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About This Presentation
for teaching only
Slide Content
ENGINEERING THERMODYNAMICS
AEU 07101
LECTURER: DR. E. KASEMBE
1
AEU 07101
LECTURER: DR. E. KASEMBE
1
Recommended textbooks
•[1]Rogers, G and Mayhew (1992): Engineering
Thermodynamics, work and Heat Transfer Long man
•[2]Easton T.D. and McConkey A. (1993): Applied
Thermodynamics for Engineers and Technologies. (5th ed.),
Longman
•[3]Kirillin, V.A. et al (1981) Engineering Thermodynamics
(2nd Ed) Mir
2
•[1]Rogers, G and Mayhew (1992): Engineering
Thermodynamics, work and Heat Transfer Long man
•[2]Easton T.D. and McConkey A. (1993): Applied
Thermodynamics for Engineers and Technologies. (5th ed.),
Longman
•[3]Kirillin, V.A. et al (1981) Engineering Thermodynamics
(2nd Ed) Mir
2
Assessment
• Written tests, assignments and experiments
forming continuous assessment (CA) is 40%
•End of module examination is 60% .
•Pass mark is 40%
Course Lecturer
Eng. Dr. E. Kasembe
3
• Written tests, assignments and experiments
forming continuous assessment (CA) is 40%
•End of module examination is 60% .
•Pass mark is 40%
Course Lecturer
Eng. Dr. E. Kasembe
3
Thermodynamics
•In Greek, Thermo means hot or heat; dynamics means motion of matter.
So, the word thermodynamic means the study of heat flow in relation to
the motion of matter
•Thermodynamics is therefore the study of energy transformation as well
as devices and materials used to accomplish these transformations
Or
•Thermodynamics is the science dealing with transference of heat and
work between a system and its surroundings and the associated changes
in the properties of the system
Or
•A science dealing with
relations between the properties of a substance
quantities of work and heat which causes a change of state.
4
•In Greek, Thermo means hot or heat; dynamics means motion of matter.
So, the word thermodynamic means the study of heat flow in relation to
the motion of matter
•Thermodynamics is therefore the study of energy transformation as well
as devices and materials used to accomplish these transformations
Or
•Thermodynamics is the science dealing with transference of heat and
work between a system and its surroundings and the associated changes
in the properties of the system
Or
•A science dealing with
relations between the properties of a substance
quantities of work and heat which causes a change of state.
4
Examples of Engineering systems/Energy transformation devices
•The majority of Engineering systems e.g boiler, condenser system, rotary
and reciprocating compressors, nozzles and diffusers, gas and steam
turbines, steam engines, I.C engines, refrigerators, air conditioners, heat
pumps in which thermodynamics analysis is important are systems in
which the working substance is fluid. Consequently the study of fluid
motion (called fluid mechanics) is also important
•Spark-ignition internal-combustion engine (known as Otto cycle in
thermodynamics) in one example of an energy-transformation device.
•Basically the engine is designed to convert chemical energy available in
fuel-air mixture into mechanical work that can be supplied to the
transmission of an automobile, the cuter or any other device.
•This is accomplished by a device consisting of a cylinder/cylinders
equipped with two or more valves and a piston that moves up and down
inside the cylinder shown in Fig below
5
•The majority of Engineering systems e.g boiler, condenser system, rotary
and reciprocating compressors, nozzles and diffusers, gas and steam
turbines, steam engines, I.C engines, refrigerators, air conditioners, heat
pumps in which thermodynamics analysis is important are systems in
which the working substance is fluid. Consequently the study of fluid
motion (called fluid mechanics) is also important
•Spark-ignition internal-combustion engine (known as Otto cycle in
thermodynamics) in one example of an energy-transformation device.
•Basically the engine is designed to convert chemical energy available in
fuel-air mixture into mechanical work that can be supplied to the
transmission of an automobile, the cuter or any other device.
•This is accomplished by a device consisting of a cylinder/cylinders
equipped with two or more valves and a piston that moves up and down
inside the cylinder shown in Fig below
5
Figure 1.1 Typical Internal-combustion Engine Piston-cylinder
Apparatus
6
Apparatus
6
Cont.
•The valves open and close at the appropriate times
to allow a fresh fuel-air charge to enter the volume
formed by the piston-cylinder through intake
manifold and to allow spent combustion gases to
leave the piston cylinder through the exhaust
manifold
•When the valves are not open, piston cylinder is
sealed so that mass cannot enter or leave the
enclosed volume.
•A spark device (spark plug) is located inside the
piston cylinder. At appropriate time, a spark
generated by this device initiates the combustion of
the fuel-air charge
7
•The valves open and close at the appropriate times
to allow a fresh fuel-air charge to enter the volume
formed by the piston-cylinder through intake
manifold and to allow spent combustion gases to
leave the piston cylinder through the exhaust
manifold
•When the valves are not open, piston cylinder is
sealed so that mass cannot enter or leave the
enclosed volume.
•A spark device (spark plug) is located inside the
piston cylinder. At appropriate time, a spark
generated by this device initiates the combustion of
the fuel-air charge
7
Cont.
•The reciprocating motion of the piston within the cylinder is generated by
a crankshaft-connecting rod mechanism. The position of the piston in the
cylinder is then directly related to the rotational position of the
crankshaft.
•At its uppermost position, the piston is said to be at top dead center
(TDC), and the volume of the space confined by the piston and cylinder in
this position is known as the clearance volume-minimum volume that the
substance occupying the piston-cylinder chamber can have.
•At its lower position, the piston is said to be at bottom dead center (BDC)-
in this position, the substance filling the chamber is at its maximum
volume.
•The difference between the confined volume at BDC and TDC is known as
the cylinder displacement volume-engine displacement.
•For example, a 6-cylinder automobile engine with a 3-liter displacement
implies that each piston displaces 0.5 liter as it moves from BDC to TDC
8
•The reciprocating motion of the piston within the cylinder is generated by
a crankshaft-connecting rod mechanism. The position of the piston in the
cylinder is then directly related to the rotational position of the
crankshaft.
•At its uppermost position, the piston is said to be at top dead center
(TDC), and the volume of the space confined by the piston and cylinder in
this position is known as the clearance volume-minimum volume that the
substance occupying the piston-cylinder chamber can have.
•At its lower position, the piston is said to be at bottom dead center (BDC)-
in this position, the substance filling the chamber is at its maximum
volume.
•The difference between the confined volume at BDC and TDC is known as
the cylinder displacement volume-engine displacement.
•For example, a 6-cylinder automobile engine with a 3-liter displacement
implies that each piston displaces 0.5 liter as it moves from BDC to TDC
8
Figure 2:Internal combustion engine with the piston at its extreme
positions
In the four stroke version, sequence of events that form a complete cycle is:
•Stroke 1(TDC to BDC) Intake of a fresh fuel-air charge while the inlet valve
is open
•Stroke 2 (BDC to TDC) Compression of the charge while both values are
closed with ignition occurring near the end of the stroke (T and P
increases)
•Stroke 3 (TDC to BDC) Expansion of the combustion gases with both valves
closed (T and P decreases, gas expands, work is delivered to crankshaft)
•Stroke 2 (BDC to TDC) Removal of the spent combustion gases through the
open exhaust valve before the entire process is repeated
9
positions
In the four stroke version, sequence of events that form a complete cycle is:
•Stroke 1(TDC to BDC) Intake of a fresh fuel-air charge while the inlet valve
is open
•Stroke 2 (BDC to TDC) Compression of the charge while both values are
closed with ignition occurring near the end of the stroke (T and P
increases)
•Stroke 3 (TDC to BDC) Expansion of the combustion gases with both valves
closed (T and P decreases, gas expands, work is delivered to crankshaft)
•Stroke 2 (BDC to TDC) Removal of the spent combustion gases through the
open exhaust valve before the entire process is repeated
9
Cont.
•This engine illustrates major features of
thermodynamic devices
•First, it transform energy from one form to
another (chemical to mechanical)
•Second, it uses working fluid whose conditions
are constantly changing
•Third it periodically returns to its starting
conditioner to begin the process once again
10
•This engine illustrates major features of
thermodynamic devices
•First, it transform energy from one form to
another (chemical to mechanical)
•Second, it uses working fluid whose conditions
are constantly changing
•Third it periodically returns to its starting
conditioner to begin the process once again
10
Example Figure 3:Basic Steam powered electrical generation station
11
11
Cont.
•Has four major components: boiler, turbine generator, condenser, and
feed water pump.
•Water in liquid and vapour forms circulates through these devices to
produce mechanical work.
•The generator converts this work into electrical power
•The boiler is a device that transfer the heat generated by the combustion
of fuel-air mixture to the water being circulated through the system. This
water enters the boiler in a high-pressure liquid form. The heat from the
combustion of the fuel-air converts this water into vapour form (steam)
before it leaves the boiler. High-pressure, high-temperature steam leaves
the boiler and enters the turbine
•In the turbine , its pressure and temperature are reduced and steam
expands and the turbine produces work to drive the electrical generator.
The turbine exhausts a mixture of low-pressure steam and
•a small amount of liquid into the condenser, where heat is removed to
convert the working fluid entirely into liquid form. The heat is removed by
environmental cooling source such as water.
•Liquid water from condenser is re-pressurized by the feedback pump
before it re-enters the boiler to repeat the process
•Note : Boiler can be replace other devices such as solar Concentrator,
nuclear reactor etc
12
•Has four major components: boiler, turbine generator, condenser, and
feed water pump.
•Water in liquid and vapour forms circulates through these devices to
produce mechanical work.
•The generator converts this work into electrical power
•The boiler is a device that transfer the heat generated by the combustion
of fuel-air mixture to the water being circulated through the system. This
water enters the boiler in a high-pressure liquid form. The heat from the
combustion of the fuel-air converts this water into vapour form (steam)
before it leaves the boiler. High-pressure, high-temperature steam leaves
the boiler and enters the turbine
•In the turbine , its pressure and temperature are reduced and steam
expands and the turbine produces work to drive the electrical generator.
The turbine exhausts a mixture of low-pressure steam and
•a small amount of liquid into the condenser, where heat is removed to
convert the working fluid entirely into liquid form. The heat is removed by
environmental cooling source such as water.
