.
The entropy of a control
volume changes as a result of
mass flow as well as heat
transfer.
Example: Entropy generation
during a throttling process
Determine the rate of entropy generation in a steady-state throttling process of steam
shown in the diagram.
Use the tables to determine the entropy at the inlet and the exit states:
State 1 :
P1  7 MPa 
 h1  3288.3 kJ/kg, s1  6.6353 kJ/kg.K
T1  450C 
State 2 :
P2  3 MPa 
 s2  7.0046 kJ/kg.K
h2  h1

0 (negligible heat transfer)
0 (steady flow process)
d S sys
m s1  m s2  S gen  0
dt

S gen  m ( s2  s1 )
Q k
  dS CV



m
s

m
s

S
 T  i i  e e gen dt
k
Sin  Sout 



Rate of entropy
transfer by mass
flow
S gen

Rate of entropy
generation

Rate of change
in entropy in the
control volume
Dividing by mass flow rate :
s gen  s2  s1  7.0046  6.6353  0.3693 kJ/kg.K
Example: Entropy generation in
a compressor
Sin  Sout 



Rate of net entropy
transfer by heat
and mass
S gen

Rate of entropy
generation

d S sys
𝑃1 = 1 MPa
𝑇1 = 327℃
0 (steady flow process)
s2o  2.40902 kJ/kg.K
dt

Rate of
change in
entropy
Q out
 S gen  0
Tb , surr
Q out

S gen  m ( s2  s1 ) 
Tb , surr
m s1  m s2 
For ideal gases : s2  s1  s20  s10  R ln
300 kW
Compressor
P2
P1
kJ
1000 kPa
m ( s 2  s1 ) air  0.853 kg/s (2.40902  1.66802)
 0.287 ln
kg.K
100 kPa
 0.0684 kW/K
25 kW
S gen  0.0684 kW/K 
 0.155 kW/K
290 K
25 kW
Air
𝑚1 = 0.853 kg/s
𝑃1 = 100 kPa
𝑇1 = 𝑇𝑎𝑚𝑏 = 17℃
s1o  1.66802 kJ/kg.K
Example: Entropy transfer
associated with heat transfer
A frictionless piston-cylinder contains saturated liquid
vapor mixture at 100oC. 600kJ is lost to the environment at
constant pressure leading to condensation of some vapor.
The entropy change of water :
 S sys 
Q
 600 kJ

 1.61 kJ/K
Tsys (100  273)K
S in  S out  S gen 




Net entropy transfer
by heat and mass
Entropy
generation
 S sys



Entropy change
of the system
The extended system includes the
water, the piston-cylinder device
and the surroundings just outside
the system that undergoes a
temperature change. The
boundary of the extended system
is at Tsurr.
Considering the extended system for
total entropy change :
Q
Q
 out  S gen   S sys  S gen  out   S sys
Tb
Tb
600 kJ

