 So, welcome to the 12th lecture of cryogenic engineering, we are talking basically on the gas liquefaction and refrigeration at very low temperature. Just to take a overview of the earlier lecture, we had studied earlier the Lindy-Hamson system for gas liquefaction and we studied the effect of heat exchanger effectiveness that is epsilon on the performance of this cycle which is called as Lindy-Hamson system. Before this we had studied ideal thermodynamic cycle and then we had the first cycle called Lindy-Hamson system in which we studied the effect of heat exchanger effectiveness. Mathematically, the heat exchanger effectiveness epsilon is actual heat transfer divided by maximum possible heat transfer thereby the epsilon value will be between 0 to 1 that is 100 percent effectiveness when epsilon is equal to 1. We also saw that in a Lindy-Hamson cycle the heat exchanger effectiveness epsilon is h 1 dash minus h g upon h 1 minus h g meaning that h 1 dash minus h g is the enthalpy difference which represent actual heat transfer that is occurring in the heat exchanger divided by h 1 minus h g this is the enthalpy difference between 0.1 and g and this is the ideal or maximum possible heat transfer that can take place when the effectiveness is 100 percent. This is we are talking at a lower pressure region or the suction pressure of the compressor side and if we talk about the discharge pressure of the compressor side that means high pressure side the same value can be interpreted as the actual heat transfer in this case now becomes h 3 dash minus h 2 at 0.3 dash and 2 divided by maximum possible heat transfer which is h 3 minus h 2 this being the enthalpy values at 0.3 and 2 respectively. At the same time the liquid yield y for a Lindy-Hamson system is given by y is nothing but the ratio of m dot f upon m that is the mass fraction of the m dot which gets liquefied which is called as m dot f. So, y is equal to h 1 minus h 2 minus 1 minus epsilon into h 1 minus h g divided by h 1 minus h f minus 1 minus epsilon h into h 1 minus h g. So, you can see in this term that the epsilon represents the effectiveness of the heat exchanger and when epsilon is equal to 1 percent is equal to 1 or 100 percent y is getting reduced to h 1 minus h 2 divided by h 1 minus h f all this right hand side which is a negative side to this numerator and denominator becomes equal to 0. So, here you can understand that how sensitive the value of y is to the value of epsilon in this case. We have also seen the effectiveness should be more than 85 percent in order to have a liquid yield in the Lindy-Hamson cycle. We are seen with the help of a numerical that as the value of epsilon decreases and when it decreases below 85 percent the value of y is very very close to 0 or almost 0 meaning which that the heat exchanger effectiveness should be very high it should be much higher than 85 percent if you want to have a good value of y and normally this value will be around 95 percent or above in order that you get a high value of y or a significant value of y in this case. With this background the outline of today's lecture which is the topic of which is gas liquefaction and refrigeration systems. We will talk about now a new system which is a pre cooled Lindy-Hamson cycle. In the earlier time we had seen Lindy-Hamson system. Now we will see pre cooled Lindy-Hamson system under which we will study the expressions or we will talk about how much y is could be obtained with the pre cooled Lindy-Hamson system that is liquid yield. What is the work requirement for this cycle and what is the maximum liquid yield for this cycle and then we will compare the results obtained using this cycle with simple Lindy-Hamson system. So, what we will do is comparison between the simple and pre cooled Lindy-Hamson systems. We have seen earlier that as the compression temperature decreases the yield y increases for a Lindy-Hamson system. This we had again seen with an example that if the compression temperature instead of happening at room temperature of 300 Kelvin, if it happens at 200 Kelvin the value of y increases or the yield increases. The method of cooling of the gas after the compression or before the entrance to the heat exchanger is called as pre cooling. So, when we are pre cooling basically what we are doing is we are pre cooling the gas and we are pre cooling the gas at the entrance to the heat exchanger or the first heat exchanger after the compressor. So, one way of doing this is basically we can reduce the compression temperature which is normally not possible because compressor cannot work at lower temperature. So, normally the process of compression would occur at room temperature or at 300 Kelvin and if you want to pre cool the gas this has to be done after the compressor and therefore, the gas would enter the first heat exchanger at a lower temperature than room temperature and this is what we call as pre cooling the gas. The Lindy-Hamson cycle with a pre cooling arrangement is called as pre cooled Lindy-Hamson cycle. So, normally we will call Lindy-Hamson cycle that means a normal or a simple Lindy-Hamson cycle, but whenever it has got a pre cooling arrangement it will always be called as pre cooled Lindy-Hamson cycle. Here after we refer to cycles as simple Lindy-Hamson system and pre cooled Lindy-Hamson system respectively. So, let us see now what is included in the pre cooled Lindy-Hamson cycle. The simple Lindy-Hamson system is shown in the figure and this is what he can see. So, here you can see the simple Lindy-Hamson cycle with a compressor then the compressed gas comes to the heat exchanger after getting the through from the heat exchanger the gas expands what you get here is m dot f liquid. The return gas goes to this heat exchanger pre cooling the gas which is coming at high pressure and then the gas goes back to the compressor m dot f is obtained at this point the return gas is m dot minus m dot f and again m dot f is added at this point as replenishment and the cycle continues. So, this heat exchanger is a part of Lindy-Hamson system in addition to this Lindy-Hamson heat