This document provides an overview of thermodynamics concepts for a diploma in mechanical engineering. It defines thermodynamics as the science dealing with thermal energy and its conversion between other forms of energy. The document outlines several key concepts in thermodynamics including systems, properties, the first law of thermodynamics, and conversions between thermal, mechanical, and electrical energy. Examples of thermodynamic processes in common appliances like fans, refrigerators, and pressure cookers are provided to illustrate applications of basic thermodynamic principles.

1. Diploma in Engineering
Mechanical Engineering Division, Ngee Ann Polytechnic

2. o Contents
Section Sub-Section Description
1.1 Define Thermodynamics
1.1.1 Basic Thermodynamics and Applied Thermodynamics
1.2 Basic Concepts of Thermodynamics
- System, Surroundings, Boundary, Properties of a
System, Thermal Equilibrium, Work Transfer, Heat
Transfer
1.3 First Law of Thermodynamics
1.4 Derive the Non-flow Energy Equation
1.5 Derive the Steady Flow Energy Equation

3. o 1.1 What is Thermodynamics
Simple definitions:
 A science which deals with the relationship between
thermal energy (heat) and all other forms of energy
(e.g. mechanical, electrical, etc.).
 A study of thermal engineering.

4. o 1.1 What is Thermodynamics (continued)
 Human activity requires energy
 Energy considered as something which “gives life” to
otherwise inanimate objects
Where does this
energy come from?

5. o 1.1 What is Thermodynamics (continued)
Energy source:
 The major source of energy for humanity is the
thermal energy received directly from the sun in the
form of solar radiation
 For engineering applications, the most common
source of thermal energy is from combustion, that is,
the burning of fuel such as wood, coal or oil.

6. o 1.1 What is Thermodynamics (continued)
 Primitive man discovered that fire provided him with
thermal energy to keep warm and to cook food
 Modern man often makes direct use of thermal
energy, it is a fact that energy in this form is less
useful, and must first be converted into one of the
other energy forms before it can be used to power
many of our modern machines

7. o 1.1 What is Thermodynamics (continued)
 Similarly, it is often necessary to convert between all
the other forms of energy
 Some of these energy conversion processes might
require several steps to produce energy in the form
it is needed.
 Common energy conversion processes

8. o 1.1 What is Thermodynamics (continued)
(a) Thermal energy to mechanical energy
 An example of this energy conversion process is the
motor car engine.
 Fuel is burned to produce thermal energy which the
engine then converts into the mechanical energy
needed to move the car along the road

9. o 1.1 What is Thermodynamics (continued)
(b) Mechanical energy to thermal energy
 The reverse process occurs when a moving car is
stopped by applying the brakes
 The mechanical energy of the car is absorbed by the
brakes and converted into thermal energy,
evidenced by the fact that the brakes become hot
after application

10. o 1.1 What is Thermodynamics (continued)
(c) Thermal energy to electrical energy
 In a power station fuel such as coal or oil is burned
to produce thermal energy. This is used to produce
steam which drives the machinery that produces
electrical energy

11. o 1.1 What is Thermodynamics (continued)
(d) Electrical energy to thermal energy
 Electrical energy can be turned into thermal energy,
as in the case of a domestic electric kettle
 An interesting comparison is the use of electrical
energy to remove thermal energy, as in the case of a
refrigerator (or air-conditioning unit). A refrigerator
removes the thermal energy from the food stored
inside the insulated box

12. o 1.1 What is Thermodynamics (continued)
(e) Electrical energy to mechanical energy
 Electrical energy supplied to a motor causes the
motor shaft to rotate. The resulting mechanical
energy can, for example, be used to drive a machine
tool or propel a vehicle

13. o 1.1 What is Thermodynamics (continued)
(f) Mechanical energy to electrical energy
 The rotating shaft of an internal combustion engine
can be used to drive an electrical generator, thus
providing a portable power source for use in remote
areas
 Regenerative braking

14. o 1.1 What is Thermodynamics (continued)
 Many more involving conversions between all the
different forms of energy
 Thermodynamics is the science that deals with these
processes
 It is concerned with how they occur, with the laws which
control their occurrence, as well as with the engineering
of the machinery needed to convert between the various
energy forms
 It is also concerned with what actually happens within
this machinery, both microscopically and macroscopically
 As such, Thermodynamics is a subject at the very core of
engineering practice

