 So, welcome to the eleventh lecture of cryogenic engineering under the NPTEL program. Just to take over view of the earlier lecture, we talked about ideal thermodynamic cycle for gas liquefaction in the earlier lecture and we found that the implementation of ideal thermodynamic cycle for gas liquefaction is impractical and hence modified cycles are proposed. The impracticality came due to the fact that the pressures required to liquefy all the gas that is compressed are very very high and therefore, this pressures cannot be obtained using conventional compressors which are available and therefore, these cycles cannot be realized in practice. However, these cycles form a very good basis of comparison for other cycles which is what the next conclusion was an ideal cycle is used as a benchmark and in effect different ratios and functions are defined to compare the various liquefaction cycles. So, if I am having any other cycle for liquefaction, the results obtained from this cycle or the liquefaction obtained from this cycle can be compared with the liquefaction that could be obtained from an ideal cycle. It becomes a basis for comparison for various cycles and therefore, the performance of various cycles can be obtained. The performance parameters can be obtained when we compare those parameters with the ideal thermodynamic cycle. Then we talked about Lindy-Hamson cycle and it consists of a compressor heat exchanger and a Jules-Thompson expansion device. An ideal cycle had only compressor and J T expansion device while a heat exchanger is added to the ideal cycle and in this system only a part of the gas that is compressed gets liquefied as against the fact in the ideal cycle whatever amount of gas is compressed it gets liquefied. In fact, all the gas gets liquefied while in this case only a part of the gas this part of the gas could be only 5 percent or 10 percent or 15 percent. So, only a part of the gas that is compressed gets liquefied. It is a very important difference between the two cycles and that is why this cycles becomes a kind of practical cycle. There is a heat exchange process occurring in heat exchanger and this process occurs at constant pressure and that is why you call it isobaric heat exchange process. So, the isobaric heat exchange process occurring in the heat exchanger used to conserve the cold and J T expansion device is used for producing lower temperature. So, the heat exchanger which was a change from the earlier ideal cycle it basically conserve the cold that is produced at the end. The cold gas which goes back to the compressor pre-cools the incoming hot gas and that is why we say that whatever cold is produced is conserved in this cycle and J T expansion wall of course is used for reducing the temperature with the J T expansion process. The work required for unit mass of gas compressed for a Linde-Hamson cycle is given by the formula minus W c upon m dot m dot is mass flow rate of the gas which is being compressed is equal to T 1 into S 1 minus S 2 minus in bracket H 1 minus S 2. The 1 and 2 basically denote the state of the points before compression and after compression and this process of compression is carried out isothermally at a temperature T 1. So, this is the work required for unit mass of gas which is compressed which is m dot. Now, we found that the yield y the yield is nothing but ratio of the gas which is liquefied divided by a gas which is compressed that is m dot f upon m dot. The yield y is maximum when the state 2 after the compression lies on the inversion curve at the temperature of the compression process. This particular part has been discussed in detail in the last lecture and we found that the yield y is maximum when the state 2 lies on the inversion curve at the temperature of the compression process. For a Linde-Hamson cycle following whole true as the compression pressure increases the liquid yield y increases for a given compression temperature. So, we had seen that as the pressure increases as the pressure P 2 the gas which is compressed from state 1 to state 2 if you compress instead of 100 bar to 200 bar that means the pressure ratio is going to be very very high the compression work also is going to be very very high but when you compress at a very higher pressure the liquid yield y increases for a compression temperature. So, compression temperature could be kept constant at 300 Kelvin which is room temperature and if we increase the pressure at the end of the compression y increases. Similarly, as the compression temperature decreases so instead of carrying the compression process at 300 Kelvin if I carry out the process at 200 Kelvin then the liquid yield y increases for a given compression pressure and both these conclusions we had seen last time using some problems or tutorials in the last lecture. So, if you understand this now we will go to the outline at today's lecture. So, today's lecture is a continuation of the earlier lecture which is gas liquefaction and refrigeration system and what we are going to learn today is the basics of heat exchangers as you know in Linde-Hamson cycle the heat exchanger is the only addition to the ideal thermodynamic cycle and therefore heat exchanger plays a very very crucial role in determining the output from this cycle. So, we will study what is the heat exchanger what are the basics related to that what are the performance parameter and then you proceed to understand the effect of heat exchanger effectiveness on Linde-Hamson system. So, today we will continue the study of the earlier cycle the Linde-Hamson cycle. However, what