 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 a particular gas, you have to lower 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 refrigerator could be calculated by COP or the coefficient of performance parameter and the importance of Carnot COP was also mentioned, which serves as a ideal COP or the maximum possible COP to reach a particular temperature. We talked about various mechanism 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 normally a closed cycle process that means working fluid never leaves the cycle while liquefier can be called as 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 counter the deficit. We talked that J T expansion is an isenthalpy 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 an ideal gas mu J T or del T by del P at constant enthalpy is equal to 0. That means if the ideal 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 talked in the earlier lecture was J T expansion of ideal gas and let us 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 used for gas liquefaction alright. 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 have 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 liquefy 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 liquefy particular gas. Now, let us understand what is Joule Thomson coefficient is for the real gas and let us start with the definition of enthalpy which is nothing but sum of internal energy u plus pv work and mathematically I can write h is equal to u plus pv. So, by definition mu Jt or a Jt 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 these 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 Jt coefficient what we will get is this. Mu Jt is equal to 1 by Cp because of delta t by delta h is nothing but 1 by Cp 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 pv at constant temperature here. Now, this is the bracket which will decide whether mu Jt 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. So in order that Jt 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 Jt 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 p by del p of pv will decide what is the net sign of this bracket. The first term is the variation of internal energy with pressure at constant temperature. Now this is nothing but this term represents departure from Joule's law. The Joule's 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. The ideal gas will not accept this 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 high 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 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 of the 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. Then you can see now this is del u by del p of p v at constant temperature. The Boyle's law says that the product p v remains constant for a given temperature. Now we say that we are talking about changes in the value of p v 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 p v variations versus pressure. And as you can see here the p v variation for high temperature p v value decreases when the pressure decreases and therefore p v del p v 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 a variation in the curve and for some part you can find the p v decreases for some part the p v increases and here comes the p v variation with pressure the sign changes. So here in this case the value of p v decreases with the decreasing pressure saying that del p v by del p in this case is positive. Well if you come in this region we understand that p v value increases when pressure decreases and in this region therefore del p v by del p in this case is going to be negative. Because there is a p v increase happens although the pressure decreases. So in this case is going to be negative and if you want to have cooling in this case because we want mu j t to be positive we expect that del p v by del p should be negative which should help us in having mu j t as a positive quantity. So if we see at this variation we can have two parts one is 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 high pressure and therefore they cannot be compressed further. And what we can say now hence they are less compressible than Poil's law prediction in this case. So when the compressibility of the gas is very well less in this case when they cannot be compressed further the p v 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 decreased the value of p v also shows decrease in this case. And therefore in this region if you expand the gas this del p v by del p term is going to be positive in this case that means pressure reduction happens del p v 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 Boil's law predictions right. Why you see here because now we are talking about lower pressure values 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 p v product increases because the volume increase will be there and the 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 p v value increases and therefore del p v 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 that 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 these 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 R T where a and b are constants and this a and b gives the major of intermolecular forces which is what a is while and size of the particle respect to 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 p v is equal to R T for ideal gas. If we rearrange this term what you get is a p is equal to R T upon v minus b minus a by v square you can do this existence 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 R T 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. So, 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 is 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 this if v is specific volume is very very large as compared 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 while this also tend to be equal to 0 and therefore, ultimately my expression for mu J T comes like this 1 by C p 2 a upon R T 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 2 a upon R T minus b as we know 1 by C p 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 2 a by R B. 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 2 a upon R B and if I want mu J T to be equal to 0 in that case my temperature is going to be 2 a upon R B. In the sense this is nothing but the inversion temperature for the gas at a given pressure. So, if my temperature is less than 2 a upon R B my mu J T is going to be positive and when expanded it is going to resulting lowering of temperature. Well, here if the temperature is more than 2 a upon R B 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 Waal equation basically would decide what are the values of the inversion temperature for a particular gas at different pressures etcetera. One can based on these values one can construct the inversion curve also. So, if I want to plot isenthalpic line on a TS chart you can see this is a 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. So, if I expand the gas from this high pressure isenthalpically 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 isenthalpically 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 obtained 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 isenthalpically. 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 JT 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 isenthalpically alright. 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 JT expansions are mostly the JT expansion would be for a two phase system when you are completely away from ideal gas. So, you got a mixture of let us say gas plus liquid or complete liquid also which is what is normally preferred in order to have more and more cooling effect if the gases or the mixture or the two phase mixture is subjected to JT expansion. Why is 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 ideal gas nature in this 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 us say at room temperature now it is not resulting in cooling but it is resulting 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 the state 2 is lying below the inversion temperature before it takes off for isenthalpic expansion. So, in the fluid expands from state 2 to state 3 the temperature rises this occurs because the initial temperature at state 2 is above the inversion temperature. This is a very important parameter one has to understand. Now there are various gases and they have got their own inversion temperature. What you can see from here is the gases like helium hydrogen neon have got their inversion temperature as 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 JT expansion they would result in cooling. So, for gases like helium hydrogen and neon in order to experience JT cooling they have to be pre cooled below T inversion temperature first and then subject to the JT 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 pre cool 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 where the entropy remains constant. Enthalpy and entropy are two thermodynamic state properties of matter which are functions of pressure and temperature. When the gases are expanded at constant enthalpy as in JT expansion it is called as isenthalpic expansion. On the similar line when the high pressure gases are expanded at constant entropy it is called as isentropic expansion or a reversible adiabatic expansion. This is all what we know from thermodynamics. The commonly used expansion devices are turbo expanders which uses 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 the difference between a JT expansion and isentropic expansion. Here in this case when expansion occurs there is some kind of work gets produced and you will get some work from this expansion. So this is a turbo expander and you will get some kind of a WE or a shaft work done when the gas expands from high pressure to low pressure or similarly if you have got a reciprocating expander what you get ultimately is a work output WE 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 us see how we derive the value of delta T by delta P at constant entropy. The entropy S 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 a 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 these 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 had 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 D S 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 D T plus delta S by delta P keeping temperature constant into D P. If I multiply by T and this is again in the same line as what we have done earlier we find that T D S is nothing but multiplied by T to this term multiplied by T to this terms. 