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Welcome to the 10th lecture of cryogenic engineering, under the NPTEL program. Take an overview
of what we learnt in the last class - we found that for liquefaction of real gas, J - T coefficient,
that is, del T by del p at constant enthalpy, depends on inversion temperature. So, we learnt
the concept of inversion temperature, which is nothing, but del T upon del p, at constant
enthalpy. We found that, in order to get J - T equaling
effect for a real gas, the temperature in the initial state of the gas should be less
than T inversion temperature. Then we went away from isenthalpic process, to isentropic
process, using reciprocating expansion engine, or Turbo-expanders. We understood that, the
isentropic expansion of a gas always results in cooling, irrespective of its initial state.
So, as it was in J - T expansion, which depends on the initial state, the isentropic expansion
does not depend on the initial state. The J - T expansion is normally used where
phase changes are required. This is a very important thing that, during J - T expansion,
normally there is phase change, while in an isentropic expansion it is used for single
phase fluids. So, the inlet is gas; the outlet also should be gas, while it is not true with
J - T expansion. In J - T expansion, inlet could be gas, or a two phase mixture, and
outlet will be liquid, or a two phase mixture. The isentropic expansion coefficient is nothing
but, delta T by delta p, at constant enthalpy. So, del T by del p at constant s. This was
del T by del p at constant h, for J - T expansion.
We also studied that, the gases like air, and nitrogen show J - T cooling, when expanded
at room temperature, while other gasses like helium, hydrogen, and neon require to be pre-cooled
to result in J - T cooling. This is going to the fact that, the inversion temperature
of helium, hydrogen, and neon is bellow ambient temperature, and therefore, if you want to
have J - T cooling for these gases, the initial state of the gas should be brought down below
its inversion temperature; and therefore, they need to be pre-cooled, and only then
they can result in J - T cooling. We studied thermodynamic ideal system for
gas liquefaction where, in a thermodynamic ideal system, all the gas that is compressed
gets liquefied. Using the ideal thermodynamic cycle, one can calculate the ideal work requirement
for liquefaction of unit mass for a given gas. So, the ideal thermodynamic cycle basically
teaches us, or basically gives us, the ideal work requirement to liquefy the unit mass
of a given gas. This ideal work requirement depends on the initial condition of the gas,
which is nothing but the initial pressure, and temperature condition of the given gas.
Extending whatever we have learnt in the earlier lecture, the outline for today’s lecture
again holds around the topic of ‘gas liquefaction and refrigeration systems’; and what we
study under this today are parameters of gas liquefaction systems, and then we come to
the first practical system, which is Linde - Hampson system, where we talk about - liquid
yield, work requirement, and optimization of liquid yield. All these are calculations;
and based firmly on their derivation, we will try to understand how Linde - Hampson system
behaves.
This is an ideal system. As seen earlier, the schematic diagram of an ideal system,
and its T - s diagram is given here. This is the ideal system, wherein you have got
a gas input, you have got a compressor, and you have got an expander. A corresponding
T - s diagram is given here; where, 1 to 2 is the compression process, and 2 to f is
an isentropic expansion process. This is the ideal thermodynamic cycle. The process of
compression and expansion are from, 1 to 2, and 2 to f, respectively. So, 1 to 2 is the
compression; 2 to f is the expansion process. The initial condition of gas at point 1, point
1 is the initial condition, and it could be ambient. 300 Kelvin, and 1 bar could be the
standard operating condition, for the initial condition, and this determines the position
of f. As you can see that, the point 1 determines the location of point f because, this is a
1 bar line, and this is corresponding to the boiling point of the gas at 1 bar. For nitrogen
it will be – 77 Kelvin. As soon as your point 1 gets determined, your
point f is always known. Therefore, we say, the initial condition 1 of the gas, determines
the position of point f. The point 2 determines the final state of the gas after the compression
process. So, point 2 comes over here, which is nothing but a point, which is perpendicular
to this line. So, this is an isentropic expansion, the entropy remains constant. So, if you extend
this line further up, and if you draw a horizontal line, temperature remaining constant, the
intersection of these two lines will be nothing but the point 2, which is the final state,
after compression process from 1 to 2.
Let us take an example of nitrogen. The initial condition at point 1, which is 1 bar, and
300 Kelvin, we can find out now, as soon as your 1 bar and 300 Kelvin is determined, we
know the point f; and if you draw a vertical line here, and if I draw a horizontal one
at this point, the intersection at this point will give the final condition after the compression
process - 1 to 2. Here, we can understand the required pressure at point 2 to follow
an ideal cycle, is more than 70,000 bar. This is a very, very high pressure; absolutely
impractical to attempt in practice. So, what you want to conclude from here is that, such
high pressures are impractical; and hence, there is a need to modify the system to lower,
maximum pressures. What we understand from here is, if you want
to have ideal thermodynamic cycle for gas liquefaction, you have to liquefy all the
gas that is getting compressed, and if you were to do this, the point 2 lies at a very
high pressure, and it is as high as 70,000 bar, which is absolutely impractical to attain;
and therefore, what we say as ideal thermodynamic cycle for liquefaction is impractical, and
hence, we have to modify this system. So, we lower the maximum pressures, which are
normally attainable using the compressors available.