 (1.61 kJ/K)  0.40 kJ/K
(25  273) K
Entropy generation associated with
a heat transfer process
Pinpointing the location of entropy generation: Be more precise about
the system, the b0undary and the surroundings.
Homework
 Steam expands in a turbine steadily at a rate of 40,000
kg/h, entering at 8 MPa and 500oC and leaving at 40
kPa as saturated vapor. If the power generated by the
turbine is 8.2 MW, determine the rate of entropy
generation for this process. Assume the surrounding
medium is at 25oC.
8 MPa
500oC
Steam
turbine
8.2 MW
40 kPa
Sat. vapor
CHAPTER 4 – EXERGY AND EXERGY ANALYSIS
Instructor:
Prof. Dr. Uğur Atikol
Chapter 4
Exergy and Exergy Analysis
Outline
 Fundamentals on Exergy
 Exergy Associated with KE and PE
 Irreversibility (Exergy Destruction)
 Second Law Efficiency
 Nonflow Exergy
 Exergy of a Flow stream
 Exergy by Heat, Work and Mass
 Exergy Balance
Exergy: Work Potential of Energy
 The exergy of a system is defined as the maximum
shaft work that can be achieved by both the system
and a specified reference environment
 Therefore exergy is a property of both the system and
the environment
Heat Source at T
Q
Carnot
Heat
Engine
Exergy transfer by heat
𝑋ℎ𝑒𝑎𝑡 𝑜𝑟 𝑋𝑄
𝑇0
= 1−
𝑄
𝑇
Xheat = Wmax = (1 - T0/T ) x Q
Q0
Dead State
Environment
at T0 at T0
Revision of Fundamentals
 Work = f (initial state, process path, final state)
 The specified initial state is constant
 Maximum work is obtained from reversible process
 To maximize the work output, final state = dead state
 Dead state means thermodynamic equilibrium of the
system with the environment
 Exergy is destroyed whenever an irreversible process occurs
 Exergy transfer associated with shaft work is equal to the
shaft work
 Exergy transfer associated with heat transfer is dependent
on the temperature of process in relation to the
temperature of the environment
Exergy Associated with KE and PE
 Kinetic and potential energies are forms of mechanical
energy
 Hence they can be converted to work entirely, i.e. The
work potential or exergy are themselves:
exergy of kinetic energy : xke  k e 
2
2
exergy of potential energy : x pe  p e  g z
V2
xke  ke 
2
x pe  pe  g z
Exergy Associated with Electricity
 Just like shaft work, exergy associated with electricity is
equal to electric energy itself.
 Hence, electric energy 𝑊𝑒𝑙 and power 𝑊𝑒𝑙 can be
converted directly to 𝑋𝑒𝑙 and 𝑋𝑒𝑙 respectively:
exergy of electric energy : xel  wel
exergy power : xel  w el
V2
xke  ke 
2
x pe  pe  g z
Surroundings Work
 Work produced by a work producing device (that involve
moving boundary) is not always completely usable
 Work done by or against the surroundings is known as
surroundings work, Wsurr
 In a piston-cylinder device some work is used to push the
atmospheric air out of the way
Atmospheric air
Atmospheric air
 In this example:
P0
P0
Wsurr  P0 (V2  V1 )
 Useful work:
Wu  W  Wsurr
 W  P0 (V2  V1 )
System
V1
System
V2
Irreversibility (exergy destruction)
 Reversible work (Wrev) is defined as the maximum
useful work that can be generated (or the minimum work
that needs to be supplied) during a process
 When the final state of the process is the dead state then
Wrev = Exergy = X
 The useful work (Wu) obtained in work producing
devices is less than Wrev due to the irreversibilities
 Irreversibility is viewed as the lost opportunity to do
work
 Irreversibilities (I) cause exergy destruction
 I = Xdes = Wrev,out – Wu,out or Wu,in – Wrev, in
I or Xdes from a Heat Source
High Temperature Reservoir at TH
QH
Carnot
Heat
Engine
Q0
Irreversibilities
Dead State (Environment) at T0
Lost available work, 𝑊𝑙𝑜𝑠𝑡
or
Exergy destruction, 𝑋𝑑𝑒𝑠
Example:
I or Xdes of a Heat Engine
 TL  


 Qin
Wrev  th ,rev Qin  1 
 TH 
300K 

W rev  1 
 (500kW )  375 kW
 1200K 
1200K
500-kJ/s
The rate of irreversibility or exergy destruction:
180-kW
X des  I  W rev  W u  375  180  195 kW
300K
Q L ,total  Q H  Wu  500  180  320 kW
Q
 Q  W  500  375  125 kW
L , rev
H
I or X des
rev
I  320  125  195 kW
Example taken from Çengel 7th Ed. p.425
Q L ,total
This is not available for
converting to work
Q L ,rev
Example:
Xdes from a Hot Water Tank
 When the water is not used the work potential is
completely wasted
0
Exergy stored in the tank is completely destroyed, I = Xdes = Wrev – Wu = Wrev
Hot Water Tank at TH = 80oC
Vout= 0
QH
Outlet valve
closed
Hot water
80oC
10oC
Carnot
Heat
Engine
Wu = 0
Q0
Inlet
Environment at T0 = 10oC
Example:
Xdes from a Hot Water Tank
 When the water is used the Xdes can be expressed as:
Exergy destroyed, I = Xdes = Wrev – Wu = Xst – Xout
Hot Water Tank at TH = 80oC
Xout
QH
Outlet
Xst
Carnot
Heat
Engine
Q0
Inlet
Environment at T0 = 10oC
Second-Law Efficiency, ηII
 Second-law efficiency is defined as the ratio of the actual
thermal efficiency to the maximum possible (reversible)
thermal efficiency under the same conditions:
th
 For heat engines:  II 
th,rev
Wu
 For work producing devices:  II  W
rev
Wrev


