Good morning I welcome you all to this session
last class, we have discussed the first
law of thermodynamics we started discussing
the first law of thermodynamics, and then
we define a property enthalpy h which is the
combination of intermolecular energy a part
of the internal energy or in a stationery
or closed system. That is the internal energy
plus
the product of pressure, and volume, and we
also recognize that this internal this
enthalpy is a property of this system bears
is physical significance relating to energy
transfer an open system or a flow process.
So, today therefore, we will apply this first
law for a flow system or a flow process or
an open system, and we will see how does
enthalpy bear, a physical significance with
the energy transfer across the boundary of
such an open system.
.
So, for that what we do? We just consider
a flow system, like this let us 
let us consider a
device a flow device this is known as a flow
device flow device well which has an inflow
where there is continuously an inflow of fluid,
and continuously an outflow of fluid
inflow, and outflow of fluid, and this flow
device is connected with a shaft. So, that
a
.shaft work is coming out in the form of the
rotation of the shaft let us consider an infinite
small amount of work d w d cut, w as I explained
earlier is coming over a time d t s that
we can represent that d w d t as the rate
at which the work is coming out from a flow
device in the form of shaft work.
And also let us consider, similar way infinite
small amount of it d cut q is added
continuously over an interval of time d t.
So, that we can represent that d q d t as
a
continuous rate of heat flowing or heat transfer
into this flow device. Now this flow
device where there is a continuous flow of
fluid at 1 section, and continuous flow of
fluid
out in other section it is flow in, and flow
out, and developing both work, and heat; that
means, transferring in terms of work, and
heat in the surrounding is termed as an open
system or a controlled volume. In fact, we
consider a close a boundary circumscribing
this flow device this way if we consider a
boundary this way.
This this system within this fixed boundary
is known as an open system or a control
volume or a control volume which characterized
by this fixed boundary as I have already
told you that a an open system is a control
volume system or simply a control volume,
where the boundary where the system or the
control volume is characterized by the fixed
boundary, and across which the mass transfer
takes place there is continuous inflow of
fluid, and there is continuous outflow of
fluid now, if we want to make the conservation
of energy statement for this open system or
control volume. It is very simple that we
can
write it like this the rate of rate of energy
in flow to the control volume I write simply
c
v.
Minus the rate of energy well outflow from,
c v must be equal to the rate of change of
rate of change of energy within the control
volume or the open system in in consideration
within the c v this is the very simple, and
a preliminary statement of conservation of
energy that the rate of energy in flow to
control volume minus the out flow rate of
energy, outflow from control volume is equal
to the rate of change of energy with the
control volume now, you see therefore, we
have find out what are the inflow energy
inflow quantities into the control volume,
and what are the outflow quantities.
In the very first case we see that we have
considered, the work is coming out in the
form
of the shaft work from the control volume
this is 1 way the energy is coming out in
the
form of work, and similarly the energy in
the form of heat is going into the control
.volume, because we have considered these
heat energy to be going into the control
volume for the general deduction apart from
this the energy quantities coming to the
control volume due to the inflow of fluid
mass across this section. Similarly energy
which is going out of the control volume are
associated with the f flux o fluid mass from
the control volume across this boundary.
]so therefore, we have to now, recognize to
know the different energies, which are
coming in to the control volume or coming
out of the control volume, what are the
energy? Quantities that are associated with
the flow of a fluid stream. So, we have to
first
recognize that. So, what are the different
quantities or different energies different
forms
of energies that are associated with the flow
of a fluid stream with the flow of a fluid
stream the energy quantities or the different
energies, those are associated are first is
intermolecular energy that is the kinetic,
and potential energy of the molecule
intermolecular energy exist in all substances
under all conditions if it is at a temperature
above that the absolute 0.
So, this intermolecular energy number 2 is
the kinetic energy this is, because of the
flow
velocity though a fluid flow or fluid stream
lowing stream possesses kinetic energy
another energy is a very well known energy
is the potential energy. This is by virtue
of
its placement or virtue of its position in
the conservative force field if there is no
such
external conservative force field. So, gravitational
force field is there. So, gravitation
potential energy is there apart from that
there is another flow energy, which is there
in a
flow of streams or fluid is the pressure energy
that you have read in your fundamental
fluid mechanics class.