•Liquid water from condenser is re-pressurized by the feedback pump
before it re-enters the boiler to repeat the process
•Note : Boiler can be replace other devices such as solar Concentrator,
nuclear reactor etc
12
Why study thermodynamics
•Exponential growth of population and modernization causes
the increase in use of energy in 21
st
century
•Beginning of industrial revolution was the use of coal
•Progressing through the use of petroleum, natural gas,
uranium, hydro power, geothermal power
•Easily exploited resources of fossil fuels/uranium are not only
limited but also approaching exhaustion in the beginning of
21
st
century
Consequences
•Ways have to be found of using energy resources more
efficient
•Other/alternative sources of energy have to be developed
13
•Exponential growth of population and modernization causes
the increase in use of energy in 21
st
century
•Beginning of industrial revolution was the use of coal
•Progressing through the use of petroleum, natural gas,
uranium, hydro power, geothermal power
•Easily exploited resources of fossil fuels/uranium are not only
limited but also approaching exhaustion in the beginning of
21
st
century
Consequences
•Ways have to be found of using energy resources more
efficient
•Other/alternative sources of energy have to be developed
13
Science of thermodynamics
•Aims at analysing power machines or schemes which convert
chemical/nuclear energy into useful work (i.e thermodynamic analysis)
•Aims at improving the effectiveness of using energy resources i.e
minimizing a waste of energy (consequence of the second law of
thermodynamics and heat transfer analysis)
•Aims at assisting the development of new sources of power. For example
proposal for more effective ways of using solar energy such as converting
it directly to electricity or proposals for conversion of energy from
renewable energy sources to a usable way, e.g conversion of energy in
biomass into gas-chemical energy , conversion of nuclear energy into
electricity. Such alternative sources of power have to be safe and non-
polluting
14
•Aims at analysing power machines or schemes which convert
chemical/nuclear energy into useful work (i.e thermodynamic analysis)
•Aims at improving the effectiveness of using energy resources i.e
minimizing a waste of energy (consequence of the second law of
thermodynamics and heat transfer analysis)
•Aims at assisting the development of new sources of power. For example
proposal for more effective ways of using solar energy such as converting
it directly to electricity or proposals for conversion of energy from
renewable energy sources to a usable way, e.g conversion of energy in
biomass into gas-chemical energy , conversion of nuclear energy into
electricity. Such alternative sources of power have to be safe and non-
polluting
14
Some concepts and definitions
•Thermodynamic system: a quantity of fixed mass
under investigation
•Surroundings: everything/matter external to the
system,
•System boundary: interface separating system and
surroundings
•Universe: combination of system and surroundings.
15
•Thermodynamic system: a quantity of fixed mass
under investigation
•Surroundings: everything/matter external to the
system,
•System boundary: interface separating system and
surroundings
•Universe: combination of system and surroundings.
15
•The system, surroundings, system-boundary for a universe
are shown for a potato-shaped system in Fig. 1.
•In here two important interactions between the system
and its surroundings are allowed:
•heat can cross into the system (our potato can get hot), and
•work can cross out of the system (our potato can expand).
Figure 1: Sketch of a universe composed of a system, its surroundings, and the
system boundary.
16
are shown for a potato-shaped system in Fig. 1.
•In here two important interactions between the system
and its surroundings are allowed:
•heat can cross into the system (our potato can get hot), and
•work can cross out of the system (our potato can expand).
Figure 1: Sketch of a universe composed of a system, its surroundings, and the
system boundary.
16
17
Cont. Thermodynamics concepts
Closed system - same matter remains within the region
throughout the process under examination/ consideration
Only work and heat cross the boundary, no material flow
across the boundary.
Fluid in a cylinder as a closed system
17
Cont. Thermodynamics concepts
Closed system - same matter remains within the region
throughout the process under examination/ consideration
Only work and heat cross the boundary, no material flow
across the boundary.
Fluid in a cylinder as a closed system
17
18
Cont. Thermodynamics concepts
Open system - material flow across the boundary
in addition to energy flow.
Fluid in a turbine as an open system
18
Cont. Thermodynamics concepts
Open system - material flow across the boundary
in addition to energy flow.
Fluid in a turbine as an open system
18
•Materials within the boundary - ‘working fluid’,
Materials which do not react chemically with one
another – simple fluids.
•The boundary separating a closed system from the
surroundings may expand or contract to
accommodate any change of volume undergone by
the fixed quantity of fluid.
The fluid processes in a closed system are non-
flow processes.
The fluid processes in an open are flow processes.
19
Materials which do not react chemically with one
another – simple fluids.
•The boundary separating a closed system from the
surroundings may expand or contract to
accommodate any change of volume undergone by
the fixed quantity of fluid.
The fluid processes in a closed system are non-
flow processes.
The fluid processes in an open are flow processes.
19
Properties of a system and State of the working fluid
•State: The state of a system is simply the condition of the
system as described by its properties
•In practice the matter contained with the boundaries of a
system can be a liquid or gas and is known as a working fluid.
•At any instant the state of the working fluid may be defined by
its properties. Some thermodynamic properties are pressure,
temperature, specific volume specific internal energy,
specific enthalpy, and specific enthalpy.
• The value of a property must be independent of the process.
The changes in the values of the property depends on the final
state of the systems
•Recall State Postulates: For any pure working fluid only two
independent properties are necessary to define completely
the state of the fluid
20
•State: The state of a system is simply the condition of the
system as described by its properties
•In practice the matter contained with the boundaries of a
system can be a liquid or gas and is known as a working fluid.
•At any instant the state of the working fluid may be defined by
its properties. Some thermodynamic properties are pressure,
temperature, specific volume specific internal energy,
specific enthalpy, and specific enthalpy.
• The value of a property must be independent of the process.
The changes in the values of the property depends on the final
state of the systems
•Recall State Postulates: For any pure working fluid only two
independent properties are necessary to define completely
the state of the fluid
20
Paths Between Thermodynamic States
•A path is a line connecting the series of equilibrium states of a system
during the process
•Thermodynamic equilibrium occurs within a system when there are no
further changes occurring in the system when isolated from the
surroundings (no heat and work cross the boundary)
•In equilibrium, the properties are uniform
•Process: As a system transforms from one state to another, it passes
through a series of intermediate states. This sequence of states is known
as a process
Figure pg 17:A process on a general state diagram
21
•A path is a line connecting the series of equilibrium states of a system
during the process
•Thermodynamic equilibrium occurs within a system when there are no
further changes occurring in the system when isolated from the
surroundings (no heat and work cross the boundary)
•In equilibrium, the properties are uniform
•Process: As a system transforms from one state to another, it passes
through a series of intermediate states. This sequence of states is known
as a process
Figure pg 17:A process on a general state diagram
21
Heat and Work
•The first law of thermodynamics is simply an
expression of the conservation of energy principle,
and it asserts that energy is a thermodynamic
property.
•Energy can cross the boundary of a closed system in
two distinct forms: heat and work. It is important to
distinguish between these two forms of energy.
•Therefore, they will be discussed first, to form a
sound basis for the development of the first law of
thermodynamics.
22
•The first law of thermodynamics is simply an
expression of the conservation of energy principle,
and it asserts that energy is a thermodynamic
property.
•Energy can cross the boundary of a closed system in
two distinct forms: heat and work. It is important to
distinguish between these two forms of energy.
•Therefore, they will be discussed first, to form a
sound basis for the development of the first law of
thermodynamics.
22
Heat Transfer
•Heat is defined as the form of energy that is transferred between
two systems (or a system and its surroundings) by virtue of a
temperature difference.
•That is, an energy interaction is heat only if it takes place because
of a temperature difference. Then it follows that there cannot be
any heat transfer between two systems that are at the same
temperature
•Heat is energy in transition. It is recognized only as it crosses the
boundary of a system. Consider the hot baked potato. The potato
contains energy, but this energy is heat transfer only as it passes
through the skin of the potato (the system boundary) to reach the
air
23
•Heat is defined as the form of energy that is transferred between
two systems (or a system and its surroundings) by virtue of a
temperature difference.
•That is, an energy interaction is heat only if it takes place because
of a temperature difference. Then it follows that there cannot be
any heat transfer between two systems that are at the same
temperature
•Heat is energy in transition. It is recognized only as it crosses the
boundary of a system. Consider the hot baked potato. The potato
contains energy, but this energy is heat transfer only as it passes
through the skin of the potato (the system boundary) to reach the
air
23
Cont.
•Once in the surroundings, the transferred heat becomes part of the internal
energy of the surroundings. Thus, in thermodynamics, the term heat simply
means heat transfer.
•A process during which there is no heat transfer is called an adiabatic process.
There are two ways a process can be adiabatic:
–Either the system is well insulated so that only a negligible amount of heat can
pass through the boundary, or
–both the system and the surroundings are at the same temperature and
therefore there is no driving force (temperature difference) for heat transfer.
•An adiabatic process should not be confused with an isothermal process.
•Even though there is no heat transfer during an adiabatic process, the energy
content and thus the temperature of a system can still be changed by other means
such as work.
•The amount of heat transferred during the process between two states (states 1
and 2) is denoted by Q12, or just Q. Heat transfer per unit mass of a system is
denoted q and is determined from
•q = Q/m
24
•Once in the surroundings, the transferred heat becomes part of the internal
energy of the surroundings. Thus, in thermodynamics, the term heat simply
means heat transfer.
•A process during which there is no heat transfer is called an adiabatic process.
There are two ways a process can be adiabatic:
–Either the system is well insulated so that only a negligible amount of heat can
pass through the boundary, or
–both the system and the surroundings are at the same temperature and
therefore there is no driving force (temperature difference) for heat transfer.
•An adiabatic process should not be confused with an isothermal process.
•Even though there is no heat transfer during an adiabatic process, the energy
content and thus the temperature of a system can still be changed by other means
such as work.
•The amount of heat transferred during the process between two states (states 1
and 2) is denoted by Q12, or just Q. Heat transfer per unit mass of a system is
denoted q and is determined from
•q = Q/m
24
Modes of Heat transfer
•Conduction
-Fourier's law of heat conduction.
•Convection
-Newton's law of cooling
•Radiation
-Kirchhoff's law of radiation
25
•Conduction
-Fourier's law of heat conduction.
•Convection
-Newton's law of cooling
•Radiation
-Kirchhoff's law of radiation
25
Work
26
26
Cont.
27
27
Cont.
28
28
Heat transfer and work are interactions between a system and its surroundings, and there are
many similarities between the two:
1. Both are recognized at the boundaries of the system as they cross them. That is, both heat and
work are boundary phenomena.
2. Systems possess energy, but not heat or work. That is, heat and work are transfer phenomena.
3. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning
at a state.
4. Both are path functions (i.e., their magnitudes depend on the path followed during a process
as well as the end states).
•Heat
Heat is energy in transit at the boundary when a system changes its state due to
differences in temperatures between the system and its surrounding
-Heat supplied to a system = + Q
-Heat rejected by the system = -Q
•Work
Work is energy in transit at the boundary when a system changes its state due to the
movement of a part of a boundary under the action of force
-Work done by the system = - W
-Work input to a system = + W
29
many similarities between the two:
1. Both are recognized at the boundaries of the system as they cross them. That is, both heat and
work are boundary phenomena.
2. Systems possess energy, but not heat or work. That is, heat and work are transfer phenomena.
3. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning
at a state.
4. Both are path functions (i.e., their magnitudes depend on the path followed during a process
as well as the end states).
•Heat
Heat is energy in transit at the boundary when a system changes its state due to
differences in temperatures between the system and its surrounding
-Heat supplied to a system = + Q
-Heat rejected by the system = -Q
•Work
Work is energy in transit at the boundary when a system changes its state due to the
movement of a part of a boundary under the action of force
-Work done by the system = - W
-Work input to a system = + W
29
Example 1.1 pg 21
•A steel ingot is pushed to the right and up
along a very low friction plane by a forklifts.
This forklift exerts a force of 3 N acting at an
angle of 30o with respect to the line of motion
followed by an ingot. Determine the total
work expended on the ingot to move it a
distance of 2m along the plane
•Ans. W=5.20Nm
30
•A steel ingot is pushed to the right and up
along a very low friction plane by a forklifts.
