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So, welcome to the ninth lecture of cryogenic engineering of this NPTEL program. My earlier
lecture, I will just take a review of my earlier lecture; what we talked about in the earlier
lecture, it was regarding gas refrigeration and gas liquefaction introduction.
So, what we talked about was basics of refrigeration, liquefaction. In order to liquefy particular
gas, you have to low the temperature to boiling point of this particular gas. And therefore
the importance or the procedure how to reach those lower and lower temperatures is very
important. And the performance of the refrigerator could be calculated by COP or the coefficient
of performance parameter, and importance of Carnot COP was also mentioned. It serves as
an ideal COP or the maximum possible COP to reach a particular temperature.
We talked about various mechanisms to get lower and lower temperatures. These temperatures
can be obtained by throttling mechanism, heat exchangers or compression expansion systems.
We talked about all these three possibilities and we also talked that a complete system
may come out as a combination of all these parameters. We talked about a definition of
a refrigerator, liquefier and a combination of these two systems. We talked that refrigeration
is a normally a close cycle process; that it is working fluid never leave the cycle.
Well liquefier can be called as a open cycle process, because the working fluid itself
gets liquefied. And whatever liquid we collect the same amount of the gas has to be added
to count of the deficit. We talked that J - T expansion is an isenthalpic
process; enthalpy gets conserved during this process. And the initial temperature of the
gas should be less than the T inversion temperature to have a cooling effect. We talked about
that, every gas has got a characteristic inversion temperature. And in order that after expansion
by J - T expansion after isenthalpic expansion if it has to result in lowering of temperature,
then the initial state of the gas should have temperature less than inversion temperature
of that particular gas. We also saw that for ideal gas mu J - T or del T by del p at constant
enthalpy is equal to 0. That means if the gas is subjected to isenthalpic expansion,
it should not result in any lowering of temperature or even in heating of the gas, because mu
J - T is equal to 0 for ideal gas.
In this particular lecture, I am continuing with gas liquefaction and refrigeration systems.
What we are going to talk about now is J - T expansion of real gas; what we talk in the
earlier lecture was J - T expansion of ideal gas. Let see how the real gas behaves when
it is subjected to J - T expansion. Then we will talk about isentropic expansion; then
we will compare the J - T expansion and isentropic expansion. And then we will see the first
cycle which is ideal thermodynamic cycle which is use for gas liquefaction all right, this
is what we will try to cover in this particular lecture.
We have seen that work input is needed to generate and maintain low temperatures. We
have seen that the first law and the second law tells us that we have to put in some work,
we have to put in some efforts in order to generate low temperature and to maintain the
low temperature. We have seen that as T L decreases as lower and lower temperature if
we want to reach, the Carnot COP decreases stating that more work input is required to
maintain very low temperature. If you remember the last lecture, we had talked
about watt of power input per watt of refrigeration, and we found that the watt of power required
per watt of cooling effect required increases as we go down lower and lower temperatures;
if you want to reach lower and lower temperature your work input increases. Hence, a knowledge
of performance of various refrigeration liquefaction cycle is necessary to design the system with
maximum COP. See ultimately important thing is to our maximum COP or maximum efficiency,
and therefore one has to understand what cycle one should use in order to reach particular
temperature or in order to liquefier particular gas. This is very, very important considering
the fact that efficiency and COP are very important functions of what cycle you choose
in order to refrigerate in order to reach lower temperatures or to liquefier particular
gas.
Now, let us understand what is Joule - Thompson coefficient is for the real gas, and let us
start with the definition of enthalpy which is nothing but some of internal energy u plus
pv work, and mathematically I can write h is equal to u plus pv. So, by definition mu
J - T or J - T expansion coefficient is nothing but del T by del p at constant enthalpy. And
if I split that it becomes del T by del h into del h by del p with a negative sign over
here. This is what we have seen earlier and therefore, I will not go in the details of
all this things. So, when I know now del h by del p I have
to find at constant temperature and del T by del h at constant pressure. Substituting
enthalpy h, we know h is equal to u plus pv and if I substitute this l h over here in
the above expression of J - T coefficient what we will get is this mu J - T is equal
to 1 by c p, because delta T by delta h is nothing but 1 by c p which we have seen earlier.
And now I am putting the value of h as u plus pv, and therefore this expression becomes
del u by del p at constant temperature plus del by del p of p v at constant temperature
here. Now, this is the bracket will will decide
whether mu J - T is positive which is required for cooling or negative which result in heating
or is equal to 0 which will not result in cooling or even heating. See, in order that
J - T expansion results in cooling, the bracket should always be negative. So, net result
of this bracket should be negative. So that taking care of this negative, the whole thing
becomes positive and in that case mu J - T is going to be positive.
