Thermodynamics chapter one lecture of wrc

PDF of wrc lecture

1. INTRODUCTION
Shacheendra Kishor Labh
Assistant Professor
Department of Mechanical and Automobile Engineering
IOE, Pashchimanchal Campus
ME 104 Engineering Thermodynamics I BME I/II IOE, Pashchimanchal Campus
1

2. Course Outline
2
Course Objective: To develop the laws of thermodynamics and their practice with real-
world engineering examples.
Unit 1 Introduction [3 hrs]
Unit 2 Energy and Energy Transfer [4 hrs]
Unit 3 Properties of Pure Substances [6 hrs]
Unit 4 First law of Thermodynamics [4 hrs]
Unit 5 First-Law Analysis for a control volume [4 hrs]
Unit 6 Second law of Thermodynamics [3 hrs]
Unit 7 Entropy [4 hrs]
Unit 8 Second-Law Analysis for a control volume [4 hrs]
Unit 9 Irreversibility and Availability [4 hrs]
Unit 10 Thermodynamic Relations [6 hrs]
Unit 11 Gas Mixtures [3 hrs]
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

3. Course Outline
3
References:
•V. Wylen, Sonntag & Borgnakke, “Fundamentals of Thermodynamics”, John Wiley & Sons,
Inc
•M. J. Moran & H. N. Shapiro, “Fundamentals of Engineering Thermodynamics", John Wiley
& Sons, Inc.
• Y. A. Cengel & M.A. Boles, "Thermodynamics: An Engineering Approach", McGraw-Hill.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

4. Evaluation
4
Internal:
•Theory [40 marks]
i.Assignments
ii.Assessments
iii.Attendance
iv.Classroom behaviour
•Practical [25 marks]
Final:
•Exam [60 marks]
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

5. Introduction
5
Thermodynamics
therme dynamis
HEATPOWER OR
MOTION
Literally, thermodynamics means ‘Heat Power’ or ‘Heat in motion’
The concise oxford dictionary defines Thermodynamics as
“the science of relation between heat and mechanical energy”
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

6. Introduction
6
Definition:
▪Thermodynamics is the study of energy interactions between systems and the effect
of these interactions on the system properties. Energy transfer between systems
takes place in the form of heat and/or work.
▪Thermodynamics is concerned with
•the concept of energy,
•the laws that govern the conversion of one form of energy into another,
•the properties of the working substance or the media used to obtain the energy
conversion.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

7. Brief history
7
Discovered Rubbing of two stones produces fire
Time Event
Probably when humans
started playing with
stones.
50 BC Hero of Alexandria developed his own engine (which was actually a
turbine)
1620 AD Francis Bacon concluded that heat is motion and nothing else
1631 AD Jean Rey constructed the first liquid expansion thermometer
1680 AD Huygens developed the first gunpowder engine
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

8. Brief history
8
Newcomen devised the first steam engine with cylinder and piston
Time Event
1824 AD Sadi Carnot published the basics of second law of thermodynamics
1842 AD Mayer and Joule published experimental works on mechanical
equivalent of heat
1847 AD Hemholtz Formulated fist law of thermodynamics
1848 AD Kelvin developed the thermodynamics scale of temperature
1705 AD
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

9. Brief history
9
Clausius and Kelvin stated their forms of the second law of
thermodynamics
Time Event
1897 AD Maxwell stated the zeroth law of thermodynamics
1907 AD Nernst enunciated the third law of thermodynamics
1850-55 AD
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

10. Macroscopic vs microscopic approach
10
•The description of a system or matter using a few of its measurable bulk properties
constitutes a point of view called macroscopic.
•The state or condition of the system is completely described by means of the large scale
characteristics or properties of the system
•Some common attributes of such description are:
a.These characteristics do not involve any assumptions about the structure of the
material,
b.these are readily measured, and
c.Only few such properties are required to completely describe the system.
•A more elaborate approach, based on the average behaviour of large groups of
individual particles is called microscopic approach or statistical thermodynamics.
•The properties like velocity, momentum, impulse etc. are studied by this approach.
•It requires advanced statistical and mathematical method.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

11. Foundation of Thermodynamics
11
Four laws are the foundation stone of thermodynamics
0
th
law Defines temperature
1
st
law Defines internal energy
2
nd
law Defines Entropy
3
rd
law Gives numerical value to entropy
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