This forklift exerts a force of 3 N acting at an
angle of 30o with respect to the line of motion
followed by an ingot. Determine the total
work expended on the ingot to move it a
distance of 2m along the plane
•Ans. W=5.20Nm
30
cycle
•A cycle is a series of processes that form a
closed figure on a state diagram. Cycles have
characteristics that the system always return
to the same state at which it started
Figure A typical cycle
31
•A cycle is a series of processes that form a
closed figure on a state diagram. Cycles have
characteristics that the system always return
to the same state at which it started
Figure A typical cycle
31
The first Law of thermodynamics
•One of the most fundamental laws of nature is the conservation of energy
principle. It simply states that during an interaction, energy can change
from one form to another but the total amount of energy remains
constant. That is, energy cannot be created or destroyed.
• Or,
•during an interaction between a system and its surroundings, the amount
of energy gained by the system must be exactly equal to the amount of
energy lost by the surroundings.
•Mathematically
•The statement of the first law: For a closed system unergoing cyclic non-
flow processes. The sum of the net quantities of heat and work absorbed by
the system is zero.
•In other words, the net work output during the cycle is equal to the net
heat supplied
dWdQ
0
dWdQ
32
•One of the most fundamental laws of nature is the conservation of energy
principle. It simply states that during an interaction, energy can change
from one form to another but the total amount of energy remains
constant. That is, energy cannot be created or destroyed.
• Or,
•during an interaction between a system and its surroundings, the amount
of energy gained by the system must be exactly equal to the amount of
energy lost by the surroundings.
•Mathematically
•The statement of the first law: For a closed system unergoing cyclic non-
flow processes. The sum of the net quantities of heat and work absorbed by
the system is zero.
•In other words, the net work output during the cycle is equal to the net
heat supplied
dWdQ
0
dWdQ
32
Corollaries of the first law
•Corollary 1: there exists a property of a closed system such that a change in its
value is equal to the difference between the heat supplied and the work done
during a change of state
•Suppose that a closed system of unit mass takes in a certain quantity of thermal
energy q, which it can receive by thermal conduction and/or radiation. As a result
the system may do a certain amount of external work w.
•The excess of the energy supplied to the body over and above the external work
done by the body is q − w.
•It follows from the principle of conservation of energy that the internal energy of
the system must increase by q − w.
•That is, Δu = q − w
•where u is the change in internal energy of the system which is the property.
–It is the energy associated with the motion, position and internal state of
atoms or molecules of the substance contained in the system
–A mass of gas possesses internal energy due to the kinetic and potential
energy of its molecules or atoms.
– Changes in internal energy are manifested as changes in the temperature of
the system.
33
•Corollary 1: there exists a property of a closed system such that a change in its
value is equal to the difference between the heat supplied and the work done
during a change of state
•Suppose that a closed system of unit mass takes in a certain quantity of thermal
energy q, which it can receive by thermal conduction and/or radiation. As a result
the system may do a certain amount of external work w.
•The excess of the energy supplied to the body over and above the external work
done by the body is q − w.
•It follows from the principle of conservation of energy that the internal energy of
the system must increase by q − w.
•That is, Δu = q − w
•where u is the change in internal energy of the system which is the property.
–It is the energy associated with the motion, position and internal state of
atoms or molecules of the substance contained in the system
–A mass of gas possesses internal energy due to the kinetic and potential
energy of its molecules or atoms.
– Changes in internal energy are manifested as changes in the temperature of
the system.
33
Corollaries
•Corollary 2: Internal energy of a closed system remains unchanged if the
system is isolated from the surroundings.
•From the equation Δu = q − w, if the system is isolated from the
surroundings Δu is zero and hence q and w are also equal to zero
•Corollary 3: A perpetual motion machine of the first kind is impossible
34
•Corollary 2: Internal energy of a closed system remains unchanged if the
system is isolated from the surroundings.
•From the equation Δu = q − w, if the system is isolated from the
surroundings Δu is zero and hence q and w are also equal to zero
•Corollary 3: A perpetual motion machine of the first kind is impossible
34
forms of first law using internal energy
•One form of mechanical work frequently encountered in practice is
associated with the expansion or compression of a gas in a piston-cylinder
device. During this process, part of the boundary (the inner face of the
piston) moves back and forth. Therefore, the expansion and compression
work is often called moving boundary work, or simply boundary work
•Consider a substance, the working substance, contained in a cylinder of
fixed cross-sectional area that is fitted with a movable, frictionless piston.
•The volume of the substance is proportional to the distance from the base
of the cylinder to the face of the piston, and can be represented on the
horizontal axis of the graph shown in the following figure. The pressure of
the substance in the cylinder can be represented on the vertical axis of
this graph.
•Therefore, every state of the substance, corresponding to a given position
of the piston, is represented by a point on this pressure-volume (p–V )
diagram.
35
•One form of mechanical work frequently encountered in practice is
associated with the expansion or compression of a gas in a piston-cylinder
device. During this process, part of the boundary (the inner face of the
piston) moves back and forth. Therefore, the expansion and compression
work is often called moving boundary work, or simply boundary work
•Consider a substance, the working substance, contained in a cylinder of
fixed cross-sectional area that is fitted with a movable, frictionless piston.
•The volume of the substance is proportional to the distance from the base
of the cylinder to the face of the piston, and can be represented on the
horizontal axis of the graph shown in the following figure. The pressure of
the substance in the cylinder can be represented on the vertical axis of
this graph.
•Therefore, every state of the substance, corresponding to a given position
of the piston, is represented by a point on this pressure-volume (p–V )
diagram.
35
Cont.
•Representation of the state of a working
substance in a cylinder on a p–V diagram.
36
•Representation of the state of a working
substance in a cylinder on a p–V diagram.
36
Cont.
•If V2 > V1, then W is positive, indicating that the substance does work on
its environment. If V2 < V1, then W is negative, which indicates that the
environment does work on the substance.
•The p–V diagram is an example of a thermodynamic diagram, in which the
physical state of a substance is represented by two thermodynamic
variables.
•If we are dealing with a unit mass of a substance, the volume V is replaced
by the specific volume v, i.e v=V/m and the work w that is done when the
specific volume increases by dw is
•dw = pdv
•The thermodynamic equation may be written
•dq = du - dw
•Using this with the equation above, we get
•dq = du - pdv
•which is an alternative statement of the First Law of Thermodynamics.
37
•If V2 > V1, then W is positive, indicating that the substance does work on
its environment. If V2 < V1, then W is negative, which indicates that the
environment does work on the substance.
•The p–V diagram is an example of a thermodynamic diagram, in which the
physical state of a substance is represented by two thermodynamic
variables.
•If we are dealing with a unit mass of a substance, the volume V is replaced
by the specific volume v, i.e v=V/m and the work w that is done when the
specific volume increases by dw is
•dw = pdv
•The thermodynamic equation may be written
•dq = du - dw
•Using this with the equation above, we get
•dq = du - pdv
•which is an alternative statement of the First Law of Thermodynamics.
37
Specific Heats
•heat capacity as the amount of heat required to raise the temperature of a mass
of a system by 1°C. It is denoted by C.
•Q = C ΔT 2-37
•Where Q is in kJ, C in kJ/kg.0C.
•Specific heat capacity. Heat absorbed by 1 kg of material that undergoes a rise in
temperature of 1°C.
•When the heat capacity is essentially constant between the two temperatures T1
and T2,
•Q = C ΔT = C ( T2 - T1)
•Suppose a small quantity of heat dq is given to a unit mass of a material and, as a
consequence, the temperature of the material increases from T to T + dT without
any changes in phase occurring within the material.
•The ratio dq/dT is called the specific heat of the material.
38
•heat capacity as the amount of heat required to raise the temperature of a mass
of a system by 1°C. It is denoted by C.
•Q = C ΔT 2-37
•Where Q is in kJ, C in kJ/kg.0C.
•Specific heat capacity. Heat absorbed by 1 kg of material that undergoes a rise in
temperature of 1°C.
•When the heat capacity is essentially constant between the two temperatures T1
and T2,
•Q = C ΔT = C ( T2 - T1)
•Suppose a small quantity of heat dq is given to a unit mass of a material and, as a
consequence, the temperature of the material increases from T to T + dT without
any changes in phase occurring within the material.
•The ratio dq/dT is called the specific heat of the material.
38
forms of the of the First law using cv and cp
39
39
Various forms of the of the First law
40
40
Cont.
•The First Law of Thermodynamics for an ideal gas can now be written in
the form
•dq = cv dT + pdv
•We can also define a specific heat at constant pressure
•cp =(dq/dT)p
•where the material is allowed to expand as heat is added to it and its
temperature rises, but its pressure remains constant.
•In this case, some of the heat added to the material will have to be
expended to do work as the system expands against the constant
pressure of its environment.
•Therefore, a larger quantity of heat must be added to the material to raise
its temperature by a given amount than if the volume of the material
were kept constant.
41
•The First Law of Thermodynamics for an ideal gas can now be written in
the form
•dq = cv dT + pdv
•We can also define a specific heat at constant pressure
•cp =(dq/dT)p
•where the material is allowed to expand as heat is added to it and its
temperature rises, but its pressure remains constant.
•In this case, some of the heat added to the material will have to be
expended to do work as the system expands against the constant
pressure of its environment.
•Therefore, a larger quantity of heat must be added to the material to raise
its temperature by a given amount than if the volume of the material
were kept constant.
41
Cont.
•We write the thermodynamic equation as
•dq = cv dT + pdv = cv dT + d(pv) − vdp
•From the equation of state, d(pv) = d(RT) = RdT .
•Therefore,
•dq = (cv + R)dT − vdp
•At constant pressure, the last term vanishes; therefore,
•cp =(dq/dT)p= cv + R
•Using this in the equation above it, we obtain an alternative form of the
First Law of Thermodynamics:
•dq = cp dT − vdp .
42
•We write the thermodynamic equation as
•dq = cv dT + pdv = cv dT + d(pv) − vdp
•From the equation of state, d(pv) = d(RT) = RdT .
•Therefore,
•dq = (cv + R)dT − vdp
•At constant pressure, the last term vanishes; therefore,
•cp =(dq/dT)p= cv + R
•Using this in the equation above it, we obtain an alternative form of the
First Law of Thermodynamics:
•dq = cp dT − vdp .
42
Table of CV and CP
43
43
Enthalpy
44
44
Cont.
•specific enthalpy is defined as:
•h = u + p
•Since u, p, and are functions of state, h is also a function of state.
•From above it follows that, at constant pressure,
•Δq= (h2 − h1)
•or, in other words,
•Δq = Δh .
•Differentiating the defining equation, h = u + pv, we obtain
•dh = du + d(pv)
•= du + pdv + vdp
•= dq + vdp
•Transferring terms to the other side, we get:
•dq = dh − vdp .
•This is another form of the First Law of Thermodynamics
45
•specific enthalpy is defined as:
•h = u + p
•Since u, p, and are functions of state, h is also a function of state.
•From above it follows that, at constant pressure,
•Δq= (h2 − h1)
•or, in other words,
•Δq = Δh .
•Differentiating the defining equation, h = u + pv, we obtain
•dh = du + d(pv)
•= du + pdv + vdp
•= dq + vdp
•Transferring terms to the other side, we get:
•dq = dh − vdp .