So, this is the expression I am talking about, and what it has? It has got two parameters;
del u by del p at constant temperature and del pv by del p at constant temperature. Now,
let us talk about the first parameter that is del u by del p, because essentially the
values what we get from del u by del p at constant temperature and del pi by del p of
pv will decide what is the net sign of this bracket. The first term is the variation of
internal energy which pressure at constant temperature. Now, this is nothing but this
term represents departure from joules law. The joules law states that the internal energy
is constant for a given temperature. And here we are talking about changes in internal energy
with pressure when the temperature is maintained constant. So, this basically talks about how
you are going away from the ideal gas right. The ideal gas will not accept these. That
means, we are going away from ideality which means that we are taking into consideration
the real effects of the gas and we are talking about real gas now.
At lower pressures, when you expand the gas from higher pressure to low pressure and when
we are talking about low pressures, the molecules are pulled apart. This is what happens in
expansion process. This results in increase in the potential energy of the gas. So, whenever
the molecules go apart, the potential energy increases, as a result the kinetic energy
decreases, and as you know the kinetic energy decreases, because the total energy remains
constant alright. So, when the kinetic energy decreases, kinetic
energy directly proportional to temperature. So, as soon as the kinetic energy decreases,
the temperature of the system is also decreases. That means, del u by del p when the pressure
decreases, the del u by del p should result in increase of u value, because the potential
energy of the system increases, and therefore this term del u by del p always becomes negative
which helps in the reducing the temperature gas when the gas expands. Please understand
that del u by del p therefore will always be negative and will always help in reduction
of temperature when the gas is subjected to expansion, when the pressure of the gas reduces.
This is true for the real gas also.
Now, we will talk about the second term. So now, if we talk about the second term, this
second term is the most important thing and this term will decide if this bracket is going
to be positive, negative or 0. The second term now it can be positive, negative or 0
and this represents now departure from Boyle’s law. Again you can see now, this is del by
del p of pv at constant temperature. The Boyle’s law says that the product pv remains constant
for a given temperature. Now, we say that we are talking about changes in the value
of pv although the temperature remains constant with pressure changes. So, this is basically
departure from the Boyle’s law where the non-ideality or the real gas comes into picture.
Please look at this graph and here it talks about the pv variations verses pressure. And
as you can see here, the pv variation for higher temperature, pv value decreases when
the pressure decreases, and therefore, pv del pv by del p in this case is going to be
positive. Well you can see as the temperature goes on lowering down, you can see that there
is the variation in the curve and for some part you can find pv decreases, for some part
the pv increases, and here comes the pv variation with pressure the sign changes. So, here in
this case, the value pv decreases with a decrease in pressure saying that del pv by del p in
this case is positive. Well if you come in this region, we understand
that pv value increases when pressure decreases and in this region therefore, del pv by del
p in this case is going to be negative, because there is a pv increase happens although the
pressure decreases. So, in this case is going to be negative. And if we want to have cooling
in this case, because you want mu J - T to be positive, we expect that del pv by del
p should be negative which should help us it having mu J - T has a positive quantity.
So, if we see at this variation, we can have two parts; one at the higher pressure and
one is at a lower pressure. We can see at a higher pressure, the molecules are very,
very squeezed. The molecule is already subjected different molecules are already subjected
to very higher pressure, and therefore, they cannot be compress further.
And what we can see now, hence, they are less compressible than Boyle’s law prediction
in this case. When the compressibility of the gas is very, very less in this case, when
they cannot be compressed further, the pv should not show an increase in this case.
Because the pressure decreases, but the volume will not increase in this case in a proportional
manner, because the gases are less compressible in the higher pressure region; that means
even if the pressure is decrease, the value of the pv also shows decrease in this case,
and therefore, in this region, if you expand the gas, this del pv by del p term is going
to be positive in this case. That in pressure reduction happens del pv also reduces, and
this is not going to help you in reduction of temperature or J - T cooling effect.