we will see and study here is the role of heat exchanger the heat exchanger performance is very very important and we will give some quantification as to what exactly happens with the deterioration in the heat exchanger effectiveness. And finally, we will understand what is the figure of merit and this is a very important term which is obtained to compare various cycle what is FOM figure of merit and all these things we will try to understand in today's lecture. So, let us come to the basics of heat exchanger. So, what is the heat exchanger? A heat exchanger is a device in which the cooling effect from the cold fluid is transferred to pre-cool the hot fluid. So, basically the hot fluid while travelling to the heat exchanger will become cold and the cold fluid while travelling through the heat exchanger will become hot because this cold is taken by the hot fluid and the hot fluid gets pre-cooled. Now, this heat exchanger could be a two fluid that means we can have one hot and one cold fluid or we can have three fluid type depending upon the number of inlets and outlets attached to the heat exchanger. Normally the heat exchangers are having two fluids, but there are various cases which we will see in the next lectures also that a heat exchanger can have two inlets and one outgoing stream or thing like that. So, we can have or one inlet and two outgoing fluids any combination that is possible depending on the different working fluids that are passing through the heat exchanger and what is the end usage of this particular heat exchanger is. The process of heat exchanger occurs at a constant pressure and hence it is an isobaric process this is an assumption. In fact, there will be some pressure drop across the length of the heat exchanger, but in an ideal heat exchange process we can assume that the pressure remains constant and therefore, we can assume that the process is isobaric. So, this is a heat exchanger or this is schematic shown which we are going to use in this particular lecture and about this all the schematics that are going to be used in this gas liquefaction. We have already shown that this is a heat exchanger and here I have got two schematics here this is a heat exchanger schematic and here I am talking about the temperature variations across the length of the heat exchanger. What you can see that the fluid A which is a hot fluid in fluid A in hot and fluid A out cold. So, while travelling through the heat exchanger this hot fluid becomes cold and how does it become cold because there is a other fluid coming from the opposite direction meaning which this could be a counter flow heat exchanger. In the counter flow heat exchanger you will have fluids coming in opposite direction. So, you got a fluid A and fluid B, the fluid B is cold fluid entering from this side while the hot fluid fluid A enters from the other side and during the travel across the length of the heat exchanger the temperature of the hot fluid A comes down. So, you can see that the temperature has come from A in hot to A out cold while the temperature of the fluid B increases when it travels across the length of the heat exchanger to this because it takes heat from the hot fluid and the temperature of the cold fluid now increases in this way the B in cold and it goes to B out hot. So, you can see that the hot fluid becomes cold while the cold fluid becomes hot during the travel alright. Lot depends on now how much heat it getting transferred and what is the A out temperature what is the B out temperature let us see those things now. So, this is basically the hot side of the heat exchanger and this we can call as a cold side of the heat exchanger. Now in order to specify the performance of the heat exchanger in terms of actual heat exchange occurring a term epsilon is defined which is called as heat exchanger effectiveness. Now in ideal case whatever heat is given by the hot fluid should be taken by the cold fluid, but no it does not happen like that. So, whatever heat is given by the hot fluid only a part of that will be taken by cold fluid and therefore, the actual effectiveness of the heat exchanger is going to be different than what it is ideally. In order to understand what are the losses in the system why the heat exchanger does not performance to 100 percent capacity a term comes into picture which is called as effectiveness of heat exchanger a very commonly known term and it defined as a ratio of actual heat transfer that is occurring in the heat exchanger to the maximum possible heat transfer that can occur theoretically. So, actual heat exchange upon maximum possible heat exchanger. So, mathematically one can write epsilon or heat exchanger effectiveness is equal to q actual the actual heat transfer divided by q max the maximum possible in heat exchanger. Naturally the maximum value of q actual could be q max value and therefore, the maximum value of epsilon also can be 1 the epsilon parameter is definitely dimensionless and therefore, it lies between 0 and 1. Here is the parameter which is q where the heat is transferred from the hot fluid to the cold fluid the actual heat exchange now in mathematical term calculation will be done by how much heat is given by hot fluid and how much heat is taken by the cold fluid. So, amount of heat that is given by hot fluid is m into C p into delta t of that particular fluid. So, if I talk about the B fluid we talk about m dot b which is mass flow rate of the B fluid into the specific capacity at constant pressure of the B fluid into the temperature difference of the B fluid which is T b in