•This is another form of the First Law of Thermodynamics
45
Cont.
•Repeating,
dq = dh − vdp .
•But we already had the equation
dq = cp dT − vdp .
•Comparing these last two equations, we conclude that
dh = cp dT
•or, in integrated form,
•h = cpT
•where h is taken as zero when T = 0.
•In view of this, h is sometimes called the heat at constant pressure,
because it corresponds to the heat given to a material to raise its
temperature from 0 to T Kelvins at constant pressure.
46
•Repeating,
dq = dh − vdp .
•But we already had the equation
dq = cp dT − vdp .
•Comparing these last two equations, we conclude that
dh = cp dT
•or, in integrated form,
•h = cpT
•where h is taken as zero when T = 0.
•In view of this, h is sometimes called the heat at constant pressure,
because it corresponds to the heat given to a material to raise its
temperature from 0 to T Kelvins at constant pressure.
46
The second law of thermodynamics
The first law shortcomings
• Statement is based on principle of energy
conservation
•Not clear whether the net heat supplied is equal to
the total heat supplied
•It does not tell us whether any of the heat supplied is
not converted into work i.e rejected
The second law is therefore an expression of the fact
that some of the heat supplied during the cycle is
rejected
47
The first law shortcomings
• Statement is based on principle of energy
conservation
•Not clear whether the net heat supplied is equal to
the total heat supplied
•It does not tell us whether any of the heat supplied is
not converted into work i.e rejected
The second law is therefore an expression of the fact
that some of the heat supplied during the cycle is
rejected
47
Statement of the Second Law of Thermodynamics
•It is impossible for a system operating in a
cycle to extract heat from a reservoir and
produce an equivalent amount of work i.e
Q>W
•Or
•Whenever a system operates in a cycle the
net work output is less than the quantity of
heat supplied
48
•It is impossible for a system operating in a
cycle to extract heat from a reservoir and
produce an equivalent amount of work i.e
Q>W
•Or
•Whenever a system operates in a cycle the
net work output is less than the quantity of
heat supplied
48
Cont-Heat Engines
•For a heat engine operating on a cycle of processes giving a
net work output, the second law implies that the engine must
operate between the two reservoirs of heat-the hot reservoir
at higher temperature supplying heat and the cold reservoir
at a lower temperature receiving the rejected heat. It is
assumed that the capacity of the hot reservoir is large enough
so as not to be affected in the temperature by the extraction
of heat by the system or the surrounding
•Whereas a heat engine absorbs a net quantity of heat and
produces a net work output, a heat pump absorbs a net
quantity of work and produces a net heat output
•The primary function of a heat pump is to maintain a high
temperature in the hot reservoir
•The primary function of a refrigerator is to maintain a low
temperature in the cold reservoir
49
•For a heat engine operating on a cycle of processes giving a
net work output, the second law implies that the engine must
operate between the two reservoirs of heat-the hot reservoir
at higher temperature supplying heat and the cold reservoir
at a lower temperature receiving the rejected heat. It is
assumed that the capacity of the hot reservoir is large enough
so as not to be affected in the temperature by the extraction
of heat by the system or the surrounding
•Whereas a heat engine absorbs a net quantity of heat and
produces a net work output, a heat pump absorbs a net
quantity of work and produces a net heat output
•The primary function of a heat pump is to maintain a high
temperature in the hot reservoir
•The primary function of a refrigerator is to maintain a low
temperature in the cold reservoir
49
Consequences of the Second Law
•let Qh=Q1=heat supplied during the cycle,
Qc=Q2=Heat rejected during the cycle
•2
nd
law states that: work output, W=Qh-Qc=Q1-Q2
•The cycle efficiency is the ratio of the heat
converted into useful work to the quantity of heat
supplied
•The quantity Qh and Qc may be expressed per unit
mass or per mass flow rate or per unit time or per
cycle
1
2
1
21
1
1
Q
Q
Q
QQ
Q
w
50
•let Qh=Q1=heat supplied during the cycle,
Qc=Q2=Heat rejected during the cycle
•2
nd
law states that: work output, W=Qh-Qc=Q1-Q2
•The cycle efficiency is the ratio of the heat
converted into useful work to the quantity of heat
supplied
•The quantity Qh and Qc may be expressed per unit
mass or per mass flow rate or per unit time or per
cycle
1
2
1
21
1
1
Q
Q
Q
Q
w
50
Lord Kelvin and Max Planck
The first scientists to state the
Second Law of Thermodynamics
1.Kelvin – Planck Statement of the Second laws of Thermodynamics:
A cyclical heat engine cannot operate while receiving heat from
a single thermal-energy reservoir, i.e Heat engines must always
reject heat to a thermal-energy reservoir
In other words, maximum possible efficiency is less than 100%
A heat engine that violates
the Kelvin-Planck
Statement of Second law
51
The first scientists to state the
Second Law of Thermodynamics
1.Kelvin – Planck Statement of the Second laws of Thermodynamics:
A cyclical heat engine cannot operate while receiving heat from
a single thermal-energy reservoir, i.e Heat engines must always
reject heat to a thermal-energy reservoir
In other words, maximum possible efficiency is less than 100%
A heat engine that violates
the Kelvin-Planck
Statement of Second law
51
2.Clausius Statement of the second law
No cyclical device can cause heat to transfer from thermal-energy
reservoirs at low temperature to reservoirs at high temperature with no
other effect
Work is therefore required to manifest the effect
It is impossible to construct a system which will operate in a
cycle, extract heat from a reservoir and do an equivalent
amount of work on the surroundings
A machine which will produce work continuously while
exchanging heat with only a single reservoir is known as
Perpetual motion machine. This would contradict the second
law
PMM is a device that violets the first or second law of
thermodynamics
52
No cyclical device can cause heat to transfer from thermal-energy
reservoirs at low temperature to reservoirs at high temperature with no
other effect
Work is therefore required to manifest the effect
It is impossible to construct a system which will operate in a
cycle, extract heat from a reservoir and do an equivalent
amount of work on the surroundings
A machine which will produce work continuously while
exchanging heat with only a single reservoir is known as
Perpetual motion machine. This would contradict the second
law
PMM is a device that violets the first or second law of
thermodynamics
52
3.The third statement of the Second law
All Reversible engines operating
between the same two reservoirs have
the same efficiency
η=f(T
h, T
c)
Thermodynamic Temperature scale.
(read and understand)
53
All Reversible engines operating
between the same two reservoirs have
the same efficiency
η=f(T
h, T
c)
Thermodynamic Temperature scale.
(read and understand)
53
Reversible processes
•A reversible process is one in which everything involved with the process (system
and surroundings) can be returned to its original condition after the process has
been executed.
•This implies that for a non flow process to be reversible it must be an infinitely
slow process. As a consequence of the second Law of Thermodynamics the
reversibility concept can be given a more precise form which is more relevant to
flow processes than non-flow closed systems.
•The reversible process obviously has to be a quasi-equilibrium process; additional
requirements are:
•No friction is involved in the process.
•Heat transfer occurs due to an infinitesimal temperature difference only.
•Unrestrained expansion does not occur.
54
•A reversible process is one in which everything involved with the process (system
and surroundings) can be returned to its original condition after the process has
been executed.
•This implies that for a non flow process to be reversible it must be an infinitely
slow process. As a consequence of the second Law of Thermodynamics the
reversibility concept can be given a more precise form which is more relevant to
flow processes than non-flow closed systems.
•The reversible process obviously has to be a quasi-equilibrium process; additional
requirements are:
•No friction is involved in the process.
•Heat transfer occurs due to an infinitesimal temperature difference only.
•Unrestrained expansion does not occur.
54
Irreversible processes
•A process which involves the spontaneous movement of a system from a non-
equilibrium state to an equilibrium state is spontaneous or natural process.
•As such a process cannot be reversed without the application of an external
agency, thus, it is an irreversible process.
•Feature found in irreversible process
1. Friction
2. unstrained expansion
3.Heat transfer across a finite temperature difference
4. Rapid mixing
5. Rapid chemical reaction
6. Other factors that create losses
55
•A process which involves the spontaneous movement of a system from a non-
equilibrium state to an equilibrium state is spontaneous or natural process.
•As such a process cannot be reversed without the application of an external
agency, thus, it is an irreversible process.
•Feature found in irreversible process
1. Friction
2. unstrained expansion
3.Heat transfer across a finite temperature difference
4. Rapid mixing
5. Rapid chemical reaction
6. Other factors that create losses
55
b. Reversible and irreversible processes
•Reversible process represented by solid lines- Both the fluid and surrounding can
always be restored to their original state
•Irreversible process represented by dotted lines
General statement:
•When a fluid undergoes a reversible process, both the fluid and its surroundings can always
be restored to their original state.
Reversible process
(Fig. a) is always
represented by a
solid line
Irreversible process
(Fig. b) is always
represented by a dotted
line
Figure : Reversible and irreversible processes
56
•Reversible process represented by solid lines- Both the fluid and surrounding can
always be restored to their original state
•Irreversible process represented by dotted lines
General statement:
•When a fluid undergoes a reversible process, both the fluid and its surroundings can always
be restored to their original state.
Reversible process
(Fig. a) is always
represented by a
solid line
Irreversible process
(Fig. b) is always
represented by a dotted
line
Figure : Reversible and irreversible processes
56
Entropy
•Entropy is a measure of the unavailability of thermal energy
to do work in a closed system.
•Heat transferred during a process may affect work transfer
and also the end state after the process
•Theoretically, the amount of heat transferred during a
process is determined by calculations if it is transferred
reversibly.
•Let us plot a graph of suitable axes so that the area
underneath a process plotted on the graph gave the quantity
of heat transferred reversibly during the process. Such a
graph would be as useful as a pressure-volume graph in
determining the quantity of work done during a reversible
process
pdVdWW
57
•Entropy is a measure of the unavailability of thermal energy
to do work in a closed system.
•Heat transferred during a process may affect work transfer
and also the end state after the process
•Theoretically, the amount of heat transferred during a
process is determined by calculations if it is transferred
reversibly.
•Let us plot a graph of suitable axes so that the area
underneath a process plotted on the graph gave the quantity
of heat transferred reversibly during the process. Such a
graph would be as useful as a pressure-volume graph in
determining the quantity of work done during a reversible
process
pdVdWW
57
Cont.
•Let on this new graph one axis be absolute
temperature (T) and the other axis be some
new function (s) as shown. The absolute
temperature is selected because it is closely
related to energy level of a substance, notable
internal energy and enthalpy
•Figure for T-s
58
•Let on this new graph one axis be absolute
temperature (T) and the other axis be some
new function (s) as shown. The absolute
temperature is selected because it is closely
related to energy level of a substance, notable
internal energy and enthalpy
•Figure for T-s
58
Cont.
•Remember that heat is that energy transfer
which is associated with temperature
difference. Further the internal energy of a
substance is zero at the absolute zero
temperature. Consider an elemental strip of
the T-s graph of thickness ds at a temperature
T during a process between states 1 and 2.
thus ds=elemental change in s at a point of
graph where temperature=T and entropy=s
59
•Remember that heat is that energy transfer
which is associated with temperature
difference. Further the internal energy of a
substance is zero at the absolute zero
temperature. Consider an elemental strip of
the T-s graph of thickness ds at a temperature
T during a process between states 1 and 2.
thus ds=elemental change in s at a point of
graph where temperature=T and entropy=s
59
Cont.