exchanger what we will have is one more heat exchanger and which is called as pre cooling heat exchanger. So, this is the second heat exchanger which is added over here and now you can see that the high pressure gas which is coming out of the compressor is getting pre cooled in this heat exchanger. Now, how is it getting pre cooled it is getting pre cooled by another gas which is passing through heat and this gas is going to pass through this heat exchanger at a much lower temperature or at a temperature at which you want to pre cool this gas. So, this gas which is coming out at room temperature at this particular point will get pre cooled because of other refrigerant or other pre cooling circuit thereby making this heat exchanger as a three fluid heat exchanger. You can see already there are two fluids passing through it we have got a third fluid coming through this heat exchanger and therefore, this is called as a three fluid heat exchanger. So, for pre cooling what we are having in addition to the Lindy-Hamson system is we got a one more heat exchanger which is now a three fluid heat exchanger. So, a three fluid heat exchanger is used to thermally couple the pre cooling and the Lindy-Hamson system. So, this is now a Lindy-Hamson system with a pre cooling heat exchanger and in addition to this now how is this pre coolant or how this refrigerant coming it has got one more circuit and this is called as a pre cooling circuit. Now, this pre cooling circuit will have its own refrigerant depending on the temperature of our interest a refrigerant has to be selected in such a way that this refrigerant with pre cool the high pressure gas at this point. Now, as soon as you have got a one more refrigerant you got one more compressor and this four this will be called as a refrigerant compressor and this follows a simple vapor compression kind of a cycle and therefore, it has got a condenser at this point and this condenser could be cooled it could be air cooled or could be water cooled. So, we got a one more arrangement which is shown over here this basically does the condensing of this refrigerant and it could be air cooled condenser or it could be a water cooled condenser also. So, what you have got now here two circuits one is a pre cooling circuit and one is a Lindy-Hamson system or a simple Lindy-Hamson system with a pre cooling heat exchanger and this whole arrangement is called as pre cooled Lindy-Hamson cycle. Hence, because of this three fluid heat exchanger the temperature is lower after compression or before the entry to the heat exchanger. So, why are we doing all this exercise? We are doing all this exercise so that the working fluid at this point it could be nitrogen oxygen or air whatever you want to liquefy is entering this heat exchanger at much lower temperature than what it used to be or basically it was coming at room temperature earlier. Now, with this pre cooling arrangement the gas enters this heat exchanger at lower temperature than the ambient temperature here alright. So, the whole exercise is basically being done in order that the gas enters this heat exchanger at lower temperature than it used to be and this is what a pre cooling circuit is doing. So, this is a pre cooling circuit which is comprising of this compressor, condenser, a JT wall and a pre cooling heat exchanger over here. Now, this pre cooling heat exchanger is basically pre cooling the Lindy-Hamson cycle. The feature of the pre cooling system are as follows. It is a closed cycle refrigerator with the cold heat exchanger thermally coupled to the simple Lindy-Hamson system. So, this is basically the connecting point and this is thermally coupled with the simple Lindy-Hamson cycle. Now, what you can see here is m dot r is the flow of refrigerant at this point. So, you got a compressor which is compressing refrigerant of mass flow rate m dot r and the power naturally when you got a compressor you got to supply the power to it and the power to this compressor is WC2 while power to the main Lindy-Hamson cycle will be WC1. So, nothing is coming free here you got to have more power input to be done for this compressor and we are compressing m dot r which is a mass flow rate in this circuit or the pre cooling circuit. Now, you can see like this system is basically the Lindy-Hamson system or this is a load on this pre cooling circuit alright. So, here you can say that the heat exchanger is thermally coupled to the simple Lindy-Hamson system. In other words, the cooling object for this refrigerator is Lindy-Hamson cycle. What is cooling? It is cooling basically the Lindy-Hamson cycle. The heat exchanger of the pre cooling system is cooled by water. So, this could be cooled by water sometimes it could be by air also and a JT device is used to attend lower temperature. So, what you have is a compressor, the heat exchanger, a JT expansion wall and a pre cooling heat exchanger. The process of compression is assumed to be adiabatic in this case. So, we are not shown QR value over here. So, we can assume that this compressor is adiabatic while as you know in the Lindy-Hamson cycle we can assume that the process of compression is close to isothermal. So, in this case QR will be will be equal to 0 because it is assumed to be adiabatic in this case. What could this pre coolants be? These pre coolants are basically refrigerant and depending on your choice or depending on your pre cooling temperature whatever you want to consider depending on what you want to liquefy in the main circuit we can have different refrigerants R134A, ammonia, CO2 etcetera are the common refrigerants in the pre cooling systems. Now, depending on this refrigerant one has to choose what kind of compressor, condenser, expansion, expander should be used and depending on these refrigerant also you will get the pre cooling temperature at this point. So, this is the whole arrangement with the pre cooling circuit and Lindy-Hamson cycle. The salient features of a pre cooled Lindy-Hamson systems are as follows. The system consists of a compressor this is a compressor heat exchanger which is a two fluid and a three