Engineering Thermodynamics Notes for Quick Reference

1
Engineering Thermodynamics
By
Dr.P.N.Kadiresh
Professor/Aerospace Engineering Dept.,
B.S.Abdur Rahman Crescent Institute of Science and Technology

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CONTENTS
1. Basic Concepts and First Law of Thermodynamics
2. Second Law of Thermodynamics
3. Pure Substance and Vapour Power Cycle
4. Gas Power Cycles
5. Isentropic Flow Through Nozzle
6. Refrigeration and Air-conditioning
7. Air Compressor

3
Units and Parameters
Fundamental units:
Mass kilograms kg
Length meter m
Time seconds s
Temperature Celcius/Kelvin o
C / K
Derived units:
Newton’s Law states: Force = mass x acceleration
F = m  a
[N] = [kg] . [m/s2
]  1 N = 1 kgm/s2
P = F / A
[Pa] [kgm/s2
] / [m2
]  1 Pa = 1 N/m2
W = F  x
[J] [kgm/s2
] . [m]  1 J = 1 N.m
Force (F) N
Pressure (p) Pa, bar, atm, psi
1 Pa = N/m2
; 1atm = 1.01325 bar
1 bar = 1 x 105
N/m2
= 0.1 MPa
Work (W) / Heat (Q) J
Power (P) J/s
Sp. internal energy, (u) J/kg
Sp. enthalpy, (h) J/kg
Sp. entropy, (s) J/kg.K
Sp. heat at constant pressure, cp J/kg.K
Sp. heat at constant volume, cv J/kg.K
Universal gas constant, 8.314 kJ/kgmol.K
ℎ , = =
8.314
28.96
= 0.287 /( . )

Why do we need to study thermodynamics?
Knowledge of thermodynamics is required to design any device involving the
between heat and work.
Examples of practical thermodynamic devices:
Thermal Power Plant
Rocket Nozzle
What is thermodynamics?
 The study of the relationship between work, heat, and energy.
 Deals with the conversion of energy from one form to another.
 Deals with the interaction of a system and it surroundings.
Chapter 1 - Introduction
Why do we need to study thermodynamics?
Knowledge of thermodynamics is required to design any device involving the
Examples of practical thermodynamic devices:
Air Conditioner
Turbojet Engine
The study of the relationship between work, heat, and energy.
Deals with the conversion of energy from one form to another.
Deals with the interaction of a system and it surroundings.
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Introduction
Knowledge of thermodynamics is required to design any device involving the interchange
Air Conditioner
Turbojet Engine

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Macroscopic and Microscopic viewpoint of Thermodynamics:
Macroscopic (or Classical Thermodynamics):
 In this approach, a certain quantity of matter is considered, without taking into the
account the events occurring at the molecular level.
 In macroscopic thermodynamics, the properties of system are assigned to the system as a
whole and that are based on observable, measurable quantities and these effects can be
perceived by the human senses.
 Example: A moving car.
Microscopic (or Statistical Thermodynamics):
 From the microscopic view point, matter is composed of large number of small molecules
and atoms.
 Microscopic thermodynamics is concerned with the effects of action of many molecules
and these effects cannot be perceived by the human senses.
 Example: individual molecules present in the air.
Energy:
Energy is a convenient generic term for “something” which is transferred in the doing of Work.
Energy of most interest:
 potential energy (gravity or “spring”)
 kinetic energy
 chemical energy
 internal energy
Above forms can “store” energy, whereas work can’t. It is only transient manifestation process
of transferring energy. Heat can’t either; it is also only transient manifestation process of
transferring energy. So heat and work are other forms of energies in transition.
System: Identifies the subject of the analysis by defining a boundary. Either “a region in space”
or “a particular collection of matter” must be treated consistently.
Surroundings: Everything outside the system boundary.

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Open System: Fixed volume in space, mass and energy exchange permitted across the system
boundary.
Example: jet engine
Closed System: Fixed non-changing mass of fluid within the system, i.e., no mass transfer
across the system boundary but can have energy exchange with the surroundings.
Example: piston-cylinder assembly.
Isolated System: a system that does not interact at all with the surroundings, e.g., neither mass
nor energy transfer across system boundary.
Properties of a system:
A property of a system is a measurable characteristic of a system that is in equilibrium.
Properties may be intensive or extensive.
Intensive property is a physical property of a system that does not depend on the system size or
the amount of material in the system.
e.g: Temperature, Pressure, and Density
Extensive property of a system does depend on the system size or the amount of material in the
system.
e.g: mass, volume, energy, enthalpy
State, Equilibrium and Process:
 State – a set of properties that describes the conditions of a system. Eg. Mass m,
Temperature T, volume V.

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 Change of State: When system interacts with another system or with surroundings, the
system is said to be undergoing change of state.
 Path – Locus of change of state.
 Process – If path is specified, process can be defined (based on which property is held
constant between initial state and end state).
……………………………………………….
Process Property held Constant
--------------------------------------------------------
isobaric pressure
isothermal temperature
isochoric volume
isentropic entropy
………………………………………………....
 Thermodynamic Equilibrium – A system that maintains thermal, mechanical, phase
and chemical equilibriums is said to be in thermodynamic equilibrium.
 Quasi-static Process – When a process is carried out in
such a way that the system passes through infinite number
of equilibrium states. It is a sufficiently slow process in
which, locus of all state points are equilibrium points.
A quasi-static process is a hypothetical process and
essentially a reversible process.
 Cyclic process - when a system in a given initial state goes
through various processes and finally returns to its initial
state, the system has undergone a cyclic process or cycle.
 Reversible process - it is defined as a process that, once
having take place it can be reversed. In doing so, it leaves no
change in the system or boundary.
 Irreversible process - a process that cannot return both the system and surrounding to
their original conditions
 Adiabatic process - a process that has no heat transfer into or out of the system. It can be
considered to be perfectly insulated.
 Isentropic process - a process where the entropy of the fluid remains constant.

