 So, welcome to the 13th lecture of cryogenic engineering under the NPTEL program. To take an overview of the earlier lecture, in the earlier lecture, we have seen that the pre cooling of a simple Linde-Hamson system improved the liquid yield. Before that we had talked about simple Linde-Hamson and then we are talking about the pre cooling of the simple Linde-Hamson cycle and we found that it improved the liquid yield. In a pre cooled Linde-Hamson system, a closed cycle refrigerator is thermally coupled to a simple Linde-Hamson system through a three fluid heat exchanger. So, you got a pre cooling arrangement through a three fluid heat exchanger. The pre cooling limit of the pre cooling cycle is governed by the boiling point of the refrigerant at its suction pressure. So, you got a refrigerant moving in a closed cycle manner in a pre cooling circuit and the boiling point of this refrigerant at a suction pressure at a low pressure, it will decide the lowest possible pre cooling temperature that can be obtained for the Linde-Hamson cycle. From the tutorial in the past lecture, we saw that the yield of a pre cooled cycle was more than that of the simple system and this is why we do pre cooling basically. The maximum liquid yield in the pre cooled system occurs when the effectiveness of the three fluid heat exchanger is 100 percent, but as you know in practical system the heat exchanger effectiveness will not be 100 percent. So, the effect of this also has to be considered. The maximum yield or a Y max is obtained by the enthalpy difference what you get after the first heat exchanger and that is h6 minus h3 upon h6 minus hf when the properties the enthalpy are 0.6 and 0.3 are evaluated at temperature which is at the boiling point of the refrigerant. So, in the above equation the values of h6 and h3 are evaluated at boiling point temperature of the refrigerant. Then only whatever yield you get is Y max. With this background the outline of current lecture is continuing on the same topic of gas liquefaction and refrigeration system. We will take this topic ahead the pre cooled Linde-Hamson system wherein we will study the effect of different parameters that affect the performance of this system. What are these different parameters? So the effect of mass ratio R this is the ratio of mass flow rate of the refrigerant to the mass flow rate of the gas to be liquefied. So m dot R upon m dot is what we call as mass ratio R. Then yield versus mass flow ratio mass ratio R yield is Y whatever liquefaction yield you get that as a function of this R which is a very important parameter and we will try to understand what happens to this ratio. Then different work requirement compressor work requirement per gram of mass which is compressed and also the work requirement per gram or kg of gas which is liquefied. So you got a work requirement versus mass ratio R and finally figure of merit versus mass ratio R which depends on the above two parameters. So in effect we will study the all this effect as a function of R which is a very very important parameter. If R is equal to 0 it reduce down to a simple Lindy-Hamson cycle. So R is the only parameter which changes the simple Lindy-Hamson cycle to pre-cooled Lindy-Hamson cycle and therefore to study the effect of this R parameters on all these different performance parameters is a very important task. To introduce what we are going to learn the work requirement of a pre-cooled Lindy-Hamson cycle is given by this formula minus Wc upon m dot is equal to the first bracket and there is a second bracket. The first term or the first bracket is a work requirement in a simple Lindy-Hamson system. We have seen the derivative of this derivation of this formula also and to this first term or the first bracket what you add is a second term when we go from a simple Lindy-Hamson system to a pre-cooled Lindy-Hamson system. So when you go from a simple Lindy-Hamson to a pre-cooled Lindy-Hamson this is the additional work that the compressor has to do and this is the work done by the refrigeration system or the pre-cooling circuit compressor. Second term is the additional work required to pre-cool the system. So the first term remain the same as long as the suction pressure and the compressor pressure at the end of compression remain the same but this is the additional amount of work that has to be done and that comes due to the pre-cooling circuit in the pre-cooled Lindy-Hamson system. The yield for a pre-cooled Lindy-Hamson system is as given below y is equal to m dot upon m dot is equal to h1 minus h2 upon h1 minus hf which is same as simple Lindy-Hamson cycle plus the additional yield which comes because of the pre-cooling cycle where the mass ratio r is given by m dot r upon m dot is equal to r. The first term in the above expression is the yield from