In the first lecture, I had talked about heat exchangers like, joule Thomson valve, and
turbo expanders, all these systems could be used to modify the system, and now that we
talked about changing the ideal thermodynamic cycle to get a relatively practical cycle,
we will have to use all these components as a two-way heat exchanger - one gas coming
in, and one gas coming in from this direction. These are the turbo expanders, and other is
the J - T valve. The heat exchangers are used to conserve cold
and J - T devices are used to achieve lower temperatures. This is what all of you know.
Now, the following slides in this lecture will explain various cycles, that are used
for gas liquefaction.
So now, we will come to the imported cycles. Before we go into various liquefaction cycles,
let us find out what the different gas liquefaction parameters are. So, we have got different
cycles, and if I were to compare to this different cycles, I will have to compare different parameters.
What are these parameters? In the refrigeration system, for example, the Carnot COP is often
used as a benchmark to compare the performances. On a similar line, there is a need to compare
different liquefaction systems. There are different cycles, and therefore, we will have
some parameters so that, we can compare the performance of cycle A to cycle B, to cycle
C. In liquefaction systems, an ideal cycle is used as a benchmark to compare the performances.
So, the way of comparison is to compare a cycle’s performance with an ideal cycle.
Different rations and functions are defined to give qualitative and quantitative information
on different liquefaction cycles.
What are these different ratios and functions? If you take a typical example of this cycle,
you have got a compressor and an expander, the gas m dot is compressed at this point,
out of which, m dot f gets liquefied, and therefore, the gas which returns to the cycle
is m minus m F dot. This gas goes in here, and replenishment is done at this point, which
is m dot F. So, whatever mass goes out is replenished over here, and the cycle continues.
So, from here, m dot gets compressed, and m dot f gets liquefied. So, what are the different
parameters? At the same time, what you can see here, the
power input to the compressor is W C, and the heat ejected to the atmosphere is Q R,
or the heat of compression. The first parameter is work required per unit mass of gas compressed.
So, the work required is W C, and the gas which is compressed m one dot. So, minus W
upon m one dot is the work per unit mass of gas compressed. The next is, work per unit
mass of gas liquefied. So, the gas liquefied is m dot f, and therefore, the work required
is minus W upon m dot f. The minus stand is given for work done on the system, which is
in this case, the work done on the compressor, and therefore, we are calculating the work
done per unit mass of gas liquefied, and work done for a mass of gas compressed.
The compressor isothermal efficiency, we are assuming that the compressor is isothermal,
however, in the actual case the compressor may not show isothermal compression, and therefore,
we have an efficiency, which is called ‘compressor isothermal efficiency’.
Similarly, we have got compressor mechanical efficiency, which amounts to all the irreversibility
or the frictional losses in a system and it is called as ‘eta c mechanical’ or ‘efficiency
mechanical’. Then we define one more parameter called ‘Figure of Merit’ or FOM. This
is the ratio of ideal work input, based on the ideal thermodynamics cycle, divided by
actual work input of the system. So naturally, the ideal work input is going to be less than
the actual work input of the system. So, this ratio is going to be less than 1, and this
is called as the ‘Figure of Merit’ of any cycle.
Then is the fraction of total gas liquefied. So, you can see here that, y represents the
ratio of m dot f divided by m dot. When the m dot gas is getting compressed, the amount
of gas, which is liquefied, is only m dot f. So, m dot f upon m dot is nothing but,
the fraction of the total gas liquefied.
With these parameters, now let us see some fundamentals, which would help us to finalize
the liquefaction cycle. The sign convention that I just talked about is that, the work
done by the system is taken as positive. Similarly, the work done on the system will be taken
as negative. The heat transferred to the system is taken as positive, and the heat leaving
the system will be taken as negative. In all these cases, the pressures we are using
is bar or Pascal; the conversion table is - 1 Pascal is equal to 1 Newton per meter
square, and 1 bar is equal to 10 to the power 5 Pascal, and 1 atmosphere is nothing but
- 1.01325 bar. In actual cases, I will assume that 1 atmosphere is equal to 1 bar, as you
can see from this approximate expression.
With this background, let us come to the Linde - Hampson cycle, which is the first practical
cycle taking us away from the ideal thermodynamics cycle. This is the cycle; this is the schematic
representation of the Linde - Hampson cycle. What it has is, you can see over here that
it has got a compressor; it has got a heat exchanger, it has got a J - T wall, and it
has got a container. The gas goes back, and the gas is replenished. So, in the sense,
if you see the thermodynamics cycle to this cycle, what we have added is a heat exchanger.
