Good morning. Welcome to this session, last
class we were discussing the first law of thermodynamics. We have started the first
law of thermodynamics applied to a finite process and to a cyclic process, what was
it that we recognized the algebraic sum of heat interactions between a system and the
surrounding in a cyclic process becomes equal to the algebraic sum of the work interactions
between a system and its surrounding in a cyclic process, which means that the cyclic
integral of heat interactions minus the work interactions between the system and the surrounding
is 0 and the immediate consequence of heat is that the difference between heat and work
interaction executed by a system in a finite process becomes equal to the change in a property
which we definite as internal energy of the system. Now, let us write this, therefore if we just
recognize which we discussed earlier that in a process connecting these two steps points
1-2 the heat added you want to is it’s change in internal energy E2 minus E1 plus the work
coming out of this system during the process from state 1 to state 2 and our usual sign
convention is the heat added to a system is positive while the work done by the system
is positive. If you write it for an infinite small process
with differential amount of the quantity, that means, if deltaQ is the infinite small
heat added for an infinite small process executed by a infinite small system, then this can
be written in terms of this dE here, E is a point function it’s a change in internal
energy which can be written in terms of its perfect differential that’s why I used d
and delta W. So, this is the first law that if we take
the cyclic integral, then we get automatically cyclic integral of del Q is cyclic integral
on both this sides dE plus this cyclic integral of delta W.
You know the E is a point function, so that cyclic integral is 0 the cyclic integral of
delta Q, this is the first law. In fact we came this way that we first recognize this
and then we define that this minus, this is a change in a property which is the internal
energy E2 minus E1. So, this is the precise form of the first law. Now, if we write this
in terms of per unit mass this will be internal energy per unit mass, but before that I should
tell this way. Now, you concentrate on these two equations,
if a closed system now this first law first of all you must know this first law that heat
added is change in internal energy plus work done is applied for both closed system and
open system, but the version of this law in a open system is little different which I
will tell afterwards. So, when we deal basically with closed system,
we write in this form that delta Q is equal to dE plus delta W. So henceforth, whatever
we will be referring to a closed system. So, let us now think of a closed system, and then
we can write this internal energy we have already recognized for a closed system is
composed of intermolecular energy plus kinetic energy plus potential energy.
All types of energy stored within this system apart from the intermolecular energy which
is because of the molecular kinetic and potential energy can be described in terms of microscopic
potential energy. For example the gravitational potential energy
and kinetic energy because of the movement or motions of the particles of the system
macroscopic particles of the system not the molecular motion, so therefore d of this comprises
d of u plus d of this quantities kinetic energy plus potential energy.
Now, for a closed system, when it is at an equilibrium state it is at rest, there is
no motion within in it and the properties are uniform throughout, so therefore the kinetic
energy within the matter is 0. So, therefore usually for a closed system
at equilibrium state because we are interested between the equilibrium states from one equilibrium
states, one to other equilibrium states, so at any equilibrium state these kinetic energy
has to be 0, but potential energy may not be 0 because any system for example in the
gravitational field of the earth it has some potential energy, but the change of potential
energy in all practical cases are very negligible or very small compare to the change in its
intermolecular energy. So that we can neglect this change of potential
energy, because of the fact there is no kinetic energy its change is also 0, so that finally
for a closed system at equilibrium state we can write the change of its internal energy
is the change of its intermolecular energy u only.
This is the usual sign convention, sign is used in nomenclature, u has the intermolecular
energy and E has the internal energy. so therefore we will denote the internal energy for a closed
system henceforth by the term u is actually the intermolecular energy.
so when we will refer the internal energy of a stationary closed system at equilibrium
states we will refer to u only. so with this in mind, we can write then for a closed system
Q 1-2 is equal to u2 minus u1 plus W1-2 in differential form delta Q is du plus delta
W. Now, we see that these are the first law when no restrictions about the system were
given. It is valid for a closed system, whatever
maybe it’s internal energy, but when you write in this fashion we consider the internal
energy is composed of only intermolecular energy, that means, closed system will be
at rest there will be no kinetic energy and more over that changes in the potential energy
during a process has to be a small as compare to the changes in its intermolecular energy
so that we can write this form of the first law for a closed system.