 For work consuming devices: II
Wu
COP
 For refrigerators and heat pumps:  II 
COPrev
Example:
Hot Water Usage from a Tank
Xout
X out
 II 
X st
Outlet
where Xout is the useful exergy extracted from the tank
and Xst is the exergy stored in the tank
Xst
Inlet
Also note that:
X des
 II  1 
X st
If all the stored exergy is destroyed, then ηII = 0
If no exergy destruction takes place (reversible case)
then ηII = 1 (maximum). This means that Wu = Wrev
Nonflow Exergy: Exergy of a fixed mass
Any useful work is due to pressure above
atmospheric pressure :

   W  P dV

 ( P  P0 ) dV  P0 dV



 Wb ,useful
Work potential due to heat transfer :

   WHE  1  T0   Q

 T


T
Q  0 Q

T


 T0 dS

  Q   WHE  T0 d S
Note that :  Wtotal useful   WHE   Wb ,useful
Substitute  Q and  W in the energy equation :
  Q   W  dU
 ( WHE  T0 d S )  ( Wb ,useful  P0 dV )  dU
  Wtotal useful  T0 d S  P0 dV  dU
  Wtotal useful   dU  P0 dV  T0 d S
Nonflow Exergy: Exergy of a fixed mass
Equation obtained :
  Wtotal useful   dU  P0 dV  T0 d S
Integrating from given state to dead state (0 subscript) :
 Wtotal useful  (U  U 0 )  P0 (V  V0 )  T0 ( S  S0 )



Availability or Exergy
On a unit mass basis the nonflow exergy can be expressed as :
   (u  u0 )  P0 (v  v0 )  T0 ( s  s0 )
Including the kinetic energy and potential energy terms :
1
   (u  u0 )  P0 (v  v0 )  T0 ( s  s0 )   2  g z
2
where g is gravitational acceleration,  is velocity and z is elevation.
The exergy change of a nonflow system (from state 1 to 2) :
1
   2  1  (u2  u1 )  P0 (v2  v1 )  T0 ( s2  s1 )  ( 22   12 )  g ( z2  z1 )
2
 (e2  e1 )  P0 (v2  v1 )  T0 ( s2  s1 )
where e is (u   2 2  gz )
Nonflow Exergy: Exergy of a fixed mass
For incompressible substances it is recalled that :
c
dT
T
For example, the nonflow exergy of a full tank of hot water
du  cdT , dv  0 and ds 
can be evaluated from :
0
   (u  u0 )  P0 (v  v0 )  T0 ( s  s0 )
 (u  u0 )  T0 ( s  s0 )
where u is the total specific internal energy
and s is the total specific entropy in the tank.
Note 1 : Suffix "0" denotes the dead state.
Note 2 : Nonflow exergy is the exergy stored in the
tank, therefore X st  
𝑋𝑠𝑡 = 𝜙
Flow Exergy: Exergy of a flow stream
For flowing fluids flow energy or flow work was defined before.
This is the energy needed to maintain flow in a control volume,
such that wflow = Pv.
The flow work is done against the fluid upstream in excess of the
boundary work against the atmosphere such that exergy associated
with this flow work:
xflow = Pv – P0v = (P – P0)v
The exergy associated with
flow energy is the useful
work that would be
delivered by an imaginary
piston in the flow section.
Flow Exergy: Exergy of a flow stream
Exergy of a flow stream :
xflowing fluid  xnonflowing fluid  xflow
1
 (u  u0 )  P0 (v  v0 )  T0 ( s  s0 )   2  g z  ( P  P0 )v
2
1
 (u  Pv)  (u0  Pv0 )  T0 ( s  s0 )   2  g z
2
1
 (h  h0 )  T0 ( s  s0 )   2  g z
2
Therefore exergy for a flow stream :
1
   (h  h0 )  T0 ( s  s 0 )   2  g z
2
The exergy change of a fluid stream (from state 1 to 2) :
1
    2   1  (h2  h1 )  T0 ( s 2  s1 )  ( 22   12 )  g ( z 2  z1 )
2
Example: Exergy change during a
compression process
Refrigerant 134a is to be compressed from 0.14 MPa and - 10  C to 0.8 MPa and 50  C.
P2 = 0.8 MPa
T2 = 50oC
Environment conditions are 20  C and 95 kPa.
Inlet state :
P1  0.14 MPA 
h1  246.36 kJ/kg and s1  0.9724 kJ/kg  K
T1  10  C