15. o 1.1.1 Basic and Applied Thermodynamics
Thermodynamics is all about energy, and the various
transformations between the different forms of energy

16. o 1.1.1 Basic and Applied Thermodynamics (continued)
 In Basic Thermodynamics, we are concerned not only
with the different forms of energy, but also with various
process by which energy is changed from one form to
another.
 We are also concerned with the fundamental physical
laws which must be obeyed during these transforming
processes.
 It is essential that we understand the important
properties of the substances which are involved in the
process, for it is by the changes in these properties that
we measure the changes of energy.

17. o 1.1.1 Basic and Applied Thermodynamics (continued)
 In Applied Thermodynamics, we are concerned with the
application of basic thermodynamics principles to actual
engineering problems.
 We are not only concerned with the physical details of
the hardware, but also with how the thermodynamic
processes actually occur inside the machinery, as well as
with the factors which affect the performance and
effectiveness of such machinery.
 This is clearly a very practical area of study.

18. o 1.1.1 Basic and Applied Thermodynamics
(continued)
 Some typical domestic examples to elaborate how
Basic Thermodynamics applies to the machinery

19. o 1.1.1 Basic and Applied
Thermodynamics (continued)
(a) Electric Fan
 Electrical energy is supplied to
a motor causing the shaft to
rotate, and with it the fan
blades, which in turn create a
draft of air. Here electrical
energy is converted into
mechanical energy

20. o 1.1.1 Basic and Applied
Thermodynamics (continued)
(b) Refrigerator
 Electrical energy is supplied to a
compressor which produces a flow of
working fluid known as a refrigerant.
This fluid is used to remove thermal
energy from inside an insulated box
to make a cold space where we can
preserve our food. Here an input of
electrical energy results in the
removal of thermal energy. However,
in making one place cold, we have to
make another place hot, which means
we can actually heat water using a
refrigerator.

21. o 1.1.1 Basic and Applied
Thermodynamics (continued)
(c) Air conditioner
 It works in much the same way as a
refrigerator, except that the
refrigerant is used to remove the
thermal energy from the air inside a
room. This cools the air to a
comfortable temperature for the
occupants. Here again, an input of
electrical energy results in the
removal of thermal energy. When we
cool a room with an air conditioner,
we sometimes see that the windows
in the room are covered with droplets
of dew. Here we are actually
extracting water from air.

22. o 1.1.1 Basic and Applied
Thermodynamics (continued)
(d) Pressure cooker
 A little water is placed inside a sealed
container along with the food to be
cooked. The cooker is placed over a
gas burner where the chemical energy
of the fuel is converted into thermal
energy. This energy is supplied to the
cooker to boil the water and produce
pressurized steam inside the
container, enabling us to cook the
food quickly.
 One result of using a pressure cooker
is that we save fuel, and hence fuel
cost at the same time.

23. o 1.1.1 Basic and Applied Thermodynamics
(continued)
 We can see from all these examples that a study of
Basic Thermodynamics can help us to understand
how many basic domestic appliances work.
 The principles of thermodynamics are evident all
around us and are not merely confined to the world
of engineers.

24. o 1.2 Basic Concepts of Thermodynamics
Basic units of the S. I. System:
Quantity Unit
Mass kg
Length m
Time s

25. o 1.2 Basic Concepts of Thermodynamics (continued)
Derived units for thermodynamics in the S. I. System:
 Force
A measure of the “push” or “pull” which is often
exerted on a body.
F=m∙a
F: force, 1 N = 1 kg∙m/s2
m: mass (kg); a: acceleration (m/s2)

26. o 1.2 Basic Concepts of Thermodynamics (continued)
Derived units for thermodynamics in the S. I. System:
 Weight
The gravitational attractive force which the earth exerts on
a mass. It depends on the acceleration due to gravity and
varies with height and location on the earth.
w=m∙g
w: weight, 1 N = 1 kg∙m/s2
m: mass (kg); g: gravity (assumption: 9.81m/s2)