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 Polytropic process - when a gas undergoes a reversible process in which there is heat
transfer. It is represented as PVn
= constant.
 Throttling process - a process in which there is no change in enthalpy, no work is done
and the process is adiabatic.
Property Definitions
In order to speak of an intrinsic property “at a point” we must treat matter as a continuum,
i.e., matter is distributed continuously in space
 In classical thermodynamics a point represents the smallest volume ‘V’ for which
matter can be considered a continuum.
 The value of the property represents an average over this volume ‘V’.
At any instant the density, ρ, at a point is defined as
 
 








lim
V V
M
V
unit: kg/m3
Mass, M, of the system with volume, V, is





 V
dV
dM
M
dV
dM


Note: if ρ is uniform over the volume M = V
Specific volume, , defined as
 
 1 unit: m3
/kg
The pressure, P, at a point is defined as
P F
A
A A

 








lim
units: 1 Pa = 1 N/m2
1 standard atmosphere = 101,325 Pa
1 bar = 100,000 Pa = 100 kPa = 0.1 MPa
Absolute pressure , Pabs, measured relative to a perfect vacuum
Gauge pressure , Pg, measured relative to the local atmospheric pressure, Patm.

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Note: Pg = Pabs - Patm
Gauge pressure measurement via a manometer
Pg = Pabs – Patm = ρgL where, g = 9.81 m/s2
Temperature, T, in units of degrees celcius, o
C, is a measure of “hotness” relative to the freezing
and boiling point of water. A thermometer is based on the thermal expansion of mercury
Temperature in units of degrees kelvin, K, is measured relative to this absolute zero temperature,
so 0 K = - 273o
C; in general, T in K = T in o
C + 273.13.
 Zeroth Law: “If two bodies are separately in thermal equilibrium with a third body, then
they must be in thermal equilibrium with each other.” (Same temperature!).
 First Law and Corollaries
Work: If Boundary of System moved against a force, Work is done on or by System
(Note: Work is transient, so not a Property)
Heat: Heat is energy transferred from one body to another due to temperature difference
(Note: Heat is transient, so not a property)

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 First Law of Thermodynamics:
When any closed system is undergoing a cycle, the net work delivered by the system is
proportional to the net heat transferred to the system,
i.e., (∑ ) ∑
∑ = ∑
− = 0
Where J is Joule’s of Mechanical equivalent of Heat
With same units for Q & W; Net heat transfer = Net work.
=
 First Law corollary 1:
There exists a property of a closed system such that a change in its value is equal to the
difference between the heat supplied and the workdone during any change of state.
Proof:
Let us consider 1-A-2 & 2-B-1 two processes constituting a cycle, 1-2-1.
From FLTD,
( ) + ( ) = +
Simply,
+ = +
− = −( − )------------------------(1)
Let us imagine, the cycle 1-2-1 is completed though the process C instead of B.
Applying FLTD for the cycle 1-2-1,
=
+ = +
− = −( − )------------------------(2)
Comparing equation 1 & 2
Change (Q-W) is same for the process B or C, it is independent of path it followed, so it
is a point function and a property. Thus based on FLTD, (dQ – dW) is a property, we call
it as internal energy (in kJ) or (‘u’ – specific internal energy in kJ/kg)
. .̇ − = = ( − )
− = =>
( − ) = ( − )
= +

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 First Law corollary 2:
If a system is isolated from its surroundings (i.e., Q = 0; W = 0) then its internal energy
remains unchanged
du = 0; u = const.
 Work Done in Vacuum
Rupture of membrane allows the gas to expand into vacuum. The system boundary
moves but no force resisting the movement
= 0 (OR) = 0
 Perpetual Motion Machine of first kind (PMMI)
A machine which produces work without absorbing energy from the surroundings is not
possible. It must be true if First Law is true.
 Energy Transport by Heat and Work and the Classical Sign Convention
Energy may cross the boundary of a closed system only by heat or work. Energy transfer
across a system boundary due solely to the temperature difference between a system and
its surroundings is called heat.
Heat and work are energy transport mechanisms between a system and its surroundings.
The similarities between heat and work are as follows:
a) Both are recognized at the boundaries of a system as they cross the
boundaries. They are both boundary phenomena.
b) Systems possess energy, but not heat or work.
c) Both are associated with a process, not a state. Unlike properties, heat or work
has no meaning at a state.
d) Both are path functions (i.e., their magnitudes depends on the path followed
during a process as well as the end states.
Since heat and work are path dependent functions, they have inexact differentials
designated by the symbol, . The differentials of heat and work are expressed as Q and
W. The integral of the differentials of heat and work over the process path gives the
amount of heat or work transfer that occurred at the system boundary during a process.
∫ = (not ΔQ)
∫ = (not ΔW)
That is, the total heat transfer or work is obtained by following the process path and
adding the differential amounts of heat (Q) or work (W) along the way. The integrals
of Q and W are not Q2 – Q1 and W2 – W1, respectively, which are meaningless since
both heat and work are not properties and systems do not possess heat or work at a state.