fluid. So, it has got two compressor one is a three fluid compressor and one of the two fluid compressor and a J T expansion device. So, what you have is a in the complete circuit what you have is a one two and three heat exchangers, but if you talk about only pre cooled circuit in the Lindy-Hamson we have two heat exchangers. Now, one is a three fluid other one is a two fluid and we have got a one J T expansion wall. The compression process is isothermal here in this case and that is why you got a value of q r and w c 1 associated over here. While as I said earlier that the process of compression here in this pre cooling circuit could be assumed to be adiabatic in nature while the J T expansion is isenthalpic you know that the expansion through J T is always isenthalpic. All the processes are assumed to be ideal in nature and there are no pressure drops in the system. So, when we are analyzing this cycle now we will assume that all the processes are ideal. That means all the heat exchangers are ideal all the expansions are isothermal the adiabatic compressor over at this point and the isothermal compression process at this point. The heat exchangers are assumed to be 100 percent effective and the processes are isobaric in nature. The gas to be liquefied by the liquefaction system is called as primary fluid. We can call this as a primary fluid which could be nitrogen, oxygen, air whatever you want to liquefied whereas, the refrigerant in the pre cooling system is called as secondary fluid. So, you can call this as the primary fluid or a primary system and this you can call as secondary fluid or a secondary system or a pre cooling system also. So, when depending on the references you can call this is the primary circuit secondary system or a lindey-hampson cycle and a pre cooled lindey-hampson cycle etcetera lot of connotation lot of ways of addressing this system. So, now if I want to represent the whole system on a T S chart which is what a very important task is it will look like this. So, process we are talking about this now here 1 to 2 is a compression and 2 to 3 is a pre cooling. So, here the gas gets pre cooled from 2 to 3 at a temperature which could be called as a pre cooling temperature alright and then the gas gets further pre cooled from 3 to 4 in a major heat exchanger at this point as you can see then there is a isenthalpic expansion the written gas goes at g it goes up to 6 at this point and then the gas gets 1 up to 0.1 in this pre cooling heat exchanger. So, the whole processes are shown over here and this temperature at 0.3 or 6 is going to be determined based on the pre cooling circuit based on the refrigerant which is flowing in the circuit based on the pressure of this pre cooling circuit etcetera. So, this is the most important thing now is what is the pre cooling temperature and it will be decided by the pre cooling circuit and also the requirement of the main Lindy-Hamson system. So, this is what it would look like a pre cooled Lindy-Hamson system the pre cooling limit of the pre cooling cycle alright. So, how much that temperature should be is governed by the boiling point of the refrigerant at its suction pressure. So, what this temperature could be we want this temperature to be as low as possible alright and the lowest temperature that could be achieved by this is basically is going to be decided by the boiling point of the refrigerant in the pre cooling circuit alright because we cannot come below that thing and the pre cooling is the boiling point at the suction pressure is going to be the lowest limit of this temperature would be alright. So, this temperature is going to be decided by the pre cooling circuit and is also going to be decided by the type of refrigerant and its pressure that will decide this temperature. So, this temperature we can call it as TD which is the temperature that is shown in the pre cooling circuit and this is nothing but equal to the boiling point of the refrigerant at the suction pressure of the pre cooling circuit alright. This will be the lowest temperature at which it enters the heat exchanger. The boiling point of the common refrigerants at 1 bar R if we assume that the suction pressure is at 1 bar there are the boiling points of this refrigerants will be for example, for CO2 it is 216 Kelvin for ammonia NH3 it is 240 Kelvin or for R 134 A that temperature could be 247 Kelvin. So, if we assume that the return pressure or the suction pressure is around 1 bar this is what the lowest temperature they can achieve in the pre cooling circuit. So, the gas would enter the heat exchanger at this lowest temperature if this conditions are maintained. So, this is what the circuit would be. Now, if I want to analyze this circuit further I will consider a control volume for this system as shown in the figure and this is my control volume in which I am excluding the compressor and I am taking pre cooled heat exchanger main heat exchanger and the container. So, it includes as three fluid heat exchanger this is one then JT device and a liquid container. So, this is what enclose in this control volume and we apply the first law as we have done always in earlier lectures also the first law is applied to analyze the system the changes in the velocities and the datum levels are assumed to be negligible as what we have been doing. The quantities entering and leaving the control volume are as follows. So, what is entering this control volume is m dot r at point D this is a flow rate this system which is m dot r refrigerant flow rate and it is entering at this point D and also what is entering is m dot at point 2. So, these are the two quantities which are entering the control volume all of the quantities are leaving the system or leaving the control volume what are they m dot r which is leaving at point A. Similarly, m dot minus m dot f leaving at point 1 and m dot f at point f. So, we say that whatever is coming in is equal to whatever is leaving in the system and therefore, applying first law what we have is m dot r H D we are taking enthalpies at various points and various temperatures and at