Good afternoon, everybody. We will continue with what we have learnt
in our last lecture. In our last class, we have seen that whenever
a fluid is entering a control volume through a control surface, it has to do some extra
amount of work. This work we have termed as flow work or flow
energy. I will repeat this. Flow work or flow energy, you have derived
it. This is given by p into small v, where p is
pressure at the control surface through which mass is flowing, v is specific
volume. We are getting this flow work or flow energy
by the product of two quantities; one is pressure at the control surface and another is the
specific volume. If we go back to our previous diagram of the
control volume, it was like this - control volume of any arbitrary shape and we had the
control surface like this. This is our inlet, where the mass is entering
and this is the exit where mass is going out. Let us say that the mass flow rate at the
inlet is given by m dot i; the mass flow rate through the exit is m dot e, pressure here
is pi, velocity is Vi, specific volume is vi and the height from the datum level is
Zi. We have also selected a datum plane at Z = 0. This is Zi and this is Ze. The corresponding quantities at the exit are
pe - pressure, Ve - velocity, small v subscript e - specific volume and Ze that is the height
from the datum level. Over and above that, we also had some work
interaction. Let us say that the rate of work interaction
is W dot. We had some heat interaction; the rate of
heat interaction is Q dot. If we think of the energy balance for this
particular control volume, let us see what are the quantities entering here? With the incoming mass, some amount of kinetic
energy will enter the control volume, some amount of potential energy will enter the
control volume and we will also have internal energy ui plus the flow work or flow energy. The energy which will be leaving the control
volume is the kinetic energy associated with the mass, the potential energy associated
with the mass, internal energy associated with mass and the flow work or flow energy. Over and above that, we have work transfer
and heat transfer. Keeping the sign convention intact we can
write, mi into ui, ui is internal energy plus Vi square by 2, that is the kinetic energy;
gZi potential energy plus pi vi flow work or flow energy. With this the amount of heat interaction will
come. This can be equated with m dot e ue plus Ve
square by 2 plus gZe plus pe ve plus W dot that is the work transfer. This equation can be written only under the
condition of steady state steady flow process where m dot i is equal to m dot e is equal
to m dot. With this, we can finally write m dot ui plus
Vi square by 2 plus gZi plus pi vi plus Q dot this is equal to m dot ue plus Ve square
by 2 plus gZe plus pe ve plus W dot. It is customary to divide both the sides with
the mass flow rate. We will get one equation where all the quantities
are specific or per unit mass. If we do that, we will get ui plus Vi square
by 2 plus gZi plus pi vi plus q that is equal to ue plus Ve square by 2 plus gZe plus pe
ve plus w. So, small q and small w are the heat transfer
and work transfer per unit mass respectively. All the quantities in this equation are specific
quantities or quantities per unit mass. We can write the steady state steady flow
equation in this particular form. This equation I will put in a box. It is termed as steady state steady flow energy equation or
in short SSSFE. This
equation is a very useful equation for the analysis of engineering systems and different
devices. We will take up different examples in a few
minutes and we will see how we can apply this equation for different systems and devices. Before that, we will do some more simplification. Both in the right hand and in the left hand
we are getting quantities like internal energy u and flow work or flow energy pv. This is one intensive property, internal energy
is one intensive property; pressure and specific volume they are also intensive properties. We can combine these two things and we can
ultimately express this quantity with a new intensive property, which is known as enthalpy. This is denoted by h and is known as enthalpy. This is a property that is why it is a state
variable; at a particular state, we will have a particular value of enthalpy. Introducing enthalpy, we will have our equation
like this: hi plus Vi square by 2 plus gZi plus q that is equal to he plus Ve square
by 2 plus gZe plus w. This will be the form of our steady state
steady flow energy equation. Let us see how we can apply the first law
of thermodynamics for different systems and processes. First, we will take examples of some closed
systems. Let us say we are taking the example of a
cycle - steam power cycle. Schematically, steam power cycle will have
4 components. This is the boiler denoted by B. Then, we
will have turbine, this is denoted by T. After the turbine, we will have condenser denoted
by C and finally we will have feed pump which is denoted by FP. The working substance which is either water
or steam will execute a cycle between all these 4 components. In the boiler, we will have supply of heat
from outside. Let us denote this thermal energy or heat
as QB. In the condenser heat will be rejected and
let us denote this quantity as QC. Similarly, in the turbine, it will do some
external work; let us say this is WT. In the feed pump, we have to supply some amount
of work from outside to run the pump, so let us denote it as WP. Recognize that these are all the quantities