12. Definitions
12
System:
A thermodynamic system is defined as a quantity of matter of fixed mass and identity
upon which attention is focused for study. In simple terms, a system is whatever we want
to study.
The thermodynamic system may be classified into the following three groups:
(a)Closed system;
(b)Open system; and
(c)Isolated system.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

13. Definitions
13
Surrounding:
Everything external to the system is the surroundings.
Boundary:
The system is separated from the surroundings by the system boundaries. The boundary
may be real or imaginary.
Universe:
A system and it surroundings together comprises a universe which can be defined as
totality of matter that exists.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

14. Terminologies
14
1.Properties:
•A property is any characteristic or attribute of matter, which can be evaluated
quantitatively.
•A thermodynamic property depends only on the state of the system and is
independent of the path by which the system arrived at the given state.
•Some of the properties, with which we are already familiar, are: temperature, pressure,
density, specific volume, specific heat etc.
•Thermodynamic properties can be either intensive (independent of size/mass, e.g.
temperature, pressure, density) or extensive (dependent on size/mass, e.g. mass,
volume)
2.State
•It is the condition of a system as defined by the values of all its properties. It gives a
complete description of the system.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

15. Terminologies
15
•Any operation in which one or more properties of a system change is called a change
of state.
3.Path and Process
•The succession of states passed through during a change of state is called the path of
the system.
•A system is said to go through a process if it goes through a series of changes in state.
•A quasi-static process is one in which the deviation from thermodynamic equilibrium is
infinitesimal.
•All states of the system passes through are equilibrium states.
4.Equilibrium or non equilibrium system
Properties of the system do not change with time and space for a system in
equilibrium
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

16. Terminologies
16
3.Homogeneous or heterogeneous system
Glass of water with ice is heterogeneous and a cup of coffee is homogeneous
4.Components of a system
Glass of water with ice has only one component but a coffee has many components
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

17. Zeroth law of thermodynamics
17
•Temperature is a property of a system which determines the degree of hotness or
coldness. It is a relative term.
•Two systems are said to be equal in temperature, when there is no change in their
respective observable properties when they are brought together.
•Zeroth law of thermodynamics states that:
“If two systems (say A and B) are in thermal equilibrium with a third system (say C)
separately (that is A and C are in thermal equilibrium; B and C are in thermal
equilibrium) then they are in thermal equilibrium themselves (that is A and B will be in
thermal equilibrium.”
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

18. Some common properties
18
Pressure
•Pressure is defined as the normal force exerted per unit area on a real or a fictitious
surface.
•For engineering work, pressures are often measured with respect to atmospheric
pressure rather than with respect to perfect vacuum.
•The pressure relative to the atmosphere is called gauge pressure. The pressure relative
to a perfect vacuum is called absolute pressure.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

19. Some common properties
19
•For pressures below atmospheric pressure, the gauge pressure will be negative. This
negative gauge pressure is known as vacuum pressure.
P
abs = P
atm - P
vacuum
•The standard value of atmospheric pressure is taken as 101.3 kPa (or 760 mm of Hg) at
sea level.
Specific Volume
•Volume is the space occupied by a substance and is measured in m
3
. This is an extensive
property.
•The specific volume (ν) of a substance is defined as the volume per unit mass and is
measured in m
3
/kg.
ν =
??????
??????
•Specific volume is the reciprocal of the density.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

20. Example 1
20
Solution:
Given,
Specific volume of B, ν
B = 5 m
3
/kg
V
A = 15 m
3
, V
B = 5 m
3
Thus,
Mass of steam in compartment B, m
B = V
B/ν
B = 5/5 = 1kg
When the membrane breaks,
Specific volume of the container, ν = 8 m
3
/kg
Total volume of container, V = 15 + 5 = 20 m
3
Hence, mass of steam in container, m = V/ν = 20 / 8 = 2.5 kg
Mass of steam in compartment A, m
A = m – m
B = 2.5 – 1 = 1.5 kg
Therefore, specific volume of steam in container A, ν
A = V
A/m
A = 15/1.5 = 10 m
3
/kg
A container having two compartments contains steam as shown in the figure below. The specific
volume of steam in compartment B is 5 m
3
/kg. The membrane breaks and the resulting specific
volume is 8 m
3
/kg. Find the original specific volume of steam in compartment A.
A B
V
A= 15 m
3
V
B= 5 m
3
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