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Work has the units of energy and is defined as force times displacement or Newton times
meter or joule. Work per unit mass of a system is measured in kJ/kg.
 Common Types of Mechanical Work Energy
• Shaft Work
• Spring Work
• Work done of Elastic Solid Bars
• Work Associated with the Stretching of a Liquid Film
• Work Done to Raise or to Accelerate a Body
Net Work Done By a System
The net work done by a system may be in two forms: other work and boundary work.
First, work may cross a system boundary in the form of a rotating shaft work, electrical
work or other the work forms listed above. We will call these work forms “other” work,
that is, work not associated with a moving boundary. Second, the system may do work
on its surroundings because of moving boundaries due to expansion or compression
processes that a fluid may experience in a piston-cylinder device.
= − +
Here, Wout and Win are the magnitudes of the other forms of work crossing the boundary.
Wb is the work due to the moving boundary as would occur when a gas contained in a
piston cylinder device expands and does work to move the piston. The boundary work
will be positive or negative depending upon the process.
= ( ) +
 Introduction to the Basic Heat Transfer Mechanisms
Heat transfer is energy in transition due to a temperature difference. The three modes of
heat transfer are conduction, convection, and radiation.
Figure: Illustration of Modes of Heat Transfer

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Conduction: It is the mode of heat transfer particularly in solids. In this mode of heat
transfer, the heat transfers from one atom to its neighbouring atom through molecular
vibrations.
Rate of heat transfer by conduction, ̇ is governed by Fourier Law of heat
conduction:
̇ = − ,
Where k - thermal conductivity of the chosen material (W/mK).
A – Area normal to the heat flow (m2
)
dT/dx – Temperature gradient in the direction of heat flow (0
C/m)
Convection: This mode of heat transfer particularly occurs in fluids in motion. That is in
both liquids and gases that are in motion. This mode of heat transfer occurs due to the
transfer of energy through the bulk mass.
In detail whenever there is a temperature difference in a fluid, density difference occurs
and motion of fluid starts as lower density fluid attempts to reach the top of the fluid.
During this motion mass and energy transfer occurs thus heat transfer takes place.
Rate of heat transfer by convection, ̇ ̇ is governed by Newton’s Law of cooling,
̇ ̇ = ℎ , where h = Convective heat transfer coefficient (W/m2
K).
Radiation: In this mode, heat transfer takes place via electromagnetic waves.
Electromagnetic waves transport energy like other waves and travel at the speed of light.
This is the mode by which we receive solar energy from the sun.
Rate of heat transfer by Radiation, ̇ ̇ is governed by Stefan-Boltzmann law,
̇ ̇ = ( − ) , where σ – Stefan-Boltzmann constant, 5.67x10-8
W/m2
K4
 Ideal gas equation:
Equation of state is defined from Avogadro’s law, which states that, equal volumes of all
gases, at the same temperature and pressure, have the same number of molecules".

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For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly
proportional if the temperature and pressure are constant.
=
if p = 760 mm of Hg = 1.01325 bar; T = 273.15 K, it occupies 22.4 m3
/kgmol
, =
̅
=
1.01325 10 22.4
273.15
= 8.314
.
We know that ̅ = where n = number of moles
So gas equation can be written as,
= ; where =
m is the mass of the gas and µ is the molecular weight of the gas.
= = ( )
 = where R is the characteristic gas constant.
 Enthalpy:
It is a composite property. It is defined or sum of internal energy and flow energy.
Enthalpy : = +
Specific Enthalpy : ℎ = +
From FLTD,
= +
= +
= ( )
= ( + )
( ) = ℎ
For constant pressure process = (ℎ − ℎ )
 Specific heat at Constant volume : :
Defined as the rate of change with temperature of the specific internal energy of the
system when volume is held constant.
=

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From FLTD,
= +
= | = 0,
. .̇ Heat transfer for constant volume process { = −
So for a small temperature change dT,
=
But s specific heat and is specific heat at constant volume-
is a property because is only expressed in terms of other properties:
thus it must be a property.
For perfect gas,
= ;  [ ] = ( )
Heat transfer for constant volume process [ ] = ( − )
 Specific heat at constant pressure: ( ) :
Defined as the state of change with temperature of the specific enthalpy of the system when
the pressure is held constant
i.e., =
From FLTD,
= +
= ( + )
[ ] = ℎ
For small temperature change dT,
=
is specific heat at constant pressure.
. .̇ =
[ ] = Dt

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Heat transfer for constant pressure process {[ ] = ( − )}
[ ] = ℎ = ( − )
ℎ − ℎ = ( − )
 Ratio of specific heats:
=
 Relationships between , , &
From FLTD, | We know that for perfect gas =
= +
ℎ = +
ℎ = + +
For constant pressure process,
ℎ = + ( ) + 0
ℎ = + ( )
ℎ = +
÷
= +
= +  − =
Where → ℎ
= where
= 8.314 .
= ℎ .
= =
.
.
 = . .
⁄

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 Thermodynamic work or pdv work:
Let us consider a piston-cylinder
arrangement. A gas is enclosed in the
cylinder at Position 1, the pressure,
temperature and volume of gas in the
cylinder is P,T & V, if the gas is
allowed to expand to an elemental
distance dl, the force acting on the
surface of the piston is,
The pressure forces acting on the piston, = [ ∗ ]
Workdone by the expanding gas in the piston for elemental distance dl is,
=
Workdone by the gas from state 1 to 2, = ∫
Thus, the area under the process curve on a p-V diagram is equal to the workdone during a
quasi-static process of a closed system.
 Property Relation
Properties have unique values for any (equilibrium) state. Properties are state functions
and independent of path or process.
∫ = − is exact differential.
Similarly; ∫ = −
but work, not a property, because it is not independent of process.
∫ ≠ −
∫ = ( ) ; It is an inexact differential.
 Work is a path function:

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

W PdV
W PdV
V
V

 
  
shaded area
W total area under curve
1 2
1
2
1
2
Consider two processes with the same start and end state
Since the area under each curve is different the amount of work done for each path is different.
2
1 )
(
)
(
2
1
2
1
path
V
V
path
V
V
PdV
PdV 