What is that pressure energy or it may be
termed as flow work this is, because of the
existence of pressure at a particular section
in the flowing fluid as you know that it is,
because of the pressure a section a fluid
layer at a section is capable of pushing the
neighbouring layer to make its flow through
the . So, therefore, to maintain the flow
the
layer at any section the fluid layer at any
section has to continuously push the
neighbouring layer, and by virtue of its pressure
t can do. So, and it makes its way
through. So, therefore, we see to maintain
a flow each, and every section the fluid at
each, and every section does work on the neighbouring
layer.
.And that work is known as flow work whose
magnitude is pressure times, this specific
volume per unit mass well that is the work
per unit mass, and I we look from this angle
that this capability of doing this work by
a particular layer of fluid at any section
on its
neighbouring layer to make its way, through
the flow is known as the energy which is
inherent in the fluid or stored in the fluid
layer, and it is stored in the fluid layer
in the
from of pressure energy we tell. It in the
as pressure energy sometimes we refer it to
as
pressure energy or sometimes we refer it to
as low work.
So, therefore, we see what are the energy
quantities, that are associated with a flow
of
fluid stream 1 is the internal energy by virtue
of its state thermodynamics state another
is
the kinetic energy by virtue of the flow of
velocity another 1 is the potential energy
by
virtue of its position in a conservative force
field, and last 1 is the pressured energy
by
virtue of the pressure in the flow field.
So, I we recognize these energy quantities
ten we
can now, come we can now, recognize the energies
which are coming into a into the
control volume, and going out of the control
volume.
So, therefore, if we again see this let us
consider the rate of inflow of mass, as d
m on d t,
and let the outflow of mass is d m to d t
the rate of the outflow this is the rate of
inflow
of mass this is the rate of outflow of mass.
Now, to identify those energy quantities we
will have to fix the state of the fluid at
inlet let the inlet pressure is given by p
1 the
corresponding specific volume is given by
v one. Let the in let velocity is given by
v one,
and let from any arbitrary referenced atom
the elevation the vertical height at the inlet
section is given by z 1 the corresponding
quantity at out. Let is given by p 2 with
a
subscript to v 2 that is the specific volume
v 2 is the pressure velocity v two, and also
the
elevation from a referenced atom vertical
height as z two.
..
Therefore with this nomenclature we can now,
write the rate of energy coming in to the
control volume associated with the inflow
of fluid, as d m 1 d t that is the rate times
this
specific that is the per unit mass internal
energy plus the flow work we write first that
the
pressure energy plus the kinetic energy per
unit mass plus the potential energy you
consider only. The gravitational potential
energies present; that means, no other external
conservative force field is there. So, this
is the rate of energy coming into the control
volume through the inflow of fluid mass well,
plus as we have already considered in this
figure for this control volume.
This heat is being added at the arte of d
q d t. So, we write d q d t. So, therefore,
this is
the in flow of energy to the control volume
similarly, what is the outflow of energy from
the control volume minus that is the energy
quantities which are going out with the fluid
stream at the outlet; that means, the outflow
of fluid that becomes u two, that is the
internal energy at the outlet p 2 v 2 pressured
energy per unit mass at the outlet fluid
stream plus v 2 square by 2 plus g z 2 plus
the work quantity, because we have
considered the for the control volume the
shaft work is coming out.
That means controlled volume or the flow device
or the open system develops work in
the form that work is coming out in the form
of the shat work, coming out to the
surrounding. So, therefore, we can write it
this is equal to d w d t, and difference.
Minus.
.D t minus.
Sir d w by d t.
Very good minus, and this will be equal to
the rate of change of energy within the
control volume. Where e c v is the energy
in the control volume internal energy in he
control volume at any instant internal energy
in the control volume in the control volume
in this c v at any instant this internal energy
at any instant comprises all the energies
in
the control volume, that is the intermolecular
energy the kinetic energy due to the
motions of fluid particles within the control
volume plus the potential energy. So, this
is
the statement of the conservation of energy.
So, now, under a special an this you can consider
as the general steady flow in a general
sorry not steady general flow, general energy
equations for a flow process we can
consider this as the general energy equation
general you can write it general energy
equation general energy equation for a flow
process for a flow process for a flow process
or open system for an open or for an open
system. Now, for a special case when the flow
process is steady as a special case for steady
flow, process for a steady flow process
when the flow process is steady.
