Basic Knowledge about Thermodynamics

1. Thermodynamics
Md. Mostafizar Rahman
Lecturer
Dept. of FET, SUST, Sylhet

2. Definition:
The word thermodynamics derives from two
Greek ideas:
Therme meaning hot or heat,
Dynamics meaning power or motion

3. Thermodynamics is
the study of heat related to matter in motion.
Or
the science of the regularities governing
processes of energy conversion.
Or
the science that deals with the interaction
between energy and material systems.

4. Its importance's :
1. Thermodynamic engines
petrol engines,
gas turbines,
steam turbines, etc.,
2. Modern industrial society

5. Working substances:
-are fluids which are capable of deformation in
that they can readily be expanded and
compressed.
Example: Air and steam
Pure substance:
-is a single substance or mixture of substances
which has the same consistent composition
throughout.
In other words, it is a homogeneous substance and
its molecular or chemical structure does not vary.
Example: steam and water, or a mixture of steam
and water and Air its gaseous and liquid form.

6. Macroscopic analysis
If the properties of a particular mass of a
substance, such as its pressure, volume and
temperature are analysed, called macroscopic
analysis
Microscopic analysis
If an analysis is made in which the behaviour
of the individual atoms and molecules of a
substance are under investigation, called
microscopic analysis.

7. Properties and state
A property of a system is a characteristic of the
system which depends upon its state, but not upon
how the state is reached.
1. Intensive properties. These properties do not
depend on the mass of the system.
Examples : Temperature and pressure.
2. Extensive properties. These properties depend
on the mass of the system.
Example : Volume.
State -is the condition of the system at an instant of
time as described or measured by its properties.
Or, Each unique condition of a system is called a
state.

8. Specific quantity/volume
The specific volume of a substance, at some
particular state, is the volume occupied by unit
mass of the substance at that particular state.
For example: if the volume of a system of mass
m is V, then the specific volume of
matter within the system, v = V/m.
Temperature
Temperature is the degree of hotness or
coldness of a body.
Unit Degree centigrade (0C) or Fahrenheit(0F)

9. Pressure-
Force per unit area. If a force F is applied to an
area A and if the force is uniformly distributed
over the area,
then the pressure P =F/A in N/m2
Here, Force in Newtons (N) and Area in Square
Metre (m2).
Volume
-is a property associated with cubic measure(ft3
or m3).
If the volume of a substance increases or
decreases then the substance is called expanded
or compressed. Symbol V

10. Phase
When a substance is of the same nature
throughout its mass, it is said to be in a phase.
Matter can exist in three phases; solid, liquid and
vapour or gas.
If the matter exists is only one of these forms
then it is in a single phase.
If two phases exist together then the substance is
in the form of a two-phase mixture.
Examples: when a solid is being melted into a
liquid or when a liquid is being transformed into a
vapour.

11. Process
When the state of a substance is changed by
means of an operation is called process.
Processes are the expansion and compression
of a gas or the conversion of water into steam.
Cycle
Any process or series of processes whose end
states are identical is termed a cycle.
if processes are carried out on a substance such
that, at the end, the substance is returned to its
original state, then the substance is said to have
been taken through a cycle.

12. The constant temperature process
This is process carried out such that the temperature
remains constant throughout the process. It is often
referred to as an isothermalprocess.
The constant pressure process
This is a process carried out such that the pressure
remains constant throughout the process. It is often
referred to as anisobaric or isopiestic process.
The constant volume process
This is a process carried out such that the volume
remains constant throughout the process. it is often
referred to as an isometric or isochoric process.

13. Energy
Energy is defined as that capacity a body or
substance possesses which can result in the
performance of work.
Work is the application of a force over a
distance. Thus W= Fxd.
Here, W= work
F= Force
d= distance

14. Work and Pressure- volume diagram
In Figure1.3, the diagram is shown a cylinder in
which a fluid at pressure P is trapped using a
piston of area A. The fluid here is the system.
From this
Force on piston = pressure x area = PA-------(1)
Let this force PA be just sufficient to overcome
some external load.
Now let the piston move back a distance L
along the cylinder, while at the same time the
pressure of the fluid remains constant. The
force on the piston will have remained
constant.
Work done= force×distance=PA×L------------(2)
Rearranging the above equation
Work done=P×AL Fig.1.3 work and P-V diagram

15. But
AL= volume swept out by the piston, called the swept or stroke
volume
= (V2-V1)
So, work done=P(V2-V1)------------------------------(3)
The above P-V graph appears as horizontal straight line ab
whose height is at pressure P and whose length is from original
volume V1 to final volume V2.
Now consider the area abcd under this graph
Area=P(V2-V1)--------------------------------------------(4)
This equation is same as equ. 3. So the area under a P-V
diagram gives the work done.
Now, if pressure in N/m2 and volume in m3 then work done=
N/m2 × m3 =Nm=Joule--------------------------------(5)

16. Consider Fig. 1.4, It is now a
curve with original pressure
and volume P1 and V1
respectively. The final
pressure and volume are P1
and V2, respectively. Both
the pressure and the
volume have changed in
this case.
Fig.1.4 Usual P-V diagram