In the ideal cycle, all the gas which was getting compressed had got liquefied, while
here, that is not happening; only some of the gases are getting liquefied, when the
remaining gas, which is m minus m f dot is coming back to the compressor, and m dot f
gas is replenished, or the makeup gas is added here.
With this background let us see the salient features of this system. The Linde - Hampson
cycle consists of a compressor, a heat exchanger, and a J - T expansion wall. This is what we
have seen. Only a part of the gas that is compressed gets liquefied; only m dot f is
getting liquefied out of m dot which is compressed. Being an open cycle, the mass deficit occurring
is replenished by a makeup gas connection.
We have just seen that all the processes are assumed to be ideal in nature, and there are
no irreversible pressure drops in the system. This is the assumption which will be used
for analyzing this system. We assume that the compression process is
isothermal, while J - T expansion is isenthalpic, and we also assume that the system incorporates
a two-fluid heat exchanger, which is assumed to be 100% effective. So, it is a perfectly
ideal system; however, the heat exchanger here also is assumed to be 100% effective.
The heat exchanger process is an isobaric process, and it is used to conserve cold in
the system; that means, when both the fluids are travelling through the heat exchanger,
there is no pressure drop across the heat exchanger. That is, the stream 2 to 3, and
g to 1- these points are the direction of the fluids flowing through here, the stream
of the gas 2 to 3 is cooled by the stream of the gas going back.
The J - T expansion device is used for phase change of gas stream to liquid stream, by
lowering the temperature. So, once you get gas at this point, what you get at this point
is a two-phase, or the liquid gets collected and gas goes back at this point.
This is the most important thing, and what you have to learn from here is, how to draw
a corresponding temperature entropy diagram, or a T - s diagram, from a given cycle.
So, what you can see from here is, the 1 to 2 process is an isothermal compression, and
2 to 3 is an isobaric heat exchanger process. The pressure remains constant, and the temperature
gets lowered, because of the return gas stream, which is from g to 1. So, g to 1, because
of this low temperature gas, which is going back to the compressor, the temperature gets
down from 2 to 3. At the point 3, the gas gets expanded from 3 to 4; and as soon as
the expansion happens, the state of the gas lies in the doom. This is what we say that,
the state of the gas at this point 4 is a two-phase gas; that is, you got a liquid and
some gas. The liquid is collected at point f of which, some of the liquid is drawn out,
or the remaining gas will go back, and this will pre-cool the incoming gas from 2 to 3.
At point g, the gas would go back from g to 1, pre-cooling the stream from 2 to 3. These
two processes - 2 to 3 and g to 1 are basically heat exchange processes. In the heat exchanger,
while the process of 1 to 2 is a compression process, 3 to 4 is an expansion process; and
4 to g is a place where we get vapor at this point, or gas at this point. The liquid is
collected depending on where the point 4 is. Depending on where the point 4 is, amount
of liquid will be decided, and the amount of gas which has got converted to liquid will
be decided, and this is what we will calculate now.
This should be very, very clear to all of you on how to draw the T - s diagram for various
cycles. Now, having done this, let us have a control volume over here in this region.
So, we want to basically now analyze this cycle, in order to understand what the work
input to the system is, or what the fraction of gas that has got into liquid region is,
as a fraction of m dot how much m f has come in a liquid form. So, consider a control volume
for this system as shown in the figure. It encloses the heat exchangers; the control
basically encloses the heat exchanger, J - T wall, and the liquid container. As you can
see from this, let us apply the first law of thermodynamics to this system. The changes
in the velocities and the datum levels are assumed to be negligible, as we had done in
the last lecture also, for ideal cycle.
The quantities entering and leaving the control volume are as given below. What is entering
is only the gas which is coming at m dot 2, the mass of gas which is coming at point 2,
while, what is leaving the system is m minus m f dot, at this point. At the same time,
what is leaving the system is the m dot f figure.
Using the first law, the energy m is equal to the energy out. Therefore, what you get
is m dot at this point, multiplied by enthalpy at this point. So, m dot is equal to m dot
minus m f dot, and corresponding to this point is nothing but point 1. So, m dot minus m
f dot into h1, which is enthalpy at point 1 plus, whatever is leaving. This control
volume is m f dot into h f.
With this background if I get a ratio, it is, m dot f upon m dot is equal to h 1 minus
h 2, divided h 1 minus h f. So, h 1 minus h 2 gives you the enthalpy difference across
the compressor, divided by h 1 minus h f. Here, the fraction of gas liquefied, or the
liquid yield, is defined as m dot f, divided by m dot equal to y. This is nothing but,
y is equal to h 1 minus h 2, divided by h 1 minus h f. It means that, y depends on the
initial condition, and the compression pressure. So, h 1, h 1 and h f depend on the initial
condition, while the point h 2 depends on the compression pressure.
