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Transcript
1
THERMODYNAMICS
Thermodynamics is the branch of science which deals with energy transfer and its effect on the
state or condition of a system.
Essentially, thermodynamics refers to the study of:
Interaction of system and surroundings.
Energy and its transformation.
Relationship between heat, work and physical properties such as pressure, volume and
temperature of the working substance employed to obtain energy conversion.
Feasibility of a process.
Thermodynamics is essentially based upon experimental results and observations of common
experience; there is no mathematical proof to laws of thermodynamics. The laws, however, lay
down the general restrictions within which energy transformations are observed to occur.
Macroscopic approach and Microscopic approach
For analyzing any system there are basically two approaches available in engineering
thermodynamics.
Sl.No.
Macroscopic approach
1.
In this approach a certain quantity of matter is
considered without taking into account the events
occurring at a molecular level. In other words,
this approach to thermodynamics is concerned
with gross or overall behavior. This is known as
classical thermodynamics.
2.
The analysis of macroscopic system requires
simple mathematical formulae.
3.
4.
The values of the properties of the system are the
average values of millions of individual
molecules. These properties, like pressure and
temperature can be measured very easily. The
changes in properties can be felt by our senses.
Microscopic approach
The approach considers that the system is made
up of a very large number of discrete particles
known as molecules. These molecules have
different velocities and energies. The values of
these energies are constantly changing with
time. This approach to thermodynamics, which
is concerned directly with the structure of the
matter, is known as statistical thermodynamics.
The behavior of the system is found by using
statistical methods as the number of molecules
is very large. So advanced statistical and
mathematical methods are needed to explain
the changes in the system.
The properties, like velocity, momentum,
impulse, kinetic energy, force of impact, etc.,
which describe the molecule cannot be easily
measured by instruments. Our senses cannot
feel them.
Large numbers of variables are needed to
In order to describe a system only a fewdescribe a system. So the approach is
properties are needed.
complicated.
Thermodynamic system
A thermodynamic system represents a fixed quantity of matter bounded by a closed surface or
control volume, upon which the attention is focused to study the changes in its properties due to
exchange of energy in the form of heat and work.
MECHANICAL DEPARTMENT, IIET
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The system may be a quantity of steam, a mixture of vapour and gas or a piston-cylinder
assembly of an I.C. engine and its contents.
Surroundings
The combination of matter and space, external to the system that may be influenced by changes
in the system is called surroundings or environment.
Boundary
The thermodynamic system and surroundings are separated by an envelope called boundary of
the system. The boundary represents the limit of the system and may be either real or imaginary
Classification of systems
(1) Closed System
A closed system can exchange energy in the form of heat and work with its environment but
there is no mass transfer across the system boundary. The mass within the system remains the
same and constant, though its volume can change against a flexible boundary. Further, the
physical nature and chemical composition of the mass may change. Thus a liquid may
evaporate, a gas may condense or a chemical reaction may occur between two or more
components of the system.
Cylinder fitted with a moveable piston: The gas can receive or reject heat, expand or contract
according to whether the piston is moved outward or inward but no matter (gas) crosses the
system boundary.
(2) Open System
An open system has mass exchange with the surroundings along with transfer of energy in the
form of heat and work. The mass within the system does not necessarily remain constant; it may
change depending upon mass inflow and mass outflow.
Water Wheel: It is a device that converts potential energy of water into mechanical work. Water
enters the wheel from head race side and leaves it to the tail race from the other end, and as such
the mass crosses the system (wheel) boundary. The work output due to rotation of the wheel also
crosses the system boundary.
MECHANICAL DEPARTMENT, IIET
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(3) Isolated System
An isolated system is of fixed mass and energy; it exchanges neither mass nor energy with
another system or with the surroundings. An isolated system has no interaction with the
surroundings; it neither influences the surroundings nor is influenced by it. When a system and
its surroundings are taken together, they constitute an isolated system.
The universe can be considered as an isolated system and so is the fluid enclosed in a perfectly
insulated closed vessel (thermos flask).
Control Volume
A control region or control volume is defined as any region in space which is separated from its
environment by a control surface which may be either physical or imaginary.
The control system is normally taken to be fixed in shape, position and orientation relative to the
observer. Heat and work interactions are present across the control surface and matter flows
continuously in and out of the control volume.
Consider a steam generator with water being fed to it at one side and steam being taken out of it
from the other side. During the continuous operation of steam formation, the region has a
constant volume; its boundary neither expands nor contracts, i.e., does not change even though
there is mass flow across the control surface.
Thermodynamic property
A thermodynamic property refers to the observable characteristics which can be used to describe
the condition or state of a system, e.g., temperature, pressure, chemical composition, colour,
volume, energy etc.
The salient aspects of a thermodynamic property are:
o It is a measurable characteristic describing a system and helps to distinguish one system
from another,
o It has a definite unique value when the system is in a particular state,
o It is dependent only on the state of the system; it does not depend on the path or route the
system follows to attain that particular state,
o Its differential is exact.
MECHANICAL DEPARTMENT, IIET
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Since a thermodynamic property is a function of the state of a system, it is referred to as a point
function or a state function.
There are two kinds of thermodynamic properties namely intensive and extensive.
Intensive property
Intensive property is independent of the extent or mass of the system. Its value remains the same
whether one considers the whole system or only a part of it.
Examples are pressure, temperature, density, composition, viscosity, thermal conductivity,
electrical potential etc.
Extensive property
Extensive property depends on mass or extent of the system.
Examples include energy, enthalpy, entropy, volume etc.
Let a system with value of a certain characteristic / property P be divided into a number of parts.
If P1 P2, P3 etc denote the value of that property for various parts of the system, then
For an intensive property P = P1= P2 = P3. For an extensive property: P = P1 + P2 + P3
Specific property
It is an extensive property expressed per unit mass of the system. Examples include specific
volume, specific energy, specific entropy, specific enthalpy etc
Specific volume v = V/m
Specific energy e = E/m, where V, E, m are volume, energy and mass of the system respectively.
Like extensive properties, specific properties too, are additive.
Entropy
Entropy is the measure of the heat Q exchanged by the system with its surroundings at
constant temperature T, in an ideal reversible process. The change in entropy of the system is:
Q
dS
T
It is a function of a quantity of heat that shows the possibility of conversion of that heat into
work. The increase in entropy is small when heat is added at a high temperature and is greater
when heat addition is made at a lower temperature. Thus, for maximum entropy, there is
minimum availability for conversion into work and for minimum entropy there is a maximum
availability for conversion into work.
Entropy is a function of the initial and final states of the system and is independent of the path of
the process. Hence, it represents a property of the system.
The unit of measurement for entropy is kJ/K.
Specific entropy(s) is the entropy of 1 kg of the substance and it is measured in kJ/kg-K,
Work
Under the concept of mechanics, work is defined as the product of the force and the distance
moved in the direction of force.
In S.I. systems, the unit of work is N-m. The work of 1 N-m is called as Joule.
MECHANICAL DEPARTMENT, IIET
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In thermodynamics, work transfer is considered as an event between the system and the
surroundings. “The work is said to be done by a system during a process if the sole effect of the
system on surroundings can be reduced to the raising of a weight”.
Let us explain the above definition by a simple example. Consider a battery and a motor as a
system. External to the system is a fan as shown in figure. The motor drives the fan by getting
the power from the battery. In this case, the system is doing work upon the surroundings.
The fan is replaced by a pulley-weight arrangement. When the motor operates, the weight will
raise since the pulley is operated by the motor. The sole effect on things external to the system
(surroundings) is then raising a weight.
Work transfer from the system (work done by the system) is assigned with positive (+) sign
while the work transfer to the system (work done on the system) is assigned with negative (-)
sign.
Mechanical Equilibrium (Equality of pressure)
A system is said to be in mechanical equilibrium, when there is no unbalanced forces within the
system or between the surroundings. The pressure in the system is the same at all points and
does not change with respect to time.
Chemical equilibrium (Equality of chemical potential)
A system is said to be in chemical equilibrium, when there is no chemical reaction within the
system or diffusion (transfer of matter)
Thermal equilibrium (Equality of temperature)
A system is said to be in thermal equilibrium, when there is no temperature difference between
the parts of the system or between the system and the surrounding.
Thermodynamic equilibrium
A system which is simultaneously in a state of mechanical equilibrium, thermal equilibrium and
chemical equilibrium is said to be in a state of thermodynamic equilibrium.
State
State is the condition of the system identified by thermodynamic properties so that one state is
distinguished with other.
When all such properties have a definite value, the system is said to exist at a definite state. With
properties as co-ordinates, the state of the system in thermodynamic equilibrium can be
represented by a point.
Consider a system constituted by gas enclosed in the piston cylinder assembly of a reciprocating
machine. Corresponding to position of the piston at any instant, the condition of the system will
MECHANICAL DEPARTMENT, IIET
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be prescribed by pressure, volume and temperature of the gas.
Path
Any operation in which properties of the system change is called a
change of state.
The locus of the series of states through which a system passes in
going from initial state to its final state constitutes the path.
Thermodynamic process
When a system changes its state from one equilibrium state to another equilibrium state, then the
path of successive states through which the system has passed is known as thermodynamic
process.
The thermodynamic process consists of enough information about the thermodynamic
coordinates at successive state points in the thermodynamic equilibrium to be plot a path of a
change of state on the thermodynamic plane.
As a process is simply a change in the state of the system, there are infinite ways for a system to
change from one state to another state.
Certain processes are identified by special names, e.g., an isochoric process is a constant volume
process, an isobaric process is a constant pressure process, and an isothermal process is a
constant temperature process.
Thermodynamic cycle
When a system in a given state undergoes through a series of
processes such that the final state is identical with the initial state, a
cyclic process or a cycle is said to have been executed.
An essential feature of the cycle is that the initial condition of the
system is restored after the processes. The change in the value of any
property of the system for a cyclic process is zero.
Quasi-static process
When the departure of the state of system from the
thermodynamic equilibrium state will be infinitesimally small
then, every state passed by the system will be an equilibrium
state. The locus of a series of such equilibrium states is called
a quasi-static or quasi-equilibrium process and can be
represented graphically as a continuous line on a state
diagram. A quasi-static process is thus a succession of
equilibrium states.
A quasi-static process can be viewed as a sufficiently slow process which allows the system to
adjust itself internally so that properties in one part of the system do not change any faster than
those at other parts.
Reversible processes
A thermodynamic process is said to be reversible if, when the process is carried out in the
reverse direction using the same amount of work and heat transferred during the forward
process, the system passes through the same states as it does in the forward direction the system
passes through a continuous series of equilibrium states. If the process is reversed, the system as
well as surroundings will restore back to their respective initial states.