associated with different components of the steam power plant and we can also have some
sort of a control volume identified as shown by this dotted line. If this is our control volume, we can see
that there are a number of interactions in terms of heat and work. There is work input in the pump. The control volume is doing some work on the
surrounding, that is the turbine work; there is heat supply to this control volume and
there is heat rejection from this control volume. If we write down now the first law of thermodynamics,
for a closed system, we have cyclic integral of dQ is equal to cyclic integral of dW. Basically, for this steam power cycle, considering
all this work transfer and heat transfer, we can write QB, this is a positive quantity
as heat is supplied to the control volume, minus QC, this is being rejected by the control
volume; this is equal to WT, work done by the control volume on the surroundings, so
it is positive and then minus WP, work which is required to run the pump, so which will
be a negative quantity. I have taken a very simple example. This is how we can apply first law of thermodynamics
taking the steam power plant as a whole to be our control volume. This is the application of the first law of
thermodynamics for a closed system because the way we have defined our control volume,
if we take the steam and water to be the working fluid, then no mass is entering the control
volume or no mass is leaving the control volume because steam or water is the working fluid
for our example. Next, let us take some examples where we have
got flow process and we have got mass entering the system and leaving the system. Let us take the example of a turbine. Either it can be a steam turbine or it can
be a gas turbine. Schematically, it is represented by an inverted
cone like this because in a turbine there is expansion. There is a change in specific volume in the
direction of flow. This is the inlet, this is the outlet. Again, there could be heat transfer and work
transfer. This is Q dot; it can have any direction,
I have shown an arbitrary direction. Assuming
steady operation, if we assume that it is operating under steady state, one can write,
this is inlet and this is exit; hi plus Vi square by 2 plus gZi plus q that is equal
to he plus Ve square by 2 plus gZe plus w. This is the equation we can write for the
turbine. This is a very generalized equation. For a turbine used in practice, we can see
the number of terms which we have written here. Some of them are significant while contributions
of some of the other terms are not that significant. If we repeat this equation hi plus Vi square
by 2 plus gZi plus q that is equal to he plus Ve square by 2 plus gZe plus w. Generally, in a turbine a compressible fluid
is entering and leaving the turbine. The change in potential energy is small. So, these two terms can be neglected. From the turbine, there could be some heat
loss but generally, this is small compared to other quantities, so this term also can
be neglected. The changes in kinetic energy or contribution
of the kinetic energy term is relatively small compared to the change in enthalpy or contribution
of enthalpy, so one can neglect the change in kinetic energy also. We will have hi is equal to he plus W, or
in other words we can have hi minus he is equal to W. Basically when the fluid is entering
a turbine, its enthalpy will reduce and that will produce the work or the mechanical work
which is supplied by the turbine, which is done by the control volume on the surrounding
that is coming due to the change of enthalpy. The enthalpy at the inlet minus the enthalpy
at the outlet is the specific work done by the turbine. Finally, we can write WT is approximately
equal to hi minus he. If we take the example of a compressor, what
will we get? The compressor is a device where the fluid
will be compressed. There could be different design of compressor
either it could be a reciprocating compressor or it could be a roto-dynamic compressor. But, in both the cases there will be a compression
of the fluid or the pressure of the fluid will increase while it will pass through the
compressor. The fluid will be energized and we have to
supply certain amount of work from outside. In general, one can represent it like this. There is an inlet through which mass is entering,
there is an exit through which mass is going out. Again, one can have work interaction and heat
interaction and one can easily recognize that it is a device to some extent reverse of turbine. Here, we are supplying work from outside and
we are energizing the fluid whereas in the turbine we have extracted energy from the
fluid and produced certain external work. Here, I will not go into the details of it. One can do the analysis just like in the previous
case. Some of the terms like kinetic energy, potential
energy are negligible compared to others. One can write down the work of compression
or compressor work again in terms of enthalpies. What will it be? It is he minus hi. These two examples that I have shown, one
should remember that in these cases we have neglected the heat transfer, the contribution
of kinetic energy and potential energy. There are certain cases where it may be mentioned
that the compressor or the turbine is having certain amount of heat transfer or having
certain amount of change in kinetic energy. In those cases we need to consider those terms,
otherwise in general those are negligible. We can think of another mechanical device
which is also used for the production of power; that is a nozzle. Schematically, one can show a nozzle like
