Download Mechanical Engineering (Electrical Branch)

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Subject
name
Semester
Name
Electrical
Mechanical
Engg
Programme
BTech 3rd
Subject code
MECH-ELE
3RD Credits
Ajaz Ali
4
Mobile no.
Teacher incharge / mentor C6
9469720362
[email protected]
Lecture 01
Topic: Engineering Thermodynamics
Links: https://www.youtube.com/watch?v=9GMBpZZtjXM

Book: Engineering Thermodynamics by P K Nag Fifth Edition, Page no. 9-13.
Lecture note:
Introduction to engineering thermodynamics
Thermodynamics is the science of energy transfer ad its causes and effects.
 In microscopic thermodynamics, the behaviour of the gas is described by summing up the
behaviour of each molecule.
 In macroscopic thermodynamics, the behaviour of the gas is described by the net effect of action
of all the molecules, which can be perceived by human senses.
System: A system is a matte or region on which analysis is done. System is separated from the
surrounding by boundary. Everything external to the system is called surroundings. System and
surrounding together is called a universe.
Mass Transfer
Energy Transfer
Example
Open System
Yes
Yes
Compressor,
etc.
Closed System
No
Yes
Piston
cylinder
arrangement, gas in a
closed container
Isolated System
No
No
Universe
Turbine
Control Volume: Volume surrounding an open system on which study is focussed.
Properties:
Properties ar point function and are exact or perfect differentials. For example; internal energy,
enthalpy, entropy.
Two types of properties: intensive properties and extensive properties
Intensive properties: Properties are independent of mass. For example; pressure, temperature,
density, specific volume, Specific heat (Cp and Cv) etc.
Extensive properties: properties are related to the mass. For example; volume, energy, Heat capacity
Specific Extensive Properties

 Extensive properties per unit mass is specific extensive properties
 It is an intensive properties. For example; Specific volume, Specific energy

It is independent of mass.
Reversible Process
  The process which can be reversed without leaving any effect on system and surrounding
 All reversible processes can be shown on diagrams. For example; P-V, T-S, P-T diagrams.
Irreversible process





All spontaneous process are irreversible process.
Irreversible process cannot be shown on diagrams. They are shown as dotted lines. For example; heat
transfer through finite temperature difference, free expansion.
A system will be in a state of thermodynamic equilibrium if the conditions for the following
three types of equilibrium are satisfied.
(i)
mechanical equilibrium
(ii)
Chemical equilibrium
(iii)
Thermal equilibrium
Pure Substance: A substance homogenous in chemical composition and homognous in
chemical aggregation.
Gibb’s Phase Rule
P+F=C+2
P= No. of phases
F= Degree of freedom
C= No. of component
Thermodynamic Cycle
 It is a series of processes when initial and final points are same.
 There is no change in property of system
Lecture 02
Temperature
Topic : Zeroth law of thermodynamics
Link: https://www.youtube.com/watch?v=xQwi9fveGTQ
Book: Engineering Thermodynamics by P.K. Nag fifth Edition, Page no.33-40
Lecture note:
Zeroth law of thermodynamics:
When a body A is in thermal equilibrium with a body B & also separately with a body C then body B
and C will be in thermal equilibrium with each other.
Zeroth law of thermodynamics is the basis of temperature measurement.
Measurement of Temperature – The Reference Points
Thermometer
Thermometric property
Constant volume gas thermometer
Pressure (P)
Constant pressure gas thermometer
Volume (V)
Electrical Resistance thermometer
Resistance (R)
Thermo couple
EMF (E)
Mercury in glass thermometer
Length (L)
Thermometers and their temperature range
Thermometers
Temperature range
o
o
Platinum resistance thermometers
-200 C to 1200 C
o
o
-200 C to 1600 C
Thermoelectric thermometers
o
Radiation pyrometers
Above 400 C
o
o
600 C to 2000 C
Segar cone
o
Optical Pyrometers
Above 650 C
o
o
Gas Thermometers
-200 C to 1200 C
Thermocouple uses copper-constantan, platinum-rhodium, chromel-alumel combinations.
Gas Thermometers
A small amount of gas is enclosed in bulb B which is in communication via the capillary
tube C with 1 limb of the mercury manometer M. the other limb of the mercury
manometer is open to adjust the mercury levels so that the mercury just touches lip L of
the capillary. The pressure in the bulb is used as a thermometric property and is given by
p  p0  M Zg
Fig. 1 Constant volume gas thermometer
Where p0 is the atmospheric pressure, M is the density of mercury.
When the bulb is brought in contact with the system whose temperature is to be measured, the bulb,
in course of time, comes in thermal equilibrium with the system. The difference in mercury level Z is
recorded and the pressure p of the gas in the bulb is estimated. Since the volume of the trapped gas is
Constant, from the ideal gas equation, T  V p
(i)
R
i.e., the temperature increase is proportional to the pressure increase.
In a constant pressure gas thermometer, the mercury levels have to be adjusted to keep Z
constant, and the volume of gas V, which would vary with the temperature of the
system, becomes the thermometric property.

