Unit 2: BASIC MECHANICAL ENGINEERING by varun pratap singh

BASIC MECHANICAL ENGINEERING
UNIT-2
THERMODYNAMICS
NOTES
By
Mr. Varun Pratap Singh
Assistant Professor
Department of Mechanical Engineering
College of Engineering Roorkee

UNIT-2:
Zeroth law: Zeroth law, Different temperature scales and temperature measurement
First law: First law of thermodynamics. Processes - flow and non-flow, Control volume, Flow
work and non-flow work, Steady flow energy equation, Unsteady flow systems and their
analysis.
Second law: Limitations of first law of thermodynamics, Essence of second law, Thermal
reservoir, Heat engines. COP of heat pump and refrigerator. Statements of second law and their
equivalence, Carnot cycle, Carnot theorem, Thermodynamic temperature scale, Clausius
inequality. Concept of entropy.
Video Reference:
1. YouTube Channel: BEST MECHANICAL ENGINEERING
https://www.youtube.com/channel/UC4vN8jWyjDlyRfryqQ3izsQ/featured
2. YouTube Channel: All About Mechanical Engineering
https://www.youtube.com/channel/UCaI6gazNIAsclpelpAuCynw
3. YouTube Channel: Yantriki: The power of Machines
https://www.youtube.com/channel/UC4YmXloYijid3FKvLoXMDCA
4. YouTube Channel: LEARN AND GROW
https://www.youtube.com/channel/UCCqGTvGZgWw8mFX5KYTHCkw
5. YouTube Channel: Learn Engineering
https://www.youtube.com/c/LearnEngineering/featured
6. YouTube Channel: Mech Learner
https://www.youtube.com/channel/UChg4N8Rkxqy49cd1c91fwoQ
7. YouTube Channel: EnggOnline
https://www.youtube.com/channel/UCuKdhLcpIlsy5yKCMxDXJiQ
8. YouTube Channel: Fizică
https://www.youtube.com/channel/UCw1H787zqIj9ZtTQK2eat3Q/featured
9. YouTube Channel: Khan Academy
https://www.youtube.com/c/khanacademy/featured
10. YouTube Channel: Professor Dave Explains
https://www.youtube.com/c/ProfessorDaveExplains/featured

Laws of Thermodynamics
The science of thermodynamics is based on the fourth law of thermodynamics namely:
1. Zeroth law of thermodynamics
2. First law of thermodynamics
3. Second law of thermodynamics
4. Third law of thermodynamics
These all laws are deducted from experimental observations and are based on logical
reasoning. There are no mathematical proof for these laws.
Equality of Temperature
https://www.youtube.com/watch?v=tEwqhdmpLus&list=PLdoIhVhbPQV6nSJ-jhlsJ9b6gwfd4tsGx&index=15
Any two bodies are said to have 'equality of temperature' when no change in any observable
property occurs when they are brought in for thermal communication. The equality of
temperature is also termed 'thermal equilibrium'.
Zeroth law of thermodynamics:
https://www.youtube.com/watch?v=Xkna_fMzer0&list=PLdoIhVhbPQV6nSJ-jhlsJ9b6gwfd4tsGx&index=16
It state that, “if two systems are each in thermal equilibrium with third system, then they are
also in thermal equilibrium with each other,"
First law of thermodynamics:
Joule's Experiment - First Law of Thermodynamic
https://www.youtube.com/watch?v=VqUAhrrW6UA&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV
First law of thermodynamics is based on law of conservation of energy, which state that
"energy can neither be created nor destroyed but it can only be transferred from one form to
another form of energy".
Statement of the first law of thermodynamics for the closed system
undergoing a cycle.
" when a system undergoes a thermodynamic close and reversible cyclic process then the
net heat supplied to the system from the surrounding is equal to net work done be the system
on its surrounding".
Its mean "for a closed, and reversible cyclic process integration of energy will always be
equal to the integration of work.

A closed system can exchange energy with its surroundings through heat and work transfer.
In other words, work and heat are the forms that energy can be transferred across the system
boundary.
Both heat transfer and work transfer may cause the same effect on a system. They both are
different forms of energy in transit. Energy that enters a system as heat may leave as work or
vice versa.
(∑W) cycle = J(∑Q) cycle
or, ∮ δW = ∮ δQ
Where, J is called Joule’s equivalent. When heat and work both are measured in same unit,
value of J will be 1.
∮ δW = ∮ δQ
Sign convention The work done by a system on the surroundings is treated as a positive
quantity.
Similarly, energy transfer as heat to the system from the surroundings is assigned a positive
sign. With the sign convention one can write,
∫dQ = ∫dW
Corollary
A device or machine is impossible, which can produce work continuously without absorbing
energy from its surroundings.

Consequences of the first law:
Suppose a system is taken from state 1 to state 2 by the path 1-a-2 and is restored to the initial
state by the path 2-b-1, then the system has undergone a cyclic process 1-a-2-b-1. If the system
is restored to the initial state by path 2-c-1, then the system has undergone the cyclic change 1-
a-2-c-1. Let us apply the first law of thermodynamics to the cyclic processes 1-a-2-b-1 and 1-
a-2-c-1 to obtain
∫1-a-2dQ+ ∫2-b-1dQ - ∫1-a-2dW - ∫2-b-1dW =0
∫1-a-2dQ+ ∫2-c-1dQ - ∫1-a-2dW - ∫2-c-1dW=0
Subtracting, we get
∫2b1dQ- ∫2c1dQ – (∫2b1dW - ∫2c1dW) =0
We know that the work is a path function and hence the term in the bracket is non-zero. Hence
we find
∫2b1dQ = ∫2c1dQ
That is heat is also a path function.
Energy is a property of the system:
https://www.youtube.com/watch?v=NgvCSQB0XcY&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=6
By rearranging we can have
∫2b1 (dQ - dW) = ∫2c1 (dQ - dW)
It shows that the integral is the same for the paths 2-b-1 and 2-c-1, connecting the states 2 and
1. That is, the quantity ∫ (dQ - dW) does not depend on the path followed by a system, but
depends only on the initial and the final states of the system. That is ∫ (dQ - dW) is an exact
differential of a property. This property is called energy (E). It is given by
dE = dQ-dW
E = KE + PE +U
where U is the internal energy. Therefore,
dE = d(KE) + d(PE) + dU = dQ-dW
Quit often in many situations the KE or PE changes are negligible.
dU = dQ – dW
An isolated system does not exchange energy with the surroundings in the form of work as well
as heat. Hence dQ = 0 and dW = 0. Then the first law of thermodynamics reduces to dE = 0 or
E2 = E1 that is energy of an isolated system remains constant.