minus T b out. Similarly, we assume that the heat which is given by the hot fluid is m dot a into C p a into delta t of the A fluid. So, m dot a into C p a into T a in minus T a out. So, this is the q actual. So, this is the amount of heat which is taken by the B fluid and this is the amount of heat which is given by the A fluid. Now, the maximum possible heat exchange is the B fluid can reach maximum temperature equal to a in hot. If the heat exchange is 100 percent if there are no problems then this B fluid in principle could have reached up to a hot or the A stream can reach the lowest temperature of B in cold. So, the maximum heat exchange is going to occur between these two temperatures a in hot minus B in cold and therefore, q max is equal to m dot C p minimum and this maximum change of temperature will happen for the fluid which has got minimum heat capacity. The heat capacity is equal to m dot into C p minimum of the two fluids multiplied by maximum temperature difference that is T b in minus T a in. So, this is a very important thing that this is a maximum temperature drop for any stream that is possible multiplied by m dot C p or the heat capacity of the fluid which has got minimum heat capacity m dot C minimum because the one which has got minimum m dot C p will experience maximum delta T. So, effectiveness is epsilon is equal to q actual upon q max. So, q actual is given by this formula and the q max can be calculated by this formula and this will give what is the epsilon or what is the effectiveness of the heat exchanger is. Now, if I want to understand because of this ineffectiveness or effectiveness being a finite value or not 100 percent how exactly it looks on a T s diagram for a Lindy-Hamsen system. So, we know that the isothermal compression process of a Lindy-Hamsen system is 1 to 2 is a compression process at any temperature which is constant. Similarly, the isenthalpic process occurs from 3 to 4 like this and this is a J T expansion process. So, the gas gets expanded by J Hamsen effect from 3 to 4 where this length 4 to g represents the amount of liquid that is generated because of this expansion while the length f to 4 or 4 to f denotes the amount of gas which is going back. So, if this length is more you will get more liquid if this length is small less gas goes back. So, the gas and the liquid sets are given by g and f the gas goes back from this and the liquid is collected at point f. Now, if the heat exchanger effectiveness or epsilon is assumed to be 100 percent the isobaric heat exchange would be from 2 to 3 process the hot fluid is getting pre cooled from 2 to 3 and after that it is getting subjected to J T expansion. Similarly, the low pressure sides of the cold fluid experiences g to 1 process the temperature will increase from point g to point 1 because of which the temperatures of the hot fluid will decrease from 2 to 3 and this cycle exist when the epsilon is 100 percent. Now, in actual case the heat exchange is not a perfect process and hence the processes are from 2 to 3 dash instead of having heat exchanger process from 2 to 3 we will have it from 2 to 3 dash it will not come up to 3 point for a hot fluid. Similarly, for the cold fluid or the low pressure fluid instead of reaching from temperature at point g up to 1 it will go up to only 1 dash. So, it will not go up to 1, but it will go up to 1 dash this is what is happening because of the heat exchanger effectiveness term. Now, the gas is leaving the heat exchanger on the low pressure side at 1 dash and it enters the compressor at temperature 1 dash. So, naturally the compressor have to do a process of compression from 1 dash to 2. So, the compression occurs from 1 dash to 2 a compressor has to do additional work for bringing its temperature to 1 dash to 1 first and then isothermally compress the gas from 1 to 2. So, in essence what you can understand is that the compressor has to do more work equal to the length from 1 to 1 dash alright. So, in such a case the gas is always compress from the state 1 dash to state 2 and similarly the J T expansion instead of happening from 3 to 4 pluses it happens from 3 dash to 4 dash and as I said that this length from 4 dash to g denotes the amount of liquid or amount of liquefaction that is occurring in this cycle. We can see from here that this length got reduced from 4 to g to 4 dash to g it means that the liquefaction has reduced because of the ineffectiveness of the heat exchanger. So, the J T expansion is occurring from 3 dash to 4 dash as shown in the figure. It is clear that the actual heat exchange is from g to 1 dash alright whereas, the maximum possible heat exchange is from g to 1 in principle one should have come from g to 1 because the temperature of the point 2 could have attained on this pressure line, but no the fluid could not reach up to 1 it has reached up to 1 dash. On the other pressure line on the high pressure line the actual heat exchange process is from 2 to 3 dash whereas, the maximum possible heat exchange should have been from 2 to 3. So, maximum possible heat exchange should have been from 2 to 3 while it has reached up to 3 dash. In an ideal system change in enthalpy on these two isobaric lines is equal. Therefore, the effectiveness is given by epsilon is equal to actual heat transfer that is h 1 dash minus h g the enthalpy at these two point which is the actual heat transfer divided by maximum possible heat transfer which is h 1 minus h g. So, maximum enthalpy difference that could have obtained if the heat exchanger effectiveness 100 percent is h 1 minus h