17. Fig.1.5 Area division on P-V diagram Fig.1.6 Area calculation on P-V diagram

18. Consider Fig. 1.5 and 1.6. This is the same P-V diagram.
Consider some point in the expansion, X say, where the pressure is P and the
volume is V. Let there be an elemental expansion δV from this volume V.
Then work done during elemental expansion=PδV----------------------(6)
The total work done will be obtained by summing all the elemental strips of
width δV from the volume V1 to volume V2 .
Therefore, Total work done=
Now, if an infinite number of strips are taken, in which case δV becomes
infinitely small (this is written as δV 0), then
Work done = ------------------------------------------------------- ---(7)

19. Heat
During an energy transfer process which results from the
temperature difference between one body and another, the
energy so transferred is called heat.
Also heat is transient quantity; it describes the energy transfer
process through a system boundary resulting from temperature
different. If there is no temperature difference, there is no heat
transfer.
Heat energy is given the symbol Q. To indicate a rate of heat
transfer, a dot placed over the symbol, thus
Q= heat transfer/ unit time

20. Specific heat capacity
For unit mass of a particular substance at a temperature t, let there be a change
temperature δt brought about by a transfer of heat δQ.
The specific heat capacity,c of the substance at temperature t is defined by the
ratio δQlδt.
Thus, c= δQlδt-------------------------------1
In the limit, as δt — 0, then
c= dQldt--------------------------------------2
It is common practice to use an average value of specific heat capacity
within a temperature range.
this average value is then used as being constant within the temperature range,
so equation (2) can be rewritten
c=Q/Δt----------------------------------------3
Where, Q= heat transfer/ unit mass, J/kg
Δt= change in temperature, k
Unit: joules/kilogram kelvin or J/kg k or KJ/kg k

22. Example 1.5: 5 kg of steel, specific heat capacity 450 Jlkg K, is
heated from 15˚Cto 100 ˚C. Determine the heat transfer.
Example 1.6: A copper vessel of mass 2 kg contains 6 kg of
water. If the initial temperature of the vessel plus water is 20
°C and the final temperature is 90 °C, how much heat is
transferred accomplish this change, assuming there is no
heat loss? Take the specific heat capacity of water to be 4.19
kJ/kg K.
Example 1.7: An iron casting of mass 10 kg has an original
temperature of 200 ˚C. If the casting loses heat to the value
715.5 KJ, determine the final temperature.
Example 1.8: A liquid of 4 kg has its temperature increased
from 15 °C to 100 °C. Heat transfer into the liquid to the value
714 kJ is required to accomplishthe increase in temperature.
Determine the specific heat capacity of the liquid.

23. The principle of the thermodynamic engine
The thermodynamic engine is a device
in which energy is supplied in the form
of heat and some of this energy is
transformed into work.
It would be ideal if all the energy
supplied was transformed into work.
The usual process in the engine can be
followed by Fig. 1.12

24. With all engines there must be a source of supply of heat
and, with any quantity heat Q supplied from the source to
the engine, an amount Q will successfully converted into
work, W. This will leave a quantity of heat (Q-W)to be
rejected by engine into the sink.
The ratio
W/Q= Work done/Heat received , is called the thermal
efficiency.---------------------1
The process in the engine is that of receiving heat, converting
some of it into work and then rejecting the remainder. So it
appears that, neglecting losses, the difference between the
heat received and the heat rejected is equal to the work
done.
Heat received-Heat rejected=Work done----------------2

25. Now Thermal ɳ= Work done/Heat received-----------------3
Using equation 2 & 3
Thermal ɳ= (Heat received-Heat rejected)/Heat received
-------------------------------4
=1-(Heat rejected/Heat received)-----------------------------5
From equation 3
Work done=Heat received×Thermal ɳ----------------------6
Heat received=Work done/Thermal ɳ------------------------7
From equation 4
Thermal ɳ=(Heat received-Heat rejected)/Heat received
Heat received×Thermal ɳ= Heat received-Heat rejected
From which
Heat rejected=Heat received-(Heat received×Thermal ɳ)
= (1-Thermal ɳ)×Heat received-----------------------------8

26. Equations 1 to 8 will apply to a system in which heat and work
transfer across the system boundary, a heat engine in fact.
For other systems, such as IC engine the thermal efficiency
may be defined as
Thermal ɳ= Work done/ Energy received---------------------9
Calorific value: is defined as the amount of energy liberated
by burning unit mass or volume of the fuel.
Example- By burning 1 kg of petrol, 43 MJ/kg are liberated,
then the calorific value of the petrol is 43MJ/kg.

27. Example 1.9: A petrol engine uses 20.4kg of petrol per hour
of calorific value 43 MJ/kg. The thermal efficiency of the
engine is 20 percent. Determine the power output of the
engine and the energy rejected/min.
Example 1.10: A steam plant uses 3.045 tonne of coal per
hour. The steam is fed to a turbine whose output is 4.1 MW.
The calorific value of the coal is 28 MJ/kg. Determine the
thermal efficiency of the plant.