27. o 1.2 Basic Concepts of Thermodynamics (continued)
 System
A collection of matter within a prescribed region
System
Cylinder
Boundary
Piston
Surroundings
Surrounding: the region enclosing the system
Boundary: the surface that separates the system from the surroundings

28. o 1.2 Basic Concepts of Thermodynamics (continued)
Properties of a system ( to define its condition)
 Most common:
Pressure
Temperature
Volume and specific volume
Density and specific gravity
 Additional:
Internal energy
Enthalpy
Entropy
At least 2
independent
properties of a fluid
known, then the state
of the system is
known

29. Pressure
p=
F
𝛢
p: pressure, the force exerted by a fluid on unit area
(N/m2) 1 bar = 105 N/m2
F: force (N)
A: area (m2)

30. Piston
(mass=0)
p
water
Force = 500 N
A = 0.01 m2
Example:

31. Atmospheric pressure, patm
 A pressure due to the atmosphere at the surface of
the earth. It depends on the weight of air above the
surface.
 Sea level ≈ 1.01325 × 105 N/m2 ≈ 1.01325 bar
Gauge pressure, pgauge
 A pressure measured with respect to the
atmospheric pressure.
 E.g Pressure= 2atm; Gauge Pressure= 1atm
patm
+
-

32. Absolute pressure, pabs
 A pressure measured above the absolute zero
pressure, which is a perfect vacuum.
Relationship:
pabs = pgauge + patm

33. Example:
The gauge pressure of the air in a vessel is 10 kN/m2.
Determine the absolute pressure of the air. Assume
the atmospheric pressure is 100 kN/m2.
Solution:
pabs = pgauge + patm
pabs = 10 + 100
= 110 kN/m2

34. Temperature, T (K)
For the degree of hotness or coldness of anything.
 The same temperature interval
 Conversion between the two
temperature scales:
T(K) = t(˚C) + 273
Celsius Kelvin
boiling pt*
freezing pt*
*atmospheric pressure
100˚C
0˚C
-273˚C
373 K
273 K
0 K (absolute zero)

35. A thermodynamics property may be defined as
any quantity that describes the state of a system
and, conversely, as any quantity, the value of
which depends solely on the state of the system.
In other words, the thermodynamic property is
independent of how the system reaches a given
state.

36. Properties are broadly classified into two
categories, namely extensive properties and
intensive properties.

37. In a given state, the value of an extensive property
depends on the amount of mass in the system.
Volume is an example of this type of property. The
greater the mass, the greater is the volume.
Extensive properties are denoted by upper case
letters. For example, volume is denoted by V.

38. A property that is independent of the mass of the
system is called an intensive property. Pressure and
temperature come into this category.
Specific Property: At a given state, the volume (V) is
directly proportional to the mass (m) of the system.
The quantity V/m is a specific property known as
specific volume and is denoted by v (lower case).
Similarly, all extensive properties expressed per unit
mass become specific properties, and are denoted by
lower case letters.

39. Volume, V(m3)
 The amount of space which it occupies.
Specific volume, v(m3/kg)
 The volume occupied by unit mass of the
substance.

40. Example:
1.29 kg of the compressed air is contained in a rigid
vessel of volume 1.0 m3. Determine the specific
volume of the air.
Solution:

41. Density, ρ(kg/m3)
 Mass per unit volume
 The density is also the reciprocal of specific volume

42. Solution:
Volume = Length × Width
× height
Vair = 6.0×4.0×2.9
= 69.6 m3
Example:
4.0 m
2.9m
A living room: density of air
ρ=1.2 kg/m3
What is the mass of air?

43. Specific gravity, s.g.sub (-)
 A ratio of the density of a substance over the
density of water at 4˚C (1000 kg/m3)
Example:
The density of sea water is 1025kg/m3. What is
the specific gravity of sea water?
Solution:

44. System equilibrium
 If a system is completely stable when one
particular value of any property (e.g. pressure)
is the same at all points throughout the system
at that particular instant.
 In practice, the properties of a system changes
and it is usually assumed that the initial and
final conditions are in states of equilibrium.