T  P V
R
(ii)
i.e., the temperature increase is proportional ti the observed volume increase. The constant
volume gas thermometer is simpler in construction and easier to operate.
Ideal Gas Temperature
1. Surround the bulb with steam condensing at 1 atm, determine the gas pressure p
and calculate
  273.16
p
1000
2. Remove some gas from the bulb so that when it is surrounded by water at its triple point,
the pressure pt is 500 mmHg. Determine the new value of p and then  fro steam
condensing at 1 atm.
p

  273.16 500 
3. Continue reducing the amount of gas in the bulb so that pt and p have smaller and
smaller values, e.g., pt having say, 250 mmHg, 100 mmHg, and so on. At each value of
pt calculate the corresponding  .
4. Plot  vs. pt and extrapolate the curve to the axis where pt=0. Read from the graph

lim
pt o
The graph indicates that although the readings of a constant volume gas thermometer
depend upon the nature of the gas, all gases indicate the same temperature as pt is lowered
and made to approach zero.
Fig. 1. Ideal gas temperature for steam point
At constant pressure when the bulb is surrounded by steam condensing at 1atm and triple point of
water respectively.
  273.16 V
V
t
And  vs. p may be plotted, similar to fig. 1. It is found from the experiments that all gases
indicate the same value of  as p approaches zero.
The ideal gas temperature T is defined by either of the two equations:
T  273.16 lim p
pt 0
pt
T  273.16 lim V
vt 0
(iii)
Vt
Where  has been replaced by T to denote this particular temperature scale, the ideal gas
temperature scale.
If ps and pt are the measured pressures at the steam point and the triple point respectively,
one gets the value of the steam point temperature Ts as
V
T 273.16 lim
s
vt 0
(iv)
Vt
Which is equal to 373.15 K.
Similarly, the temperature of Ti at the ice point is
T  273.16 lim
i
pt 0
p
(v)
pt
However when extrapolated to zero pressure all curves converge, and the ratio ps/pi tends to a
constant value giving
T
p
s
Ti
 lim
pt 0
 1.366099
i
pt
(vi)
This value may be considered as a universal constant.
One may now decide to hbe a certain number of divisions between the steam point and ice point,
say 100 as in the Kelvin and Celsius scales so that
Ts  Ti  100
(vii)
Ts =373.15 K and Ti = 273.15 K
Celsius Temperature Scale
The Celsius temperature scale employs a degree of the same magnitude as that of the ideal gas scale,
but its zero point is shifted, so that the Celsius temperature of the triple point of water is 0.01 degree
o
Celsius or 0.01 C. if t denotes the Celsius temperature, then
o
t = T - 273.15
thus the Celsius temperature ts at which steam condenses at 1atm. Pressure
o
o
ts = Ts – 273.15 = 373.15−273.15= 100.00 C
o
Similar measurements for ice points show this temperature on the Celsius scale to be 0.00 C. the
only Celsius temperature which only fixed by definition is that of the triple point.
Lecture 03
Topic: Energy Interactions
Link: https://www.youtube.com/watch?v=0jXeNaSM5Xc&index=3&list=PLA17CE108925DC94F
Book: Engineering Thermodynamics by P K Nag Fifth Edition, Page no. 48 – 77
Lecture note:
Energy Interactions (Heat And Work)
Thermodynamic work: Work is said to be done by the system when the sole effect on things external
to the systems can be equated to raising of weights (weights may not be actually raised but the effect can
be equated to raising of weights).
Fig. 1. Battery motor system driving a fan
Fig. 2 Work transfer from a system
Note: work transfer occurs only when it crosses the boundary and hence work transfer is a
boundary phenonmenon.
Convention of Work Transfer
work done by the system is +ve and work done on the system is taken as negative.
Generalised equation for closed system Work: (Non-flow work)
P =F/A
F =PA
Work = Force x Distance
dW= PA x dx
dW  PdV