Work done depends on the path taken and not just the value of the end states. Work is not a
property!
 PdV work for different thermodynamics processes:
 Constant pressure process (Isobaric process) :
= ∫
= ∫
( ) = ( − )
Constant Volume process (Isochoric Process)
( ) = ∫
( ) =
Because = ; = = 0

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 Constant Temperature Process (Isothermal Process)
= ∫
We know that gas equation, =
When, =
. .̇ = = =
= … … … … … … … … … . (1)
( ) = ∫ = ∫
( ) = ln
 Polytrophic Process
= = =
 =
( ) = ∫ = ∫
 ( ) =
=
=
−
1 −
=
−
1 −
( ) =
−
− 1
 Property relations for different thermodynamic processes:
 Constant pressure process (Isobaric process, p1 = p2)
=  =
 Constant volume process (Isochoric process, v1 = v2)
=  =

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 Constant temperature process (Isothermal process, T1 = T2)
=  =
 For polytropic process ( = )
=  =
=  =  =
 =
=  =  =
 =
/
 =

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Summary of Thermodynamic Processes
Process
Constant
Pressure
(Isobaric)
Constant
Volume
(Isochoric)
Constant
Temp.
(Isothermal)
Rev.
Adiabatic
(Isentropic)
Polytropic
Index ( n ) n = 0 n =  n = 1 n =  n = n
p v T
Relations 1
2
1
2
v
v
T
T

Charle's Law
2
1
1
2
p
p
T
T

2
2
1
1 v
p
v
p 
Boyle's Law

 1
1
2
1
2










p
p
T
T
1
2
1
1
2











v
v
T
T
n
n
p
p
T
T
1
1
2
1
2










1
2
1
1
2










n
v
v
T
T
Specific heat
C p
c v
c  0









n
n
c
c v
n
1

Heat added, Q  
1
2 T
T
cp   
1
2 T
T
cv 
1
2
1
1 ln
v
v
v
p 0 WD
n
*
1 











Work Done,
W
 
1
2 v
v
p  0
1
2
1
1 ln
v
v
v
p
1
2
2
1
1



v
p
v
p
1
2
2
1
1


n
v
p
v
p
Entropy
Change
s2 - s1 1
2
ln
T
T
cp
1
2
ln
T
T
cv
1
2
ln
v
v
R 0 















1
2
ln
*
1 T
T
n
n
cv

Different Thermodynamic Processes:

22
Problem #1.1: A piston-cylinder assembly contains 1 kg of nitrogen gas (N2). The gas expands
from an initial state where T1= 700K and P1= 5 bar to a final state where P2= 2 bar. During the
process the pressure and specific volume are related by Pv1.3
= const. Assuming ideal gas
behaviour and neglecting KE and PE effects, determine the heat transfer during the process, in
kJ.
From FLTD,
W
U
Q
W
Q
U






2
2
1
2
1
2
2
1
1
2
1
T
need
1
mRT
PV
gas
ideal
an
for
Recall,
1









n
)
-T
mR(T
pdV
n
V
P
V
P
pdV
W
For polytropic process,
1
2
1
1
2
1
1
2
1
2






















n
n
n
v
v
T
T
P
P
T
T
567K
(700K)
bar
5
bar
2 1.3
0.3
2 







T
132kJ
1
3
.
1
567)
-
0
(0.287)(70
1
1
2
1






W
n
)
-T
mR(T
W
Note: work is positive  work done by the system

23
The average value V
c
~ = 0.786 kJ/kg.K is assumed for air.
kJ
104.5
-
700)K
-
K)(567
g
(0.786kJ/k
kg
1
)
(
)
( 1
2
1
2







 T
T
mc
u
u
m
U v
Q = ΔU + W => Q = (-104.5 kJ) + (132 kJ) = +27.5 kJ
Note: heat transfer is positive  So, heat is transferred into the system.
Problem #1.2: 0.1 m3
of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8 bar. It is
then cooled at constant volume and further expanded isothermally so as to reach the condition
from where it started. Calculate a) pressure at the end of constant volume cooling, b) change in
internal energy during constant volume process, c) net work done and heat transferred during the
cycle. Assume cp = 14.3 kJ/kgK and cv = 10.2 kJ/kgK.
Solution:
= =
14.3
10.2
= .
For Process 1-2 (Adiabatic Compression):
= ----- V2 = 0.0227 m3
= ---- T2 = 544.5 K
= W1-2 = - 20.27 kJ
For Process 2-3( Constant Volume Cooling):
Change in internal energy, u3 – u2 = m cv(T3 – T2) = - 20.2 kJ
Work done, W2-3 = 0
For Process 3-1( Isothermal Expansion):
= ln = ln ; Because, p3v3 = p1v1 for isothermal Process
Therefore, W3-1 = 14.82 kJ
Net work done, Wnet = W1-2 + W2-3 + W3-1 = - 5.45 kJ
-‘ ve sign indicates work has been done on the system.
From FLTD, for a cyclic process;
(∑ ) = (∑ )  Qnet = - 5.45 kJ -‘ ve sign indicates heat is lost from the system.