this. A nozzle is used for increasing the velocity
of the fluid. Let us write down. This is the inlet of the nozzle and this is
the exit of the nozzle. Let us write down the steady state steady
flow energy equation for the nozzle. We will have hi plus Vi square by 2 plus gZi
plus Q that is equal to he plus Ve square by 2 plus gZe plus W. Generally, the fluid
enters with a small velocity, with a very low velocity into the nozzle. The kinetic energy of the incoming fluid can
be neglected. Generally, the nozzle is a small device and
a fluid which is passing through it that experiences a very small change in elevation, so change
of potential energy is very small. We neglect these two quantities. This is a stationary device and there is no
work interaction with the surroundings. The work done is equal to zero. The way I have drawn it, the nozzle is insulated,
so there is no heat transfer. In certain cases, there could be a certain
amount of heat transfer. But, generally, it is small compared to the
other terms so we are also neglecting the heat transfer term. The exit velocity Ve square by 2 is equal
to hi minus he or Ve is equal to 2 hi minus he and within a square root. This is how we can determine what the exit
velocity of the nozzle will be or let us say we need to have certain amount of exit velocity
of the nozzle. So, at what condition will the fluid be supplied
to the nozzle that can be designed from the first principle. Let us take the example of a pump. It is slightly different compared to a compressor
because in a compressor the working fluid is a compressible fluid, whereas in a pump
the working fluid is incompressible. The fluid is entering the pump. This is our inlet and this is our exit. The fluid is going out of the pump. Again, let us write down the steady state
steady flow energy equation. This is the potential energy term plus q that
is equal to he plus Ve square by 2 plus gZe plus W. Now, just like before if we try to analyze
the process, there is no heat transfer. A very small change of temperature will occur
when a fluid will pass through a pump. As the fluid is being churned inside the pump,
there could be certain amount of temperature rise but that is negligibly small, so heat
transfer is neglected here. Again, between the inlet and outlet of the
pump one can neglect the change of potential energy. The change in kinetic energy can also be neglected. It can be like this; if we are having the
same diameter of the pump suction pipe and discharge pipe, the kinetic energies will
be the same here because, it is an incompressible fluid, the mass flow rate is remaining same. The kinetic energy at the inlet and kinetic
energy at the outlet or exit are the same and they will cancel each other. There can be a variation in the suction pipeline
diameter and discharge pipeline diameter. There will be some contribution from the change
of kinetic energy but generally, that is small. So, we can neglect these two terms. Basically, what we are getting W is equal
to hi minus he. Again, let us write this down in terms of
internal energy and flow work. If we do that, we will have ui plus pi vi
minus ue plus pe ve. We can see this is the internal energy at
the inlet; this is the internal energy at the outlet. The internal energy is a property of the fluid
and is dependent to a very great extent on the temperature of the fluid. We know that between inlet and outlet there
is very small change in temperature. Basically, there will not be a very appreciable
amount of change in internal energy in the case of a pump. We can neglect the quantity ui minus ue. Let me write it in two steps: ui minus ue
plus pi vi minus pe ve. The first term is very small, we can neglect
it. Ultimately, we are writing Wpump is equal
to pi vi minus pe ve. Again, this specific volume as the fluid is
incompressible fluid there will not be appreciable change in specific volume. We can write vi is equal to ve and this is
denoted by v; so Wpump is pi minus pe into v. What is Wpump? This is the work transfer associated with
unit mass, which is going through the control volume which is now our pump. What are we getting here? This quantity is equal to inlet pressure minus
exit pressure multiplied by this specific volume. In this quantity pi inlet pressure, which
is called suction pressure, this is less; exit pressure is higher compared to the inlet
pressure. This quantity will be a negative quantity,
which means that we have to supply work to run the pump for pressurizing the liquid. Basically, if we are interested in the magnitude
only one can again reformat the expression and one can write p exit minus p inlet into
v. W dot pump one can write m dot into pi minus pe into v or Q into delta p. We are familiar with this formula. Is there any difficulty? This is the specific volume per unit mass,
and this is multiplied by m dot, so we will get this expression. Let us take some other example, which is again
interesting. Let us take the example of a throttling process. A throttling process is a process in which
a compressible fluid is made to pass through a narrow aperture. This narrow aperture could be a porous plug,
could be a very small opening like orifice or could be a small opening of a valve. Basically, we are having a pipeline. In the pipeline, we are having a very narrow
opening and this change in cross sectional area, the pipeline has got a larger area;
we are having a narrow opening so it is having a small area. This change in cross section is abrupt. There is an abrupt change in cross section,
