Unit-I :Definitions and Concepts of Thermodyanmics
Thermodynamic System : In the study of thermodynamics, we always focus our attention on certain mass of a substance. Whenever this substance under consideration passes through some process, its properties are changed. For observing the changes in the properties, the substance is separated out from its surrounding effects. The substance under consideration is enclosed in a space having definite boundary. This boundary may be real or imaginary. Everything within this boundary is called thermodynamic system or simply a system.
System Boundary Surrounding
Thermodynamic
}
Energy Mass{
System
The boundary itself called System Boundary. Everything outside the system boundary is known as Surroundings. The system and surroundings may interact each other across the system boundary.
Following are the possible interactions between the system and surrounding;
(a) Mass Interaction : The substance (i.e. mass) may enter or leave the system through the boundary. This indicates the flow of mass through the system boundary.
(b) Energy Interaction : The energy may transfer through the system boundary in either direction i.e. from system to surrounding or vice-versa. The dominating forms of energy in the thermodynamics systems are Heat and Mechanical Work. So the heat may enter the system or leave it and also the mechanical work may be done on the system (by the surrounding) or may be delivered by the system (to the surrounding).
Types of Systems
(1) Open System : When the mass as well as energy both cross the system boundary then the system is called an open System.
(2) Closed System : When only the energy crosses the system boundary and no mass. It means there is no mass flow across the system boundary. Such type of the system is called a closed system.
(3) Isolated System : When neither mass nor energy cross the system boundary then it is called an isolated system.
Sign Convention
Heat: heat supplied to the system (substance) or heat absorbed by the system (substance) is taken as +ve
System
Heat : heat rejected by the system (substance) is taken as -ve
Work : Work done by the system on the surrounding is taken as +ve Work : Work done on the system by the surrounding is taken as -ve
Thermodynamic Properties : A number of parameters or variables are associated with the substance under consideration. The particular values of these parameters decide the condition of substance. In general these variables are known as thermodynamic properties of the substance (or system).
Pressure (p), Volume (V), Temperature (T), Internal Energy (U), Enthalpy (H), Entropy (S) Specific Values The value of a property per unit mass (m) is called its specific value.
Sp. Volume=V/m=v m3/kg; Sp. Internal Energy =U/m=u J/kg; Sp. enthalpy (h), Sp. Entropy (s) Types of Properties
(i) Extensive Properties : When the value of a thermodynamic property depends upon the mass of the substance, it is known as extensive property.
e.g. Volume, Internal Energy, Enthalpy and Entropy
(ii) Intensive Properties : When the value of a thermodynamic property do not depend upon the mass of the substance, it is known as intensive property.
e.g. Pressure and Temperature
Thermodynamic State : The set of particular values of the thermodynamic properties at any time instant determines the condition of the substance. This particular condition of the substance at any instant of time is called the thermodynamic state or simply state.
A thermodynamic state of a substance is shown by a point on a property diagram.
Equation of State
P For a perfect (or ideal) gas at any state,
P1 1 referred as Equation of State
V1 V Where P= absolute pressure, V= total volume
T=absolute temperature, R=Gas constant
Gas constant (R) of a particular gas is obtained from universal gas constant (Ru) from the following PV = mRT
relation,
M
R Ru
,
where Ru=8314 J/kmol K, and ‘M’ is the molecular mass of the gas under consideration.Other Forms of Equation of State
1. Dividing the above equation both side by the mass of the gas, we have m RT
m m
P V
Pv RT, here v is the specific volume of the gas
2. Dividing the above equation both side by the total volume (V) of the gas, we have V RT
m V
P V
PRT, here ‘ρ' is the density of the gas under consideration 3. Dividing and multiplying the RHS of Eqn. of state by the molecular mass of the gas;
MR
TM PV m
PVNRuT, where ‘N’ is the no. of moles of the gas
Specific Heat of a Substance : The specific heat of a substance is the amount of the heat absorbed (on heating) or rejected (on cooling) by the unit mass of that substance for unit degree change in its temperature. It is denoted by ‘C’ and its SI unit is J/kgK. Every solid and liquid has only one value of specific heat.
Two Specific Heats of a Gas : Since the heating or cooling of a gas may take place either on constant pressure or constant volume. Therefore every gas has two values of specific heat as given below;
Specific Heat at Constant Pressure(Cp) : Cp of a gas is the amount of the heat absorbed (on heating) or rejected (on cooling) by the unit mass of that gas for unit degree change in its temperature at constant pressure.