However, if you see the other half that is the second term is negative at lower pressures
and lower temperatures. So, if we talk about this region now, we find in this region which
is low pressure and low temperature; the gases are more compressible than Boyle’s law predictions
right. Why see here, because now we are talking about lower pressure value. This is not a
higher pressure value as it was in earlier case. We are talking about lower pressure
values, and when I expand the gas from this pressure let us do up to this pressure. The
pv product increases, because the volume increase will be there, and volume increase will be
more, because the gases are more compressible in this region. And therefore, as soon as
I expand the gas from higher pressure to low pressure in this case, the pv value increases,
and therefore, del pv by del p in this case would be negative which will help in making
mu J - T as positive. And therefore, if you expand the gas in lower temperature and low
pressure region, it is always beneficial or it will always result in cooling when subjected
to isenthalpic expansion. And this is where we found there are the temperature
or the pressure above which it will not result in cooling when expanded; while in this case,
the gas will result in cooling when expanded in a J - T expansion manner. So, for a real
gas Joule - Thompson coefficient depends upon the relative magnitude of both this terms.
So, what is this magnitude positive, negative or 0. This is always going to be negative,
but what is the net result of this bracket will decide, if mu J - T is going to be positive
or negative or when subjected to Joule - Thompson expansion, is it going to result in cooling
or is it going to result in heating will be decided by this parameters.
Now, this was true when we talked about real gas and we talked about the enthalpy equation.
Now, same thing could be understood when I am using a real gas equation or van der Waals
equation. So, van der Waals equation of state for a real gas is given as this, which is
p plus a by v square into v minus b is equal to RT where a and b are constants, and this
a and b gives the measure of intermolecular forces which is what a is while, and size
of the particle respectively the value of b denotes basically the size of the particles,
which we say that the finite volume of value to this b and a value in this case. For ideal
gas, we assume that both a and b is equal to 0, and therefore what we get is pv is equal
to RT for ideal gas. If we rearrange this term, what you get is a p is equal to RT upon
v minus b minus a by b square. You can do this exercise by yourself to derive this particular
term.
Now, if I want to differentiate this term, real gas equation, van der Waals equation;
upon differentiating the following equation at constant pressure, this is my equation,
because the pressure remains constant in this case. I will get p is equal to RT v minus
b and minus a by v square. If I differentiate I will get 0 here, because I am keeping pressure
as constant and then I get this particular term. From where if I rearrange the term what
I ultimately want to get is a value of del v by del T at constant pressure. This is the
most important thing, because I would like to put this value in my earlier equation.
See if I get del v by del T at constant pressure at this value, I will substitute this del
v by del T at constant pressure in the J - T coefficient equation, this is the value and
this my J - T coefficient equation.
So, here this del v by del T at constant pressure, I will put from here into this equation. And
if I do that for real gas, what I get is this. Now, I will try to approximate these, if v
is specific volume is very, very large as compare to the values of a and b. If I assume
this and this is a very realistic assumption, if I do that then b by v will tend to be equal
to 0, well this also tend to be equal to 0, and therefore, ultimately my expression for
mu J - T comes like this - 1by c p 2a upon RT minus b. So, here the value of mu J - T
is now determined by this particular equation, and the sign of mu J - T will be decided by
this bracket - 2a upon RT minus b. As we know 1 by cp is always going to be positive, because
c p is always a positive term.
So, what I understand now? For a real gas with large specific volume, I will have this
equation, and if I want to have mu J - T to be more than 0, this bracket should be more
than 0, and therefore, temperature should be less than 2a by Rb. If I want to be mu
J - T to be less than 0 which means it is going to be result in heating in this case,
I would like the temperature of the gas - initial temperature of the gas to be more than 2a
upon Rb. And if I want mu J - T to be equal to 0 in that case, my temperature is going
to be 2a upon Rb. In this sense, this is nothing but the inversion temperature for the gas
at a given pressure. See, if my temperature is less than 2a upon Rb, my mu J - T is going
to be positive and when expanded it going to be result in lowering of temperature.
Well here, if the temperature is more than 2a upon Rb which is nothing but the inversion
temperature for a particular gas; my expansion is going to result in heating in this case.
Well this is going to be point where nothing will happen. And if a is equal to b is equal
to 0, ultimately everything gets reduced to the ideal gas, again in which case mu J - T
is equal to 0. So, this essentially talks about what the values of inversion temperature
would be for a particular gas depending on the values of a, b and r for those gases.
This is the van der Waals equation basically would decide, what are the values of the inversion
temperature for a particular gas at different pressures etcetera. One can based on this
values, one can construct the inversion curve also.
See, if I want to plot isenthalpic line on a T – S chart, you can see this is the dome
in which the liquefaction happens, temperature and entropy here. And you can see, suppose
this is a room temperature at this point, the isenthalpic lines are like this. See,
if I expand the gas from this high pressure isenthalpicaly, I will get this as delta T
in this case, this minus this. But as you go down the temperature here, and if I expand
the gas isenthalpicaly from this pressure to this pressure, my delta T in this case
is much larger than what I get near room temperature. What does it tell? It shows that the drop
in temperature obtain after isenthalpic expansion at lower temperature is very, very high.