Now, if we go little ahead now you see put little more constant now if we consider a
closed system which perform only reversible displacement work which we discussed earlier,
that may quasistatic displacement work which is nothing but pdv work we have recognize
this earlier So, any form of reversible or quasistatic
displacement work that is the work interaction between the closed system and the surrounding
because of the displacement of its system boundary that is the reversible or quasistatic
displacement work is given by the expression pdv infinite small amount of work is given
by pdv, that means delta W equals that’s why this work will be refer henceforth as
pdv work, so if we consider the closed system only performs this type of work that is pdv
work, then we can write for that occasion Q1-2 is u2 minus u1 plus integration of pdv
between the state 1-2 or deltaQ is equal to du plus pdv.
Now, we see that there are three different forms, now, this is valid there is no restriction
input even the restriction of closed and open systems are not imposed here, this is the
general statement of the first law usually we use this equation for closed system.
When the restriction of closed system at equilibrium state comes with the further assumption that
change in potential energies are negligible compare to the change in intermolecular energy
then this becomes these two, one is for a finite process another is for an infinite
small process in differential form for closed system at stationary equilibrium states.
Then if we again further put another constant that closed system only interacts with the
surrounding in the form of pdv work then the first law for a finite process becomes like
that, now if we write this expression, sometimes it is written for a closed system conventionally
per unit mass. That means if you write for a finite process
let m is the mass of the closed system which remains fixed per unit mass then this becomes
the change in the specific internal energy and this is expressed as small u script is
same but the size is small u and this will be definitely p specific volume this is denoted
by small v. So, therefore here also del Q by del m, if
del m is the mass of an infinite small system which receives infinite small amount of heat
del Q here also it is the differential of small u that means small u refers to specific
internal energy per unit mass plus p into dv specific volume, that means this is a point
function in terms of specific values; these becomes an intensive property, this is also
specific internal energy and this is the heat addition per unit mass after this I will go
to the definition of a new property enthalpy. Now if you are asked at any time, what is
the definition of enthalpy the very first line of this definition comes from the mathematical
statement that enthalpy is a property which is defined as H is the denoted which is defined
as summation of internal energy plus pressure and volume for this is the enthalpy, very
first line of its definition there is no physical concept at the beginning.
Let us start with this definition that enthalpy is a property of a system with equals to u,
the internal energy for a closed system, internal energy for a open system also it is the same
internal energy u plus the product of p into v that is the definition of enthalpy.
So enthalpy is a property of a system it refers to both closed and open system in such a property
which is defined by this. Now, the question comes that, if we define enthalpy why such
a particular combinations of internal energy u pressure and volume has been defined as
enthalpy, this will be made clear afterwards and which will give the physicals significance
of enthalpy. So, at the present moment first we start with
this definition mathematically without understating much as if this particular combination is
defined as a property which is known as enthalpy of a system which refers to both closed and
open system, because it is the definition of a property which does not depend upon the
type of the system or the type of the process executed by a system it is a property of the
system. So, it’s very important sometimes it is
asked that whether enthalpy is defined for a closed system or open system, for any system
it’s a property just like a internal energy pressure volume temperature like that enthalpy,
now you see u is an extensive property v is an extensive property so, therefore H is a
extensive property so we can defined the intensive property corresponding
intensive property that is enthalpy per unit mass which is specific enthalpy as specific
internal energy plus pressure into specific volume, so this is the definition of specific
enthalpy. Now, with this definition of enthalpy, let
us now find out one small thing or a small problem to solve one small problem that a
closed system that a closed system receive some heat from outside and performing a process
from state 1 to 2 and is doing pdv work at constant pressure understand
So, what is the amount of heat added? Now, let us find out these as a problem. Now, Q1-2
we can write as what is this u2 minus u1 plus integration of pdv. Now, earlier also we recognize
that, integral cannot be found out or evaluated until and unless we know the process constant
that what is the relationship between p in terms of v. Now, if we know that process constant
is such that pressure is constant, then we can write this u2 minus u1 plus a constant
pressure that is p times v2 minus v1. so you can write this we can use p2 when you multiply
with v2 and or p1 multiplied with v1 since p2 is equal to p1 pressure remains constant
that means that constant pressure the displacement is taking place.