Then the elemental heat transfer reversibly=area of
elemental strip
dQ=Tds
From this,
Heat transfer reversibly from state 1 to 2=total area
of the graph under points 1,2
i.e
In limit as ds goes to 0
Heat transferred reversibly from state 1 to 2
2
1
2
1
ss
ss
TdsdQ
2
1
2
1
2
1
s
s
rev
s
s
rev
TdsQ
TdsdQ
60
Then the elemental heat transfer reversibly=area of
elemental strip
dQ=Tds
From this,
Heat transfer reversibly from state 1 to 2=total area
of the graph under points 1,2
i.e
In limit as ds goes to 0
Heat transferred reversibly from state 1 to 2
2
1
2
1
ss
ss
TdsdQ
2
1
2
1
2
1
s
s
rev
s
s
rev
TdsQ
TdsdQ
60
Entropy
Differentianting, dQrev=Tds or
Entropy (s) is a system property as a
consequence of the second law,
Useful for conducting first and second law analysis of
thermodynamic devices and systems
e.g dq=du+pdv
Tds=du+pdv
if ds=(dq/T)
reversible
Tds=du+pdv
if h=u+pvdh=du+pdv+vdp Tds=du+pdv
Tds=dh-vdp:
reversibleT
dq
ds
61
Differentianting, dQrev=Tds or
Entropy (s) is a system property as a
consequence of the second law,
Useful for conducting first and second law analysis of
thermodynamic devices and systems
e.g dq=du+pdv
Tds=du+pdv
if ds=(dq/T)
reversible
Tds=du+pdv
if h=u+pvdh=du+pdv+vdp Tds=du+pdv
Tds=dh-vdp:
reversibleT
dq
ds
61
Entropy
Also
Hence
•Similarly, Eq. Tds=du+pdv is transformed using the
substitutions P = RTv and du =cvdT to yield
•To evaluate the entropy change in going from state 1
to state 2 state, we integrate above Eqs., that is
v = RT/P dh =cpdT
62
Also
Hence
•Similarly, Eq. Tds=du+pdv is transformed using the
substitutions P = RTv and du =cvdT to yield
•To evaluate the entropy change in going from state 1
to state 2 state, we integrate above Eqs., that is
v = RT/P dh =cpdT
62
Entropy
•The second term on the right-hand side of each of
these expressions can be easily evaluated since R is a
constant, and so
•In many engineering applications, an average value
of cp (or cv) over the temperature range of interest
can be used to evaluate s2-s1 with reasonable
accuracy:
63
•The second term on the right-hand side of each of
these expressions can be easily evaluated since R is a
constant, and so
•In many engineering applications, an average value
of cp (or cv) over the temperature range of interest
can be used to evaluate s2-s1 with reasonable
accuracy:
63
•If heat is received then Qrev is positive and
the substance entropy has increased
•If heat is rejected, Qrev is negative and the
entropy of the substance has decreases
•Thus +ve and –ve changes of entropy show
whether heat has been received or rejected
during a process under consideration
64
the substance entropy has increased
•If heat is rejected, Qrev is negative and the
entropy of the substance has decreases
•Thus +ve and –ve changes of entropy show
whether heat has been received or rejected
during a process under consideration
64
Non flow Processes
•Meaning of non-flow processes;- All processes associated with closed systems are called non-
flow processes-.
•NFEE; Due to first law of thermodynamics,
Q-W=u2-u1
dQ-dW=du
•Isothermal process: a process undergone of constant temperature
Q- ∫pdv=u2-u1
dq-pdv=du
•Since temperature is constant, there is a definite relationship between p and v. Thus, for the
given end states, ∫pdv has a definite value independent of the process. Since heat is not a
property, Q is dependent on the path of the process. Assuming no frictional effects and no work
is done by mechanical agitation the isothermal process may be termed reversible. It follows
that there is only one possible reversible isothermal process between the two equilibrium end
states which will satisfy the equation, Q=∫pdv + (u2-u1)
•Estimating the amount of heat transferred for a perfect gas,
•W= ∫pdv=pv∫ (dv/v)=pvln(v2/v1)=pvln(p1/p2)=-pvln(p2/p1)(Joules Law).
•Also for an isothermal process, du=cvdT=0
•Therefore, NFEE: Q= Q=p1vln(v2/v1)=p2v2ln(v2/v1)
65
•Meaning of non-flow processes;- All processes associated with closed systems are called non-
flow processes-.
•NFEE; Due to first law of thermodynamics,
Q-W=u2-u1
dQ-dW=du
•Isothermal process: a process undergone of constant temperature
Q- ∫pdv=u2-u1
dq-pdv=du
•Since temperature is constant, there is a definite relationship between p and v. Thus, for the
given end states, ∫pdv has a definite value independent of the process. Since heat is not a
property, Q is dependent on the path of the process. Assuming no frictional effects and no work
is done by mechanical agitation the isothermal process may be termed reversible. It follows
that there is only one possible reversible isothermal process between the two equilibrium end
states which will satisfy the equation, Q=∫pdv + (u2-u1)
•Estimating the amount of heat transferred for a perfect gas,
•W= ∫pdv=pv∫ (dv/v)=pvln(v2/v1)=pvln(p1/p2)=-pvln(p2/p1)(Joules Law).
•Also for an isothermal process, du=cvdT=0
•Therefore, NFEE: Q= Q=p1vln(v2/v1)=p2v2ln(v2/v1)
65
•Constant Pressure (Isobaric) process: A process undergone at constant pressure
•A process undergone at constant pressure. It is clear that for a piston cylinder
device enclosing a fluid, it the fluid compressed at constant pressure heat must be
extracted; and conversely if the fluid expands at constant pressure heat must be
added.
•If the process is carried out infinitely slowly without internal frictional effects and if
no work is done by mechanical agitation, the forces may be termed reversible. For
constant pressure process, NFEE becomes
•Q= ∫pdv + (u2-u1)
•=p (v2-v1)+ (u2-u1)
•=(pv2+u2)-(pv1+u1)
• where enthalpy, h=u + pv
•Or Q= (h2-h1)
•Ie Heat supplied = Change of enthalpy
•In the differential form, dQ=dh
•Since enthalpy is a property, the change of enthalpy h2-h1 is independent of the
path of the process. Since heat is not a property, the quantity Q is dependent on
the path of the forces. Consequently, there is only one possible reversible
constant pressure process between two specified end state which will satisfy
equation Q=h2-h1. It is now possible to predict the heat quantity transferred
66
•A process undergone at constant pressure. It is clear that for a piston cylinder
device enclosing a fluid, it the fluid compressed at constant pressure heat must be
extracted; and conversely if the fluid expands at constant pressure heat must be
added.
•If the process is carried out infinitely slowly without internal frictional effects and if
no work is done by mechanical agitation, the forces may be termed reversible. For
constant pressure process, NFEE becomes
•Q= ∫pdv + (u2-u1)
•=p (v2-v1)+ (u2-u1)
•=(pv2+u2)-(pv1+u1)
• where enthalpy, h=u + pv
•Or Q= (h2-h1)
•Ie Heat supplied = Change of enthalpy
•In the differential form, dQ=dh
•Since enthalpy is a property, the change of enthalpy h2-h1 is independent of the
path of the process. Since heat is not a property, the quantity Q is dependent on
the path of the forces. Consequently, there is only one possible reversible
constant pressure process between two specified end state which will satisfy
equation Q=h2-h1. It is now possible to predict the heat quantity transferred
66
•Constant volume (Isochoric) process: A process done at constant
volume. Assuming no work is done on the stationery fluid by
mechanical agitation a Constant volume process has work term =O.
Thus,
•NFEE; Q=u2-u1 or Heat supplied=Change of internal energy.
dq-dw=du
dq-pdv=du
dq=du ( ∫pdv as dv=0)
•Assuming the process is an infinitely slow one, we may regard it as a
reversible process and the differential form of NFEE becomes: dQ=du to
be applicable to any infinitesimal part of the system.
•Since internal energy is a property, u2-u1 is independent of the path of
the process. Since heat is not a property, the quantity Q is dependent
on the path of the process. Hence there is only one possible constant
volume process joining the specified end states which satisfy the
equation: Q=u2-u1 from which the quantity of heat transferred during
the process may be estimated.
67
volume. Assuming no work is done on the stationery fluid by
mechanical agitation a Constant volume process has work term =O.
Thus,
•NFEE; Q=u2-u1 or Heat supplied=Change of internal energy.
dq-dw=du
dq-pdv=du
dq=du ( ∫pdv as dv=0)
•Assuming the process is an infinitely slow one, we may regard it as a
reversible process and the differential form of NFEE becomes: dQ=du to
be applicable to any infinitesimal part of the system.
•Since internal energy is a property, u2-u1 is independent of the path of
the process. Since heat is not a property, the quantity Q is dependent
on the path of the process. Hence there is only one possible constant
volume process joining the specified end states which satisfy the
equation: Q=u2-u1 from which the quantity of heat transferred during
the process may be estimated.
67
•Adiabatic process: A process undergone without the transfer of heat
across the system boundary
dq=0
-dw=du
Since internal energy is a property, Change of internal energy u2-u1 is
independent of the path of the process. Since work is not a property, the
quantity W is dependent on the path of the process. Hence there is only
one possible adiabatic process joining two specified equilibrium end
states which will satisfy the equation W=u2-u1 from which the work done
or/by the system may be estimated.
68
across the system boundary
dq=0
-dw=du
Since internal energy is a property, Change of internal energy u2-u1 is
independent of the path of the process. Since work is not a property, the
quantity W is dependent on the path of the process. Hence there is only
one possible adiabatic process joining two specified equilibrium end
states which will satisfy the equation W=u2-u1 from which the work done
or/by the system may be estimated.
68
•Polytropic process: A general non flow process of
the form pvn=constant, where p=pressure,
v=specific volume and n=index of compression or
expression. The previous four processes are special
cases of this general process.
When n=0, we have a constant pressure process
When n= α, we have a constant volume process
When n=1, we have an isothermal process
•Work of expansion done for unit mass for a
reversible process between two end states can be
evaluated thus:
•For a perfect gas undergoing an adiabatic process,
Q=0, the work done reduces to:
•W=PV-PV but pv=RT
•Therefore, when a perfect gas undergoes an
adiabatic process the index of compression or
expansion is equal to the ratio of specific heats at
constant pressure and constant volume.
11
1
1
1
22111122
1
111
1
222
2
1
111
2
1
2211
2
1
n
vpvp
n
vpvp
vvpvvp
n
v
n
vp
v
dv
c
vpvpcpvwherepdvW
nnnn
v
v
n
n
n
nnv
v
p
vvvp
vpv
v
c
c
n
cnccc
ccRwhereTTcnTTR
TTcuuandRTpvbut
uu
n
vpvp
uupdvQbecomesNFEE
n
vpvp
W
)()1()(
)(
)(
1
0
)(
1
1212
1212
12
2211
12
1
1
2211
69
the form pvn=constant, where p=pressure,
v=specific volume and n=index of compression or
expression. The previous four processes are special
cases of this general process.
