Subject: ME8391 Engineering Thermodynamics Topic: Basic Concepts & First law of Thermodynamics B.E. Mechanical Engineering Second year, III Semester. [Anna University R-2017]

1. ME 8391-ENGINEERING THERMODYNAMICS
S.Thirumalvalavan
Assistant Professor
Department of Mechanical Engineering
E-mail : thirumalbemech@gmail.com
BASIC CONCEPTS AND
FIRST LAW OF THERMODYNAMICS

2. UNIT I BASIC CONCEPTS AND FIRST LAW 9+6
• Basic concepts - concept of continuum, comparison of microscopic and macroscopic
approach. Path and point functions. Intensive and extensive, total and specific
quantities. System and their types. Thermodynamic Equilibrium State, path and
process. Quasi-static, reversible and irreversible processes. Heat and work transfer,
definition and comparison, sign convention. Displacement work and other modes of
work .P-V diagram. Zeroth law of thermodynamics – concept of temperature and
thermal equilibrium– relationship between temperature scales –new temperature
scales. First law of thermodynamics –application to closed and open systems – steady
and unsteady flow processes.
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

3. UNIT – 1 BASIC CONCEPTS AND FIRST LAW
• Thermodynamics is the science of energy transfer which deals with the
relations among heat, work and properties of systems.
• Thermodynamics = Thermo + dynamics
Heat + Motion
Heat Energy into POWER.
Thus, thermodynamics is basically the study of heat and power.
Today conservation of heat energy into power plays important role in power
generation, refrigeration etc.,
Ex: Burning of Coal (Water get heated to produce steam) – Turbine – Generator
– POWER output in the form of Electricity ;
Ex: Motorcycle = Fuel (Burned inside the Engine) – Heat – POWER output in the
form of Wheel rotation
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

4. Application Area of Thermodynamics:
Energy transfer is present in almost all the engineering activities. Hence,
the principles of thermodynamics are playing vital role in designing all the
engineering equipments such as internal combustion engines, rockets, jet
engines, thermal and nuclear power plants, refrigerators etc.
Energy  Capacity to do work
• Force = Mass x Acceleration (N)
F = m x a
• Weight = Mass x Acceleration due to gravity (g = 9.81 m/s2)
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5. Law of conservation of ENERGY Principle
Energy can be changed from one form to another from without losses
during energy interaction.
So, First law of thermodynamics is simply an expression of conservation
of energy. It considers only Quantity.
But, the Second law of thermodynamics considers both Quantity as well
as Quality.
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6. BASIC CONCEPTS
• Density (ρ) = mass / volume
m / v (kg / m3)
Wkt, Air density = 1 kg / m3 ; Water density = 1000 kg / m3
Mass – Scalar Quantity – Not depend on gravity
• Specific Weight (w) = weight / volume
W / V (N/m3 or kN/m3)
Weight – Vector quantity – Depend on gravity
• Specific volume (v) = volume / mass (m3/kg)
• Specific gravity (s) =
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7. BASIC CONCEPTS (Contd..)
• Pressure (p) = Force / Area
= F / A (N/m2)
It is defined as the force per unit area.
If we are dealing with, Liquid and gas – termed as Pressure ;
In solid – termed as Stress (Stress = Load/Area)
Various units of pressure,
1 pascal = 1 N/m2
1 Bar = 1x105 N/m2 = 100 kN/m2 = 100 000 N/m2
1 mm of Mercury (Hg) = 1 Torr = 133.3 N/m2
1 mm of water (H2O) = 9.80665 N/m2
Pressure, p --> 0.1 Mpa = 100 kpa = 105 pa = 105 N/m2 = 1 bar
1 Pa = 1 N/m2; 1 kpa = 1 kN/m2; 1 Mpa = 1000 kN/m2; 1 bar = 100 kN/m2 or kPa
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

8. Figure 1. Pressure Relations
Temperature (T) : It is defined as the measure of velocity of fluid particles. It is a property which is used to
determine the degree of hotness or coldness or the level of heat intensity of a body.
T = t + 273 K
Where,
T – Temp. on Kelvin scale
t – Temp. on Celsius scale
Figure 2. Temperature Relationship
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9. HEAT (Q)
• Heat is defined as the energy crossing the boundary of a system due to the
temperature difference between system and surrounding. It is usually represented by
Q and expressed in Joule or kJ.
• Let, m be the mass of the substance heated from temperature T1 to T2, then the heat
transfer is given by
Q = m C (T2 – T1) = m C dT (Joule)
Where, C = specific heat of the substance
Note:
If Q is positive, heat is supplied to the system
If Q is negative, heat is rejected from the system
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10. • Specific heat capacity (C) : It is defined as “the Quantity of heat transfer required for
raising or lowering the temperature of unit mass of the substance through one degree”.
Denoted by C and is expressed in  J/kg-K or kJ/kg-K
• Specific heat capacity at Constant volume (Cv) : It is defined as “the Quantity of heat
transfer required for raising or lowering the temperature of unit mass of the substance
through one degree when the volume remains constant”.
Heat transfer, Q = m Cv (T2 – T1) kJ
• Specific heat capacity at Constant Pressure (Cp) : It is defined as “the Quantity of heat
transfer required for raising or lowering the temperature of unit mass of the substance
through one degree when the pressure remains constant”.
Heat transfer, Q = m Cp (T2 – T1) kJ
• For any gas, Cp is always greater than Cv. (i.e., Cp > Cv)
• The ratio of two specific heats remains constant and is denoted by gamma (γ).
γ = Cp / Cv
For air, Cp = 1.005 kJ/kg K; Cv = 0.718 kJ/kg-K; γ = 1.4
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