That means, heat is added in such a way that whenever it is displaced it is at constant
pressure then we get u2 plus p2 v2 minus u1 plus p1 v1 is equal to H2 minus H1. So, therefore
we see the heat added is given by the difference is enthalpy at the two states enthalpy at
state 2 minus enthalpy at states. This can be more easily found out from the differential
form that delta Q is equal to those who are little smatter then du plus pdv then you can
use it like this d of u plus pv why because p of dv is d of pv since p is constant it
can go within this differential operator that means it is dH
So, from here one can conclude Q1-2 means if you integrate H2 minus H1, therefore we
see that the heat interactions between a system closed system and the surrounding when closed
system performs displacement that pdv work at constant pressure is giving by its the
difference of enthalpy between the two states. That means, if it goes from state 1 to state
2, if the enthalpy at state 2 is more than the state 1 then heat is being added because
Q is positive in that case if the enthalpy of state 2 is less than the enthalpy of state
1, the heat is being rejected by the system that is Q is negative.
That means the different between the enthalpy at the two states point gives the heat interactions
between the system and the surrounding provided the system is a closed system and performs
pdv work at constant pressure. How the pdv work at constant pressure is perform when
the displacement work is being obtained, that means the system boundary expand the pressure
will reduce, then heat is added in such a way that the pressure remains constant that
means always expansion is taking place at constant pressure.
Similarly, you can conceive physically that is a compression type of thing, that means
if the system boundary collapses then pressure should increase so that heat should be rejected
in such a way that it counter balances that increase in pressure that means that compression
takes places with constant. Next is we will define specific heats. Specific heats
we have already we know there are two specific heats Specific heats at constant volume and
specific heats at constant pressure which we have already read at school level specific
heat at constant volume and specific heat at constant pressure. How do you define the
specific heat? We define the specific heats in terms of the heat quantity but specific
heat is a property of a system. Specific heat at constant volume or constant pressure, these
are the process constant, we know the specific heat at constant volume is defined as the
infinite small amount of heat added to a closed system at constant volume to raise its temperature
by an infinite small amount delta T, then this ratio del Q by del t is the specific
heat. Now we will define these in terms of this
system property rather than the path function and Q and W how do you do it? Let us start
the definitions of specific heat, let us consider the similar way that a system of small mass
delta m receives an infinite small amount of the delta Q for which there is a rise in
a temperature delta T. Then, as you know that definition of specific heat is delta Q per
unit mass, delta m divided by delta T, limit of this quantity as delta T approach a 0.
So, this is the basic definition of specific heat at constant volume provided, this is
done at constant volume v is equal to constant that means heat has to be added at constant
volume of the system. Now, if we look to the first law what is first law here, deltaQ is
du plus pdv now in defining this specific heat one restriction is there you always forget
that there should not be any other work transfer except the displacement work of the system.
So, we will not allow any other work transfer without the displacement work understand,
so only displacement work that is pdv work will be allowed. If we write delta Q by delta
m what is this is capital U I have just told this is small u that is means specific internal
energy plus p into dv. Now, when we make this at constant volume,
that means heat added per unit mass at volume constant, then we can write is as du v is
equal to constant, this is 0. So, therefore del Q by del m is du at constant volume.
