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Hello and welcome to lecture 9 of this lecture
series on Introduction to Aerospace Propulsion.
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In the last few lectures, we have been discussing
some of the basic concepts of thermodynamics.
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In the last lecture that was lecture 8, we
discussed about the first law of thermodynamics
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applied to closed systems that is those systems,
where there is no mass interaction between
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the system and the surroundings.
In this lecture, we are going to discuss about
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first law of thermodynamics applied to open
systems; that is, how do we apply first law
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of thermodynamics to a system, which has mass
as well as energy interaction with the surroundings.
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In this lecture, what we are going to discuss
are basically the following: we shall be discussing
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about first law of thermodynamics as applied
to open systems. Before we actually define
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it for open systems, we need to look at what
is meant by flow work and the energy associated
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with a flowing fluid. This is basically applied
only for those systems, where there is mass
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interaction between the system and the surroundings.
We shall then define what is meant by total
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energy of a flowing fluid. We shall subsequently
define energy transport by mass and we shall
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carryout energy analysis of steady-flow systems.
We will also define, what are steady-flow
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systems. We shall take up some examples of
some steady-flow engineering devices; some
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commonly observed engineering devices. We
shall derive the steady-flow energy equation
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for such devices.
Now, before we look into the first law of
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thermodynamics applied to open systems, we
need to understand that there are 2 ways or
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2 methods in which you can approach a particular
problem. One is known as the Lagrangian approach
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and the other is known as the Eulerian approach.
Depending upon how you would like to analyze
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a particular system, you can use either of
these approaches. In the case of Lagrangian
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approach or system approach as it is called
in the usual practice. You would track a particular
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particle and see what are the changes that
particular particle or a fluid element undergoes
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as it moves within the system boundaries.
In the case of the second approach, that is
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the Eulerian approach, or it is also known
as the control volume approach. We only look
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at the system boundaries as a whole. We are
not really interested about what is actually
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happening to certain fluid element or a group
of particles. So, we are going to use the
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Eulerian approach for majority of discussions
that we are going to have during this course.
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In some problems, we may also look at the
Lagrangian approach as it might simplify the
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analysis in some sense.
Now, this is regarding the different approaches
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that you can have for analyzing a particular
system. Now, besides this, a particular flow
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process could also be steady or it could be
unsteady. The system you are looking at may
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have certain steady-flow process that is undergone
by the system or it might have an unsteady-flow
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process. So, we shall look at what we mean
by steady and unsteady-flow processes.
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As I mentioned, we are going to discuss control
volume approach for majority of the discussion.
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There are 2 types of processes - steady and
unsteady. Now, in a steady-flow process, the
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rates of flow of mass and energy are constant
across the system boundary. Some examples
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of steady-flow processes are turbines, compressors,
heat exchangers and so on. We shall take up
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detailed analysis of some of these systems
when we explicitly state the steady-flow energy
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equation or the first law of thermodynamics
as applied to these systems little later in
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this particular lecture.
Now, a process could also be unsteady, if
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the rates of mass or energy are not constant
across system boundaries. Some examples are
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like charging and discharging processes. If
you are charging a tank with compressed air
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or some other compressed gas or you are charging
a tank from a pipeline, which carries a certain
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compressed air or a certain fluid, it is basically
an unsteady-flow process because the rates
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of mass or energy are not constant across
the system boundaries.
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We will not be covering any unsteady-flow
process in this particular course as it is
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beyond the scope of the syllabus that we have.
We shall definitely be analyzing the steady-flow
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processes and steady-flow process systems
in this lecture because majority of the processes
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that we are interested as aerospace engineers
can be approximated to be a steady-flow process.
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That is the reason why we shall not really
go into details of the unsteady-flow process
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in this course.
Before we take up the first law of thermodynamics
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for open systems, we need to understand a
very fundamental law of nature known as the
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conservation of mass principle. If you recollect,
we had already discussed about the first law
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of thermodynamics for close systems. We stated
that it is basically the conservation of energy
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principles stated in a different way to define
the first law of thermodynamics for open systems.
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We also need to understand what is mean by
conservation of mass.
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Conservation of mass law states that the total
mass of a particular system will be equal
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to the difference between the total mass entering
the system minus the total mass leaving the
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system and it will be equal to the net change
in mass within the system. So, this is what
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basically the conservation of mass principle
states. The total mass, which enters a system
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minus the total mass that leaves the system
should be equal to the net change in mass
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within the system. It means that there is
no creation or generation of mass within the
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system boundaries. It is basically the mass
contained in the system; it is the net difference
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of the mass that is entering the system and
the mass leaving the system.
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So, stated in equation, we would have m subscript
in minus m subscript out should be equal to
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delta m cv, where cv stands for the control
volume. So, the mass in minus mass out is
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equal to delta m that is net change in mass
of the control volume or rate of mass flow
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in minus rate of mass flow out is equal to
dm cv by dt. Therefore, you can calculate
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the total mass within the control volume m
cv as the integral of rho times dv, where
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row is density and v is the volume.
In the rate form, the rate of change of mass
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within the control volume is dm cv by dt,
which is d by dt of rho dv and this is for
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a control volume. Basically, the conservation
of mass states that there is a change in the
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mass of the system and that is attributed
to the mass coming in and the mass that is
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going out.
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Now to understand flow process, we also need
to understand or we should be familiar with
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the concept of what is known as the flow work
or flow energy. If you are looking at a certain
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flow process, there is a certain mass entering
a system and a certain mass leaving a system.
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It means that there is certain amount of work
that is required to push a certain mass into
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a control volume or out of the control volume.
This mass or this work that is required to
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either push in the mass or push out the mass
of a control volume is known as the flow work
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or flow energy. This is the major difference
between a closed system and an open system.
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It makes a lot of difference in the analysis
of closed and open systems and that is because
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of the certain of work that is required for
pushing in a certain mass of fluid into a
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system or pushing out certain mass out of
the system. So, there is a certain work required.
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Now to define or to understand this better,
let us consider a certain system, which has
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a certain volume V. Let us say, the pressure
acting on that particular control volume or
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particular fluid volume is P through an area
A. We know that the work done for any particular
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process is equal to force times the distance
.To understand this better, let us consider
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that there is an imaginary piston, which is
pushing this mass flow into the control volume.