MECHANICAL DEPARTMENT, IIET
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A reversible process between two states can be drawn as a continuous line on any diagram of
properties as shown in the figure.
The following conditions need to the satisfied for a process to be
reversible
o There should be no friction; solid or fluid
o The heat exchange to or from the system, if any, should be only
through small temperature difference
o The process should be quasi-static; it should proceed at infinitely
slow speed. For this the pressure difference between the system and surrounding must be
infinitely small.
Irreversible processes
A process, which cannot be completely reversed with out living a
change in either in the system or surrounding is called irreversible
process.
All actual process is irreversible. Work during an irreversible
process will be less than the work during the same process if the
process is assumed to be reversible.
A reversible process between two states can be drawn as a dotted
line on any diagram of properties as shown in the figure.
LAWS OF PERFECT GASES
A perfect gas or an ideal gas may be defined as a gas having no forces of intermolecular
attraction / repulsion and obeys all gas laws at all rages of pressures and temperatures.
Real gases follow gas laws at low pressures or high temperatures or both. This is because the
forces of attraction between molecules tend to be very small at reduced pressures and elevated
temperatures.
An ideal gas obeys the law pv RT , the specific heat capacities are not constant but are
functions of temperature. A perfect gas obeys the law pv RT and has constant specific heat
capacities. A perfect gas is well suited to mathematical manipulation and is therefore a most
useful model to use for the analysis of practical machinery, which uses real gases as a working
substance.
In reality there is no ideal or perfect gas. At a very low pressure and at a very high temperature,
real gases like hydrogen, oxygen, nitrogen, helium, etc., behave nearly the same way as perfect
gases.
The behaviour of a perfect gas is governed by the following laws (1) Boyle's law (2) Charle's
law (3) Gay-Lussac law.
Boyle's Law
Boyle's law states that volume of a given mass of a perfect gas varies inversely as the absolute
pressure when temperature is constant.
If p is the absolute pressure of the gas and v is the volume occupied by the a unit mass of perfect
gas at constant temperature T oK, then
MECHANICAL DEPARTMENT, IIET
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1
p
i.e pv C, when T is cons tan t
v
p1v1 p 2 v 2 p3 v3 C
Charle's Law
Charle's law states that the volume of a given mass of a perfect gas varies directly as its absolute
temperature, when the absolute pressure remains constant.
v T when p is cons tan t
i.e
v
C,
T
v1
v 2 v3
C
T1
T2 T3
Gay-Lussac Law
The law states “The absolute pressure of a given mass of a perfect gas varies directly as its
absolute temperature, when the volume remains constant”
p T when V is cons tan t
i.e
p
C,
T
p1
p 2 p3
C
T1
T2 T3
Joule's Law
Joule's law states that the change in internal energy of a perfect gas is directly proportional to the
change of temperature.
dU dT
i.e dU m C dT, where m is the mass of the gas
and C is the proportionality cons tan t and is the specific heat.
Avogadro's Law
Avogadro’s law states equal volume of all gases, at the same temperature and pressure contain
equal number of molecules.
According to this law, 1 m3 of O2 will contain same number of molecules as 1 m3 of H: when the
temperature and pressure of O2 and H2 are same.
Characteristic gas equation (Equation of state of an ideal Gas)
The characteristic gas equation represents the relation between p, v, and T and can be set up by
applying Boyle’s law and Charle’s law to a perfect gas undergoing a change of state.
Consider a unit mass of an ideal gas that passes from state p1, v1, T1 to another state identified by
p2 ,v2 ,T2. Let this change be (1) first at constant pressure p1 to some intermediate volume vi, and
(2) then at constant temperature T2 to final volume v2
MECHANICAL DEPARTMENT, IIET
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Process 1-i
It is at constant pressure and, therefore, state of the gas
changes in conformity with Charle’s law. Then
v1
vi
T1
Ti
(1)
T
i.e vi v1 i
T1
Process i-2
It is at constant temperature and, therefore, state of the gas changes in conformity with Boyle’s
law. Then
pi vi p 2 v 2
i.e vi v 2
p2
pi
(2)
From equ (1) and equ (2)
v1
Ti
p
v2 2
T1
pi
Since pi p1 and Ti T2
v1
T2
p
v2 2
T1
p1
p1v1
T1
p2 v2
T2
pv
R (a cons tan t)
T
pv = RT is the equation of the state or the characteristic gas equation for unit mass and R
V
is Characteristic gas constant. Since v
where V is the volume of the gas and m is its
m
mass the above equation can be rewritten as pV = mRT .
Unit of Characteristic gas constant is N-m / Kg-K or J / Kg-K
Universal gas constant
For molar volume of a gas, the characteristic gas equation can be written as
p Vmol = M R T
= R mol T
Where R mol , Universal gas cons tan t = M R
= Molecular weight Characteristic gas constant
8314 J K mole K
MECHANICAL DEPARTMENT, IIET
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R mol
M
Univeras l g as consatnt
Chateristic gas cons tan t
=
Molecular weight
The Universal gas constant is independent of the nature of gas.
R
=
A standard condition of t = 0 o C = 273.15 K and p = 101325 N/m 2 ,
3
the volume of one kilo of all gases is equal to 22.413 m
Temperature
The temperature of a system is an intensive thermodynamic property that determines whether or
not a system is in thermal equilibrium with other systems. It can also be defined as the degree of
hotness or the level of heat intensity of the body.
Heat
In thermodynamics, heat is defined as energy in transition flowing by virtue of temperature
difference between two systems or between a system and its surroundings. Heat is visible only at
the system boundary. It is not contained in the system. It is therefore a transient form of energy.
Heat is not a function of the thermodynamic coordinates, that is, not a state function, so the
calculation of the heat depends on the path of integration.
Heat transfer to the system (heat absorbed) is assigned with positive (+) sign while the heat
transfer from the system (heat rejected) is assigned with negative (-) sign. Its units are Calories.
Point function
When two properties locate, a point on the graph {co-ordinate axes) then those properties are
called a point function.
Eg:- Pressure, temperature, volume, etc.
Volume is an exact differential. The change in volume can be written as difference between their
end states.
2
dV
V2 V1
1
Path function
There are certain quantities that cannot be located on a graph by a point but are given by the area
on that graph. In that case, the area on the graph, pertaining to the particular process, is a
function of the path of the process. Such quantities are called path function.
Eg:- Heat, work etc.
Heat and work are inexact differentials. Their change cannot be written as difference between
their end states.
2
2
Q 1Q2 or Q Q12
1
1
2
.
But not Q Q 2 Q1
1
MECHANICAL DEPARTMENT, IIET
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Work and Heat transfer-a path function
The amount of work/heat transferred from state 1 to state 2, when a system changes its position,
depends on the intermediate positions taken by the system during that process. This means that
work/heat transfer is a path function. Therefore work ( W )/heat transfer ( Q ) is an inexact
differential. Therefore
2
Q
2
Q12 and Q Q2 Q1 ;
1
1
2
W
2
W12 and W W2 W1
1
1
Comparison of Work and Heat
Similarities:
•
Both work and heat are the forms of transient energy
•
Both are path functions and inexact differentiate.
•
Both are boundary phenomenon i.e. both are observed at the boundaries of the system as
they cross them.
•
Systems possess energy, but not work or heat
Dissimilarities:
•
In heat transfer temperature difference is required.
•
In a stable system there cannot be work transfer, however, there is no restriction for the
transfer of heat.
Internal energy
Internal energy of a thermodynamic system is the total amount of energy contained in the
system.
In general, the internal energy of a substance is made up of two parts: kinetic energy due to
molecular motion and potential energy arising from molecular dispersion.
Changes in kinetic internal energy are indicated by changes in temperature of the system. Since
kinetic energy of a molecule is associated with its motion - the faster the molecule moves, the
more kinetic energy it possesses Changes in potential internal energy are indicated by changes in
the phase of the system.
If the expansion of the system is without any transfer of heat, the work will be done at the
expense of internal energy. Further the system receives heat without any change in boundaries;
the entire heat added will go in for increasing the internal energy of the system.
It is an extensive property as it value depends upon the mass. As the absolute value of internal
energy is difficult to measure, the change in internal energy is considered in most of the
thermodynamic applications.
The symbol for internal energy is U and in SI unit it is measured in joules (J) or kilojoules (kJ.
Enthalpy
It is defined as the energy content of a system and is given by the sum of internal energy; the
energy required to create the system and the product of pressure and volume; the amount of
energy required to make room for it by displacing its environment and establishing its volume
and pressure.
It is state function and extensive property. As absolute value of enthalpy can not be obtained,
only change in enthalpy of substance is considered.
MECHANICAL DEPARTMENT, IIET
12
Then H = U pV
The change in enthalpy is given by
dH = dU d pV
H 2 H1 U 2 U1 p 2 V2 p1V1
m C v T2 T1 mRT2 mRT1
pV mRT
m C v T2 T1 mR T2 T1
m C v R T2 T1
C p C v R
m C p T2 T1
m C p dT
dH
Specific enthalpy, dh Cp dT
Specific heat of a gas
Specific heat of a gas is defined as the amount of heat required to raise a unit mass of the gas
through a unit rise in temperature.
C
Q
m T2 T1
Where Q Heat trasfer in J
C Specific heat in J kg K
m mass in kg
Since a gas can be heated under constant pressure and constant volume, it has two specific heats.
Specific heat at Constant volume
Specific heat at constant volume is defined as the amount of heat required to raise a unit mass of
the gas through a unit rise in temperature, when the gas is heated at constant volume.
Cv
Q
J kg K
m T2 T1
The specific heat at constant volume is also be defined as the rate of change of internal energy
with respect to the temperature when the volume remains at constant.
Cv
du
du
dT v
Cv dT
u1 u 2 Cv T2 T1
Specific heat at Constant pressure
Specific heat at constant pressure is defined as the amount of heat required to raise a unit mass
of the gas through a unit rise in temperature, when the gas is heated at constant pressure.
Cp
Q
J kg K
m T2 T1
The specific heat at constant volume is also be defined as the rate of change of enthalpy with
respect to the temperature when the pressure remains at constant.