So, if I expand the gas at lower and lower temperatures, I am going to get bigger and
bigger temperature drop if expanded isenthalpicaly. Why, because as I am going down and down I
am reaching in a real gas state, I am going away from ideality. And therefore, my J - T
expansion is going to be more and more prominent, I am going to get more and more real gas expansion,
and in this case I am getting very large cooling when expanded isenthalpicaly all right. One
should always have isenthalpic expansion when one goes away from ideal gas, and best way
to go away from ideal gas is to have liquid; liquid is nothing but completely away from
ideality. So, normally you find that J - T expansions
are mostly the J - T expansion would be for a two phase system; when you are completely
away from ideal gas. So, you have got a mixture of let say gas plus liquid or complete liquid
also which is what is normally prepared in order to have more and more cooling effect,
if the gas is or the mixture or the two phase mixture is subjected to J - T expansion. Why
this happening? This is because the gases are imperfect at very low temperature, and
therefore, they show the real gas behavior and they are completely away from the gas
nature in these cases.
So, if you want to see what happens near room temperature, you can see now. If I expand
the gas from 2 to 3 which is let say at room temperature now, it is not result in in cooling,
but it is result in heating in this case. Why, because this initial state of the gas
is above the inversion temperature, and therefore, instead of getting into cooling zone it is
going into heating way. And therefore, one should ensure that this state 2 is lying below
the inversion temperature before it takes off for isenthalpic expansion. So, in the
fluid expands from state 2 to 3, the temperature rises. This occurs because the initial temperature
at state 2 is above the inversion temperature. This is the very important parameter one has
to understand.
Now, there are various gases and there got the one inversion temperature. What you can
see from here is the gases like Helium, Hydrogen, Neon have got their inversion temperature
has 45, 205 and 250 kelvin which are below the room temperature, while all other gases
over here having their inversion temperature above the room temperature; the room temperature
being 300 kelvin. It means that if I want to have isenthalpic expansion to be resulting
into cooling for these gases, their initial state has to be brought down below their inversion
temperature. Then and then only when they are subjected to isenthalpic expansion or
a J - T expansion, they would result in cooling right.
So, for gases like Helium, Hydrogen and Neon, in order to experience J - T cooling, they
have to be precooled below T inversion temperature first and then subject to the J - T expansion.
However, the other gases if they are subjected to isenthalpic expansion at room temperature
only, they will result in cooling in this case, because their inversion temperature
lie above the room temperature. This is very important for cryogenic temperature for these
gases, we have to precool these gases first below their inversion temperature and then
get cooling effect. Now, we talked about isenthalpic expansion.
Now, we will talk about isentropic expansion, whether entropy remains constant. Enthalpy
and entropy are two thermodynamics state properties of matter which are functions of pressure
and temperature. When the gases are expanded at constant enthalpy as in J - T expansion
it is called as isenthalpic expansion. On the similar line, when the high pressure gases
are expanded at constant entropy it is called an isentropic expansion or a reversible adiabatic
expansion. This is all what we know from thermodynamics.
The commonly used expansion devices are turbo expanders which is a turbine for isentropic
expansion or reciprocating expanders where cylinder and piston arrangement could be used
to cause isentropic expansion. This is a work producing device; now, this is a difference
between a J - T expansion and an isentropic expansion. Here in this case, when expansion
occurs, there is some kind of work gets produced and we will get some work from this expansion.
So, this is a turbo expander and you will get some kind of a W e or a shaft work done,
when the gas expand from high pressure to low pressure. Or similarly, if we have got
a reciprocating expanders, what you get ultimately is a work output W e when the expansion occurs.
Now, similar as what you saw earlier, delta T by delta p at constant entropy is what we
will consider now and this ratio is called as isentropic expansion coefficient.
Now, let see how we derive the value of delta T by delta p at constant entropy. The entropy
is a function of both pressure p and temperature T, and therefore, what we can write is s is
equal to function of pressure and temperature. Using the calculus as what we have done earlier,
we derive following expression, where we say now, delta s by delta p at constant temperature
into delta p at delta T at constant entropy into delta T by delta s at constant pressure
is equal to minus 1. And therefore, if I rearrange this what ultimately I want to have is a mu
s which is nothing but delta T by delta p here, delta T by delta p at constant entropy
which is what this value is. If I rearrange this term what I get is equal to minus sign
delta T by delta s at constant pressure into delta s by delta p at constant temperature.