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The piston moves by, let us say, a distance
of L and this requires a certain force, which
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is let us say, F. So, F times L will give
you the work required. We also know that force
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is nothing but pressure times the area. Since
we already know the pressure or we have defined
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the pressure as P and area as A. So, P times
A will give you the force acting on the piston,
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multiplied by L will be equal to work. So,
F into A into L will be the work and since
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A times L is volume, we get the net work done
as equal to P times the volume PV.
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Basically, the flow energy will come out to
be the product of the pressure times the volume.
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This was some way obvious to you because in
the last lecture or in the 2nd lecture, we
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had discussed about work and heat transfer.
One of the major forms of work was displacement
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work. This is basically displacement work,
but we are defining it in a different way
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here and that is basically known as the flow
work or flow energy. So, if we were to look
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at it in a different way as I mentioned, we
consider some fluid element of volume V. If
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the fluid pressure acting is P through a cross
sectional area A, let us say, L is the distance
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through which an imaginary piston must moved.
So, the work required or work done in pushing
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this fluid element across the system boundary
would be F times L, where F is pressure times
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the area and therefore it is PA times L. Therefore,
it is equal to PV and so work done during
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this process is PV.
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I have illustrated this in a in a small example
here. What I was mentioning about this fluid
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element, which you can see here is - it has
a certain volume of V, pressure acting is
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P and a mass m. This is the control volume,
which has been indicated by dotted lines and
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the piston that is shown here is an imaginary
piston. This is just to indicate that there
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is a certain force, which has to be acted
upon this fluid element to push it into the
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control volume. So, this is the force acting,
F and it is acting through an area A. As the
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piston moves by a distance of L, this much
amount of fluid, which has a volume of V,
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pressure of P and mass of m goes into the
control volume. Similarly, there is a certain
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amount of mass flow leaving the system.
The work done for this process is this force
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acting on the imaginary piston multiplied
by the distance L through which it was displaced.
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So, force times L and force is pressure times
area and PA into L and A into L is volume;
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area times the length is volume. Therefore,
the work done for this process or the flow
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energy or the flow work is equal to P times
V pressure times the volume.
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Now, we shall look at what is the total energy.
Now, you have understood what is flow energy
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and we will now look at what is the total
energy that is associated with a fluid that
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is flowing. Let us say, entering a system
or leaving a control volume, it has a certain
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amount of additional energy. If you recall,
we had defined total energy for closed systems
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as the sum of the internal energy plus kinetic
energy plus the potential energy. Now, in
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the case of fluid, which is flowing there
is an additional work or energy associated
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with it. It is given by the flow work or flow
energy, which is the product of pressure times
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the volume.
So, the total energy of a fluid, which is
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flowing will be the sum of the internal energy
u plus the kinetic energy plus the potential
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energy and the flow energy that is Pv. So,
the total energy, which we normally represent
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for flow processes by theta will be equal
to u plus ke plus pe plus Pv, which is the
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flow energy. Now, in one of the earlier lectures,
I had defined a term, which is known as a
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combination property as enthalpy. We had defined
enthalpy as the sum of internal energy u and
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the flow energy Pv.
The total energy is now equal to the sum of
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the enthalpy, the kinetic energy and the potential
energy. So, for a flowing fluid, there are
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the total energy consists of 4 different terms
- it is the sum of internal energy- u, kinetic
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energy, potential energy and the flow energy.
For closed system, the total energy consists
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of 3 terms - the internal energy, kinetic
energy and the potential energy. So, there
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is no flow energy associated with closed systems.
It makes sense to define this term, combination
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property as enthalpy because enthalpy actually
takes care of the internal energy as well
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as the flow energy. So, you do not have to
worry about these 2 terms separately. Enthalpy
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has already taken into account for the internal
energy and the flow energy. So that is the
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convenience of defining this combination property
of enthalpy.
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The total energy of an open system or a flow
process will be the sum of enthalpy, which
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in turn is equal to u plus Pv, enthalpy plus
kinetic energy plus the potential energy pe.
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So, the total energy associated with the flowing
fluid is normally associated as h plus ke
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plus pe.
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If you were define it in terms of an equation,
the total energy of a flowing fluid is e and
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it is equal to the total energy associated
with a non-flowing fluid. It is u plus ke
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plus pe and which is in turn equal to u plus
V square by 2 and that is kinetic energy per
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unit mass plus g times z, which is the potential
energy per unit mass.
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For a flowing fluid on the other hand, theta
is the total energy and it is equal to h plus
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ke plus pe and which is in turn equal to u
plus Pv plus V square by 2 plus gz. So, you
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can see that the total energy consists of
3 parts for a non-flowing fluid. For a flowing
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fluid, the total energy consists of 4 components
or 4 terms and that is the basic difference
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between a flowing fluid and a non-flowing
fluid.
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I have mentioned that for a process, which
involves a fluid element crossing the system
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boundary. If there is a certain flow that
is taking place, we can define the total energy
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per unit mass equal to the mass, which the
flowing fluid multiplied by the energy itself.
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Therefore, the energy transport associated
with the mass will be equal to m times of
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theta that is mass times the total energy.
We have already defined total energy for the
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flowing process, which was the sum of h plus
kinetic energy plus potential energy. The
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amount of energy transported is E subscript
mass is equal to mass times theta. Theta is
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the flow energy equal to h plus V square plus
gz. If you were to look at this in the rate
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form, the rate of energy transport E dot mass
is equal to m dot times theta, where m dot
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is the mass flow rate. So, the amount of energy
transport E mass is in kilo joules and the
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rate of energy transport will be in kilo watts
because you are looking at mass flow rate.
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So, you would get kilo joules per second,
which is equal to kilowatts.
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The amount of energy transport here will be
the product of the mass. The total energy
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associated with this particular process and
the amount of or the rate of energy transport
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will be equal to the product of the mass flow
rate times the total energy. So, you would
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get m dot multiplied by h plus V square by
2 plus gz, where h is the enthalpy for the
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process, V square by 2 is the kinetic energy
and gz is the potential energy per unit mass.
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This multiplied by the mass flow rate would
give you the total energy associated or energy
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transport associated with that mass in kilowatts.
The amount of energy that a certain fluid
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flow process carries depends on 4 terms -internal
energy; which is u, the flow work or flow
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energy; which is P, the kinetic energy and
the potential energy. So, this is the total
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energy associated with a flow process and
this multiplied by the mass will give you
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the total energy transport associated with
this particular mass flow process.