21. Solution:
Given,
Pressure gauge A reading, (Pgauge)
A = 180 kPa
Pressure gauge B reading (Pgauge)
B = 120 kPa
Atmospheric pressure, P
atm =100 kPa
For gauge B,
(Pabs)
X = P
atm + (Pgauge)
B = 100 kPa + 120 kPa = 220 kPa
For gauge A,
(Pabs)
X = (Pabs)
Y + (Pgauge)
A
220 kPa = (Pabs)
Y + 180 kPa
(Pabs)
Y = 40 kPa
Example 2
21
A large chamber is separated into two compartments which are maintained at different pressures,
as shown in the figure. Pressure gauge A reads 180 kPa, and pressure gauge B reads 120 kPa. If the
barometric pressure is 100 kPa, determine the absolute pressure existing in the compartments and
the reading of gauge C.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

22. Contd…
For gauge C,
(Pabs)
Y = P
atm + (Pgauge)
C
40 kPa = 100 kPa + (Pgauge)
C
(Pgauge)
C = -60 kPa
Example 2
22
A large chamber is separated into two compartments which are maintained at different pressures,
as shown in the figure. Pressure gauge A reads 180 kPa, and pressure gauge B reads 120 kPa. If the
barometric pressure is 100 kPa, determine the absolute pressure existing in the compartments and
the reading of gauge C.
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

23. Solution:
Given:
P
air = 200 kPa
P
atm = 100 kPa
mass (m) = 20 kg
So,
Weight = mg = 20 x 9.81 = 196.2 N
From free body diagram,
P
air x A = P
atm x A + mg
A = 1.962 x 10
-3
m
2
Hence,
A =
π??????
2
4
D = 0.05 m
Example 3
23
A piston cylinder arrangement is shown in the figure. If the mass of piston is 20 kg and the pressure
of air inside the cylinder is 200 kPa, find the diameter of piston. [Take atmospheric pressure to be
100 kPa]
P
atm x A
W
P
air x A
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

24. Solution:
Given:
Area of piston (A) = 0.1m
2
mass of piston (m) = 80 kg
Atmospheric pressure (P
atm) = 100 kPa
spring constant (k) = 50 kN/m
g = 9.81 m/s
2
; initial volume (V
i) = 0.06 m
3
Final condition:
Volume (V
f) = 2 x V
i = 2 x 0.06 = 0.12 m
3
Compression of the spring,
x =
??????
�
??????
-
??????
??????
??????
= 0.12/0.1 – 0.06/0.1 = 0.6 m
Example 4
24
A piston cylinder arrangement loaded with a linear spring as shown in figure has cross sectional
area of 0.1 m
2
, contains gas piston mass of 80 kg. Initially, the spring touches the piston but exerts
no pressure on it. Heat is supplied to the system until its volume doubles. Determine the final
pressure. [Take g= 9.81 m/s
2
, Atmospheric pressure = 100 kPa, spring constant k = 50 kN/m and
initial volume = 0.06 m
3
]
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh

25. Contd..
From free body diagram,
(P
gas)
f x A = P
atm x A + F
spring + mg
(P
gas)
f = P
atm + F
spring /A + mg/A
= P
atm +
????????????
??????
+
??????�
??????
= 100 +
50 ?????? 0.6
0.1
+
80 ?????? 9.81
0.1 ?????? 103
= 407.8 kPa
Example 4
25
A piston cylinder arrangement loaded with a linear spring as shown in figure has cross sectional
area of 0.1 m
2
, contains gas piston mass of 80 kg. Initially, the spring touches the piston but exerts
no pressure on it. Heat is supplied to the system until its volume doubles. Determine the final
pressure. [Take g= 9.81 m/s
2
, Atmospheric pressure = 100 kPa, spring constant k = 50 kN/m and
initial volume = 0.06 m
3
]
P
atm x A
W
(P
gas)
f x A
F
spring
ME 104 Engineering Thermodynamics I IOE, Pashchimanchal Campus Shacheendra K Labh