So, here I have got an expression now for mu s; in a similar way, what I have earlier
got expression for mu h or mu J – T.
Now, for the same variables entropy s, temperature T and pressure p, we use again calculus and
now I can write ds as, basically I am writing this expression by using partial derivative
in terms of T and p - temperature and pressure. So, delta s by delta T keeping pressure constant
into dT plus delta s by delta p keeping temperature constant into dp. If I multiply by T and this
is again in the same line as what we have done earlier. We find that Tds is nothing
but multiplied by T to this term, multiplied by T to this term. Again using the same approach,
this particular thing is nothing but c p. While this delta s by delta p gives minus
del v by del T at constant pressure which is what we obtained from Maxwell’s equation.
If I use these two terms now in my original equation over here, I will put the values
and what I put the values as delta T by delta s as c p, because if I compare this with this,
I get this is equal to this and I get this is equal to this all right. So, I put the
value of delta T by delta s as 1 by c p, and I get delta s by delta p as nothing but delta
v by delta T at constant pressure. This is a negative sign over here and del 1 negative
sign here, over here. If I put them together, I get mu s is equal to T by c p into del v
by del T at constant pressure.
So, now I apply the ideal gas and I know for ideal gas pv is equal to RT, and therefore,
v is equal to RT by p. If I differentiate that with respect to temperature keeping pressure
p constant, I will get delta p by delta T is equal to R by p, because a T will go. And
R by p is nothing but equal to v by T. See if I put these values in this equation, and
if I replace del v by del T by is equal to v by T, I will get T and temperature constant,
and my mu s ultimately gets reduce to the value of v by c p. So, mu s is equal to v
by c p by using ideal gas law, what does it mean? It means that for an ideal gas mu s
is non-zero quantity. In earlier case for ideal gas, mu J - T was
0. Unlike the case, in the J - T expansion when the enthalpy was constant we found that
mu J - T was equal to 0. It means that the ideal gas does exhibit a cooling effect when
it undergoes an isentropic expansion. This is the very important thing to understand
that whatever happen when the gas is expanded isentropicaly, it would always result in cooling,
and it is true for the gas also; which was not true in earlier case when the J - T expansion
happened. Now, what we will see? We just found that for ideal gas mu s is positive non-zero,
and what we will see now is what happens mu s when it is subjected to real gas.
So, here mu s is equal to this particular quantity. The derivative term represents the
variation of volume with temperature; del v by del T is nothing but volume changes with
temperature at constant pressure. This term is called as volumetric coefficient and is
always positive, and hence isentropic expansion coefficient will always be positive in this
case. Whatever happens mu s is going to be always positive, because volumetric coefficient
will always positive in this case. It is clear that the isentropic expansion results in cooling
irrespective of its initial state unlike J - T expansion, because this term is always
going to be positive in this case all right.
Now, if I apply the same for real gas using van der Waals equation, we get the similar
term as what it was in the earlier case. And if I substitute that in the same value as
mu s is equal to this term which is what we got from earlier state.
What you can find, again with approximation, I get mu J - T is equal to v by c p which
is means that for real gas also mu J - T is always going to be positive here. We found
that whether is a gas or real gas mu J - T is always going to be positive; that means,
when expanded isentropic, it will always result in lowering of temperature.
If I want to compare isenthalpic expansion with isentropic expansion; so, J - T expansion
with isentropic expansion, I should get the J - T expansion as a condition of T inversion
or inversion temperature. I do not have any such condition in this case. Similarly, when
I have a J - T expansion, I get no work output and this is called therefore as internal work
process. While here I get some work output and therefore it is called as external work
process. The J - T expansion is a very simple device is a very simple in construction. While
in this case, you will have a lot of rotary mechanism, reciprocating mechanism, lot of
linkages, frictional losses, etcetera, and therefore the device is a complex device in
this case. It is normally used for phase change of fluids that is what we talked about, because
J - T expansion always bucks for real gas, and therefore, we always prepared to have
phase change of fluid during the expansion. While here we should work with single phase
fluid, because it have got two phase fluid, it will not be acceptable to the turbo expander
or reciprocating expander. And lastly, because the J - T expansion is
subjected through small wall constriction, the constriction can get subjected to clogging
with the dust or the particles coming with the gas. And the clogging of the constriction
is a disadvantage, and many times you have to clean it. While here one has to do regular
maintenance and periodic checks are required, some time the lubrication has to be taken
care of anything like that. So, this is what we will compare a J - T expansion on isentropic
expansion. And one has to prepare depending on what you want, whether one should go for
J - T expansion or one should go for isentropic expansion.