Exit state :
win
P1  0.8 MPA 
h1  286.69 kJ/kg and s1  0.9802 kJ/kg  K
T1  50  C

The exergy change of the refrigerant is determined from :
1
   2   1  (h2  h1 )  T0 ( s 2  s1 )  ( 22   12 )  g ( z 2  z1 )
2
 (h2  h1 )  T0 ( s 2  s1 )
Compressor
P0 = 95 kPa
T0 = 20oC
Refrigerant 134a
P1 = 0.14 MPa
T1 = - 10oC
 (286.69  246.36) kJ/kg  (293 K )(0.9802  0.9724) kJ/kg  K
 38.0 kJ/kg
This represents the minimum work input (win,min) required to compress the refrigerant
to the specified state.
Exergy transfer by heat, XQ
Heat source
Temperature: T
Q
T0
The maximum work can be obtained by a carnot engine :
Exergy transfer
 T 
X Q  1  0  Q 
T 

by heat


Carnot efficiency
When temperature is not constant :
 T0 
X Q   1    Q
T 

Exergy transfer by work, XW
 W  Wsurr (for boundary work)
XW 
(for other forms of work)
W
Note that Wsurr  P0 (V2  V1 )
Exergy transfer by mass, Xmass
There is no
useful work
transfer
associated
with boundary
work when the
pressure of the
system is
maintained
constant at
atmospheric
pressure.
When mass, m, enters or leaves a system the amount of exergy that accompanies it:
X mass  m
Mechanisms of Exergy Balance
 Total exergy   Total exergy   Total   Change in the 

 
 
 

entering
the

leaving
the

exergy

total
exergy
of

 
 
 

 system   system   destroyed   the system 

 
 
 

X in

Exergy entering
the system by mass flow,
heat and work transfers


X out

Exergy exiting
the system by mass flow,
heat and work transfers
X des

 X system
Exergy destroyed
during the process
Xmass,in
XQ,in
Xmass,out
System
Xsystem
XW,in
XQ,out
XW,out
Xdes
Also defined as
lost available work, Wlost
Exergy Balance: Closed Systems
 Total exergy   Total exergy   Total   Change in the 

 
 
 

entering
the

leaving
the

exergy

total
exergy
of

 
 
 

 system   system   destroyed   the system 

 
 
 


X in

Exergy entering
the system by mass flow,
heat and work transfers

X out

Exergy exiting
the system by mass flow,
heat and work transfers
X des

 X system
Exergy destroyed
during the process
A closed system does not involve any mass flow
Xmass,out
Xmass,in
XQ,in
XW,in
System
Xsystem
Xdes
XQ,out
XW,out
Exergy Balance: Control Volumes
 Total exergy   Total exergy   Total   Change in the 

 
 
 

entering
the

leaving
the

exergy

total
exergy
of

 
 
 

 system   system   destroyed   the system 

 
 
 


X in

Exergy entering
the system by mass flow,
heat and work transfers

X out

Exergy exiting
the system by mass flow,
heat and work transfers
Mass
entering
X des

Exergy destroyed
during the process
Control
Volume
XCV
Xdes
 X system
Procedure for Exergy Analysis
 Subdivide the process under consideration into sections as
desired
 Conduct conventional energy analysis
 Select a reference environment
 Evaluate energy and exergy values relative to the
environment
 Set up the exergy balance and determine exergy
destruction
 Define first and second law efficiencies of the system
 Interpretation of results and conclusions
Example: Solar Water Heating
System from Hepbasli*
Tave
Tw,out
Hot water outlet
Storage
Tank
Cold water inlet
Tw,in
X output
X output
X des
 II  
1


X
X
X
input
input
sun
*Hepbasli A. Renewable and Sustainable Energy Reviews 2008;12
Solar Collector
The instantaneous exergy efficiency of solar collector :
Increased exergy of water
X u
 II,col 

Exergy of the solar radiation X
col
where
X  m [(h
u
w
w , out
Tw,out
 hw,in )  T0 ( sw, out  sw,in )]
(Note that s  dq T  CdT T  C ln T )