g. However, the actual enthalpy difference that occurred is basically h 1 dash minus h g. So, effectiveness could be written as h 1 dash minus h g upon h 1 minus h g. This is on the low pressure side if I am to give the same definition on this side then the epsilon in this case is equal to h 3 dash minus h 2 actual enthalpy difference divided by h 3 minus h 2 which is the maximum possible heat transfer that is possible for the high pressure line. So, both these terms will indicate the heat exchanger effectiveness for a given heat exchanger. Now, if I want to understand all these points here on this cycle the gas enters the heat exchanger at 0.2 which is this and instead of coming at 3 it comes at 3 dash and instead of expanding from 3 to 4 it expands from 3 dash to 4 dash across the J T expansion wall. So, this is the very important to understand the gas does enter the heat exchanger at point g, but instead of coming at 0.1 it is coming out at 1 dash alright. So, this is what is shown over here. Now, if I were to do the heat balance in this cycle as we have done earlier I will consider this control volume. So, consider a control volume for this system as shown in the figure the quantities entering and leaving this control volume are as given below. So, what is entering this control volume is the mass flow rate m dot at 0.2 and what is leaving the system are m dot minus m dot f at 1 dash and at same time what is leaving is m dot f that is liquid here which is m dot f at point f and if I have the first law of energy balance across this control volume I can write that whatever enters is equal to whatever leaving the system or whatever leaving the control volume. So, whatever is entering is m dot h 2 and whatever is leaving is m dot minus m dot f h 1 dash plus m dot f h f this is what you see from this in out table. If I rearrange this I can now develop a correlation for y which is nothing, but m dot f upon m dot which is equal to h 1 dash minus h 2 upon h 1 dash minus h f. So, you can see now in earlier case it was h 1 minus h 2 divided by h 1 minus h f when the effectiveness is 100 percent, but because the effectiveness is not 100 percent we have got h 1 replaced by h 1 dash and therefore, y is defined like this now. As seen earlier the effectiveness is epsilon is equal to h 1 dash minus h g upon h 1 minus h g this is what we have calculated earlier we have developed a correlation for this. If I rearrange this I will get h 1 dash if I want to have an expression in terms of h 1 dash. So, I can write h 1 dash is equal to epsilon into h 1 minus h g plus h g on this side and what I am going to do now is replace this h 1 dash here by this particular value and rearrange the equation for y. This is what my expression for y is and this is what my expression for h 1 dash is. So, if I replace h 1 dash over here you can get an expression like this. So, y is equal to h 1 minus h 2 minus 1 minus epsilon h 1 minus h g divided by h 1 minus h f minus 1 minus epsilon h 1 minus h g. So, what we can see from here is the earlier expression for y which is h 1 minus h 2 upon h 1 minus h f as numerators and denominators both the numerators and denominators are getting lessened by the value equivalent to this. So, naturally as a result of which y as compared to what it was is going to be less in this case algebraically. So, expression for y is this and what you can see now the second term being negative it should be minimum to maximize the y. If I want to have maximum y this negative term should be as minimum as possible and what you can see now from this all the parameters are constant once my 1 and 2 parameters are fixed state 1 and state 2 pressures are fixed the only variables in the whole expression is only epsilon which is the effectiveness of heat exchanger. So, all other parameters being constant for a given cycle the effectiveness epsilon should be very close to 1. So, everything depends on what is the value of epsilon if the value of epsilon is equal to 1 you get 1 minus 1 is equal to 0 and the whole expression reduces down to what it was earlier and therefore, one has to ensure that this epsilon value should be as close to 1 as possible. We will have several tutorials to understand the effect of heat exchanger effectiveness on the liquid yield. So, based on whatever we have learnt till now these tutorials will help you understand all those theoretical aspects which I just talked about and therefore, I have planned for almost four tutorials. So, kindly go through those tutorials in detail to understand every part of it and what is most important is to interpret the results it is just not mathematical but the understanding should not be only mathematical terms while what is important is to understand the output of these tutorials the results of these tutorials or the conclusions that are drawn from these tutorials that is the most important thing of these tutorials. Then only you can sort of give some quantitative understanding you can have some number fill for all these cycles and their results. So, this is the first tutorial and again read the tutorial problem definition properly. Determine the liquid yield for a Lindy-Hamson cycle with air as working fluid the moment I say air as working fluid you should have temperature entropy diagram or a T s diagram of air available with you when the system is operated between 1.013 bar or 1 atmosphere and 202.6 bar or 200 atmosphere at 300 Kelvin this is a compression process state 1 and state 2 approximately 1 bar and 200 bar and the compression temperature is 300 Kelvin. The effectiveness of heat