That means we know the definition of a steady
process or a steady flow fill; that means,
each, and every point in this system the properties
are in variant with time; that means, as
a whole the properties within this control
volume or the open system would be invariant
with time this will happen when the mass flow
rate at inlet, and mass flow rate at outlet
will be same. So, there will not be any mass
accumulation within the control volume, and
at the same time the energy inflow to the
control volume must be equal to the energy
outflow from the control volume. So, that
there is not change of energy within the
control volume with time.
So, that the energy quantity remain same the
mass remain same. So, all the
thermodynamic properties at each, and every
point within the control volume will remain
same. So, this situation refers to a steady
flow process. So, therefore, we can write
in
those process first of all from mass point
of view d m 1 d t is equal to d m 2 d t is
equal
to zero; that means, the control volume mass,
will not increase sorry sorry sorry sorry
I
am sorry its constant let d m d t I am sorry
this is constant otherwise for an general
case
.we can write, similar to energy conservation
d m 1 d 2 minus d m 2 d 2 very correct d m
c v.
Which means the general conservation of mass
the rate of mass inflow to the control
volume minus rate of mass outflow from the
control volume is equal to the rate of
change of mass in the control volume. So,
in n c v is the mass of the control volume
at
any instant. So, for a steady flow this becomes
0 this is equal to d m d t similarly this
becomes zero. So, if we now, this is the general
equation for a steady flow process, but
conventionally this is written in a form like
this if we divide the terms each, and every
term by the d m 1 d t or d m to d t which
is equal to each other at let it be d m d
t, and
divide each, and every term by d m d t, then
we get an equation if we divide the terms
on
the the left hand side all the terms by d
m d t.
.
We get an equation in this form u 1 plus p
1 v 1 plus v 1 square by 2 plus g z 1 plus
d
cube d m well is equal to what we can write,
u 2 plus p 2 v 2 plus v 2 squarer by 2 plus
g
j 2 plus d w d difference is that here, we
what is what is implied by this equation the
rate
of energy inflow is equal to the rate of outflow
from the control volume each, and every
term represents the energy per unit mass.
It is more conventional to represents the
energy
quantities per unit mass, this is the heat
flowing to the control volume per unit mass
this
is the work flow out of the control volume
per unit mass.
.In this equation these were the energy quantities
on time bases the rate of energy, coming
in rate of heat coming in. So, rate of energy
quantity going out rate of heat going rate
o
work coming out or going out, but in general
the for a steady flow process. It is
conventional to express this quantities in
terms of mass bases. So, therefore, each,
and
every quantity is the energy per unit mass.
So, energy coming in per unit mass is equal
to
energy going out per unit mass in the control
volume, and it is out of the control volume.
So, this is known as steady flow energy equation
the special case steady flow energy
equation.
Now, you see that we have defined this, summation
of u plus p v as the enthalpy. So,
specific internal energy plus pressure times
specific volume can be defined as specific
enthalpy since the enthalpy is defined as
u plus p v you known u is an extensive property
volume is an extensive property. So, if you
define the corresponding intensive property
by dividing by the mass of the system. We
can define the specific enthalpy the sum of
specific internal energy plus the product
of pressure, and specific volume. So, therefore,
we can, write the equation in terms of the
enthalpy quantities v 1 square by 2 plus g
z 1
plus d q d m is equal to u 2 plus p 2 v 2
plus sorry we are writing h 2 plus v 2 square
by 2
plus g z 2 plus d w d x.
This can be manipulated algebraically in this
fashion that you can write this way that d
q
rather here you can write this way, d w d
m minus d q d m is equal to h 1 minus h 2
plus
v 1 square minus v; that means, we are taking
this here, and we are writing this in the
left
hand side a simple algebraic manipulation
h 1 minus h 2 d w m all right g z 1 minus
z 2
divided by oh this is g z 1 minus z two. So,
you see that this is the final outcome of
the
steady flow energy equation, but incidentally
it happens that in all engineering devices
all these things that heat transfer work transfer
do not take place simultaneously. And
moreover the changes in potential energies,
and kinetic energies are sometimes neglected
for example, let us consider only the devices
which are only work interactions which are
only work interactions without any heat interactions.
..