 Tw, out 

  m wCw (T w, outTw,in )  T0  ln


 Tw,in 

 Tw, out 
T0

 ln

  Qu 1 

 Tw, out  Tw,in  Tw,in 
and
X col  
A IT rsrad , max
 
Area
Total The maximum
global
exergy - to irradiance energy ratio
for radiation
Tw,in
According to Petela 
4
1T  4 T
rsrad, max  1   0   0
3 T  3 T
*Petela R. Exergy of undiluted thermal radiation. Solar Energy 2003;74
Storage Tank
Tw,out
Tave
Ttop
Hot water outlet
Storage
Tank
Exergy from the storage tank to the end-user as
presented by Xiaowu et al*:
Tbottom
Cold water inlet
 Ttop
 TbottomT0 m wCw  Ttop 


X output  m wCw (Tave  T0 )  m wCwT0  ln
 1 
ln
 T0
 Ttop  Tbottom  Tbottom 
Exergy from the collector to the storage tank as
presented by Xiaowu et al*:
 Tw, out 


X col  tank  m wCw (Tw, out  T0 )  T0  ln
T0 

*Xiaowu et al. Exergy analysis of domestic-scale solar water heatersRenewable and Sustainable Energy Reviews
2005;9
CHAPTER 5 – ENTROPY GENERATION
Instructor:
Prof. Dr. Uğur Atikol
Chapter 5
Entropy Generation (Exergy Destruction)
Outline
 Lost Available Work
 Cycles
 Heat engine cycles
 Refrigeration cycles
 Heat pump cycles
 Nonflow Processes
 Steady-Flow Processes
 Exergy wheel diagrams
Lost Available Work W  P dVdt  W

Atmospheric temperature
and pressure reservoir
at (To,Po)
Atmosphere
P0
Wu
m in
System
m out
Q n
Q 2
...
Reservoir
at T1
Reservoir
at T2
dV
dt
P0
Q 0
Q1
 W shear  Wmagnetic
Work done
against the
atmosphere
dV
dt
in
electrical
Reservoir
at Tn
out
W
(All modes of
work transfer)
dV
Wu  W  P0
dt
Lost Available Work
T0
First law:
n
dE
  Q i  W   m h o   m h o
dt i 0
in
out
Q 0
m in
System
Q1
Q 2
Q n
T1
T2
Tn
Note: h o is known as methalpy, such that
V2
h  h
 gz
2
o
Second law:
n

Q
dS
S gen 
  i   m s   m s  0
dt i 0 Ti in
out
m out
W
Lost Available Work
n
dE
  Q i  W   m h o   m h o
dt i 0
in
out
dS n Q i

S gen 
    m s   m s  0
dt i 0 Ti in
out
Eliminate Q 0 between the two equations :
n
 T 
d

W   ( E  T0 S )   1  0 Q i   m (h o  T0 s )   m (h o  T0 s )  T0 S gen
dt
Ti 
i 1 
in
out
When reversible S gen is zero, hence :
n
 T 
d

Wrev   ( E  T0 S )   1  0 Q i   m (h o  T0 s )   m (h o  T0 s )
dt
Ti 
i 1 
in
out
Therefore generally :
W  W rev  T0 S gen however we know that Wlost  W rev  W
Hence
Wlost  T0 S gen
Also known as «exergy destruction Xdes»
or «Irreversibility»
Lost Available Work
Wlost is always positive although W and W rev can be either positive or negative
(remember Wlost  W rev  W )
0
Wout
W rev
Work producing devices
Wlost is  ve
Wlost is  ve
Work absorbing devices
Win
W rev
0
The main purpose of studying the lost available work is to diagnose the areas
where irreversibilities are taking place in a prosess so that thermodynamic
improvements can be made.
Lost Available Work
When the system is doing work against the atmosphere
that has pressure P0 then the atmosphere consumes a work
rate of P0 dV dt such that :
X W