W  PdV
Work on P-V diagram
Area = PdV
Work = Area
Area under the curve when projected on volume axis gives closed system work.
Wa  Wb
Though the end points are same . Work transfer is not same, therefore work transfer is not a property,
it depends on path and hence it is a path function and inexact differential.
 PdV
Conditions for applying the equation W 
1. The system must be a closed system.
2. Work should cross the boundary.
3. The process must be a reversible process.
Closed system work for various processes.
1. Process 1st: Constant Volume or Isochoric or Isometric Process
V= constant
 dV  0

W  PdV
W0
Constant volume closed system work = 0.
nd
2. Process 2 : Constant pressure process or Isobaric Process or Iso-prestic Process
P  W P
atm
A
gas
V2
W  PdV
V1
W  P(V2 V1 )
This equation is valid for a closed system undergoing reversible constant pressure process.
rd
3. Process 3 : Constant Temperture or Isothermal Process
T = Const.
PV = mRT
PV = mRT = C
PV = C = P1V1 = P2V2
V
1
V
P
C dV
W  2
V
V
C
V
1
V
dV
W  C 2
V
V
1
V2

W  PdV
W  ClnV V
2
V
W  mRT ln
1

V 
2




V1 
This equation is valid when an ideal gas undergoes reversible isothermal process in a closed system.
th
4. Process 4 : Adiabatic Process
A process in which there is no heat transfer from the system or to the system is known as
adiabatic process.

For adiabatic process PV  C
Where  is adiabatic index and it is greater than 1.
PV
C

 P  CV

V2
W  PdV
V1


W= CV
dV
V2
W  C V

V1
dV
W
W
W
W

 V  1
C

V
2

  1V
1

C
V  1 V1 1
  1 2
1
  1
P V
2
P V P V
1
1
2

 PV 
2
1 1
2
 1
This equation is valid for a closed system undergoimg reversible adiabatic process.
th
5. Process 5 : Polytropic Process
n
A process in which PV  Constant is known as polytropic process, where n is known as Polytroic
Index. Generally, 1  n   in polytropic process, there is both heat transfer and work transfer, but
whereas in adiabatic process, there is only work transfer.
Polytropic Work, W 
P1V1  P2V2
n 1
Ideal Gas Equation for various Processes
1. Constant Volume Process
V= Const.
PV = mRT
P T
P
2
P1

2
T1
2. Constant Pressure Process
P = Constant, PV = mRT
V
2
V1
T
 2
T1
3. Isothermal Process
T = Const.
PV = mRT
PV = Const
P1V1 = P2V2
4. Adiabatic Process
PV   C
PV = mRT
P = mRT/V
mRT
V
C

V
TV  1  mR
TV
 1
 Constant


PV
PV
1 1
V
2
2
1
 P 
2
1
V2
C


 P1 
TV
 T V  1
 1
1
1
2
2
1
 T   1
V
 
1
V

2

2


T

1
1
 T   1


2


T



 P 
2


1

1
 1


T

P


2

2

T 



1

1
 P 



P
1

5. Polytropic Process
n
PV
PV
1 1
TV
2
n1
1 1
T 

2

T


1

n
2
 T V n1
2
2
P 


P 
1 
2
n 1
n
Heat: the energy transfer due to temperature difference is known as Heat Transfer.
Specific heat: It is the amount of heat required to raise the temperature of unit mass of
substance through a unit degree temperature difference.


