ful thermodynamics lecture.pptx

CHAPTER
1
Thermodynamics
Fundamental
Concepts
1

What is Thermodynamics?
The science of energy, that
in
concerned with the ways
which energy is stored within a
body.
Energy transformations – mostly
involve heat and work movements.
The Fundamental law is the
conservation of energy principle:
energy cannot be created or
be
to
destroyed,
transformed
another.
but can only
from one form
2

System, surroundings and boundary
❖ System: A quantity of matter or a
region in space chosen for study.
❖ Surroundings: The mass or region
outside the system
❖ Boundary: The real or imaginary
surface that separates the system
from its surroundings.
3

Type of system
(isolated system)
Isolated system – neither
mass nor energy can cross
the selected boundary
Example (approximate): coffee in
a closed, well-insulated thermos
bottle
4

Type of system
(Closed system)
Closed system – only energy
can cross the selected
boundary
Examples: a tightly capped cup of
coffee
5

Type of system
(Open system)
Open system – both mass and
energy can cross the selected
boundary
Example: an open cup of coffee
6

7
Properties of a system
Properties of a system is a measurable characteristic of a system that is
in equilibrium.
Properties may be intensive or extensive.
Intensive – Are independent of the amount of mass:
e.g: Temperature, Pressure, and Density,
Extensive – varies directly with the mass
e.g: mass, volume, energy, enthalpy

Specific properties – The ratio of any extensive property of a system to that
of the mass of the system is called an average specific value of that property
(also known as intensives property)
Properties of a system

State, Equilibrium and Process
State – a set of properties that describes the conditions of a
system. Eg. Mass m, Temperature T, volume V
Thermodynamic equilibrium -
system that maintains thermal,
mechanical, phase and chemical
equilibriums.
9

Process – change from one
equilibrium state to another.
Process
10
Property held
constant
pressure
temperature
volume
entropy
isobaric
isothermal
isochoric
isentropic

The prefix iso- is often used to designate a process for which a particular property
remains constant.
Isobaric process: A process during which the pressure P remains constant.
Pressure is Constant (ΔP = 0)

Isothermal process: A process during
which the temperature T remains
constant.
.
Isochoric (or isometric) process: A process during which the specific volume v
remains constant
Process Property held
constant
pressure
temperature
volume
entropy
isobaric
isothermal
isochoric
isentropic

Types of Thermodynamics Processes
Cyclic process - when a system in a given initial
state goes through various processes and finally
return 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
13

Types of Thermodynamics Processes
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.
Polytropic process - when a gas undergoes a reversible
process in which there is heat transfer, it is represented
with a straight line, PVn = constant.
Throttling process - a process in which there is no
change in enthalpy, no work is done and the process is
adiabatic.
14

Zeroth Law of Thermodynamics
“ If two bodies are in thermal equilibrium with a third
body, there are also in thermal equilibrium with each
other.”
15

Application Areas of Thermodynamics
16

CHAPTER
2
Thermodynamics
Properties of
Pure Substances
1

What is Pure Substances?
A substance that has a fixed
chemical composition throughout
is called a pure substance.
A pure substance does not have
to be of a single chemical
element or compound, however.
A mixture of various chemical
elements or compounds also
qualifies as a pure substance as
long as the mixture is
homogeneous.
2

Faculty of Mechanical Engineering, UiTM
3
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A mixture of liquid and water vapor is a pure substance, but a
mixture of liquid and gaseous air is not.
Examples:
❖ Water (solid, liquid, and vapor phases)
❖ Mixture of liquid water and water vapor
❖ Carbon dioxide, CO2
❖ Nitrogen, N2
❖ Mixtures of gases, such as air, as long as there is no
change of phase.

Phases of A Pure Substance
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The substances exist in different phases, e.g. at
room temperature and pressure, copper is solid
and mercury is a liquid.
It can exist in different phases under variations
of condition.
There are 3 Principal phases
• solid
• Liquid
• gas
Each with different molecular structures.

Phase-change Processes of Pure Substances
Solid: strong intermolecular bond
Liquid: intermediate intermolecular bonds
Gas: weak intermolecular bond
There are many practical situations where two phases of a pure
substances coexist in equilibrium.
E.g. water exists as a mixture of liquid and vapor in the boiler and etc.
5
MEC 451 –
Solid Liquid Gas

Phase-change Processes
6
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s
7
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This constant
pressure heating
process can be
illustrated as:

Property Diagram
8
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Saturation
9
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Saturation is defined as a condition in which a mixture
of vapor and liquid can exist together at a given
temperature and pressure.
Saturation pressure is the pressure at which the liquid
and vapor phases are in equilibrium at a given
temperature
For a pure substance there is a definite relationship
between saturation pressure and saturation
temperature. The higher the pressure, the higher the
saturation temperature

The graphical representation of this relationship between temperature
and pressure at saturated conditions is called the vapor pressure curve
10
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Saturated and Sub-cooled Liquids
If a substance exists as a liquid at the
saturation temperature and pressure,
it is called a saturated liquid
If the temperature of the liquid is
lower than
temperature
pressure, it
the
for the
is called
saturation
existing
either a
subcooled liquid or a compressed
liquid
11
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If a substance exists entirely as
vapor at saturation temperature, it
is called saturated vapor.
When the vapor is at a temperature
greater than the saturation
temperature, it is said to exist as
superheated vapor.
The pressure and temperature of
superheated vapor are independent
properties, since the temperature
may increase while the pressure
remains constant
Saturated and Superheated Vapors
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Latent Heat
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Latent heat: The amount of energy absorbed or
released during a phase-change process.
Latent heat of fusion: The amount of energy absorbed
during melting. It is equivalent to the amount of energy
released during freezing.
Latent heat of vaporization: The amount of energy
absorbed during vaporization and it is equivalent to the
energy released during condensation.
At 1 atm pressure, the latent heat of fusion of water
is 333.7 kJ/kg and the latent heat of vaporization is
2256.5 kJ/kg.

Quality
❖ When a substance exists as part liquid and part vapor at
saturation conditions, its quality (x) is defined as the
ratio of the mass of the vapor to the total mass of both
vapor and liquid.
❖ The quality is zero for the saturated liquid and one for
the saturated vapor (0 ≤ x ≤ 1)
❖ For example, if the mass of vapor is 0.2 g and the mass
of the liquid is 0.8 g, then the quality is 0.2 or 20%.
x 
14
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masssaturated vapor
masstotal
mg
mf  mg


Quality
Mixture of liquid and vapor
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Moisture Content
The moisture content
opposite
of a
of its
substance is the
quality. Moisture is defined as the
ratio of the mass of the liquid to
the total mass of both liquid and
vapor
Recall the definition of quality x
Then
x 
mg mg
m mf  mg

m m m
16
MEC 451 –
m m
 1 x
f g


Moisture Content
Take specific volume as an example. The specific volume of the
saturated mixture becomes
v  (1 x)vf  xvg
The form that is most often used
v  vf  x(vg  vf )
Let Y be any extensive property and let y be the corresponding
intensive property, Y/m, then
y 
Y
 y
where yfg  yg yf 17
MEC 451 –
m
 x(yg  yf )
 yf  x yfg
f

Property Table
For example if the pressure
and specific volume are
specified, three questions are
asked: For the given pressure,
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Property Table
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If the answer to the first question is yes,
the state is in the compressed liquid
region, and the compressed liquid table is
used to find the properties. (or using
saturation temperature table)
If the answer to the second question is
yes, the state is in the saturation region,
and either the saturation temperature table
or the saturation pressure table is used.
If the answer to the third question is yes,
the state is in the superheated region and
the superheated table is used.
v  vf
vf  v  vg
vg  v

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Example 2.1
Determine the saturated pressure, specific volume, internal energy
and enthalpy for saturated water vapor at 45oC and 50oC.
21
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Example 2.2
Determine the saturated pressure, specific volume, internal energy and
enthalpy for saturated water vapor at 47⁰ C .
22
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Solution:
Extract data from steam table
Interpolation Scheme for Psat
Interpolation for Psat
sat@47
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Psat 9.5953

47  45
12.3529.5953 50  45
10.698kPa
P 
Do the same principal to
others!!!!
T Psat v u h
45 9.5953 15.251 2436.1 2582.4
47 Psat v u h
50 12.352 12.026 2442.7 2591.3

Exercises
2. Determine the saturated temperature, saturated pressure and
enthalpy for water at specific volume of saturated vapor at
10.02 m3/kg .
1. Fill in the blank using R-134a
24
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Example 2.3
Determine the enthalpy of 1.5
kg of water contained in a
volume of 1.2 m3 at 200 kPa.
Solution:
Specific volume for water
v 
Volume

1.2m
mass 1.5kg
 0.8
m
kg
3 3
From table A-5:
v  0.001061m3
m3
vg  0.8858
kg
f
kg
Is v  v ? No
Is vf  v  vg ? Yes
Is vg  v ? No
Find the quality
v  vf  x(vg  vf )
f
0.80.001061

0.88580.001061
 0.903 (What does this mean?)
f
vg vf
v v
x 
h  hf  xhfg
 504.7  (0.903)(2201.6)
 2492.7
kJ
kg 25
MEC 451 –
The enthalpy

Example 2.4
Determine the internal energy of refrigerant-134a at a temperature
of 0C and a quality of 60%.
Solution:
From table A-5:
uf  51.63
kg
ug  230.16
kg
kJ
kJ
The internal energy of
at given condition:
u  uf  x(ug uf )
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 51.63 (0.6)(230.1651.63)
158.75
kJ
kg
R 134a

Example 2.5
Consider the closed, rigid container of
water as shown. The pressure is 700
kPa, the mass of the saturated liquid is
1.78 kg, and the mass of the saturated
vapor is 0.22 kg. Heat is added to the
water until the pressure increases to 8
MPa.
enthalpy,
water
Find the final temperature,
and internal energy of the
mg, Vg
Sat. Vapor
mf, Vf
Sat. Liquid
27
MEC 451 –

Solution:
Theoretically:
v2  v1
The quality before pressure
increased (state 1).
mg1
1
mf 1  mg1
0.22kg

(1.78 0.22)kg
 0.11
x 
Specific volume at state 1
v1  vf 1  x1 (vg1 vf 1)
 0.001108 (0.11)(0.27280.001108)
m3
 0.031
kg
State 2:
Information :
P2  8MPa v2  0.031
From table A-5:
m3
kg
vg 2  v2
Since that it is in superheated
region, use table A-6:
2
2
u2  2776
T  361.8o
C
h  3024 kJ
kg
kJ
kg
28
MEC 451 –

1. Four kg of water is placed in an enclosed volume of 1m3.
Heat is added until the temperature is 150°C. Find ( a )
the pressure, ( b )the mass of vapor, and ( c ) the volume
of the vapor.
2. A piston-cylinder device contains 0.1 m3 of liquid water and
9.m3 of water vapor in equilibrium at 800 kPa. Heat is
transferred at constant pressure until the temperature reaches
350°C.
(a)what is the initial temperature of the water,
(b) determine the total mass of the water,
(c) calculate the final volume, and
(d)show the process on a P-v diagram with respect to
saturation lines.
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MEC 451 –
Exercises

Exercises
30
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3. For a specific volume of 0.2 m3/kg, find the quality of steam
if the absolute pressure is (a) 40 kPa and ( b ) 630 kPa. What
is the temperature of each case?
4. Water is contained in a rigid vessel of 5 m3 at a quality of
0.8 and a pressure of 2 MPa. If the a pressure is reduced to
400 kPa by cooling the vessel, find the final mass of vapor
mg and mass of liquid mf.