11. CHARACTERISTIC GAS EQUATION
General gas equation for ideal gas,
Where,
p – pressure in N/m2
V – volume in m3
T – temperature in oC
Taking R as constant,
PV = RT
If we consider mass ‘m’, then the Equ. becomes pV = mRT. This equation is known as
characteristic gas equation.
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12. TYPES OF THERMODYNAMICS
1. Statistical Thermodynamics is microscopic approach in which, the matter is
assumed to be made of numerous individual molecules. Hence, it can be regarded
as a branch of statistical mechanics dealing with the average behaviour of a large
number of molecules.
2. Classical thermodynamics is macroscopic approach. Here, the matter is considered
to be a continuum without any concern to its atomic structure.
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13. THERMODYNAMIC SYSTEM
1. System : A thermodynamic system is defined as a defined as a definite space or area
on which the study of energy transfer and energy conversions is made.
2. Boundary : The system and surroundings are separated by boundary. It may be fixed
or movable and real or imaginary.
3. Surroundings : Anything outside the system which affects the behavior of the system
is known as surroundings or the environment.
4. Control volume : A specified large number thermal device has mass flow in and out
of a system called as control volume.
5. Universe : A system and surroundings together comprise a universe.
Mass Mass
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14. CLASSIFICATION OF THERMODYNAMIC SYSTEMS
1. In Open System both the mass and energy transfer takes place. The open system is
often called as control volume.
2. A Closed System does not permit any mass transfer. But, only the energy transfer
takes place.
3. Isolated System : A system which is not affected by the surrounding, simply there is
no heat, work and mass transfer. It is an imaginary system. Ex: Entire universe.
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15. PROPERTIES
• It is defined as any measurable or observable characteristics of the substance when
the system remains in equilibrium state. Ex: Pressure, Temperature Volume, Entropy
etc.,
Properties are classified into two types,
i) Intensive or Intrinsic property : These properties are independent on the mass of
the system. If we consider a part of the system, these properties remain same.
Ex: Pr, Temp, Specific volume, density etc.,
ii) Extensive or Extrinsic property : These properties are dependent upon the mass of
the system. If we consider a part of the system, these properties have lesser value.
Ex: Mass, Volume, Total energy, weight etc.,
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16. STATE OF A SYSTEM
• STATE : It is the condition of the system at any particular moment. The state is
identified by the properties of the system such as pressure, volume, temperature etc.,
If the value of one property changes, the state will change to a different state
called a change of state. Ex: ICE Solid state  WATER liquid state (Due to
temperature drop)
• PATH : The succession of state crossed through the control volume during change of
state in the thermodynamic system is called path.
THERMODYNAMIC PROCESS
1. Quasi-Static process
2. Reversible and Irreversible process
3. Flow and Non flow process
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17. CYCLE
• A series of state changes such that the final state is identical with the initial state is
known as cycle.
• If a thermodynamic system undergoes a series of processes and returns to its initial
position, then the process is called cyclic process.
• There are two types of cyclic processes,
(i) Closed cycle the working substance is recirculated again and again within the system
itself without taking any mass transfer.
(ii) Open cycle the working substance is exhausted to atmosphere after completing the
process. So, here both the mass and energy transfer take place.
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18. EQUILIBRIUM
• A system is said to be in equilibrium, if it does not tend to undergo any change of
state on its own accord.
• A system is said to be in thermodynamic equilibrium, then it should satisfy the
following three conditions of equilibrium.
(i) Mechanical Equilibrium A system is said to be in mechanical equilibrium, when
there is no unbalanced forces acting on it.
(ii) Thermal equilibrium A system is said to be in thermal equilibrium, when there is no
temperature difference throughout the system.
(iii) Chemical Equilibrium A system is said to be in chemical equilibrium, when there is
no chemical reaction throughout the system.
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

19. WORK TRANSFER (W)
• Work is an energy interaction between system and surroundings.
• Work can be defined as the energy interaction which is not caused by the
temperature difference between system and surroundings.
Specifically, it is the energy transfer associated with force acting through a distance.
Work = force x distance moved
W = F x x N-m or kJ/kg
POWER : The Work done per unit time is called power unit is kJ/sec or kW.
• Work done BY the system is denoted as Positive work
• Work done ON the system is denoted as Negative work
Modes of Work transfer : (i) Mechanical work (ii) Non-mechanical work
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