Rate of
available work
dV
 W  P0
dt
In most flow systems P0 dV/dt = 0,
therefore ẊW = Ẇ (i.e., exergy transfer
by work is simply the work itself)
n
 T0  
d
  ( E  P0V  T0 S )   1   Qi
Ti 
dt
i 1 
  m (h o  T0 s )   m (h o  T0 s )  T0 S gen
in
out
Lost Available Work
In the reversible limit :
X   W  P dV
dt
W rev
rev
0
n
 T 
d
  ( E  P0V  T0 S )   1  0  Q i   m (h o  T0 s )   m (h o  T0 s )
Ti 
i 1 
in
out
dt
 


 


Maximum delivery






Accumulation of nonflow
X
W rev

of available power
Intake of flow exergy
with mass flow
Ψ  m 
Exergy tansfer
with heat transfer
exergy dΦ dt
Release of flow exergy
with mass flow
Ψ  m 
 
 
in
out
in
out
Environment (T0 , P0 )
in
Intake of
flow exergy
with mass flow
 m 
Accumulation of
nonflow exergy
dΦ
dt
Maximum delivery of
available mechanical power
X W rev
in
out
T1
T2
X Q1
Tn
X Q2 ....................... X Qn
Release of flow exergy
with mass flow  m 
out
Lost Available Work
Lost available work is defined as the difference between the maximum available
work Wrev and the actual work W. Alternatively it can be defined as:
X lost  ( X W ) rev  X W
Same as Ẇlost
Same as Ẋdes
Environment (T0 , P0 )
Actual work
in
X 
 m 
W rev
dΦ
dt
in
X W
X 
W lost
Lost available work
Lost exergy
Exergy destruction
Irreversibilities
out
T1
T2
X Q1
Tn
X Q2 ....................... X Qn
 m 
out
Lost Available Work
Exergy balance of the open system discussed can be shown on a flow diagram
as follows:
Exergy inflow
Exergy outflow
 m 
X 
in
W rev
n
 X
i 1
Qn
dΦ
dt
X W
X 
W lost
 m 
out
Lost Exergy in Cycles
Consider as closed systems that operate in an integral number of cycles.
The ceiling value for available power (maximum available power) is
n
 T0  

( X W ) rev   1   Qi
Ti 
i 1 
Exergy content of heat transfer (Q , T , T0 ) can be expressed as
 T0 


X Q  Q 1  
 T 
T
T
T
Therefore the lost available work for
closed systems operating in cycles :
n
Wlost   ( X Q ) i  X W
i 1



( X W ) rev
1
2
Q1
Q2
n
Closed
system
Qn
W
Cycles
T0
Heat Engine Cycles
First and second laws state that :
QH  QL  W  0
QL QH
S gen 

0
TL TH
Obtained by applying the
definition of entropy to the
2 reservoirs. QH is -ve
Wlost can be expressed as follows if
temperature TL is assumed to be T0
 TL 
  W
Wlost  X QH  X W  QH 1 
 TH 
Also can be expressed as :
Wlost  TL S gen
Source
TH
Q H
Ẇ
Heat
Engine
Q L
Sink
TL
Heat Engine Cycles
Temperature -energy diagram for a heat engine cycle proposed by Adrian Bejan
Reversible
Irreversible
QH
QH
TH
TH
tan  
Wlost
TL
since Wlost  TL S gen ,
tan  S gen or   tan 1 S gen
QL ,rev
TL
QL
Wrev
W
Wlost
TL
Wlost
0
0
tan 1 S gen
Heat Engine Cycles
Comparison between the first- and second-law efficiency of a heatengine cycle
High Temperature Reservoir at TH
Exergy transfer by
heat transfer
QH
X
EQQHH
tio
no
f Exe
c
Heat
Engine
Destr u
Heat
Engine
W = I x Q H
QL
Low Temperature Reservoir at TL
 II 
II IIxEXQQHH
rgy
XW
X QH
 T 
where X QH  ( X W ) rev  QH 1  L 
 TH 
Heat Engine Cycles
Second-law efficiency of a heat-engine cycle can also be expressed
as follows:
TL S gen
XW
( X W ) rev  Wlost
 II 

 1
( X W ) rev
( X W ) rev
( X W ) rev
Relationship between first and second law efficiencies:
I 
W
QH
and
 II 
XW
X QH
We know that work transfer is the same as the exergy transfer associated with it
(i.e., W  X W ) Therefore,
X
QH