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It is important for us to understand these
flow processes because many of the engineering
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systems that we shall be discussing soon are
basically involving flow processes and a certain
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mass associated with these flow processes.
So, we need to understand the importance of
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flow processes and also how do you apply the
first law of thermodynamics to different flow
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processes. We will take up some examples towards
the end of this lecture on how you can derive
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first law of thermodynamics. So, energy equation
for different steady-flow processes like turbines
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and nozzles etc, which we will take up towards
the end of the lecture.
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I mentioned in the beginning of this lecture
that the flow process can be of different
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types. One of them is known as a steady-flow
process and the other is an unsteady-flow
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process. I mentioned that steady-flow processes
are those in which the mass and energy do
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not change across the system boundaries. So,
what is the importance of understanding or
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analyzing steady-flow processes? Well, there
is a lot of importance in understanding steady-flow
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processes because several engineering devices
or systems can be can be well approximated
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00:21:06,120 --> 00:21:13,120
as steady-flow systems. Some examples are
turbine, compressors, nozzles etc. During
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a steady-flow process, the basic assumption
is that no intensive or extensive property
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with in a control volume change with time.
The element of time does not really appear
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in a steady-flow process. If there is a d
by dt term in the steady-flow equation of
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the energy equation, it will become 0 because
the rate of change of any of any intensive
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00:21:40,640 --> 00:21:47,640
property or any extensive property with time
will be equal to 0 and so that is the basic
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definition of a steady-flow process.
200
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As I mentioned, there are many engineering
devices, which we can approximate as steady-flow
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00:22:02,049 --> 00:22:09,049
devices like turbines, compressors, nozzles
etc. During a steady-flow process, we may
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discuss that the none of these properties
like extensive or intensive properties would
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actually change with time. Therefore, the
boundary work associated with a steady-flow
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system will be 0 because the volume of the
control volume is fixed or constant and so
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boundary work associated with this process
will be 0.
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00:22:33,590 --> 00:22:38,690
The other property with a steady-flow system
is that the total mass or energy entering
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the control volume must be equal to the total
mass or energy leaving the control volume.
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So, there is no accumulation of mass or energy
within the control volume. If that was the
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case, it could become an unsteady process.
I mentioned that we are not going to discuss
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about unsteady-flow process in this lecture.
We shall be talking only about steady-flow
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processes during this lecture.
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What are the properties of steady-flow process?
Well, the basic properties of a steady-flow
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process are - none of the properties, whether
they are extensive properties or intensive
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properties, they do not change within the
control volume with time. Of course, the properties
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00:23:25,350 --> 00:23:32,350
can change within the control volume itself,
but there is no rate of change of the control
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properties within the control volume with
time. It is also a fact that no properties
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00:23:38,980 --> 00:23:44,820
will actually change at the boundaries of
the control volume with time. The rate of
218
00:23:44,820 --> 00:23:51,350
change of mass or energy with time across
the control surface is actually a constant,
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00:23:51,350 --> 00:23:57,150
if the process is to be approximated as a
steady-flow process.
220
00:23:57,150 --> 00:24:02,500
The different thermodynamic properties like
pressure, temperature etc has fixed values
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00:24:02,500 --> 00:24:09,500
at a particular location. They do not change
with time that is if you look at a system,
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00:24:09,600 --> 00:24:14,260
it has to be approximated as a steady-flow
system or a steady-flow process. Then the
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00:24:14,260 --> 00:24:19,100
different thermodynamic properties of the
system will have fixed values at different
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00:24:19,100 --> 00:24:25,210
locations within the system. They do not change
with time. So, these are some basic properties
225
00:24:25,210 --> 00:24:31,850
that need to be satisfied, if a particular
process has to be approximated or if a process
226
00:24:31,850 --> 00:24:36,000
has to qualify as a steady-flow device or
a steady-flow process.
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00:24:36,000 --> 00:24:43,000
Very shortly, we shall define the first law
of thermodynamics for steady-flow devices.
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00:24:45,120 --> 00:24:51,340
We shall apply the first law for some examples
of steady-flow engineering devices. So, we
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00:24:51,340 --> 00:24:56,150
need to keep these properties in mind, if
we are going to approximate a particular process
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or a system as a steady-flow process.
231
00:25:02,250 --> 00:25:08,780
This is an example or just an illustration
showing, what are the properties associated
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00:25:08,780 --> 00:25:14,980
with steady-flow systems. For a steady-flow
system, which has certain mass entering and
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00:25:14,980 --> 00:25:20,090
a certain mass leaving the system, the net
mass of the control volume is a constant.
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00:25:20,090 --> 00:25:25,510
The net energy associated with this control
volume is also a constant.
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00:25:25,510 --> 00:25:29,910
Now, the first one on the left hand side,
what you see is an example of a single entry
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00:25:29,910 --> 00:25:35,419
and single exit system. You could also have
a system, which has multiple entries and multiple
237
00:25:35,419 --> 00:25:40,890
exits. The basic definition remains the same
that if the net mass within the control volume
238
00:25:40,890 --> 00:25:47,890
is constant and the net energy of the control
volume remains a constant, then we can approximate
239
00:25:49,400 --> 00:25:56,400
this process as a steady-flow process. Under
steady-flow conditions, the fluid properties
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00:25:56,600 --> 00:26:02,410
at the inlet or exit remain a constant. They
do not change with time and so the element
241
00:26:02,410 --> 00:26:09,410
of time in the steady-flow energy equation
would become equal to 0 because we are assuming
242
00:26:10,150 --> 00:26:15,390
that the properties do not change with time.
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00:26:15,390 --> 00:26:22,390
Let us now look how we can derive the energy
equation for a steady-flow process. For steady-flow
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00:26:22,760 --> 00:26:29,760
system, the amount of energy entering a control
volume in all its forms, it could be energy
245
00:26:30,090 --> 00:26:34,679
transfer by heat, it could be energy transfer
by work or it could be energy transfer by
246
00:26:34,679 --> 00:26:40,350
virtue of the mass flow itself. So, whatever
energy is entering a control volume, it should
247
00:26:40,350 --> 00:26:45,600
be equal to the energy leaving the control
volume because we have seen that energy of
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00:26:45,600 --> 00:26:52,600
a control volume, whether it is single entry
or multi entry, the net energy of the system
249
00:26:53,250 --> 00:26:58,770
within the control volume is a constant.