When n=0, we have a constant pressure process
When n= α, we have a constant volume process
When n=1, we have an isothermal process
•Work of expansion done for unit mass for a
reversible process between two end states can be
evaluated thus:
•For a perfect gas undergoing an adiabatic process,
Q=0, the work done reduces to:
•W=PV-PV but pv=RT
•Therefore, when a perfect gas undergoes an
adiabatic process the index of compression or
expansion is equal to the ratio of specific heats at
constant pressure and constant volume.
11
1
1
1
22111122
1
111
1
222
2
1
111
2
1
2211
2
1
n
vpvp
n
vpvp
vvpvvp
n
v
n
vp
v
dv
c
vpvpcpvwherepdvW
nnnn
v
v
n
n
n
nnv
v
p
vvvp
vpv
v
c
c
n
cnccc
ccRwhereTTcnTTR
TTcuuandRTpvbut
uu
n
vpvp
uupdvQbecomesNFEE
n
vpvp
W
)()1()(
)(
)(
1
0
)(
1
1212
1212
12
2211
12
1
1
2211
69
Some Properties of a Perfect gas
–Full treatment latter
–pv=RT, R= the gas constant=R
o
/M where R
o
is the universal
gas constant=8.314kJ/kmol-K
–c
p-c
v=R where c
p and c
v are gas specific heat capacities at
constant pressure and volume respectively
•Isothermal Process
For a reversible process
dq=Tds, s=S/m; m=mass, s=entropy
q=T(s
2
-s
1
)
70
–Full treatment latter
–pv=RT, R= the gas constant=R
o
/M where R
o
is the universal
gas constant=8.314kJ/kmol-K
–c
p-c
v=R where c
p and c
v are gas specific heat capacities at
constant pressure and volume respectively
•Isothermal Process
For a reversible process
dq=Tds, s=S/m; m=mass, s=entropy
q=T(s
2
-s
1
)
70
Flow Process
•2.2 A Flow process: a process undergone by an open system.
• Steady flow process: a flow process in which the fluid mass
flow rate is uniform (constant); the work and heat energy
transfers across the boundary occurring also at constant
rates; and the values of the properties of the fluid at any
point in the system are independent of time (steady).
•Unsteady (non-steady) flow process: a flow process in which
the fluid mass in the system, the energy E contained in the
system and the fluid properties at inlet and or outlet, all vary
with time. Thus, the general problem is that for E to be
calculated, the values of the fluid properties at any instant of
time throughout the system must be known
71
•2.2 A Flow process: a process undergone by an open system.
• Steady flow process: a flow process in which the fluid mass
flow rate is uniform (constant); the work and heat energy
transfers across the boundary occurring also at constant
rates; and the values of the properties of the fluid at any
point in the system are independent of time (steady).
•Unsteady (non-steady) flow process: a flow process in which
the fluid mass in the system, the energy E contained in the
system and the fluid properties at inlet and or outlet, all vary
with time. Thus, the general problem is that for E to be
calculated, the values of the fluid properties at any instant of
time throughout the system must be known
71
Prof. G.R. John-March 2012 72
•The diagram
Apparatus for a flow process
72
•The diagram
Apparatus for a flow process
72
•For this process, we may now apply the first
law of thermodynamics or principal of energy
conservation as follows:
••At station 1 or initial instant of time, Internal
energy of the imaginary closed system
•=E1 + (U1 + ½C² + gz1)
••At station 2 or final instant of time, Internal
energy of the imaginary closed system
•=E2 + (u2 + ½C² + gz2)
•Thus, the change of internal energy of the
imaginary closed system during the elemental
non-flow process is
•But E2=E1 =constant for a steady flow process
•Total work done during the elemental process
•recall: compression at 1, expansion at 2
•Since the sum of the net heat and work
supplied to the system is equal to the change
of internal energy.
[Recall NFEE: Q=W+(u2-u1)
)]()(2/1)[(
12
2
1
2
21212
zzgccuuEE
)]()(2/1)[(
12
2
1
2
212
zzgccuu
2211
vpvpdw
)]()(2/1)()(
12
2
1
2
2122211 zzgccuuvpvpdwdQ
73
law of thermodynamics or principal of energy
conservation as follows:
••At station 1 or initial instant of time, Internal
energy of the imaginary closed system
•=E1 + (U1 + ½C² + gz1)
••At station 2 or final instant of time, Internal
energy of the imaginary closed system
•=E2 + (u2 + ½C² + gz2)
•Thus, the change of internal energy of the
imaginary closed system during the elemental
non-flow process is
•But E2=E1 =constant for a steady flow process
•Total work done during the elemental process
•recall: compression at 1, expansion at 2
•Since the sum of the net heat and work
supplied to the system is equal to the change
of internal energy.
[Recall NFEE: Q=W+(u2-u1)
)]()(2/1)[(
12
2
1
2
21212
zzgccuuEE
)]()(2/1)[(
12
2
1
2
212
zzgccuu
2211
vpvpdw
)]()(2/1)()(
12
2
1
2
2122211 zzgccuuvpvpdwdQ
73
•Recall: enthalpy h=u+pv and so
•Summing up a succession of such
elemental processes, we obtain
the SFEE:
•(Sum or Difference of net heat
and work transferred for unit
mass of fluid in a steady flow
process)= (Change of specific
enthalpy) + (Change of specific
kinetic energy) + (Change of
specific potential energy)
)]()(2/1)[(
)]()(2/1)()(
12
2
1
2
212
12
2
1
2
2111222
zzgcchhdwdQ
zzgccvpuvpudwdQ
)]()(2/1)[(
12
2
1
2
212 zzgcchhwQ
74
Steady flow energy equation
dq-dw=∆h+∆KE+∆PE
•Summing up a succession of such
elemental processes, we obtain
the SFEE:
•(Sum or Difference of net heat
and work transferred for unit
mass of fluid in a steady flow
process)= (Change of specific
enthalpy) + (Change of specific
kinetic energy) + (Change of
specific potential energy)
)]()(2/1)[(
)]()(2/1)()(
12
2
1
2
212
12
2
1
2
2111222
zzgcchhdwdQ
zzgccvpuvpudwdQ
)]()(2/1)[(
12
2
1
2
212 zzgcchhwQ
74
Steady flow energy equation
dq-dw=∆h+∆KE+∆PE
Applications of SFEE
•We will assume that the fluid properties are the uniform
over the cross sections at entry to and exit from the
system.
•There are many cases of systems containing gases in which
the potential energy term g( z2-z1) is negligible.
•There are also examples in which velocity changes are
small and hence the kinetic energy can be neglected
compared with enthalpy changes.
Typical engineering systems can be listed below:
•Boilers are steam generators, condensers.
•Heaters and coolers.
•Turbo machinery e.g. axial flow (rotary) compressors, gas
turbine, steam turbine.
•Reciprocating machinery e.g. reciprocating air compressor,
reciprocating internal combustion engines.
•Flow devices e.g. diffusers, nozzles, throttles.
75
•We will assume that the fluid properties are the uniform
over the cross sections at entry to and exit from the
system.
•There are many cases of systems containing gases in which
the potential energy term g( z2-z1) is negligible.
•There are also examples in which velocity changes are
small and hence the kinetic energy can be neglected
compared with enthalpy changes.
Typical engineering systems can be listed below:
•Boilers are steam generators, condensers.
•Heaters and coolers.
•Turbo machinery e.g. axial flow (rotary) compressors, gas
turbine, steam turbine.
•Reciprocating machinery e.g. reciprocating air compressor,
reciprocating internal combustion engines.
•Flow devices e.g. diffusers, nozzles, throttles.
75
Typical Engineering Systems Consideration
(a) Boiler or Steam generator/Condenser
•Layout is shown. Heat is supplied by the combustion of a fuel
in a furnace. Furnace gases then flow through exchanger tubes
located at the lower portion of a boiler. Water is fed by pump
and on exchanging heat the vaporized steam escapes through
an outlet pipe at the top of the water level. So, there are two
systems in the boiler: the water steam system and the furnace
gas system which supplies the heat.
Analysis:
• There is no work done in a boiler
• Potential and kinetic energy changes are small compared with
enthalpy changes
• We’ve to know the states of water at inlet 1 and of the steam
at outlet 2.
W=0, ∆KE=0, PE=0
76
(a) Boiler or Steam generator/Condenser
•Layout is shown. Heat is supplied by the combustion of a fuel
in a furnace. Furnace gases then flow through exchanger tubes
located at the lower portion of a boiler. Water is fed by pump
and on exchanging heat the vaporized steam escapes through
an outlet pipe at the top of the water level. So, there are two
systems in the boiler: the water steam system and the furnace
gas system which supplies the heat.
Analysis:
• There is no work done in a boiler
• Potential and kinetic energy changes are small compared with
enthalpy changes
• We’ve to know the states of water at inlet 1 and of the steam
at outlet 2.
W=0, ∆KE=0, PE=0
76
•Then the SFEE reduces to:
•Q= h2 – h1, change of specific enthalpy
•Total heat transfer
•Q= m (h2 – h1), m= mass flow rate of steam
•h1=hf, liquid enthalpy of feed water
•h2= enthalpy of steam produced
•The evaporative capacity of the boiler is the quantity of steam it can
produce per hour. This boiler capacity depends upon its design quality
and type of fuel and furnace design. Since boilers can operate at different
running conditions, it is necessary to compare them by converting
capacities to some common standard.
•The equivalent evaporation of a boiler from and at 100ºc, is its
evaporative capacity per enthalpy of evaporation from and at 100ºc i.e. ni
(h2 –h1)/22256,9 kg of water.
77
•Q= h2 – h1, change of specific enthalpy
•Total heat transfer
•Q= m (h2 – h1), m= mass flow rate of steam
•h1=hf, liquid enthalpy of feed water
•h2= enthalpy of steam produced
•The evaporative capacity of the boiler is the quantity of steam it can
produce per hour. This boiler capacity depends upon its design quality
and type of fuel and furnace design. Since boilers can operate at different
running conditions, it is necessary to compare them by converting
capacities to some common standard.
•The equivalent evaporation of a boiler from and at 100ºc, is its
evaporative capacity per enthalpy of evaporation from and at 100ºc i.e. ni
(h2 –h1)/22256,9 kg of water.
77
Prof. G.R. John-March 2012 78
•Steady flow energy equation
dq-dw=∆h+∆KE+∆PE
•Turbine
Neglect ∆PE
Gas turbine
78
•Steady flow energy equation
dq-dw=∆h+∆KE+∆PE
•Turbine
Neglect ∆PE
Gas turbine
78
Prof. G.R. John-March 2012 79
Cont - FLOW PROCESSES
•Nozzle and Diffuser
∆Q=0, ∆W=0
∆KE=∆h
•Turbine and Compressors
∆Q=0, ∆PE=0, ∆KE=0 ∆W=∆h
•Throttling
∆Q=0, ∆W=0, ∆KE=0, ∆PE=0
∆h=0→ h
2-h
1=0 for throttling
Nozzle and Diffuser
79
Cont - FLOW PROCESSES
•Nozzle and Diffuser
∆Q=0, ∆W=0
∆KE=∆h
•Turbine and Compressors
∆Q=0, ∆PE=0, ∆KE=0 ∆W=∆h
•Throttling
∆Q=0, ∆W=0, ∆KE=0, ∆PE=0
∆h=0→ h
2-h
1=0 for throttling
Nozzle and Diffuser
79
CONCEPT OF STATE RELATIONSHIPS
State Principle
•An important concept in thermodynamics is the
state principle:
•In dealing with a simple compressible substance,
the thermodynamic state is completely defined by
specifying two independent intensive properties.