So, let us talk about the gas liquefaction systems here. I will understood in what with
the expansions happen. We talk about now various gas liquefaction systems. And there are various
gas liquefaction systems now . And first one is Thermodynamically Ideal system, Linde Hampson
system, Precooled Linde Hampson system, Linde Dual Pressure system, Claude system, Kapitza
system, Heylandt system and Collins system. Let us talk first about the thermodynamically
ideal system, because we will have to compare all other systems with this ideal system as
normally happens in engineering. So, this we this system serves as a comparison basis
and let see how does this system work.
So, thermodynamically ideal system will have following salient features. Here this is a
compressor, and this is a expander, and this is a container whether liquid after liquefaction
is stored and the liquid is transferred from this. So, you can see that process 1 to 2
is a compression process, process 2 to f is a expansion process, and some work output
W e is obtained over here. While in compression process W c is supplied, this is the work
of compression and this is heat which is generated Q r during compression. So, 1 to 2 process
is a compression process, while 2 to f is a expansion process all right.
So, here what we can see is whatever gas is compressed is expanded, and it is converted
to liquid and this gas is collected as a liquid form from this container. So, there is no
return path for this gas. So, in a thermodynamically ideal system, all the gas that is compressed
gets liquefied. So, all the processes are ideal processes in this and there are no irreversible
pressure drops in a system. The process of compression and expansions are isothermal
and isentropic respectively.
So, if I were to represent this system on a T - S diagram, this would look like this.
These are two pressure lines and this is starting point at point 1 over here. So, 1 to 2 process
is a isothermal compression process, temperature remains constant. And after pressurization
at point 2, here the gas is expanded isentropically, therefore entropy remains constant. And this
gas after expansion hits this end which is a such as a liquid curve at this point. So,
whatever gas is compressed completely gets liquefied at this point; this is what the
liquid varying, we get at this particular point all right.
So, this point is shown as here and this is going to lie at a 1 bar pressure line, because
this is my constant pressure, gas is expanded from point number 2 to f from very high pressure
to 1 bar which is my starting point. So, this is ambient point 300 kelvin and 1 bar, and
this is 1 bar line, and this is what my final state would be. What does it mean? The initial
state 1 and final state f can be determined as soon as I know the pressure at this point.
So, pressure is known, temperature is known, I know the f condition. So, initial condition
1 of the gas, this condition determines the final position f of the gas after liquefaction.
This is an open thermodynamic cycle and that is why we get the mass transferred gets transferred
from the working fluid only, because the working fluid flows across the system. My consider
this as a control volume, I can apply the 1 st law of thermodynamics to this system
and I can analyze this system. And the changes in velocities and datum levels are assumed
to be negligible and they are neglected while doing this analysis.
The quantities entering and leaving the systems are given as below. So, if you see this table,
I have got what is entering this control volume and I have to see what is leaving this control
volume. So, what is coming in the control volume is the mass at point 1 which is m 1
at point 1, and W c that is the work which is being supplied to the compressor W c. So,
what is coming out of the system is the heat of compression or Q r, W e is the work of
expansion. And what again coming out is m 1, because whatever compress is m 1; same
thing is liquefied m 1 at point f. See if I apply the 1 st law, I will say energy
coming in the control volume is equal to energy leaving the control volume. So therefore,
m 1 h 1 is the energy associated with this gas which is entering at point 1, which is
h 1 is nothing but the enthalpy of the gas at point 1, W c which is work done is equal
to Q r which is leaving the system plus W e plus m f and h f associated enthalpy at
point f all right. So, m 1 dot and m m f dot are nothing but the flow rates or the mass
flow rate of the system. In fact as you can see in this case, m 1 dot is equal to m f
dot.
If I further rearrange this system, the work W e produced by the expander actually is negligible,
and therefore can be neglected as compare to the systems. And therefore, if I neglect
W e, my equation gets reduced to Q r minus W c is equal to m 1 in bracket h f minus h
1, because m 1 dot is equal to m dot f. The compression process is assumed to be isothermal
that means the temperature remains constant and this is what we talked about.
Hence, from second law of thermodynamics, we can write that Q r is equal to m 1 dot
T 1 into difference between two entropy between the process 1 to 2 which is the isothermal
compression process. So, heat generated during compression is nothing but m dot into T 1
into d s that is s 2 minus s 1. Also the expansion process is isentropic process, therefore s
2 is equal to s f, and therefore, I can replace this s 2 by s f. So, by substitution what
you get is W c is a work of compression is equal to m 1 dot T 1 into s f minus s 1 minus
m 1 dot into h f minus h 1. So, entropy difference and enthalpy difference between the point
1 and final point most important parameters to be use for calculation of W c.