C
p
1 Cv
  1

Specific heat iat constant pressure Cp is greater than specific heat at constant volume Cv, because Cp
includes internal energy ‘U’ and external work ‘W’. Whereas Cv includes internal energy ‘U’ only.
NOTE: In case of solid and liquid, as the expansion work is negligible, therefore, Cp and CV are
almost same.
Questions:
1. When is work said to be done on a system?
2. Under what conditions is the work doen equal to

2
PdV ?
1
3. What is the difference between work transfer and heat transfer?
4. Show that heat is a path function and not a property.
5. A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1 MPa to 0.7MPa, 0.03
3
n
m . Assume that the pressure and volume are realted by pv = constant, find the work done
by the gas system.
6. At the beginning of the compression stroke of a teo-cylinder internal combustion engine the
air is at a pressure of 101.325 kPa. Compression reduces the volume to 1/5 of its original
1.2
volume, and the law of compression is given by pv =constant. If the bore and stroke of
each cylinder is 0.15 m and 0.25 m, respectively, determine the power absorbed in kW by
compression strokes when the engine speed is such that each cylinder undergoes 500
compression strokesper minute.
7. A system undergoes 3 processes as shown in figure. 1-2 is isobaric process, 2-3 is
adiabatic process (  1.4) , 3-1 is isothermal process in which PV = constant. Find
i)
V2
ii)
Net Work transfer
Lecture 4
Topic: First law of Thermodynamics (Conservation of Energy)
Link:
https://www.youtube.com/watch?v=sUDfpFD0xX4&index=4&list=PLA17CE108925DC94F&spfreload
=10
https://www.youtube.com/watch?v=bCToK4_dmbU&index=5&list=PLA17CE108925DC94F
Book: Engineering Thermodynamics by P K Nag Fifth Edition, Page no. 78 – 97
Lecture Notes:
First law of thermodynamics
For a closed system undergoing a cycle, net heat transfer is equal to net work
transfer. i.e., Q  w
the above equation valid for any cycle (reversible or irreversible)
Consequences of First law of thermodynamics
Ist Consequence: Heat transfer is a path function
dQ dQ
1a2