So, if you substitute this, we get limit du at constant volume v is equal to constant
divided by delta T as delta T tends to 0, which becomes is equal to del U del t at constant
volume usually it is written like this in terms of the partial difference equation,
U is function of two variables, in case of single phase single component system u may
be expressed as a function of T and v small v small u.
So, therefore we can write partial differential, therefore you see that is specific heat at
constant volume becomes equal to partial differential of specific internal energy with temperature
at constant volume, it is clear cv is del U del t at constant volume. Similar way we can prove that cp is del h
and del t at constant pressure. How? Let us define small cp is equal to, as we know the
pressure constant has to be change only that means is the same thing at constant pressure
that means with the same nomenclature delta Q by delta m delta Q is the small amount of
heat added to a system of mass delta m, where p is equal to constant well divided by delta
T and limit of this quantity as delta T tends to 0.
Now, we will have to look to the first law, delta Q is du, I have told only we will relax
this work transfer no other work transfer only pdv work, if it has to be there at constant
pressure what it will be delta Q, first of all per unit mass equal to small u, plus pdv,
that means at constant pressure. Just now we have recognized this; so it is a repetition
at p is equal to constant it is the change in enthalpy and it will be change in specific
enthalpy because we are representing Q per unit mass.
We write it at pressure is equal to constant, that means p simply I am giving p is equal
to or you can write p is equal to constant, that means here this things will be limit
of change in specific enthalpy at constant p at constant p divided by deltaT and the
limit when deltaT tends to 0, so this becomes delh delt, so it will be delh delt at constant
pressure. So, therefore specific heat at constant volume
is the partial differential of specific internal energy with respect to temperature at constant
volume. Similarly, the specific heat at constant pressure of a system is the partial differential
of specific enthalpy with temperature at constant pressure.
So, this is now clear, with this I will close this lecture regarding first law applied to
a closed system, now I will come to the first law to this section, first law applied to
an open system or control volume. Now, we may ask, first law, why it is not
applicable? First law is the same, but how the different version when you use the control
volume the version changes little bit, because of the fact that control volume and closed
system that is open system and closed system, control volume system and control mass system.
Usually, when we tell system it means, I have told earlier that is the control mass system
we usually do not use this word control mass system means control mass system on the other
hand control volume means, control volume system or open system.
So, their basic definition, say the basic requirement of control volume system or control
volume and open system and closed system are different, what is the control volume system.
Let us recapitulate it is the fixed region in the space upon which the concentration
is put and there is a continuous flow of matter into the control volume and continuous flow
of matter coming out of the control volume and during this what happens the properties
of the matter which comings in and going out coming in and going out changes and the control
volume interacts with the surrounding in terms of work and heat.
For an example, we have appreciated a compressor is a control volume just an example where
continuously air comes at certain properties and goes out at certain property. Whereas,
the interactions with the surrounding in terms of taking work, work is being added to the
compressor, so that a desired change in the properties of the air flow in through it is
made. Similarly the turbine continuously some gas
or air is coming into heat and it is going out and this unit as a control volume delivers
work to the surrounding, heat exchanger continuously a fluid is coming in and going out and heat
is being added from the surrounding to this control volume.
So, these are the examples of control volume, let us discuss how the first law of thermodynamics,
that is basically the conservation of energy principle is written for this type of control
volume. Now, let us define a control volume like this, we are not much interested what
happens exactly within that we are only recognizing these as well control volume.
Now, let us identify a section where the fluid or the material comes in switches is usually
fluid it and the another sections where from the material which is usually fluid in practice
goes out it is usually a fluid goes out there may be multiple such inflow streams. There
will be multiple such outflow streams but it doesn’t matter much and the analysis
will remain same, but for our simplicity we consider only one inflow stream and one outlet
stream. Let us define a section here just near to
the control volume 1 1, let us define a section here ah just at the outlet of the control
volume 2 2. Now, this is the control volume, this is the situation, now how do you make
this analysis of the control volume and let us now consider as if the control volume is
receiving some heat, always we make analysis with the positive quantity of the energy indirections,
since control volume operates on the principle that always there is an continuous inflow
and there is a continuous outflow, this is the definition of control volume.