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 alright. 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 del V by del T at constant pressure. There is a negative sign over here and there was 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 P V is equal to R T and therefore V is equal to R T by P. If I differentiate that with respect to temperature keeping pressure P constant I will get delta V by delta T is equal to R by P because the T will go and nothing R by P is nothing but equal to V by T. If I put these values in this equation and if I replace del V by del T is equal to V by T I will get T and temperature constant and my mu S ultimately gets reduced 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 zero unlike the case in the J T expansion when the enthalpy was constant we found that mu J T was equal to zero. It means that the ideal gas does exhibit a cooling effect when it undergoes an isentropic expansion. This is a very important thing to understand that whatever happen when the gas is expanded isentropically it would always result in cooling and this is true for the ideal 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 represent 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 is 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. Now if I apply the same for real gas using Van der Waals equation we get the similar term as what it was 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 it is ideal gas or real gas mu J T is always going to be positive that means when expanded isentropically it will always result in lowering of temperature. If I want to compare isentropic expansion with isentropic expansion so J T expansion with isentropic expansion I should get the J T expansion has 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 works for real gas and therefore we always prefer to have phase change of fluid during the expansion while here we should work with single phase fluid because if you 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 valve 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 sometimes the lubrication has to be taken care of and thing like that. So, this is what we will compare a J T expansion on isentropic expansion and one has to prefer depending on what you want whether one should go for a J T expansion or one should go for isentropic expansion. So, let us talk about the gas liquefaction systems here having understood in what way the expansions happen we talk about now various gas liquefaction systems and there are a various gas liquefaction systems now and the first one is thermodynamically ideal system, Lindy-Hamson system, Precool Lindy-Hamson system, Lindy dual pressure system, Cloudy system, Capitza system, Heland 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 system serves as a comparison basis and let us see how does this system work. So, thermodynamically ideal system will have falling salient features here this is a compressor and this is expander and this is a container where the 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 e c is supplied this is a 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 an expansion process alright. 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 in 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 0.1 over here. So, 1 to 2 process is a isothermal compression process temperature remains constant and after pressurization at 0.2 here the gas is expanded isentropically therefore entropy remains constant and this gas after expansion hits this end which is a saturated liquid curve at this point. So, whatever gas is compressed completely gets liquefied at this point this is what the liquid we get at this particular point alright. 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, it 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 the 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 expansion liquefaction. This is an open thermodynamics 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. Now, if I consider this as a control volume I can apply the first 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 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 a work of expansion and what again coming out is m 1 because what was compress is m 1 same thing is liquefied m 1 at point f. See, if I apply the first 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 alright. So, m 1 dot and 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 compared to other 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 chef 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 used 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 represented 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 compress 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. So, 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 compressing this case. Since in an ideal system the mass of gas compress 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 for 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 1 is obtained and the pressure remain the same and therefore, s f at saturated liquid condition can be obtained at the same pressure and h f is an 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 liquefy gas. So, here 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 as shown. This is my T s diagram and this is what is available in various books this is available on the net also for nitrogen we should get this diagram and find out a condition as soon as I know the point 1 I will look at the point f because I will come down at constant pressure go the constant pressure line and look at 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 got a pressure of 1 bar at f also I got a pressure of 1 bar here I got a temperature of 377 Kelvin 377 Kelvin at 1 bar 77 nothing 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 it 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 an ideal 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 hit 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 bother to look at this point 2 because this point 2 will be at a very high pressure very high pressure. So, what I need to look at 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 1 bar pressure which is the room temperature is 300 Kelvin in both the cases while the point f is at the boiling point of respective 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. So, if I put these values in this equation these 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 zero 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 requirement for liquefying helium gas. So, it is not although the boiling part of helium is just 4.2 Kelvin which is less than that of hydrogen still the work done to liquefy 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 liquefy hydrogen and work required to liquefy helium. So, this table gives you ideal work requirement for various gases and you can see the boiling point of helium is 4.2 upon 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 this ideal work requirement and you can compare these 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 should 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 Joule's law and Boyle's law. We saw the two parameters and 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 pre cooled to bring down their initial temperature below the inversion temperature and then only they will experience J T effect or J T cooling. In expansion devices like turbo expanders and expansion engines the expansion process is isentropic or reversal 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 such 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 made clear right because J T expansion normally we want to have real behavior we want to have non-ideal behavior and therefore, we prefer to have a phase changes or we would prefer 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.