respective pressures. So, m dot r H D r this is the energy at this point which is entering the control volume plus m dot H 2 which is entering at this point. So, whatever is coming in are these two energies and whatever are leaving are m dot r H A r which is at this point m dot minus m dot f at H 1 which is at this point and m dot f H f at this point. So, the energy which is coming in is equal to the energy which is leaving the control volume. So, if I could do the same thing here rearrange the terms what we get ultimately is m dot upon m dot m dot f upon m dot which is nothing but y is equal to H 1 minus H 2 upon H 1 minus H f plus m dot r upon m dot into H A r minus H D r that is enthalpy difference across this heat exchanger divided by H 1 minus H f. So, what you can see we have got an expression now from this rearranging this term is the yield that is y is equal to the term which is normally common to all the which is simple Lindy-Hamson cycle yield plus an additional yield over here which is coming because of the pre cooling circuit. So, if we denoting the ratio m dot r upon m dot is equal to r that is the ratio of how much mass flow through refrigerant divided by the mass flow in the main circuit. So, if we have a relationship between the m dot r in the pre cooling circuit to the m dot in the main circuit as r the expression could be now written as y is equal to m dot f upon m dot is equal to H 1 minus H 2 upon H 1 minus H f plus r times H A r minus H D r upon H 1 minus H f. The first term in the above expression is the yield for a simple Lindy-Hamson system. So, what you get normally in a Lindy-Hamson system will be this in addition to that there is a second term now and this second term is coming as an additional yield because of the pre cooling arrangement. The second term is the additional yield occurring due to the pre cooling of the simple system alright. So, this is an additional yield which we are getting because the gas after the compressor here is getting pre cooled to temperature T 3 or T 6. Basically getting pre cooled to T 3 because of the refrigerant coming at this point in this three fluid heat exchanger. So, this is the role which is played by the pre cooling circuit. So, the expression is now y is equal to m dot f upon m dot is this. This is the increment in the yield is now dependent on what. So, what is this y additional increment in y it depending on what we can see from this is the change in enthalpy of the values H D and H A. So, H A minus H D is going to basically determine the additional value of y or additional increment in the value of y while H 1 minus H F is same in both the cases and the second point is the refrigerant flow rate m dot r. So, these two things what is the enthalpy change across this three fluid heat exchanger in the pre cooling circuit that is H D minus H A and the value of r that is what is the ratio of m dot r upon m dot. So, from here you can see that if ratio of r is high and if this enthalpy difference is high you will have more and more increment to the value of y as compared to the simple Lindy-Hamson cycle. Since the three fluid heat exchanger is assumed to be 100 percent effective the following conditions hold good. The minimum value of T 3 would be equal to T D the value of T 3 at this point will be equal to the value of T D because this the gas is getting pre cooled depending on the what is the temperature at point D is. So, the minimum value of point 3 at point 3 which is T 3 would be equal to T D which is the boiling point of the refrigerant. So, in this circuit the minimum temperature at which the gas enters the heat exchanger is going to be equal to the boiling point of the refrigerant in this circuit. So, boiling point of the refrigerant corresponding to the suction pressure of this compressor which is at this point and this is going to decide what is the lowest temperature at which this gas can enter the heat exchanger. The maximum value of T 6 similarly if we assume that this heat exchanger is a perfect on ideal heat exchanger the maximum value of T 6 will be equal to T 3 the minimum value of T 3 equal to T D the maximum value of T 6 is also equal to T 3 or equal to T D. So, the maximum value of T 6 would be equal to T D which is the boiling point of the refrigerant. So, we are talking in a case when the heat exchanger effectiveness is 100 percent. So, at this condition the system produces the maximum yield for a given refrigerant. So, as we were talking earlier that if the temperature at which the gas enters the heat exchanger is lowest you will get maximum yield. So, in this case the lowest temperature at which the gas can enter this heat exchanger is going to be the boiling point of this refrigerant at its suction temperature at the suction pressure. So, in this case if the value of temperature 3 at point T 3 is going to be equal to the boiling point of this refrigerant then the yield what you get y is going to be equal to y max or the maximum yield. So, at this condition when the gas enters the heat exchanger at the boiling point of this refrigerant whatever yield you get is going to the maximum yield for a given refrigerant maximum yield for a given liquefier also. So, mathematically if we want to have y is equal to y max if you want to have maximum yield then T 3 equal to T 6 is equal to T D and T D is nothing but equal to the boiling point of the refrigerant in the this circuit or at this particular pressure. Now, if I consider this as a control volume. So, consider a control volume enclosing the heat exchanger JT device and the liquid container as shown in the figure earlier we had taken a bigger enclosure, but if I take this as a enclosure now and do the energy balance. So, again we find the quantity is entering and leaving the control volume. So, what is entering is mass at point 3 that is m dot at 3 and what are leaving is m dot f at point f and m minus m dot minus m dot f at point 6. So, applying the first law again what you get is m dot h 3 is equal to m dot f h f plus m dot minus m dot f h 6 and if we rearrange this terms what you get is this and if I write