This work of compression is done on the system and therefore, hence, the value is expressed
as a negative quantity. So, work done on the system can be represent as negative quantity.
So, minus W c is the work of compression which is equal to now m 1 dot T 1 into s 1 minus
s f minus m 1 dot into h 1 minus h f. So, work required per unit mass of gas compressed
which is m 1 dot is equal to minus W c upon m 1 dot is equal to T 1 into s 1 minus s f
minus h 1 minus h f. This is the most important thing that if I know the point 1, I get T
1 into s 1 minus s f and then I can know the enthalpy difference h 1 minus h f also. And
this determines the work required for compressing this gas and this is work required per unit
mass of gas compressed this gas.
Since in an ideal system, the mass of gas compressed is same as mass of gas liquefied,
and therefore we got m 1 is equal to m f or m 1 dot is equal to m f dot. So, work required
per unit mass of gas that is liquefied, this is now W c by m dot f. So, I am just replacing
m 1 dot by m f dot. So, W c by m f dot with a negative sign here is equal to T 1 into
s 1 minus s f minus h 1 minus h f. So, here again we can understand that - the work required
per unit mass of gas liquefied depends on the initial condition of the gas, because
the moment initial condition is T 1 and s 1 are obtained or h f obtained, and the pressure
remains the same, and therefore s f at saturated liquid condition can be obtained at the same
pressure, and h f is the enthalpy at the of the saturated liquid condition obtained at
the same pressure. Therefore, moment your initial condition gets decided, this value
gets decided immediately. Now, based on this we can have a simple tutorial to understand
how do we calculate the work of compression to liquefier gas.
So, here there is a problem. Determine the ideal work requirement for liquefaction of
nitrogen beginning at 1 bar pressure and 300 kelvin. So, what I know is my initial condition
is at 1 bar and 300 kelvin. So, my first step is basically get a T - s diagram for ideal
thermodynamic cycle, I shown. This is my T-s diagram and this is what is available in various
books; this is available in net also. For nitrogen, we should get this diagram. And
find out a condition, as soon as I know the point 1, I will locate the point f. Because
I will come down at constant pressure, go the constant pressure line and locate the
point f. I draw a horizontal line from 1 to 2, I draw a vertical line from over here,
and the intersection would otherwise give me the point 2 at this. So, this is 1, 2 and
f is obtained or basically I should know what is my 1 point and what is my f point. This
is the most basic requirement to calculate the work ideal work requirement for liquefaction
of any gas.
Step 2 now; the state properties at different points are as given below. So, what are the
state properties at 1 and f? At 1 I have got the pressure of 1 bar, at f also I have got
the pressure of 1 bar. Here I got a temperature of 300 and 77 kelvin 300 and 77 kelvin at
1 bar; 77 is the boiling point of nitrogen. Corresponding enthalpy values have to be taken
from this chart now. So, a T-s chart has to be made available, you can see from different
books. And the read enthalpy at point f and read enthalpy at point 1; these are the two
values. Similarly, read entropy at point 1 and read entropy at point f.
And here put those values in the equation; substitute in the equation these values. So,
this is my equation now. W i we can call ideal work done per mass of the gas which is liquefied.
So, W i by m f dot with a minus sign into T 1 into s 1 minus s f minus h 1 minus h f,
if I put these values what I get is a 767 joule per gram. So, my compressor requirement,
per gram of mass which is liquefied, I will get I will have to put in 767 joules. So,
my work requirement, therefore ideal work requirement for nitrogen gas to be liquefied
is 767 joules per gram.
If I go to the second tutorial now; calculate the ideal work requirement for liquefaction
of helium and hydrogen beginning at 1 bar pressure and 300 K, and compare the results.
I will follow the same technique as what I did earlier for nitrogen gas. My initial state
at 1 bar and 300 K will give me the final position also. So, the T - s diagram for a
thermodynamic cycle is as shown exactly same thing. Now, again I will go for helium and
hydrogen T-s chart. Look at the point 1 at 1 bar and 300 kelvin, this is the 1 bar line.
And for hydrogen, this will heat at 20 kelvin, for helium it will be at 4.2 kelvin. So, the
temperature will remain actually 4.2 kelvin. One need not be bothered to locate this point
2, because this point 2 will be at a very high, high pressure - very high pressure.