Important Definition
Critical point - the temperature and pressure above which there
is no distinction between the liquid and vapor phases.
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o
o Triple point - the temperature and pressure at which all three
phases can exist in equilibrium.
o Sublimation - change of phase from solid to vapor.
o Vaporization - change of phase from liquid to vapor.
o Condensation - change of phase from vapor to liquid.
o Fusion or melting - change of phase from solid to liquid.

32
MEC 451 –

Ideal Gas Law
Robert Boyle formulates a well-known law that states the pressure of a
gas expanding at constant temperature varies inversely to the volume,
or
P
1V1  P2V2  constant
As the result of experimentation, Charles concluded that the pressure of
a gas varies directly with temperature when the volume is held
constant, and the volume varies directly with temperature when the
pressure is held constant, or
V2 T2 P2 T2
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MEC 451 –
V1

T1 P1

T1
or

By combining the results of
and Boyle's
the following
Charles'
experiments,
relationship can be obtained
The constant in the above
equation is called the ideal gas
constant and is designated by
R; thus the ideal gas equation
becomes
In order to make the equation
applicable to all ideal gas, a
universal gas constant
introduced
RU is
 constant
T
Pv
or PV  mRT
Pv  RT
M
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MEC 451 –
R
R  U

For example the ideal gas constant for air, Rair
8.3144
  0.2871kJ / kg.K
(M )air
(R )
 U air
Rair
28.96
The amount of energy needed to raise the temperature of a unit of
mass of a substance by one degree is called the specific heat at
constant volume Cv for a constant-volume process and the specific
heat at constant pressure Cp for a constant pressure process. They
are defined as
P
35
MEC 451 –
P
v
v
T 
 h

and C  

T
 u

C  
 



The specific heat ratio, k is defined as
Using the definition of enthalpy (h = u + Pv) and writing the
differential of enthalpy, the relationship between the specific heats
for ideal gases is
h  u  Pv
dh  du  RT
CPdt  CV dt  RdT
CP  CV  R
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MEC 451 –
Cv
C
k  P

➢ For ideal gases u, h, Cv, and Cp are functions of temperature alone.
The Δu and Δh of ideal gases can be expressed as
u  u2  u1  Cv (T2 T1 )
h  h2  h1  CP (T2 T1 )
37
MEC 451 –

Example 2.6
An ideal gas is contained in a
closed assembly with an initial
pressure and temperature of 220
kPa and 700C respectively. If
the volume of the system is
the
increased 1.5 times and
temperature drops to 150C,
determine the final pressure of
the gas.
Solution:
given
P
1  220kPa
T1  70  273K  343K
T2 15  273  288K
V2 1.5V1
From ideal-gas law:
state1
state2

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MEC 451 –

P
1V1

P2V2
T1 T2
V1 288 3
2
1.5V1  343 
22010
123.15kPa
P 
 
 

Example 2.7
A closed assembly contains 2 kg
of air at an initial pressure and
temperature of 140
2100C respectively.
kPa and
If the
volume of the system is doubled
and temperature drops to 370C,
determine the final pressure of
the air. Air can be modeled as an
ideal gas.
Solution:
given
P
1 140kPa
1
T2  37  273  310K
V2  2V1
From ideal-gas law:
state1
T  210  273K  483K
state2

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MEC 451 –

P
1V1

P2V2
T1 T2
3
1
2
2V1  483
V 310
14010
 44.93kPa
P   
 

Example 2.8
An automobile tire with a volume
of 0.6 m3 is inflated to a gage
pressure of 200 kPa. Calculate the
mass of air in the tire if the
temperature is 20°C.
Solution:
given
state1
P  200 100 kPa
T  20  273K  293K
From ideal-gas law:
3  
40
MEC 451 –
293K
3 2
kg.K
m 
PV
RT
30010 0.6m
287
 2.14kg
N
m
Nm


Supplementary Problems
41
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42
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CHAPTER
3
Thermodynamics
First Law of
Thermodynamics
1

FIRST LAW OF THERMODYNAMICS
6
0
ENERGY ANALYSIS
OF CLOSED SYSTEM

First Law of Thermodynamics
6
1
The First Law is usually referred to as the Law of Conservation
of Energy, i.e. energy can neither be created nor destroyed, but
rather transformed from one state to another.
The energy balance is maintained within the system being
studied/defined boundary.
The various energies associated are then being observed as
they cross the boundaries of the system.

Energy Balance for Closed System
Heat
Work
z
Closed
System
Reference Plane, z = 0

V
or
6
2
MEC 451 –
Ein  Eout  Esystem

6
3
MEC 451 –
According to classical thermodynamics
Qnet Wnet  Esystem
The total energy of the system, Esystem, is given as
E = Internal energy + Kinetic energy + Potential energy
E = U + KE + PE
The change in stored energy for the system is
E  U  KE  PE
The first law of thermodynamics for closed systems then can be
written as
Qnet Wnet  U  KE  PE

6
If the system does not move with a velocity and has no change in
elevation, the conservation of energy equation is reduced to
Qnet Wnet  U
The first law of thermodynamics can be in the form of
(kJ )
1000
g(z  z )
2000
1
2
2000
2 2
V V
 2 1
 u1
2
  m u
W net
Qnet









(kJ / kg)
1000
V 2
V 2
g(z  z )
2 1 1
2
1
2  u 
 u
qnet  wnet









For a constant volume process,



V V g(z  z ) 

 mu




 u 
W net
1000
2000
2
2 1
2
2 1
2
1
2
Qnet

 

V V g(z  z ) 

Q  mu 
 u 
1000
2000
2 1
2
2 1
1
2
net

For a constant pressure process,



V V g(z  z ) 

 mu



 u 
W net
1000
2000
2 1
2
2 1
2
1
2
Qnet



V V g(z  z ) 

Q  P(V V )  mu  u 



1000
2000
1
2
2
2 1
2
1
2
2 1
net



V g(z  z ) 

Q  mu



 u1  P(V2 V1) 
1000
2000
2 1
2
2 1
2
2
V
net



6
5
MEC 451 –

V V g(z  z ) 

Q  mh  h 


1000
2000
2 1
2
2 1
2
2 1
net

Rigid tank
6
6
MEC 451 –
Piston cylinder
Example of Closed Systems

A closed system of mass 2 kg
undergoes an adiabatic process.
The work done on the system is
30 kJ. The velocity of the system
changes from 3 m/s to 15 m/s.
During the process, the elevation
of the system increases 45 meters.
Determine the change in internal
energy of the system.
Example 3.1
Solution:
Energy balance,



g(z  z ) 

 mu  u 



V V
W net
1000
2000
2 1
2
2 1
2
2 1
Qnet
Rearrange the equation
Qnet
30 2u  2
6
7
MEC 451 –
 9.8145
V 2
V 2
2 1
2000
2 1
1000
 mu2  u1 
V 2
V 2
2 1
2000
2 1
1000
 mu2  u1 

152
 32

g(z  z ) 
g(z  z ) 
 2
2000 1000
u  14.451 kJ Ans..
W net
Wnet

 



 



   

Steam at 1100 kPa and 92 percent
quality is heated in a rigid container
until the pressure is 2000 kPa. For a
mass of 0.05 kg, calculate the amount
of heat supply (in kJ) and the total
entropy change (in kJ/kg.K).
Example 3.2 Solution:
v1  vf 1  x1vfg1
 0.00113  0.920.17753  0.001133
kg
u1  uf 1  x1ufg1
 780.09  0.921806.3
kg
s1  sf 1  x1sfg1
 2.1792  0.924.3744
3
 0.1634 m
at P
1  1100 kPa, x1
kJ
kg.K
State1
 0.92
 2441.9 kL
 6.204
68
MEC 451 –

3116.9  2945.9
m3
kg
at P2  2000kPa,v2
u2  2945.9  
2
State2
 0.1634
 0.1634  0.15122 
 0.17568  0.15122 
 3030.42 kJ
s  7.1292   0.1634  0.15122 
7.4337  7.1292
 0.17568  0.15122 
kg

kJ
kg.K

 7.2790
For a rigid container,
v2=v1=0.1634 m3/kg
superheated
69
MEC 451 –
v u s
0.15122 2945.9 7.1292
0.1634 u2 s2
0.17568 3116.9 7.4337

Amount of heat supplied, Q
Q  mu2  u1 
 0.053030.42  2441.9
 29.43kJ
The change in entropy, Δs
s  s2  s1
 7.2790  6.204
70
MEC 451 –
kJ
kg.K
 1.075

Example 3.3
A rigid tank is divided into two equal
parts by a partition. Initially one side
of the tank contains 5 kg water at 200
kPa and 25°C, and the other side is
evacuated. The partition is then
removed, and the water expands into
the entire tank. The water is allowed to
exchange heat with its surroundings
until the temperature in the tank
returns to the initial value of 25°C.
Determine (a) the volume of the tank
(b) the final pressure (c) the heat
transfer for this process.
Solution:
initial volumeof half resevoir
 50.001003
; 0.005m3
The initial volume for entire tank
 20.005
 0.01m3
m3
kg
1
v1  vf @ 25
C
1
1
State1
P  200kPa,
 0.001003
T  25o
C
V  mv


resevoir
V
Comp. liquid
71
MEC 451 –

 mu  ke  Pe
 mu  ke  Pe 
Qnet Wnet
Qnet  Wnet
 mu  mu2  u1 
Qnet
The final pressure
The heat transfer for this process
3
 0.001003 m
m3
kg  vg  43.34 kg
v 
0.01
 0.002 m3 
2
2
then :P2
State 2
T  25
C
5 
check region!
vf  v  vg  saturated mixture
 3.169kPa
 f kg
 Psat
v
2304.9
104.93 104.88
1 f @ 25
C
u2  uf  x2u fg
2 f
x2 
5
5
104.88
104.93
Then :
u2  104.83  2.310
 104.88 kJ
 2.310
Q  5 (104.88-104.83)
 0.25 kJ
kg
fg
kJ
kg
net
u  u
v  v
v
+ve sign indicates heat transfer
into the system.
 104.88
72
MEC 451 –
(2304.3)