We always prescribe the energy interaction quantities with respect to time on the time
basis, because it is a continuous flow process, understand the continuous flow comes in and
flows go out that is the control volume concept, it may be steady, may be unsteady but it is
the continuous flow process. So, therefore because of this continuous process with time
things are expressed in terms of time rate, so delQ delt let us consider the rate at which
heat is added to the control volume. At the same time, we consider as if the control volume
is delivering work to the surrounding by this amount delta which may be coming in terms
of shaft work usually a control volume compressor turbine which I discussed there is a shaft
mounted on it rotates against a resisting torque gives the work done but we are not
much interested that mechanism by which work is transferred in thermodynamic. Let us consider
some work is continuously being transferred to the surrounding at the rate of del w delt
while the control volume receives heat by the amount delQ delt.
Now, first write the conservation of mass for the control volume which is very simple
we have done it fluid mechanics class, let us denote the mass flow rate because it is
rate is very important d m1 and d t at the inlet and let us write d m2 and dt is the
rate of mass out and at any instant let us define mcv is the mass of the control volume.
So, conservation of mass is very simple from commonsense, that is the mass rate flow coming
in inflow rate of mass minus mass flow rate reflects dm to dt, that means rate of change
of mass that means mass flow rate that is the mass flow rate mass flow rate coming into
the control volume minus the mass flow rate going out of the control volume is equal to
dm, that is the rate of change of mass within the control, this simple conservation of mass
which was exploited in fluid mechanics to derive the equation in terms of the velocity
components which we are not doing here that is known as popular continuity equation.
So, basically conservation of mass tells like that the rate of mass inflow minus rate of
mass outflow is the rate of increase of control volume mass mcv dt, now what about energy,
that is the first law conservation of energy also will come from the same commonsense that
if we denote Ecv as the internal energy of the control volume at any instant, the rate
of change of this internal energy within the control volume will be the difference between
the rate at which energy comes into the control volume minus the rate at which energy goes
out of the control volume, but here we will have to recognize one thing as unlike the
closed system, that in closed system energy comes in and goes out only in the form of
the work and heat, because no matter no material no mass of the system goes out or comes in,
but here since the masses are coming in and going out so the energy quantities associated
with this mass also comes within the control volume and going out of the control volume.
So, we will have to recognize these two forms of energy which are coming into the control
volume. For example here along with this mass the energy is coming into the control volume
apart from that heat coming into the control volume. Similarly work is going out of the
control volume energy same time the energy associated with this stream of mass going
out of the control volume causes an energy going out of the control volume, so to recognize
that one has to know what is the energy? So, now to know what is the energy coming
into the control volume associated with this stream of mass or what is the energy going
out of the control volume associated with the stream of mass, one has to no first what
are the different forms of energy which are being associated with a stream of mass as
a stored energy within that stream. If we defined e then we have to know, that
means what is the stored energy in a stream of fluid or mass in a stream of fluid, so
therefore we have to know this thing if we know that then we can find out what is the
energy that is s coming into the control volume per unit time, so either to know then what
is the different form of stored or what is the total stored energy in a stream of fluid.
Now, with the stream of fluid, what are the forms of energy that it can have stored in
as an internal energy of the stream of fluid if we tell that internal energy of a stream
of fluid. The Kinetic energy, potential energy, intermolecular energy because of its temperature,
kinetic energy because of the velocity potential energy because of its position in a conservative
force field intermolecular energy because of its temperature absolute above absolute
0 always as an internal energy intermolecular energy, then another one is pressure energy
that is the pressure energy of flow work which we discussed in detail at the beginning of
this class that pressure energy pv or the flow work.