The total energy is entering in all its forms
250
00:26:58,770 --> 00:27:04,380
of heat, work and mass because these are the
3 modes of energy interactions, which a system
251
00:27:04,380 --> 00:27:09,600
can have with the surroundings. The net energy
entering in all these forms should be equal
252
00:27:09,600 --> 00:27:15,280
to the net energy leaving the system. Energy
balance for a steady-flow system would be
253
00:27:15,280 --> 00:27:22,280
equal to energy in minus energy out. The net
is equal to rate of change of energy of a
254
00:27:23,730 --> 00:27:29,040
system. In the case of steady-flow processes,
the rate of change of energy is equal to 0
255
00:27:29,040 --> 00:27:35,370
because it is a steady-flow process. Therefore,
energy in will be equal to energy out that
256
00:27:35,370 --> 00:27:42,160
is e in is equal to the rate of change of
energy leaving the control volume.
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00:27:42,160 --> 00:27:48,090
Now, as I mentioned, energy in could be in
different forms. It could be in the form of
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00:27:48,090 --> 00:27:55,090
heat work or mass, but whatever the form of
energy interaction, which the system has with
259
00:27:55,309 --> 00:28:02,260
the surroundings. The net rate of change of
energy transfer by heat mass or work should
260
00:28:02,260 --> 00:28:09,260
be equal to rate of change of energy transfer
out of the system by heat work or mass. This
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00:28:10,600 --> 00:28:16,870
is basically the energy balance or the energy
equation for a steady-flow process.
262
00:28:16,870 --> 00:28:22,429
We can write this more explicitly. If you
look at the steady-flow process, the energy
263
00:28:22,429 --> 00:28:29,429
equation is expressed in more explicit form.
As I mentioned, energy transfer can take place
264
00:28:30,350 --> 00:28:37,270
in 3 modes - heat, work and mass flow. So,
these are the 3 modes, which have been written
265
00:28:37,270 --> 00:28:44,270
here. Q in dot, which is rate of input of
heat minus W dot in, which is rate of work
266
00:28:47,020 --> 00:28:54,020
input plus the mass flow input, which is equal
to sigma in m dot theta. This sigma is for
267
00:28:56,059 --> 00:29:03,059
accounting the different multi input flow
rates into the system. Since E in is equal
268
00:29:03,789 --> 00:29:10,070
to E out, this should be equal to Q dot out,
which is rate of heat transfer out of the
269
00:29:10,070 --> 00:29:17,070
system minus W out dot, which is rate of work
out of the system plus sigma out of m dot
270
00:29:21,309 --> 00:29:27,080
theta.
As we now discussed, the third term is the
271
00:29:27,080 --> 00:29:34,080
total energy associated with the flow process.
It should be equal to Q in dot, which is heat
272
00:29:35,169 --> 00:29:42,150
transfer in minus W in dot, which is work
input plus sigma in times mass flow rate multiplied
273
00:29:42,150 --> 00:29:49,150
by the enthalpy, the sum of enthalpy plus
kinetic energy plus the potential energy.
274
00:29:49,200 --> 00:29:56,200
This is applicable for each inlet and this
should be equal Q dot out, which is heat transfer
275
00:29:56,280 --> 00:30:03,280
out of the system minus W dot out, which is
work done by the system, plus m dot out multiplied
276
00:30:05,260 --> 00:30:12,260
by the flow energy, which is h plus V square
by 2 plus gz. So, this defines the general
277
00:30:15,220 --> 00:30:20,700
energy equation for steady-flow processes,
which consists of 3 terms.
278
00:30:20,700 --> 00:30:27,700
In a steady-flow energy equation, you have
3 distinct terms. One is the heat transfer
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00:30:28,630 --> 00:30:35,630
term, the work done term and the flow work
term. So, Q in minus W in plus mass flow rate
280
00:30:36,429 --> 00:30:41,590
times the flow work or flow energy should
be equal to the same terms leaving out of
281
00:30:41,590 --> 00:30:48,480
the system. If the system consists of multiple
entries and multiple exits, it will affect
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00:30:48,480 --> 00:30:52,350
the third term that is the flow energy term.
You have to calculate the flow energy for
283
00:30:52,350 --> 00:30:57,030
each of these entries and each of these exits
from the system.
284
00:30:57,030 --> 00:31:03,580
In general, steady-flow energy equation will
consist of these 3 different terms, which
285
00:31:03,580 --> 00:31:08,039
have to be applied at the entry as well as
the exit. You can now compare this steady-flow
286
00:31:08,039 --> 00:31:12,530
energy equation with the energy equation,
which we had discussed in the last lecture.
287
00:31:12,530 --> 00:31:19,130
It was applicable for closed systems or closed
processes. So, you might recall that we did
288
00:31:19,130 --> 00:31:24,419
not have this flow energy term in the first
law of thermodynamics as applied to closed
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00:31:24,419 --> 00:31:30,419
systems. For the first law as applied to closed
systems, it was Q minus W is equal to delta
290
00:31:30,419 --> 00:31:37,100
e or delta u.
There was no flow energy in a closed system
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00:31:37,100 --> 00:31:44,100
or closed process. So, energy equation can
be written in the generalized form of energy
292
00:31:44,809 --> 00:31:51,630
equation. It can be expressed as Q dot minus
W dot is equal to the difference between the
293
00:31:51,630 --> 00:31:58,630
mass flows entering the system and mass flows
leaving the system or the flow energy associated
294
00:31:58,980 --> 00:32:05,960
with the mass leaving the system minus the
energy associated with mass entering the system.
295
00:32:05,960 --> 00:32:12,960
So, Q dot would refer to the net heat transfer
into the system and W dot would refer to the
296
00:32:13,340 --> 00:32:16,130
net work transfer from the system.
297
00:32:16,130 --> 00:32:23,020
Q dot minus W dot is equal to sigma out m
dot times the flow energy term. Energy flow
298
00:32:23,020 --> 00:32:30,020
term is is h plus V square by 2 plus gz for
each exit minus sigma in into m dot times
299
00:32:33,299 --> 00:32:40,299
h plus V square by 2 plus gz for each inlet.