Supplementary Problems 1
73
MEC 451 –
1. Two tanks are connected by a valve. Tank A contains 2 kg of carbon
monoxide gas at 77°C and 0.7 bar. Tank B holds 8 kg of the same gas
at 27°C and 1.2 bar. Then the valve is opened and the gases are
allowed to mix while receiving energy via heat transfer from the
surrounding. The final equilibrium temperature is found to be 42°C.
Determine (a) the final pressure (b) the amount of heat transfer. Also
state your assumption.
+37.25 kJ]
[P2=105 kPa, Q =
2. A piston cylinder device contains 0.2 kg of water initially at 800 kPa
and 0.06 m3. Now 200 kJ of heat is transferred to the water while its
pressure is held constant. Determine the final temperature of the
water. Also, show the process on a T-V diagram with respect to
saturation lines.
[ 721.1oC]

[50°C, 400 kPa] 74
MEC 451 –
3. A piston-cylinder device contains 6 kg of refrigerant-134a at 800 kPa
and 50oC. The refrigerant is now cooled at constant pressure until it
exist as a liquid at 24oC. Show the process on T-v diagram and
determine the heat loss from the system. State any assumption made.
[1210.26
4
. k
A
J]
0.5 m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40
percent quality. Heat is now transferred to the refrigerant until the
pressure reaches 800 kPa. Determine (a) the mass of the refrigerant in
the tank and (b) the amount of heat transferred. Also, show the process
on a P-v diagram with respect to saturation lines.
[12.3 kg, 2956.2
5
. k
A
Jn
] insulated tank is divided into two parts by a partition. One part of
the tank contains 6 kg of an ideal gas at 50°C and 800 kPa while the
other part is evacuated. The partition is now removed, and the gas
expands to fill the entire tank. Determine the final temperature and the
pressure in the tank.

75
MEC 451 –
Some thermodynamic cycle composes of processes in which
the working fluid undergoes a series of state changes such
that the final and initial states are identical.
For such system the change in internal energy of the
working fluid is zero.
The first law for a closed system operating in a
thermodynamic cycle becomes
Closed System First Law of a Cycle
Qnet Wnet  Ucycle
Qnet  Wnet

Boundary Works
3
4
5
V
76
MEC 451 –
2
1
P

No Value of n Process Description Result of IGL
77
MEC 451 –
3 1 isothermal constant temperature
(T1 = T2)
4 1<n< γ polytropic -none-
5 γ isentropic
1
constant entropy (S1 = S2) P2
According to a law of PV n
 constant
P1V1  P2V2
 T n1
T2 
1
 V1 
2



  


 

P V
n
n
1 ∞ isochoric constant volume (V1 = V2) P1 P2
2 0 isobaric constant pressure (P1 = P2) T1
V1
 T2
V2
T1 T2

Various forms of work are expressed as follows
Process Boundary Work
isochoric
isobaric
isothermal
polytropic
isentropic
W12  P(V2 V1)  0
W12  P(V2 V )
1
V1
78
MEC 451 –
V 2
 P1V1 ln
W 12
1 n
P V  PV
W  2 2 1 1
12

Example 3.4
Sketch a P-V diagram showing the following processes in a cycle
79
MEC 451 –
Process 1-2: isobaric work output of 10.5 kJ from an initial volume of 0.028
m3 and pressure 1.4 bar,
isothermal compression, and
isochoric heat transfer to its original volume of 0.028 m3 and
pressure 1.4 bar.
Process 2-3:
Process 3-1:
Calculate (a) the maximum volume in the cycle, in m3, (b) the isothermal work,
in kJ, (c) the net work, in kJ, and (d) the heat transfer during isobaric expansion,
in kJ.

Solution:
Process by process analysis,
Section 1 2isobaric
W12  PV2 V1  10.5
140V2  0.02810.5
V  0.103m3
2
The isothermal work
 
Section 2  3isothermal 
 1400.103ln  0.028 
P2V2  P3V3
3  
3
 W23  P2V2 ln
V2
 0.103 
140  515kPa
0.028
 0.103 
 18.78kJ
P 
V
 
 
80
MEC 451 –

The net work
Section 31isochoric
81
MEC 451 –
W31  0
 W12 W23 W31
10.5 18.78
 8.28kJ
Wnet

Example 3.5
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.
82
MEC 451 –

Solution:
Process by process analysis,
Section 1 2isothermal 
0.220.12
 4150.00732ln
0.0264
P
1V1  P2V2
V1  
 0.00732 m3
V2
V1
12 1 1
 115 
 415 
W  PV ln
0.00732
 3.895kJ
83
MEC 451 –

Section 2  3isobaric
W23  PV3 V2  0.525kJ
3
115
 0.03097 m3
V 
0.525
 0.0264
Section 3 4isochoric
W34  0
Section 4 1PolytroPic
4150.0072 620.03097
11.3182
4
  1

P
1
62
V4 
PV  PV
 1 1 4 4
41
  0.00732 
415  0.03097 
ln 0.1494  nln 0.2364
n  1.3182
1 n
 3.5124kJ
n
n
P  V 
W
 

The net work of the cycle
 W12 W23 W34 W41
 0.9076kJ
84
MEC 451 –
Wnet

85
MEC 451 –
1. A mass of 0.15 kg of air is initially exists at 2 MPa and 350oC. The air
is first expanded isothermally to 500 kPa, then compressed
polytropically with a polytropic exponent of 1.2 to the initial state.
Determine the boundary work for each process and the net work of the
cycle.
2. 0.078 kg of a carbon monoxide initially exists at 130 kPa and 120oC. The
gas is then expanded polytropically to a state of 100 kPa and 100oC.
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. Two kg of air experiences the three-
process cycle shown in Fig. 3-14.
Calculate the net work.
4. A system contains 0.15 m3 of air pressure of 3.8 bars and 150⁰ 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
86
MEC 451 –

87
MEC 451 –
MASS & ENERGY ANALYSIS
OF CONTROL VOLUME

Conservation of Mass
88
MEC 451 –
Conservation of mass is one of the most fundamental
principles in nature. We are all familiar with this
principle, and it is not difficult to understand it!
For closed system, the conservation of mass principle is
implicitly used since the mass of the system remain
constant during a process.
However, for control volume, mass can cross the
boundaries. So the amount of mass entering and leaving
the control volume must be considered.

Mass and Volume Flow Rates
Mass flow through a cross-sectional area per unit time is called the
mass flow rate. Note the dot over the mass symbol indicates a time
rate of change. It is expressed as
m
  V .dA
If the fluid density and velocity are constant over the flow cross-
sectional area, the mass flow rate is
where 
1

 iscalled specificvoulme
89
MEC 451 –
AV
m
  AV 


Principal of Conservation of Mass
90
MEC 451 –
The conservation of mass principle for a control volume can be
expressed as
m
in  m
out  m
CV
For a steady state, steady flow process the conservation of mass
principle becomes
(kg/s)
m
in  m
out

As the fluid upstream pushes mass across the control volume, work
done on that unit of mass is
flow
91
MEC 451 –
Wflow
w 
flow
W  FdL  FdL
A
 PdV  Pvm
A
 Pv
m
Flow Work & The Energy of a Flowing Fluid

Total Energy of a Flowing Fluid
92
MEC 451 –
The total energy carried by a unit of mass as it crosses the control
surface is the sum of the internal energy + flow work + potential
energy + kinetic energy
V 2
V 2
energy  u  P 
2
 gz  h 
2
 gz
The first law for a control volume can be written as
 in
  mout hout 






 
 gz
h 
 gzout 








. .
Qnet  W net
in
in
in
in
out
out V
m
V
2
2
2
.
2
.

The steady state, steady flow conservation of mass and first law of
thermodynamics can be expressed in the following forms
(kW )
1000
V 2
V 2
g(z  z )
2 1
2000
1
2
1 
Qnet W net  m 2  h
. . .
h







(kJ )
1000
g(z  z )
2000
1
2
2 2
V V
 2 1
 h1
2
 m h
W net
Qnet










(kJ / kg)
93
MEC 451 –
1000
V 2
V 2
g(z  z )
2 1
2000
1
2
1
2 h 
 h 
qnet  wnet










Steady-flow Engineering Devices
94
MEC 451 –

Nozzle & Diffuser
Nozzle - device that increases
the velocity fluid at the expense
of pressure.
Diffuser - device that increases
pressure of a fluid by slowing it
down.
Commonly utilized in jet
engines, rockets, space-craft
and even garden hoses.
Q = 0 (heat transfer from the
95
MEC 451 –
fluid to surroundings
small
W = 0 and ΔPE = 0
very

Energy balance (nozzle & diffuser):
 out out
out
 in in
in








gz












gz

.
W  out
out
out
out
in
in
in
in
V
h
m
.
W
.
Q
V
m h
.
Q
2
2
2
.
2
.



2

out  out
 h



  m


2

in  in
m  h 


2
2
.
.
out
in V
V

96
MEC 451 –


2


   h 
 

2 
V 2


 h 


2
2
2
1
1
V

Example 3.6
Steam at 0.4 MPa, 300ºC,
enters an adiabatic nozzle with
a low velocity and leaves at 0.2
MPa with a quality of 90%.
Find the exit velocity.
Solution:
Simplified energy balance:
V 2
1
1
h 
 
2
2
 2

1
1
1
State 2
P2  0.2 MPa h2  hf  x2hfg
 h2  2486.1
x2  0.9
2 2
State1
h  3067.1
P  0.4 MPa
T  300 C sup erheated
kJ
kg
o
kJ
kg
V
 h 




 


 
 




P1  0.4 MPa
o
P2  0.2 MPa
x2  0.9
1
V1 ; 0
T  300 C
State1 State 2
Exit velocity:
V2  20003067.1 2486.1
 1078 m / s
97
MEC 451 –

Example 3.7
Air at 10°C and 80 kPa enters the
diffuser of a jet engine steadily
with a velocity of 200 m/s. The
inlet area of the diffuser is 0.4
m2. 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 and (b) the temperature
of the air leaving the diffuser.
1 2
1
1
A  0.4 m2
1
Solution:
P  80 kPa V ; 0
T  10o
C
State1
V  200m / s
State 2
2
1
1 2
 
 2 
h 
0
V 2 
2
V 
 h 

2






Simplified energy balance:
From Ideal Gas Law:
3
1.015 m
98
MEC 451 –
1
1
1
kg
RT
v 
P

Mass flow rate

2000.4
m
  V1 A1
v
1
1
1
1.015 
 78.8 kg
s
 
 
Enthalpy at state 1
h1  CpT1 1.005283
 284.42 kJ
kg
From energy balance:
2
99
MEC 451 –
V1
h2  h1 
2002
2
2
2000
 284.42 
2000
 304.42
Cp
304.42
1.005
 302.9 K
kJ
kg
h
T 


Turbine & Compressor
10
0
MEC 451 –
Turbine – a work producing device through the expansion of a
fluid.
Compressor (as well as pump and fan) - device used to increase
pressure of a fluid and involves work input.
Q = 0 (well insulated), ΔPE = 0, ΔKE = 0 (very small compare
to Δenthalpy).