 II X QH
 II  QH (1  TL TH )
I 

QH
QH
 TL 
 I   II 1  
 TH 
Refrigeration Cycles
 They are closed systems in communication with two heat
reservoirs
(1) the cold space (at TL) from which refrigeration load
QL is extracted
 (2) the ambient (at TH) to which heat QH is rejected

First and second laws state that :
Ambient
TH
Q H
QL  QH  W  0
Ẇ
Q
Q
S gen  L  H  0
TL TH
Obtained by applying the
definition of entropy to the
2 reservoirs. QL is -ve
Here dead state - temperature T0 is the temperature of the
Refrigerator
Q L
ambient, which is TH . Wlost can be expressed as follows :
 T 
Wlost  X QL  X W  QL 1  H   (W )



TL  Work

will
be
W
input
 is a negative
 T 
itself with a
Q  1 H 
L

TL 
(  ) ve sign
This term will be
negative
T

Rearranging  W  QL  H  1  Wlost
 TL

number
Refrigerated space
TL
Refrigeration Cycles
Temperature -energy diagram for a refrigeration cycle proposed by Adrian Bejan
TH
QL
TL
Reversible
Irreversible
QH
QH
TH
QL
Wrev
Wlost
Wlost
W
TL
tan 1 S gen
tan  
0
0
Wlost
TL
since Wlost  TH S gen ,
tan  S gen or   tan 1 S gen
Refrigeration Cycles
Energy conversion vs exergy destruction during a refrigeration cycle
High Temperature Reservoir at TH
Refrigerator
-XW
W
QL
Low Temperature Reservoir at TL
Exergy destruction
Q
COP  L
W
fr igerato r
Re
QH
Minimum work input
when Wloss  0 or when
 X QL  (  X W ) rev
 II 
-XQL
( X W ) rev

 XW
( X QL )  TH S gen
COP
COPrev
1
Noting that COPrev 
TH TL  1
T

 II
 II  COP H  1 or COP 
TH TL  1
 TL

COP 
QL
W

 X QL
and
 II 
Heat-Pump Cycles
Energy conversion vs exergy destruction during a heat-pump cycle
High Temperature Reservoir at TH
-XQH
QH
W
at -pump
He
QH
COP 
Heat-pump
 T 
 X QH  1  L  QH
 TH 
W   XW
W
QL
-XW
Low Temperature Reservoir at TL
 T 
Wlost  1  L (QH )  (W )

TH 

Exergy

destruction
TL S gen
W rev or  X QH
 T 
or re - arranging  W  1  L  QH  Wlost
 TH 
Heat-Pump Cycles
The second - law efficiency of the heat - pump cycle is calculated by dividing the minimum work
requirement by the actual work :
 X QH
( X W ) rev
 II 

 XW
( X QH )  TL S gen

Heat-pump
-XQH
Exergy
destruction
 T 
 X QH  1  L  QH
 TH 
W   XW
-XW
QH
W
COP
COPrev
1
Noting that COPrev 
1  TL TH
 T 
 II
 II  COP1  L  or COP 
1  TL TH
 TH 
COP 
and
 II 
Nonflow Processes
General equation for available work :
dV
X W  W  P0

dt
Rate of
available work
n
 T 
d
  ( E  P0V  T0 S )   1  0  Q i   m (h o  T0 s )   m (h o  T0 s )  T0 S gen
dt
Ti 
i 1 
in
out
For the closed system shown consider a
process 1  2 and integrate the above equation
T1
T2
Tn
from t  t1 to t  t2 :
n
X W  A1  A2   ( X Q )i  T0 S gen
i 1
where A  E  T0 S  P0 V 
Nonflow availability
or a  e  T0 s  P0 v 
A is a thermodynamic property of the system
as long as T0 and P0 are fixed.
Q1
Cycles
Q2
Closed
system
Qn
W
T0
Nonflow Processes
n
X W  A1  A2   ( X Q )i  T0 S gen
i 1
When the atmosphere is the only reservoir, the max work a closed system
delivers can be expressed as :
( X W ) rev 
A A
0
This is known as
the nonflow exergy
Note that the last two terms in the original equation drop out.
The nonflow exergy in full :
Φ  A  A0  E  E0  T0 ( S  S0 )  P0 (V  V0 )
  a  a0  e  e0 T 0( s  s0 )  P0 (v  v0 )
The nonflow exergy is
the reversible work
delivered by a fixed-mass
system during a process in
which the atmosphere is
the only reservoir.
Steady-flow Processes
General equation for available work :
dV
X W  W  P0