MECHANICAL DEPARTMENT, IIET
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Cp
dh
dT p
dh
C p dT
h1 h 2 C p T2 T1
Relation between C p , C v and R
Specific enthalpy for perfect gas is given by
h
u pv
We know pv RT
u RT
dh du RdT
dh
du
R
dT
dT
Cp C v R
Cp C v R
The ratio of specific heat is given by
Cp
Cv
C
Cp 1 v R
Cp
1
Cp 1 R
1
Cp
R
Cp
R
1
Cp
Cv
1 R
Cv
C v 1 R
Cv
1
1
R
1. A gas of certain mass is expanded from an initial state of 4 bar and 0.04 m 3 to another
condition of 1.2 bar and 0.1 m3. The temperature fall is observed to be 140. If the value of Cp
and Cv are 1.0216 kJ/kg-K and 0.7243 kJ/kg-K, calculate the change in internal energy and
enthalpy of the gas.
Given
p1= 4 bar = 4×105 N/m2= 400 kN/m2,
p2= 1.2 bar = 1.2×105 N/m2= 120 kN/m2
Cp = 1.0216 kJ/kg-K
Cv = 0.7243 kJ/kg-K and T1- T2 = 140
MECHANICAL DEPARTMENT, IIET
14
p1V1 p 2 V2
T
pV
400 .04
R; 1 1 1
1.33
T1
T2
T2
p 2 V2
120 .1
T1 1.33 T2 and T1 T2 140
0.33 T2 140
140
424.24 424 K
0.33
T1 424 140 564 K
T2
Cp C v R;
R 1.0216 0.7243 0.2973 kJ kg K ;
p1V1
m R;
T1
m
p1V1
400 .04
0.09542 kg
RT1
0.2973 564
U m C v T2 T1
0.09542 0.7243 140
9.675 kJ
H m Cp T2 T1
0.09542 1.0216 140
13.647 kJ
2. 1 kg of ideal gas is heated from 20 oC to 100 oC. Assuming R = 0.264 kJ/kg-K and γ = 1.18
for the gas. Find out 1) Specific heats 2) Change in internal energy 3) Change in enthalpy
Given T1= 20 oC = 293 K,
T2= 100 oC = 373 K
m = 1 kg
R = 0.264 kJ/kg-K and γ = 1.18
R
1
1.18
0.264
1.18 1
1.731 kJ kg K
Cp
1
R
1
1
0.264
1.18 1
1.467 kJ kg K
Cv
MECHANICAL DEPARTMENT, IIET
15
dU m C v T2 T1
11.467 373 293
117.36 kJ
dH m C p T2 T1
11.731 373 293
138.48 kJ
LAW OF THERMODYNAMICS
Zeroth law of thermodynamics
Zeroth law of thermodynamics states that if the bodies A and B are in thermal equilibrium with a
third body C separately, then the two bodies A and B shall also be in thermal equilibrium with
each other. This is the principle of temperature measurement.
The zeroth law provides the basis for the measurement of temperature. It enables us to compare
temperatures of two bodies A and B with the help of a third body C and say that the temperature
of A is the same as the temperature of B without actually bringing A and B in thermal contact.
In practice, body C in the zeroth law is called the thermometer. It is brought into thermal
equilibrium with a set of standard temperatures of a body B, and is thus calibrated. Later, when
any other body A is brought in thermal communication with the thermometer, we say that the
body A has attained equality of temperature with the thermometer, and hence with body B. This
way, the body A has the temperature of body B given for example by, say, the height of mercury
column in the thermometer C.
First law of thermodynamics
First law of thermodynamics states that in a closed system undergoing a cyclic process, the net
heat transfer is equals to the net work done.
First law of thermodynamics can't be proved but it is supported by a large number of
experiments and no exceptions have been observed. It is therefore termed as the law of nature.
Mathematical expression for the first law of thermodynamics can be,
Q W
First law for a process -Corollary I
When a system executes a process, the net heat interaction equals the net work interaction plus
change in stored energy.
Q W dE
The change in energy depends only on the end states and hence E is a property. Internal energy,
kinetic energy, potential energy, electrical energy, chemical energy and magnetic energy are
some of the important modes of energy which add to give the total energy E of the system
MECHANICAL DEPARTMENT, IIET
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First law for an isolated system - Corollary II
For an isolated system, both heat and work interactions are absent.
i.e. Q 0 and W 0
dE 0
E Cons tan t
The above identity indicates that the energy of a system remains unchanged if the system is
isolated from its surroundings as regards heat and work interactions. This fact is often referred to
as the principle of conservation of energy and may be stated as “Energy can neither be created
nor destroyed; however, it can be converted from one form to another”.
Perpetual Motion Machine of First Kind (PMM1) - Corollary III
A perpetual motion machine of first kind is an imaginary device which produces a continuous
supply of work without absorbing any energy from the surroundings or from other systems.
Such a machine, in effect, creates energy from nothing and violates the first law of
thermodynamics.
Corollary 3 may then be stated as “A perpetual motion machine of the first kind is impossible”
Limitation of First law of thermodynamics
1. First law of thermodynamics does not differentiate between heat and work and assures full
convertibility of one into other whereas full conversion of work into heat is possible but the
vice-versa is not possible.
2. First law of thermodynamics does not explain the direction of a process. Such as
theoretically it shall permit even heat transfer from low temperature body to high
temperature body which is not practically feasible.
SECOND LAW OF THERMODYNAMICS
Second law came up as a concrete form of real happenings keeping the basic nature of first law
of thermodynamics.
Feasibility of process, direction of process and grades of energy such as low and high are the
potential answers provided by the second law.
Second law of thermodynamics is capable of indicating the maximum possible efficiencies of
heat engines, coefficient of performance of heat pumps & refrigerators, defining a temperature
scale independent of physical properties etc.
Heat reservoir
Heat reservoir is the system having very large heat capacity i.e. it is a body capable of absorbing
or rejecting any finite amount of energy without any appreciable change in its temperature.
Large river, sea etc. can also be considered as reservoir as dumping of heat to it shall not cause
appreciable change in temperature.
Heat reservoirs can be of two types depending upon nature of heat interaction i.e. heat rejection
or heat absorption from it. Heat reservoir which rejects heat from it is called source. While the
heat reservoir which absorbs heat is called sink.
MECHANICAL DEPARTMENT, IIET
17
Heat engine
Heat engine may be defined as "a device operating in cycle
between high temperature source and low temperature sink and
producing work". Heat engine receives heat from source,
transforms some portion of heat into work and rejects balance
heat to sink. All the processes occurring in heat engine
constitute cycle.
Heat and work have been categorized as two forms of energy
of low grade and high grade type. Conversion of high grade of
energy to low grade of energy may be complete (100%), and
can occur directly whereas complete conversion of low grade
of energy into high grade of energy is not possible. For
converting low grade of energy (heat) into high grade of
energy (work) some device called heat engine is required.
The performance of any machine is expressed as the ratio of “what we want” to “what we have
to pay for”.
In the context of an engine, work is obtained at the expense of heat input. Accordingly, the
performance of a heat engine is given by
th
net work output
total heat supplied
This ratio is called thermal efficiency. Thermal efficiency is a measure of the degree of useful
utilization of heat received in a heat engine.
th
Q1 - Q2
Q1
1
Q2
Q1
1
T2
T1
Heat pump and refrigerator
Refrigerators and heat pumps are reversed heat engines i.e. devices used for extracting heat from
a low temperature surroundings and sending it to high temperature body, while operating in a
cycle. In other words heat pump maintains a body or system at temperature higher than
temperature of surroundings, while operating in cycle.
The performance of a refrigerator and heat pump is expressed in terms of coefficient of
performance (COP) which represents the ratio of desired effect to work input.
COP
desired effect
work input
MECHANICAL DEPARTMENT, IIET
18
In a refrigerator, the desired effect is the amount of heat Q2 extracted from the space being
cooled, i.e. the space at low temperature
CO P r ef
Q2
Q2
heat extracted at low temperature
work input
W
Q1 - Q 2
In a heat pump, the desired effect is the amount of heat Q1 supplied to the space being heated.
CO P heat pump
Q1
Q1 - Q 2
1
Q1
Q1 - Q 2
1 CO P ref
Statements for Second Law of Thermodynamics
1. Clausius statement of second law of thermodynamics:
"It is impossible to construct a device that operates in a cycle and produces no effect other than
the transfer of heat from a system at low temperature to another system at high temperature".
The statement implies that heat cannot flow of itself from a system at low temperature to a
system at high temperature. The schematic arrangement that is prohibited by Clausius statement
is shown in figure. The coefficient of performance of such an arrangement equals: COP=
COP = Q/W=Q/0 .Obviously the Clausius statement tells that COP of a heat
pump/refrigerator cannot be equal to infinity.
The only alternative for the transfer of heat from low temperature to high temperature level is
that some external work must be supplied to the machine as shown in figure
MECHANICAL DEPARTMENT, IIET
19
2. Kelvin-Planck statement of second law of thermodynamics:
It is impossible to construct an engine that operates in a cycle and produces no effect other than
work output and exchange of heat with a single heat reservoir
The statement implies that no heat engine can be developed that receives a certain amount of
heat from a high temperature source and converts that into an equivalent amount of work. i.e,
W= Q. The thermal efficiency of such an engine η = W/Q =1 or 100 %.
Figure represents the schematic arrangement of a heat engine that exchanges heat with a single
heat source and is 100 percent efficient. Such a system satisfies the principle of energy
conservation (1st law) but violates the Kelvin statement of second law. Obviously Kelvin-Planck
statement tells that no heat engine can have thermal efficiency equal to 100 percent
The only alternative for continuous power output from a heat engine is that a portion of the heat
received must be rejected to a heat reservoir at low temperature (heat sink). This engine receives
Q1, units of heat, rejects Q2 units of heat and converts (Q1 - Q2) units of heat into work per cycle.
All possible heat engines conform to this representation.
Whereas the Kelvin-Planck statement is applied to heat engines, the Clausius statement concerns
heat pumps and refrigerators. Both the Kelvin-Planck and Clausius statements are negative
statements, they have no mathematical proof. The law is based on experimental observations,
and to-date no observation has been made that contradicts the law and this aspect is taken as
sufficient evidence of its validity.
MECHANICAL DEPARTMENT, IIET
20
ANALYSIS OF THERMODYNAMIC PROCESSES
1. Constant volume process (Isochoric process)
A change in the state of a system at constant volume is called isochoric process. An isochoric
process results when the gas system is heated or cooled in an enclosed space.