Here, Q dot refers to the net heat input into
300
00:32:40,360 --> 00:32:47,360
the system, W dot is the net work output of
the system and this again is an assumption.
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00:32:47,400 --> 00:32:53,380
As I mentioned in the last lecture, we normally
assume that for a given process, there is
302
00:32:53,380 --> 00:32:57,559
heat input into the system and work output
from the system.
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00:32:57,559 --> 00:33:02,789
After any of the calculations that you are
carrying out, if these numbers come out to
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00:33:02,789 --> 00:33:07,940
be negative, it just means that the net heat
was not input, but there was a net heat output
305
00:33:07,940 --> 00:33:14,830
from a system. If W comes out to be negative,
it means that there was net work done on the
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00:33:14,830 --> 00:33:21,830
system because we normally have systems, which
generate work and requires heat input. This
307
00:33:22,070 --> 00:33:28,020
is why, as a common practice, we assume net
heat input to the system and net work output
308
00:33:28,020 --> 00:33:29,059
from the system.
309
00:33:29,059 --> 00:33:35,250
Now, if we look at this particular equation
for a single entry system, which is how most
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00:33:35,250 --> 00:33:42,250
of the engineering devices would be. For example,
a turbine or a compressor or nozzle and diffusers,
311
00:33:42,299 --> 00:33:49,299
all these are all devices, which have single
entry and single exit. The energy equation
312
00:33:49,559 --> 00:33:56,559
becomes simpler; you have only one particular
term for the flow work. The energy equation
313
00:33:58,470 --> 00:34:05,470
is modified as Q minus W is equal to the difference
in the enthalpies at the exit and the entry
314
00:34:08,649 --> 00:34:12,849
plus the difference in the kinetic energies
plus the difference in potential energies
315
00:34:12,849 --> 00:34:17,720
at the inlet and exit.
If you were to write down the steady-flow
316
00:34:17,720 --> 00:34:24,720
energy equation for single entry device, then
the equation would look like Q dot minus W
317
00:34:28,229 --> 00:34:35,229
dot is equal to m dot times h 2 minus h 1
plus V 2 square minus V 1 square by 2 plus
318
00:34:35,509 --> 00:34:42,509
g times z 2 minus z 1. This h 2 minus h 1
is the net enthalpy, V 2 square minus V 1
319
00:34:44,289 --> 00:34:50,200
square by 2 is a net kinetic energy, g times
z 2 minus z 1 is the net potential energy.
320
00:34:50,200 --> 00:34:57,200
The same equation per unit mass would be Q
dot minus W dot is equal to h 2 minus h 1
321
00:34:58,549 --> 00:35:05,239
plus V 2 square minus V 1 square by 2 plus
g times z 2 minus z 1. This is the general
322
00:35:05,239 --> 00:35:11,410
energy equation for steady-flow processes.
In some textbooks, you might see that this
323
00:35:11,410 --> 00:35:17,839
equation is referred to as the steady-flow
energy equation. So, steady-flow energy equation
324
00:35:17,839 --> 00:35:24,839
for single entry and single exit devices is
Q dot minus W dot is delta h plus delta ke
325
00:35:25,410 --> 00:35:27,930
plus delta pe.
326
00:35:27,930 --> 00:35:34,930
We shall now discuss about some common steady-flow
devices, which we see in daily life. Some
327
00:35:35,440 --> 00:35:42,440
of these commonly used steady-flow energy
devices are nozzles and diffusers, compressors
328
00:35:42,599 --> 00:35:49,599
and turbines, throttling devices, mixing chambers,
heat exchangers etc. We shall now derive the
329
00:35:49,969 --> 00:35:55,729
steady-flow energy equation for some of these
engineering devices, which are commonly used.
330
00:35:55,729 --> 00:36:02,729
We shall see how the general form of steady-flow
energy equation can be used by applying appropriate
331
00:36:04,430 --> 00:36:10,450
boundary conditions for these devices like
nozzles and diffusers, compressors, turbines
332
00:36:10,450 --> 00:36:15,589
or certain mixing chambers, throttling devices
etc.
333
00:36:15,589 --> 00:36:22,589
The first such device, which we shall discuss
are nozzles and diffusers. A nozzle is a device,
334
00:36:23,710 --> 00:36:30,029
which increases the velocity of a fluid at
the expense of pressure. Diffuser on the other
335
00:36:30,029 --> 00:36:36,999
hand is a device, which increases the pressure
of a fluid by slowing it down. A nozzle and
336
00:36:36,999 --> 00:36:42,779
a diffuser in some sense are devices, which
are opposite to that of each other in terms
337
00:36:42,779 --> 00:36:49,479
of their basic function.
The cross-sectional area of a nozzle decreases
338
00:36:49,479 --> 00:36:55,229
in the flow direction for subsonic flows and
it increases for supersonic flows. Well, you
339
00:36:55,229 --> 00:37:00,440
might wonder what subsonic and supersonic
flows are. At the moment, we will just define
340
00:37:00,440 --> 00:37:07,239
them as those flows, which have a mach number
less than 1 are known as subsonic flows and
341
00:37:07,239 --> 00:37:12,839
those flows, which have a mach number greater
than 1 are known as supersonic flows. I think
342
00:37:12,839 --> 00:37:19,309
we will define this little later in the course,
when we discuss about compressible flows.
343
00:37:19,309 --> 00:37:26,309
A nozzle in a subsonic flow has decreasing
area in the direction of the flow and increasing
344
00:37:27,589 --> 00:37:32,670
area in direction of flow in supersonic flows.
Reverse of this is true for diffuser, because
345
00:37:32,670 --> 00:37:37,719
I mentioned that nozzles and diffusers are
2 devices, which have opposite functions.
346
00:37:37,719 --> 00:37:43,239
So, their geometry also in some sense will
be opposite to that of each other.
347
00:37:43,239 --> 00:37:50,239
We will derive the energy equation or we will
simplify the basic energy equation. Let us
348
00:37:50,680 --> 00:37:57,099
say, a nozzle will be very similar to that
for a diffuser with appropriate boundary conditions
349
00:37:57,099 --> 00:38:03,699
applied. Now, we know that in the case of
an energy equation, we have 3 different terms
350
00:38:03,699 --> 00:38:09,140
- the heat transfer, work and the mass on
the flow energy terms. Some of these terms
351
00:38:09,140 --> 00:38:14,190
will become 0 as you apply it for different
engineering devices.