an abrupt constriction is there. Let us say, that this is your inlet plane
and this is your exit plane. This is i and e; flow is taking place in this
direction. Our control volume is something like this. We have taken a sufficiently long or large
control volume and we have encompassed this narrow aperture or narrow opening inside this
control volume. Now, we can apply our first law of thermodynamics
for this particular process. We can have hi plus Vi square by 2 plus gZi
plus q that is equal to he plus Ve square by 2 plus gZe plus W. Let us concentrate only
in the inlet plane and outlet plane. This is a constant area term or constant area
pipeline. It is a steady state steady flow process;
mass flow rate remains constant. There will not be any change in kinetic energy. So from both the sides this kinetic energy
term can be cancelled. There will not be any change in potential
energy, so this can also be cancelled from both the sides. There is no work interaction with the surroundings. So W, this term, can be made equal to 0. The duct or pipeline must be well insulated
so there is no heat transfer and we can make q is equal to 0. So, for the throttling process we can write
hi is equal to he. If we want to give a verbal description, we
can write: enthalpy before throttling is equal to enthalpy after throttling. The
throttling process is such that enthalpy before throttling is equal to enthalpy after throttling. This we get from the first law of thermodynamics
but it has got some far-reaching significance. The throttling process is used for production
of low temperature. We know that in our domestic refrigerator
or air conditioner, we use a throttled valve or a throttling device. The throttling device could be some sort of
expansion valve or some sort of a capillary tube through which the fluid is throttled. If certain conditions are maintained, then
we can see that due to throttling, low temperature can be produced or low temperature effect
can be produced. That is why the throttling process is very
important. But, there we will have this derivation, and
that means, what we have derived is valid in throttling process - the enthalpy before
throttling and enthalpy after throttling will be identical. If I elaborate this, actually hi is equal
to he. Let us elaborate it a bit further. Though hi and he are identical, they are not
at the same state. They are at different states. That means for hi, the pressure and temperature
are not identical with the pressure and temperature of the fluid after throttling. There is a fall in pressure, because the fluid
is experiencing a sudden expansion and due to this sudden expansion there is a fall in
temperature. We are interested in those situations. One can see that if it is an ideal gas, then
there will not be any change in temperature. In case of real gases there are cases when
there will be a fall in temperature. We are interested in those applications because
using those applications we can have our refrigeration cycle. This process is known as Joule-Thompson expansion
process. The throttling process when we have, enthalpy
before throttling and after throttling identical; this process is known as Joule-Thompson expansion
process. I will take another example where we can apply
the first law of thermodynamics. So far, we have considered that there is only
one inlet and one outlet but the control volume need not have only one inlet and one outlet. There could be multiple inlets and multiple
outlets. We can take the example of a heat exchanger. Let us have a very simple heat exchanger which
is known as double pipe heat exchanger or tube in tube
heat exchanger. Basically, there is a central tube through
which one fluid flows and over this central tube, there is another tube which makes an
annular passage and through this, another fluid flows. This is fluid 1 and this is the second fluid,
fluid 2. If we take this device to be our control volume,
we can see that the control volume has two inlets and two outlets. Let us say this is i1, one inlet; this is
i2, second inlet. This is exit one - e1 and this is e2, exit
two. There are two exits and two inlets. One can select or one can take example of
some other control volume where there could be more number of inlets and more number of
exits. How can we analyse this particular system? There are different ways of analyzing it. What one can do is, first one can take only
fluid 1 to be the control volume and then, one can see its interaction with fluid 2 because
if we take fluid 1 to be our control volume then fluid 2 becomes the surroundings. One can arrive at the final equation or one
can do the analysis the way I have drawn the control volume taking both the fluids inside
the heat exchanger to be our control volume. If we adopt the second method then one can
write sigma hi plus Vi square by 2 plus gZi. This is j is equal to 1 to n, there are n
number of inlets, plus q that is equal to sigma; let us say there are p number of outlets,
p is equal to 1 to m; p is any arbitrary outlet and there are m number of outlets, then he
plus Ve square by 2 plus gZe plus w. We can write it in this particular form. For this particular heat exchanger, we have
two inlets and two outlets. We have taken fluid 1 and fluid 2, so we can
write hi1 plus Vi1 square by 2 plus gZi1 plus hi2 plus Vi2 square by 2 plus gZi2 plus q
that is equal to he1 plus Ve1 square by 2 plus gZe2 plus he2 plus Ve2 square by 2 plus
gZe2 plus w. So one can write it like this. I think we will take a break here and we will
continue from this in our next lecture.