Let us consider turbine or compressor, let
us consider a turbine for example, the simple
case the turbine you known the turbine this
is insulated. Let this is 1 this is 2 let
this is
insulated. So, in this case d w d t is there
which is coming out whereas, d q d t is 0
or you
can tell d w d m is there, and d q d m is
zero. And if we neglect the change in the
kinetic
energy at the inlet, and outlet of this turbine,
and also the changes in the potential energy,
then what we get we get from this equation,
you see that this becomes 0 this becomes
zero. So, simply, and also this is zero; that
means, the work coming out from the turbine
in that case is h 1 minus h two, and therefore,
we see that for only work interacting
devices or example in a turbine. When work
is coming out where the changes in kinetic,
and potential energies are negligible compared
to the changes in the enthalpy.
It is the property enthalpy whose difference
between the inlet, and outlet state
straightaway give the work interaction quantity.
Similarly in case of compressor if you
see that the reverse in case of a compressor,
if you see in case of a compressor c let us
in
case of a compressor if you see it is the
same thing that if it is insulated let this
is one,
and this is two, and if it is insulated similarly
way, d q d m is zero, but
it absorbs
energy; that means, shaft work is filled into
the compressor; that means, it has d w d m,,,
but in this direction.
In that case also we can express the same
equation that is in this case, also d w d
m is h 1
minus h the same equation here h 1 is greater
than h 2 always. So, that this quantity is
.positive work is coming out in this case
h 1 is less than h two; that means, the enthalpy
of the outlet in is outlet section is more
than that at inlet. So, that d w d m is negative,
which implies physically the work is gong
in similarly there are devices like heat
exchange are where there is only heat input
to a system for example, a fluid flowing
through a pipe flow, and there is no work
transfer; that means, d w d m is zero.
And if the flow area is uniform we can neglect
the kinetic energy change; that means, v 1
is almost equal to v two, and if we change
the potential energies neglect the changes
in
the potential energy here also we can write
from this equation, this same this equation
again we can write here making this 0 minus
d q d m is h 1 minus h 2 or d q d m is h 2
minus h one; that means, it is the difference
of enthalpy which gives this heat transfer;
that means, this is positive; that means,
heat is added when h 2 is more than h 1 if
h 2 is
less than h 1 the heat is rejected.
So, therefore, we see now, physical significance
of enthalpy is such that in flow devices
steady flow devices the difference between
the enthalpy at inlet, and outlet section
gives
the net work, and heat interactions by the
device with the surroundings. So, for only
work interacting devices. It is say the work
transfer is simply coming out as the
difference in enthalpy provided the change
in kinetic, and potential energies are
neglected, and in all practical applications
for those devices which interact in terms
of
either work or heat the changes in kinetic,
and potential energies are usually very small
as compared to the changes in enthalpy.
So, therefore, it is the change in the enthalpy
quantity, which is straightaway giving the
work. So, here you see the enthalpy in such
system bears more or less the synonymous
role as the internal energy as if it is, because
of the release of enthalpy or absorption of
enthalpy the work transfer takes place in
a turbine the work is done, because of the
change in enthalpy. So, enthalpy bears the
same physical significance or same physical
entity as the internal energy does for a closed
systems, similarly in a compressor the
work is absorbed by virtue of which its enthalpy
is raised, because it is the difference of
enthalpy which is given the work quantity.
Similarly, in case of heat interacting devices
where only heat interaction takes place it
is
by virtue of the change in enthalpy the heat
energy is coming out this equated with the
change in enthalpy. So, therefore, in this
way we can tell that enthalpy bears a similar
.thing that is the internal as internal, but
truly speaking it is not the internal energy
it is a
property of this system whose definition comes
only by the mathematical expression h is
equal to u plus p v. Now, we see some more
examples which may come in your
compressible flow calculations that there
is no heat, and work interaction.
.
That means the flow takes place through a
nozzle, but there is a change in velocity
for
example, the flow through a nozzle or a diffuser
flow through a nozzle or a diffuser this
is a nozzle this is a nozzle one, and two,
and this is a diffuser flow through a nozzle,
and
diffuser here what happens v 2 is greater
than v one, and here v 2 is less than v 1
that is
one, and two, similarly the p 1 here is greater
than p 2 here here v 2 is more less than v
1
p is less than p two.
So, pressure increases velocity decreases
pressure decreases velocity increases in this
case of you see the energy equation, here
again the common these 2 terms are zero. So,
this terms we will take care of and if we
neglect the potential energy changes to be
0,
because the there is no change or the change
in the vertical heights at the inlet, and
outlet
are small, then here the equation tells like
that, h 1 plus v 1 square by 2 is equal to
h 2
plus v 2 square by two. Now, if you explicitly
see u 1 plus p 1 v 1 plus v 1 square by
two; that means, this is the conservation
of energy when heat, and work quantities are
not
there the flow takes place in such a way the
total energy comprising the enthalpy, and
the
kinetic energies remain same.