exchanger is 100 percent 95 percent 90 percent and 85 percent comment on the results. So, what basically I want you to do is comment on the liquefaction yield you get when the heat exchanger effectiveness 100 percent then 95 percent then 90 percent and 85 percent and then comment on the results. So, there are actually four problems in this solve these problems to get liquid yield when this is the ideal Lindy-Hamson cycle then we have got effectiveness of 95 percent 90 percent and 85 percent of the heat exchanger. So, the first thing is to get a T s diagram. So, T s diagram of Lindy-Hamson system if assumed the heat exchanger to be of 100 percent effective then it is as shown here. So, this is as you now know 1, 2 to the compression process 2, 2, 3 is a heat exchanger process 3 to 4 is a J T expansion process G 2 1 is a heat exchanger process J 2 1 prequels 2 to 3 alright. So, this is what is happening when the heat exchanger effectiveness is 100 percent. This is the state 1 at 1.013 bar or 1 atmosphere this is the state 2 the process 1 2 is a compression process occurring at 300 Kelvin. So, first is the formula which we want to apply which is y is equal to h 1 minus h 2 minus 1 minus epsilon h 1 minus h g divided by h 1 minus h f minus 1 minus epsilon h 1 minus h g. This is a general formula and if you put the first case now this is the formula which we get from a Lindy-Hamson cycle having an heat exchanger which has effectiveness value epsilon. The first case is when we want to have epsilon 1 is equal to 1 or 100 percent if you put this value you should get the value of y for the first case when epsilon is 100 percent. The most important thing is now look at the point 1, 2, f and g on the T s diagram. The one is at 1 bar approximately and 300 Kelvin the point 2 is at 200 bar and 300 Kelvin the point f is 1 bar and 78.8 Kelvin which is the boiling point of air because air is the working fluid and the point g is on the gas side and the point f is on the liquid side. So, point g and f lie on the same temperature which is 78.8 Kelvin which is nothing but the boiling point of air. Get the corresponding enthalpy and entropy and the first step to do any problem is to make this table so that all these values are available to you. Now with my experience I can tell you if you make mistake in getting even one value the whole result will change and you will not be able to get a correct conclusion drawn from this. So, the most important thing is keeping calm go to the T s chart, look at this point and get correct enthalpy and entropy value. Inaccuracy of plus minus 5 percent is acceptable if you write instead of 28.47 as 30 no problems or 26 no problem but do not go from 28 to 40 or 35 that far. So, having got these values from this T s diagram the important thing is now put these values of h 1, h 2, h g and h f in this formula for y. So, the first case is when epsilon is 100 percent or epsilon 1 in this case is 1. So, if I put those values h 1, h 2, h 1, h f and here I got epsilon is equal to 1. So, in principle the whole this side gets equal to 0. So, 1 minus 1 is equal to 0 on the numerator as well as denominator and the expression gets reduced to h 1 minus h 2 upon h 1 minus h f which is what we know when the heat exchanger if it is 100 percent if you put these values y 1 you get is equal to 0.085. So, first case for the epsilon 1 is equal to 1 you get y 1 is equal to 0.085. Now let us go to the second case when epsilon 2 is equal to 0.95 or 95 percent is the effectiveness of the heat exchanger again go to the same table the values will remain the same here put the values correctly calculate these values and now here you get y 2 is equal to 0.060. So, if you compare with earlier values you can suddenly find that the value of y 2 is less than what you got in the earlier case as soon as you came down on the effectiveness from 100 percent to 95 percent. Now let us go to the third case where the epsilon 3 value is 90 percent that means the heat exchanger effectiveness in this case is now 90 percent apply the same formula again get the same values again for 1, 2, f and g put the values for y 3. See if I put those value and put the value of effectiveness of heat exchanger as 90 percent or 0.9 in this case calculate the whole thing and y 3 has come out to be 0.034 which is again less than what you got for 95 and 100 percent effectiveness of the heat exchanger. So, in this case y 3 is equal to 0.034 the fourth case now we are going is for epsilon 4 is equal to 0.85 or 85 percent effectiveness of the heat exchanger again I put the values from this table and I get the values of y 4 and if I put all those values with effectiveness of the heat exchanger equal to 0.85 in this expression I get the value of y 4 to be 0.006 very close to 0. So, here we have got 4 cases and we went on reducing the heat exchanger effectiveness from 100 percent to 85 percent. So, in summary we want to show the schematic in the graphical form the y values for different effectiveness of heat exchanger. So, if I do that for working fluid of air keeping the pressure from 1 bar to 200 bar of compression process the temperature of compression is 300 Kelvin for a Linde-Hampson cycle the 0.1 is effectiveness of 1 the y value is 0.85 which I have plot over here. So, 0.1 for 1 what you get is 0.085, 0.2 for 95 percent effectiveness of heat exchanger what is you get is 0.06 then 0.3 is for 0.9 and what you get is 0.034 as y and 0.4 is effectiveness of 0.85 what you get is 0.006 which is very close to 0 plot and connect by line what you get is this. So, joining these points we