Specific Heat at Constant Voulme(CV) : CV of a gas is the amount of the heat absorbed (on heating) or rejected (on cooling) by the unit mass of that gas for unit degree change in its temperature at constant Volume.
For an ideal or perfect gas, CP > CV and CP - CV =R i.e. the gas constant
Thermodynamic Process : When the state of a substance changes from one to another in some define manner, it is said to be that the substance undergoes a thermodynamic process. A thermodynamic process is shown by a line (or curve) on a property diagram.
P
P2 2 State1: Pressure-P1, Volume-V1, Temperature-T1 State2: Pressure-P2, Volume-V2, Temperature-T2
P1 1 Indicates the direction of process
i.e. from state 1 to state 2 and is called process1-2 V1 V2 V
This line or curve is called path of the process and it is obtained by joining the intermediate equilibrium states between 1 & 2.
General Gas Equation (Eqn. of a Process) :
Let, m=mass of the gas under consideration and ‘R’ is the gas constant of the gas
From eqn. of state, for state1 P1V1 mRT1---(i) and for state2 P2V2 mRT2---(ii) Dividing eqn.(i) by eqn. (ii)
2 1 2
2 1 1
mRT mRT V
P V P
2 1 2 2
1 1
T T V P
V
P ---(iii)
The above equation (iii) is referred as General Gas eqn.
Important Definitions
Work Done : The work done during a thermodynamic process is given by the area enclosed by the process line with the volume axis i.e. the area 1-2-V2-V1 as given in the following figure;
P
P2 2 Taking an elementary strip of thickness dv at pressure, p p The strip is so small that the pressure remains constant over dv P1 1 Area of element or elemental work
Wpdv
V1 dv V2 V Work done during the whole process 1-2
2 1
2 1
pdv W
Therefore the work done during the process 1-2,
Internal Energy : The internal energy of a substance is the energy associated with the intermolecular structure of the substance. Internal energy of a substance includes various type energies of molecules, atoms and particles of the substance. The various types of energies which provide the internal energy to a substance are potential, kinetic and vibrational energy etc.
The internal energy of a substance directly depends upon its absolute temperature.
The elemental change of internal energy of a substance during a thermodynamic process;
dT mC
dU v , for whole process 1-2 2
1 2 1
vdT mC
dU U2U1mCvT2T1 Therefore change of internal energy during a thermodynamic process 1-2,
Flow Work : The product of pressure (absolute) and total volume of a gas at any state is called flow work at that state.
Flow work at state1=P1V1, similarly, Flow work at state2=P2V2
Change of flow work during a thermodynamic process 1-2, =
P2V2 P1V1
From eqn. of State,
P2V2P1V1
=
mRT2mRT1
=2 2 2 1
1 1
T V P T
V P
2 2 2 1
1 1
T V P T
V P
21 12 pdv W
2 1
v 12 mC T T
U
T2 T1
mR
Enthalpy : The sum of internal energy and flow work of a perfect gas at any state is called enthalpy of the gas at that state. It is denoted by ‘H’.
Enthalpy at state1, H1 U1P1V1, similarly, Enthapy at state2, H2 U2P2V2
Change of enthalpy during a thermodynamic process 1-2, =H2H1
U2 U1
P2V2P1V1
H12 U12mR
T2T1
H12 mCv
T2T1
mR
T2T1
H12 m
CvR
T2T1
or (as Cp-Cv=R) Entropy : Entropy of a substance is a thermodynamic property such that it changes when substance absorbs or rejects the heat. In general when the substance absorbs the heat its entropy increases and vice-versa. It is denoted by ‘S’.For analyzing the change of entropy during a thermodynamic process, a T-S diagram is drawn.
T
T2 2 Taking an elementary strip of thickness dS at temperature,T T Stip is so small that the temperature remains constant over dS T1 1 ∂Q=elemental heat transfer during the element
T = Constant temperature during the element S1 dS S2 S dS=elemental change of entropy during the element
Then,
T
dS Q, for whole process 2
1 2 1 T
dS Q
2
1 12 1
2 T
S Q S
S
General expression for Change of entropy during a thermodynamic process;
State1: Pressure-P1, Volume-V1, Temperature-T1 and entropy –S1
State2: Pressure-P2, Volume-V2, Temperature-T2 and entropy –S2
Change of entropy during the process 1-2,
Flow Process : An open system allows the flow of substance across the boundary i.e. it enters the system at one section and leaves at another. In other words there is a continuous flow of substance through the system.