During an isochoric process, both the pressure and temperature will change. The pressure and
temperature increase when heat is supplied to the system and decrease when heat is rejected by
the system. Since there is no expansion of gas in constant volume process, no mechanical work
is done on or by the system. In this process, all the heat is utilized to change the internal energy
of the system. The isochoric process can be depicted in p - V co-ordinates by a vertical line and
curve in T-S diagram.
a) p-V-T relationship
For perfect gas,
p1V1
T1
p 2 V2
T2
Since V1 V2
p1
T1
p2
T2
b) Work done
W1-2
pdV
Since V C, dV 0
W1-2 0
c) Change in internal energy
Since there is temperature rise from T1 to T2
dU mCv T2 T1
d) Heat supplied
MECHANICAL DEPARTMENT, IIET
21
From 1st law of thermodynamics, Q1-2 dU + W1-2
Since W1-2 0;Q1-2 dU
Q1-2 mCv T2 T1
e) Change in Entropy
Heat transfer, Q mC v dT
Q
dT
mC v
T
T
Divide by T ;
i.e. dS mC v
2
dT
T
2
dT
T
1
dS mCv
1
T
S2 S1 mC v ln 2
T1
Since
T2
T1
p2
p1
p
S2 S1 mC v ln 2
p1
1.
A closed insulated chamber of capacity 3 m3 contains 15 kg of nitrogen. Paddle work is
done on the gas still the pressure inside the chamber increases from 5 bar to 10 bar.
Calculate 1) change internal energy 2) Work done 3) Heat transfer 4) Change in entropy.
Assume CP = 1.04 kJ/kg-K and CV = 0.7432 kJ/kg-K.
Given
p1 = 5 bar = 5×105 N/m2= 500 kN/m2,
p2 = 10 bar = 10×105 N/m2= 1000 kN/m2
V1 = 3 m3
m = 15 kg
CP = 1.04 kJ/kg-K
CV = 0.7432 kJ/kg-K
MECHANICAL DEPARTMENT, IIET
22
T2 p 2
T1 p1
We know p1V1 mRT1 , C p C v R and
p1V1
mR
p1V1
500 3
m Cp Cv 15 1.04 0.7432
T1
T2
336.9 K
p2
T1
p1
1000
336.9 673.8 K
500
We know dU m C v T2 T1
dU
15 0.7432 673.8 336.9
3755.8 kJ
We know Q dU W
Since the container is insulated, Heat transfer Q 0
W dU
3755.8 kJ
ie. Work is done on the system
T
m C v ln 2
T1
673.8
S2 S1 15 0.7432 ln
336.9
7.7 kJ
We know dS
2. Constant pressure process (Isobaric process)
A change in the state of a system at constant pressure is called isobaric process. An isobaric
process results when the gas system confined by a piston in a cylinder is heated or cooled slowly
in a way such that pressure is kept constant.
MECHANICAL DEPARTMENT, IIET
23
Both the volume and temperature will change during constant pressure process. Since there is a
change in volume in constant pressure process, mechanical work is done on or by the system. In
this process, a part of the heat supplied is utilized to change the internal energy of the system
and remaining part is used to do external work. The isobaric process can be depicted in p - V coordinates by a horizontal line and curve in T-S diagram.
a) p-V-T relationship
For perfect gas,
p1V1
T1
p 2 V2
T2
Since p1 p 2
V1
T1
V2
T2
b) Work done
W1-2
pdV
Since p C
W1-2
p dV
W1-2 p V2 V1
c) Change in internal energy
Since there is temperature rise from T1 to T2
dU mCv T2 T1
d) Heat supplied
MECHANICAL DEPARTMENT, IIET
24
From 1st law of thermodynamics, Q1-2 dU + W1-2
Q1-2 mC v T2 T1 p V2 V1
mC v T2 T1 mR T2 T1
m C v R T2 T1
Q1-2 mCp T2 T1
e) Change in Entropy
Heat transfer, Q mC p dT
Q
dT
mC p
T
T
Divide by T ;
i.e. dS mC p
2
dT
T
2
dT
T
1
dS mCp
1
T
S2 S1 mC p ln 2
T1
Since
T2
T1
V2
V1
V
S2 S1 mC p ln 2
V1
1. 2 kg of an ideal gas is heated at constant pressure from 25 oC to 200 oC. The values of CP
and CV are 0.984 kJ/ kg-K and 0.728 kJ/ kg-K. If the initial volume was 3 m3, find 1) Final
volume and pressure 2) Heat added 3) Change in internal energy 4) Work done and 3)
Change in entropy energy.
Given
m = 2 kg
T1 = 25 oC = 298 K
T2 = 200 oC = 473 K
V1 = 3 m3
CP = 0.984 kJ/kg-K
CV = 0.728 kJ/kg-K
MECHANICAL DEPARTMENT, IIET
25
We know p1V1 mRT1 , Cp C v R and
p1 p 2
V2 T2
V1 T1
mRT1
V1
2 0.984 0.728 298
50.86 kN m 2 0.5086 bar
3
T
V2 2 V1
T1
473
3 4.76 m3
298
We know Q mC p T2 T1
Q 2 0.984 473 298
344.4 kJ
We know dU m C v T2 T1
dU
2 0.728 473 298
254.8kJ
We know W p V2 V1
W 50.86 4.76 3
89.51 kJ
or
W Q dU
344.4 254.8
89.6 kJ
T
We know dS m C p ln 2
T1
473
S2 S1 2 0.984 ln
298
0.909 kJ
3. Constant temperature process (Isothermal process)
A change in the state of a system at constant temperature is called isothermal process. An
isothermal process can be performed in a piston–cylinder assembly which is surrounded by a
constant temperature reservoir.
MECHANICAL DEPARTMENT, IIET
26
The pressure and volume are the changing variables during constant temperature process. Since
the gas is at constant temperature, all the heat is utilized to accomplish mechanical work done.
During isothermal expansion heat is received by the working fluid, and in isothermal
compression the working fluid rejects heat to the surroundings. The isothermal process can be
represented in p - V co-ordinates by a rectangular hyperbola and horizontal line in T-S diagram.
a) p-V-T relationship
For perfect gas,
p1V1
T1
p 2 V2
T2
Since T1 T2
p1V1 p 2 V2
b) Work done
W1-2
pdV
Since pV C and p
W1-2
C
C
; C p1V1 p 2 V2
V
dV
V
V
W1-2 p1V1 ln 2
V1
V2 p1
V1 p 2
p
W1-2 p1V1 ln 1
p2
c) Change in internal energy
Since T1 T2 ;
dU mC v T2 T1
dU 0
d) Heat supplied
From 1st law of thermodynamics, Q1-2 dU + W1-2
dU
0
Q1-2 W1-2
V
Q1-2 p1V1 ln 2
V1
p
or Q1-2 p1V1 ln 1
p2
e) Change in Entropy
MECHANICAL DEPARTMENT, IIET
27
Heat transfer, Q dU W
Since T=constant, dU = 0;and W = pdV
Q pdV
dV
V
dV
mRT
V
pV mRT
pV
Q
dV
mR
T
V
Divide by T ;
i.e. dS mR
2
dV
V
2
dV
V
1
dS mR
1
V
S2 S1 mR ln 2
V1
Since
V2
V1
p1
p2
p
S2 S1 mR ln 1
p2
1. Air initially at 1 bar (100 kPa), 80 oC and occupying a volume of 0.4 m3 is compressed
isothermally until the volume is 0.1 m3. Find 1) Final pressure 2) Heat rejected 3) Change in
internal energy 4) Work done and 3) Change in entropy energy.
p1 = 1 bar = 100 kPa = 100 kN/m2
Given
T1 = 80 oC = 353 K
V1 = 0.4 m3
V2 = 0.1 m3
Assume
R = 0.287 kJ/kg-K
We know p1V1 mRT1 , p1V1 p 2 V2 and
p2
V2 p1
V1 p 2
p1V1
V2
100 0.4
400 kN m 2 4 bar
0.1
p1V1
m
RT1
100 0.4
0.395 kg
0.287 353
MECHANICAL DEPARTMENT, IIET
28
V
p
We know Q p1V1 ln 2 or Q p1V1 ln 1
V1
p2
0.1
Q 100 0.4 ln
-55.45kJ
0.4
Heat rejected 55.45kJ
We know dU 0
W Q -dU dU
T2 T1
0
-55.45kJ
Work done on the gas 55.45kJ
V
We know dS mR ln 2
V1
0.1
S2 -S1 0.395 0.287 ln
0.4
-0.157 kJ
Decrease in entropy 0.157 kJ
2. 1 kg of a gas is compressed reversibly according to a law pv = 0.2 where ‘p’ is in bar and ‘v’
is in m3/kg. Final volume is ¼th of the initial volume. Find work done on the gas.
Given
pv = 0.2 (bar × m3/kg) = 0.2 (100 kN/m2× m3/kg) = 20 kJ/kg
V1 = 4 V2
V
V
W p1V1 ln 2 20 ln 2 20 ln 1
4
V
1
4V2
27.73 kJ
Work done on the gas 27.73kJ
4. Adiabatic process
A change in the state of the system without heat exchange with the surrounding medium is
called an adiabatic process; here changing variables are pressure, volume and temperature. An
adiabatic process can be carried out by the expansion or compression of gas in a cylinder whose
walls are insulated or carried out the process very quickly so that there is no time for appreciable
heat flow. A reversible adiabatic process is called isentropic process.
MECHANICAL DEPARTMENT, IIET
29
Work is done by the gas at the expense of the internal energy of the system. The adiabatic
process can be represented in p - V co-ordinates by a curve as shown in figure and vertical line
in T-S diagram.