•The state principle allows us to define state
relationships among the various thermodynamic
properties.
80
State Principle
•An important concept in thermodynamics is the
state principle:
•In dealing with a simple compressible substance,
the thermodynamic state is completely defined by
specifying two independent intensive properties.
•The state principle allows us to define state
relationships among the various thermodynamic
properties.
80
IDEAL GASES:
Processes in P–v –T Space
•Plotting thermodynamic processes on P–v or
other thermodynamic property coordinates is
very useful in analyzing many thermal systems.
This section introduce this topic by illustrating
common processes on P–v and T–v
coordinates, attention is to ideal gases.
•We begin by examining P–v coordinates.
Rearranging the ideal-gas equation of state
•to a form in which P is a function of v yields
the hyperbolic relationship 81
Processes in P–v –T Space
•Plotting thermodynamic processes on P–v or
other thermodynamic property coordinates is
very useful in analyzing many thermal systems.
This section introduce this topic by illustrating
common processes on P–v and T–v
coordinates, attention is to ideal gases.
•We begin by examining P–v coordinates.
Rearranging the ideal-gas equation of state
•to a form in which P is a function of v yields
the hyperbolic relationship 81
•By considering the temperature to be a fixed parameter, can be
used to create a family of hyperbolas for various values of T, as shown in below.
Increasing temperature moves the isotherms further from the origin.
•The usefulness of graphs such as Fig. below is that one can immediately
visualize how properties must vary for a particular thermodynamic process.
•For example, consider the constant-pressure expansion process shown in
below, where the initial and final states are designated as points 1 and 2,
respectively. Knowing the arrangement of constant-temperature lines
allows us to see that the temperature must increase in the process 1–2.
•For the values given, the temperature increases from 300 to 600 K. The
important point here, however, is not this quantitative result, but the
qualitative information available from plotting processes on P–v
coordinates.
•Also shown in Fig. below is a constant-volume process (assuming that we
are dealing with a system of fixed mass).
•In going from state 2 to state 3 at constant volume, we immediately see that
both the temperature and pressure must fall.
82
used to create a family of hyperbolas for various values of T, as shown in below.
Increasing temperature moves the isotherms further from the origin.
•The usefulness of graphs such as Fig. below is that one can immediately
visualize how properties must vary for a particular thermodynamic process.
•For example, consider the constant-pressure expansion process shown in
below, where the initial and final states are designated as points 1 and 2,
respectively. Knowing the arrangement of constant-temperature lines
allows us to see that the temperature must increase in the process 1–2.
•For the values given, the temperature increases from 300 to 600 K. The
important point here, however, is not this quantitative result, but the
qualitative information available from plotting processes on P–v
coordinates.
•Also shown in Fig. below is a constant-volume process (assuming that we
are dealing with a system of fixed mass).
•In going from state 2 to state 3 at constant volume, we immediately see that
both the temperature and pressure must fall.
82
FIGURE Constant-pressure (1–2) and constant-volume (2–3) processes are shown on Pv
coordinates for an ideal gas (N2). Lines of constant temperature are hyperbolic, following the
ideal-gas equation of state, P= (RT)v .
83
coordinates for an ideal gas (N2). Lines of constant temperature are hyperbolic, following the
ideal-gas equation of state, P= (RT)v .
83
•Treating pressure as a fixed parameter, this
relationship yields straight lines with slopes of PR.
•Higher pressures result in steeper slopes.
•Figure below illustrates the ideal-gas T–v
relationship. Consider a constant temperature
compression process going from state 1 to state 2.
•For this process, we immediately see from the graph
that the pressure must increase.
•Also shown on Fig. below is a constant- (specific)
volume process going from state 2 to state 3 for
conditions of decreasing temperature.
•Again, we immediately see that the pressure falls
during this process.
84
relationship yields straight lines with slopes of PR.
•Higher pressures result in steeper slopes.
•Figure below illustrates the ideal-gas T–v
relationship. Consider a constant temperature
compression process going from state 1 to state 2.
•For this process, we immediately see from the graph
that the pressure must increase.
•Also shown on Fig. below is a constant- (specific)
volume process going from state 2 to state 3 for
conditions of decreasing temperature.
•Again, we immediately see that the pressure falls
during this process.
84
FIGURE 2.8 Constant-temperature (1–2) and constant-volume (2–3) processes are shown on Tv
coordinates for an ideal gas (N2). Lines of constant pressure are straight lines, following the
ideal-gas equation of state, T = (PR)v.
85
coordinates for an ideal gas (N2). Lines of constant pressure are straight lines, following the
ideal-gas equation of state, T = (PR)v.
85
example
•An ideal gas system undergoes a
thermodynamic cycle composed of the
•following processes:
•1–2: constant-pressure expansion,
•2–3: constant-temperature expansion,
•3–4: constant-volume return to the state-1
temperature, and
•4–1: constant-temperature compression.
•Sketch these processes (a) on P–v coordinates
and (b) on T–v coordinates.
86
•An ideal gas system undergoes a
thermodynamic cycle composed of the
•following processes:
•1–2: constant-pressure expansion,
•2–3: constant-temperature expansion,
•3–4: constant-volume return to the state-1
temperature, and
•4–1: constant-temperature compression.
•Sketch these processes (a) on P–v coordinates
and (b) on T–v coordinates.
86
87
•Looking at the sketches of Example, state whether
the pressure P, temperature T, and specific volume
v increase, decrease, or remain the same for each of
the four processes.
•(Answer: 1–2: P constant, T increases, v increases; 2–
3: P decreases,
•T constant, v increases; 3–4: P decreases, T
decreases, v constant; 4–1:
•P increases, T constant, v decreases)
88
the pressure P, temperature T, and specific volume
v increase, decrease, or remain the same for each of
the four processes.
•(Answer: 1–2: P constant, T increases, v increases; 2–
3: P decreases,
•T constant, v increases; 3–4: P decreases, T
decreases, v constant; 4–1:
•P increases, T constant, v decreases)
88
Example
•Determine the changes in the mass-specific
enthalpy and the mass-specific
•internal energy for air for a process that starts
at 300 K and ends at 1000 K.
•Also show that dh-du= RT, and compare a
numerical evaluation of this
•with tabulated data.
89
•Determine the changes in the mass-specific
enthalpy and the mass-specific
•internal energy for air for a process that starts
at 300 K and ends at 1000 K.
•Also show that dh-du= RT, and compare a
numerical evaluation of this
•with tabulated data.
89
90
•Assumption
•Air behaves as a single-component ideal gas.
•Analysis From Table, we obtain the following
values for h and u:
91
•Air behaves as a single-component ideal gas.
•Analysis From Table, we obtain the following
values for h and u:
91
92
93
Ideal-Gas Isentropic Process Relationships
•For an isentropic process, a process in which
the initial and final entropies are identical,
Eqs. and
•can be used to develop some useful
engineering relationships. Setting s2 - s1 equal
to zero
•Regarding
94
•For an isentropic process, a process in which
the initial and final entropies are identical,
Eqs. and
•can be used to develop some useful
engineering relationships. Setting s2 - s1 equal
to zero
•Regarding
94
•This equation can be simplified by introducing
the specific heat ratio γ and
•the fact that cp - cv = R
95
the specific heat ratio γ and
•the fact that cp - cv = R
95
•Regarding using similar
manipulations, we obtain
•Applying the ideal-gas equation of state
Or
obtain a third and final ideal-gas isentropic-
process relationship
96
manipulations, we obtain
•Applying the ideal-gas equation of state
Or
obtain a third and final ideal-gas isentropic-
process relationship
96
Processes in T–s and P–v Space
•With the addition of entropy, we now have
four properties (P, v, T, and s) to define states
and describe processes. We now investigate
T–s and P–v space
Ideal-Gas Isentropic-Process Relationships
97
•With the addition of entropy, we now have
four properties (P, v, T, and s) to define states
and describe processes. We now investigate
T–s and P–v space
Ideal-Gas Isentropic-Process Relationships
97
•T–s space, isothermal and isentropic processes are by definition horizontal
•and vertical lines, respectively.
•Less obvious are the lines of constant pressure and specific volume shown in Fig
below
•On a Ts diagram, lines of constant pressure (isobars) and lines of constant
•specific volume (isochors) both exhibit positive slopes. At a particular state, a
•constant-volume line has a greater slope than a constant-pressure line.
98
•and vertical lines, respectively.
•Less obvious are the lines of constant pressure and specific volume shown in Fig
below
•On a Ts diagram, lines of constant pressure (isobars) and lines of constant
•specific volume (isochors) both exhibit positive slopes. At a particular state, a
•constant-volume line has a greater slope than a constant-pressure line.
98
•In P–v space, constant-pressure and constant-volume processes are, again
by definition, horizontal and vertical lines, whereas constant-entropy and
constant-temperature processes follow curved paths as shown in below.
•On a Pv diagram, lines of constant temperature (isotherms) and lines of
constant entropy (isentropes) have negative slopes. At a particular state,
an isentrope is steeper (has a greater negative slope) than an isotherm.
99
by definition, horizontal and vertical lines, whereas constant-entropy and
constant-temperature processes follow curved paths as shown in below.
•On a Pv diagram, lines of constant temperature (isotherms) and lines of
constant entropy (isentropes) have negative slopes. At a particular state,
an isentrope is steeper (has a greater negative slope) than an isotherm.
99
Example
•Sketch the following set of processes on T–s
and P–v diagrams:
•1–2: isothermal expansion,
•2–3: isentropic expansion,
•3–4: isothermal compression, and
•4–1: isentropic compression
100
•Sketch the following set of processes on T–s
and P–v diagrams:
•1–2: isothermal expansion,
•2–3: isentropic expansion,
•3–4: isothermal compression, and
•4–1: isentropic compression
100
•We begin with the T–s diagram since the temperature is fixed for two of the four
processes whereas the entropy is fixed for the remaining two.
•Recognizing that in an expansion process v2 is greater than v1 and, therefore,
from Fig. that state 2 must lie to the right of state 1, we draw process 1–2 as a
horizontal line from left to right.
•Because the volume is continuing to increase in process 2–3, our isentrope is a
vertical line downward from state 2 to state 3. In the compression process from
state 3 to state 4, the volume decreases; thus, we draw the state-3 to state- 4
isotherm from right to left, stopping when s4 is equal to s1.
•An upward vertical line concludes the cycle and completes a rectangle on T–s
coordinates.
101
processes whereas the entropy is fixed for the remaining two.
•Recognizing that in an expansion process v2 is greater than v1 and, therefore,
from Fig. that state 2 must lie to the right of state 1, we draw process 1–2 as a
horizontal line from left to right.