Energy balance: for turbine
 out out
out
10
1
MEC 451 –
 in in
in








gz
h 
. .
Q W 








gz

. .
Q W  out
out
out
out
in
in
in
in
V
m
V
m h
2
2
2
.
2
.
. .
min hin  W out  mout hout 
.
. .
W out  m h1  h2 

Energy balance: for compressor, pump and fan
 out out
out
10
2
MEC 451 –
 in in
in








gz
h 
. .
Q W 


 




gz

. .
Q W  out
out
out
out
in
in
in
in
V
m
V
m h
2
2
2
.
2
.
. . .
W in  min hin  mout hout 
 m h2  h1 
.
.
W in

Example 3.8
The power output of an adiabatic steam turbine is 5 MW. Compare
the magnitudes of Δh, Δke, and Δpe. Then determine the work done
per unit mass of the steam flowing through the turbine and calculate
the mass flow rate of the steam.
Data :
10
3
MEC 451 –
Inlet (P = 2 MPa, T = 400ºC,v = 50 m/s, z = 10 m)
Exit (P = 15 kPa, x = 90%, v = 180 m/s, z = 6m)

Solution:
2373.1
1
1
1
2
x2  0.9
h2  hf 2  x2hfg2
 225.94  0.9 (2372.3)
State1
p  2 MPa superheated
3247.6
T  400 C
State2
P 15kPa
sat.mixture
2361.73
o kJ
kg
kJ
kg
h  3248.4





 2361.01
.
Qin
. .
 Win
From energy balance:
2
gzin 

2
. . .
2
2
in
min hin 
in
out
Qout Wout mout hout  gzout


out
V
V
 

 

g z2  z1 
PE   0.04 kJ
h  h2  h1  -887.39
10
4
MEC 451 –
V 2
V 2
2 1
2000
885.87
14.95 kJ
1000
kJ
kg
kg
kg
KE 
Solve the equation:

the work done per unit mass
 V
W  h  h 
  g z  z 

2 2
1 2
1 2
1 2
2000 1000
885.87
 887.39 14.95 0.04
870.96
out
kJ
kg
V
 


 

  
 
 872.48
The mass flow rate
5000
 5.74
870.96
kg
s
W

m
  out
out
W
 5.73
10
5
MEC 451 –
872.48

Example 3.9
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
necessary power input to the
compressor.
Solution:
simplified energy balance:
W
in m
 h2 h1Q
out
m
 h2 h1m
qout
1
10
6
MEC 451 –
1
1
2
2
2
State1
P 100kPa
h 280.13
T 280K
State2
P 600kPa
h 400.98
T 400K
kJ
kg
kJ
kg
air
air




changes in kinetic and
potential energies are
negligible, determine the

Thus
W
in 0.02
400.98280.1316

2.74kW
10
7
MEC 451 –

Throttling Valve
Flow-restricting devices that
cause a significant pressure drop
in the fluid.
10
8
MEC 451 –
Some familiar examples are
and
ordinary adjustable
capillary tubes.
valves

Steam enters a throttling valve at
8000 kPa and 300°C and leaves
at a pressure of 1600 kPa.
Determine the final temperature
and specific volume of the
steam.
Example 3.10
10
9
MEC 451 –
P
1 8000kPa superheated

1
1
2
P 1600kPa
h2  h1
State1
h  2786.5
T  300o
C
State2
makeinterpolation
kJ
kg




P kPa T  o
C vf vg hf hg
1500 198.29 0.001154 0.131710 844.55 2791
1600 T2
vf 2 vg2 hf 2 hg2
1750 205.72 0.001166 0.113440 878.16 2795.2

T2 Tsat  201.3 C
o
At state 2, the region is sat.
mixture
Getting the quality at state 2
h2 hf 2
hg2 hf 2
2
2786.5857.994

2792.68857.994
 0.997
x 
Specific volume at state 2
v2  vf 2  x2vfg2
 0.0011588
0.9970.1244020.0011588
3
11
0
MEC 451 –
 0.1240 m
kg

The section where the mixing process
takes place.
An ordinary T-elbow or a Y-elbow in
a shower, for example, serves as the
mixing chamber for the cold- and
hot-water streams.
11
1
MEC 451 –
Mixing Chamber

Mixing Chamber
Energy Balance:
m
1h1  m
 2h2  m
 3h3
m
1h1  m
 3  m
1 h2  m
 3h3
m
1 h1  h2  m
3 h3  h2 
11
2
MEC 451 –
 h3  h2 
1 3 
 h1  h2
m
  m
 


Devices where two moving fluid
streams exchange heat without
mixing.
Heat exchangers typically involve
no work interactions (w = 0) and
negligible kinetic and potential
energy changes for each fluid
stream.
Heat Exchanger
11
3
MEC 451 –

Liquid sodium, flowing at 100
kg/s, enters a heat exchanger at
450°C and exits at 350°C. The
specific heat of sodium is 1.25
kJ/kg.oC. Water enters at 5000
kPa and 20oC. Determine the
minimum mass flux of the water
so that the water does not
completely vaporize. Neglect the
pressure drop through the
exchanger. Also, calculate the
rate of heat transfer.
simplified energy balance:
m
s h1s  h2s  m
 w h2w  h1w 
m
sCp,s T1s T2s  m
w h2w  h1w 
m
sh1s  m
 wh1w  m
sh2s  m
wh2w
1
11
4
MEC 451 –
h1w
1
P2  5000kPa
kJ
h2w
State1: water
P  5000kPa comp.liquid
 88.61
T  20 C
State2 : water
 2794.2
o kJ
kg
kg

Assume a sat. vapor
state to obtain the
max. allowable exiting
enthalpy.

the minimum mass flux of the water
completely vaporize
msCp,s T1s T2s 
 
1001.25450  350
2w 1w
2794.2 88.61
 4.62 kg
m
w
s
h  h


the rate of heat transfer
 
 4.622794.2 88.61
12.5MW
11
5
MEC 451 –
h2w  h1w
w  m
w
Q


11
6
MEC 451 –
1. Air flows through the supersonic nozzle . The inlet conditions are 7 kPa
and 420°C. The nozzle exit diameter is adjusted such that the exiting
velocity is 700 m/s. Calculate ( a ) the exit temperature, ( b )the mass flux,
and ( c ) the exit diameter. Assume an adiabatic quasiequilibrium flow.
2. Steam at 5 MPa and 400°C enters a nozzle steadily velocity of 80 m/s,
and it leaves at 2 MPa and 300°C. The inlet area of the nozzle is 50 cm2,
and heat is being lost at a rate of 120 kJ/s. Determine (a) the mass flow
rate of the steam, (b) the exit velocity of the steam, and (c) the exit area
nozzle.
3. Steam enters a turbine at 4000 kPa and 500oC and leaves as shown in Fig
A below. For an inlet velocity of 200 m/s, calculate the turbine power
output. ( a )Neglect any heat transfer and kinetic energy change ( b )Show
that the kinetic energy change is negligible.

Figure A
59
MEC 451 –
4. Consider an ordinary shower where hot water at 60°C is mixed with cold
water at 10°C. If it is desired that a steady stream of warm water at 45°C
be supplied, determine the ratio of the mass flow rates of the hot to cold
water. Assume the heat losses from the mixing chamber to be negligible
and the mixing to take place at a pressure of 150 kPa.
5. Refrigerant-134a is to be cooled by water in a condenser. The refrigerant
enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70ºC
and leaves at 35°C. The cooling water enters at 300 kPa and 15°C and
leaves at 25ºC. Neglecting any pressure drops, determine (a) the mass
flow rate of the cooling water required and (b) the heat transfer rate from
the refrigerant to water.

CHAPTER
4
Thermodynamics
First Law of
Thermodynamics
1

School of Mechanical Industrial Engineering,
1
1
MEng 2101 –
ENERGY ANALYSIS
OF CLOSED SYSTEM

First Law of Thermodynamics
School of Mechanical Industrial Engineering,
 The First Law is usually referred to as the Law of Conservation
of Energy, i.e. energy can neither be created nor destroyed, but
rather transformed from one state to another.
 The energy balance is maintained within the system being
studied/defined boundary.
 The various energies associated are then being observed as
they cross the boundaries of the system.
1
2
MEng 2101 –

School of Mechanical Industrial Engineering
Energy Balance for Closed System
Heat
Work
z
Closed
System
Reference Plane, z = 0

V
or
1
2
MEng 2101 –
Ein  Eout  Esystem

 According to classical thermodynamics
Qnet Wnet  Esystem
 The total energy of the system, Esystem, is given as
E = Internal energy + Kinetic energy + Potential energy
E = U + KE + PE
 The change in stored energy for the system is
E  U  KE  PE
 The first law of thermodynamics for closed systems then can be
written as
1
2
MEng 2101 –
Qnet Wnet  U  KE  PE

6
 If the system does not move with a velocity and has no change in
elevation, the conservation of energy equation is reduced to
Qnet Wnet  U
 The first law of thermodynamics can be in the form of
(kJ )
g(z2  z1 )
1000
2000

V2 V1
 u1
2
  m u
W net
Qnet









2000
2 2
(kJ / kg)
V 2
V 2
g(z  z )
2 1 2 1
1000
1
2  u 
 u
qnet  wnet









 For a constant volume process,



g(z  z ) 

 mu




2
V 2
 u 
W net
1000
2000
2 1
2 1
1
2
V
Qnet


MEng 2101 –


g(z  z ) 

Q  mu 
2
V 2
 u 
1000
2000
2 1
2 1
1
2
V
net

 For a constant pressure process,



g(z  z ) 

 mu



2
V 2
 u 
W net
1000
2000
2 1
2 1
1
2
V
Qnet



g(z  z ) 

Q  P(V V )  mu



2
V 2
 u 
1000
2000
1
2
2 1
1
2
net 2 1
V



g(z  z ) 

Q  mu



2
V 2
 u1  P(V2 V1) 
1000
2000
2 1
2 1
2
V
net

2000 1000
 
7
MEng 2101 –

g(z  z ) 