So, a stream of fluid can be designated by all these sorts of energy at any section,
that means when a fluid flows in a stream a section of this fluid possess all this energy
if you describe in per unit of mass. So, therefore denoting that stored energy per unit mass
as e we can write e therefore becomes equal to in general u first I write u that is per
unit mass that is specific internal energy plus the flow you should write after work,
that is the pressure energy p by rho in fluid mechanics we deal it at p by rho plus the
kinetic energy as you told v square by 2 plus potential energy.
Here, we consider only the earth gravitational fields as the body force field there is no
extra body force field imposed on the fluid, so that we tell d dash gz that means this
total sum of these four quantities represents the energy stored per unit mass in a stream
of fluid and designate it as e. Now, if we designate the d here as e1 and
designate these as section e2 then we can write the energy coming in with this stream
is d m1 dt into e1, similarly energy going out with this stream is dm2 dt e2..
Here one assumption is made what is that the properties that means all these properties
not only this properties all properties of stream inlet and outlet stream remains invariant
with time they do not change time that means, the properties of the fluid stream entering
into the control volume are going out of the control volume is invariant with time there
is a change of this properties while flowing through the control volume, but at this section
all the properties are invariant so that we can write the energy quantities entering into
the control volume per unit rate is the mass flow rate into energy per unit mass flow rate
into energy per unit mass. now we can write the conservation of energy
like this that the total energy coming into the control volume then what it will be it
will be d m1 dt into e1 plus in the form of heat it is coming delQ by delt minus the energy
going out with the flow of mass outlet stream into e2 minus del W delt is equal to there
is no other statement it has to be like that, means, energy quantity coming into the control
volume minus energy quantity going out of the control volume is the rate of change of
internal energy of the control volume. So, this is precisely the conservation of energy
statement of the control volume. Now if we write this in terms of if we split
this in terms of this then we get here itself we can write d m1 dt you can see this one
know u1 plus p1 v1 plus v1 square by 2 plus g z1 plus delQ delt well minus d m2 dt.
So we just splitting this u1 u2 terms u2 plus p2 v2 it will be small v because it is per
unit plus v2 square by 2 plus g z2 is equal to is equal to dEcv by dt so the energy coming
in minus energy going out is the rate of change of energy within the control volume so this
is the energy coming in with the inflow of the stream.
This is what specific internal energy internal energy per unit mass this is the flow work
or pressure energy per unit must this is the specific volume this is the kinetic energy
per unit volume this is a potential energy per unit this is total is this stored energy
per unit mass with this flowing stream times the mass flow rate it is the energy flow rate
or energy flow rate coming into the control volume plus the rate of heat coming into the
control volume which we defined for the problem minus this is the similar way rate of energy
coming out of the control volume minus. Here minus the rate of work coming out of
the control volume which we first defined for the problem must be equal to the rate
of change of internal energy of the control volume and this is precisely the thermodynamic
to a control volume, it is very important this is first law of thermodynamics to a control
volume. Here, what we have assume we have assume that
the properties at the inlet stream and the outlet streams are invariant with time there
may be multiple of inlets streams which will come here which will we go on adding all the
energy quantities and there may be a multiple of outlet streams which will go on adding
at the outlet stream So only simplification is that in this analysis
of thermodynamic that this is the invariant with time and this is uniform across a cross
section obvious but here you always assume that all properties not only this velocity
v specific value internal energy all this things are uniform across a cross section
and invariant with time then it becomes as simple as this is only the accountability
of energy for the control volume. Now at steady state, what is the definition of steady state?