expression for y what you get here y max is equal to m dot f upon m dot is equal to h 6 minus h 3 divided by h 6 minus h f. So, basically now I am taking a control volume at this point and enthalpy difference at this point is going to decide what is my y max value will be. So, in order that I get y max now the properties of h 6 and h 3 are evaluated at the lowest temperature which is possible at this point. So, the quantities h 3 and h 6 are evaluated at the boiling point of the refrigerant T D. So, if we do this then whatever we get as y is going to be the y max value or the maximum yield that is possible from this particular circuit. So, this was as far as the liquefaction was concerned. Now, we will talk about the compressors and do not forget that there are two compressors now one is a adiabatic compressor one is a isothermal compressor. So, consider a control volume for a compressor in the liquefaction cycle as shown in this figure the quantity is entering and leaving the control volume are as given here. So, again we have got a m dot entering at 0.1 and m dot leaving at 0.2 the work is given at this point which is as we earlier know we have pointed out earlier whatever is entering, but whenever the work is done on the system it is written as minus W c 1 is the work done on the system and whatever is leaving the heat which is leaving the heat of compression is also being written negative minus q r. And if we write the heat balance now using the first law for the following table what we are getting is this energy in is equal to energy out and therefore, we have got a m dot h 1 minus W c 1 is equal to m dot h 2 minus q r rearing this term what we get is q r minus W c 1 is equal to m dot into h 2 minus h 1. The heat of compression q r can be obtained by using second law for an isothermal compressor it is given by q r is equal to m dot into t 1 into s 2 minus s 1. If you put this value of q r over here combining these two equations what we get is an expression for minus W c 1 or the work of compression done on this circuit. So, minus W c 1 is equal to m dot t 1 into s 1 minus s 2 minus m dot into h 1 minus h 2 this is the amount of work done on this compressor at this point. In addition to this we have got one more compressor and which also does we have to supply the work to this compressor also. However, we have assumed that this compressor works in a adiabatic fashion that means q r for this compressor is equal to 0. Similarly, now let us have a control volume for this compressor a control volume is taken including the refrigerating compressor at this point in this circuit the quantity is entering and leaving this control volume r as given below. What is entering is m dot r at a and the work done on the compressor is minus W c 2 what are leaving is m dot r at point b, but there is no q r leaving. So, I have just written 0 in order to match earlier tables. The heat of compression is 0 because the process is assumed to be adiabatic in this case. So, using the first law for the following table what we get is whatever entering is leaving and therefore, E in is equal to E out and therefore, m dot r into h a r enthalpy at point a minus W c 2 is equal to m dot r into h b r. So, rearranging this terms as we have done earlier what you get now is the work to be done on the pre cooled circuit or on this compressor which is minus W c 2 is equal to m dot r into h b r minus h a r this is nothing, but the enthalpy difference across this compressor. So, the total work requirement of the system is work done on this compressor plus work done on this compressor. When you are adding a pre cooling circuit no doubt we get advantage of the pre cooling, but at the same time we have to do some work on this compressor also. So, the total advantage of this pre cooling circuit should be weighed with respect to this work additional work which is required to be done on this compressor also. So, the total work done is equal to W c is equal to W c 1 plus W c 2 if we add these values of minus W c 1 for the work done for this compressor plus minus W c 2 which is the work done on the refrigeration compressor. If we put them together we get this expression. So, minus W c is equal to m dot t 1 into s 1 minus s 2 minus m dot into h 1 minus h 2 plus m dot r into h b r minus h a r the work required for a unit pass of primary gas compressed is given by. So, if I want to reduce and find out the work done per unit mass of gas compressed in the primary circuit or in the Lindhamson cycle which is m dot then I get minus W c upon m dot is equal to t 1 into s 1 minus h 2 minus h 1 minus h 2 plus m dot r upon m dot into h b r minus h a r and if you recollect we have called m dot r upon m dot at r value as r this is the ratio of the 2 mass flow rates. So, denoting the ratio of m dot r upon m dot as r the expression now gets reduced to minus W c upon m dot is equal to t 1 into s 1 minus h 2 minus h 1 minus h 2 plus r which is over here into h b r minus h r. So, this is the expression which I get for the total work done on this pre cooled Lindhamson cycle. So, here you can understand that the first two terms are coming from this primary cycle or the primary circuit the first and the second terms are that work requirement in the simple Lindhamson system this was our expression earlier. This expression has come because of the additional circuit or the pre cooling circuit and the third term is the additional work required to pre cool the system. So, one can really find what is the work done on this cycle from the pre cooling cycle and from the primary cycle. So, just find out two work done add them together and this should justify the additional circuit or the pre cooling circuit that has been incorporated in a pre cooled Lindhamson cycle. Now, with this background of the pre cooling circuit what we now do is a tutorial here and this tutorial will help you to understand what we have learnt till now. This tutorial have got all the components that one needs to understand from a pre coolant circuit and a pre cool Lindhamson cycle. So, please read the problem again correctly and as I said every time