.That means, here the kinetic energy is created
out of the enthalpies similarly here the
kinetic energies are destroyed or kinetic
energies are reduced at the pressured energy
the
enthalpies are increased by virtue of the
kinetic energy, you see that this is the
expression. So, this expression holds good
for this type of now, I will deduce the
Bernoulli’s equation which we have already
read at you fluid mechanics the well known,
Bernoulli’s equation from this principle
or from this equation of energy equation for
a
steady flow process for this bernoulli’s
equation. If you recall bernoulli’s equation
is
basically the equation for conservation of
energy.
This is a mathematical statement for the conservation
of energy for the flow of fluid, but
this is not the Bernoulli’s equation perfectly,
this is the broad statement now, what type
of energy conservation in Bernoulli’s equation
we consider only the mechanical energy.
We see how the total mechanical energy is
conserved that is why it is known, as
mechanical energy equation this is the conservation
of mechanical energy, and we a
simplified form of energy in certain special
cases of fluid flow; that means, in case of
flow of an inviscid steady flow of fluid inviscid
steady flow of fluid.
.
What we know that in Bernoulli’s equation,
if you recapitulate the your Bernoulli’s
equation Bernoulli’s equation Bernoulli’s
equation we know that the pressured energy
p
by rho sometimes, we can write it in terms
of this specific volume or 1 by rho these
are
synonymous, but usual convention this that
in fluid mechanics we use is as 1 by rho we
.use density rather instead of specific volume,
but
in thermodynamics we prefer this a
convention the specific volume. So, p by rho
plus v square by 2 we know these
expression g z is equal to constant what does
it mean.
So, this quantity as we have seen is the pressured
energy or the flow work per unit mass
this quantity is the kinetic energy per unit
mass in a flow of fluid, and this quantity
is the
potential energy. So, sum of these three quantities
give the total mechanical energy per
unit mass in the flow of the fluid which are
associated with the fluid stream, and
Bernoulli’s equation tell, that for an inviscid
incompressible 1 of the conditions inviscid
incompressible steady flow of a fluid these
three things are constant, and this is constant
only along a stream line if the fluid is rotational.
And if the fluid is irrotational constant
is constant throughout the flow field this
very
important again I tell you this is the recapitulation
in many times, this question is asked
in an interview or in viva quize what is Bernoulli’s
equation, and what are kits
restriction. So, if you tell the Bernoulli’s
equation it is basically the conservation
of
mechanical energy. So, that those mechanical
energies associated with the flow of a fluid
stream. This is the pressure energy of the
flow work kinetic energy, and the potential
energy usually we will consider the gravitational
potential energy. So, sum of these
three energy will be constant provided there
is no energy interactions from outside.
So, sum of these three energy will be constant
provided there is no energy interactions
from outside. So, first condition there is
no heat, and work interactions in the fluid
flow
from outside number 2 there is no change from
mechanical energy to other form of
energy or intermolecular energy, which we
physically call it as dissipation of energy
or
degradation energy of energy from the second
law of thermodynamics, because energy is
converted from a higher grade to a lower grade.
So, there is no such conversion which is
only possible if the friction of the fluid
is as same; that means, fluid is non viscous
or
inviscid.
So, inviscid fluid, and fluid is incompressible
pressure remains same sorry I am sorry
density remains same density does not change
in the flow. So, for an incompressible
inviscid, and steady flow fluid without any
work, and heat interactions these three
quantities p by rho represents the pressured
energy per unit mass, the v square by 2
represents the kinetic energy per unit mass,
and g z represents the gravitational potential
.energy per unit mass sum of these three quantities
remain constant, but this constant
value remains same only along the stream line
only along the stream line.
But for a rotational flow, but for any rotational
flow this is constant throughout now, I
would like to tell you I do not know whether
you known, these things from basic fluid
mechanics class that why this constant along
stream line why in a rotational flow what
isa the difference between rotational irrotational
flow from the energy from the energy
view point from the view point of energy transfer
you know the rotational, and
irrotational flow definition is that a flow
is said to be irrotational when the rotation
is 0
rotation of fluid element that is the curve
of the velocity vector is 0 how do you define
the rotation of a fluid element. It is the
average angular velocity of the 2 adjacent
linear
sides of a fluid element which were initially
perpendicular.