352
00:38:14,190 --> 00:38:20,410
In this particular example, we are discussing
for nozzle and diffuser. I mentioned that
353
00:38:20,410 --> 00:38:26,640
a nozzle is a device, which has increase in
velocity along the flow direction for subsonic
354
00:38:26,640 --> 00:38:33,640
flows. Let us say V 1 is the velocity at the
inlet and V 2 is velocity at outlet. So, nozzle
355
00:38:33,979 --> 00:38:39,130
increases the velocity at the expense of pressure
and therefore V 2 is usually greater than
356
00:38:39,130 --> 00:38:43,729
V 1.
For a diffuser, it is the opposite. If V 1
357
00:38:43,729 --> 00:38:49,019
is the inlet velocity, V 2 is the exit velocity.
V 2 will be less than V 1. You can also notice
358
00:38:49,019 --> 00:38:56,019
that a nozzle and diffuser have opposite shapes.
A nozzle in the reverse way would look like
359
00:38:57,380 --> 00:39:04,380
a diffuser. So, nozzles and diffusers are
those devices, which can cause large changes
360
00:39:05,529 --> 00:39:10,069
in fluid velocities. Therefore, there is a
large change in the kinetic energy of the
361
00:39:10,069 --> 00:39:16,319
fluid as they pass through a nozzle and a
diffuser.
362
00:39:16,319 --> 00:39:23,319
The general form of energy equation was E
in is equal to E out for steady-flow systems.
363
00:39:23,440 --> 00:39:29,099
In the case of nozzle and diffuser, there
is no net change in heat transfer. Therefore,
364
00:39:29,099 --> 00:39:34,569
Q dot can be approximated to be 0. There is
no work done by a nozzle or a diffuser and
365
00:39:34,569 --> 00:39:40,640
therefore W dot is 0. We can also approximate
the change in potential energy to be 0 as
366
00:39:40,640 --> 00:39:47,640
long as the nozzle is horizontal or the diffuser
is horizontal. Even, if nozzles and diffusers
367
00:39:48,989 --> 00:39:54,700
do not have a long length, the net change
in potential energy across the nozzle or diffuser
368
00:39:54,700 --> 00:40:00,619
can also be assumed to be 0.
If you apply all these boundary conditions
369
00:40:00,619 --> 00:40:07,619
on the energy equation, net heat transfer
as 0, work done as 0 and delta pe is equal
370
00:40:07,920 --> 00:40:14,920
to 0. The energy equation reduces to m dot
into h 1 plus V 1 square by 2 is equal to
371
00:40:15,979 --> 00:40:22,979
m dot times h 2 plus V 2 square by 2. Mass
entering in and mass leaving the system are
372
00:40:23,150 --> 00:40:30,150
the same, m dot will get canceled out. Therefore,
the energy equation for a nozzle or a diffuser
373
00:40:30,359 --> 00:40:37,359
would be h 2 is equal to h 1 minus V 2 square
by minus V 1 square by 2.
374
00:40:38,819 --> 00:40:44,890
You could also write it as delta h is equal
to the delta k E across the system. The net
375
00:40:44,890 --> 00:40:50,890
change in the enthalpy is equal to the net
change in kinetic energy. There is no net
376
00:40:50,890 --> 00:40:56,450
change in potential energy, heat transfer
is 0, work done is also 0. This would be the
377
00:40:56,450 --> 00:41:01,410
basic energy equation for nozzles and diffusers.
378
00:41:01,410 --> 00:41:08,410
Let us now look at the second set of devices,
turbines and compressors. I think you are
379
00:41:08,859 --> 00:41:13,789
already aware that pumps compressors and fans
are devices, which are used to increase the
380
00:41:13,789 --> 00:41:19,809
pressure of a fluid. Therefore, they require
work input. Turbines on the other hand, generate
381
00:41:19,809 --> 00:41:26,809
work. So, pumps and turbines in some sense
are like nozzles and diffusers. They are opposite
382
00:41:27,619 --> 00:41:34,619
in function in some way or the other.
In the case of the energy equation, the heat
383
00:41:35,190 --> 00:41:40,890
transfer, kinetic energy and potential energy
may or may not be 0. It depends upon the type
384
00:41:40,890 --> 00:41:47,410
of device that you are looking at. Usually,
it is practice to assume heat transfer across
385
00:41:47,410 --> 00:41:53,079
the system boundary to be 0. If we consider
that the turbine boundaries or casing is well
386
00:41:53,079 --> 00:41:59,920
insulated or compressor boundaries are insulated,
then the process can be considered as adiabatic.
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00:41:59,920 --> 00:42:06,079
So, heat transfer is 0, kinetic energy and
potential energy may or may not be 0. Kinetic
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00:42:06,079 --> 00:42:11,059
energy is usually not really assumed to be
0 because there is a change in velocities
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00:42:11,059 --> 00:42:16,029
across the system boundaries. They may or
may not be 0 and potential energy depending
390
00:42:16,029 --> 00:42:21,609
upon the type of system, we may assume it
to be 0 and sometimes it is not taken as 0.
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00:42:21,609 --> 00:42:28,609
In fact, when we solve some example problems
later on in a lecture, we will see how much
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00:42:29,499 --> 00:42:36,259
is the error introduced, if you were to neglect
kinetic energies and potential energies for
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00:42:36,259 --> 00:42:41,200
such systems.
If you look at the energy equation, for turbine
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00:42:41,200 --> 00:42:46,140
and which is what we are going to derive today.
In the case of turbine, we are going to assume
395
00:42:46,140 --> 00:42:53,140
that the changes in kinetic energy and potential
energy are close to 0. So, Q is equal to 0
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00:42:54,890 --> 00:42:58,890
because we are going to assume that the process
is adiabatic in some sense.
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00:42:58,890 --> 00:43:05,459
Let us look at a schematic diagram of a turbine
here. What you see here is a turbine, which
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00:43:05,459 --> 00:43:10,410
has a certain insulation. It means that there
is no heat transfer across the system boundaries,
399
00:43:10,410 --> 00:43:17,309
across the control surface. The turbine generates
a net work output and there is a certain mass
400
00:43:17,309 --> 00:43:21,910
flow entering the system and mass flow leaving
the system. So, the turbine generates a net
401
00:43:21,910 --> 00:43:26,160
work across the system boundaries.