a simple Lindy-Hamson system and the second term is the additional yield occurring due to the pre-cooling of the simple system. The increment in the yield is related to this bracket we are talking about what is it related to the additional increment the additional yield which comes because of pre-cooling what you can see from this formula is basically dependent on h a and h d or the difference between h a and h d the change in enthalpy from the value h d to h a of the refrigerant. So, this is the very important thing to understand that h a and h d value will determine or the difference in this enthalpy value will determine how much y increment will be occurring when we go from a simple Lindy-Hamson cycle to a pre-cooled Lindy-Hamson cycle. At the same time the other parameter is r which is the refrigerant flow rate m dot r rate across the three fluid heat exchanger. So, r is a parameter which is the ratio of m dot r to m dot and this will decide this basically decide decided by the m dot r value. So, m dot r upon m dot because m dot remain the same and if we are going to change the value of m dot r because of which the r changes. So, r and the enthalpy difference across the heat exchanger will decide how much increment will occur when we go from a simple system to a pre-cooled system. From the above equations it is clear that the liquid yield and the work requirement are dependent on the parameters like refrigerant flow rate m dot r, compression pressure and pre-cooling temperature. By varying these parameters the performance of the system can be optimized. So, we got various parameters and we got to worry about which parameter we should change so that we get more yield. Should we change m dot r? Should we change the suction pressure? Should we change the m dot? All this will decide how much increase or decrease would there be in terms of in parameters like liquid yield, work requirement etcetera. Hence, there is a need to study the effect of various parameters on the performance of the system for the proper design. So, one need not worry, one need not straight away go and increase the refrigerant m dot r. One should not really play with the suction pressure or a compression pressure unless one understands what is the effect of these parameters on the cycle in TOTO, I mean pre-cooled cycle plus the real simple Linde-Hampson cycle. The effect of these two parameters, these two cycles are actually related. Now, this is the cycle, this is the pre-cooling circuit and this is the simple Linde-Hampson cycle and this is the three fluid heat exchanger. One such parameter which is of great importance is the refrigerant flow rate r, this is what we talked about and r depends on this m dot r to m dot. The ratio of m dot r to m dot will decide the value of r. The state of the working fluid entering the refrigerant compressor is very important. So, m dot r the value of m dot r also determines at what state m dot r enters the refrigeration compressor alright. So, this is very important as you know in a reciprocating compressor the entry to the compressor should be in gaseous phase and this is what will the value of m dot r will decide or will ensure that gas which is entering the compressor should is in proper state. Let the heat exchange or let the heat change of the refrigerant be represented by Q R E F. So, Q R E F is basically the load on this system or the pre-cooling requirement of this heat exchanger and what you get across is R into the enthalpy difference across this heat exchanger. So, R into H R D minus H R A. This will determine how much cooling load is offered by the cooling circuit alright. Similarly, the required heat exchange for the Lindy-Hampson cycle be denoted by Q L H S alright. So, Q Lindy-Hampson cycle system load will be determined by the whole of three fluid heat exchanger. Now what we want exactly is Q L H S and what this system offers or the what this pre-cooled circuit offers is Q R E F. These two values actually should match if we have got a optimized value of Q L H S. The relative values of Q R E F and Q L H S determine the state of the refrigerant at the point A. Now, we want for a optimized design of the pre-cooled Lindy-Hampson system what we want is a Q L H S while the pre-cooled circuit offers you Q R E F. Now, depending on the magnitude of Q R E F and the requirement of Q L H S this will decide what is the state of the fluid when it leaves this three fluid heat exchanger. What is the state of the fluid at point A which is the point at which it enters the refrigeration compressor or the pre-cooling circuit compressor alright. So, the relative magnitude of these two parameters will decide what is the state of the refrigerant when it enters the refrigerant compressor. So, there are three possibilities Q R E F less than Q L H S, Q R E F is equal