When a control volume will be at steady state all its properties will remain invariant with
time that means both the mass and the energy being two of its important properties will
remain invariant with time which means at steady state a control volume is specified
by this equation that the rate of change of mass dmcv dt is 0 with time similarly rate
of change of energy with time is 0 that means there will be in variant with time that means,
the conservation of mass will tell that this will be 0 that means dm one dt is equal to
d m 2 d t that means the mass inflow is equal to the mass outflow . Similarly, dEcbdt dEcv
dt will be 0. Now if we make this first condition then we
get d m1 dt is equal to dm 2 dt because the mass inflow is equal to mass outflow there
are will be no mass change in the control volume let this becomes is equal to dm and
dt and if we substitute this in this energy equation that means if we write d m1 dt and
dn 2 d t as dm dt. Then we can write, dm dt into this quantity
u1 plus p1 v1 plus v1 square by 2 plus g z1 just simply a repetition delQ by delt minus
again dm dt into u2 small u specific internal energy plus v2 square by 2 plus g z2 minus
del W is equal to 0 because there is dcb is 0 d Ecv dt is 0.
Now this equation at steady state can be expressed in terms of the quantities per unit mass here
all the quantities are energy per unit time but we will now express energy per unit mass
to do that we will have to divide all the terms by dm dt.
So that we get u1 plus p1 v1 plus v1 square by 2 plus g z1 then this quantity will appear
as delQ del m whose meaning is that the heat transferred ah or heat added per unit mass
heat added per unit mass u2 plus p2 v2 plus v2 square by 2 plus g z2 minus del w del m
is equal to 0. That means, we transfer this equation from
the time rate bases to mass rate bases that means this quantity contains internal energy
per unit mass flow work per unit mass kinetic energy per unit mass potential energy per
unit mass this is the stored energy per unit mass with the inflow stream, that means at
the inlet section this is the heat added per unit mass.
Similarly, these are the stored energy at unit mass at the outlet sections of the fluid
stream and this is the work done per unit mass. So, this can be written now in a fashion
now by you want u1 plus p1v1 by your definition h then I can write now little this way delQ
del M minus del w del m is equal to h1 plus v1 square by 2 plus z g1 minus h2 minus h1
plus v1 square by 2 plus g z1 this will be 2. This is known as steady flow energy equation
any one of this either this or this or this is the time rate bases this is per unit mass
bases and this is written in a different arrangement. So, any one of this three are known a, steady
flow energy equation, that means in steady flow energy equation we see that the heat
added per unit mass minus the work done per unit mass is given by the difference of this
stored energy per unit mass between the two system. Now you will find out the immense
physical significant of the term enthalpy in almost all the practical appliances in
engineering use the difference in kinetic energies and potential energies are negligible
compare to the change in enthalpy. We are interested in these two quantities
as engineers so therefore we are interested in the changes not in the absolute value because
engineer want what is the work what is the heating interactions and they are always described
by the first law in terms of the changes of the properties, changes of enthalpy, changes
of internal energy even changes of kinetic energy, changes of potential energy.
We are not bothered about the absolute potential energy if you want to know what is the work
done in replacing a system from one point one to other point in a conservative force
field you are not bothered about the absolute potential energy at a particular point where
from it was taken and the particular point where it has been kept but you were only bothered
about the change in the potential energy. So that they should be defined from a conventional
and same reference data. So, therefore it is the change, so these changes are small.
In certain cases, the changes in kinetic energy in case of nozzles and diffusers, we come
across afterwards, changes in kinetic energy may be appreciable compare to the change in
enthalpies, but changing in potential energies are always negligible because change in potential
energy means what? It is with respect to the gravitational force speed. So, we have to
make a appliance very tall so that changes in potential energy becomes appreciable to
that of its change in enthalpy it is not so. Therefore, if we consider those cases where
changes in kinetic and potential energies are small with respect to change in enthalpy,
then we can write this one that delQ by delm minus delW by delm is simply h2 minus h1.
See the important implications that means, for a control volume the difference between
the heat and work interaction comes out to be its changing enthalpy at the outlet and
inlet stream of fluid, provided, its change of kinetic energy and change of potential
energies are negligible to understand. For a device which is only work interacting
type for a work interacting device for which, delQ by delm is 0. I can write this equation
as delW by delm is h1 minus h2 that mean, dealing with a compressor. I have a compressor.