have the trend as shown in the figure which you see that as the effectiveness of heat exchanger decreases the value of y decreases it is clear that as the effectiveness decreases the yield y decreases drastically and what you also can see from here is if the heat exchanger is now less than 85 percent the value of the y is almost 0 or there is no yield what does it mean? It means that if the heat exchanger effectiveness is less than 85 percent the J T expansion process will never fall in the dome it will fall outside the dome meaning which there is no liquefaction. When I am having J T expansion process happening outside the dome it means that before the J T expansion there is a gas and after the J T expansion when we want to have two phase fluid what you get at the end of the expansion is again gas and there is no liquid available. So, the point is that the heat exchanger effectiveness if it is less than 85 percent the J T expansion process will never fall in the dome the point before the J T expansion will be far away from the dome and therefore there will not be any liquefaction in this case. So, one has to be sure that the effectiveness of the heat exchanger should be much higher than 85 percent in fact in order to get proper value of y or the liquefaction for a Linde-Hamson cycle. Furthermore as we can see from this that the effectiveness should be much more than 85 percent in order to have liquid yield. So, this tutorial basically shows how important is the effectiveness of heat exchanger is in order to implement the Linde-Hamson cycle for gas liquefaction. It is not a very simple process to have a epsilon of 85 percent and more to realize in practice it is not very simple and therefore one has to have a very very important task to design a heat exchanger to have higher values of epsilon and to have a very high values close to 95 percent epsilon values are to be realized in practice it is not a very simple thing and therefore in cryogenesis what we have is very special kind of heat exchangers of which we will talk about in the coming lectures. So, this tutorial one basically shows how important the heat exchanger effectiveness is from gas liquefaction point of view. The second term now I want to introduce here is figure of merit or FOM and we will then solve some more tutorials on this. So, what is figure of merit? In the earlier lecture we have seen that the ideal cycle is used as a benchmark to compare a various liquefaction cycle. We talked about that the ideal thermodynamics cycle is used as a benchmark to compare various liquefaction cycle. In effect a parameter called figure of merit is defined and what is it? It is a ratio of the ideal work input to the actual work input to the system per unit mass of gas which is liquefied. So, what is the work input per unit mass of gas which is liquefied ideally and in actual case? So, ideal work is going to be always less than the actual case and therefore mathematically we can write figure of merit is equal to W i that is ideal work input divided by W c which is actual work input system and based on this now we got a second tutorial which is this. Determine the falling for a Linde-Hamson system with nitrogen as a working fluid. When the system is operated between again 1 bar or 1 atmosphere to 200 atmosphere at 300 Kelvin the effectiveness of heat exchanger is 100 percent. This is my tutorial number 2. So, working fluid is nitrogen again the state 1 and state 2 are at 1 atmosphere to 200 atmosphere and the compression temperature is 300 Kelvin. The heat exchanger effectiveness for this particular purpose is 100 percent. What you want to compare or what you want to get is what is the ideal work requirement, what is the liquid yield, what is the work per unit mass of gas which is compressed, what is the work per unit mass of gas which is liquefied and finally what is the figure of merit for this particular case. So, 1 2 3 4 and 5 are required to be answered for this particular tutorial using nitrogen as a working fluid. So, as I said the first thing to be done is to get a T S diagram for nitrogen. Now, when we want to calculate the ideal work requirement we have got this cycle. In the ideal cycle we say whatever is compressed gets liquefied and therefore, you are a 0.1 only determines what the point F is alright. So, we need to know only the state 1 or the starting point to get the ideal work requirement and we have done the previous calculation. The ideal work requirement is W i upon m dot is equal to T 1 into S 1 minus S f minus H 1 minus H f. So, now there are 2 points which is 1 and F. So, if I get a value of entropy at 1 and F and if I get value of enthalpy at 1 and F I can calculate the ideal work requirement per unit mass of gas which is compressed. So, again as I said go to the T S diagram temperature entropy diagram of nitrogen and gets point 1 and F in this particular case for the ideal cycle point 2 is actually not required in this case. So, the point 1 is 1 bar and 300 Kelvin, point F is 1 bar and 77 Kelvin which is the boiling point of nitrogen get corresponding enthalpy and entropy for this. Apply the formula this formula and put the values T 1 is equal to 300 Kelvin S 1 minus S f this bracket H 1 minus H f this bracket and if you calculate this the ideal work is 761 joule per gram. So, the ideal work requirement for nitrogen for this particular condition is 761 joule per gram when it is compressed at 300 Kelvin from 1 bar. The liquid yield is H 1 minus H 2 upon H 1 minus H f this is for the actual cycle now the state 2 is at this point which is 200 bar or 