Governing equation
Consider unit mass of gas
q du + w ; where q 0, du CvdT and w pdv
Cv dT + pdv 0
(1)
Consider gas equation pv RT for unit mass
On differentiating,
pdv vdp RdT
dT
pdv vdp
R
(2)
On substituting equ (2) in (1)
pdv vdp
Cv
+ pdv 0
R
C v pdv vdp + pRdv 0
R C p C v
C v pdv C v vdp + p Cp C v dv 0
C v pdv C v vdp + pCp dv pC v dv 0
C v vdp + pCp dv = 0
Divide
by C v pv
dp Cp dv
+
= 0
p
Cv v
dp
dv
+
= 0
p
v
dp
dv
p + v = C
ln p + ln v C
pv C
Hence the adiabatic process follows the law
p1v1 p 2 v 2 C
a) p-V-T relationship
MECHANICAL DEPARTMENT, IIET
30
1. Relation between p and V
For a adiabatic process
p1V1 p 2 V2
V
p1
2
p2
V1
2. Relation between p and T
For a adiabatic process
p1V1 p 2 V2
V2
V1
p
1
p2
1
(1)
For perfect gas
p1V1 p 2 V2
T1
T2
p T
1 2
(2)
p 2 T1
From equ(1) and equ(2)
V2
V1
p1
p2
1
p1 T2
p 2 T1
1
p
T2
1
T1
p2
By rewriting
p
T2
2
T1
p1
1
T 1
p2
and
2
p1
T1
3. Relation between V and T
MECHANICAL DEPARTMENT, IIET
31
V
T 1
p
p
We know 1 2 and 1 1
p2
p2
V1
T2
V2
T1 1
V1
T2
T 1
1
T2
V2
V1
1
1
1
T
1
T2
V2
V1
b) Work done
2
W1-2
pdV
1
Since pV C and p
C
; p1V 1 p 2 V 2 C
V
2
W1-2
dV
1 V
C
2
C V dV
1
C 1
V2
V11 C = p1V 1 p 2 V 2
1
1
CV2 1 CV11
1
1
p 2 V 2 V2 1 p1V 1V11
1
1
p2 V2 p1V1
1
MECHANICAL DEPARTMENT, IIET
32
1
p1V1 p2 V2
1
1
p1V1 p 2 V2
1
W1-2
For perfect gas p1V1 mRT1 and p 2 V2 mRT2
1
mRT1 mRT2
1
mR
T1 T2
1
W1-2
W1-2
c) Change in internal energy
dU
mC v T2 T1 and Q 12 0
dU
W12
dU
1
p 2V2 p1V1
1
mR
T2 T1
1
or
dU
d) Heat supplied
Q1-2 0
e) Change in Entropy
2
Q
T
1
dS
Since Q 0
S2 S1 0
S2 S1 i.e.Entropy at finalstate is equal to that of the initial state
1. 2 kg of an ideal gas is compressed adiabatically from pressure 100 kPa and temperature 220
K to a final pressure of 400 kPa. Find 1) initial volume 2) Final volume and temperature
3) Work performed 4) Heat transferred 5) Change in internal energy. 6) Change in entropy
energy. Assume CP = 1 kJ/kg-K and CV = 0.707 kJ/kg-K
Given
p1 = 100 kPa = 100 kN/m2
p2 = 400 kPa = 400 kN/m2
T1 = 220 K
m = 2 kg
MECHANICAL DEPARTMENT, IIET
33
CP = 1 kJ/kg-K
CV = 0.707 kJ/kg-K
R = CP - CV =1 - 0.707 = 0.293 kJ/kg-K and
For perfect gas p1V1 mRT1
V1
mRT1 2 0.293 220
p1
100
V1 1.29 m3
For a adiabatic process
p1V1 p 2 V2
V
p1
2
p2
V1
p
V2 V1 1
p2
1
100
1.29
400
1
1.414
V2 0.484 m3
We kmow p 2 V2 mRT2
T2
p 2 V2 400 0.484
mR
2 0.293
T2 330.4 K
Wok done W
1
p1V1 p 2V2
1
1
100 1.29 400 0.484 156.04 kJ
1.414 1
W
Wok done on the gas 156.04 kJ
mR
Also W 1 T1 T2
2 0.293
W 1.414 1 220 330.4 156.27 kJ
Q 0
dU
W
W 156.04 kJ
dU
156.04 kJ
dS S2 S1 0
S2 S1 i.e.Entropy at finalstate is equal to that of theinitial state
MECHANICAL DEPARTMENT, IIET
Cp
Cv
1
1.414
0.707
34
2. The temperature of 1 kg of ideal gas is fallen from 240 oC to 115 oC when it is expanded
adiabatically to twice of its initial volume. The gas does 90 kJ of work in the process. Find
1) CP and CV
T1 = 220 oC = 513 K
Given
T2 = 115 oC = 388 K
m = 1 kg
W 90 kJ
For adiabatic process Q = 0 i.e dU = - W
dU mC v T2 T1 90
Cv
90
90
m T2 T1
1 388 513
C v 0.72 kJ kg K
1
T 1
1
T2
Taking logarithm on both side
V
We know 2
V1
V
ln 2
V1
1 T1
ln
1 T2
T
ln 1
T2
1
V
ln 2
V1
T
513
ln 1
ln
T2 1
388
1
ln 2
V
ln 2
V1
1.4
We know
Cp
Cv
Cp C v 1.4 0.72
Cp 1.008 kJ kg K
3. A certain perfect gas is contained in a perfectly insulated piston cylinder arrangement at 100
kPa and 17 oC temperature. When the gas undergoes a reversible adiabatic compression
process to 500 kPa, its temperature rises to 77 oC. If the work done on the gas during the
compression process is 50 kJ, evaluate the adiabatic exponent, CV, the characteristic gas
constant and molecular mass of the gas.
Given
T1 = 17 oC = 290 K
MECHANICAL DEPARTMENT, IIET
35
T2 = 77 oC = 350 K
p1 = 100 kPa = 100 kN/m2
p2 = 500 kPa = 500 kN/m2
m = 1 kg
W 50 kJ
For adiabatic process Q = 0 i.e dU = - W
dU mC v T2 T1 50
Cv
50
50
m T2 T1
1 350 290
C v 0.833 kJ kg K
1
p
2
p1
Taking logarithm on both side
We know
T2
T1
T
ln 2
T1
1 p1
ln
p2
T
ln 2 ln 350
1
T1 290 0.1168
p
500
ln 1 ln
100
p2
1 0.1168
1
1.132
1 0.1168
Cp
We know
Cv
C p C v 1.132 0.833
Cp 0.943 kJ kg K
R 1 C v 1.132 1 0.833 0.1099 kJ kg K
Molecular mass
Ru
8.314
75.65
R
0.1099
4. Air is compressed from 1 bar and 27 oC to 5 bar isothermally. Then it receives heat at
constant pressure and finally returns to its original state following constant volume path.
Determine 1) Work transfer, 2) Change in internal energy, 3) Heat transfer and 4) Change in
entropy per kg for each process and also for the entire cycle.
Given
p1 = 1 bar = 100 kN/m2
p2 = p3= 5 bar = 500 kN/m2
MECHANICAL DEPARTMENT, IIET
36
T1 = 27 oC = 300 K
m = 1 kg
v1= v3
For Air
1.4
Cp 1.008
kJ kg K
C v 0.721 kJ kg K
R 0.287 kJ kg K
For unit mass , p1v1 RT1
v1
R T1 0.287 300
0.861m3 kg
p1
100
1. Process 1-2 (Constant temperature process)
Work done
p
100
w1-2 p1v1 ln 1 100 0.861 ln
138.57 kJ kg
500
p2
Change in internal enrgy
u 2 u1 0 as T2 T1
Heat Transfer
q1-2 w1-2 138.57 kJ kg
Change in entropy
p
s 2 s1 R ln 1
p2
R T2
v2
and
p2
v2
100
0.287 ln
0.462 kJ kg K
500
Since T1 T2
0.287 300
0.172 m3 kg
500
2. Process 2-3 (Constant pressure process)
v
v2
3 and Since v3 v1 and T2 T1
T2
T3
T3 T2
v3
v
0.861
T1 1 300
1501.7 K
v2
v2
0.172
Work done
w 2-3 p 2 v3 v 2
p2 v1 v2 500 0.861 0.172 344.5
MECHANICAL DEPARTMENT, IIET
kJ kg
37
Changein internal enrgy
u 3 u 2 C v T3 T2 0.7211501.7 300 866.43 kJ kg
Heat Transfer
q 2-3 u 3 u 2 w 2-3 866.43 344.5 1210.93 kJ kg
Changein entropy
T
1501.7
s3 s 2 Cp ln 3 1.008 ln
1.62 kJ kg K
300
T2
3. Process 3-1 (Constant volume process)
Work done
w 3-1 0 as v3 v1
Changein internal enrgy
u1 u 3 C v T1 T3 0.721 300 1501.7 866.43 kJ kg
Heat Transfer
q 3-1 u1 u 3 w 3-1 866.43 0 866.43 kJ kg
Changein entropy
T
300
s3 s1 C v ln 1 0.721 ln
1.16 kJ kg K
1501.7
T3
5. Air is compressed from 1 bar and 27 oC to 10 bar adiabatically. Then it is expanded
isothermally up to initial pressure and finally returns to its initial conditions under constant
pressure. Determine 1) Work transfer, 2) Change in internal energy, 3) Heat transfer and 4)
Change in entropy 5) Change in enthalpy per kg for each process and also for the entire
cycle.
Given
p1 = p3 =1 bar = 100 kN/m2
p2 = 10 bar = 1000 kN/m2
T1 = 27 oC = 300 K
m = 1 kg
p1= p3
T2 = T3
For Air
1.4
Cp 1.008
kJ kg K
C v 0.721 kJ kg K
R 0.287 kJ kg K
MECHANICAL DEPARTMENT, IIET
38
1. Process 1-2 (Adiabatic process)
Work done
p
T2
2
T1
p1
1
1
1.41
p
1000 1.4
T2 T1 2 300
579.2 K
100
p1
R
0.287
w1-2
300 579.2 200.33 kJ kg
T1 T2
1
1.4 1
Change in internal energy
u 2 u1 C v T2 T1 0.721 579.2 300 201.3 kJ kg w
Change in enthalpy
h 2 h1 C p T2 T1 1.008 579.2 300 281.43 kJ kg
Heat Transfer
q1-2 0
Change in entropy
Since q 0
s 2 s1 0
s 2 s1
2. Process 2-3 (Constant temperature process)
Work done
p
p
1000
w 2-3 RT2 ln 2 RT2 ln 2 0.287 579.2 ln
382.76 kJ kg
100
p1
p3
Change in internal energy
u 3 u 2 0 as T2 T3
Change in enthalpy
h 3 h 2 0 as T2 T3
Heat Transfer
q 2-3 w 2-3 382.76 kJ kg
Change in entropy
p
1000
s3 s 2 R ln 1 0.287 ln
0.661 kJ kg K
100
p2
MECHANICAL DEPARTMENT, IIET
39
3. Process 3-1 (Constant pressure process)
Work done
v1
R T1 0.287 300
0.861 m3 kg
p1
100
v3
R T3 R T2 0.287 579.2
1.662 m3 kg
p3
p1
100
w 3-1 p1 v1 v3
100 0.861 1.662 80.1
kJ kg
Change in internal energy
u1 u 3 C v T1 T3 C v T1 T2 0.721 300 579.2 201.30 kJ kg
Change in enthalpy
h1 h 3 Cp T1 T3 Cp T1 T2 1.008 300 579.2 281.4 kJ kg
Heat Transfer
q 3-1 u1 u 3 w 3-1 201.30 80.1 281.4 kJ kg
Change in entropy
T
T
300
s3 s 2 Cp ln 1 C p ln 1 1.008 ln
0.663 kJ kg K
T
T
579.2
2
3
4. Polytropic process
Each of the isochoric, isobaric, isothermal and adiabatic processes are quasi-static in operation
under ideal working conditions. For all these processes, the relation between pressure and
volume can be described by a single equation pV n C where n is a positive number. The value
of index n remains constant throughout in any particular process. The equation pV n C is
called polytropic law, and the process in which the state of the system changes in accordance
with this correlation is called a polytropic process.