Here we can see that the values - h 1 and h f, are governed by the initial conditions,
which are often ambient condition. So, h 1 is corresponding to 1 bar, and 300 k; corresponding
to 1 bar, h f dot will be at 77 Kelvin, which is the boiling point of liquid nitrogen, at
1 bar pressure. In order to maximize y, the value of h 2 should
be as small as possible, or h 1 minus h 2 should be as high as possible. So, if you
want to get maximum yield, or m dot f upon m dot, the numerator should be as high as
possible, while the denominator is fixed because, h1 minus h f is immediately fixed, once your
initial conditions are fixed. In order to maximize y, h 1 minus h 2 should be as high
as possible, or h 2 should be as small as possible. What is h2? It is enthalpy at a
point after the compressor. So, it depends on what the pressure of the gas after the
compression is, and the corresponding enthalpy at that particular point.
To have a minimum h 2, the change in enthalpy for a given change in pressure, should be
zero, at temperature T 1. H 1 minus h 2 is a compression process, or the process 1 to
2 is a compression process. It is happening at constant temperature, so it is a isothermal
compression process. Now, in order that, the value of h 2 should be minimum, the change
in enthalpy, or del h 2 by del p 2, change in enthalpy for a given change in pressure
should be zero, at a given temperature T 1. This can be better understood later. What
I want is, del h 2 by del p 2. If I go on changing the pressure p 2, at a certain point,
or a certain pressure, the value of h 2 is going to be minimum. It also has to take into
consideration that, during this process, the temperature remains constant.
Mathematically I will say that, del h by del p, at temperature T 1, is equal to T 2, at
point 1 is equal to point 2, this should be equal to 0. So, I get the minimum value of
h2 here, or I get maximum value of h 1 minus h 2, in that case.
Using calculus h, p, and T, we have seen earlier that, del h by del p at constant temperature,
into del p by del T at constant enthalpy, into del T by del h at constant pressure,
is equal to minus 1. So, what is this? Del T by del p is nothing but 1 upon, and del
T by del h is nothing but 1 upon c p. If I take these points on this side, what is this
point, del h by del p is equal to 0; it means that, mu J – T, into c p is equal to 0.
See, if I say mu J - T into c p is equal to 0, the quantity c p is cannot be 0, because
c p is always a finite quantity. Therefore, what you learn from this is, mu J - T is equal
to 0. Point 2 is in such a way that mu J - T should be equal to 0. In that case, for that
particular temperature, it implies that, in order to maximize y, the state 2 should lie
on the inversion curve. As you remember that, on inversion curve, mu J - T is equal to 0.
So, point 2 should lie in such a way that, the point 2 lies on the inversion curve for
a particular gas, at a temperature of compression. The temperature remains constant, and if you
were to get h 2 as a minimum value, in that case, the point should lie on the inversion
curve. This is the most important requirement that one can understand, in order to maximize
the value of y, or in order to minimize the value of h 2.
What do you understand from this? If you see at curve, which is the temperature, pressure
diagram, and this is your inversion curve, and there are different enthalpy lines. If
I want to compress the gas from point 1 to point 2 isothermally, that means, the temperature
remaining constant, you can understand from here that, I can have various infinite values
of different enthalpies, but my h 1 minus h 2 is going to be maximum only when the point
2 lies on the inversion curve. All other points in between will have enthalpy
more than point h 2. Therefore, h 1 minus h 2 will not be at maximum in that case. The
h 1 minus h 2 is at maximum, only when point 2 lies on the inversion curve. One can easily
understand from this curve. So, consider 3 constant enthalpy lines here as, h 1, a intermediate
values over here, and h 2. What you can see from here is, h 1 is more than any intermediate
value over here, and more than h 2. Intermediate value lies between h 1 and h
2. From the figure it is clear that, h 1 minus h 2 is at maximum when the point 2 lies on
the inversion curve, so that, y is maximum. This is very clear from this figure that,
if you want to have y as maximum, h 1 minus h 2 has to be maximum; and then, h 2 has to
lie, or the point 2 has to lie on the inversion curve.
Let us find out the work requirement for this. The work requirement for a Linde - Hampson
cycle can be derived by considering a control volume, enclosing the compressor. Let us have
a compressor over here, and again do the energy balance at this point. The quantities entering,
and leaving this control volume are as given below. So, what is entering is mass at point
1, and what is also entering is W c, or the work done on the system, which is written
as negative minus W c. What is leaving the system is mass at m 2, and also the q r, because
it is leaving the system, it is given as minus q r.