24
Problem #1.3:: A fluid at 4.15 bar is expanded reversibly according to a law PV = constant to a
pressure of 1.15 bar until it has a specific volume of 0.12 m3
/kg. It is then cooled reversibly at a
constant pressure, then is cooled at constant volume until the pressure is 0.62 bar; and is then
allowed to compress reversibly according to a law PVn
= constant back to the initial conditions.
The work done in the constant pressure is 0.525 kJ, and the mass of fluid present is 0.22 kg.
Calculate the value of n in the fourth process, the net work of the cycle and sketch the cycle on a
P-V diagram.
Solution:
Given Data:
p1 = 4.15 bar; p2 = 1.15 bar; v2 = 0.12 m3
/kg
p4 = 0.62 bar; W(p=c) = 0.525 kJ; m = 0.22 kg
Process 1-2 (Isothermal):
=
= = .
= ln = .
Process 2-3 (Constant Pressure): Process 3-4 (Constant Volume):
= ( − ) = . W3-4 = 0
=
0.525
115
+ 0.0264 = .
Process 4-1 (Polytropic):
= = =
.
.
 ln (0.1494) = n* ln (0.2364)  n = 1.3182
= = − .
Wnet = W1-2+ W2-3+ W3-4 +W4-1  Wnet = 0.9076 kJ

25
 STEADY FLOW PROCESS
Many engineering devices such as nozzles, compressors and turbines operate a steady state
conditions i.e., properties are not changing with time. Let us consider an open system where
system boundary is fixed. The confined region where attention is focused can be selected as a
control volume.
Applying the principle of conservation of energy for a general steady flow system as shown
in fig.
[ ℎ ℎ ]
= [ ℎ ]
− [ ℎ ℎ ]
The flowing mass is having the following energies at inlet and exit i.e., molecular internal
energy(mu),flow energy(pv),kinetic energy( /2),and potential energy (mgz)
(The total energy of flowing
Fluid per unit mass at entry) ∑ = + + +
= ℎ + + /
At exit, ∑ = (ℎ + + ) /
Applying energy principle
̇ ℎ + + + = ̇ ℎ + + + …… (2)
( − )= ̇ (ℎ − ℎ ) + + ( − ) ……. (3)
Where ̇ = ̇ law of conservation of mass.
Eqn (3) is the general form of steady flow energy equation where Q & W and rate of heat
and work transfer respectively in .
Dividing the eqn (3) by ̇ we obtain
̇
−
̇
= (ℎ − ℎ ) + + ( − )
− = (ℎ − ℎ ) +
−
2
+ ( − )
Where the subscript 1 & 2 stands for inlet and outlet conditions.

26
 STEADY FLOW DEVICES:
Applying steady flow energy equation for devices like nozzle, compressor and turbine, the
equation can be simplified as follows;
A nozzle is a device which increases the velocity of a fluid at the expense of pressure,
Applying SFEE for nozzle
− = (ℎ − ℎ )+ + ( 2 − 1)
In nozzle flow, the rate of heat transfer between the fluid and the surrounding is usually
insignificant because of the fact that the fluid will be flowing at high velocity and they pass
through a nozzle quickly without any significant heat transfer taking place ( = 0); The shaft
work ( = 0) for a nozzle flow PE change is also negligible.
0 = ℎ − ℎ + 2
2
− 1
2
2
= = 2(ℎ − ℎ ) , ≫
= 2(ℎ − ℎ ) in m/s where ℎ & ℎ in J/kg
Applying SFE principle to flow through compressor
∆ = 0; ∆ = 0 & ≠ 0
There is shaft work involved. Compressor is driven by external power; work is done on the
system. = 0 (heat rate is small compared to the shaft work). Therefore, ,
− = ℎ − ℎ ;
= −(ℎ − ℎ ) [-ve sign indicates, work is done on the system for compressor]
For turbine, all conditions are similar to the compressor except that work done is positive
W = (ℎ − ℎ ) in kJ/kg

27
For throttling valve, there is a significant pressure drop in the fluid where it is passed through a
restricted passage. No work and heat transfer considered. ∆ & ∆ is also insignificant.
Therefore, the SFEE becomes ℎ = ℎ
Problem # 1.4: Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity
of 200 / . The inlet area of the diffuser is 0.4 . The air leaves the diffuser with a velocity
that is very small compared with the inlet velocity. Determine, a) The mass flow rate of the air,
b) The temperature of the air leaving the diffuser.
Given Data:
= 80
= 200 /
= 10℃ = 283
= 0.4
= 0
̇ =?
=?
Solution:
Simplified SFEE for diffuser flow is
ℎ + = ℎ + [ = = 0]
Enthalpy at state 1, ℎ = = 1.005 ∗ 283 = 284.42 /
Mass flow rate, ̇ =
= 0.9852 ∗ 200 ∗ 0.4
̇ = . /
From gas equation =
=
= = 1.015
=
1
= 0.9852 /
Enthalpy at state 2, can be obtained by,
Substituting ℎ & values in SFEE,
ℎ +
2
= ℎ +
2
ℎ = ℎ +
2
= (284.42 ∗ 10 ) +
200
2
= . /
ℎ =
=
304.42
1.005
= .

28
Problem # 1.5: Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The
Mass flow rate of the air is 0.02 kg/s and a heat loss of 16 kJ/kg occurs during the process.
Assuming the changes in kinetic and potential energies are negligible, determine the necessary
process input to the compressor.
Given Data:
= 100
= 280
̇ = 0.02 /
= 600
= 400
= −16 /
∆ = ∆ = 0
SOLUTION:
Simplified energy balance equation is
− = ̇ (ℎ − ℎ )
= − ̇ (ℎ − ℎ )
= ̇ − ̇ (ℎ − ℎ )
= ̇ − (ℎ − ℎ )
ℎ = = 400.98 / ; ℎ = = 280.13 /
= 0.02[(−16) − (400.98 − 280.13)]
= −2.74 Negative (-‘ve) sign indicates, work is done on the system.
Problem #1.6: In a gas turbine installation, the gases enter the turbine at the rate of 5 kg/s with a
velocity of 50 m/s and enthalpy of 900 kJ/kg and leave the turbine with 150 m/s and enthalpy of
400 kJ/kg. The loss of heat from the gases to the surroundings is 25 kJ/kg. Assume R = 0.285
kJ/kgK, cp = 1.004 kJ/kgK and inlet conditions to be 100 kPa and 27 ◦C. Determine the diameter
of the inlet pipe.
Solution:
= −′
(ℎ ℎ )
= − 25 ∗ 5 = 125 10 /
̇ = ̇ = 5 /
Applying SFEE,