20. POINT AND PATH FUNCTIONS
From above figure, the properties like pressure and volume do not depend upon the
path [ 1 – C – 2 or 1 – B – 2 ] followed by the gas. It requires end points only. Therefore,
these properties are called as point function.
Some properties like work transfer, heat transfer etc., are dependent upon the path
followed by a gas. These properties are called as path function.
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

21. ZEROTH LAW OF THERMODYNAMICS
• It states that “When two systems are in thermal equilibrium with the third system
separately, then they themselves are in thermal equilibrium with each other”.
• A system A in thermal equilibrium with another system B. Also let another system C is
thermal equilibrium with the system B. Then from Zeroth law of thermodynamics, the
system A is in thermal equilibrium with the system C. Hence, A and C are at same
temperature.
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

22. FIRST LAW OF THERMODYNAMICS
• It states that “When a system undergoes a cyclic process, then the net heat transfer is
equal to the net work transfer”.
• Mathematically,
• It may be stated, the heat and work are mutually convertible.
• In general, for any thermodynamic systems, the first law of thermodynamics can be
written in the form of following equation.
Heat transfer = Work done + Change in internal energy
Q = W + ∆U
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23. LIMITATIONS OF FIRST LAW OF THERMODYNAMICS
1. First law of thermodynamics does not specify the direction of flow of heat and
work. i.e., whether the heat flows from hot body to cold body or cold body to hot
body.
2. The heat and work are mutually convertible. The work can be converted fully into
heat energy. But heat cannot be converted fully into mechanical work.
This violates the foresaid statements. A machine which violates the First law of
thermodynamics is known as Perpetual Motion Machine (PMM-1) of the first kind
which is impossible.
PMM-1 is a machine which delivers work continuously without any input.
Thus, the machine violates first law of thermodynamics.
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

24. Unit Conversion (To Remember)
We know that,
• Kilo – 103 ; Mega – 106 ; Giga – 109 ; Terra – 1012 ;
• 1 watt = 1 J/sec
• 1 kW = 1 kJ/sec
• 1 J/s = 1 W
• 1 tonne = 3.5 kW
• 1 Hr = 3600 sec
• m  x103  mm; mm  ÷103  m
• km  x103  m; m  ÷103  km
• km/s x103  m/s; m/s  ÷103  km/s
• min  x60  sec; sec  ÷60 min
• kJ  x103  J; J  ÷103  kJ
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

25. Problems on Basics
1. During a flow process 5kW paddle wheel work in supplied while the internal energy of the
system increases in one minute in 200kJ. Find the heat transfer when there is no other form
of energy transfer.
Given Date:
Work done, W = -5 kW (-ve sign; Work is supplied)
Internal energy, ∆U = 200 kJ/m = 3.33 kJ/s
To find:
Heat transfer, Q = ?
Solution: By first law of thermodynamics,
Q = W + ∆U
= -5 + 3.33
Q = -1.67 kW
Result: Heat transfer, Q = -1.67 kW [ -ve sign indicates that the heat is transferred from the
system] @ S. Thirumalvalavan, AP/Mech, 5104 - AEC.
Unit conversion:
(min  Sec ; ÷ by 60)
1 kW = 1 kJ/sec

26. Problems on Basics
2. A liquid of mass 18kg is heated from 25oC to 85oC. how much heat transfer is required?
Assume Cp for water is 4.2 kJ/kg-K.
Given Date:
Mass, m = 18 kg
Initial temp, T1 = 25 oC = 25 + 273 = 298 K (oC  K ; +273)
Final temp, T2 = 85 oC = 85 + 273 = 358 K (oC  K ; +273)
Cp for water = 4.2 kJ/kg-K (Cp = Specific heat capacity at Const. Pressure)
To find: Heat transfer, Q = ?
Solution: From Specific heat capacity at Constant Pressure (Cp) definition wkt,
Q = m Cp ( T2 – T1)
= 18 x 4.2 x (358 - 298)
Q = 4536 kJ
Result: Heat transferred, Q = 4536 kJ
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.
Unit conversion:
So, Balance : kJ

27. Problems on Basics
3. A closed system receives an input heat of 450 kJ and increases the internal energy of the
system for 325 kJ. Determine the work done by the system.
Given Date:
Heat received, Q = 450 kJ
Change in Internal energy, ∆U = 325 kJ
To find: Work done, W = ?
Solution: By first law of thermodynamics,
Q = W + ∆U
W = Q - ∆U
= 450 – 325
Q = 125 kJ
Result: Work done, W = 125 kJ
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.

28. THANK YOU
@ S. Thirumalvalavan, AP/Mech, 5104 - AEC.