one has to really understand the language of this problem. So, let us see the tutorial now. Determine the Y that is yield Y max that is the maximum yield the work per unit mass of gas compressed work per unit mass liquefied and FOM which is nothing, but figure of merit for the simple and pre cooled Lindhamson systems with nitrogen as working fluid. Now, this is very important to understand what is asked in the problem. So, what is your working fluid which is nitrogen what are we talking about we want to talk about both the cycles that is simple Lindhamson cycle as well as pre cooled Lindhamson cycle. The R 134 A is the refrigerant for the pre cooling system with ratio R is equal to 0.08 that is m dot R upon m dot is 0.08. The liquefaction system is operated between 1 atmosphere and 100 atmosphere or 1.013 bar and 101.3 bar and this compression is carried out at 300 Kelvin. So, this pressure change is happening at 300 Kelvin which is isothermal compression process. The following is the data for R 134 A and what ultimately what we want is your comment on the results. So, the data for R 134 A is given in terms of point A B and C. The A B C are referring to the conditions what you saw earlier in a pre cooled circuit that could be called as 1 2 3 4 whatever you want to have, but this is in the reference to the circuit which I have already shown to you. What you can see from here is the point A is at the entrance of the refrigeration compressor which is at 1 bar or 1.013 bar. The point B is after compression which is 10.13 bar that means a pressure ratio of 10 is ok at this point. The points is after the condenser or after the heat exchanger and corresponding to these 3 points what you have is a temperatures and what you have is an enthalpy at this point. So, this information regarding pre cooling circuit is sufficient enough for you to carry out different calculations required to find out all these parameters as specified in this problem. So, as I said again and again from this language convert the information as to what exactly you want. So, what is given what is the data which is known to you and what is asked in this problem this is very important to understand. So, if I could write all this information properly we see the second slide which is this. So, what is given to you what is given we have been asked to do a simple and pre cooled Linde-Hamson cycle. We have to consider simple and pre cooled Linde-Hamson system which there are 2 system that means the working fluid is nitrogen. The pressure for this nitrogen refrigerant is 1 atmosphere and 100 atmosphere. The temperature is 300 Kelvin for this compressor process then the pre cooling circuit has a refrigerant R134A which is compressed from 1 atmosphere to 10 atmosphere and the mass ratio R that is the ratio of the mass flow rate of the refrigerant to the mass flow rate of nitrogen in the primary circuit is m dot R is R is equal to 0.08. For the above cycles now these 2 cycles calculate and comment on various values you get from the system that is liquid yield y and y max work per unit mass of gas compressed work per unit mass of gas which is liquefied and figure of merit FOM. So, this is what is asked in this problem and this is what the data is for this problem. So, as you can see this is our circuit for a pre cooled Linde-Hamson cycle this is the pre cooling circuit and this is the primary circuit or a Linde-Hamson cycle m dot R comes in this compressor m dot comes in this compressor. So, if I now get the enthalpy entropy values for different temperatures and pressures in this cycle that are points are 1 2 3 1 2 and F are the points which are required for calculations of all the parameters which are asked for in the problem. So, you got a point 1 2 and F corresponding to pressures of 100 and 1 respectively the temperatures at these 3 points are 300 300 and point F what you have is a boiling point of nitrogen which is 77 Kelvin corresponding to these temperatures and pressures you got enthalpy values which are given as this 462 445 and 29 joule per gram and similarly we have got entropy values for these 3 expressions. Now, these values are either taken from chart and many times it has taken from temperature entropy chart. So, whenever we got this problem we should have a temperature entropy chart of nitrogen air oxygen whatever use of primary working fluid in this case. So, first make this kind of a table so that you will never make mistakes. So, point 1 is at this here point 2 is after the compression which is at this point and point F is the point which is at this point alright. Similarly, now let us go to the pre cooling circuit or the refrigeration R 134 A circuit and let us locate these points again which are A B and C. So, where is A point we can see the point A to be over here which is at the entrance to the refrigeration compressor where is point B the point B is located after the compressor and therefore it is at high pressure of 10 bar 10.13 at this point. Then what you have is a point C which is at this point that is before the expansion before the J T expansion is point C. So, again similar to what we did for nitrogen as a working fluid we similarly make a table for the properties of enthalpy and entropy for different value of pressure and temperature for the pre cooling circuit and this is done for R 134 A. So, what is required here is mostly the enthalpy value. So, you can see that we are not written the entropy values because these values are not required in the calculations. What is to be noted here is the enthalpy at point D it is what is required for us now H D value is the value which is used in the calculation over here. The point to be noted here the enthalpy at point C is equal to enthalpy at point D because this is a isenthalpic expansion process and therefore many times you find that the enthalpy may be given at point C and you might be wondering what do I do about the enthalpy at point D, but one has to understand that the enthalpy at point D is nothing but enthalpy at point C since the expansion is isenthalpic process. This is a very twist many time this is