First law for a closed system undergoing a change of state
https://www.youtube.com/watch?v=COuWZTMj-8o&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=5
Consider a system undergoing a cycle, in which it changes from state 1 to state 2 through a
process A and returns from state 2 to state 1 through process
Now for a cycle, we have
For a cycle made up of two process A and B,
Now consider another cycle using process C and B, then
Similarly,
Now subtracting the second from the first equation, we get
On rearranging,
Since, A and C represents arbitrary process between states 1 and 2, the quantity is
same for all processes between states 1 and 2. Therefore, depends upon only
initial and final states, and not on the path followed, so it is a point function and differential
of the property of the mass. This property is the Energy of the mass, E.
On integration we get,
1Q2 = E2-E1 + 1W2.
This is the general expression of first law for a process.
The property E represents all the energy contained by the system at given state. If we separate
the bulk kinetic energy and potential energy for the property E, remaining all the energy is
called the internal energy U,
So, E = U + KE + PE
In differential form,
dE = dU + d (KE) + d(PE)
So, first law for change of state may be written as
dQ = dU + d(KE) + d(PE) + dW

Internal energy is a property of the system
https://www.youtube.com/watch?v=NgvCSQB0XcY
[BEST MECHANICAL ENGINEERING]
Internal Energy (U)
U is an extensive property and so is KE and PE. Whereas, KE and PE are the stored macroscopic
energy, U is the microscopic energy composed of translational, rotational, vibrational,
chemical, electronic, nuclear, etc. energy possessed by the molecules of a substance.
It has already been discussed in the first chapter that the internal energy is a property of the
system and depends on temperature only. From the first law of thermodynamics as discussed
in previous article
Q = W + dU
If there is a case when heat is supplied to a fix volume of gas (thermodynamic system) confined
in fix boundary of the system i.e. there is no change in volume of gas during supplying of heat,
then there will not be any work. So
W = 0 and dU = Q
As Q is heat supplied to gas keeping it at constant volume, it can be calculated as mCvdt. Thus
change in internal energy is heat exchanged at constant volume and can be calculated as
Change in internal energy per unit mass i.e.
Thus internal energy is a property derived from first law of thermodynamics. It accounts for
the difference between heat exchange and work exchange and shows that energy cannot be
created nor be destroyed. Change in specific internal energy of a system is always calculated
as Cvdt i.e. as a function of temperature change.
Enthalpy (H)
It is also a property of thermodynamic system which is calculated in terms of other properties.
It is defined as the sum of internal energy and product of pressure and volume of a
thermodynamic system. Thus it is a calculated property and loosely defined as total heat content
of the system. It is denoted by H.

H = U + PV
And dH = dU + d (PV)
dH = dU + P.dV + V. dP
First law analysis for a control volume
A control volume is a volume in space in which one has interest for a particular study or
analysis. Mass, heat and work can cross the control surface and the mass and its properties can
change with time in the control volume.
Conservation of mass in control volume
The rate of change of mass inside a control volume can be different from zero if we add or take
a flow of mass out as
Rate of change =( +)in and (–) out
For several possible flows,
This equation is termed as continuity equation.

Non-Flow Reversible Processes (Heating/Cooling and Expansion/
Compression of Gases)
Non Flow Process and Flow Process
https://www.youtube.com/watch?v=DDmhcMLU6b8&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=13
Until now we have studied that thermodynamic system of a gas is used to convert heat energy
into work energy or vice versa which is required in a number of practical applications.
Thermodynamic system is also of many types. Here we will study a closed system which can
exchange energy with surroundings but not the mass. Thus mass does not flow in or out of the
system and so the processes of heating/cooling/compression/expansion etc are undergone by
this fix mass of gas confined in continuous closed boundary are called Non-flow processes. In
these processes some property of the gas may change and some may not change based on which
a particular process is characterized. By applying first law of thermodynamics, the various
forms of energy exchange can be calculated considering the processes as reversible i.e. taking
all internal/external losses due to friction etc as nil.
For this let us consider a fix quantity of an ideal gas filled in a metallic cylinder as shown in
below mentioned, one side of which is covered or fixed by a solid end plate and other side is
covered by a moving piston. The outer wall of piston matches with inner wall of cylinder such
that it makes a leak proof sliding joint. This moving piston makes one of the boundaries of
system as moving or flexible.
Fig.: Closed system of gas filled in metallic cylinder
When infinitely small heat 𝛅Q is supplied to gas from outside through the wall of cylinder, the
gas tends to expand and forces the piston weight F to move up. Let piston moves by a short
distance, dl. Then the infinitesimal work done can be calculated as
= Pressure x Change in volume
Total work during a non-flow process 1-2 i.e. compression or expansion of gases can be
calculated as taking integral of P.dV