Q  mh  h 


2
V 2
2 1
1
2
2 1
V
net

Example of Closed Systems
Rigid tank Piston cylinder
1
2
MEng 2101 –

A closed system of mass 2 kg
undergoes an adiabatic process.
The work done on the system is
30 kJ. The velocity of the system
changes from 3 m/s to 15 m/s.
During the process, the elevation
of the system increases 45 meters.
Determine the change in internal
energy of the system.
Example 4.1
Solution:
 Energy balance,



V 2
V 2
g(z  z ) 
2 1
2000

 mu  u 



W net
1000
2 1
2 1
Qnet
 Rearrange the equation
Qnet
30 2u  2
12
6
MEng 2101 –
 9.8145
V 2
V 2
2 1
2000
2 1
1000

 mu2  u1 

V 2
V 2
2 1
2000
2 1
1000

 mu2  u1 

152
 32

g(z  z ) 
g(z  z ) 
  2

2000 1000
u  14.451 kJ Ans..
W net
Wnet
 

 



   

Steam at 1100 kPa and 92 percent
quality is heated in a rigid container
until the pressure is 2000 kPa. For a
mass of 0.05 kg, calculate the amount
of heat supply (in kJ) and the total
entropy change (in kJ/kg.K).
12
7
MEng 2101 –

 For a rigid container,
v2=v1=0.1634 m3/kg
12
8
MEng 2101 –

 Amount of heat supplied, Q
Q  mu2  u1 
 0.053030.42  2441.9
 29.43kJ
 The change in entropy,Δs
s  s2  s1
 7.2790  6.204
12
9
MEng 2101 –
kg.K
 1.075 kJ

Example 4.3
A rigid tank is divided into two equal
parts by a partition. Initially one side
of the tank contains 5 kg water at 200
kPa and 25°C, and the other side is
evacuated. The partition is then
removed, and the water expands into
the entire tank. The water is allowed to
exchange heat with its surroundings
until the temperature in the tank
returns to the initial value of 25°C.
Determine (a) the volume of the tank
(b) the final pressure (c) the heat
transfer for this process.
Solution:
 The initial volume for entire tank
 20.005
3
 0.01m
Vresevoir
13
0
MEng 2101 –

 The final pressure
 The heat transfer for this process
 +ve sign indicates heat transfer
into the system.
13
1
MEng 2101 –

13
2
MEng 2101 –
1. Two tanks are connected by a valve. Tank A contains 2 kg of carbon
monoxide gas at 77°C and 0.7 bar. Tank B holds 8 kg of the same gas
at 27°C and 1.2 bar. Then the valve is opened and the gases are allowed
to mix while receiving energy via heat transfer from the surrounding.
The final equilibrium temperature is found to be 42°C. Determine (a)
the final pressure (b) the amount of heat transfer. Also state your
assumption. [P2=105 kPa, Q = +37.25 kJ]
2. A piston cylinder device contains 0.2 kg of water initially at 800 kPa
and 0.06 m3. Now 200 kJ of heat is transferred to the water while its
pressure is held constant. Determine the final temperature of the water.
Also, show the process on a T-V diagram with respect to saturation
lines. [ 721.1oC]

[50°C, 400 kPa] 13
3
MEng 2101 –
3. A piston-cylinder device contains 6 kg of refrigerant-134a at 800 kPa
and 50oC. The refrigerant is now cooled at constant pressure until it
exist as a liquid at 24oC. Show the process on T-v diagram and
determine the heat loss from the system. State any assumption made.
[1210.26 kJ]
4. A 0.5 m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40
percent quality. Heat is now transferred to the refrigerant until the
pressure reaches 800 kPa. Determine (a) the mass of the refrigerant in
the tank and (b) the amount of heat transferred. Also, show the process
on a P-v diagram with respect to saturation lines.
[12.3 kg, 2956.2 kJ]
5. An insulated tank is divided into two parts by a partition. One part of
the tank contains 6 kg of an ideal gas at 50°C and 800 kPa while the
other part is evacuated. The partition is now removed, and the gas
expands to fill the entire tank. Determine the final temperature and the
pressure in the tank.

 Some thermodynamic cycle composes of processes in which
the working fluid undergoes a series of state changes such
that the final and initial states are identical.
 For such system the change in internal energy of the
working fluid is zero.
 The first law for a closed system operating in a
thermodynamic cycle becomes
Qnet Wnet  Ucycle
Qnet  Wnet
13
4
MEng 2101 –
Closed System First Law of a Cycle

 constant
13
5
MEng 2101 –
No Value of n Process Description Result of IGL
1 ∞ isochoric constant volume (V1 = V2) P1

P2
T1 T2
2 0 isobaric constant pressure (P1 = P2) V1

V2
T1 T2
3 1 isothermal constant temperature
(T1 = T2)
P1V1  P2V2
4 1<n< γ polytropic -none-
n n
P1

V2 

 T1 n1
P 

V 
 

T 

2  1   2 
5 γ isentropic constant entropy (S1 = S2)

 V
arious forms of work are expressed as follows
13
6
MEng 2101 –
Process Boundary Work
isochoric W12  P(V2 V1)  0
isobaric W12  P(V2 V1 )
isothermal
W  PV ln
V2
12 1
1
V 1
polytropic W 
P2V2  P1V1
12
1 n
isentropic

Example 4.4
Sketch a P-V diagram showing the following processes in a cycle
13
7
MEng 2101 –
Process 1-2: isobaric work output of 10.5 kJ from an initial volume of 0.028
m3 and pressure 1.4 bar,
isothermal compression, and
isochoric heat transfer to its original volume of 0.028 m3 and
pressure 1.4 bar.
Process 2-3:
Process 3-1:
Calculate (a) the maximum volume in the cycle, in m3, (b) the isothermal work,
in kJ, (c) the net work, in kJ, and (d) the heat transfer during isobaric expansion,
in kJ.

Solution:
 Process by process analysis,
 The isothermal work
 The net work
13
8
MEng 2101 –

Example 4.5
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.
13
9
MEng 2101 –

Solution:
 Process by process analysis,
 The net work of the cycle
 W12 W23 W34 W41
 0.9076 kJ
14
0
MEng 2101 –
Wnet

14
1
MEng 2101 –
1. A mass of 0.15 kg of air is initially exists at 2 MPa and 350oC. The air is
first expanded isothermally to 500 kPa, then compressed polytropically
with a polytropic exponent of 1.2 to the initial state. Determine the
boundary work for each process and the net work of the cycle.
2. 0.078 kg of a carbon monoxide initially exists at 130 kPa and 120oC. The
gas is then expanded polytropically to a state of 100 kPa and 100oC.
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. Two kg of air experiences the three-
process cycle shown in Fig. 3-14.
Calculate the net work.
4. A system contains 0.15 m3 of air pressure of 3.8 bars and 150⁰ 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
14
2
MEng 2101 –

14
3
MEng 2101 –
MASS & ENERGY ANALYSIS
OF CONTROL VOLUME

• Reminder of an open System.
–Open system = Control volume
–It is a properly selected region in space.
–Mass and energy can cross its boundary.
First low of thermodynamics for open
Systems
14
4
MEng 2101 –

Control volume involves two main
processes
• Steady flow processes.
– Fluid flows through the control volume
steadily.
– Its properties are experiencing no change with
time at a fixed position.
• Unsteady flow processes.
– Fluid properties are changing with time.
14
5
MEng 2101 –

Mass and Volume Flow Rates
 Mass flow through a cross-sectional area per unit time is called the
mass flow rate. Note the dot over the mass symbol indicates a time
rate of change. It is expressed as
m
  V .dA
 If the fluid density and velocity are constant over the flow cross-
sectional area, the mass flow rate is
where 
1

 iscalled specificvoulme
m
  AV 
AV

14
6
MEng 2101 –

(kg/s)
m
in  m
out
Principal of Conservation of Mass
 The conservation of mass principle for a control volume can be
expressed as
m
in  m
out  m
CV
 For a steady state, steady flow process the conservation of mass
principle becomes
14
7
MEng 2101 –

 As the fluid upstream pushes mass across the control volume, work
done on that unit of mass is
flow
W
w  flow
 Pv
flow
W  FdL  FdL
A
 PdV  Pvm
A
m
Flow Work & The Energy of a Flowing Fluid
14
8
MEng 2101 –

energy  u  P 
V
 gz  h 
V
 gz
2 2
2 2
 The first law for a control volume can be written as
 in
 out 






 
 gz
h 








 gz
h 

. .
Qnet  W net
in
in
in
in
out
out
out
out
V
m
V
m
2
2
2
.
2
.
 The total energy carried by a unit of mass as it crosses the control
surface is the sum of the internal energy + flow work + potential
energy + kinetic energy
14
9
MEng 2101 –

 The steady state, steady flow conservation of mass and first law of
thermodynamics can be expressed in the following forms
(kW )
g(z2  z1)
1000
2000
2 2
V V1
2
2 1
.
.
W net
.
Qnet  mh  h  








(kJ )
1000
g(z  z )
2000
1
2
2 2

V2 V1
 m h  h1
2
W net
Qnet










(kJ / kg)
1000
V 2
V 2
g(z  z )
2 1
2000
1
2
2 1
 h  h 
qnet  wnet









15
0
MEng 2101 –

Steady-flow Engineering Devices
Only one in and one out More than one inlet and
exit
15
1
MEng 2101 –

Nozzle & Diffuser
 Nozzle - device that increases
the velocity fluid at the expense
of pressure.
 Diffuser - device that increases
pressure of a fluid by slowing it
down.
 Commonly utilized in jet
engines, rockets, space-craft
and even garden hoses.
 Q = 0 (heat transfer from the
fluid to surroundings
small)
 W = 0 andΔPE = 0
very
15
2
MEng 2101 –

 gz  zi 





QW  m

he

 V Vi
 hi 
 
e
e
2
2 2
Is there work in this system?NO
Is there heat transfer?
In fact, it depends on the problem!
Does the fluid change elevation?NO
0  he  hi 
2
 
2 2
Vi
e
V Q :What happened to the m
 ?
Ans: It is divided out
let us say: NO
Nozzles and Diffusers
(1st low analysis)

  
36
MEng 2101 –

37
 Energy balance (nozzle & diffuser):
 out out
out
 in in
in








gz
h 
. .
Q W 








gz

. .
Q W  out
out
out
out
in
in
in
in
V
m
V
m h
2
2
2
.
2
.