considered as a twist in the problem, but one should note that the point D has a enthalpy which is same as point C. So, first our calculation starts with the ideal thermodynamic cycle because we want to calculate the figure of merit and therefore we have to consider about ideal work requirement to the cycle. So, the ideal work requirement for the cycle is assumed on the fact that whatever is compressed is expanded and is liquefied alright. So, the ideal work of requirement is minus w i upon m dot is equal to t 1 into s 1 minus s f minus h 1 minus h f and the point f is decided by the point 1 only. So, as soon as the point 1 is fixed we get a point f also fixed taking the values at 1, 2 and f from the earlier table we have to consider only 1 and f point in this case. So, get the value of enthalpy at 1 and f and entropy at 1 and f we can calculate minus w c upon m dot is equal to 300 into s 1 minus s f which is s 1 minus s f minus h 1 minus h f which is h 1 minus h f and this is equal to 767 joule per gram. This actually is now a known values to us because we have solved various problems for this. So, this is the ideal work requirement which is working on a ideal thermodynamic cycle. Now, this is the liquid yield for the simple Linde-Hamson cycle which is h 1 minus h 2 upon h 1 minus h f. Again take the value of enthalpies at 1 and point 2 we can calculate the value of y is equal to h 1 minus h 2 upon h 1 minus h f which is equal to 462 minus 445 upon 462 minus 29. These are the enthalpy values 1, 2 and f respectively putting those values you get y is equal to 0.04. So, m dot f upon m dot for a simple Linde-Hamson cycle comes out to be 0.04. So, if I want to calculate work per unit mass of gas which is compressed is minus w c upon m dot is equal to t 1 into s 1 minus s 2 minus h 1 minus h 2. Now, we are talking about 1 and 2 and not 1 and f as we have done earlier in a ideal thermodynamic cycle. So, putting those values again for enthalpies and entropies what you get is a 379 joule per gram as work done per unit mass of gas compressed in a simple Linde-Hamson cycle. If I want to convert that to work done per unit mass of gas liquefied from w c upon m dot is equal to 379, I have to know the value of y. If I divide this by y what I get is work per unit mass of gas liquefied. We have done solved these problems in the earlier lectures. So, minus w c upon m dot f is equal to minus w c upon m dot divided by y which is equal to 379 upon 0.04 which is equal to 9 4 7 5 joule per gram. So, this is the work done per unit mass of gas which is liquefied in this case. So, the figure of merit if I want to calculate is equal to what you get w c upon m dot f is equal to 9 4 7 5 what you got as w i for ideal thermodynamics cycle is w i upon m dot f is equal to 767. And so the figure of merit in this case is equal to 767 upon 9 4 7 5 is equal to 0.081. This is my figure of merit for the simple Linde-Hamson cycle. What I am now consider is a pre cooled Linde-Hamson system and corresponding temperature entropy diagram for the pre cooled Linde-Hamson cycle is this. So, you can see now 1 2 and a point at point 2 is getting pre cooled up to point 3 and at this point the pre cooled gas enters the heat exchanger. So, point 3 is basically now pre cooling temperature. The state properties are as tabulated below and here you can understand now 1 2 and f are as we saw earlier corresponding enthalpy and entropy values are given over here. The liquid yield now in this case is given by this formula which we have derived earlier is equal to y is equal to m dot f upon m dot is equal to h 1 minus h 2 upon h 1 minus h f plus and this is the additional increment in the value of y which is coming because of the pre cooling. So, into R into h a minus h d upon h 1 minus h f well as you remember this h a and h d are the enthalpy of the refrigerant at point a and d respectively which is nothing but the enthalpy across the three fluid heat exchanger or across the pre cooling heat exchanger of the pre cooled Linde-Hamson system. From here what we know is the value of R is equal to 0.08. So, if I put these two tables together 1 table is for 1 2 f which is for nitrogen and point a b c are meant for R 134 a pre cooling refrigerant and corresponding to those pressures which is 1 bar and 10 bar what you have is a enthalpy values. So, at point a b and c we have got enthalpy values over here and as you know that the value of h d in this case is equal to the enthalpy at point c as it is undergoing isenthalpic expansion from point c to point d. So, putting these values in this expression now y is equal to h 1 minus h 2 upon h 1 minus h f which is a standard value plus the additional increment that is going to come because of pre cooling circuit is 0.08 into h a minus h d which is 390 minus 260 divided by h 1 minus h f again 462 minus 29 thereby giving the value equal to 0.063. So, the yield now has increased to 0.063 for the pre cooling circuit and again I will do the same exercise of calculating work per unit mass of gas compressed and work per unit mass of gas liquefied. So, if I want to calculate work per unit mass of gas compressed I have to use the formula for this is the additional increment in the compressor which is going to come because of the refrigeration compressor. So, what you get is a normal compressor which is first two terms in addition to that what you have is plus r into h b r minus h r this is nothing the additional work that has to be done for the pre cooling circuit for which the value of r is 0.08 and if I add those enthalpy again from this table what I get is 386.3 joule per gram. So, work done per unit mass of gas which is compressed is 386.3. Now, I want to calculate work per unit mass of gas nitrogen which is liquefied. So, we will calculate for the unit mass of gas which is liquefied w c by m dot is 386.3 y is equal to 0.063 and therefore, w c upon m dot f is equal to w c upon m dot divided by y is equal to 386.3 divided by 0.063 is equal to 6131.7 joule per gram. So, as soon as I have I want to calculate and I want to compare the value of work done per unit mass of