When the substance in an open system undergoes a thermodynamic process, it is known as a
flow process. Open System
Substance
State1 Process Substance State 2
Non-Flow Process : A fixed mass of substance remains in a closed system. When a substance in a closed system under goes a thermodynamic process, it is known as non-flow process.
Reversible Process : When a substance under goes a thermodynamic process in such a way that the process can be reversed completely i.e. the substance regains its initial state. In such
2 1
p
12 mC T T
H
1 p 2 1
v 2
12 V
ln V P mC
ln P mC S
situations the properties of substance attain the initial values and the process is called Reversible Process.
2 A thermodynamic process will be reversible only when no internal friction is present during the process.
1 A reversible process is shown by a continuous line on the property diagram.
Irreversible Process : When a substance under goes a thermodynamic process in such a way that the process cannot be reversed completely i.e. the substance does no regain its initial state. In such situations the properties of substance do not attain the initial values and the process is called irreversible Process.
2 The cause of irreversibility in a process is the presence of internal friction during the process. The net result of internal
1 friction is to generate the heat which then absorbed by the substance. As a result, the entropy of the substance increases
during the process. On reversing the process, the internal friction will produced the same effect again i.e. to increase the entropy again and there is a net increase of entropy. It means the entropy of substance does not attain the initial value on reversing and hence the process is irreversible.
An irreversible process is shown by a dotted line on the property diagram as shown above.
Laws of Thermodynamics
(i) Zeroth Law (ii) First Law (iii) Second Law (iv) Third Law
(i) Zeroth Law of Thermodynamics : If there is no heat exchange between two bodies when they come in contact with each other, the bodies are called in thermal equilibrium.
Zeroth law of thermodynamic states that when a body ‘A’ is in thermal equilibrium with another body ‘B’. The body ‘B’ is separately
in thermal equilibrium with a third body ‘C’, then the bodies
‘A’ and C’ ’will also be in thermal equilibrium.
(ii) First Law of Thermodynamics : The law of conservation of energy states that the total energy of a system remains constant during a process. When this law of conservation of energy is applied to a thermodynamic system undergoing a thermodynamic process, it becomes the First Law of Thermodynamics.
First law of thermodynamics states that the heat transfer during a thermodynamic process is balanced by the work done and change of internal energy during the process.
B
A C
Mathematically,
When a substance undergoes a series of thermodynamic processes in 3 sequence in such a way that they form a closed figure on the
4 property diagram, then it is known as a Cyclic process.
2 Applying first law of thermodynamics for a cyclic process;
1 Qcyclic Wcyclic
U cyclicFor a cyclic process
Ucyclic0, Then, QcyclicWcyclicIn the mathematical notations, for a cyclic process the first law of thermodynamics takes the form;
The first law of thermodynamics established the fact that the heat energy can be converted into mechanical work.
The conversion of thermal energy into work is achieved when a substance undergoes a cyclic process. At one part of this process the heat is supplied to the substance and the mechanical work is given out at another part of the process. The device which converts the heat into mechanical work is called a ‘Heat Engine’.
Source: high temperature heat reservoir Sink : low temperature heat reservoir Qs HE : heat engine
W Qs= heat supplied (from source) to the heat engine Qr= heat rejected by the heat engine to the sink Qr W= mechanical work out put
Energy balance at HE, Qs Qr W
The mechanical work out put, WQs Qr, i.e. work out put=Heat supplied –Heat rejected
The thermal performance of a heat engine is expressed by a term called ‘Thermal Efficiency’ (ηth).
s r s
r s s
th Q
1 Q Q
Q Q Q
W
Limitations of First Law of Thermodynamics
(i) The first law of thermodynamics does not specify the directions of heat transfer and work done with respect to the system and surrounding.
(ii) The first law of thermodynamics is unable to impose any restriction on the amount of heat which converts into the work. Therefore we cannot determine that what portion of the heat supplied can be converted into mechanical work. These limitations of first law of thermodynamics
12 12
12 W U
Q
QW
Source
HE
Sink
s th r
Q 1 Q
open the scope of some other law that can explain the quantitative analysis of conversion of heat into work.