Using the first law, for the following tables, whatever is coming in, is equal to whatever
is going out; as E in is equal to E out. We can find out that, therefore, m dot h 1 minus
w c is equal to m dot h 2 minus q r. If we rearrange these terms, what we get is q r
minus w c, is equal to m dot into h 2 minus h 1.
We are taking energy balance across the compressor. What you get is q r minus W c is equal to
m dot into h 2 minus h 1.
The expression for q r can be obtained by using the second law of isothermal compression,
and it is given by q r into m dot into T 1, S 2 minus S 1, T D S is what you know this
as. So, q r is equal to m dot into T 1, into change of entropy across the process, which
is S 2 minus S 1. If you put this value over here, and rearrange the terms, combining the
above equations, the work required for a unit mass of gas compressed is minus w c, upon
m dot is equal to T 1 into S 1 minus S 2, minus, h 1 minus h 1. This is the work of
compression per mass of gas, which is compressed; m dot is nothing but, per mass of gas which
is compressed. This is the expression for that, which one as to calculate if you want
to calculate the work of compression, if the gases compressed from point 1 to point 2 isothermally,
at temperature T 1.
The liquid yield y, which is already calculated, is given by m dot by m is equal to h1 minus
h 2, upon h 1 minus h f. Combining the above equations, the work required for a unit mass
of gas liquefied, if I want to calculate now W c by m dot f, w c by m dot f is nothing
but w c upon m dot, divided by y, which is nothing but, w c upon y m dot. So, if I know
what my w c upon m dot is, I just divide by y, and what you get therefore, is w c upon
m f, or work of compression per unit mass of gas, which is liquefied.
One has to be very sure about what is the question that is being asked. Is it the work
done per unit mass of gas compressed? In which case, we have to use this expression. If the
question is, what is the work done per mass of gas, which is liquefied? Then, whatever
you have calculated here should be divided by y. Therefore, what you get is work of compression
for unit mass of gas, which is liquefied. With this background, now I would like to
solve a few problems, so that, whatever we have learnt till now could be first used,
and you will understand how to do this calculation, and importantly, how to interpret the results.
Let us have a small tutorial of one or two lectures. Under the first Tutorial now, the
problem statement is, ‘determine the liquid yield, which is nothing but y, the work per
unit mass of gas compressed, which is nothing but W c upon m dot, and the work for unit
mass of gas liquefied, which is nothing but w upon m dot f, for a Linde - Hampson cycle
with air as working fluid’. It is very important to understand what has been asked in the problem.
Sometimes, the problem statements here are very, very large; and you have to really split
those into small sub-problems, and then solve the problem. So, here you are to understand
first what has been asked. So, what is asked is y, which is m dot f upon m dot; work per
unit mass of gas compressed, which is w upon m dot; and work per unit mass of gas which
is liquefied, which is w upon m dot f for a Linde - Hampson cycle, and your working
fluid is air. Therefore, what you should have the T - s diagram for air, that is the basic
requirement that, you should have a temperature entropy diagram of air. This can be obtained
from NIST website, or various books. The system is operated between 1.013 bar,
which is nothing but 1 atmosphere. This is the point 1, and 200 atmospheres, or 202.6
bar at 300 Kelvin. So, your point 1 is going to be one atmosphere and 300 Kelvin, pressure
and temperature respectively. The point 2 is equal to 200 atmosphere, and 300 K, respectively.
So, what is step 1? Step 1 is basically the T - s diagram for Linde - Hamson cycle as
shown. So, first plot T - s diagram, and get the different enthalpy and entropy values,
associated with those points from the chart, which is available with you. So, here I am
drawing a small schematic representation, and you should also draw a small schematic
representation showing the cycle 1, 2, 3, and 4, or the Linde - Hamson cycle as given
over here. The point 1 is at 300 Kelvin and 1 bar. So, locate the pressure line at 1 bar;
locate the temperature at 300 Kelvin, and locate the point 1.
The process 1 to 2 is an isothermal compression, and the point 2 should be located in such
a way that, this pressure is 200 atmospheres, or 200 bar, at this point, temperature remaining
constant because, 1 to 2 is an isothermal compression process. Now, as soon as your
point 1 is fixed, we know where the point f is. What we need to know is the state 1,
and state 2. You may not know state 3 immediately, which may not be required for solving this
problem for the data, which has to be calculated over here.
What you understand from here is, the state properties at different points are as shown
over here. What are the state points? These properties are taken from NIST. I have talked
about NIST in the first lecture - National Institute for Standards and Technology. If
you go to this site, you will have all the T - s diagrams for various gases, which could
be downloaded, and could be used; or, there are other books on cryogenics which give these
T - s diagrams. So, if you see the format which we want to use now, what properties
does it require? The property at point 1, the properties at point 2, and the properties
at point f, is what is required. We have got a small table, and I will request
you to also, while solving such problems to make a small table of all the points, and
get all the property data, like, pressure, temperature, enthalpy, and entropy at those
points. Why only these 3 points? It is because, only the data of these 3 points is required
to solve all the problems associated with whatever question has been asked in the problem.