gas liquefied from the pre cooling circuit from the pre cooled Lindy-Hampson cycle to the simple Lindy-Hampson these are the values to be compared in order to justify the usage of pre cooling circuit or the pre cooled Lindy-Hampson cycle. The figure of merit is this divided by this the ideal work input divided by the actual work and therefore, figure of merit is this divided by this which is equal to 0.1251. So, this is the figure of merit for this case for the pre cool cycle. So, what we have done is we have considered ideal thermodynamic cycle we have considered simple Lindy-Hampson cycle and we have also considered pre cooled Lindy-Hampson cycle and we would like to compare all the values obtained from these 3 cycles. The problem also wanted to understand the Ymax value also. The Ymax value or the maximum liquid yield is going to come when the pre cooling temperature is equal to the boiling point of the refrigerant which is nothing but 247 Kelvin. So, if I get Y is equal to Ymax when T 3 is equal to T 6 is equal to T d is equal to 247 Kelvin we have studied this earlier and from this formula Ymax is equal to S 6 minus H 3 upon S 6 minus H f for R is equal to 0.08 we get 0.074. The Ymax value happens to be 0.074 temperature of pre cooling equal to 247 Kelvin and therefore, if I want to have a comparison between the simple pre cooled circuit you can see that for a simple cycle it was 0.04 Y value was 0.04 which got increased to 0.063. If I talk about the work done per unit mass of gas compressed this has increased from 379 to 386. This work done has been increased because there is additional work which has come from the refrigeration compressor in this case for the pre cooling circuit. However, if I want to understand what is the work done per unit mass of gas liquefied and that is what is very critical we can see that it has reduced down from 9475 to 6131 which justifies the usage of pre cooling circuit and the figure of merit is has increased from 0.081 to 0.1251. If I want to compare the same thing for the Ymax value so that the pre cooling is done to 247 Kelvin corresponding to that you can see that the Y value increased to 0.074 provided that pre cooling temperature is now 247 Kelvin. The work of comparison remains the same but the work done per unit mass of gas which is liquefied has still further decreased justifying the lowering of temperature further and the figure of merit has increased to 0.147 in this case. And this is the most important table which justifies the usage of pre cooling circuit or a pre cooled Linde-Hamsen cycle. Based on this we have given this assignment for the simple and pre cooled Linde-Hamsen system with air as a working fluid and again find out these values for r is equal to 0.1 in this. Please do this assignment and the answers to these questions will be given. In order to summarize what we have learnt in this particular lecture the method of cooling of the gas after the compression or before the entrance to the heat exchanger is what is called as pre cooling. The Linde-Hamsen cycle with a pre cooling arrangement is called as pre cooled Linde-Hamsen cycle. In a pre cooled Linde-Hamsen system a closed cycle refrigerator is thermally coupled to a simple Linde-Hamsen system through a three fluid heat exchanger. So, what is important is a three fluid heat exchanger existence in a case of a pre cooled Linde-Hamsen system. The compression process is isothermal in liquefaction cycle but is adiabatic in pre cooling system of a pre cooled Linde-Hamsen cycle. These are basically the assumption or a realistic assumption I should say for a pre cooled Linde-Hamsen system. In actual case they will not be isothermal they will not be adiabatic and we have to consider these effects through efficiency of this particular compresses. And this is the normally what we consider for calculation purposes. The pre cooling limit of a pre cooling cycle is governed by the boiling point of the refrigerant at its suction temperature. So, how much can we pre cool? We want to pre cool as below as possible as at low temperature as possible but this is going to be governed by the boiling point of the refrigerant. So, we would definitely like to pre cool to the extent possible but the refrigerant whether R 134A or ammonia this will govern that what is the lowest temperature at which the the gas can enter the heat exchanger it is going to be governed by the boiling point of this refrigerant at its suction pressure alright. So, this is basically very important. The yield for a pre cooled Linde-Hamsen cycle is given by this formula and this right hand side gives the additional increment that is going to come from the pre cooling circuit and at m dot R upon m dot is what we call as R in this case that the ratio of the mass flow ratios for the refrigerant to the mass flow of the nitrogen or the primary fluid. And the Y max that is the maximum value of Y or the yield what you can get is H 6 minus H 3 upon H 6 minus H F when this H 6 and H 3 are evaluated at the boiling point of the refrigerant. The maximum liquid yield is given by the above expression the enthalpy values are evaluated at the boiling point of the refrigerant and this will give you the value of Y max. The work requirement for the unit mass of primary fluid compress is this these two terms are for simple Linde-Hamsen while this additional term comes because of the refrigerating compressor and based on the tutorial what we did we have understood the yield of the pre cooled system is more than that of a simple system. So, what is important is to understand whether the pre cooling is justified or not because pre cooling adds upon an additional circuit and therefore, we have to study the effect of additional circuit or additional arrangement of this pre cooling on the entire system and this is what we will do in the next lecture as to understand the effect of various parameters in terms of how much refrigerant should flow through what should be its pressure what is the value of R should be so that we justify the usage of pre cooling circuit simple Linde-Hamsen cycle. Thank you very much.