All these reversible processes of heat exchange (Heating or Cooling) and work exchange
(compression or expansion) by a system of ideal gas with its surroundings can take place in
various ways as discussed below:
Constant volume heating/cooling process
Let the piston is fixed at one point in the walls of cylinder such that volume V = Constant
So the ideal gas law PV/T = Constant will reduce to
Now let a small increment of heat 𝛅Q is supplied to the gas. As per 1st
law of thermodynamics
As V = Constant, 𝛅W = P.dV = 0
Thus knowing the values of 𝛅Q and Cv, ∆T (Increase in temperature) can be calculated and
so ∆P can also be calculated from above equeation. By measuring the small increment in
temperature and pressure of system, the quantity 𝛅Q & ∆U can also be calculated.
Constant pressure process
Let the piston is free to move in the cylinder and force, F or pressure, P on the piston remains
constant. So now the boundaries of the system can move and so the gas can expand or can be
compressed i.e. the system is able to exchange work.
Now let a small quantity of heat, 𝛅Q is supplied to the gas through the walls of cylinder.
On absorbing this heat, the temperature of gas will tend to increase and simultaneously the gas
will tend to expand against force F on the piston. Applying 1st
law of thermodynamic to this
process.

Or δQ = P.dV + mCvdT (Eq. 4.7)
or Cp = R + Cv
or Cp − Cv = R
Thus characteristic gas constant of an ideal gas is the difference between specific heat at
constant pressure (Cp) and at constant volume (Cv).
The ratio of specific heats Cp/Cv is denoted as γ.
The physical meaning of R or γ can be taken as the characteristic of an ideal gas to expand
under the influence of heat or we can say the increase in product of pressure and volume PV
with increase in T.
Also by ideal gas law,
As pressure, P = Constant, the gas law reduces to V/T = Constant
Thus the increase in volume of gas due to expansion and increase in temperature are interrelated
and if one can be measured the other can be found and the quantities W, ∆U and so Q can be
calculated from equation 4.7 as
Constant temperature process (Isothermal Process)
The beauty of isothermal expansion or compression process is that the internal energy remains
constant and so whole of the heat exchange by system is converted to work and vice versa.
However, the process is difficult to visualize while thinking that when the system (gas filled
behind piston in the cylinder) absorbs heat, its temperature should always increase. But what
happens in the isothermal process that while absorbing heat, simultaneously the gas expands
thus decrease in its pressure and temperature takes place and the net change in temperature is
zero. While expanding, the gas gives positive work equal to the heat supplied. Conversely also
if the gas is compressed by doing extra work on the gas, its pressure and temperature tends to
increase, but simultaneously if the gas is cooled in such proportion that the net change in
temperature remains zero, the compression process becomes isothermal. In this isothermal
compression process, again the work supplied to the gas is given away by the system in the
form of heat.

Applying first law of thermodynamics to this isothermal process in
which dU=0, because dT=0,
𝛅Q = 𝛅W = P.dV
From gas law
At T= constant
Gas law reduces to PV = constant or P1V1 = P2V2 = PV
Putting this in equation 4.8
By using this equation, we can calculate the work exchange or heat exchange during isothermal
process if we know the change in volume.
Also in case of Isothermal process we know that
P1V1 = P2V2
Putting in eq. (4.9)
By using this equation, we can calculate the work exchange or heat exchange if we know the
change in pressure.
Adiabatic process
Adiabatic process is one in which there is no exchange of heat between system and
surroundings
i.e. Q= 0.
Applying first law of thermodynamics to this process
0 = δW + dU or δW = -dU
So, dU = −δW = −P.dV
Or in specific terms du = - δw = - P.dv
Also from the definition of enthalpy, the change in specific enthalpy is

Putting above eq. in each othr
dh = vdP
Thus in an adiabatic process
And dh = CpdT = v.dP
Dividing above eq
or
or
Integrating on both sides
or
or
=Constant
or
or
This represents a reversible adiabatic process: 1-2
From gas law
or
or
Putting the value of P1/P2 from equation

Thus the overall relation between initial and final properties (Pressure, volume and
temperature) in an adiabatic process 1-2 is
Work exchange
We know that work exchange during a non-flow process is given as
W=
Thus work during non-flow reversible adiabatic process 1-2 is
Specific work i.e. work per unit mass will be
If work exchange comes as positive that means gas is doing work on its surrounding during the
process and if it comes as negative, the work is being done on the gas.
Also in the reversible adiabatic process 1-2, Q = 0 (no heat exchange) so as per first law of
thermodynamics:
∆U = -W
Or
Polytrophic process

In real practice it is found that an ideal gas while undergoing a non-flow process which may be
any one or combination of two of the heating / cooling and compression / expansion processes,
follows the law.
PVn
= Constant
Where n is known as index of compression or expansion.
It is a general form of any non-flow process and the value of n decides the particular type of
process. For example
If n = 0 ↔ then PV0
= constant Or P = constant
↔ constant pressure process.
If n = ∞ then PV∞
= constant
or P1/∞
.V = constant or P0
V = constant or V = constant
↔ constant volume process
If n = 1 then PV1
= PV = Constant.
Mixing it with ideal gas law
If PV = Constant, then T = constant
⟷ constant temperature process
If n = γ then P = Constant
↔ adiabatic process
If n has any other value except 0, 1, γ and ∞
↔ polytrophic process
Depending on the value of n, all these processes can be represented on the PV- diagram as
follows.
Combining the polytrophic process law PVn
= Constant with the ideal gas law

the relation between initial and final properties (pressure, volume and temperature) in a
polytrophic process1-2 can be derived as
Heat / work exchange
The difference in mathematical law governing an adiabatic and polytrophic process is only of
γ and n. So work exchange in a polytrophic process can also be similarly derived as in case of
adiabatic process and it will be
And specific work,
Applying first law of thermodynamics to the polytrophic process 1-2
Q = W + ∆U