So, we have got a point 1, which is the pressure, around 1 bar, temperature is 300 Kelvin. The
corresponding enthalpy and entropy are this. If I go to point 2, the pressure is 202 bar,
the temperature remains constant which is 300 Kelvin, corresponding enthalpy and entropy
values are taken over here. Similarly, at point f, which is this, pressure is 1 bar
or atmospheric pressure. The temperature at this point is the boiling point of air, associated
with this pressure. So, what is the boiling point of air at 1 bar? It is 78.1 Kelvin.
Corresponding enthalpy and entropy values are over here, and this is the most important
thing. If you make mistakes in finding out these properties, it will go wrong, and you
will get very absurd values, which will be very difficult for you to explain or interpret
later. Therefore, you should get this point two or
three times. Some error is always allowed, because you are reading the chart, and therefore,
some mistakes can happen. However, you should see that 28.47T should not become 35 or 40;
it can be 30 or 26, or 27, on this side. Always 10% error could be allowed in that case.
Now, the first problem is to solve the liquid yield. The formula for this is, y is equal
to h1 minus h 2 upon h 1 minus h f. So, what do you understand from here? The properties
are required form point 1, point 2 and f. We know all the enthalpy values, which have
been given in the earlier table. This is the table; I have just produced it for you, you
can read now - what is h 1, what h 2 is, and what h f is. This is the third row of this
table; y is equal to h 1 minus h 2 upon h 1 minus h f. If I put these values - 28.47
minus of 8.37, so, I will get plus 8.37 over here.
Similarly, h 1 minus h f, again minus of 406, which is going to be kept over here, if you
solve this, what you get is, 36.84 divided by 434.47, which is equal 0.085. What does
it mean that, y is equal to 0.085? What does it mean? It means that, m dot f upon m dot
is equal to 0.085, which also means that m dot f is only 8.5% of whatever m dot is compressed.
This is a very important thing that, whatever amount of gas which is getting compressed
in the compressor at point 1, only 8.5% of that gas gets liquefied. That is what the
product is. The liquefaction product is only 8.5% of whatever is getting compressed at
point 1.
Now, let us calculate the work per unit mass of gas, which is compressed. Again understand
that, this is w upon m dot. So, apply the formula which is minus W c upon m dot is equal
to T 1 into s 1 minus s 2, minus, h 1 minus h 2. Again take the property data from this
table. Put the value of s 1, and s 2 which is this; S 1 and S 2 and h 1 minus h 2. So,
if I put those values, W c by m dot is equal to 300, which is nothing but, T 1 into s 1
minus s 2, which is this. Point 1 minus of minus 1.5, minus, h 1 minus h 2; h 1 minus
h 2 over here, which is nothing but 443.16 joules per gram.
The amount of compressor power required per unit mass of gas compressed is 443.16 joule
per gram.
Now, the third requirement of the problem is to calculate the work required per unit
mass of the gas, which is liquefied. What we calculated earlier, is work per unit mass
of gas which is compressed. So, what we have earlier found is minus w c upon m dot is equal
to 443.16 joule per gram. We know is y is equal to 0.085. Choosing this,
we can find out what is work done per unit mass of gas, which is liquefied, which is
nothing but W c upon m dot f, which is nothing but equal to minus W c upon m dot, divided
by y, or minus w c upon y into m dot. So, what is w c upon m dot? It is 443. If you
divide that by y, what you get is w c upon m dot f.
So, work required per unit mass of gas which liquefied is 5213.64 joule per gram. This
is way higher as compared to the work required per unit mass of gas, which is compressed.
This is quite understandable because, m dot f value is very, very small; therefore, work
done per unit of mass, which is liquefied, is going to be very, very high. It is just
a way of representing the values. This is the work done per unit mass of gas, which
is compressed, and this is the work done per unit mass of gas, which is liquefied. Therefore,
m dot f being a very small quantity, this quantity is always going to be much higher
as compared to the work done per unit mass of gas, which is compressed.
Now, this was the problem wherein, we have got all the three values. We have understood
now how to calculate these values, and now I will take you to the next tutorial, where
we will understand the effects of pressure of compression, and also the temperature at
which, the gas compression happens. Let us see the next tutorial. Again, as I
said, it is very important to understand what the problem is, because, the problems are
normally statements which are very, very big; and therefore, you have to understand them
in parts.
In this case now, determined the liquid yield for a Linde - Hampson cycle, with nitrogen
as the working fluid, for the following operating conditions; and ultimately what is important
is to comment on the results, or interpret the results. So, again we use the Linde - Hampson
cycle, with the working fluid in this case is nitrogen, and not air. This is the schematic
representation of the Linde - Hampson cycle, and these are the different operating conditions.