2

out  out
h



  m


2

in  in
m h 


2
2
.
.
out
in V
V



2

 
   h 


2 
V 2


 h 


2
2
2
1
1
V
MEng 2101 –

38
2 2
V 2
V 2
e
e
i
i  h 
h 
How can you find the mass flow rate in a nozzle?
m
  1V1 A1  2V2 A2
In a nozzle, enthalpy is converted into kinetic energy
v2
V A V A
m
  1 1
 2 2
v1
which can be rearranged to
MEng 2101 –

Example 5.1
Steam at 0.4 MPa, 300ºC,
enters an adiabatic nozzle with
a low velocity and leaves at 0.2
MPa with a quality of 90%.
Find the exit velocity.
Solution:
 Simplified energy balance:
 Exit velocity:
V2  2000 3067.1 2486.1
 1078 m / s
15
6
MEng 2101 –

Example 5.2
Air at 10°C and 80 kPa enters the
diffuser of a jet engine steadily
with a velocity of 200 m/s. The
inlet area of the diffuser is 0.4 m2.
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 and (b) the temperature
of the air leaving the diffuser.
Solution:
2

0
V 2 
2
2
1
1

 
 2
V
h 

 h 
2






 Simplified energy balance:
 From Ideal Gas Law:
3
1.015 m
v 
RT1
1
1
kg
P
15
7
MEng 2101 –

 Mass flow rate
 Enthalpy at state 1
h1  CpT1 1.005283
 284.42 kJ
kg
 From energy balance:
15
8
MEng 2101 –

 Turbine – a work producing device through the expansion of a
fluid.
 Compressor (as well as pump and fan) - device used to increase
pressure of a fluid and involves work input.
 Q = 0 (well insulated),ΔPE = 0,ΔKE = 0 (very small compare to
Δenthalpy).
42
Turbine & Compressor
MEng 2101 –

43
Turbines
A turbine is a device that produces work at the
expense of temperature and pressure.
As the fluid passes through the turbine, work is done
against the blades, which are attached to a shaft. As a
result, the shaft rotates, and the turbine produces work.
MEng 2101 –

44
Compressors
A compressor is a device that increases the
pressure of a fluid by adding work to the system.
Work is supplied from an external source through a
rotating shaft.
Compressor
Inlet
Exit
Win
MEng 2101 –

45
  

2 2
 


 
Ve Vi
  
Q W m

he hi
2
g ze zi


 
Turbines and Compressors
Is there work in this system? Y
es!
Is there heat transfer? Negligible because of insulation. Exception: Internal
cooling in some compressors.
Does the fluid change elevation? NO
Does the kinetic energy change? Usually it can be ignored
 
W mh h (W)
e i
wh h (kJ/kg)
e i
MEng 2101 –

Example 5.3
The power output of an adiabatic steam turbine is 5 MW
. Compare
the magnitudes of Δh, Δke, and Δpe. Then determine the work done
per unit mass of the steam flowing through the turbine and calculate
the mass flow rate of the steam.
Data : Inlet (P = 2 MPa, T = 400ºC,v = 50 m/s, z = 10 m)
Exit (P = 15 kPa, x = 90%, v = 180 m/s, z = 6m)
16
3
MEng 2101 –

 From energy balance:
. 2
Qin
. .
 Win
2
. . .
2
2
in
min hin  gzin 

in
out
Qout Wout mout hout  gzout


out
V
V
 

 

 Solve the equation:
Solution:
16
4
MEng 2101 –

5.74
870.96
out
W 872.48
 the work done per unit mass
 The mass flow rate

5000
 5.73 kg
s
W

m
  out
16
5
MEng 2101 –

Example 5.4
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
process.
changes
occurs during the
the
and
are
the
Assuming
in kinetic
energies
determine
potential
negligible,
necessary power input to the
compressor.
Solution:
 simplified energy balance:
W
in m
 h2 h1Q
out
m
 h2 h1m
qout
 Thus
W
in 0.02
400.98280.1316

2.74kW 16
6
MEng 2101 –

Throttling Valve
Flow-restricting devices that
cause a significant pressure drop
in the fluid.
Some familiar examples are
and
ordinary adjustable
capillary tubes.
valves
16
7
MEng 2101 –

51
  

2 2
 


 
Ve Vi
  
Q W m

he hi
2
g ze zi


 
Throttling Valve
Is there work in this system?
Is there heat transfer?
NO
Usually it can be ignored
Does the fluid change elevation? NO
Does the fluid change velocity? Usually it can be ignored
0he hi   he hi
isenthalpicdevice
MEng 2101 –

Steam enters a throttling valve at
8000 kPa and 300°C and leaves
at a pressure of 1600 kPa.
Determine the final temperature
and specific volume of the
steam.
Example 5.5
P
1 8000kPa superheated
1 kg
1
2
P 1600kPa
h2  h1
State1
h  2786.5kJ
T  300o
C
State2
makeinterpolation




16
9
MEng 2101 –
PkPa T o
C vf vg hf hg
1500 198.29 0.001154 0.131710 844.55 2791
1600 T2
vf 2 vg2 hf 2 hg 2
1750 205.72 0.001166 0.113440 878.16 2795.2

T2 Tsat  201.3 C
o
 At state 2, the region is sat.
mixture
 Getting the quality at state 2
h h
x  2 f 2
2
hg2 hf 2

2786.5857.994
2792.68857.994
 0.997
 Specific volume at state 2
v2  vf 2  x2vfg2
 0.0011588
0.9970.1244020.0011588
3
 0.1240 m
kg
17
0
MEng 2101 –

The section where the mixing process
takes place.
An ordinary T-elbow or a Y
-elbow in
a shower, for example, serves as the
mixing chamber for the cold- and
hot-water streams.
Mixing Chamber
54
MEng 2101 –

Mixing Chamber
 
  

    V 2
   V 2

Qnet Wnet  m  h   gz  m  h  gzi 
e i
e e e i i
 2   2 
   
We no longer have only one inlet and one exit stream
Is there any work done? No
Is there any heat transferred? No
Is there a velocity change? No
Is there an elevation change? No
 
0  me
he  mi hi 
17
2
MEng 2101 –

Mixing Chamber
• Material Balance
 
mi
 me
  
m1
 m2
 m3
• Energy balance
 
mi
hi  me he
  
m1 h1
 m2 h2
 m3 h3
17
3
MEng 2101 –

Devices where two moving fluid
streams exchange heat without
mixing.
Heat exchangers typically involve
no work interactions (w = 0) and
negligible kinetic and potential
energy changes for each fluid
stream.
Heat Exchanger
17
4
MEng 2101 –

 simplified energy balance:
Liquid sodium, flowing at 100
kg/s, enters a heat exchanger at
450°C and exits at 350°C. The
specific heat of sodium is 1.25
kJ/kg.oC. Water enters at 5000
kPa and 20oC. Determine the
minimum mass flux of the water
completely vaporize. Neglect the
pressure
exchanger.
drop
Also,
through the
calculate the
rate of heat transfer.
17
5
MEng 2101 –

 the minimum mass flux of the water
completely vaporize
 the rate of heat transfer
17
6
MEng 2101 –

1. Air flows through the supersonic nozzle . The inlet conditions are 7 kPa
and 420°C. The nozzle exit diameter is adjusted such that the exiting
velocity is 700 m/s. Calculate ( a ) the exit temperature, ( b )the mass flux,
and ( c ) the exit diameter.Assume an adiabatic quasiequilibrium flow.
2. Steam at 5 MPa and 400°C enters a nozzle steadily velocity of 80 m/s,
and it leaves at 2 MPa and 300°C. The inlet area of the nozzle is 50 cm2,
and heat is being lost at a rate of 120 kJ/s. Determine (a) the mass flow
rate of the steam, (b) the exit velocity of the steam, and (c) the exit area
nozzle.
3. Steam enters a turbine at 4000 kPa and 500oC and leaves as shown in Fig
A below. For an inlet velocity of 200 m/s, calculate the turbine power
output. ( a )Neglect any heat transfer and kinetic energy change ( b )Show
that the kinetic energy change is negligible.
17
7
MEng 2101 –

FigureA
4. Consider an ordinary shower where hot water at 60°C is mixed with cold
water at 10°C. If it is desired that a steady stream of warm water at 45°C
be supplied, determine the ratio of the mass flow rates of the hot to cold
water. Assume the heat losses from the mixing chamber to be negligible
and the mixing to take place at a pressure of 150 kPa.
5. Refrigerant-134a is to be cooled by water in a condenser. The refrigerant
enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70ºC
and leaves at 35°C. The cooling water enters at 300 kPa and 15°C and
leaves at 25ºC. Neglecting any pressure drops, determine (a) the mass
flow rate of the cooling water required and (b) the heat transfer rate from
the refrigerant to water.
61
MEng 2101 –

CHAPTER
4
Thermodynamics
Second Law of
Thermodynamics
1

Introduction
2
MEC 451 –
A process must satisfy the first law in order to occur.
Satisfying the first law alone does not ensure that the process will take
place.
Second law is useful:
provide means for predicting the direction of processes,
establishing conditions for equilibrium,
determining the best theoretical performance of cycles, engines
and other devices.

A cup of hot coffee does
not get hotter in a cooler
room.
Transferring heat to a wire
will not generate electricity.
Transferring
heat to a paddle
wheel will not
cause it to
rotate.
3
MEC 451 –
These processes cannot occur
even though they are not in
violation of the first law.

Second Law of Thermodynamics
Kelvin-Planck statement
As for a power plant to
operate, the working fluid
must exchange heat with the
environment as well as the
furnace.
4
MEC 451 –
No heat engine can have a
thermal efficiency 100
percent.

Heat Engines
5
MEC 451 –
Work can easily be converted to other forms of
energy, but?
Heat engine differ considerably from one another,
but all can be characterized :
o they receive heat from a high-temperature
source
they convert part of this heat to work
o
o they reject the remaining waste heat to a low-
temperature sink atmosphere
they operate on a cycle
o

The work-producing
device that best fit into
the definition of a heat
engine is the steam
power plant, which is
an external combustion
engine.
6
MEC 451 –

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
th
the cycle operate.
MEC 451 –
7

Required Input
Desired Result
For a heat engine the desired result is the net work
done and the input is the heat supplied to make

The thermal efficiency is always less than 1 or less than
100 percent.
th
8
MEC 451 –
Wnet, out
Qin

Wnet, out
Qin
 Wout Win
 Qnet
where

Applying the first law to the cyclic heat engine
Qnet, in Wnet, out
Wnet, out
Wnet, out
 U
 Qnet, in
 Qin  Qout
The cycle thermal efficiency may be written as
th
Qin
MEC 451 –
9
Wnet, out
Qin
Q  Q
 in out
Qin
 1
Qout


A thermodynamic temperature scale related to the heat
transfers between a reversible device and the high and low-
temperature reservoirs by
QL

TL
QH TH
The heat engine that operates on the reversible Carnot
cycle is called the Carnot Heat Engine in which its
efficiency is
th, rev
10
MEC 451 –
TL
TH
 1

Heat Pumps and Refrigerators
11
MEC 451 –
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
the
refrigerators are expressed in terms of
coefficient of performance.