First law of thermodynamics for control volume
For control mass, we have the first law as,
as rate equation, we have,
Consider a control volume that involves rate of heat transfer, rates of work transfer and mass
flows. Since we cannot create or destroy energy, so any rate of change of energy must be caused
by rates of energy in or out of the control volume.
Since heat and work transfer are already included we need explanation about the energy
associated with the mass flow rates.
The fluid flowing across the control surface (which envelops the control volume) enters or
leaves with an amount of energy per unit mass as,
Relating to the state and position of the fluid.
Whenever a fluid mass enters a control volume at state i, or exits at state e, there is a boundary
movement work associated with that process. Fluid mass enters the control volume as it is
pushed by the surrounding against the local pressure with a velocity, giving the control volume
a rate of work in the process.
So, the flow work,
So, the flow work per unit mass is Pv and total ener4gy associated with the flow of mass is,
So, first law of thermodynamics for control volume becomes,
For, general control volume we may have several (n) entering or (m) leaving mass flow rates
so, for that case, the final form or the first law becomes,
Steady state steady flow process
https://www.youtube.com/watch?v=chqkWcP51R4&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=14

Control Volume
A control volume may involve one or more forms of work at the same time. If the boundary of
the control volume is stationary, the moving boundary work is zero, and the work terms
involved are shaft work and electric work. Another work form with the fluid is flow work.
Flow Work
Work is needed to push the fluid into or out of the boundaries of a control volume if mass flow
is involved. This work is called the flow work (flow energy). Flow work is necessary for
maintaining a continuous flow through a control volume.
Consider a fluid element of volume V, pressure P, and cross-sectional area A as shown left.
The flow immediately upstream will force this fluid element to enter the control volume, and
it can be regarded as an imaginary piston. The force applied on the fluid element by the
imaginary piston is:
F = PA
The work done due to pushing the entire fluid element across the boundary into the control
volume is
Wflow = FL = PAL = PV
For unit mass,
wflow = Pv
The work done due to pushing the fluid element out of the control volume is the same as the
work needed to push the fluid element into the control volume.

STEADY FLOW ENERGY EQUATION
https://www.youtube.com/watch?v=6VA6A_qJjCs&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=15
The system through which the mass flow rate is constant i.e.
Mass Input = Mass Output
The state of working substance at any point in this system remains constant. Examples: Most
actual thermodynamic equipment work as steady flow system under steady state conditions.
Examples are IC engine, Compressors, steam turbines etc.
Analysis of steady flow system
Under steady state, Total Inlet Energy = Total Outlet Energy
Now Total Energy = I.E. + F.E + K.E + P.E
Where
I.E. = Internal Energy = mu
F.E. = Flow Energy = mPv
K.E. = Kinetic Energy = 1/2 mV2
P.E. = Potential Energy = mgZ
So,
In steady flow system the rate of mass flow is constant.
i.e. Mass Input = Mass output
Or m1 = m2
Figure: Open system
Also let the height of working substance at any point in the system remains constant. i.e.
z1 = z2.
And the kinetic energy is usually so small w.r.t. heat, work & enthalpy term, it can also be
neglected.
Also taking U1 + P1V1 = h1 (enthalpy)
H1 + Q = H2 + W

Or Q = dH + W
Or dH = δQ - δW
Or dU + P.dV + V.dP = dU + P.dV−dW
Or V.dP = −dW
Here we establish the first law for control volume for the long-term steady operation of devices
like turbines, compressors, nozzles, boilers and condensers.
SSSF Assumptions
1. The control volume doesn’t move relative to the co-ordinate frame. (No work
associated with the acceleration of the control volume.)
2. The state of mass at each point in the control volume doesn’t vary with time. (This
implies,
3. The mass flux and its state remain constant with time at the inlets and outlets.
4. The rate of heat and work transfer across the system boundary remains constant.
So, according to the assumptions the first law of SSSF devices will be,
For models having only one inlet and outlet, the continuity equation becomes,
And the first law will be,
or,

First law applied to SSSF devices:
1. Heat Exchanger
Theory:
https://www.youtube.com/watch?v=A8SLR33Xe0Q&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=31
Numerical:
https://www.youtube.com/watch?v=fJ0X0BX9dYU&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=33
It is a Device in which heat from one flowing fluid is transferred to another flowing fluid
inside a control volume.
In the ideal heat exchanger
1. The fluids flowing will have very low pressure drop
2. There is no means of doing any shaft, electrical etc work through the control volume.
3. Change in K.E. and P.E. of the fluids is very small.
4. No heat transfers to or from the surrounding.
So, for the heat exchanger the continuity equation will be,
And the energy equation will be,
2. Nozzle:
https://www.youtube.com/watch?v=AFhAfaJY3ac&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=16
Numerical:
https://www.youtube.com/watch?v=587-rus_0UQ&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=17
It is a device to generate a high velocity fluid stream at the expense of its pressure.
In an ideal nozzle,
1. The fluid pressure will be dropped and velocity increases significantly.
2. Neither Work nor Heat crosses the system boundary.
3. No change in P.E. of the fluid takes place.
4. K.E. at the inlet is usually small and can be neglected.
So, the continuity equation will be,

And the energy equation will be,
3. Diffuser:
Theory:
https://www.youtube.com/watch?v=noodV9cbHbw&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=20
It is a device constructed to decelerate a high velocity fluid in a manner that results in a increase
in pressure of the fluid (opposite to Nozzle). So, its continuity equation and the energy equation
will be same as of the Nozzle.
4. Throttle
Theory: https://www.youtube.com/watch?v=5nIwX5zR6Wo&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=34
It is a device in which a fluid flowing in a line suddenly encounters a restriction in the flow
passage (may be plate with a hole, capillary tube).
In an ideal throttling device,
1. Abrupt pressure drop in the fluid occurs
2. Some increase in velocity will be there but can be neglected because of its very low
value.
3. No work, no change in P.E. no heat transfers to or from the surrounding.
So, the first law reduces to
hi= he
So, throttling is a constant enthalpy process.
5. Turbine
Water Turbine:
Theory: https://www.youtube.com/watch?v=_bGs7-VVOPA&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=23
Numerical: https://www.youtube.com/watch?v=6CINDY-X3uc&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=25
https://www.youtube.com/watch?v=ncXfTFoa-Ao&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=26
Gas Turbine:
Theory: https://www.youtube.com/watch?v=kDLZpbCHA3A&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=24
Numerical: https://www.youtube.com/watch?v=6CINDY-X3uc&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=25
It is a rotary SSSF Machine whose purpose is the production of shaft work in expense of
pressure of the working fluid.