What do you understand from this? There are 4 cases - case 1, case 2, case 3,
and case 4. In the first case, the point 1, and point 2 are at same temperature, where
the temperature is 300 Kelvin, and the pressures are at point 1 is 1 bar. So, 1 bar, and 300
Kelvin is the condition at point 1 over here, while the point 2 condition is 300 Kelvin,
and 50 bar. The gas is getting compressed from 1 bar to 40 bar.
In the case 2, the temperature is now not 300 Kelvin, but it is at 200 Kelvin. So now,
the compression is happening at 200 Kelvin, which is way below the ambient temperature,
while the pressure is same as 1 bar compressed to 50 bar; pressure after the compression
is 50 bar. In the case three, here, the pressure is now
changed from 50 bar to 100 again. The cases are taken at 300 Kelvin, and 200 Kelvin respectively.
In both the cases, the compression is happening from 1 bar to 100 bar. The objective of this
tutorial is to understand what happens, if you have a compression at 300 Kelvin, or at
200 Kelvin. This is the isothermal compression from point 1 to point 2 here. This can happen
at 300 Kelvin, which is the room temperature, or we can have the compression at reduced
temperature, which is at 200 Kelvin, and you will have to understand how it affects the
yield, or work done, etcetera. Similarly now, instead of compressing the
gas up to 50 bar, if I went ahead, and if I compress it up to 100 bar, what are the
corresponding changes? We will like to understand these changes even when the temperature of
compression is reduced from 300 to 200 Kelvin. So, this is a very important, and interesting
problem, where you understand the effect of compression pressure, as well as, you can
understand the effect of compression temperature as thermal compression temperature. Let us
study the effect of this parameter in this problem.
So, step one, as I said, the assumption is that, the heat exchanger is effective; 100%
effectiveness. The T - s diagram from Linde - Hampson cycle is as shown. What is important?
What I wanted to say was that, first of all, get hold of the T - s diagram of nitrogen.
Again, it could be downloaded from various places. In this tutorial, we assume 1 atmosphere
is equal to 1 bar. I mean, I do not want to write every time that, 1 atmosphere is 1.013
bar. We can fairly assume that 1 atmosphere is 1 bar, and take the values accordingly.
The point one in the first case, the temperature at 300 Kelvin, and the compression is happening
from 1 bar to 50 bar. The T - s diagram is here. 1 bar to 50 bar, 1 to 2 the compression
pressure happening at 300 Kelvin. The properties at 1, 2, and f are as given over here. As
soon as I say 1 bar nitrogen, the point f will be at 77 Kelvin. So, the point 1 is 1
bar, 300 Kelvin, and corresponding enthalpy and entropy is given over here.
The point 2 is 50 bar, and 300 Kelvin, corresponding enthalpy and entropy are given here. The point
f, which is again at 1 bar, 77 Kelvin, and enthalpy and entropy values are given over
here. So, I can straight away start calculations now, which is y for the first case, is equal
to h 1 minus h 2 upon h 1 minus h f. Put all the values over here, what you get ultimately
is 0.023.
So, for the first case I am getting yield as 0.023. Now I go to the second case, which
is the compression process happening at 200 Kelvin. The pressure remains the same, that
is, 1 bar to 50 bar compression. What I am doing is now, instead of 300 Kelvin I am carrying
out the process of compression at 200 Kelvin. This is schematically shown. 1 to 50 bar,
at 200 Kelvin now. Again I go up to T - s diagram, and I get values. The point 1 is
at 1 bar, 200 Kelvin, corresponding enthalpy and entropy values are here.
The point 2 is 50 bar, 200 Kelvin, corresponding enthalpy and entropy over here. The point
f is at 1 bar, and 77 Kelvin; this will not change. If I calculate now y, for case number
2, same formula, putting the values of enthalpies, at this point, I get 0.076. So, as compared
to 0.023, what we got in case one, I am now getting is 0.076. What is the reason? The
reason is now the process of compression is carried out at a lower temperature. At 1 to
2 is happening at 200 Kelvin, while the earlier process was happening at 300 Kelvin, and you
see correspondingly, I have got an increase in the value of y in this case.
Let us go to case number 3, which is happening at 300K. The process of compression is happening
at 300K only; but the pressure at the end of compression is now 100 bar, and not 50
bar anymore. Correspondingly, now I have got point 1 at 1 bar, and point 2 is at 100 bar,
and 300 Kelvin. Again go back to T - s diagram, and get the
properties. The point 1 is now at 1 bar, and 300 Kelvin; the point 2 is now 100 bar, and
300 Kelvin; the point f is remains at the same condition. If I put these values, y for
case 3 is going to be, putting all the respective values of the enthalpy in this formula, what
you get here is 0.044. If you again compare it with the first value,
which was at 50 bar, this value is going to be higher as compared to what it was for 50
bar, and 300 Kelvin compression temperature. As soon as your pressure after the compression
at this point has increased, what you get is a higher yield, or higher ‘y’ value.