29
̇ ℎ + + + = ̇ ℎ + + +
= 5[(900 10 − 400 10 ) + ] − 125 10
Turbine Work, = = 2325 /
To get Dia of Inlet Pipe:
Volume of gas entering the turbine per sec,
From gas equation, = = = 4.275
From continuity equation, ̇ =
ℎ , = = 0.4275 ==> = 73

30
Supplementary Problems for Practice
1. A mass of 0.15 kg of air is initially exists at 2 MPa and 350o
C. The air is first expanded
isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of
1.2 to the initial pressure and finally compressed at the constant pressure to the initial
state. Determine the boundary work for each process and the net work of the cycle.
[-4.75 kJ]
2. 0.078 kg of a carbon monoxide initially exists at 130 kPa and 120o
C. The gas is then
expanded polytropically to a state of 100 kPa and 100o
C. Sketch the P-V diagram for this
process. Also determine the value of n (index) and the boundary work done during this
process. [1.248,1.855 kJ]
3. A gas contained within a piston-cylinder assembly undergoes three processes in series:
Process 1-2: Compression with pV=C from p1 = 1 bar, V1= 1m3
to V2 = 0.2 m3
. Process
2-3: Constant pressure expansion to V3 = 1 m3
. Process 3-1: Constant volume. Sketch the
processes in series on a p-V diagram labeled with pressure and volume values at each
numbered state.
4. A gas undergoes a thermodynamic cycle consisting of the following processes:
Process 1-2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W1-2 = 10.5 kJ; Process 2-3:
Compression with pV = constant, U3 = U2; Process 3-1: Constant volume, U1 – U3 = -
26.4 kJ. There are no significant changes in KE and PE. a) Sketch the cycle on a p-V
diagram, b) Calculate the net work for the cycle in kJ.
5. A system contains 0.15 m3
of air pressure of 3.8 bars and 150o
ͦC. It is expanded
adiabatically till the pressure falls to 1.0 bar. The air is then heated at a constant pressure
till its enthalpy increases by 70 kJ. Sketch the process on a P-V diagram and determine
the total work done. Use cp=1.005 kJ/kg.K and cv=0.714 kJ/kg.K [54.9 kJ]
6. If a gas of volume 6000 cm3
and at a pressure of 100 kPa is compressed quasi-statically
according to pv1.2
= constant until the volume becomes 2000 cm3
, determine the final
pressure and the work transfer. [382 kPa, 0.82 kJ]
7. A certain gas of mass 4 kg is contained within a piston-cylinder assembly. The gas
undergoes a process for which pv1.5
= constant. The initial state is given by 3 bar, 0.1 m3
.
The change in internal energy of the gas in the process is, u2 – u1 = - 4.6 kJ/kg. Find the
net heat transfer for the process when the final volume is 0.2 m3
. Neglect the changes in
KE and PE. [-0.8 kJ]
8. A turbine operates under steady flow conditions, receiving steam at the following state:
pressure 1.2 MPa, temperature 188 ᵒC, enthalpy 2785 kJ/kg, velocity 33.3 m/s and
elevation 3m. The steam leaves the turbine at the following state: pressure 20kPa,
enthalpy 2512 kJ/kg, velocity 100 m/s and elevation 0 m. Heat is lost to the surroundings
at the rate of 0.29 kJ/s. If the rate of steam flow through the turbine is 0.42 kg/s, what is
the power output of the turbine in kW? [112.51kW]
9. A reciprocating air compressor takes in 2 m3
/min at 0.11 MPa, 20 ᵒC, which it delivers at
1.5 MPa, 111 ᵒC to an after-cooler where the air is cooled at constant pressure to 25 ᵒC.
The power absorbed by the compressor is 4.15 kW. Determine the heat transfer in
a) the compressor and b) the cooler. [-0.16 kW, 3.768 kW]

31
CHAPTER 2 - SECOND LAW OF THERMODYNAMICS
Limitations of First Law:
 It does not ensure that the process will take place.
 It does not provide means for predicting the direction of processes
Second law
 provide means for predicting the direction of processes,
 establishing conditions for equilibrium,
 determining the best theoretical performance of cycles, engines and other devices.
Kelvin-Planck statement
No heat engine can have a thermal efficiency 100 percent. (QR ≠ 0).
Clausius Statement
Heat cannot flow from body at lower temperature to body at higher temperature
spontaneously but it can be made possible with an external work.
Heat Engine:
o they receive heat from a high-temperature source
o they convert part of this heat to work
o they reject the remaining waste heat to a low temperature
sink atmosphere
o they operate on a cycle
Thermal efficiency:
 Represent the magnitude of the energy wasted in order to
complete the cycle.
 A measure of the performance that is called the thermal
efficiency.
 Can be expressed in terms of the desired output and the required input
 For a heat engine the desired result is the net work done and the input is the heat supplied
to make the cycle operate.
The thermal efficiency is always less than 1 or less than 100 percent.
where
th 
Desired Result
Required Input
th
net out
in
W
Q