•Because the volume is continuing to increase in process 2–3, our isentrope is a
vertical line downward from state 2 to state 3. In the compression process from
state 3 to state 4, the volume decreases; thus, we draw the state-3 to state- 4
isotherm from right to left, stopping when s4 is equal to s1.
•An upward vertical line concludes the cycle and completes a rectangle on T–s
coordinates.
101
•To draw the P–v plot, we refer to Fig. below, which shows lines of constant
temperature and lines of constant entropy.
•Because the volume increases in the process 1–2, we draw an isotherm
directed downward and to the right.
• The expansion continues from state 2 to state 3, but the process line now
follows the steeper isentrope to the lower isotherm. The cycle is
completed following this lower isotherm upward and to the left until it
intercepts the upper isentrope at the initial state 1.
Comments: This sequence of processes is the famous Carnot cycle
102
temperature and lines of constant entropy.
•Because the volume increases in the process 1–2, we draw an isotherm
directed downward and to the right.
• The expansion continues from state 2 to state 3, but the process line now
follows the steeper isentrope to the lower isotherm. The cycle is
completed following this lower isotherm upward and to the left until it
intercepts the upper isentrope at the initial state 1.
Comments: This sequence of processes is the famous Carnot cycle
102
Example SI Engine application
•The following processes constitute the air-
standard Otto cycle. Plot these processes on
P–v and T–s coordinates:
•1–2: constant-volume energy addition,
•2–3: isentropic expansion,
•3–4: constant-volume energy removal, and
•4–1: isentropic compression
103
•The following processes constitute the air-
standard Otto cycle. Plot these processes on
P–v and T–s coordinates:
•1–2: constant-volume energy addition,
•2–3: isentropic expansion,
•3–4: constant-volume energy removal, and
•4–1: isentropic compression
103
•Solution
Since there are two constant-volume processes
and two constant-entropy processes, we
begin by drawing two isentropes on P–v
coordinates and two isochors on T–s
coordinates. These are shown in the sketches.
Because
104
Since there are two constant-volume processes
and two constant-entropy processes, we
begin by drawing two isentropes on P–v
coordinates and two isochors on T–s
coordinates. These are shown in the sketches.
Because
104
Polytropic Processes
•In the previous section, we saw that an
isentropic process can be described by the
equation
•By replacing g, the specific heat ratio, by an
arbitrary exponent n, we define a generalized
process or polytropic process as
•For certain values of n, we recover
relationships previously developed.
105
•In the previous section, we saw that an
isentropic process can be described by the
equation
•By replacing g, the specific heat ratio, by an
arbitrary exponent n, we define a generalized
process or polytropic process as
•For certain values of n, we recover
relationships previously developed.
105
•These are shown in Table below. For example,
when n is unity, a constant temperature
•(isothermal) process is described, that is,
•Special Cases of Polytropic Processes
106
when n is unity, a constant temperature
•(isothermal) process is described, that is,
•Special Cases of Polytropic Processes
106
Properties
•This is an experiment conducted at different pressures
•During the time that temperature remains fixed, the system
contents are converted from liquid phase to the gaseous
phase
•At this time the substance is said to be saturated
•The properties of the substance at this point are called
saturation properties
•State is known at saturation state
•A unique relation between pressure and temperature occurs
here represented by a phase state diagram .solid, liquid and
vapour
•This is an experiment conducted at different pressures
•During the time that temperature remains fixed, the system
contents are converted from liquid phase to the gaseous
phase
•At this time the substance is said to be saturated
•The properties of the substance at this point are called
saturation properties
•State is known at saturation state
•A unique relation between pressure and temperature occurs
here represented by a phase state diagram .solid, liquid and
vapour
General phase diagram
A general P-v state diagram for the
liquid and vapour phases
liquid and vapour phases
A general T-s state Diagram for liquid
and vapour phases
and vapour phases
Property Tables/Saturation Tables
•In saturated liquid-vapour tables, notation f is used
as a subscript to donate the saturated liquid state,
and the notation g as a subscript donates the
saturated vapour state.
•When consider saturated mixture of a liquid and
vapour phase, a certain fraction of the mixture will
be in saturated vapour form, while the balance will
be in the saturated liquid form. The relative
proportion of the two phases is determined by the
following quality:
The vapor mass fraction in a saturated liquid – vapor
mixture, or dryness fraction, represented by x.
•In saturated liquid-vapour tables, notation f is used
as a subscript to donate the saturated liquid state,
and the notation g as a subscript donates the
saturated vapour state.
•When consider saturated mixture of a liquid and
vapour phase, a certain fraction of the mixture will
be in saturated vapour form, while the balance will
be in the saturated liquid form. The relative
proportion of the two phases is determined by the
following quality:
The vapor mass fraction in a saturated liquid – vapor
mixture, or dryness fraction, represented by x.
•According to this definition, 65% of the mass of a saturated mixture of
liquid and vapor whose quality is 65% is in the saturated vapor form.
The remainder of this mixture (i.e. 35% of the mass) must then be in
the saturated liquid form.
•Hence, quality specifies the relative amounts of liquid and vapor in a
saturated liquid-vapor mixture. Note that this definition only applies to
mixtures of phases and not to systems that contain a single phase.
•The quality is useful for determining specific properties of saturated
mixtures.
•Consider a saturated mixture consisting of mf units of saturated liquid
mass and mg units of saturated vapor mass.
•The total mass of this mixture is then mf + mg.
• An extensive property Z (such as volume, total internal energy, total
enthalpy, or total entropy of this mixture,) then has the specific
counterpart Z (mf+mg).
•The liquid phase of the mixture has the specific property zf, and the
vapor phase of the mixture has the specific property zg.
•The total amount of Z for the entire mixture is then mfzf+mgzg.
• Consequently, the specific value of Z for the mixture is
liquid and vapor whose quality is 65% is in the saturated vapor form.
The remainder of this mixture (i.e. 35% of the mass) must then be in
the saturated liquid form.
•Hence, quality specifies the relative amounts of liquid and vapor in a
saturated liquid-vapor mixture. Note that this definition only applies to
mixtures of phases and not to systems that contain a single phase.
•The quality is useful for determining specific properties of saturated
mixtures.
•Consider a saturated mixture consisting of mf units of saturated liquid
mass and mg units of saturated vapor mass.
•The total mass of this mixture is then mf + mg.
• An extensive property Z (such as volume, total internal energy, total
enthalpy, or total entropy of this mixture,) then has the specific
counterpart Z (mf+mg).
•The liquid phase of the mixture has the specific property zf, and the
vapor phase of the mixture has the specific property zg.
•The total amount of Z for the entire mixture is then mfzf+mgzg.
• Consequently, the specific value of Z for the mixture is
gf
g
gf
g
f
gf
g
gf
gf
gf
ggff
xzzx
z
mm
m
z
mm
m
mm
mm
mm
zmzm
z
)1(
gf
gf
gf
xssxs
xhhxh
xuuxu
hence
)1(
)1(
)1(
g
gf
g
f
gf
g
gf
gf
gf
ggff
xzzx
z
mm
m
z
mm
m
mm
mm
mm
zmzm
z
)1(
gf
gf
gf
xssxs
xhhxh
xuuxu
hence
)1(
)1(
)1(
•Applying less arithmetic
application, the equations can be
expressed in the form:
•The physical interpretation of
above equations is that: the
enthalpy of unit mass of wet
vapour is equal to the enthalpy of
unit mass of saturated liquid plus
the increase of enthalpy entailed
by evaporation of a proportion x
at constant pressure. The term
containing hf can often be
neglected, in this case h=xhg
•The simplification cannot be
employed when the fraction x is
small or the pressure is high.
fgf
fgf
fgf
xsss
xhhh
xuuu
application, the equations can be
expressed in the form:
•The physical interpretation of
above equations is that: the
enthalpy of unit mass of wet
vapour is equal to the enthalpy of
unit mass of saturated liquid plus
the increase of enthalpy entailed
by evaporation of a proportion x
at constant pressure. The term
containing hf can often be
neglected, in this case h=xhg
•The simplification cannot be
employed when the fraction x is
small or the pressure is high.
fgf
fgf
fgf
xsss
xhhh
xuuu
Example
•A rigid tank contains 1 kg of N2 initially at 300
K and 1 atm. Energy is added to the gas until a
final temperature of 600 K is reached.
Calculate the entropy change of the N2
associated with this heating process.
•Solution
•Known MN2, P1, T1, T2
•Find s2-s1
116
•A rigid tank contains 1 kg of N2 initially at 300
K and 1 atm. Energy is added to the gas until a
final temperature of 600 K is reached.
Calculate the entropy change of the N2
associated with this heating process.
•Solution
•Known MN2, P1, T1, T2
•Find s2-s1
116
117
118
119
HEAT ENGINES
•In thermodynamics, a heat engine is a system that performs the conversion of heat or
thermal energy to mechanical work.
• It does this by bringing a working substance from a higher state temperature to a lower
state temperature.
•A heat "source" generates thermal energy that brings the working substance to the high
temperature state.
•The working substance generates work in the "working body" of the engine while
transferring heat to the colder "sink" until it reaches a low temperature state. During this
process some of the thermal energy is converted into work by exploiting the properties of
the working substance. The working substance can be any system with a non-zero heat
capacity, but it usually is a gas or liquid.
•In general an engine converts energy to mechanical work.
•Heat engines distinguish themselves from other types of engines by the fact that their
efficiency is fundamentally limited by Carnot's theorem.
•Although this efficiency limitation can be a drawback, an advantage of heat engines is that
most forms of energy can be easily converted to heat by processes like exothermic reactions
(such as combustion), absorption of light or energetic particles, friction, dissipation and
resistance.
•Since the heat source that supplies thermal energy to the engine can thus be powered by
virtually any kind of energy, heat engines are very versatile and have a wide range of
applicability.
•Heat engines are often confused with the cycles they attempt to mimic.
•Typically when describing the physical device the term 'engine' is used. When describing the
model the term 'cycle' is used.
•In thermodynamics, a heat engine is a system that performs the conversion of heat or
thermal energy to mechanical work.
• It does this by bringing a working substance from a higher state temperature to a lower
state temperature.
•A heat "source" generates thermal energy that brings the working substance to the high
temperature state.
•The working substance generates work in the "working body" of the engine while
transferring heat to the colder "sink" until it reaches a low temperature state. During this
process some of the thermal energy is converted into work by exploiting the properties of
the working substance. The working substance can be any system with a non-zero heat
capacity, but it usually is a gas or liquid.
•In general an engine converts energy to mechanical work.
•Heat engines distinguish themselves from other types of engines by the fact that their
efficiency is fundamentally limited by Carnot's theorem.
•Although this efficiency limitation can be a drawback, an advantage of heat engines is that
most forms of energy can be easily converted to heat by processes like exothermic reactions
(such as combustion), absorption of light or energetic particles, friction, dissipation and
resistance.
•Since the heat source that supplies thermal energy to the engine can thus be powered by
virtually any kind of energy, heat engines are very versatile and have a wide range of
applicability.
•Heat engines are often confused with the cycles they attempt to mimic.
•Typically when describing the physical device the term 'engine' is used. When describing the
model the term 'cycle' is used.
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