2b1
dQ dQ
1a2
From Eqn. (1) and (2), we get,
2c1
dQ dQ
2b1
dW
2b1

 dW
dW
1a2
2b1
dW  dW
1a2
2c1
(1)
(2)
 dW  dW
2c1
 dW
2b1
2c1
2c1
dW2b1  dW2c1  0
(3)
Though the end points are same for path b and path c, heat transfer is not same. Therefore
heat transfer is not a property and it is a path function and inexact differentials.
Important points with respect to Heat transfer and Work Transfer
a. Both are path functions.
b. Both are not properties.
c. Both are inexact differentials.
d. Both are boundary phenomenon.
e. Both are energies in transit. i.e., Heat and work are associated only with a process. They
lose their meaning once the process is over.
NOTE. It is wrong to use Heat and Work, the correct terms are Heat Transfer and Work Transfer.
2
nd
Consequence: Energy is a property
dQ  dQ
 dW  dW
2b1
2c1
 dQ  dW
2b1
2c1
2b1
dQ dW
2b1
2c1
2c1
(4)
(5)
( ( dQ  dW )1a2  (dQ  dW )2c1
(dQ  dW ) is independent of path
(dQ  dW ) is not a property
This property is energy.
(dQ  dW )2b1  (dQ  dW )2c1  dE
(dQ  dW )2b1  dE
(dQ  dE  dW)2b1
(dQ  dE  PdV )
This is the first law of thermodynamics for any process. (reversible and irreversible process).
NOTE: in equation dQ=dE+PdV is valid for a closed system undergoing reversible process.
Energy: E= KE+PE+U
KE+PE= Macroscopic function
U = Microscopic function
dE=d(KE)+d(PE)+dU
generally KE and PE changes are neglected.
Therefore dE= dU
dQ  dE  dW
dQ  dU  dW
this equation is valid for a process when KE and PE changes are neglected.
Internal Energy (U): The energy associated with molecules in known as internal energy.
Internal energy is an external property. It is generally expressed in KJ.
Specific internal Energy u=U/M =kJ/kg and it is an intensive property and is generally expressed in
KJ/Kg.
3
rd
Consequence: Energy of an isolated system is always constant.
Fig. 1. An isolated system showing no energy and mass transfer
dQ  dE  dW
0  dE 
0 dE  0
E2  E1  0
E1  E2 = Constant
th
4 Consequence: Perpetual motion machine of first kind (PMM – I ) is impossible.
There can be no machine which develops work continuously without absorbing heat energy. If such a
machine is developed, then it would violate Ist law of thermodynamics.
Fig. 2. Perpetual Motion Machine – I
Enthalpy (H):
In thermodynamics the term U+PV appears frequently and for convenience this term is taken as
enthalpy. Therefore H = U+PV
Enthalpy is an extensive property and it is generally expressed in KJ.
Specific enthalpy, h 
H
and it is an intensive property. It is generally expressed in KJ/Kg.
M
NOTE: the equation H= U + PV is applicable for any system, any working fluid and for any process.
Specific Heats
Specific heat at constant Volume Cv (kJ/kg K)
Specific heat at constant pressure Cp (kJ/kg K)
Heat transfer in various processes (Closed System)
1. Constant
volume
process: dQ  dU  PdV
dQV  dU  0 ,
dQ  dU
V= const.
Heat transfer in a constant volume closed system process is equal to change in internal energy.
2. Constant Pressure process:
dQ  dU  PdV
P = Constant
dQ  dU  d (PV )
dQP  d (U  PV
) dQ  dH
Heat transfer in a constant pressure closed system process is equal to change in enthalpy.
NOTE: For an ideal gas
1. U = mCvT
dU= mCvdT
2. H= mCpdT
dH=mCpdT
3. U= f(T) Joule’s Law
For real gas, U = f(T,P,V)
To show that Cp – Cv = R for an ideal gas
H = U + PV
dH = dU + d(PV)
Ideal gas equation, PV = mRT
mCp dT  mCv dT  d (mRT )
Cp  Cv  R
Cp  Cv  R
C
p

C
v
Cp  Cv
Cv  Cv  R
Cv ( 1)  R
Cv 
R
( 1)
C p  R
( 1)
For ideal gas (Air)
Cp = 1.005 KJ/kg K
Cv = 0.718 KJ/kg K
R = 0.287 KJ/kg K

  1.4
3. Isothermal Process:
dQ  dU  dW
T = Constant
Ideal gas U=f(T)
As T = Constant
U= Const.
dU=0
dQ  0 
dW dQ  dW
When an ideal gas undergoes isothermal process heat transfer is equal to work transfer
4. Adiabatic Process: There is no heat transfer in adiabatic process, so Q=0.
5. Polytropic Process:
dQ  dU  dW
Ideal gas undergoes polytropic process
W  P1V1  P2V2
n 1
dU = mCvdT
dQ
poly
Q
d
poly
Q
d
poly
 mC dT  P1V1  P2V2
v
n 1
 mRT2 T1   P1V1  P2V2
 1
n 1
 P2V2  P1V1  P1V1  P2V2
 1
n 1
1 
 1
dQ poly  PV  1P1 V
2 2
 PV  P V
dQ
poly
dQ
poly
dQ
poly



 n 1

 1 

  1 n 1

n 1 1
   n
 PV  P V 



1 1
2 2   1 
   n 

W





1


poly
1 1
2 2





Polytropic Specific heat (Cpoly):
dQ
poly
poly
   n  PV11  P2V2



  1 
n 1
   n  mRT1 T2 
 



  1 
n 1
   n 
T T 

mC
 n 1  v 1
2
   n 
 dT 

mC



n
1



v
n    
 
dQ
poly
dQ
poly
dQ
poly
dQ
poly
dQ
poly
C
   n 

W


  1 
 m

 n 1
n  
poly
C
poly

 n 1 
Cv dT 



C
 v
is negative as n   = -ve and n 1 = +ve and Cv = +ve
1n
Q poly
   n W poly
 1