402
00:43:26,160 --> 00:43:33,160
If you look at the energy equation for a turbine,
since Q dot is 0, we get m dot multiplied
403
00:43:33,690 --> 00:43:40,690
by h 1 plus V 1square by 2 plus gz 1. It is
equal to m dot is equal to work output that
404
00:43:41,859 --> 00:43:48,859
is W dot out plus m dot times h 2 plus V 2
square by 2 plus gz 2. If you assume kinetic
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00:43:50,469 --> 00:43:56,809
energy and potential energy to be negligible,
then the net work output is equal to m dot
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00:43:56,809 --> 00:44:03,099
times h 1 minus h 2 that is work output from
the turbine is mass flow times the difference
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00:44:03,099 --> 00:44:07,160
in enthalpies.
So, difference between the enthalpy times
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00:44:07,160 --> 00:44:12,499
the mass flow will give you, what is the work
output that this particular turbine is giving.
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00:44:12,499 --> 00:44:18,469
You can modify the same equation; the general
energy equation for a compressor. Depending
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00:44:18,469 --> 00:44:23,729
upon whether you neglect kinetic energy and
potential energy, you can derive a very similar
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00:44:23,729 --> 00:44:30,729
expression of work input required for compressors
or pumps or fans. The equation will be similar
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00:44:31,049 --> 00:44:36,329
to this, but just that the signs of h 1 and
h 2 would be different because enthalpy leaving
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00:44:36,329 --> 00:44:43,259
a compressor or a fan or a pump would be higher
than the enthalpy coming in because there
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00:44:43,259 --> 00:44:49,910
is work done on the system. You can modify
the equation appropriately for the compressors
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00:44:49,910 --> 00:44:55,380
or pumps. The equation for compressor and
pump will be the same because they are thermodynamically
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00:44:55,380 --> 00:44:57,819
same devices.
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00:44:57,819 --> 00:45:02,930
The next device that we are going to discuss
is a throttling device. Well, a throttling
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00:45:02,930 --> 00:45:09,930
device is a one, which basically cause flow
restrictions leading to significant pressure
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00:45:10,519 --> 00:45:17,519
drop in the fluid. Some examples are capillary
tubes, valves or porous plug etc. Unlike turbine,
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00:45:20,410 --> 00:45:26,089
where there is pressure drop across the turbine
and a work output. In the case of throttling
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00:45:26,089 --> 00:45:33,089
devices, they produce a pressure drop without
involving any work. A large drop in temperature
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00:45:35,190 --> 00:45:39,539
often accompanies this pressure drop. This
is the reason why throttling devices are very
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00:45:39,539 --> 00:45:46,539
commonly used in refrigeration and air-conditioning
systems. Throttling device forms one of the
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00:45:46,549 --> 00:45:53,549
components of a refrigeration cycle. So, throttling
devices do not generate any work output. They
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00:45:54,789 --> 00:46:01,059
obviously do not have any heat transfer and
we will derive the energy equation for throttling
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00:46:01,059 --> 00:46:02,150
devices.
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00:46:02,150 --> 00:46:07,829
Some example of throttling devices are shown
here, one example is a valve. As you adjust
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00:46:07,829 --> 00:46:13,489
the valve, there is a large change in pressure
across the valve. It is considered to be a
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00:46:13,489 --> 00:46:20,489
throttling device. A porous plug have a porous
substance within a pipe and as mass flow flows
430
00:46:22,519 --> 00:46:28,239
through the porous plug, it leads to significant
drop in pressure. Capillary tube is another
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00:46:28,239 --> 00:46:35,239
example of a throttling device. Thermodynamically
all the 3 devices are the same.
432
00:46:35,680 --> 00:46:40,239
For throttling devices, the net heat transfer
is 0. It can be assumed to be 0, there is
433
00:46:40,239 --> 00:46:45,819
no work output. Therefore w is 0, kinetic
energy and potential energy can again be assumed
434
00:46:45,819 --> 00:46:52,819
to close to 0. Therefore, the energy equation
will basically reduce to h 2 is equal to h
435
00:46:54,660 --> 00:47:01,660
1 that is the enthalpy across throttling device
is the same. This means that throttling processes
436
00:47:02,819 --> 00:47:08,940
are isenthalpic processes that is these are
processes where enthalpy is a constant. Since,
437
00:47:08,940 --> 00:47:15,940
h 2 is equal to h 1, we can also write them
in terms of their components that is internal
438
00:47:17,109 --> 00:47:23,049
energy and flow energy. Therefore, u 1 plus
P 1v 1 is equal to u 2 plus P 2 v 2 that is
439
00:47:23,049 --> 00:47:29,599
the sum of internal energy plus the flow energy
is a constant across a throttling device.
440
00:47:29,599 --> 00:47:36,529
Now, this means that you can have different
values of internal energy at the inlet and
441
00:47:36,529 --> 00:47:43,459
exit. That is compensated by a corresponding
change in the flow energy. Let us say, you
442
00:47:43,459 --> 00:47:49,849
have a drop in internal energy across the
throttling device. This has to be compensated
443
00:47:49,849 --> 00:47:55,839
by increase in the flow energy term across
the throttling device. This is required because
444
00:47:55,839 --> 00:48:01,130
the net change in enthalpy across the throttling
device will be the same and that is h 2 will
445
00:48:01,130 --> 00:48:08,130
be equal to h 1. Therefore, the total of u
1 plus v 1 should be conserved. Across a throttling
446
00:48:09,809 --> 00:48:16,809
device, the change in internal energy plus
P 1v 1 should be compensated by corresponding
447
00:48:18,279 --> 00:48:23,999
changes in internal energy and the flow energy
after the throttling device.
448
00:48:23,999 --> 00:48:30,859
To illustrate it, if you look at one case,
where P 2v 2 is greater than P 1v 1. This
449
00:48:30,859 --> 00:48:37,859
has to mean that u 2 should be less than u
1, so that there is h 1 equal to h 2 conserved.
450
00:48:40,549 --> 00:48:47,440
Therefore, this also means that if flow energy
increases, the temperature has to decrease
451
00:48:47,440 --> 00:48:53,569
because u is a function of temperature and
the reverse is also true. If flow energy decreases,
452
00:48:53,569 --> 00:48:57,390
you may also have increase in internal energy
and temperature.