*A System (hypothetical) which violates the first law of thermodynamics is called Perpetual Machine of First Kind i.e PPM-I
(iii) Second Law of Thermodynamics : There are two statements of second law of thermodynamics as discussed below;
(1) Kelvin-Plank’s Statements :According to this statement it is impossible to construct a heat engine working on cyclic process which can convert all the heat supplied into mechanical work.
W=Qs Not Possible
Qs Qs W=Qs-Qr Possible
W W Qr
A part of heat supplied (Qs) must be rejected (Qr) by the heat engine for its continuous working and delivering the mechanical work (W).
(2) Clausius’s Statement : According to this statement the heat cannot flow from a body at lower temperature to the body at higher temperature on its own accounts as shown below in fig.(a).
TA<TB TA<TB Possible TA Heat TB TA Heat TB
Not Possible External source of energy
Fig.(a) Fig.(b)
The flow of heat from lower temperature to higher temperature is possible with the aid of an external source of energy as shown above in fig.(b).
A device, in which the heat flows from lower temperature body to higher temperature body with the help of an external source of energy, is called a ‘Heat Pump’. A heat pump is obtained by reversing the processes of a heat engine.
Source
HE
Source
HE
Sink
A B A B
Qa=heat abstract from low temperature heat reservoir i.e. Sink Qd =heat delivered to high temperature heat reservoir i.e. Source
Qd W= external energy given.
W Energy balance at heat pump, Qd=Qa+W ---(i)
Thermal performance of a heat pump is expressed in two ways;
Qa (a) As a Refrigerator: when the purpose of this system is to abstract the Heat from a medium which is already at lower temperature ;
Coefficient of performance,
W COPref Qa
(b) As a Heat Pump: when the purpose of this system is to deliver the heat to a medium which is already at higher temperature;
Coefficient of performance,
W W Q W
COPhp Qd a
W 1 Q COPhp a
but, a COPref W
Q , therefore, COP hp 1COPref
*A System (hypothetical) which violates the second law of thermodynamics is called Perpetual Machine of second Kind i.e PPM-II
(iv) Third Law of Thermodynamics
At a temperature of absolute zero (0 K or -273.15°C), the substances posses no thermal energy or heat. At a temperature of zero Kelvin the atoms in a pure crystalline substance are aligned perfectly and do not move. The temperature of absolute zero is the reference point for determination of entropy.
The third law of thermodynamics states that ‘The entropy of all the perfect crystalline solids is zeros at absolute zero temperature”. The third law of thermodynamics is also referred to as Nernst law. It provides the basis for the calculation of absolute entropies of the substances.
Mathematically, 0 S
limT0 , where S=entropy (J/K) and T = absolute temperature (K) Experimentally, it is not possible to obtain -273.15°C (0 K). It is found that most of the gases either
liquefy or solidify before reaching such a temperature and gaseous molecules no longer remaining.
Importance of third law of thermodynamics;
1) It helps in calculating the thermodynamic properties.
2) It explains the behavior of solids at very low temperature.
3) It helps in analyzing chemical and phase equilibrium Source
HP
Sink
Standard Thermodynamic Processes
1. Isobaric or Constant Pressure Process 2. Isochoric or Constant Volume Process 3. Isothermal or Constant Temperature Process 4. Adiabatic or No Heat Transfer Process 4. Polytropic Processes
1. Isobaric or Constant Pressure Process : When a substance undergoes a thermodynamic process in such a way that its pressure remains constant then the process is called an isobaric process.
P (i) Law or Equation of the Process Since, P1=P2, from general gas eqn., P1=P2
2 2 2 1
1 1
T V P T
V P
2
2 1 1
T V T V
T C V
i.e. the ratio of volume and abs. temperature is constant V1 V2 V (ii) Work Done
P 2
1
12 pdV
W , since p1=p2 then 2
1 1 12 p dV
W
(iii) Change of Internal Energy
P1=P2 U12 mCv
T2 T1
(iv) Heat Transfer
From Ist Law of Thermodynamics, Q12 W12U12 T1 T2 T (v) Change of Enthalpy
H12mCp
T2T1
(vi) Change of Entropy : From general expression for change of entropy during a process;
1 p 2 1
v 2
12 V
ln V p mC
ln p mC S
Here p1=p2, putting in the above eqn., we have
1 p 2 1
v 1
12 V
ln V p mC
ln p mC S
1 p 2 v
12 V
ln V mC 1 ln mC S
{as ln(1)=0}
Problem-1 : A quantity of gas at pressure 2 bar, temperature 1500C and
volume=(digits of R.No./1000)-0.05 m3. The gas is compressed at constant pressure until its volume becomes 0.1 m3. Determine; (i) Mass of the gas. (ii) Work done (iii) Heat transfer, mention whether it is absorbed or rejected. (iv) Change of enthalpy (v) Change of entropy
Take for the gas, Cp=1005 J/kgK and Cv=710J/kgK
1 2
1 2
2 1
1
12 p V V
W
1 p 2
12 V
ln V mC S
2. Isochoric or Constant Volume Process : When a substance undergoes a thermodynamic process in such a way that its volume remains constant then the process is called an isochoric process.