In an ideal turbine
1. Change in P.E. and K.E. of the flow are negligible.
2. Heat rejection form the turbine is negligibly small and is undesirable
3. The turbine process is assumed to be adiabatic
Hence, the first law reduces to
6. Compressor/ Centrifugal Pump
Compressor
Theory: https://www.youtube.com/watch?v=pPnIO5txsU8&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=29
Numerical: https://www.youtube.com/watch?v=T_TzctEOcuc&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=30
Centrifugal Pump
Theory: https://www.youtube.com/watch?v=VySFOAUZVz8&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=27
Numerical: https://www.youtube.com/watch?v=TZve9CCcQtw&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=28
It can be a reciprocating or a rotary device whose purpose it to increase the pressure of the fluid
by putting in some shaft work through external means. Devices fulfilling this purpose for liquid
are termed Pumps and those for gases are termed Compressors. Theoretically, these devices
work exactly in opposite to that of turbine. So,
1. Change in P.E. and K.E. of the flow are negligible.
2. Heat rejection form the turbine is negligibly small and is undesirable
3. The turbine process is assumed to be adiabatic
Hence, the first law reduces to

7. Boiler
Theory: https://www.youtube.com/watch?v=OJwG-qPQqkI&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=21
Numerical: https://www.youtube.com/watch?v=NBC2hGzBOZw&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=19
https://www.youtube.com/watch?v=MlY0_0NJa9M&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=22
It is a device whose purpose is to increase the pressure of a fluid by putting heat through some
external means.
In an ideal boiler
1. Change in K.E. and P.E. of the flow are negligible.
2. There will be no work interaction with the surrounding.
Hence, the first law for an ideal boiler will be,

Uniform State Uniform Flow Process (Not in course; Just for knowledge)
This USUF processes occurs in the devices in which there is provision for change of state of
mass of the working fluid. An ideal bottle filling process is an example of USUF process.
Assumptions
1. The control volume remains constant relative to the co-ordinate frame.
2. The state of the mass within the control volume may change with time, but at any
instant of time the state is uniform throughout the entire control volume.
3. The state of the mass crossing each of the areas of flow on the control surface is
constant with time although the mass flow rates may be time varying.
The overall USUF process occurs within a span of time t.
Continuity equation at any instant of time is;
Now integrating over time t, we get,
This is the continuity equation for USUF process.
Similarly,
We have Energy Equation at any instant of time;
Since, in USUF process, state within the control volume is uniform at any instant of time,
we can write,
Now integrating over time t, we get
This is the first law of Uniform State Uniform Flow process for period of time t.

Perpetual Motion Machine of the First kind (PMM1)
https://www.youtube.com/watch?v=BzrSKFq8bkI&list=PLdoIhVhbPQV6aXEcTFqhPxvo8Wh7zkuTV&index=12
A perpetual motion machine of the first kind produces work without the input of energy. It thus
violates the first law of thermodynamics: the law of conservation of energy. A perpetual motion
machine of the second kind is a machine which spontaneously converts thermal energy into
mechanical work.
Limitation of the first law of thermodynamics
https://www.youtube.com/watch?v=shbsgsRKM8w
The limitation of the first law of thermodynamics is that it does not say anything about the
direction of flow of heat. It does not say anything whether the process is a spontaneous
process or not with what amount conversion will take place. The reverse process is not
possible. In actual practice, the heat doesn't convert completely into work

SECOND LAW OF THERMODYNAMICS, ENTROPY,
CARNOT CYCLE
Introduction
We studied that heat and work are different forms of energy and are convertible in to each
other. During this conversion, law of conservation of energy i.e. first law of thermodynamics
is followed. But this law has limitation in depicting the fraction of heat energy of a system or
supplied to system which can be converted to work. Also it does not specify the conditions
under which conversion of heat in to work is possible. Second law of thermodynamics removes
this limitation and tells under what conditions, in what direction of heat flow and how much of
it can maximum be converted in to work.
Thermal Reservoir
https://www.youtube.com/watch?v=OTo-U90m97I
A thermal reservoir is a large system (very high mass x specific heat value) from which a
quantity of energy can be absorbed or added as heat without changing its temperature. The
atmosphere and sea are examples of thermal reservoirs.
Any physical body whose thermal energy capacity is large relative to the amount of energy it
supplies or absorbs can be modeled as a thermal reservoir.
A reservoir that supplies energy in the form of heat is called a source and one that absorbs
energy in the form of heat is called a sink.
Heat Engine
https://www.youtube.com/watch?v=4skt_FJVgpQ&t=129s
It is a cyclically operating device which absorbs energy as heat from a high temperature
reservoir, converts part of the energy into work and rejects the rest of the energy as heat to a
thermal reservoir at low temperature.
The working fluid is a substance, which absorbs energy as heat from a source, and rejects
energy as heat to a sink.
Thermal Power Plant
Q1 = Heat received from hot gases
WT = Shaft work by turbine
Q2 = Heat rejected to cooling water in condenser
WP = Work done on the pump
Wnet=WT-WP
W = Q1 - Q2
Thermal Efficiency,