And this is given by the curl of the velocity
vector three rotational component in three
coordinate direction. So, if the rotation
is 0 flow is irrotational,,, but if the rotation
is not
0 the call of the velocity vector is not 0,
then the flow is rotational from the energy
point
of view in a rotational flow field. There
is always a work interactions work is either
given to the fluid from outside or work is
extracted from the fluid form outside from
the
inside to the outside from the inside to the
outside. So, because of this work giving to
the
fluid or taking from the fluid the constant
varies from stream line to stream line,,,
but if
there is more interaction along with no heat
interaction, then the fluid Behaviour is the
rotation.
So, it is both in visit, and irritation, but
it may be an in visit flow, but rotational
flow,
but
the work interaction may take place from outside
to the fluid or inside that mean
between the flow of fluid with the surrounding
this is very important. So, if you discuss
the work interaction; that means, flow of
an in visit fluid without any work interactions
from the surrounding also no heat interactions
from the surrounding, then the fluid
behaves as an irrotational fluid in visit
irrotational fluid, and visit irrotational
flow is
synonymous to isentropic flow in thermodynamics
isentropic flow without work
interactions.
You can write, isentropic flow without work
interaction. Now, let us immediately
recapitulate hurriedly how we can derive,
this equations now, from the steady flow
energy equations now, you see this is steady
flow energy equation general steady flow
.energy equations. Now, if we apply this to
a fluid visit irrotational flow of fluid visit
irrotational in visit steady flow of fluid.
So, irrotational flow means there is no work
or
heat interaction this is 0. So, therefore,
we can write that this steady flow energy
equation takes the from h 1 plus v 1 square
by 2 plus g z 1 is h 2 plus v 2 square by
2
plus g j 2.
This it is very simple, we can write from
a differential in a differential form as d
of we
can write h rather we write d h plus d v square
by 2 plus d of g j is equal to 0, because
if
we integrate it between section one, and 2
you get h 2 minus h 1 plus v 2 square by 2
minus v 1 square by 2 plus g j minus 0 you
can write this, now, if you recollect the
thermodynamic property relation which I will
discuss again in this class that t d s is
equal
to d u plus p d v, and we can write h is equal
to u plus p v.
So, d h is equal to d u plus p d v plus p
d v sorry. So, if you replace the d u in terms
of d
h we can write d h is t d s minus v d this
is the very useful relationship, which we
get I
will explain this in the next class also d
h is t d s plus v d p yes d h is t d s plus
v d p all
right. So, if you substitute this. So, for
an isentropic flow. So, I have told you that
in visit
irrotational flow is synonymous to isentropic
flow without work interaction. So, if I
replace it we get v d p plus d v square by
2 plus d of g z is.
.
So, I we integrate this equation well, if
we integrate this equation we get integration
of v
d p plus integration of t v square by 2 plus
integration of d g z is equal to zero. So,
far we
.have put the constant of isentropicness of
the flow isentropic that is the flow is isentropic
that is in visit without any work interaction,
but incompressibility has not put into
consideration. If we put the incompressibility
into fact that this v comes out from the
integration plus sorry it is constant after
integration it is constant very good d v square
by
2 plus integration of d d z is equal to constant
ok.
So, therefore, we can write after integration
p by rho v can be written as 1 upon rho plus
v square by 2 plus g z is constant, and this
is constant throughout the flow field, and
this
is simple the mechanical energy equation for
an in visit for an in visit invisuid steady
irrotational irrotational. These are the conditions
in compressible flow in gravitational
field in gravitation force field; that means,
if a in visit steady irrotational in compressible
flow in gravitational force field takes place
the irrotational flow means there is from
the
energy point o view no energy interaction
neither it nor work total mechanical energy
that is the pressure energy kinetic energy,
and this thing constant.
But if we draw this 
of conservation of energy for a real fluid
we 
use a term as loss which
is meant as the loss in mechanical energy.
So, some form of mechanical energy is lost;
that means, has disappeared, but it has taken
place in the other form as the intermolecular
energy. So, this conversion we term is as
a loss in the mechanical energy equation,
because this is a loss from the sense of total
mechanical energy; that means, the fluid
flow is unable to conserve its mechanical
energy as a whole well this is, because of
the
friction. So, this is all about the principle
of application of first law of conservation
of
energy to an open system any question.
Thank you.
.