P (i) Law or Equation of the Process P2 2 Since, V1=V2, from general gas eqn.,
2 2 2 1
1 1
T V P T
V
P
2
2 1 1
T P T P
T C P
P1 1 i.e. the ratio of pressure and abs. temperature is constant V1=V2 V (ii) Work Done
T 2
1
12 pdV
W , since no change in volume (dV=0) then T2 2 (iii) Change of Internal Energy
U12 mCv
T2 T1
T1 1 (iv) Heat Transfer
From Ist Law of Thermodynamics, Q12 0U12 V1=V2 V (v) Change of Enthalpy
H12mCp
T2T1
(vi) Change of Entropy : From general expression for change of entropy during a process;
1 p 2 1
v 2
12 V
ln V p mC
ln p mC S
Here V1=V2, putting in the above eqn., we have
1 p 1 1
v 2
12 V
ln V p mC
ln p mC S
1 ln P mC
ln P mC
S p
1 v 2
12
{as ln(1)=0}
Problem-2 : A gas occupies a volume of 0.25 m3 at temperature 850C and pressure as follows Pressure = (digits of R.No./50) bar. The gas is heated at constant volume until its pressure becomes 11 bar. Determine; (i) Mass of the gas. (ii) Change of internal energy (iii) Heat transfer, mention whether it is absorbed or rejected. (iv) Change of enthalpy (v) Change of entropy. Take for the gas, Cp=1000 J/kgK and Cv=700 J/kgK
3. Isothermal or Constant Temperature Process : When a substance undergoes a thermodynamic process in such a way that its temperature remains constant then the process is called an isothermal process.
P (i) Law or Equation of the Process P2 2 Since, T1=T2, from general gas eqn.,
2 2 2 1
1 1
T V P T
V
P P1V1P2V2 PVC P1 1
i.e. the product of pressure and volume is constant V2 V1 V
0 W12
1 v 2
12 P
ln P mC S
12
12 U
Q
(ii) Work Done 2
1
12 pdV
W , here PVC i.e.
V
P C then 2
1
12 dV
V
W C 2
1
12 V
C dV W
1212 C lnV W
W12 C
lnV2lnV1
1 12 2
V ln V C W
,
but CPVP1V1P2V2, therefore for an isothermal process
1 1 2 1
12 V
ln V V P W
(iii) Change of Internal Energy : Since T1=T2
U12 0
(iv) Heat Transfer : From Ist Law of Thermodynamics, Q12W12U12 Q12W120 (v) Change of Enthalpy : Since T1=T2
H120
(vi) Change of Entropy : From general expression for change of entropy during a process;
1 p 2 1
v 2
12 V
ln V p mC
ln p mC S
For an isothermal process ,
2 2 1 1V PV
P
2 1 1 2
V V P P
putting the above eqn., we have
1 p 2 2
v 1
12 V
ln V V mC
ln V mC
S
1 p 2 1
v 2
12 V
ln V V mC
ln V mC S
1 v 2
p
12 V
ln V C C m
S
(as CpCv R)
Problem-3 : A certain mass of a gas occupies a volume of 0.05 m3 and at pressure 100 kN/m2 and temperature=(digits of R.No.-100)0C. The gas is compressed isothermally to a volume of 0.02 m3. Determine; (i) Mass of the gas. (ii) Work done (iii) Heat transfer, mention whether it is absorbed or rejected. (iv) Change of entropy
Take for the gas, Cp=1005 J/kgK and Cv=715 J/kgK
2 1 1
12 2
P ln P V mR
ln V mR S
12 12 W Q
1 1 2 1
1 2 1
12 V
ln V V mRT
ln V V P
W
1 2 2 1
2 2 2
12 V
ln V V mRT
ln V V P W
(4) Adiabatic or No Heat Transfer Process : When a substance undergoes a thermodynamic process in such a way that there is no heat transfer during the process then it is called an adiabatic process.