45. Work transfer
Rectilinear motion
Work transfer of a constant force F is the product
of the force and the distance traveled by the force
measured along the line of action of the force.
F F
S
Work transfer = F × S

46. Work transfer
Angular motion
» Work transfer of a constant torque is the
product of the torque and the angular
displacement by the torque.
torque
Work transfer = r × θ
r: torque in N·m
θ: angular displacement in rad

47. Work transfer
Convention
» Positive: when work energy is transferred from the
system to the surroundings
» Negative: when work energy is transferred into the
system from the surrounding
system
surroundings
work output
(positive)
work input
(negative)
boundary

48. Heat transfer
A form of energy which crosses the boundary of a
system during a change of state produced by a
difference of temperature between the system and its
surrounding.

49. Heat transfer
Convention
» Positive: when heat energy flows into the system
from the surroundings
» Negative: when heat energy flows from the system
to the surroundings
system
surroundings
heat loss
(negative)
heat supplied
(positive)
boundary

50.  The principle of the conservation of energy states:
Energy can neither be created nor destroyed
» The First Law of Thermodynamics refers to heat
energy and work energy

51. The First Law of Thermodynamics:
 When a system undergoes a thermodynamics cycle
then the net heat supplied to the system from its
surroundings is equal to the net work done by the
system on its surroundings.
ΣQ = ΣW
Where:
ΣQ: the net heat supplied
ΣW: the net work done/work output

53. Example:
A system undergoes a complete thermodynamic cycle.
Determine the value of the work output, Wout.
system
surroundings
Qin=10 kJ boundary
Qout=3 kJ Wout
Win=2 kJ
Solution:
ΣQ = ΣW
Qin + Qout = Win + Wout
Wout = Qin + Qout - Win
=10×103 + (-3 ×103) – (-2 ×103)
=9×103 J (9 kJ)

54. Non-flow Processes
» In a closed system energy may be transferred across
the boundary in the form of work energy and heat
energy but the working fluid itself never crosses the
boundary
» Any process undergone by a closed system is
referred to as a non-flow process.

55. Non-flow Processes
Example:
i. Suction stroke:
» The working fluid
flows into the
cylinder, which is
then sealed by
closing of the inlet
valve
A cylinder of an internal combustion engine
T T
Fluid flows in

56. Non-flow Processes
Example:
ii. Compression stroke:
» Whilst the cylinder
is sealed, the fluid is
compressed by the
piston moving into
the cylinder
A cylinder of an internal combustion engine
T T
Non-flow
process

57. Non-flow Processes
Example:
iii. Working stroke:
» Heat energy is
supplied so that the
fluid possesses
sufficient energy to
force the piston
downward and
produce work output
A cylinder of an internal combustion engine
T T
Non-flow
process

58. Non-flow Processes
Example:
iv. Exhaust stroke:
» The exhaust valve is
opened for the fluid
to flow out of the
cylinder
A cylinder of an internal combustion engine
T T
Fluid flows out

59. Non-flow Energy Equation
» When the fluid in a closed system is undergoing a non-
flow process from State 1 to State 2, the internal energy
of a fluid depends on pressure and temperature.
» The non-flow energy equation:
or U2 - U1 = Q12 – W12
• U2 the internal energy of the fluid at State 2
• U1 the internal energy of the fluid at State 1
• Q12 the net heat energy transferred to the system
from the surrounding
• W12 the net work energy transferred from the
system to the surrounding

60. Non-flow Energy Equation
» The fluid in a closed system produces work
amounting to 600 kJ whilst heat energy amounting
1000 kJ is transferred into it. Determine the change
of internal energy of the fluid and state whether it is
an increase or decrease.
Solution:
U2 - U1 = Q12 – W12
=1000 – 600
=400 kJ
Since U2 > U1, the internal energy has increased.
Q=1000 kJ W=600 kJ

61. Flow Processes
» In an open system, in addition to energy transfers
taking place across the boundary, the fluid may also
cross the boundary.
» Any process undergone by an open system is called a
flow process.
» Steady flow processes, and unsteady flow processes

62. Steady Flow Processes
The conditions:
» The mass of fluid flowing past any section in the system
must be constant with respect to time
» The properties of the fluid at any particular section in
the system must be constant with respect to time
» All transfers of work energy and heat energy which take
place must do so at a uniform rate