Reversible Heat Engine
Every reversible heat engine operating between the same two temperature reservoirs have
identical efficiency. This means no matter how a reversible heat engine is constructed or what
the working fluid is, its efficiency is the same as all other heat engines working from the same
two temperatures.
Refrigerator
Refrigerator is a reversed heat engine which either cool or maintain the temperature of a body
(T1) lower than the atmospheric temperature (Ta). This is done by extracting the Heat from a
cold body and delivering it to a hot body (Q2). In doing so, work WR is required to be done on
the system. According to First law of thermodynamics,
WR = Q2 – Q1
The performance of a refrigerator is expressed by the ratio of amount of heat taken from the
cold body (Q1) to the amount of work required to be done on the system (WR). This ratio is
called coefficient of performance. Mathematically, coefficient of performance of a refrigerator,
T1 < Ta
Heat Pump
A refrigerator used for cooling in summer can be used as a heat pump for heating in winter. In
the similar way, as discussed for refrigerator, we have
Wp =- Q2 – Q1
The performance of a heat pump is expressed by the ratio of the amount of the heat delivered
to the hot body (Q2) to the amount of work required to be done on the system (Wp). This ratio
is called coefficient of performance or energy performance ratio (E.P.R.) of a heat pump.
Mathematically, coefficient of performance or energy performance ratio of a heat pump,

From above we see that the C.O.P. may be less than one or greater than one depending on the
type of refrigeration system used. But the C.O.P. of a heat pump is always greater than one.
Statements of Second Law of Thermodynamics
https://www.youtube.com/watch?v=WTtxlaeC9PY
Kelvin-Planck statement
https://www.youtube.com/watch?v=3crTTPyG-GQ
It is impossible to construct a thermodynamic system or device which operates in a cycle and
produce no effect other than the production of work by exchange of heat with a single
reservoir. Or in simple terms it states that all the heat from a single heat reservoir cannot be
converted to work.

In detail, the meaning of statement is that there is no such device possible which can
continuously take heat from heat reservoir on one side and convert all of it into work on the
other side. But only a part of heat energy while flowing from high temperature reservoir to low
temperature reservoir can be converted to work and the remaining part must be rejected to low
temperature reservoir i.e. atmosphere. Therefore, only a part of heat energy while in transition
from high temperature to low temperature is possible to be converted in to work.
Classius statement
https://www.youtube.com/watch?v=H5CSDpQCHLg
This statement is regarding the conversion of work in to heat and it states that (it is impossible
to construct a thermodynamic system or device which, while operating in cycle (i.e. working
continuously), transfers heat from low temperature reservoir to high temperature reservoir
without taking help or absorbing work from some external agency.)
In detail, the meaning of statement is that heat can be made to flow from low temperature to
high temperature only by applying external work.
Equivalence of Kelvin Planck Statement and Clausius Statement
https://www.youtube.com/watch?v=_z8vycD2oBQ
Heat Engine
It is a thermodynamic system or device which can continuously convert heat energy into work
energy or we can say thermal energy in to mechanical energy. We know that to work
continuously, anything has to operate in a cycle. Therefore, heat engine is also a thermodynamic
device operating in a cycle. The performance of a heat engine is measured in terms of its thermal
efficiency which is the ratio of work output to heat absorbed by engine, i.e.
Where W = Rate of mechanical work done by engine
Q = Heat absorbed by engine or rate of heat supplied to engine.
Reversible heat engine

A heat engine which operates through a reversible cycle is called Reversible Heat Engine. As
per second law of thermodynamics, heat engine absorbs heat Qh from a high temperature
source, converts a part of it into Mechanical work W and rejects the remaining part of heat Qc
to a low temperature heat sink as shown in below mentioned figure
If the heat engine is reversible i.e. all its processes are reversible, then it can be operated on
the reverse cycle in reverse direction with the same performance. Then it will start taking heat
Qc back from low temperature heat source by absorbing same amount of work W from some
external agency and reject the sum of heat absorbed and work absorbed in the form of heat Qh
to high temperature heat source. This reversed heat engine will be called heat pump as shown
in figure. If this heat pump is used for the purpose of extracting heat from a low temperature
body, it is called a refrigerator.
Corollary of 2nd
law of thermodynamics
https://www.youtube.com/watch?v=a7f9_F21DYA
No heat engine operating between two heat reservoirs, always operating at constant
temperature, can be more efficient than a reversible heat engine operating between the same
temperature limits. Also all types of reversible heat engines operating between same
temperature limits will have the same efficiency. It can also be proved with a simple logic. Let
us say, there is an irreversible engine having more efficiency than that of a reversible engine
operating between same temperature limits. Let irreversible engine produces work Wirr and
reversible engine produces work Wrev (such that Wirr>Wrev) by absorbing heat Qh from heat
source at temperature Th and rejecting heat Qc to heat reservoir at temperature Tc. Now if we
operate reversible engine in reverse direction like a heat pump taking work Wrev from the work
produced by irreversible engine and absorbing heat Qc back from heat reservoir at temperature
Tc and rejecting back heat Qh to heat reservoir at temperature Th as shown in fig 7.2. We will
find that a net positive work, W = Wirr - Wrev should be produced continuously without any
effect or any net heat exchanged with reservoirs which is completely opposite to the law of
conservation of energy i.e. energy cannot be produced without its expenditure.