P (i) Law or Equation of the Process
P2 2 Consider an elementary strip of the process at pressure ‘P’
P volume ‘V’ and temperature ‘T’. The strip is so small that the P1 1 dV pressure remains constant as ‘P’ over the strip.
Let, dV= change of volume across the strip V1 V V2 V dT= change of temperature across the strip
dU= elemental change of internal energy and ∂w is the elemental work done across the strip Applying First Law of thermodynamics over the strip, we have;
∂Q=∂W+dU, but the process is an adiabatic then ∂Q=0, we have, ∂W+dU=0 ∂W= -dU ----(i) We know that, ∂W=pdV and dU=mCvdT, putting in the above eqn.(i), we have
dT mC
PdV v , from eqn. of state PVmRT
V
P mRT , then we have, dV mC dT V
mRT
v
T
C dT V
RdV v , Integrating both sides between the limits, 2
1 2
1
T T v V
V T
C dT V
R dV
R
lnV2lnV1
Cv
lnT2lnT1
1 v 2 1
2
T ln T V C
ln V
R ---(ii)
From general gas eqn.,
2 2 2 1
1 1
T V P T
V P
1 1
2 2 1 2
V P
V P T T
Putting in the above eqn. (ii),
1 1
2 v 2
1 2
V P
V ln P V C
ln V
R
1 1
2 v 2
1 v 2
p PV
V ln P V C
ln V C
C
1 1
2 v 2
1 v 2 1
p 2
V P
V ln P V C
ln V V C
ln V
C
1 1
2 2 1
v 2 1
p 2
V P
V ln P V ln V V C
ln V C
2 2
1 1 1 v 2
1 p 2
V P
V P V ln V V C
ln V
C
2 v 1 1
p 2
P ln P V C
ln V
C
2 1 1
2 v
p
P ln P V ln V C
C but
v p
C
C
Therefore,
2 1 1
2
P ln P V
ln V
2 1 1
2
P ln P V
ln V
2 1 1
2
P P V
V
2 1 1 2
P P V V
Equation of Adiabatic Process
The ‘γ’ is known as ‘adiabatic index’ and it is the ratio of two specific heats (Cp/Cv) of the gas under consideration. The value of ‘γ’ is always greater than unity as Cp>Cv.
* The value of ‘γ’ is about 1.6 for monatomic gases such as Argon, Neon and Helium etc.
* The value of ‘γ’ is about 1.4 for diatomic gases such as O2, N2 and H2 etc.
* The value of ‘γ’ is about 1.3 for triatomic gases such as CO2, NO2 and SO2 etc.
2 2
1V1 P V
P or PV C
Also
1 2
2 1
V ln V
P ln P
(ii) Work Done 2
1
12 pdV
W , here PV C i.e. V
P C then 2
1
12 dV
V
W C 2
1
12 V
C dV W
VV1212 V 1
1
W C
1
1 1
2
12 V V
1
W C
1 V C V
W C
1 1 1
12 2
but CPV P1V1 P2V2, Then we have
1 V V P V
V W P
1 1 1 1 2 1
2 2
12
1
V P V
W12 P2 2 1 1
1
V P V
W12 P1 1 2 2
(iii) Change of Internal Energy :
U12 mCv
T2 T1
(iv) Heat Transfer : From Ist Law of Thermodynamics,
Q12W12U12 But here Q120 (v) Change of Enthalpy :
H12mCp
T2T1
(vi) Change of Entropy during an Adiabatic Process : The change of entropy during a thermodynamics process is because of following two reasons;
(i) Change of entropy due to the external heat transfer i.e. Q12
(i) Change of entropy on absorption of the heat which is generated internally due to internal friction i.e Qgen.
Therefore, ∆S12 = (∆S12)Q12 + (∆S12)Qgen , But for an adiabatic process Q12=0 → (∆S12)Q12=0 But still there is a possibility that the entropy of substance may change during an adiabatic process because of (∆S12)Qgen.
(a) Adiabatic process is Reversible : The adiabatic process under consideration may be reversible i.e. without internal friction. In that situation the Qgen will be zero and hence (∆S12)Qgen =0
Then for a reversible adiabatic process, ∆S12=0.