MECHANICAL DEPARTMENT, IIET
40
a) p-V-T relationship
1. Relation between p and V
For a polytropic process
p1V1n p 2 V2 n
V
p1
2
p2
V1
n
2. Relation between p and T
For a polytopic process
p1V1n p 2 V2 n
V2
V1
p
1
p2
1n
(1)
For perfect gas
p1V1 p 2 V2
T1
T2
p T
1 2
p 2 T1
From equ(1) and equ(2)
V2
V1
p1
p2
T2
T1
1n
p1 T2
p 2 T1
p
1
p2
(2)
1 n
n
By rewriting
p
T2
2
T1
p1
n 1
n
n
T n 1
p2
and
2
p1
T1
3. Relation between V and T
MECHANICAL DEPARTMENT, IIET
41
n
n
V
T n 1
p
p
We know 1 2 and 1 1
p2
p2
V1
T2
n
n
V2
T1 n 1
V1
T2
n 1
n
1 n
V2
V1
T
1
T2
V2
V1
T n 1
1
T2
1
b) Work done
2
W1-2
pdV
1
Since pV n C and p
C
; p1V n1 p 2 V n 2 C
n
V
2
W1-2
dV
n
1V
C
2
C V n dV
1
C n 1
V2
V1 n 1 C = p1V n1 p 2 V n 2
1 n
1
CV2 n 1 CV1 n 1
1 n
1
p 2 V n 2 V2 n 1 p1V n1V1 n 1
1 n
1
p2 V2 p1V1
1 n
1
p1V1 p2 V2
n 1
1
W1-2
p1V1 p2 V2
n 1
For perfect gas p1V1 mRT1 and p 2 V2 mRT2
1
mRT1 mRT2
n 1
mR
T1 T2
n 1
W1-2
W1-2
c) Change in internal energy
MECHANICAL DEPARTMENT, IIET
42
dU mCv T2 T1
d) Heat supplied
By 1st law, Q1-2 dU W1-2 ;
dU mC v T2 T1 mC v T1 T2 and
mR
W1-2
T1 T2
n 1
mR
Q1-2 mC v T2 T1
T1 T2
n 1
R
m T1 T2
Cv
(1)
n 1
We know Cp C v R
Cp
Cv
1 R
Cv
C v 1 R
Cv
R
1
Substitute this in equ (1)
Q1-2
Q1-2
R
R
m T1 T2
n 1 1
1 n 1
m T1 T2 R
1 n 1
n mR T1 T2
1
n 1
or
Q1-2
n
polytropic work done
1
e) Polytropic specific heat
For unit mass of perfect gas,
MECHANICAL DEPARTMENT, IIET
43
q 12 du w 12
p1v1 p 2 v 2
n 1
R(T1 T2 )
C v (T2 T1 )
n 1
C v (T2 T1 )
C v (T2 T1 )
p1v1 RT1; p 2 v 2 RT2
R Cp C v
Cp C v (T1 T2 )
n 1
Cp C v
Cp C v
Cv
(T2 T1 )
n 1
1 n
(T2 T1 ) C v
Cp nC v
(T2 T1 )
1
n
Cn (T2 T1 )
Cp nC v
Where Cn
is called the polytropic specific heat.
1 n
Also
Cp
Cv
n
Cn
Cv
1 n
f) Polytropic index
For a polytropic process
p1V1n p 2 V2 n
Taking log s on both side,
ln p1 n ln V1 ln p 2 n ln V2
n ln V1 n ln V2 ln p 2 ln p1
n ln
V1
p
ln 2
V2
p1
n
ln p 2 p1
ln V1 V2
g) Change in Entropy
1. Relation in terms of v and T
MECHANICAL DEPARTMENT, IIET
44
For unit mass of perfect gas, Heat transfer, q du w
q C v dT pdv
Divide by T , we get
2
2
dT p dv
q
Cv
T T v
T
2
dv
dT
q
T Cv T R v
1
1
1
p R
pv RT; T v
s 2 s1 C v ln T2 T1 R ln v 2 v1
2. Relation in terms of p and v
For unit mass, the change in entropy
s 2 s1 C v ln T2 T1 R ln v 2 v1 and p1v1 = RT1 ; T1 =
s 2 s1 C v ln p 2 v 2 p1v1 R ln v 2 v1
p1v1
p v
;T2 = 2 2 ;C p C v R
R
R
C v ln p 2 p1 C v ln v 2 v1 R ln v 2 v1
C v ln p 2 p1 C v R ln v 2 v1
s 2 s1 C v ln p 2 p1 C p ln v 2 v1
3. Relation in terms of p and T
For unit mass, the change in entropy
s 2 s1 C v ln T2 T1 R ln v 2 v1 and p1v1 = RT1 ; v1 =
RT1
RT2
; v2 =
;C p C v R
p1
p2
T p
s 2 s1 C v ln T2 T1 R ln 2 1
T1 p 2
T
p
C v ln T2 T1 R ln 2 R ln 1
T1
p2
T
p
C v R ln 2 R ln 1
T1
p2
T
p
s 2 s1 Cp ln 2 R ln 1
T1
p2
4. Relation in terms of volume ratio
MECHANICAL DEPARTMENT, IIET
45
For unit mass of gas, heat interation is given by,
n
w
1
Divided by T
q
q
n w
p R
pv RT;
T
1
T
T v
q
n Rdv
T
1
v
2
2
q
n
dv
T 1 R v
1
1
s 2 s1
n
R ln v 2 v1
1
DISTINCTION BETWEEN DIFFERENT POLYTROPIC PROCESSES
The different polytropic processes commonly
encountered in engineering practice have been shown on
the p - V plot in figure below. Starting from the initial
state, the lower right quadrant shows the expansion
processes and the upper left quadrant shows the
compression processes.
1. If n 0,
then the identity pV n C , becomes
pV 0 C; p C which indicates a constant pressure
process.
2. If n 1, then the identity pV n C , becomes
pV C which represents a constant temperature or isothermal process.
3. If n = γ, then the identity pV n C , becomes pV C which conforms to an adiabatic
process.
4. If n , then the identity pV n C , becomes V C which indicates a constant volume
1
n
th
n
process pV C; Taking n root ; p V C; If n ; p0V C; V C
1. 3 kg of air kept at an absolute pressure of 100 kPa and temperature of 300 K is compressed
polytropically until pressure and temperature become 1500 kPa and 500 K respectively.
Evaluate 1) Polytropic index 2) Final volume 3) Work of compression, 4) Heat interaction.
Take R = 0.287 kJ/kg-K
Given
p1 = 100 kPa = 100 kN/m2
MECHANICAL DEPARTMENT, IIET
46
T1 = 300 K
p2= 1500 kPa = 1500 kN/m2
T2 = 500 K
m = 3 kg
p
T
We know 2 2
T1
p1
n 1
n
T
log 2
n 1
T1
n
p
log 2
p1
p
1500
log 2
log
p1
100
n
1.23
p2
T2
1500
500
log log log 100 log 300
p1
T1
We know p1V1 mRT1 and p 2 V2 mRT2
V1
mRT1 3 0.287 300
2.583m3
p1
100
V2
mRT2 3 0.287 500
0.287 m3
p2
1500
We know W
W
mR T1 T2
n 1
3 0.287 300 500
3 0.287 300 500
748.7 kJ
0.23
1.23 1
n
We know Q
polytropic workdone
1
1.4 1.23
Q
748.7 318.2 kJ
1.4 1
2. 0.2 m3 of an ideal gas at a pressure of 2 MPa and 600 K is expanded isothermally to 5 times
the initial volume. It is then cooled to 300 K at
constant volume and then compressed back
polytropically to its initial state. Determine 1) Net
work done, 2) Net heat transfer during the cycle.
Given
p1 = 2 MPa = 2000 kN/m2
T1 = 600 K
T2 = 300 K
MECHANICAL DEPARTMENT, IIET
47
v1= 0.2 m3
v2 = 1 m3
1. Process 1-2 (Isothermal process)
We know p1V1 = p 2 V2
p2
p1V1
2000 .2
400 kPa
V2
1
Work done
V
5 0.2
We know W1-2 p1V1 ln 2 2000 0.2 ln
643.8 kJ
V
0.2
1
2. Process 2-3 (Isochoric process)
V2 = V3
p3
p 2T3
400 300
200 kPa
T2
600
Work done
W2-3 0 as volume is cons tan t
3. Process 3-1 (Polytropic process)
We know p3V3n p1V1n
n
V3 p1
V1 p3
p
ln 1 ln 2000
p
200
n 3
1.431
V3
1
ln ln
0.2
V1
Work done
W3-1
p3V3 p1V1 200 1 2000 0.2 464.04 kJ
n 1
1.431 1
Net work done Wnet = W1-2 W2-3 W3-1 643.8 0 464.04 179.76 kJ
For cyclic process
Q W
Heat transfer during complete cycle = 179.76 kJ
AIR STANDARD CYCLE
MECHANICAL DEPARTMENT, IIET
48
In most of the power developing systems, such as petrol engine and diesel engine take in either a
mixture of fuel and air as in petrol engine or air and fuel separately and mix them in the
combustion chamber as in diesel engine.
The mass of fuel used compared with the mass of air is rather small. Therefore the properties of
mixture can be approximated to the properties of air.
Exact conditions existing within the actual engine cylinder are very difficult to determine, but by
making certain simplifying assumptions, it is possible to approximate these conditions more or
less closely. The approximate engine cycles thus analysed are known as theoretical cycles.
The simplest theoretical cycle is called the air-cycle approximation. The air-cycle approximation
used for calculating conditions in internal combustion engines is called the air-standard cycle.
The analysis of all air-standard cycles is based upon the following assumptions:
1. The gas in the engine cylinder is a perfect gas, i.e., it obeys the gas law and has constant specific
heats.
2. The compression and expansion processes are adiabatic and they take place without internal
friction, i.e., these processes are isentropic.