Figure: Reversible and irreversible heat engines
In this way, our assumption that an irreversible engine is more efficient than a reversible engine
is totally wrong. Hence
Wrev > Wirrev
One more fact about a reversible heat engine is that it does not exist in reality. But the idea of
reversible heat engine is completely hypothetical in which the heat exchange process is thought
as reversible without any change in temperature. Otherwise heat exchanged in a medium is
irreversible and taken as Q = m.C.ΔT,
m = mass
C = Specific heat
ΔT = temperature change.
When some heat flows from a high temperature body to low temperature body, change in their
temperatures occurs i.e. hot body becomes somewhat cool and cool body becomes somewhat
hot, but now this heat cannot come back from cold body to hot body i.e. it is an irreversible
process.
So, with ΔT = 0 i.e. heat exchanged without change in temperature can only be visualized in a
way that the heat reservoirs and working medium in the reversible heat engine, which is
exchanging heat with heat reservoirs, both are of infinite heat capacity and there is no change
in their temperature and the amount of heat exchanged depends on the absolute temperature of
reservoirs at which heat is being exchanged i.e. Q ∝ T. So in a reversible heat exchange process
happening at constant temperature, heat exchanged is proportional to absolute temperature.
This is also named as CLASSIUS statement.

Carnot Cycle/Carnot Engine
https://www.youtube.com/watch?v=NasmLQOf30s
https://www.youtube.com/watch?v=aAfBSJObd6Y
Carnot suggested a reversible cycle comprising of two reversible isothermal heat exchange
processes and two reversible adiabatic expansion/compression processes as shown on P-V and
T-S charts in Figure.
Figure: Carnot cycle
Carnot Engine is the reversible heat engine working on Carnot cycle 1-2-3-4 as explained
below:
Process 1-2: Reversible isothermal heat addition: Heat, Qh is transferred to the working
substance from the high temperature reservoir at temperature Th = T1 =T2. The heat transfer is
reversible and isothermal. Expansion of gas takes places i.e. heat energy is converted to work
but the internal energy of system remains constant.
Process 2-3: Reversible adiabatic expansion: During the expansion process, the system is
thermally insulated so that Q = 0. The temperature of the working substance decreases from
high temperature, Th to low temperature, Tc = T3 = T4. Expansion of gas takes place at the
expense of its own internal energy.
Process 3-4: Reversible isothermal heat rejection: Heat Qc is transferred from the working
substance (gas) to low temperature heat reservoir (sink) at constant temperature Tc. Heat
transfer is isothermal & reversible. Gas is compressed by spending of external work and
equivalent heat to this work is rejected to heat sink. Internal energy remains constant.
Process 4-1: Reversible adiabatic compression: During the compression process, the system
is thermally insulated, so Q =0. Temperature of working substance increases from Tc to Th. So
internal energy of the system increases by an equal amount to the compression work done on
the system.
Carnot efficiency
An engine operating on the Carnot cycle has maximum efficiency.
Now Qh ∝ Th, Absolute temperature of hot reservoir
Qc ∝ Tc, Absolute temperature of cold reservoir
Thus,

Thus, the Carnot efficiency does not depend on the type of working substance but only on the
absolute temperature of hot & cold reservoirs.
Carnot Cycle
Carnot cycle is an ideal cycle as adopted for an ideal heat engine. It consists of two isothermal
process (expansion and compression) and two adiabatic process (expansion and compression).
The cylinder and piston of the engine are considered as perfect non-conductor of heat but the
cylinder cover head is a good conductor of heat. The hot body at a higher temperature is brought
in contact with the bottom 'B' of the cylinder. The cylinder is fitted with a weightless and a
frictionless piston.
The French engineer Nicolas Leonard Sadi Carnot was the first scientist who realize the
problem of the efficiency of heat engine and invented the Carnot cycle. The pressure-volume
(p-v) and temperature-entropy (T-S) graph are shown in the fig.
From the above two p-v and T-S graph, the horizontal axis represents volume 'v' and entropy
'S' and the vertical axis represents pressure 'p' and temperature 'T'. Let, engine cylinder contains
m kg of air at its original condition represented by point 1 on the p-v and T-S diagrams. At this
point, let p1,v1 and T1 be the pressure, temperature and volume.
First Stage (Isothermal Expansion)
Let, unit mass of perfect gas is admitted into the cylinder at the beginning of the outward
moment of the piston and the pressure, temperature, volume of the gas at a point 1 is p1, v1 and
T1 respectively. The bottom 'B' of the cylinder can be covered by an insulating cap .During the
movement of the piston, the heat Q1 is supplied to the perfect gas and the gas expands
isothermally keeping temperature T1 constant until the volume v2 and pressure p2.The heat

supplied by the hot body is fully absorbed by the air and is utilised by doing external work. So,
heat is supplied during this process is equal to the work done during this process. This
isothermal expansion is represented by the curve 1-2 on p-v and T-S diagram.
So, work done by the air
Since, there is no change of temperature from point 1 to 2, so, (T1=T2) and as per first law of
thermodynamics internal energy is also zero (E=0).
Then the heat supplied
Then change of entropy
Second Stage (Isentropic Expansion)
As the piston moves outward, the gas expands adiabatically till the pressure p3, volume v3 and
temperature T2 and the hot body is removed from the bottom of the cylinder 'B' and the
insulating cap is brought in contact. In this process, there is no interchange of heat of the
surrounding gasses (Q = 0) The reversible adiabatic expansion is represented by the curve 2-3
on PV diagram.