* The entropy of the substance remains constant during a reversible adiabatic process. Such a reversible adiabatic process is known as an ‘Isentropic Process’.
(b) Adiabatic process is Irreversible : If the internal friction during an adiabatic process is considerable then it makes the process irreversible. The effect of internal friction is to generate the
12
12 U
W 1
V P V
W12 P1 1 2 2
heat which is absorbed by the substance itself and hence its entropy increases. Such a process is called an irreversible adiabatic process and not the isentropic process.
A T-S diagram is useful for studying the irreversibility in an adiabatic process;
T
T2 2 2/ The shaded area, 1-2-2/-1, indicates the energy loss due to Internal friction.
T1 1
S1=S2 S2/ S
Problem-4 : Air at a pressure of 1.2 bar and temperature 2000C is compressed adiabatically to a pressure of 15 bar. If the mass of the air is taken as; mass=(digits of R.No./1000) kg, calculate (i) Initial and final volume of air. (ii) Work done (iii) Heat transfer, mention whether it is absorbed or rejected. (iv) Change of enthalpy (v) Change of entropy.
Take for the gas, Cp=1005 J/kgK and Cv=700 J/kgK
(5) Polytropic Processes : The standard thermodynamics processes which we have studied can be express as follows;
a. Isobaric (Constant Pressure) Process PC→ P1C→PV0C
b. Isochoric (Constant Volume) Process VC→1VC→ P0VC→ P V C
1
Raise the power by ‘∞’ both side
C V P
1
→ PV C→ PV C
c. Isothermal (Constant Temperature) Process PV C→ PV C d. Adiabatic (No heat transfer) Process PV C→ PV C The above analysis shows that all the possible thermodynamic processes can be expressed by a single equation of the form; PVn C or P1V1n P2V2n. All such processes are known as polytropic processes.
The exponent ‘n’ is called polytropic index and which remains constant throughout a particular process. The value of ‘n’ may vary from ‘0’ to ‘∞’.
(i) Law or equation of process : P1V1n P2V2n, taking log both sides, we have
n 2 2n
1
1V ln P V P
ln →lnP1nlnV1lnP2nlnV2→ nlnV1nlnV2 lnP2lnP1
→
1 2 2
1
P ln P V ln V
n →
2 1 1 2
V ln V
P ln P
n
* The adiabatic index ‘γ’ is wholly a property of the substance which undergoes the adiabatic process.
* The polytropic index ‘n’ depends upon the process only and it is independent of the substance involved.
(ii) Work done : The equation of polytropic process is identical to the equation of an adiabatic process with the difference in the exponent of V i.e. ‘n’ instead of ‘γ’.
In the expression of work done for an adiabatic process, replacing the ‘γ’ by ‘n’, we have Work done during a polytropic process,
(iii) Change of Internal Energy : U12 mCv
T2 T1
(iv) Heat Transfer : From Ist Law of Thermodynamics, Q12W12U12 Q12W12mCv
T2T1
We know that, Cp/Cv=γ → Cp=γCv, putting this value in Cp-Cv=R → γCv -Cv=R → 1 Cv R
We get, 12 12
T2 T1
1 m R W
Q
1
mRT W mRT
Q12 12 2 1
1
V P V W P
Q12 12 2 2 1 1
1
V P V W P
Q12 12 1 1 2 2
n 1
V P V P 1 1 W n
Q12 12 1 1 2 2 12 12
W12
1 1 W n
Q
1
1 1 n
W
Q12 12 For a polytropic process
(v) Change of Enthalpy :
H12 mCp
T2T1
(vi) Change of Entropy : From general expression for change of entropy during a process;
1 p 2 1
v 2
12 V
ln V p mC
ln p mC S
For a polytropic process, P1V1nP2V2n
n 1 2 n 2 1 1
2
V V V
V P
P
, putting in above eqn.
1 p 2 n 1 v 2
12 V
ln V V mC
ln V mC S
1 p 2 1
v 2
12 V
ln V V mC
ln V nmC S
n
V ln V mC S
1 v 2 12
For a polytropic process,
n
V ln V mC S
1 v 2
12
Problem-5 : In the problem-4, If the compression is polytropic (instead of adiabatic) with n=1.2, then calculate (i) to (v) for this situation.
1 n
V P V W12 P1 1 2 2
12
12 W
1 Q n