12
MEC 451 –

COP 
QH
W
QH
QH  QL
HP
net, in
 COP 
QL
W
R
net,in
13
MEC 451 –

Carnot Cycle
Proces
s
Description
1-2 Reversible isothermal heat addition at
high temperature
Reversible adiabatic expansion from high
temperature to low temperature
Reversible isothermal heat rejection at
low temperature
Reversible adiabatic compression from low
temperature to high temperature
2-3
3-4
4-1
14
MEC 451 –

Execution of Carnot cycle in a piston cylinder device
15
MEC 451 –

16
MEC 451 –

The thermal efficiencies of actual and reversible heat
engines operating between the same temperature limits
compare as follows
The coefficients of performance of actual and reversible
refrigerators operating between the same temperature limits
compare as follows
17
MEC 451 –

Example 4.1
A steam power plant
produces 50 MW of net
work while burning fuel
to produce 150 MW of
heat energy at the high
temperature. Determine
the cycle thermal
efficiency and the heat
rejected by the cycle to
the surroundings.
Solution:
Wnet , out
 
18
MEC 451 –
th
QH
50 MW
150 MW
  0.333 or 33.3%
Wnet , out
QL
 QH  QL
 QH Wnet , out
 150 MW 50 MW
 100 MW

QL
WOUT
A Carnot heat engine receives 500 kJ of heat per cycle from a high-
temperature heat reservoir at 652ºC and rejects heat to a low-
temperature heat reservoir at 30ºC. Determine :
(a) The thermal efficiency of this Carnot engine
(b) The amount of heat rejected to the low-
temperature heat reservoir
Solution:
Example 4.2
TL = 30oC
HE
th, rev
TL
H
 1
T
(30  273)K
(652  273)K
 0.672 or 67.2%
 1
QL

TL
19
MEC 451 –
Q T
(30  273)K
(652  273)K
Q  500 kJ(0.328)
 164 kJ
H H
L
  0.328
TH = 652oC
QH

An inventor claims to have developed a refrigerator that maintains the
refrigerated space at 2ºC while operating in a room where the
temperature is 25ºC and has a COP of 13.5. Is there any truth to his
claim?
Example 4.3
Solution:
Win
QH
TH = 25oC
R
COP 
20
MEC 451 –
QL
QH  QL
TL
TH  TL
(2  273)K
(25  2)K
R 

 11.96
- this claim is also false!
QL
TL = 2oC

Supplementary Problem 4.1
21
MEC 451 –
1. A 600 MW steam power plant, which is cooled by a river, has a thermal
efficiency of 40 percent. Determine the rate of heat transfer to the river
water. Will the actual heat transfer rate be higher or lower than this
value? Why?
[900
M
AW]
steam power plant receives heat from a furnace at a rate of 280
GJ/h. Heat losses to the surrounding air from the steam as it passes
through the pipes and other components are estimated to be about 8
GJ/h. If the waste heat is transferred to the cooling water at a rate of
145 GJ/h, determine (a) net power output and (b) the thermal
efficiency of this power plant.
[ 35.3 MW,
2.
45.4% ]
3. An air conditioner removes heat steadily from a house at a rate of 750
kJ/min while drawing electric power at a rate of 6 kW. Determine (a) the
COP of this air conditioner and (b) the rate of heat transfer to the outside
air.
[ 2.08, 1110 kJ/min
]

22
4. Determine the COP of a heat pump that supplies energy to a house at a
rate of 8000 kJ/h for each kW of electric power it draws. Also,
determine the rate of energy absorption from the outdoor air.
kW ]
MEC 451 –
[ 2.22, 4400
kJ/h ]
5. An inventor claims to have developed a heat engine that receives 700 kJ
of heat from a source at 500 K and produces 300 kJ of net work while
rejecting the waste heat to a sink at 290 K. Is this reasonable claim?
6. An air-conditioning system operating on the reversed Carnot cycle is
required to transfer heat from a house at a rate of 750 kJ/min to
maintain its temperature at 24oC. If the outdoor air temperature is
35oC, determine the power required to operate this air-conditioning
system.
[ 0.463
kW ]
7. A heat pump is used to heat a house and maintain it at 24oC. On a
winter day when the outdoor air temperature is -5oC, the house is
estimated to lose heat at a rate of 80,000 kJ/h. Determine the
minimum power required to operate this heat pump.
[ 2.18

Entropy
23
MEC 451 –
The 2nd law states that process occur in a certain
direction, not in any direction.
It often leads to the definition of a new property called
entropy, which is a quantitative measure of disorder for
a system.
Entropy can also be explained as a measure of the
unavailability of heat to perform work in a cycle.
This relates to the 2nd law since the 2nd law predicts
that not all heat provided to a cycle can be
transformed into an equal amount of work, some heat
rejection must take place.

Entropy Change
The entropy change during a reversible process is defined
as
24
MEC 451 –
For a reversible, adiabatic process
dS  0
S2  S1
The reversible, adiabatic proces
s
is called an isentropic
process
.

Entropy Change and Isentropic Processes
25
MEC 451 –
The entropy-change and isentropic relations for a process
can be summarized as follows:
i. Pure substances:
Any process: Δs = s2 –
s1
(kJ/kgK)
Isentropic process: s2 = s1
ii. Incompressible substances (liquids and solids):
Any process: s2 – s1 = cav T2/T1
Isentropic process: T2 = T1
(kJ/kg

s2  s1  Cv,av
T v
1 1
ln
T2
 Rln
P2
2 1 p,av
T1 P
1
s  s  C
for isentropic process
iii. Ideal gases:
a) constant specific heats (approximate treatment):
for all process
26
MEC 451 –
ln
T2
 R ln
v2
 P2  1
 P
1 s  const.  v2 
k
 v 

   

Example 4.5
Steam at 1 MPa, 600oC, expands in a turbine to 0.01 MPa. If the
process is isentropic, find the final temperature, the final enthalpy of the
steam, and the turbine work.
Solution:
 m
 h1  h2 
massbalance :m
1  m
2  m

energybalance
E
in  E
out
m
1h1  m
2h2 W
out
W
out
1
27
MEC 451 –
1
1
1 kg.K
State1
superheated
kJ
P 1MPa
h  3698.6
T  600 C
s  8.0311
kg
kJ
o



h2 191.8 0.9842392.1
 Tsat@ P
2
P2
s2  8.0311kg.K 
 x2
T2
State2
 0.01MPa 
sat.mixture
 0.984
 2545.6 kJ
 45.81 C
kJ 
kg
o
❖ Since that the process is
isentropic, s2=s1
❖ Work of turbine
 h1  h2
 3698.6  2545.6
28
MEC 451 –
1153 kJ
Wout
kg

Isentropic Efficiency for Turbine
29
MEC 451 –

Isentropic Efficiency for Compressor
30
MEC 451 –

Example 4.6
Steam at 1 MPa, 600°C,
expands in a turbine to 0.01
MPa. The isentropic work
of the turbine is 1152.2
kJ/kg. If the isentropic
efficiency of the turbine is
90 percent, calculate the
Find the
actual work.
the steam.
Solution:
 0.91153
31
MEC 451 –
w h  h
 a
 1 2a
isen,T
h1  h2s
 w
ws
isen,T s
1037.7 kJ
a
kg
w 

❖ Theoretically:
actual turbine exit
temperature or quality of

1
1
1  1
kJ
kg.K
P2  0.01MPa
x2s
s2s  s1  8.0311 kg.K 
h2s
State1
h  3698.6
P 1MPa
T  600 C s  8.0311
State 2s
sat.mixture
 0.984
 2545.6
kJ
kg
o
kJ
kJ
kg





❖ Obtain h2a from Wa
wa  h1  h2a
h2a  h1  wa
 2660.9 kJ
kg
2
32
MEC 451 –
2a kg 
 2a
State2a
P  0.01MPa  superheated
h  2660.9 kJ
T  86.85o
C



Example 4.7
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 air for a compressor
isentropic efficiency of 80
percent.
Solution:
❖ From energy balance
 m
 h2s  h1 
W
c,s
W

W  c,s
c,s  h2s  h1
m

❖ For isentropic process of IGL
k1
T  27  273 0.5 
2s 2
   
 T1   P
1 
0.4/1.4
2s
 0.1 
 475.4 K
W 1.005475.4 300
T2s T1

 CP 33
MEC 451 –
k
 T   P 
 
❖ Then
c,s
W
W  c,s
c,a
isen,c
176 kJ
 220
kg
kJ
kg


Supplementary Problems 4.2
34
MEC 451 –
1. The radiator of a steam heating system has a volume of 20 L and is
filled with the superheated water vapor at 200 kPa and 150oC. At
this moment both inlet and exit valves to the radiator are closed.
After a while the temperature of the steam drops to 40oC as a result of
heat transfer to the room air. Determine the entropy change of the
steam during this process.
[ -0.132 kJ/.K ]
2. A heavily insulated piston-cylinder device contains 0.05 m3 of steam at
300 kPa and 150oC. Steam is now compressed in a reversible
manner to a pressure of 1 MPa. Determine the work done on the
steam during this process.
[ 16 kJ ]
3. A piston –cylinder device contains 1.2 kg of nitrogen gas at 120 kPa
and 27oC. 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.
[ -0.0617 kJ/kg.K ]

4. Steam enters an adiabatic turbine at 8 MPa and 500oC with a
mass flow rate of 3 kg/s and leaves at 30 kPa. The isentropic
efficiency of the turbine is 0.90. Neglecting the kinetic energy of the
steam, determine (a) the temperature at the turbine exit and
35
MEC 451 –
(b) the power output of the turbine.
[ 69.09oC,3054 kW ]
5. Refrigerant-R134a enters an adiabatic compressor as saturated
vapor at 120 kPa at a rate of 0.3 m3/min and exits at 1 MPa
pressure. If the isentropic efficiency of the compressor is 80
percent, determine (a) the temperature of the refrigerant at the exit
of the compressor and (b) the power input, in kW. Also, show the
process on a T-s diagram with respect to the saturation lines.
[ 58.9oC,1.70 kW ]

CHAPTER
5
Thermodynamics
Air Standard
Cycle

th
Wnet
Qin

th, Carnot
TL
TH
of the real
 1
Upon derivation the performance cycle is often
measured in terms of its thermal efficiency
Review – Carnot Cycle
The Carnot cycle was introduced as the most efficient heat
engine that operate between two fixed temperatures TH and TL.
The thermal efficiency of Carnot cycle is given by
2