The change of internal energy E = - W and there is a no change of entropy, so S2 = S3.
Third Stage (Isothermal Compression)
Now the piston moves inward the gas and insulating cap I.C. is also removed from the bottom
of the cylinder and bring the cold body in its contact. The air pressure is compressed
isothermally keeping temperature constant T3 from v3 to v4.It means at a point 4 temperature
T4 is equal T3.That's why heat is rejected to the cold body is equal to the work done on the air.
The isothermal compression curve is represented by 3-4 on p-v curve.
Here change of entropy E = 0 and change of entropy
Fourth Stage (Isentropic Compression)

As the piston moves inwards the gas insides the cylinder is compressed adiabatically till the
pressure p1, volume v1, and temperature T1 such that the gas returns to its original condition to
complete the Carnot cycle and insulated cap I.C. is brought in contact with the bottom of the
cylinder B. The temperature of air increases from T4 to T1 and not heat is absorbed or rejected
by the air. The reversible adiabatic compression is represented by the curve 4-1 on P-V and T-
S diagram. So, work done by the air for adiabatic compression
From the above discussion, it is seen that the total internal energy decrease in reversible
adiabatic expansion shown in curve 2-3 is equal to the increase in internal energy during
reversible adiabatic compression 4-1. So, the net effect of the whole Carnot cycle is zero.
Net work done,
And Carnot efficiency
From the Carnot cycle efficiency equation, T1 is greater than T3. For the smaller value of T3,
the thermal efficiency will be maximum. That's why Carnot cycle has highest thermal
efficiency of all heat engine.
Why Carnot cycle can't be used in actual practice?

During the isothermal process, the piston should move very slowly within the cylinder. So that
insufficient time is available for the transfer of heat to the working medium within the cylinder
during expansion or from the working medium within the cylinder during isothermal
compression.
But during the adiabatic process, the piston moves very fast within the cylinder. So that no time
is available for the transfer of heat.
So, sudden change of speed (from low speed to high speed) of the piston is required to make
the cycle efficient. But this is not possible due to direction of the cylinder and the piston is none
conductor of heat and cover head of the cylinder is good conductor. Again there is a weightless
and frictionless piston fitted on a cylinder which is practically impossible. That's why Carnot
cycle cannot be used in actual practice.
Classius Equality & Inequality
As per corollary of 2nd
law of thermodynamics all the reversible engines operating between two
heat reservoirs at the same temperature, have the same efficiency, irrespective of the working
substance used. That means between the temperature th & tc the value of Qh & Qc are always
same for any reversible engine.
The function f proposed by Kelvin and subsequently accepted is given by the simple relation.
Where C is a constant and Th & Tc are the absolute or thermodynamic temperatures on
thermodynamic temperature scale.
Or
or
It is called Classius equality for a reversible cycle.
In case of an irreversible cycle for a heat engine,
Wirrev < Wrev
Or Qh - Qirr < Qh Qrev
(for the same amount of heat absorbed, Qh in both cases)
Or Qc irr > Qc rev
Or

,
which is called Classius inequality.
Thus Classius Equality and In- equality statements can be combined in mathematical terms as:
Concept of Entropy
https://www.youtube.com/watch?v=Eo_DnJZXh0A
As the first law of thermodynamics introduces a property named as internal energy. Second
law of thermodynamics, when applied to a process, introduces the property named as entropy,
which is of extensive type.
The physical significance of entropy is somewhat difficult to imagine, but if we start from the
very basic, it will definitely give an idea of entropy and also its importance in thermodynamic
calculations.
Now consider a reversible cycle 1A2B1 having two reversible processes A & B between the
states 1 & 2 and another reversible cycle 1A2C1 having two reversible processes A and C
between the same states 1 & 2, as shown in fig 7.4.
Figure: Reversible cycles
Applying Classius equality to reversible cycles 1A2B1 and 1A2C1
From the above two equations:
Thus, we see that for any reversible process between state 1 & 2 is same i.e. independent
of path followed B or C or any other path (process) and depends only on states 1 & 2. Thus it
is a point function and hence is a property called entropy (S), such that a change in this
property, Hence change in entropy during a process 1-2 is given as:

Thus, entropy of a system may be defined as a property such that change in it from one state to
other is always equal to integral of heat exchanged divided by absolute temperature, at which
heat is exchanged, during any reversible process between the states.
Specific entropy
Entropy per unit mass is called specific entropy. It is denoted by small letter s.
Conclusion: For a reversible process in a closed system
a) Entropy increases if heat is added
b) Entropy decrease if heat is rejected
c) Entropy remains constant if there is no heat transfer.
Third law of thermodynamics: -
https://www.youtube.com/watch?v=kuGmsnzjbpE
Unattainability of absolute zero is based on the third law of thermodynamics which states
that the entropy of a pure substance of absolute zero temperature is zero. This is also called
Nernst Theorem.

Entropy and Availability
Entropy Change for an Irreversible Process in a Closed System
First of all, it may be made clear here that, entropy is a property and any change in it between
two states is always constant irrespective of the type of process between these states.
Let us consider a reversible cycle 1A2B1 formed by two reversible processes A and B between
state 1 & 2 and an irreversible cycle 1A2C1 formed by one reversible process A and other
irreversible process C between states 1 & 2 as shown in Fig 8.1.
Figure: Irreversible cycle
Applying Classius equality & inequality to both cycle (1A2B1)rev & (1A2C1)irr
0
From (i) and (ii)
It is due to -ve sign otherwise the numerical value of is high.
Now
Thus for the irreversible process C, change in entropy
So even if Q =0 for irreversible process, the entropy will always increase. It can be taken
mathematically as
Where, δI is the rise in entropy due to irreversibility factor
By integrating the total change in entropy during an irreversible process is
The effect of this irreversible term I is always to increase some entropy whether the heat is
added or rejected or not exchanged.
Conclusion
For an Irreversible Process

1. If heat is added, entropy increases due to heat addition and irreversibility.
2. If there is no heat transfer, the entropy still increases due to irreversibility.
3. If heat is rejected, entropy decreases due to heat rejection and increases due to
irreversibly. Net effect may be +ve or -ve.
General mathematical statement for change in entropy between two states is
Where
= sign is for Reversible process between the given states.
> Sign is for Irreversible Process between the given states.
It can be understood more clearly with the help of fig as given below:
Figure: Relationship of change in entropy with heat exchange during reversible and
irreversible processes

Unit 2: BASIC MECHANICAL ENGINEERING by varun pratap singh

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Unit 2: BASIC MECHANICAL ENGINEERING by varun pratap singh