,
W W W
Q Q
net out out in
in net
,  


32
 Applying the first law to the cyclic heat engine
, − , = 0
, = ,
, = −
 The cycle thermal efficiency may be written as
 A thermodynamic temperature scale related to the heat transfers between a reversible
device and the high and low-temperature reservoirs by
 The heat engine that operates on the reversible Carnot cycle is called the Carnot Heat
Engine in which its efficiency is
Heat Pumps and Refrigerators
 A device that transfers heat from a low temperature medium to a high temperature one
is the heat pump.
 Refrigerator operates exactly like heat pump except that the desired output is the
amount of heat removed out of the system
 The index of performance of a heat pumps or refrigerators are expressed in terms of the
coefficient of performance.
th
net out
in
in out
in
out
in
W
Q
Q Q
Q
Q
Q



 
,
1
Q
Q
T
T
L
H
L
H

th rev
L
H
T
T
,  
1

33
Carnot Cycle
Process Description
1-2 Reversible isothermal heat addition at high temperature
2-3 Reversible adiabatic expansion from high temperature to low
temperature
3-4 Reversible isothermal heat rejection at low temperature
4-1 Reversible adiabatic compression from low temperature to high
temperature
COP
Q
W
Q
Q Q
HP
H
net in
H
H L
 

,
COP
Q
W
R
L
net in

,

Execution of Carnot cycle in a piston cylinder device
Execution of Carnot cycle in a piston cylinder device
34

35
Carnot’s Theorem: Among the many heat engines operating between the same temperature
limits, only reversible heat engine (Carnot Engine) will have highest efficiency.
The thermal efficiencies comparison are as follows
The coefficients of performance of actual and reversible refrigerators operating between the
same temperature limits compare as follows
Problem #2.1: A power station contains a heat engine operating between two heat reservoirs, one
consisting of steam at C and the other consisting of water at C. What is the maximum
amount of electrical energy which can be produced for every Joule of heat extracted from the
steam?
Solution:
The maximum efficiency comes from using a Carnot (reversible)
engine, for which = , and so
With TH =373 K and TH =293 K;
For every Joule of heat extracted from the hot reservoir (QH = 1 J)
= 1 1 − = 0.2144
Problem #2.2: When a fridge stands in a room at C, the motor has to extract 500W of heat
from the cabinet, at C, to compensate for less than perfect insulation. How much power must
be supplied to the motor if its efficiency is 80% of the maximum achievable?

36
Solution:
The maximum efficiency comes from using a Carnot (reversible)
fridge, for which = . Here QC is known, and so
(In this question W, QC, and QH will refer to energy transfer per
second, measured in Watts.) With TH =293 K and TC = 277 K,
QC = 500 W gives Wmax = 29 W.
However the real fridge works at 80% of the maximum efficiency,
so =
.
= 36
Problem #2.3: Two reversible engines A & B are arranged in series, EA rejecting heat directly to
engine, EB. EA receives 200 kJ at a temperature of 421 ͦ
C from a hot source, while EB is in
communication with a cold sink at a temperature of 4.4 ͦ
C. If the work output of EA is twice that
of EB, find the efficiency of each engine.
Solution:
From Thermodynamic temperature scale we know that,
= ; =
But = =
 Q3 = 79.94 kJ
Since, WA = 2 WB
− = 2( − )  Q2 = 119.96 kJ
=  T2 = 143.26 ͦ
C ;
= = % where WA = Q1 –Q2
= = . % where WB = Q2 –Q3

37
 ENTROPY – A Measure of Disorder
 Derivation of Entropy from Second Law
Consider the following combined system consisting of a Carnot heat engine and a
piston cylinder system
First Law applied to the combined system
C
rev
C W
Q
dE 
 
 ; C
rev
C dE
Q
W 


For a Carnot Engine (reversible cycle)
T
T
Q
Q
T
T
Q
Q R
rev
R
rev



 



Substituting
C
R
C dE
T
Q
W 


T


Let the system undergo a cycle while the Carnot cycle undergoes one or more cycles.
Integrate over entire cycle. (Recall, system energy is a state property)


 

 C
R
C dE
T
Q
W
T


The net work for one cycle is
 






boundary
R
C
Q
T
W
T

The combined system (cycle) draws heat from a single reservoir while involving work WC
Based on K-P statement the combined system cannot produce net work output  WC 0
0
T








boundary
Q


38
This is the Clausius Inequality which is valid for all thermodynamic cycles, reversible or
irreversible.
For a reversible cycle (no irreversibilities in the system)
0
T








rev
Int
Q

Since the cycle integral of (dQ/T) is 0, the quantity dQ/T is a state property, it does not
depend on the path (similar to   0
dE ).
We call this as new property entropy, S,
kJ/K
:
units
int
rev
T
Q
dS 







For a process where the system goes from state 1 to 2;
rev
T
Q
S
S
int
2
1
1
2  








Specific entropy s= S/m
rev
rev
T
q
T
m
Q
s
s
int
2
1
int
2
1
1
2
/

 
















The entropy change between two specified states is the same whether the process is
reversible or irreversible
To calculate S2-S1 evaluate  T
Q /
 for the reversible path. Consider a cycle made up of an
irreversible process followed by a reversible process
Applying Clausius Inequality to the cycle
1
2
Irreversible
process
T
S
Internally reversible
process

39
0
T
T
T int
1
2
2
1 
 





  







rev
b
Q
Q
Q 


 
 









 






2
1
1
2
2
1
2
1
T
0
T
b
b
Q
S
S
S
S
Q


Remove inequality sign to get the entropy balance equation for a closed system
gen
b
S
Q
S
S
T
2
1
1
2 
 
















process
impossible
0
process
reversible
0
process
le
irreversib
0
gen
S
For an isolated system (adiabatic closed system)
gen
S
Q
T
S 2
1
isol 

 