   1.4
 n 1.2
i.e., though heat is supplied, there is a decrease in temperature. This is because in such a polytropic
process, work transfer is more than heat transfer and this excess work transfer comes from the internal
energy of the system, as there is decrease in internal energy transfer. Therefore there is decrease in
temperature.
To show that PV

 C for an adiabatic process:
dQ  dU  PdV
Ideal gas undergoing adiabatic process
dU  mCv dT
d Q  0 for adiabatic process
U  mCv dT  PdV
PdV= mCVdT
(1)
H = U + PV
dH = dU + PdV +VdP
dH  d Q VdP
mCPdT = U + VdP
(2)
dividing Eqn. 2 with 1
VdP

PdV

mC p dT
mCv dT
dP
V
P dV

dP dV
P
V0
 ln P   lnV  ln C
ln P  lnV
ln PV
PV



 ln C
 ln C
 C (Reversible Adiabatic)
This equation is valid when an ideal gas undergoes reversible adiabatic process and this
equation is valid for open as well as closed systems.
Question:
1. A fluid is contained in a cylinder by a spring loaded frictionless piston, so that the pressure in
the fluid is a linear function of volume. i.e., P= a+bV, where a and b are constants. Internal
3
energy of the fluid is given by U = 34 +3.15PV, where U is in KJ, P in kPa and V in m . If
3
the fluid changes from initial state to P1 = 170 kPa, V1 = 0.03m to a final stage of P2 =
3
400kPa, V2 = .06m , then find the magnitude and direction of heat and work transfer.
3
2. A stationary mass of gas is compressed without friction from an initial state of 0.3m and
3
0.105MPa to a finalstate of 0.15m and 0.105MPa, he pressure remaining constant during
the process. There is a transfer of 37.6 KJ of heat from the gas during the process. How
much does the internal energy of the gas change?
3. In a cyclic process, heat transfers are 14.7 kJ, -25.2kJ, -3.56 kJ and 31.5 kJ. What is the net
work for this cyclic process.
4. 1.5 kg of liquid having a constant specific heat of 2.5kJ/kg Kis stirred in a well-insulated
o
chamber causing the temperature to rise by 15 C. find change in internal energy and W for
the process.
Lecture 5
Equation of state
Link: http://nptel.ac.in/courses/112103016/19
Book: Fundamental of Engineering Thermodynamics by Moran, Shapiro, Fifth Edition. (Page no. 99
– 101)
Equation of state
Ideal gas equation of state is pv = RT.
This equation was established from the postulates of the kinetic theory of gases by Maxwell with the
following assumptions:
(1) There is little or no attraction between the molecules of the gas
(2) Volume occupied by the molecules is negligibly small compared to the volume of gas
Applicability: Ideal gas equation of state is applicable to gas with very small pressure or very high
temperature. With increase in pressure the intermolecular force of attraction and repulsion increases,
and also the volume of the molecules becomes appreciable compared to the gas volume. Under such
condition real gas equations are to be used.
Following are some of the real gas equations:
1. van der Waals equation of state for real gas
(1)
2. where
is accounting for the force of cohesion and b is accounting for volume of the molecules
termed as co-volume .
3. Berthelot equation of state
(2)
4. Dieterici equation of state
(3)
5. Redlich-Kwong equation of state
(4)
6. Saha-Bose equation of the state
(5)
where a and b are constants which can be evaluated from the critical data.