3. No chemical reaction takes place in the cylinder. Heat is supplied or rejected by bringing a
hot body or a cold body in contact with cylinder at appropriate points during the process.
4. The cycle is considered closed, with the same 'air' always remaining in the cylinder to repeat the
cycle.
Because of many simplifying assumptions, it is clear that the air-cycle approximation does not
closely represent the conditions within the actual cylinder. Due to the simplicity of the air-cycle
calculation, it is often used to obtain approximate answers to complex engine problems.
THE EFFICIENCY OF AIR STANDARD CYCLE
The purpose of any cycle is to produce the maximum amount of work, out of a given amount of
energy supplied to the fluid during the cycle.
The efficiency of the (in the absence of other losses) cycle is given by,
a
Work doveloped Heat supplied Heat rejected
Heat rejected
1
Heat supplied
Heat supplied
Heat supplied
The efficiency given by the above expression is known as thermal efficiency of an air standard
cycle which is also known as air-standard efficiency. The actual efficiency of a cycle is always
less than the air-standard efficiency of that cycle under ideal conditions. This is taken into
account by introducing a new term, "Relative efficiency" which is defined as
relative
Actual thermal efficiency
Air standard efficiency
CARNOT CYCLE
A Carnot cycle is a hypothetical cycle consisting of four distinct processes: two reversible
isothermal processes and two reversible adiabatic processes. The cycle was proposed in 1824 by
a young French engineer, Sadi Carnot.
The essential elements needed for making an analysis of this cycle are:MECHANICAL DEPARTMENT, IIET
49
1. A working substance which is assumed to be a perfect gas
2. Two heat reservoirs; the hot reservoir (heat source) at temperature T1 and the cold reservoir
(heat sink) at temperature T2
3. Piston- cylinder arrangement for getting the work out of the working substance. The piston
and cylinder walls (excluding the cylinder head) are taken as perfect heat insulators. The
cylinder head is imagined to provide alternatively a diathermic cover (perfect heat
conductor) and an adiabatic cover (perfect heat insulator).
The piston-cylinder arrangement is shown in figure. There is no friction to the movement of
piston inside the cylinder.
Sequence of Operation
Process 1-2 (Isothermal expansion)
During this process, heat is supplied to the working fluid at constant temperature T1 by bringing
the heat source in contact with the cylinder head through diathermic cover. The gas expands
isothermally from state point l (p1, V1) to state point 2 (p2, V2). The heat supplied is used to do
work on the working fluid and which is represented by area under the curve 1-2 on p-V plot and
is given by
V
V
Qin W1-2 p1V1 ln 2 mRT1 ln 2
V1
V1
Process 2-3 (Adiabatic expansion)
At the end of isothermal expansion i.e. state point 2, the heat source is replaced by adiabatic
cover. The expansion continues adiabatically and reversibly up to state point 3 (p3, V3). Work is
done by the working fluid at the expense of internal energy and its temperature falls to T2 at state
point 3.
Process 3-4 (Isothermal compression)
MECHANICAL DEPARTMENT, IIET
50
After state point 3, the piston starts moving inwards and the working fluid is compressed
isothermally at temperature T2. The constant temperature T2 is maintained by removing the
adiabatic cover and bringing the heat sink in contact with the cylinder head. The expansion
continues up to state point 4. The working fluid loses heat to the sink and its amount equals the
work done on the working fluid. This work is represented by area under the curve 3-4 and its
amount is given by.
V
V
Q r W3- 4 p3V3 ln 4 mRT2 ln 3
V4
V3
Process 4-1 (Adiabatic compression)
At the end of isothermal compression (state point 4), the heat sink is removed and is replaced by
adiabatic cover. The compression now proceeds adiabatically and reversibly till the working
fluid returns back to its initial state point 1. Work is done on the working fluid, the internal
energy increases and temperature is raised to T1
Since all the processes that constitute a Carnot cycle are reversible, the Carnot Cycle is referred
to as a reversible cycle. Further, a cyclic heat engine working on the Carnot cycle is called
Carnot engine and its thermal efficiency is given by
net work output Wnet
heat input
Qin
For process 2 3
1
V
T1
3
T2
V2
For process 4 1
V
T1
4
T2
V1
1
V
3
V2
V3
V
4
V2
V1
V3 V2
V4
V1
1
V
4
V1
i.e.
1
Adiabatic expansion ratio Adiabatic compression ratio
i.e. Isothermal compression ratio Isothermal expansion ratio
Net work done Wnet = Qin Q r
V
V
Wnet = mRT1 ln 2 mRT2 ln 3
V1
V4
V
V
= mRT1 ln 2 mRT2 ln 2
V1
V1
MECHANICAL DEPARTMENT, IIET
51
V
V
mRT1 ln 2 mRT2 ln 2
V1
V1
V
mRT1 ln 2
V1
V
mRT2 ln 2
V1
1
V
mRT1 ln 2
V1
1
T2
T1
Following conclusions can be made with respect to efficiency of a Carnot engine:
1. The efficiency is independent of the working fluid and depends upon the temperatures of
source and sink.
2. If T2 = 0, the engine will have an efficiency of 100 %. However, that means absence of heat
sink which is violation of Kelvin-Planck statement of the second law.
3. The efficiency is directly proportional to temperature difference (T1 - T2) between the source
and the sink.
4. Higher the temperature difference between source and sink, the higher will be the efficiency
obtained.
5. The efficiency increases with an increase in temperature of source and a decrease in
temperature of sink.
6. If T1 = T2, no work will be done and efficiency will be zero.
Reasons for a Carnot engine to be an impracticable one
Carnot cycle gives the maximum possible thermal efficiency which can be obtained for the
given temperature limits. The Carnot engine, however, is a hypothetical device and it is not
possible to devise it due to the following reasons
a) All the four processes have to be reversible. This necessitates that the working fluid must
have no internal friction between the fluid particles and no mechanical friction between the
piston and cylinder walls.
b) For attaining isothermal operation, the piston movement is required to be very slow.
However, the piston must move fast for the adiabatic process. A variation in the speed of the
piston during different processes of a cycle is rather impossible.
c) There is insignificant difference in the slopes of isothermal and adiabatic lines. Consequently
the p- V plot is greatly extended both in the horizontal and vertical directions. The cylinder
then involves greater pressures and volumes, and as such becomes bulky and heavy.
1. The work out put of a Carnot engine operates between two heat reservoirs is 60% of the heat
rejected. The difference in temperature between the source and the sink is 200 K. Calculates
the thermal efficiency, source temperature, and sink temperature.
MECHANICAL DEPARTMENT, IIET
52
Wnet 0.6 Q r
Qin Wnet Q r 0.6 Q r Q r 1.6 Q r
net work output 0.6
0.375 37.5%
heat input
1.6
Efficiency is also given by
T1 T2
T1
But T1 T2 200 K
0.375 T1 200
200
533.3K
0.375
T2 T1 200 533.3 200 333.3K
T1
2. A Carnot engine works with adiabatic compression ratio of 5 and isothermal expansion ratio
of 2. The volume of air at the beginning of the isothermal expansion is 0.3 m 3. If the
maximum temperature and pressure is limited to 550 K and 21 bar, determine 1) minimum
temperature in the cycle 2) thermal efficiency of the cycle 3) Pressure at all important points.
Take 1.4
V1 = 0.3 m3
T1 = 550 K
p1 21bar 2100 kN m 2
V1
1
=
V4
5
V2
=2
V1
For process 4 1
1
V
T1
4
T2
V1
T1
550
T2
288.92 K
1
51.41
V4
V1
Efficiency
1
1
T2
T1
288.92
47.47 %
550
For isothermal process 1 2
p1V1 p 2 V2
p 2 p1
V1
1
2100 1050 kN m 2 10.50 bar
V2
2
MECHANICAL DEPARTMENT, IIET
53
For isotropic process 2 3
p 2 V21.4 p3V31.4
1.4
V
p3 p 2 2
V3
1.4
1
1050
5
110.31 kN m 2 1.1031 bar
For isotropic process 4 1
p 4 V41.4 p1V11.4
1.4
p4
V
p1 1
V4
1.4
1
2100
5
220.63 kN m 2 2.2063 bar
OTTO CYCLE (CONSTANT VOLUME CYCLE)
Otto cycle is a theoretical cycle of spark ignition engine. Air standard Otto cycle consists of four
reversible processes. Heat is added and rejected at constant volume. Expansion and compression
takes place adiabatically.
Sequence of Operation
Process 1-2 (Adiabatic compression)
During this process, m kg of air inside the cylinder at state 1 (p1, V1, and T1) is compressed
isentropically to state point 2 (p2, V2, and T2) by doing work on the air. In this process, no heat is
absorbed or rejected by the air. This is represented by the curve 1-2 on p-V plot and by a vertical
line 1-2 on T-S diagram.
Process 2-3 (Constant volume heating)
At the end of isentropic compression, the heat is supplied to the air at constant volume from an
external body till the state 3 (p3, V3, T3) is reached. This is represented by a vertical line 2-3 on
p-V diagram and by a curve 2-3 on T-S diagram.
Heat supplied, Qin mCv T3 T2
Process 3-4 (Adiabatic expansion)
MECHANICAL DEPARTMENT, IIET
54
After state 3, the air is expanded isentropically to state 4 (p4, V4, T4) doing external work done as
shown in curve 1-2 on p-V plot and by a vertical line 1-2 on T-S diagram. In this process, there
is no heat absorption or rejection by the air.
Process 4-1 (Constant volume cooling)
At the end of adiabatic expansion (state point 4), heat is rejected at constant volume to external
cold body till initial state point 1 is reached. This is represented by a vertical line 2-3 on p-V
diagram and by a curve 2-3 on T-S diagram. Thus the air finally returns to its original state after
completing the cycle.
Heat rejected, Qr mCv T1 T4 mCv T4 T1
Air standard efficiency of Otto cycle is given by
Qin Q r
Wnet
Qin
Qin
m C v T3 T2 m C v T4 T1
m C v T3 T2
m C v T3 T2 m C v T4 T1
1
m C v T3 T2
T4 T1
T3 T2
(1)
For process 1 2
V
T2
1
T1
V2
1
r 1 , where r
V1
, the compression ratio
V2
T2 T1 r 1
(2)
For process 3 4
V
T3
4
T4
V3
1
T3 T4 r 1
V
1
V2
1
r 1
(3)
MECHANICAL DEPARTMENT, IIET
55
Substituting equ(2) & equ(3) in equ(1)
T4 T1
T3 T2
T4 T1
1
1
T r
4
1
r
1
1
T1 r 1
T4 T1
T4 T1
1
1
r
1
From the above equation, it can be observed that the efficiency of the Otto cycle is mainly the
function of compression ratio for the given ratio of Cp and Cv and increases with increasing
compression ratio. However, practically the compression ratio of petrol engines is restricted to
maximum of 9 or 10 due to the phenomenon of knocking at high compression ratios.