63. Steady Flow Processes:
Example:
A steam boiler under constant load
» To maintain the water level in the
boiler:
» To maintain the production of team at
this rate ( )at a steady pressure, the
furnace will need to supply heat
energy at a steady rate
» Under these conditions, the
properties of the working fluid at any
section within the system must be
constant with respect to time
Boiler
Qin (from
furnace)
Feed
water in
Steam
out
Boundary
wm
sm

64. The Continuity of Mass Equation
» In steady flow, the mass flow rate of fluid is the same
across any section in the system
» Consider kg/s of fluid flowing through a system in
which all conditions are steady (i.e. under steady flow
conditions)
Specific volume, v1
Velocity, c1
Specific volume, v2
Velocity, c2
Cross-sectional area, A1 Cross-sectional area, A2
𝑚
·
=
𝑐 ⋅ 𝐴
𝑣Continuity of mass equation: in out
m m 

65. The Steady Flow Energy Equation
» The working fluid flows along the inlet pipe at a constant rate and
enter the system at point 1.
» Various energy transfers take place across the boundary of the
system.
» The fluid flows out of the system at point 2 along the outlet pipe.
system
1
2
pressure, p1
inlet velocity, c1
specific internal energy, u1
pressure, p2
outlet velocity, c2
specific internal energy, u2
inQ
outQ
outW
inWZ1
Z2

66. Energy balance:
The total amount of
energy entering the
system
The total amount of
energy leaving the
system=
or
The total energy entering the system :
• Rate of heat energy entering the system per second,
• Rate of work energy entering the system per second,
• The rate of energy of the fluid entering the system
 Internal energy,
 Potential energy,
 Kinetic energy,
• The rate of work energy required to push the fluid across the boundary and
enter the system,
inQ
inW
1 1m u
1 1m gZ
2
1 1
1
2
m c
1 1 1m p v

67. Therefore:
Similarly:
Since:
And
Hence:
Steady flow energy equation
2
1
1 1 1 1 1( )
2
in inin
c
E Q W m u p v gZ     
2
2
2 2 2 2 2( )
2
out outout
c
E Q W m u p v gZ     
in outE E
h u pv 
2 2
1 2
1 21 1 2 2( ) ( )
2 2
in outin out
c c
Q W m h gZ Q W m h gZ        
Specific enthalpy, h, is a measure of the total energy of a
thermodynamic system. It includes the internal energy,
which is the energy required to create a system, and the
amount of energy required to make room for it by
displacing its environment and establishing its volume
and pressure.

68. Example
» In an open system, the fluid flow through the system at 17
kg/s and the power developed by the system is 14 000 kW.
The specific enthalpy of the fluid at inlet and outlet are 1 200
kJ/kg and 360 kJ/kg respectively, and the velocities of the
fluid at inlet and outlet are 60 m/s and 150 m/s respectively.
Assuming the change in potential energy is negligible.
Determine:
a. the rate at which heat is rejected from the system
b. the area of inlet pipe, given that the specific volume of the
fluid at inlet is 0.5 m3/kg

69. Solution:
a. Applying the continuity of mass equation
Applying steady flow energy equation
outQ
outW
outlet
inlet
Hence:

70. Applying
Since
Hence the cross-sectional area of inlet pipe

71. Example
» Fluid flows steadily at the rate of 0.4 kg/s through an open
system, entering at 6 m/s with a pressure of 1 bar and a
specific volume of 0.85 m3/kg and leaving at 4.5 m/s with a
pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The
specific internal energy of fluid leaving is 88 kJ/kg, greater
than that of the fluid entering. 59 kJ/s of heat is rejected
from the system to its surrounding and assuming the change
in potential energy is negligible.
Determine:
a. the power required to drive the system
b. the inlet and outlet pipe cross-sectional area

72. Solution:
a. Applying the continuity of mass equation
Applying steady flow energy equation
outQ
outlet
inlet
Hence:
inW

73. Solution:
Hence:
Since
Hence:

74. » b) applying
And
Hence, the cross-sectional area of inlet pipe
The cross-sectional area of inlet pipe

75. Q & A