So, what I need to locate is point 1 only and point f only.
So, step 2 is the state property of hydrogen and helium at different points are as given
below. So, here are my properties. This is point 1 and f for hydrogen, and this is point
1 and f for helium, all are at one bar pressure which is the room temperature is 300 kelvin
in both the cases; while the point f is at the boiling point of respect to gases. So,
point f is going to be at 20 kelvin for hydrogen and at 4.2 kelvin for helium. Read the corresponding
enthalpy values and read the corresponding entropy values as what we have done earlier.
See, if I put this values in this equation, this values, what I get for hydrogen is 300
into s 1 minus s f which is this minus h 1 minus h f which is this, look at this minus
point, because this is with respect to a particular initial condition or 0 point condition, and
therefore, this is minus 75 is written over there, this will become positive in this case.
What I get is 9835 joules per gram for hydrogen and if I do it for helium put the same values
here, I get only 6864 joule per gram. What does it indicate? It indicates that the work
done for liquefaction of hydrogen is more than work done requirements for liquefier
helium gas. So, it is not although the boiling point of
helium is just 4.2 kelvin which is less than that of hydrogen. Still the work done to liquefier
1 gram of helium gas is less than that for hydrogen. So, basically it depends on what
is my h 1 minus h f values or what is my latent heat in this case. This value is much higher
for hydrogen as compared to helium. Similarly, this value also is much higher as compared
to what it is for helium. So, it basically depends on the nature of the gas, what is
the amount of work required to liquefier hydrogen and work required to liquefier helium.
So, this table give you ideal work requirement for various gases. And you can see the boiling
point of helium is 4.21, 20.27; still the ideal work requirement for hydrogen in this
case is more than that for helium. Similarly, nitrogen is only 768, but for air it is 738
argon, oxygen and ammonia. These are different values given. We will expect that you can
work out your own ways of calculating is ideal work requirement, and you can compare this
values with this table or you can refer to any literature, and you can compare your findings
with this ideal work requirement in terms of kilo joule per kg or joule per gram.
With these tutorials you should be in position to calculate using charts, the ideal work
of liquefaction for air, oxygen, helium, we have already done ammonia. Take the T - s
diagram - T-s chart for these gases, find out the enthalpy values, find out the entropy
values and calculate the ideal work requirement to liquefy this gas. Compare these values
with the values given in the table. So, we have already given the table and whatever
values you calculate, please compare, the values will be little different depending
on how good or bad, your charts you read, sometimes you can read some wrong values and
according to which your values can show some difference.
In order to summarize, whatever we have covered today, we found that for ideal gas mu J - T
is equal to 0. But for a real gas J - T coefficient depends upon the relative magnitude of departure
from joules law and Boyle’s law. We saw the two parameters and we we found out that
these two parameters in the bracket decide whether the mu J - T is going to be negative,
positive or 0. The gases like nitrogen, air, show J - T cooling when expanded at room temperature.
Because the inversion temperature for these gases is more than room temperature; so, one
need not worry about temperature in this case, because the inversion temperature for these
gases is way above the ambient temperature of 300 kelvin, and therefore, if they are
subjected to J - T expansion, they would immediately result in cooling. However, the other gases
like helium, hydrogen and neon they required to be precooled to bring down their initial
temperature below the inversion temperature, and then only the y will experience J - T
effect or J - T cooling.
In expansion devices like turbo expanders and expansion engines, the expansion process
is isentropic or reversible adiabatic. The coefficient of isentropic expansion is given
by mu s value which is T by c p into del v by del T at constant pressure. And we found
that the isentropic expansion is always positive for both real and ideal gases. Therefore,
one need not worry about inversion temperature for search cases. It results in cooling for
any initial state, unlike the J - T expansion which is dependent on what is the T inversion
temperature for that particular gas.
The J - T expansion is normally used where phase changes are required, where whereas
isentropic expansion is used for single phase fluids; this is what is met clear right. Because
J - T expansion normally we want to have real behavior, we want to have non-ideal behavior,
and therefore, we prepared to have a phase changes or we would prepare to have liquid
in this case at initial condition also. While isentropic condition expansion is normally
used for single phase fluid, because having liquid there can spoil can have problems for
the turbo expanders or reciprocating expansions. In a thermodynamic ideal system, this is what
we saw all the gas that is compressed gets liquefied. And the work required per unit
mass of gas compressed and gas liquefied are given by W c by m f dot minus sign is equal
to T 1 into entropy difference between the initial state and the final state minus the
enthalpy difference between the initial state and the final state. Thank you very much.