I have got case 4, where the same compression, instead of doing it at 300 Kelvin, now from
1 bar to 100 bar, I am doing it at 200 Kelvin. Again I will go back to the T - s diagram,
again I will get points 1, 2 and f from this chart. Here, my point 1 is at 1 bar, and 200
Kelvin; point 2 is at 100 bar, and 200 Kelvin. So, the process of compression is happening
at a lower temperature. We can say it is a pre-cooled cycle, which is happening at 200
Kelvin, and not at 300 Kelvin. Again I will take the enthalpy and entropy
values from the T - s diagram, and calculate y. So, y in the 4th case by this formula is
going to be 0.14, which is very, very high as compared to what it was for earlier cases.
What does it mean? It means that, if you carry out compression at lower temperature, you
get higher yield. At the same time, if you compress the gas at higher pressure, you get
higher yield. This is what we understood from this. So now, let us summarize the result.
Here are the 4 cases - case number 1, 2, 3, and 4. Case number 1 was compressed from 1
to 50 bar at 300 Kelvin, and it gave us 0.023 y, while I carried out this compression not
at 300 Kelvin, but at 200 Kelvin, the yield increased from 0.023 to 0.076. When I did
the compression not up to 50 bar, but up to 100 bar, the yield increased from here to
here. Again, 0.023 to 0.044 at 300 Kelvin compression temperature, while when I reduced
the temperature of compression from 1 to 100 bar at 300 Kelvin to 200 Kelvin, the yield
increased drastically, from 0.044 to 0.14. This can be graphically represented in the
slides over here. So, understand from here that, as the compression process increases,
the liquid yield increases at a given compression temperature; and as the compression temperature
decreases, the liquid yield increases at a given compression pressure. This is what we
learnt from this.
If I want to plot this now on y as a function of T2 that means, the compression temperature,
or the isothermal compression temperature you can see is, point number 1. This point
2, if I plot at 200 Kelvin and 50 bar pressure is this; and if I join this point and extend
this point further, this is the line of 50 atmospheres, and you can see as you go on
loading the temperature of the compression, the yield is going to increase in this fashion.
If I take into consideration of point 3, and point 4, or case 3 and case 4. This is the
case 3, which is 0.044 which will come at this point at 300K, and 100 bar. This 200K
and 100 bar; if I join these 2 points, this is the 100 bar case, and you can see now.
You can see how much yield has increased, if we went from 50 bar to 100 bar; or, you
can also see how much yield has increased when the process of compression was carried
instead of at 300 Kelvin, I carried it at 200 Kelvin.
If I now extend this further, for different pressures, what you get is this at 200 atmospheres;
and this at 300 atmospheres. It is fair to say that I should go for as high pressure
as possible, and as low temperature as possible for the compression process.
Summarizing, the compression pressure increases, the liquid yield y increases at a given compression
temperature; as the compression temperature decreases, the liquid yield y increases at
a given compression pressure. This is what has been summarized from this.
This is very important to understand what happens when the pressure changes, and what
happens when the temperature changes. Best of this, we have got different assignments.
Please calculate the liquid yield work done compression, and work done for unit mass liquefied.
For the above parameters, again as air as the working fluid, and nitrogen as the working
fluid, please do this assignments, and similar to what we have learnt I have got a different
assignment for you. Please carryout these assignments.
To summarize today’s lecture, the ideal cycle demands very high pressure, which is
impractical, and hence modified cycles are proposed to lower the maximum pressure. An
ideal cycle is used as a benchmark to compare the performances of different liquefaction
cycles. In a Linde - Hampson system, only a part of the gas that is compressed gets
liquefied, that is why you got m dot or m dot f, which is much less than m dot and therefore,
a fraction comes which is called y.
A heat exchanger is used in a Linde - Hampson system to conserve cold in the system. This
process is an isobaric process, and is assumed to be 100% effective.
In order to maximize y, or m dot f upon m dot, for a Linde - Hampson system, the state
2 should lie on the inversion curve, at the temperature of the compression process. This
is made fairly clear in this lecture.
The work required for a unit mass of gas compressed for a Linde - Hampson cycle is given by this
formula. We have seen the derivation of this formula also. For a Linde - Hampson cycle,
the following holds true - as the compression pressure increases, the liquid yield y increases
at a given compression temperature. This is fairly clear from the examples also, which
we have taken. At the same time, as the compression temperature decreases, the liquid yield y
increases at a given compression pressure, this is what we have understood from the last
problem. Thank you very much.