dt
Rate of
available work
n
 T0  
d
  ( E  P0V  T0 S )   1  Qi   m (h o  T0 s )   m (h o  T0 s )  T0 S gen

 out 

dt
Ti 
i 1 
in
b
b

( X Q ) i
n
  ( X Q )i   m b   m b  T0 S gen
i 1
in
out
The flow availability at each port is defined as :
B  H o  T0 S
b  h o  T0 s
Steady-flow Processes
Consider multi - stream flow through devices where the streams do not mix. The equation
obtained in the previous slide
n
X W   ( X Q )i   m b   m b  T0 S gen
i 1
in
out
can be written as
n
r
i 1
k 1
X W   ( X Q )i   (m b)in  (m b) out k  T0 S gen
where k is the number of streams between 1 and r
Most popular examples would be single - stream devices and two - stream heat exchangers.
If the flow availability evaluated at standard environmental conditions (T0 , P0 ) is b0 , such that
b0  h0o  T0 s0
then we can define flow exergy x f as :
Remember the flow exergy from Chp 4
x f  b  b0
The flow work is done against the fluid
Hence
upstream in excess of the boundary work
n
r
i 1
k 1


X W   ( X Q )i   (m x f )in  (m x f ) out k  T0 S gen
against the atmosphere such that exergy
associated with this flow work:
xflow = Pv – P0v = (P – P0)v
Steady-flow Processes
Consider a Rankine cycle operating between a high temperature TH and the
atmospheric reservoir temperature T0 . Using the equation
Boiler
n
X W   ( X Q )i   m b   m b  T0 S gen
i 1
in
out
ẊQH
it is possible to derive the following equation :
(m x f )3
n
X W   ( X Q )i   m x f   m x f  T0 S gen
i 1
in
TH
out
(m x f ) 2
Pump
Turb.
(m x f )1
Q H
Cond.
-ẊWp
3
2
Ẇp
ẊWt
(m x f ) 4
Ẇt
4
1
Q L
T0
 II X Q
H
Steady-flow Processes
n
X W   ( X Q )i   m x f   m x f  T0 S gen
i 1
in
In the case of the boiler :
0  X  m ( x )  m ( x )  T S
QH
f
2
f
3
0
ẊQH
gen , boiler
or
X QH  m ( x f ) 2  m ( x f )3  T0 S gen ,boiler
 


Exergy inflow
Boiler
out
Exergy outflow
(m x f )3
(m x f ) 2
Pump
Exergy destroyed
In the case of the turbine ( X Wt  Wt ) :
Turb.
(m x f )1
Cond.
-ẊWp
ẊWt
(m x f ) 4
X Wt  0  m ( x f )3  m ( x f ) 4  T0 S gen ,turb
or
m ( x f )3  X Wt  m ( x f ) 4  T0 S gen ,turb
 



 
Exergy inflow
Exergy outflow
Exergy destroyed
 II X Q
H
Steady-flow Processes
n
X W   ( X Q )i   m x f   m x f  T0 S gen
i 1
in
Boiler
out
In the case of the condenser :
0  m ( x f ) 4  m ( x f )1  T0 S gen , condenser

 

A significant portion
of stream exergy is
destroyed due to
heat transfer from
condenser to the ambient
ẊQH
(m x f )3
(m x f ) 2
Pump
(m x f )1
The exit temperature of the condenser is T1 , which is
greater than T0 and hence the exit exergy ( x f )1 is finite.
In the case of the pump ( X W p  W p ) :
 X W p  0  m ( x f )1  m ( x f ) 2  T0 S gen , pump
 m ( x f ) 2  T0 S gen , pump  X W p  m ( x f )


It is so small, it
is not shown on
the diagram
Turb.
Cond.
-ẊWp
ẊWt
(m x f ) 4
 II X Q
H
HOMEWORK
 Determine (by drawing an exergy wheel diagram) the
exergy flow with the associated exergy destruction
components of each component of a simple vaporcompression refrigeration cycle. Write down the
exergy balance equations for each component and
state any assumptions made.
Mechanisms of Entropy Generation
or Exergy Destruction
Heat Transfer across a Finite
Temperature Difference
Flow with Friction
Mixing