200 atmosphere again get the data for 1, 2 and F this is now Lindy-Hamson cycle. So, get all the values associated with this and calculate the value of y which is nothing but m dot f upon m dot which is the liquid yield. So, putting this value of H 1 minus H 2 upon H 1 minus H f is equal to this what you get is 0.074. So, liquid yield is y which is 0.074 in this case. So, work for unit mass of gas which is compressed is now T 1 into S 1 minus S 2 minus H 1 minus H 2 here not S 1 minus S f and not H 1 minus H f which was the case for the ideal cycle. So, if I get the state 1 and state 2 and if I put this properties in that particular case I can calculate the value and my actual work of compression per unit mass of gas which is compressed is 463 joule per gram. Now, I want to calculate work per unit mass of gas which is liquefied and not compressed what we obtain earlier is work per unit mass of gas which is compressed. So, W c by m dot is 463 which is work per unit mass of gas which is compressed the y value is 0.074 and if I want to calculate now work per unit mass of gas which is liquefied which is W c upon m dot f what I am going to do is to divide W c by m dot by y right if you put the value of y here what you get is W c upon m dot f. So, 463 upon 0.074 gives you 6265.22 joule per gram alright. So, this is the rather high value as compared to what you got ideally in ideal case work per unit mass of gas liquefied is equal to the work per unit mass of gas which is compressed. So, what we had calculated then is also work per unit mass of gas liquefied what you are getting here also is now work per unit mass of gas which is liquefied in this case. So, if I want to calculate figure of merit or FOM which is I should get W c upon m dot f for actual case and ideal case is W i upon m dot f which is 767. So, the figure of merit is the ideal work of compression per unit mass of gas which is liquefied divided by the actual work per unit mass of gas which is liquefied. So, if I put that I will get 767 upon 6265 and 0.1225 this is my figure of merit in this particular case alright. So, we have solved all the 4 required things for the tutorial number 2. Now, I go to tutorial number 3 which is the same thing, but now here I am changing the gas from nitrogen to argon. So, determine the following for a Linde-Hampson system with argon as a working fluid when the system is operated from again 1 atmosphere to 200 atmosphere at 300 Kelvin the effectiveness of heat exchanger is 100 percent. The tutorial is solved just to show the difference what happens when I go from A gas to B gas alright. So, it is just repetition of the earlier tutorial just to get you habituated with how to read a T s diagram for argon T s diagram for nitrogen and repeat of the methodologies involved. So, again I will calculate ideal work of requirement which is W i upon m dot. Now, I will go for a argon T s chart. The moment I go for argon T s chart I got different enthalpy and entropy value at 0.12 and f. As far the ideal work is concerned I do not have to depend on this state 2 I have to go for 1 and f only while for actual case I will have to go for 0.12 and f. So, if I get these values 1 and f and put in this equation I will get the ideal work requirement as 476 joule per gram for argon. The liquid yield is depending on the state 1 and state 2 h 1 minus h 2 upon h 1 minus h f again get the table and I can calculate the value of liquid yield in this case is 0.1176. So, if I want to calculate work per unit mass of gas which is compressed my formula is W c upon m dot is T 1 into s 1 minus s 2 minus h 1 minus s 2 1 and 2 are the state point for the compression processes. Again put the value of enthalpy and entropy in this equation and what you get is now 326 joule per gram. This is the work per unit mass of gas compressed. If I want to calculate the work per unit mass of gas liquefied I will divide this value by y or the liquid yield. So, work per unit mass of gas liquefied is going to come from this value and corresponding y. So, W c upon m dot f is equal to W c upon m dot divided by y and now work per unit mass of gas liquefied comes to be 2772.1. And finally, when I want to calculate the figure of merit W c upon m dot f is equal to 2772.1 W i upon m dot f is 476. So, the ideal work input per mass of gas which is liquefied actual work input per mass of gas which is liquefied and this divided by this is nothing, but figure of merit. So, figure of merit is ideal work input actual work input per unit of gas which is liquefied putting this values over here the figure of merit comes out to be 0.1717 for Aragon when compressed from 1 bar to 200 bar at 300 Kelvin. We have just solved the problem for nitrogen and Aragon and we found that the figure of merit in this case was around 0.12 and 0.17 corresponding boiling points are given corresponding y values are given liquid yield are given. What is the work of compression per unit mass of gas which is compressed? What is the work of compression per unit mass of gas which is liquefied? And then finally, what you get is the figure of merit. We have also given values for air and oxygen for which we have done calculations, but these two values I am going to give you for assignments at the end of this lecture. So, please solve those assignments and compare your values with the values of figure of merit or y or these values given in this table which we will see about in the later part of this lecture. The above table is for Linde-Hamson cycle when the pressures are from 1 bar to 200 bar at 300 Kelvin. The heat exchanger in all these