453
00:48:57,390 --> 00:49:03,569
Now, for an ideal gas, enthalpy is only a
function of temperature. Therefore, for an
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00:49:03,569 --> 00:49:09,900
ideal gas or for a system, which involves
only ideal gases, the enthalpy and the temperature
455
00:49:09,900 --> 00:49:15,449
have to remain a constant during a throttling
process. For an ideal gas, enthalpy is only
456
00:49:15,449 --> 00:49:22,449
a function of temperature. For ideal gases,
enthalpy is only a function of temperature,
457
00:49:22,769 --> 00:49:27,949
which means that enthalpy has to be a constant
and therefore temperature also has to remain
458
00:49:27,949 --> 00:49:32,789
a constant across a throttling device.
459
00:49:32,789 --> 00:49:38,699
Just an example, as I mentioned, if across
this device, at the inlet, you had u 1 is
460
00:49:38,699 --> 00:49:45,160
equal to 87 point some kilo joules. P 1v 1
as something else and total enthalpy is 88.56.
461
00:49:45,160 --> 00:49:51,589
After the throttling device, let us say, u
1 has reduced and P 1v 1 and therefore, it
462
00:49:51,589 --> 00:49:58,199
has to correspondingly increase, so that the
net enthalpy is the same. During a throttling
463
00:49:58,199 --> 00:50:04,440
process, enthalpy of a fluid has to remain
constant, but internal energy and flow energies
464
00:50:04,440 --> 00:50:08,819
are inter-convertible. So, you can covert
internal energy into flow energy partially
465
00:50:08,819 --> 00:50:13,119
and so on and vice versa across a throttling
device.
466
00:50:13,119 --> 00:50:18,239
The last device that we are going to discuss
is a mixing chamber. Mixing chamber is a section,
467
00:50:18,239 --> 00:50:23,459
where there is a mixing process taking place.
Some examples are the mixing of hot and cold
468
00:50:23,459 --> 00:50:30,390
water at a T junction of a shower. For example,
you have 2 different masses coming in m 1
469
00:50:30,390 --> 00:50:36,809
and m 2, leaving out as m 3 for energy equation.
Therefore, it will reduce to m 1 h 1 plus
470
00:50:36,809 --> 00:50:42,449
m 2 h 2 and it is equal to m 3 h 3. Since
we assume again that net heat transfer is
471
00:50:42,449 --> 00:50:47,640
0, work is 0, changes in kinetic energy and
potential energy are 0.
472
00:50:47,640 --> 00:50:54,519
You combine the energy and mass balances because
m 3 is equal to m 1 plus m 2. Energy equation
473
00:50:54,519 --> 00:51:01,519
reduces to m 1 dot h 1 plus m 2 dot h 2 is
equal m 1 plus m 2 dot times h 3. This would
474
00:51:01,680 --> 00:51:06,009
be the energy equation for a mixing chamber
process that is the net enthalpy leaving the
475
00:51:06,009 --> 00:51:10,459
system is equal to sum of the enthalpies entering
the system.
476
00:51:10,459 --> 00:51:17,459
Now, that brings us to the end of this lecture.
We have discussed about the first law as applied
477
00:51:18,839 --> 00:51:25,839
to flow processes. Let us take a look at what
we had discussed during this lecture. We discussed
478
00:51:25,849 --> 00:51:31,819
about the first law of thermodynamics applied
for open systems or flow processes. We then
479
00:51:31,819 --> 00:51:38,449
discussed about flow work and the energy associated
with a flowing fluid. Then, what is the total
480
00:51:38,449 --> 00:51:45,449
energy that is available for a flowing fluid.
As a consequence of mass flow rate what is
481
00:51:45,449 --> 00:51:51,599
the energy associated with mass flow process.
We then discussed the energy equation for
482
00:51:51,599 --> 00:51:56,420
steady-flow processes. We derived the energy
equation for the steady-flow processes and
483
00:51:56,420 --> 00:52:01,940
also we discussed about application of these
energy equation. The basic energy equation
484
00:52:01,940 --> 00:52:08,940
for certain steady-flow devices like nozzles
and diffusers, turbines, compressors, mixing
485
00:52:09,979 --> 00:52:16,150
chambers or heat exchangers. We also discussed
about how we can apply energy equation for
486
00:52:16,150 --> 00:52:23,009
throttling processes.
The idea was to help you in understanding
487
00:52:23,009 --> 00:52:28,529
what is a process involved in making certain
justifiable assumptions and applying first
488
00:52:28,529 --> 00:52:33,390
law thermodynamics for different flow processes.
If you come across a different type of flow
489
00:52:33,390 --> 00:52:40,390
processes based on what we had discussed here,
you could probably be able to extend or simplify
490
00:52:40,670 --> 00:52:47,319
the general energy equation to any other open
system or any other flow processes.
491
00:52:47,319 --> 00:52:53,430
Now, in the next lecture, what we are going
to discuss? Second law of thermodynamics,
492
00:52:53,430 --> 00:53:00,150
we shall define what is known as thermal energy
reservoirs. We shall then state the Kelvin-Planck
493
00:53:00,150 --> 00:53:05,359
statement of the second law of thermodynamics.
We shall subsequently discuss about refrigerators
494
00:53:05,359 --> 00:53:12,359
and heat pumps. As a consequence, what is
the Clausius statement of second law of thermodynamics?
495
00:53:13,209 --> 00:53:20,130
We shall then prove that the Kelvin-Planck
statement of second law and the Clausius statement
496
00:53:20,130 --> 00:53:24,880
of second law are equivalent. Towards the
end of the next lecture, we shall again discuss
497
00:53:24,880 --> 00:53:29,859
about certain devices, which violate the second
law of thermodynamics. They are known as perpetual
498
00:53:29,859 --> 00:53:34,359
motion machines of the second kind. We had
already discussed this for the first law,
499
00:53:34,359 --> 00:53:39,150
which were known as perpetual motion machines
of the first kind. Those devices, which violate
500
00:53:39,150 --> 00:53:43,259
second law, are known as perpetual motion
machines of the second kind. So that is something
501
00:53:43,259 --> 00:53:45,670
we shall discuss during the next lecture.