Note
compression ratio, r
r
V1
V
4
V2
V3
V
Total volume
Clerance volume +Stroke volume Vc + Vs
1 s
Clerance volume
Clerance volume
Vc
Vc
1. The compression ratio of an engine working on Otto cycle is 8.5:1. The temperature and
pressure at the beginning of compression is 366 K and 0.93 bar respectively. The maximum
pressure in the cycle is 38 bar. Determine the temperature and pressure at the salient points
of the cycle and air standard efficiency. Take γ = 1.4
Given T1= 366 K, p1= 0.93 bar, p3= 38 bar, compression ratio, r
MECHANICAL DEPARTMENT, IIET
V1
V
4 8.5
V2
V3
56
For process 1 2
V
T
We know 2 1
T1
V2
1
r 1
T2 T1 r 1
T2 366 8.51.41 861.5 K
V
p
We know 2 1 r
p1
V2
p 2 p1 r
p 2 0.93 8.51.4 18.6 bar
For process 2 3
We know
T3 p3
T2 p 2
p
T3 3 T2
p2
38
861.5 1760.1K
18.6
For process 3 4
V
T
We know 3 4
T4
V3
T
1760.1
T4 31
r
8.51.41
1
V
1
V2
1
r 1
747.8 K
V
p
We know 3 4 r
p4
V3
p
p 4 3
r
p4
38
1.9 bar
8.51.4
Air s tan dard efficiency 1
1
1
r
1
1
1
r
1
1
57.5%
8.51.41
DIESEL CYCLE (CONSTANT PRESSURE CYCLE)
Diesel cycle is a theoretical cycle on which a diesel engine works. Air standard diesel cycle
consists of four reversible processes. Heat is added at constant pressure and rejected at constant
volume. Expansion and compression takes place adiabatically.
MECHANICAL DEPARTMENT, IIET
57
Sequence of Operation
Process 1-2 (Adiabatic compression)
During this process, m kg of air inside the cylinder at state 1 (p1, V1, and T1) is compressed
isentropically to state point 2 (p2, V2, and T2) by doing work on the air. In this process, no heat is
absorbed or rejected by the air. This is represented by the curve 1-2 on p-V plot and by a vertical
line 1-2 on T-S diagram.
Process 2-3 (Constant pressure heating)
At the end of isentropic compression, the heat is supplied to the air at constant volume from an
external body till the state 3 (p3, V3, T3) is reached. This is represented by a vertical line 2-3 on
p-V diagram and by a curve 2-3 on T-S diagram.
Heat supplied, Qin mCp T3 T2
Process 3-4 (Adiabatic expansion)
After state 3, the air is expanded isentropically to state 4 (p4, V4, T4) doing external work done as
shown in curve 1-2 on p-V plot and by a vertical line 1-2 on T-S diagram. In this process, there
is no heat absorption or rejection by the air.
Process 4-1 (Constant volume cooling)
At the end of adiabatic expansion (state point 4), heat is rejected at constant volume to external
cold body till initial state point 1 is reached. This is represented by a vertical line 2-3 on p-V
diagram and by a curve 2-3 on T-S diagram. Thus the air finally returns to its original state after
completing the cycle.
Heat rejected, Qr mCv T1 T4 mCv T4 T1
Air standard efficiency of Otto cycle is given by
Qin Q r
Wnet
Qin
Qin
MECHANICAL DEPARTMENT, IIET
58
m Cp T3 T2 m C v T4 T1
m Cp T3 T2
m Cp T3 T2 m C v T4 T1
m Cp T3 T2
C T T
1 v 4 1
Cp T3 T2
1 T T
1 4 1
T3 T2
Compression ratio r
Cut-off ratio
(1)
V1
,
V2
V3
,
V2
Expansion ratio r1
V4
V V
V
1
1
4 2 1
r
V3
V2 V3
V2 V3
V2
For process 1 2
V
T2
1
T1
V2
1
r 1
T2 T1 r 1
(2)
For process 2 3
T3
V
3
T2
V2
T3 T2
T3 T1 r 1
(3)
For process 3 4
V
T4
3
T3
V4
1
T4 T3
r
T4 T1 r
1
T4 T1
1
V4
V3
1
1
r
1
r
1
1
r
1
(4)
MECHANICAL DEPARTMENT, IIET
59
Substituting equ(2) , equ(3) and equ(4) in equ(1)
1 T T
1 4 1
T3 T2
T1 T1
1
1
T1 r 1 T1 r 1
1 1 1
1 1
r 1
From the above equation, it can be observed that the efficiency of the Diesel cycle is function of
compression ratio, cut-off ratio and ratio of specific heats. Practically the compression ratio of
diesel engines ranges from 15 to 25.
1. In a diesel engine the compression ratio is 13:1. The volume at cut off is 19.6 m3. The
clearance volume is 1 m3. Find the standard efficiency of the engine. Take γ = 1.4
Given, Compression ratio, r
V1
13, 1.4, Volume at cut off V3 1.96 m3 ,
V2
Clearence volume V2 1m3 , Cut off ratio
V3 1.96
1.96
V2
1
1 1 1
Air s tan dard efficiency 1 1
r 1
1.4
1 1 1.96 1
1 1.41
58.3%
13
1.4 1.96 1
2. In an ideal diesel cycle, the temperatures at the beginning and end of the compression are
330 K and 876 K respectively. The temperatures at the beginning and end of the expansion
are 2223 K and 1143 K respectively. Find the ideal efficiency of the cycle. Take γ = 1.4 If
the compression ratio is 14 and the pressure at the beginning of the compression is 1 bar,
calculate the maximum pressure in the cycle.
Given, Compression ratio, r
V1
14, 1.4, p1 1bar, Clearence volume V2 1m3 ,
V2
T1 330 K, T2 876 K , T3 2223 K, T4 1143 K
1 T T
Air s tan dard efficiency 1 4 1
T3 T2
1 1143 330
1
56.9 %
1.4 2223 876
We know p1v1 p 2 v 2
v
1.4
p2 p1 1 1 14 40.23 bar
v2
MECHANICAL DEPARTMENT, IIET
60
HEAT TRANSFER
Heat transfer is defined as energy-in-transit due to temperature difference. Heat transfer takes
place whenever there is a temperature difference within a system or whenever two systems at
different temperatures are brought into thermal contact.
Generally heat transfer takes place in three different modes: conduction, convection and
radiation. In most of the engineering problems heat transfer takes place by more than one mode
simultaneously.
1. Conduction
Conduction heat transfer takes place whenever a temperature difference exists in a stationary
medium. Conduction is one of the basic modes of heat transfer. On a microscopic level,
conduction heat transfer is due to the elastic impact of molecules in fluids, due to molecular
vibration and rotation about their lattice positions and due to free electron migration in solids.
The fundamental law that governs conduction heat transfer is called Fourier's law of heat
conduction, it is an empirical statement based on experimental observations. It states that the
rate of heat flow, through a homogeneous solid is directly proportional to the area, of the
section at right angles to the direction of heat flow, and to the temperature difference along the
path of heat flow.
Qx k A
dT
dx
dT
is the
dx
temperature gradient in x-direction. A is the cross-sectional area normal to the x-direction and
k is a proportionality constant and is a property of the conduction medium, called thermal
conductivity. The '-' sign in the above equation states that in spontaneous process heat must
dT
always flow from a high temperature to a low temperature (i.e.
must be negative).
dx
In the above equation, Q x is the rate of heat transfer by conduction in x-direction.
2. Convection
Convection heat transfer takes place between a surface and a moving fluid, when they are at
different temperatures. In a strict sense, convection is not a basic mode of heat transfer as the
heat transfer from the surface to the fluid consists of two mechanisms operating simultaneously.
The first one is energy transfer due to molecular motion (conduction) through a fluid layer
adjacent to the surface, which remains stationary with respect to the solid surface. Superimposed
upon this conductive mode is energy transfer by the macroscopic motion of fluid particles by
virtue of an external force, which could be generated by forced convection or generated due to
buoyancy, caused by density difference.
The rate of heat transfer by convection between a surface and adjacent fluid is given by
Qc h c A Ts Tf
where Qc = Rate of convective heat transfer, W
h c = Convective heat transfer coefficient, W m 2 K
A = Surface area, m 2
Ts = Surface Temperature, K
Tf = Fluid Temperature, K
MECHANICAL DEPARTMENT, IIET
61
The above equation is also referred to as Newton's law of cooling. It states that the rate of heat
loss from a body is proportional to the difference in temperature between the body and its
surrounding.
The convective heat transfer coefficient (hence heat transfer by convection) depends on the
temperature difference near the surface and the thermal conductivity of the fluid. The
temperature difference near the surface depends on the rate at which the fluid near the surface
can transport energy into the mainstream. Thus the temperature difference depends on the flow
field i.e. with higher velocities, high is the temperature difference and so higher heat transfer
rates. The convective heat transfer coefficient can vary widely depending upon the type of fluid
and flow field and temperature difference.
3. Radiation
Radiation is another fundamental mode of heat transfer. Unlike conduction and convectionradiation heat transfer does not require a medium for transmission as energy transfer occurs due
to the propagation of electromagnetic waves. A body due to its temperature emits
electromagnetic radiation, and it is emitted at all temperatures. It is propagated with the speed of
light in a straight line in vacuum. Its speed decreases in a medium but it travels in a straight line
in homogenous medium.
The radiation energy emitted by a surface is given by Stefan-Boltzmann's law. The Stefan–
Boltzmann’s law, states that the total energy radiated per unit surface area of a black body per
unit time is directly proportional to the fourth power of the black body's absolute temperature T
Q r A Ts 4
where Q r = Rate of thermal energy emission, W
= Emissivity of the surface
= Stefan-Boltzmann's constant = 5.669 10-8 W/m 2 -K 4
A = Surface area, m 2
Ts = Surface Temperature,K
The emissivity is a property of the radiating surface and is defined as the emissive power
(energy radiated by the body per unit area per unit time over all the wavelengths) of the surface
to that of an ideal radiating surface. The ideal radiator is called as a "black body", whose
emissivity is 1. A black body is a hypothetical body that absorbs all the incident (all wave
lengths) radiation.
MECHANICAL DEPARTMENT, IIET