cases is assumed to be 100 percent. I am now solving a tutorial number 4 which is basically meant to understand again effectiveness of heat exchangers on all these parameters. So, for the conditions specified in tutorial number 2 for Linde-Hamson system with nitrogen as working fluid, calculate the following when the effectiveness of heat exchanger is 90 percent. So, get the liquid yield work per unit mass of gas compress, work per unit mass of gas liquefied and figure of merit. And ultimately what is most important is comment on the results. So, what you got earlier was for effectiveness of heat exchanger equal to 100 percent and what you calculate now in this problem is going to be for the epsilon value equal to 90 percent. So, get the results for this problem and compare all these values with the tutorial number 1 or 2 which you got for nitrogen as working fluid. And comment on the result is the most important thing because what has happened because the effectiveness came down from 100 to 90 percent is very important to understand. So, the process is same and the formula changes now because epsilon comes into picture. So, this is the formula I am going to use now in this case. So, epsilon is equal to 90 percent and this is the table now 1, 2, f and g what you get is y is equal to in this case 0.021. So, y has suddenly decreased now to 0.021. In order to calculate the additional work that is to be done we have to calculate what is the h 1 dash if you remember from earlier discussion h 1 dash is equal to epsilon h 1 minus h g plus h g putting those values epsilon value is 90 percent h 1 dash is equal to 438.8. So, if I want to calculate now the additional work the compressor does which is equal to h 1 minus h 1 dash is equal to 23.0 joule per gram. So, the compressor does this much additional work in 462 which is almost 5 percent of what it was earlier. So, 5 percent more work has to be done in order to get the enthalpy from h 1 dash to h 1 and then the compress the gas from h 1 to h 2 in this case when the effectiveness of heat exchanger is 90 percent. So, work per unit mass of gas compressed is the same formula I will get these values and calculate these and what you get now is 463 over here. So, if I want to get total work done is equal to this is the work done from 0.1 to 2. So, 0.1 to 2 plus additional work which is going to be done is going to give me actual work or the total work of compression per unit mass of gas which is compressed which is now 486.2. So, in earlier case it was 463 now it is 486.2 work per unit mass of gas which is liquefied it is this divided by y which is very small 0.021. So, I get now W c upon m dot f is equal to 22613.95. So, it is rather a high as compared to what it was earlier and the corresponding figure of merit now here is W i upon m f dot is equal to 761 W c upon m f dot is this if I divide this the f y m comes to be 0.03. So, you can see if I want to summarize what you can see is these are the two columns when the epsilon is 100 percent and the epsilon is 90 percent in this case. So, you can compare the y values the y value for 100 percent effectiveness was 0.074 and it has come to 0.021 the percentage decrease has been 71 percent. If I see the compressor work per mass of gas which is compressed it has increased from 463 to 486 and therefore, percentage changes minus 5 percent. If I see the compression of work done per unit mass of gas which is liquefied it has been rather high 6265 it has increased to 2200614 that means there is the increase of 261 percent this is happening because y value had decreased drastically by 71 percent and finally, the figure of merit has decreased by almost 72 percent. So, this example shows that as soon as you reduce down from 100 percent to 90 percent there are so many change drastic changes are happening in the yield value as well as figure of merit and therefore, one has to be very careful in the effectiveness calculation and heat exchange in design. This is a very very important design aspects of any liquefaction system. So, this example highlights the importance of epsilon value for any cycle in cryogenic liquefaction of gases all right. So, about if the significance of heat exchanger effectiveness for a Lindy-Hamson system and we have got assignment here based on what we have done with air as a working fluid calculate these we have done for nitrogen and argon calculate all these things please do it for air and oxygen and also repeat the same problem when the effectiveness is 90 percent and compare and comment on the results. Please do these three assignments this will really give you a feel for the values for these two particular gases. Finally, to summarize what we have learnt today a heat exchanger is a device in which the cold is transferred from cold fluid to hot fluid. Everybody knows this the effectiveness epsilon is defined as ratio of actual heat transfer that is occurring to the maximum possible heat transfer that can occur theoretically in a heat exchanger. It is a dimensionless number between 0 to 1. In a Lindy-Hamson cycle the heat exchanger effectiveness is epsilon is equal to h 1 dash upon h g divided by h 1 minus h g or on a high pressure side which is h 3 dash of minus h 2 divided by h 3 minus h 2. The liquid yield y of a Lindy-Hamson cycle is given by this expression where epsilon plays a very important role as we have just found. As the effectiveness decreases the yield y decreases drastically. Furthermore, the effectiveness should be more than 85 percent in order to have liquid yield in Lindy-Hamson cycle. Thank you very much.