to Q L H S and the third one is Q R E F is greater than Q L H S. In the first case the value of the T 3 would not be equal to T D what is the first case when Q R E F or the refrigerant effect that is given by this pre-cooling circuit is not is less than Q L H S. In this case the gas state or the gas state at point 3 will not be at the same temperature at T D because the refrigeration of the cooling effect that is available in the pre-cooling circuit is less than the desired value of Q L H S. In that case the temperature at point 3 will be will not be equal to the point T D that means it will not be equal to the boiling point of the refrigerant at this suction pressure. In the second case when Q R E F is equal to Q L H S what we achieve is the temperature at point T 3 is equal to T D that means based on the second law of thermodynamics the lowest temperature that is possible at point 3 is equal to the temperature at point D which is in that case is equal to the boiling point of the refrigerant at the suction pressure at this pressure pressure at this point pressure after the expansion of the refrigerant here and therefore in this case what we achieve is the lowest possible temperature at point 3 and therefore the yield what you get is y max we have talked about this in the earlier lecture. However what is important is the third case when the refrigeration effect that is available in the pre-cooling circuit is more than required that is Q R E F is more than Q L H S. So, in the third case now the liquid would enter the refrigeration compressor because the cooling effect that is available that is Q R E F is more than required and therefore the gas which is coming which is the refrigerant which is going to come out at this point A is going to be in two phase zone or in liquid zone because it has not got vaporized completely alright and therefore the state of the refrigerant at this point could be liquid plus gas over here which is not good from the compressor point of view. And this is a very important thing to understand that if Q R E F is more than Q L H S then the state of the refrigerant at this point when it enters the refrigeration compressor could be liquid or liquid plus gas and therefore this is not a desired condition that is any excess flow of the refrigerant then required results into an excess available heat content as a result the state of the fluid at point A would be a two phase mixture which is unfavorable which is not desired in this case and hence for a given operating condition there is an optimum value of R. We want to avoid such situation in fact we should never run a compressor in which the state of the refrigerant at this point is liquid or two phase and therefore we should never operate in such a condition and therefore we should always see that this optimum value of R which ensures that the refrigerant state at this point A is always in gaseous phase alright this is a very important limitation of this system that the corresponding value the corresponding phase at this point or the R value should be optimized value which will ensure that the refrigerant when it enters the compressor is in gaseous phase. This is better explained through a tutorial solved in the subsequent slides. So, to understand this we will solve a problem and from there we can draw conclusions in which in this problem we got a various flow ratio R values are taken both below and above the limiting value to explain the principle of the value of R or the dependence of the entire performance of this circuit on the value of R. So, a tutorial is taken here and let us understand the tutorial and this tutorial has got two parts. The part one is this the statement is like this a pre cooled Linde-Hampson system has nitrogen and R134A as primary and secondary fluids respectively this you understand nitrogen is a working fluid R134 is a pre coolant. Determine the liquid yield and liquid yield is y and figure of merit. The operating conditions and other useful data are given below the operating conditions are this we have got four cases over here and this is very important to understand where we vary the value of R which is 0.05, 0.07 when we have got a compression pressure of 101.3 bar for this two cases and we also want to study the effect of value of R for different pressures and in which case we have taken the pressure as 202.6 bar or 200 atmosphere and again here we value the value of R from 0.05 to 0.1. So, here in short we are studying the effect of pre cooling for two different pressures 100 atmosphere and 200 atmosphere and also we are studying the effect for two different R values 0.05 and 0.07 R and 0.05 and 0.1 for 200 atmosphere. The corresponding values for the pre cooling circuits are given over here where we are compressing the pre coolant of the refrigerant from 1 atmosphere