There is an inlet stream air 1 and there is an outlet stream air well 1 and 2; delW by
delm is h1 minus h2 in this equation. We see that in compressor that this h2 is
more than h1; it comes at a lower pressure and temperature; it goes out with a higher
pressure and temperature; h2 is more than h1 and this becomes negative which means the
compressor takes the 1. Reverse is the case in turbine; turbine work is coming out, h1
is more than h2. Similarly, if we consider a pipe, just i am
giving some example only heating interacting devices. You write this only for heat interacting
device. If flow takes place like this and we give heat per unit mass, so if one wants
to know what is the heat added, if i know the enthalpy at this two point i can write
this steady flow energy equation making this term 0, that delQ by delm is simply… that
means if heat is added h2 will be more than h1 or heat is rejected h2 will be less than
h1. Sometimes in a control volume, there is no work in interaction; there is a conversion
from one form to other form. For example, a nozzle, what you will do here in adiabatic
nozzles? There is adiabatic surface insulated at stream 1 and stream 2. At stream 1, liquid
or fluid comes with enthalpy h1 and a velocity v1 and here in fluid goes out enthalpy h2
and velocity v2 and what happens is that v2 is greater than v1, whereas, h2 is less than
h1. At the expense of this enthalpy that means internal energy plus pressure energy, we create
the kinetic energy. Then its kinetic energy is created out of the some of the internal
energy and the pressure energy. In that case, we can write the energy equation not by neglecting
this v1 square by 2 terms that means we can neglect the change in the potential energy
term. In that case, we can write that delQ by delm
minus delW by delm is equal to h2 plus v2 square by 2 minus... Since this is the adiabatic,
no heat transfer is taking place, no work is performed by the nozzle with this surrounding
and this term will be zero. We can write that v2 square minus v1 square by 2 is h1 minus
h2. It is because of this enthalpy difference, the change in kinetic mass. This way we can
apply the steady flow energy equations in different control volume in practices.
[Conversation between professor and student.] Try to understand volume is changed, pressure
is changed everything is change but gross energy balance will be an increase in enthalpy
because of the work done. Now, how we get it in increase in enthalpy within because
increase in enthalpy knows enthalpy composed of two parts u plus pv. So whether there is
a change in u or not i do not know so far that will come from the property relations
if it is an air. So this is an ideal gas; internal energy is a function of temperature.
Only if temperature increases, the internal energy will increase. So if there is an increase
in internal energy we will find out there is an increase in temperature and there is
an increase in internal energy. Similarly, there will be an increase in pressure energy
also because of the product of p and v whether volume decrease and pressure increase, probably,
their product is increased. We do not know these things, conclusively, that what will
be there with this two counter acting effects. So specific volume is decrease at the same
time pressure is increase, but the product will be increased. At the present moment we
are not even going into that detail because we do not know how much pv is increased? How
much u is increased so that there is a decrease in pv or increase in u? But we are not reading
the air compressor. We are analyzing the thermodynamically that there is an increase in enthalpy.
We will find out whether this increases is because of increase in u or increase in pv
or how much for u or how much for pv or a combination of increase decrease total increase
that we are not going through at the present moment, that will be found out if we know
the property relations and how do they change but only thing is that the enthalpy is increased.
Think it from thermodynamic point of view, do not go into that detail, that will come
afterwards, but in fact I tell you for your information, the pressure energy increased,
because increase in pressure is much more than the decrease in
the specific volume. So, first law for a closed system and first
law for an open system or a control volume that we have finished. Next class, what I
will do, I will solve some problems in this class here, so you come prepared, you go through
this lecture that first law and second low analysis to closed and open system, we will
solve, some simple problems and some complicated problem also.
Thank you