Since Sgen 0  Sisol 0
This is the Increase in Entropy Principle which simply stated says that for an isolated
system the entropy always increases or remains the same
A process that is both adiabatic and reversible is referred to as isentropic, and for a closed
system, 0
T
2
1 



 gen
isentropic S
Q
S
 .
Therefore, entropy is constant, S1=S2.
Change in entropy
of the system Entropy generated in the
system due to irreversibilities
Entropy transfer to the
system by heat transfer

40
Evaluation of Entropy Change in a Closed System
Apply First Law neglecting KE and PE effects, no shaft work
In differential form Pdv
q
w
q
du 


 


Pdv
du
q 


For an internally reversible process
T
q
ds

 so
Pdv
du
Tds 
 ----------------- (1)
recall vdP
Pdv
dh
Pv
h
d
du
Pv
u
h 






 )
(
substitute du into the above equation
vdP
dh
Tds
Pdv
vdP
Pdv
dh
Tds





 )
( ------------------- (2)
These are the Gibb’s equations (or simply Tds equations)
For a system undergoing a process from state 1 to state 2














2
1
2
1
2
1
1
2
2
1
2
1
2
1
1
2
dP
T
v
T
dh
ds
s
s
dv
T
P
T
du
ds
s
s
W
Q
System

41
 Entropy Change for an Ideal Gas
For an ideal gas RT
Pv
dT
c
dh
dT
c
du P
V 

 ,
,



 




2
1
2
1
2
1
2
1
1
2
v
dv
R
T
dT
c
dv
T
P
T
du
s
s V
and
2
1
2
1
2
1
2
1
1
2 


 




P
dP
R
T
dT
c
dP
T
v
T
dh
s
s P
If the specific heats cP and cV are taken as constant






























































1
2
1
2
1
2
1
2
2
1
1
2
1
2
1
2
1
2
1
2
2
1
1
2
ln
ln
ln
ln
ln
ln
P
P
R
T
T
c
s
s
P
P
R
T
dT
c
s
s
v
v
R
T
T
c
s
s
v
v
R
T
dT
c
s
s
P
P
V
V
 Isentropic Efficiency for Turbine

42
 Isentropic Efficiency for Compressor:
Problem#2.4: Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27 ͦ C to a
final state of 0.5 MPa. Find the work done on the compressor with an isentropic efficiency of
80 %.
Solution:
Applying SFEE for compressor,
̇ = ̇ (ℎ − ℎ )  = ̇
̇
= (ℎ − ℎ )
= ( − )
For isentropic process; =  T2s = 475.4 K
Then,
= 1.005(475.4 − 300) = 176 /
( ) = = /

43
Problem #2.5: A piston–cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27°C.
The gas is now compressed slowly in a polytropic process during which PV1.3
= constant. The
process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen
during this process.
Solution:
Entropy Change, 


















1
2
1
2
1
2 ln
ln
v
v
R
T
T
c
s
s V
For a Polytropic process,
1
2
1
1
2


















k
v
v
T
T
So, T2 = T1 * (2)0.3 = 369 K
 
5
.
0
ln
*
287
.
0
300
369
ln
*
718
.
0
1
2 







 s
s
s2 – s1 = 0.0596 kJ/kgK
Problem #2.6: Air at 20 ͦ C and 1.05 bar occupies 0.025 m3
. The air is heated at constant
volume until the pressure is 4.5 bar, and then cooled at constant pressure back to original
temperature. Calculate a) the net heat flow from the air, b) the net entropy change.
Solution:
cp = 1.005kJ/kgK; cv = 0.718 kJ/kgK ------- Assumed
From gas equation, = =
. .
= .
For Process 1-2 (Constant Volume Process):
= == T2 = 1255.7 K (Q)v=c = m cv (T2 – T1) = 21.56 kJ

44
For Process 2-3 (Constant Pressure Process):
(Q)p=c = m cp (T3 – T2) = - 30.18 kJ where T3 = T1 and –‘ve sign indicates heat loss
Entropy change in process 1-2, − = ln = 0.0326 /
Entropy change in process 2-3, − = ln = −0.0456 /
Net entropy change, (∆s) = s1 –s3 = - 0.013 kJ/K
Supplementary Problems for Practice
1. A reversible heat engine operates between two reservoirs at temperatures of 600 ͦ
C and
40 ͦ C. The engine drives a reversible refrigerator which operates between reservoirs at
temperatures of 40 ͦ C and -20 ͦ C. The heat transfer to the engine is 2 MJ and the net
output of the combined engine and refrigerator plant is 360 kJ. Find the heat transfer to
the refrigerant and the net heat transfer to the reservoir at 40 C. Also find these values if
the efficiency of the heat engine and COP of the refrigerator are each 40% of their
maximum possible values.
2. A reversible engine as shown in Figure below
during a cycle of operation draws 5 MJ from
the 400 K reservoir and does 840 kJ of work.
Find the amount and direction of heat
interaction with other reservoirs.
3. The scales are so chosen that a reversible cycle plots clockwise as a circle on T-s plane,.
The minimum and maximum values of the temperature are 305 and 627 K and the
entropy 1.23 and 2.85 kJ/k respectively. Find the cycle work and efficiency.
4. A gas initially at 603 K expands until its volume is 5.2 times the initial volume according
to pvn
= constant. If the initial and final pressures are observed to be 8.5 bar and 1 bar,
determine i) index of expansion, ii) workdone per kg of gas, iii) heat exchange per kg of
gas, and iv) entropy change per kg of gas. Assume cv = 0.712 kJ/kgK and γ = 1.4.
5. One kg of water at 273 K is brought into contact with a heat reservoir at 373 K. When the
water has reached 373 K, find the entropy change of the water, the heat reservoir and the
universe.

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