The ideal gas equation is defined as
Pv  RT or PV  mRT
where P = pressure in kPa
v = specific volume in m3/kg (or V = volume in
m3)
R = ideal gas constant in kJ/kg.K
m = mass in kg
T = temperature in K
Review – Ideal Gas Law
3 3

The Δu and Δh of ideal gases can be expressed as
21
7
u  u2  u1  Cv (T2 T1 )
h  h2  h1  CP (T2 T1)
Δu - constant volume process
Δh - constant pressure process

Process Description Result of IGL
isochoric constant volume (V1 = V2)
isobaric
constant pressure (P1 =
P )
2
isothermal
constant temperature
(T = T )
1 2
polytropic -none-
isentropic constant entropy (S1 = S2)
 constant
T1 T2
V1 V2
P1 P2

1 2

T T
 P2V2
P1V1
 T n1
T2 
1
 V1 
2
1
P2



  


 

P V
n
n
Review – Thermodynamics Processes
21
8

R = 0.2871 kJ/kg.K
Cp = 1.005 kJ/kg.K
Cv = 0.718 kJ/kg.K
k = 1.4
where R = ideal gas constant
Cp = specific heat at constant pressure
Cv = specific heat at constant volume k
= specific heat ratio
Review – Properties of Air
21
9

IC Engine – combustion of fuel takes place inside an engine’s
cylinder.
Introduction
22
0

Air continuously circulates in a closed loop.
Always behaves as an ideal gas.
All the processes that make up the cycle are internally
reversible.
The combustion process is replaced by a heat-addition
process from an external source.
Air-Standard Assumptions
22
1

A heat rejection process that restores the working fluid to its
initial state replaces the exhaust process.
The cold-air-standard assumptions apply when the
working fluid is air and has constant specific heat
evaluated at room temperature (25o
C or 77o
F).
No chemical reaction takes place in the engine.
Air-Standard Assumptions
22
2

Top dead center (TDC), bottom dead center (BDC), stroke,
bore, intake valve, exhaust valve, clearance volume,
displacement volume, compression ratio, and mean
effective pressure
Terminology for Reciprocating Devices
22
3

The compression ratio r of an
engine is defined as
r 
V max

VBDC
V min VTDC
The mean effective pressure
(MEP) is a fictitious pressure
that, if it operated on the piston
during the entire power stroke,
would
amount
produced
cycle.
produce the
work
a
s
sam
e
that
of net
during the actual
MEP 
Wnet
V V
wnet
v  v

max min max min11

Otto Cycle
The Ideal Cycle for Spark-Ignition Engines
22
5

22
6
The processes in the Otto cycle are as per following:
Process Description
1-2
2-3
3-4
4-1
Isentropic compression
Constant volume heat addition
Isentropic expansion
Constant volume heat rejection

Related formula based on basic thermodynamics:
 T n1
22
7
V1  T2 
1
2
1
P2

  
 


 

P V
n
n
 T n1
V1  T2 
1
2
1
P2

  
 


 

P V
n
n
Qin  mCv T3 T2 
 mCv T4 T1 
Qout

Thermal efficiency of the Otto cycle:
th
Wnet Qnet Qin  Qout
Qin Qin Qin Qin
Qout
    1
Apply first law closed system to process 2-3, V = constant.
3
Wnet,23 Wother,23 Wb,23  0  PdV  0
2
Thus, for constant specific heats
Qnet , 23  U23
Qnet , 23  Qin  mCv (T3  T2 )
Qnet,23 Wnet,23  U23
22
8

Apply first law closed system to process 4-1, V = constant.
Qnet,41 Wnet,41  U41
1
4
Thus, for constant specific heats,
Qnet, 41  U41
Qnet, 41  Qout  mCv (T1  T4 )
Qout  mCv (T1  T4 )  mCv (T4  T1)
The thermal efficiency becomes
th, Otto
Qout
Qin
 1
mCv (T4  T1)
mCv (T3  T2 )
 1
16

th, Otto
(T4  T1)
 1
(T  T )
T (T / T 1)
 1 1 4 1
T2 (T3 / T2 1)
Recall processes 1-2 and 3-4 are isentropic, so
3 2
Since V3 = V2 and V4 = V1,
T2

T3
T1 T4
T4

T3
T1 T2
or
k1 k1
23
0
T3
T2 1 4
T1 V2  T4 V3 
 V  V 
and 
  
 

The Otto cycle efficiency becomes
th, Otto
T1
 1
T2
Since process 1-2 is isentropic,
where the compression ratio is
r = V1/V2 and
th, Otto  1
rk1
1
k1
23
1
T2 1
T1
T1
V2 
k1 k1
2
T2  V1 
  1 
 V 
V 
 r 
  
    

An Otto cycle having a compression ratio of 9:1 uses air as the
working fluid. Initially P1 = 95 kPa, T1 = 17°C, and V1 = 3.8
liters. During the heat addition process, 7.5 kJ of heat are
added. Determine all T's, P's, th, the back work ratio and the
mean effective pressure.
Example 5.1
Solution:
Data given:
T1  290K
V1
V2
Q23  7.5kJ
P
1  95kPa
V1  3.8Litres
 9
23
2

Example 5.1
 
Process1 2isentropiccompression
 P2  959
Process 2 3Const.volumeheat addition
k1
0.4
T2 1
2
T1
P2
V2 
k1
1.4
 V1 
P
1 V2 
T  290 9  698.4K
 2059kPa
1st
law:Q  W  U
net net
 V 
  
  
Q23  mCv T3 T2 
  3
0
P
1v1  RT1  v1 
Q
 23
v1
23
23
V1
0.2871 290
IGL:  0.875
95
1727 kJ
m
kg
kg
q  Q
m 20

Example 5.1
 
q23  Cv T3 T2 
 0.718T3  698.4
T3  3103.7K
Pr ocess3 4isentropicexp ansion
 P4  P3 1/ 9
ButV3 V2
P3

P2
T3 T2
P3  9.15MPa
k1
0.4
3
T4
4 3
T3
P4
V4 
1.4
V3 
P3 V4 
Back to IGL :
 T  T 1/ 9 1288.8K
 422kPa
k
V 
  
  
21

Example 5.1
Pr ocess 4 1Const.volumeheat rejection
Q41  mCv T4 T1 
q41  Cv T4 T1 
 0.7181288.8 290
 717.1kJ
kg
Then:
Wnet  qin  qout
 q23  q41
 0.58558.5%
th,Otto
1009.6 kJ
kg
Wnet
qin
 
22

Example 5.1
What else?
 
v1 1v2 / v1 
0.87511/ 9
Vmax Vmin vmax  vmin
wnet
v1  v2
wnet
1
r
1
u12
wexpans u34
1009.6
1298kPa
v 1
Wnet wnet
wnet
Cv T2 T1 

wcompr
 
bw
MEP 
r

 
 
Cv T3 T4 
23
6
 0.22522.5%

Supplementary Problems 5.1
23
7
1. An ideal Otto cycle has a compression ratio of 8. At the beginning of
the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg
of heat is transferred to air during the constant-volume heat-
addition process. Taking into account the variation of specific heats
with temperature, determine (a) the pressure and temperature at
the end of the heat addition process, (b) the net work output, (c) the
thermal efficiency, and (d) the mean effective pressure for the cycle.
[(a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent,(d ) 495 kPa]
2. The compression ratio of an air-standard Otto cycle is 9.5. Prior to
the isentropic compression process, the air is at 100 kPa, 35°C, and
600 cm3. The temperature at the end of the isentropic expansion
process is 800 K. Using specific heat values at room temperature,
determine (a) the highest temperature and pressure in the cycle; (b)
the amount of heat transferred in, in kJ; (c) the thermal efficiency;
and (d) the mean effective pressure.
[(a) 1969 K, 6072 kPa,(b) 0.59 kJ, (c) 59.4 percent, (d) 652 kPa]

The processes in the Diesel cycle are as per following:
Diesel Cycle
23
8

c
v  Cut  off ratio,r
v
v3
 Compression ratio,r and
v1
v 2
2
Diesel Cycle
23
9

Related formula based on basic thermodynamics:
 T n1
24
0
V1  T2 
1
2
1
P2

  
 

P V 
 

n
n
 T n1
V1  T2 
1
2
1
P2

  
 


 

P V
n
n
Qin  mCP T3 T2 
 mCv T4 T1 
Qout

Thermal efficiency of the Diesel cycle
th, Diesel
Wnet Qout
Qin Qin
  1
Apply the first law closed system to process 2-3, P = constant.
Qnet,23 Wnet,23  U23
3
Wnet,23  Wother,23 Wb,23  0  PdV  0
2
 P2 V3 V2 
Qnet, 23  U23  P2 (V3 V2 )
Qnet, 23  Qin  mCv (T3  T2 )  mR(T3  T2 )
Qin  mCp (T3  T2 )
28

Apply the first law closed system to process 4-1, V = constant
Qnet,41 Wnet,41  U41
1
4
Qnet, 41  U41
Qnet, 41  Qout  mCv (T1  T4 )
Qout  mCv (T1  T4 )  mCv (T4  T1)
The thermal efficiency becomes
th, Diesel
Qout
Qin
 1
mCv (T4  T1)
mCp (T3  T2 )
 1
29

P4V4

P
1V1
T4 T1
where V4 V1
T4

P4
T1 P
1
Recall processes 1-2 and 3-4 are isentropic, so
PV k
 PV k
and PV k
 PV k
1 1 2 2 4 4 3 3
Therefore,
Since V4 = V1 and P3 = P2, we divide the second equation by
the first equation and obtain
k
3
P4
T4
V2 
 r k
c
V 
  
k r 1
24
3
th,Diesel
rk1
1 rc 1
1
k
c


An air-standard Diesel cycle has a compression ratio of 18 and a
cut-off ratio of 2.5. The state at the beginning of compression is
fixed by P = 0.9 bar ant T = 300K. Calculate:
i. the thermal efficiency of the cycle,
ii. the maximum pressure, Pmax, and
iii. The mean effective pressure.
Solution:
Data given:
Example 5.2
V1
V2
V3
V2
18
 2.5
24
4

Example 5.2
Pr ocess1 2isentropiccompression
T2  30018
Pr ocess 2 3Const. pressureheat addition
Pr ocess3 4isentropicexp ansion
181/ 2.5 7.2
k1
0.4
T2  V1 
T1 V2 
V V
2
 3 3
P2  P3  3 2 
T2 T3 V2 
V4

V1
.
V2
V3 V2 V3
T4
 953.3K
 2383.3K
V 
 T  T
  

 
k1
0.4
3
4
T3 V4 
 T  2383.3 1/ 7.2 1082 K
V 
  
32

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