to 8 atmosphere and expanding from 8 atmosphere to 1 atmosphere again and we got different temperature values and enthalpy values for R134A which is the refrigerant in this case. We got a part 2 of this problem where we extend this problem definition and here we would like to calculate the y max for each of this pressures mentioned and their corresponding R values. So, identify the y max for given pressures and corresponding to those values find out the R values or the limiting R values in this case plot the data graphically and comment on the nature of y work requirement FOM versus R. So, this is what a part 2 will be. So, let us understand the part 1 which is a very simple case we have solved the tutorial in the earlier class for this. Again the problems same I am just repeating the table in order to understand what is basically asked for in these two problems or these two tutorials. So, what is given as for part 1 is a we want to solve the pre cooled Lindehamsen cycle with nitrogen as a working fluid the temperature of compression is 300 Kelvin. The refrigerant is R134A where it is compressed from 1 atmosphere to 8 atmosphere. What we have to do is find out liquid yield y work per unit mass of gas compress work per unit mass of gas liquefied and FOM. The nitrogen R is having 2 values 0.05 to 0.07 for 100 atmosphere and then 0.05 to 0.1 R for 200 atmosphere. For part 2 similarly we got the same values only in that case we want to understand the dependence or R at y max value for 100 bar atmosphere. Again we want to find R at y max value for 200 atmosphere. The methodology we are going to follow to solve this tutorial is the two pressures what we have taken are 100 bar and 200 bar or 100 atmosphere and 200 atmosphere. The liquid yield and FOM are calculated only for 100 atmosphere here. So, what I am going to do is solve the problem for one pressure and we would expect that you will solve the problem for 200 atmosphere case. We will also calculate the y max value and corresponding to y max value we find the R value and we will also go beyond y max. We will go for an R value which is beyond y max condition and again we will do the same calculations for 100 atmosphere or 101.3 bar pressure condition only. Calculations pertaining to 200 atmosphere or 202.6 bar conditions are left as an exercise for the students. So, we got a circuit here we got a 0.12 and point F as given over here. Similarly, we got point A B and C as given over here. So, we got A B and C conditions given over here for R 134 A and 1 2 F as given over here. It is just to basically make you habituated to locate these points or corresponding to those values find out the temperature enthalpy and entropy respectively. Again as you know at C what you have is a h d is equal to h c the enthalpies at point C is equal to enthalpy at point D. First we do the ideal work requirement I will not go in the details of these calculations again because we have done this calculation earlier I will show these calculations. So, as you know the ideal calculation come from this fact that whatever is compressed is getting liquefied. Using this formula we get the values for point 1 and point F put these values in this formula what you get is a 761 joule per gram which is a very standard value for nitrogen as a working fluid. Now, this is a T s diagram for a pre cooling circuit and this is a pre cooling temperature. The properties are given over here as given in the data find out the liquid yield by this formula we got a y is equal to m dot F upon m dot and this is the formula which is R dependent and enthalpy difference across the three fluid heat exchanger from refrigerant side putting those values over here. In the first case we are taking R is equal to 0.05 put the enthalpy values at different temperatures and calculate the y 1 1 is depicting R is equal to 0.05 for 100 bar putting these values of enthalpy and R is equal to 0.05 what you get is a 0.055 as yield value at in the first case y 1 and then calculating work unit mass of nitrogen compressed this is the formula put the value of R and H B and H A this is the additional compressor work one has to do for the pre cooling circuit putting these values over here what you get is a 381 joule per gram as the yield for the first case and work per unit mass of nitrogen which is liquefied is coming from W C by m dot and y 1 dividing this what you get is a 6927.2 joule per gram as work per unit mass of nitrogen liquefied. The figure of merit comes by ideal work per unit mass of gas liquefied divided by the actual work per unit mass of gas liquefied W I by m dot f required W C by m dot f putting these two values the FOM figure of merit for the first case is 0.1107. Now, I want to do the same thing for next R value we have done all the calculations for R is equal to 0.05 will