The van der Walls, Berthelot and Dieterici equations of state are of limited accuracy.
The Redlich-Kwong equation gives good results at high pressures and is fairly accurate for
temperatures above the critical value.
Saha-Bose equation of state is quite accurate for densities less than about 0.8 times the critical density
One more widely used equation of state with good accuracy is the Beattie-Bridgeman Equation:
(6)
where
There are five constants, A0 , B0 , a, b, and c to be determined experimentally for each gas.
Virial expression
The relation between pv and p in a term of power series may be expressed as
(7)
For any gas
(8)
(9)
An alternative expression is
(10)
Both expressions in Eqs.(9) and (10) are known as virial expansions or virial equations of state, first
introduced by the Dutch physicist, Kammerlingh onnes B´, C´, B, C,…….. etc. are called virial
coefficients. B´ and B are called second virial coefficients, C´ and C are called third virial
coefficients, and so on. For a given gas, these coefficients are functions of temperature only.
The ratio is called the compressibility factor, Z. for an ideal gas Z=1. The magnitude of Z for a certain
gas at a particular pressure and temperature gives an indication of the extent of deviation of the gas
from the ideal gas behavior.
The virial expansions become
(11)
and
(12)
The relations between B', C' and B, C… can be derived as follows
(13)
(14)
Comparing this equation and rearranging
Therefore
(15)
The terms
etc. of the virial expansion arise on account of molecular interaction. If no
such interactions exist (at very low pressures) B=0,C=0 etc. Z=1 and
. It may be observed
from Eq (2.152) that the compressibility factor is a function of temperature and pressure of a gas.
A graph for variation of pressure versus Z with constant temperature can be plotted from which value
of Z for a pair of pressure and temperature states can be estimated. Further using the equation pv=
ZRT, the volume of the gas can be obtained.
Example 1: Express the van der Waals equation of state in virial form.
Solution:
(16)
2
The second virial coefficient the third virial coefficient c = b , etc.
From Eq. (1), on mass basis
Example 2: Determine Boyle temperature from van der Waal's equation
Solution:
The point at which B is equal to zero gives the Boyle temperature.
The second virial coefficient is the most important. Since this B is known, the behavior of the gas at
moderate pressure is completely determined.
The terms which contain higher power becomes significant only at very high pressures.
LAW OF CORRESPONDING STATES
The general shapes of the vapour dome and the constant temperature lines on the p-v plane are similar
for all substances. However, scales may be different for different substances. This similarity can be
utilized with application of dimensionless properties called reduced properties.
A reduced property is the ratio of the existing property (X) to the critical property ( Xc ) of the
substance. Thus
(17)
Where the subscript r and c refer to the reduced and critical properties, respectively.
Various primary properties in terms of reduced property are given in Table 1 .
Table 1: Reduced Properties
The relations among the reduced properties, Pr , Tr , vr and vr is known as the law of corresponding
states.
The specific volume or molal volume of different gases are different under same pressure and
temperature. However, it is found from experimental data that at the same reduced pressure and
reduced temperature, the reduced volumes of different gases are approximately the same. Therefore,
for all substances
(18)
Now,
(19)
where
is called the critical compressibility factor.
From these two equation, by rearranging and dividing,
(20)
Substituting the value of b
(21)
Substituting the values of b and R
(22)
(23)
Therefore, the value of R becomes
(24)
The values of a, b, and R have thus been expressed in terms of critical properties, substituting these in
van der Waals equation of state
(25)
or,
(26)
Using the reduced parameter
(27)
In the reduced equation of state (27) the individual coefficients a, b and R for a particular gas have
disappeared. So this equation is an expression of the law of corresponding states because it reduces
the properties of all gases to one formula.
Similarly, two different substances are considered to be in ‘corresponding states', if their pressure,
volumes and temperatures are of the same fraction or multiples of the critical pressure, volume and
temperature of those substances.
Values of constants 'a' and 'b' for van der Waals and Redlich-Kwong are given below (Table 2)
Table: 2. Values of a and b
The generalized compressibility chart in terms of reduced properties is shown in generalized
compressibility chart. It is very useful in predicting the properties of substances for which more
precise data are not available.
Other form of real gas equations in terms of reduced properties can be estimated similarly.
(2) Benedict-Webb-Rubin
The Beattie-Bridgeman equation of state is
(28)
3
3
In this equation, P is in Kpa, is in m /kmol, T is in K and Ru = 8.314Kpa.m / (kmol.K)
where
The five constants in the Beattie-Bridgeman equation are as follows (Table 3):
Table 3. Different Gas Constants
Questions
1.
2.
3.
4.
5.
6.
Write down the ideal gas equation and compare the same with other real gas equation.
What is the law of corresponding states?
What is the usedulness of the general compressibility chart?
How does van-der wall equation differ from ideal gas equation?
Using virial equations determine the Boyle temp for the van der wall equation.
What is the advantage of using virial equations for the analysis of different real gas equation?
Lecture 06
Topic: Steady Flow Processes
Link: https://www.youtube.com/watch?v=0jXeNaSM5Xc&list=PLA17CE108925DC94F&index=3
Book: Engineering Thermodynamics by P K Nag Fifth Edition, Page no. 99 – 107
Lecture note:
Steady Flow Process:


In a flow if thermodynamic properties do not change with time at different locations, the
process is steady flow process
In steady flow process there is no accumulation of mass or energy in control volume i.e.,
conservation of mass and energy occurs.
For steady flow m1 = m2 = m
Eentering = Eleaving (steady)
1 2
1
mc1  mgZ1  U 1  Q  mc22  mgZ 2  U 2  W
2
2
1 2
1
mc1  mgZ1  U 1  Q  mc22  mgZ 2  U 2  P1V1  Wcv  P2V2
2
2
1 2
1
mc1  mgZ1  U 1  Q  mc22  mgZ 2  U 2  P1V1  Wcv  P2V2
2
2
1 2
1
mc1  mgZ1  U 1  P1V1  Q  mc22  mgZ 2  U 2  P2V2  Wcv
2
2
1 2
1
mc1  mgZ1  H 1  Q  mc22  mgZ 2  H 2  Wcv
2
2
The above equation is known as Steady flow energy equation.
This is the first law of thermodynamics equation for open system under steady flow conditions.
1 2
1
c1  gZ1  h1  q  c22  gZ 2  h2  wcv
2
2
1 2
1
c1  gZ1  u1  p1v1  q  c22  gZ 2  u 2  p2 v2  wcv
2
2
u1 
p1
p 1
1
 c12  gZ1  q  u2  2  c22  gZ 2  wcv
1 2
2 2
(1)
(2)
(3)
(4)
(5)
Steady flow
If u1 = u2
q=0
w=0
incompressible (p1 = p2)
Bernoulli Equation
Complete examples of Steady Flow devices
1.
2.
3.
4.
5.
Turbine
Compressor
Nozzle
Diffuser
Throttling valve, etc
Appication of steady flow energy equation
1. Nozzle: Nozzle is a device which is used for increasing kinetic enrgy at the expense of pressure
energy, whereas in diffusers the pressure energy increase and kinetic energy decreases.in both
nozzles and diffusers the work transfer is zero.
c12
c22
h1   Z 2 g  q  h2   Z 2 g  Wcv
2
2
h1 
c12
c2
 h2  2
2
2
If c1 << c2
2. Turbine:
c12
c22
h1 
 Z 1g  h2 
 Z 2 g  Wcv
2
2
Assumptions:
1. Steady flow
2. Perfectly insulated
3. Neglect PE change
4. Neglect KE change
c12
c22
h1   Z 2 g  q  h2   Z 2 g  Wcv
2
2
h1  h2  Wcv
h1  h2  Wcv
Wturbine = h1 – h2
3. Compressor
Wturbine = h1 – h2
Wturbine = - (h2 – h1)
-Wturbine = h2 – h1
Wcompressor = h2 – h1
4. Throttling valve:
Example of Throttling valve:
1. Flow through a partially opened valve
2. Flow through a very small opening (orifice)
3. Flow through a porous plug.
Characteristics of Throttling:
1. No work transfer (loss due to friction)
2. No heat transfer (the length of the device is so small and hence the time available for heat
transfer is less)
3. It is an irreversible process
4. It is an isenthalpic process
h1 = h2
Isenthalpic Processes