 So, in this lecture, we are going to look at the vapor compression refrigeration cycle. This is the last of the cycle that we will consider in this module on thermodynamic cycles. In contrast to all the cycles that we have looked at so far, the vapor compression refrigeration cycle is a power absorbing cycle. And what we are going to discuss today is the cycle that is used in domestic refrigerators and domestic air conditioners. This was discussed in detail in the previous course. So, for those of you who have not gone through that material, I suggest that you go through this the module on vapor compression cycle in the previous lecture, but we will quickly recap whatever was said there. So, basically we start at state 1 here where the refrigerant. So, here the refrigerant is the working substance. So, similar to the Rankine cycle where water was the working substance and it remained as water throughout the cycle. Here refrigerant is the working substance and it remains as refrigerant throughout unlike the air standard cycles where the air was taken in and then usually it is mixed with fuel and burnt. So, you have air coming in and then combustion gases leaving. So, the working substance does not execute a cyclic process there. In this case, the refrigerant is the working substance and it executes a cyclic process. So, let us start with state 1. So, state 1 is entry to the compressor. So, here the working substance is at low pressure and relatively low temperature and it can be a saturated vapor or slightly superheated also. It does not matter. Let us write it down as saturated vapor. So, that is the thermodynamic state at state 1. So, the refrigerant then enters the compressor where it is compressed to high pressure, high temperature and superheated. So, the high pressure, high temperature refrigerant is then sent to the condenser where it loses heat with the ambient. So, if you look at say slightly older refrigerators, you should be able to see the cooling coils at the back of the refrigerator. So, that is the condenser where the refrigerant loses heat with the ambient. So, as it comes out of the condenser, the refrigerant is at a high pressure, reasonably high temperature, not very high, reasonably high temperature and it actually usually comes out as a saturated liquid. Although this need not exactly be the case, but for now let us take it as saturated liquid. Now, the refrigerant is then throttled in the throttling valve. So, the high pressure, reasonably high temperature refrigerant now comes out as low pressure, low temperature, saturated mixture. So, the temperature of the refrigerant drops as a result of throttling and it then enters the evaporator. So, this is the refrigerated compartment or the air conditioned room. So, the refrigerant then the cold refrigerant picks up the heat from the refrigerated compartment and as it picks up the heat, it undergoes change of phase and then it goes from being a saturated mixture to a saturated vapor and the cycle is repeated. Now, we may illustrate the cyclic process executed by the refrigerant on a T S diagram or on a P H diagram. I will come to the P H diagram next. Let us first look at the T S diagram. So, just like the Rankine cycle or the Brayton cycle, the vapor compression refrigeration cycle also operates between 2 isobars P equal to P C which is the condenser pressure and P equal to P E which is the evaporator pressure. So, state 1, we have taken it to be a saturated vapor. So, saturated vapor at evaporator pressure. So, it enters the compressor where it is compressed from state 1 to 2 is in case of an isentropic compression. In the situation where we allow for internal irreversibilities in the compressor, the end state will be state 2. So, you can see that the state corresponding to 2 is high pressure, high temperature and superheated as I mentioned before. So, it is not taken to the condenser and it loses heat to the ambient at constant pressure, undergoes change of phase starting from here and the fluid leaves the condenser as a saturated liquid at condenser pressure. So, it is not throttled in the throttling valve. So, this is the throttling process. So, it is throttled to a low temperature, low pressure saturated mixture and it then enters the evaporator where it picks up the heat. So, this is the evaporator and this is the condenser. Now, this cycle is called the ideal vapor compression refrigeration cycle. Even if the compressor is not ideal, for example, even if the compression process has internal irreversibilities, this cycle is still usually referred to as the ideal vapor compression refrigeration cycle in the refrigeration community. There are two reasons for that. First one is that T H, which is the temperature of the hot reservoir or the reservoir to which heat is rejected. So, T H usually in the case of an ideal vapor compression cycle corresponds to the saturation temperature of condenser pressure. So, T H is T sat of P C and T C the temperature of the cold reservoir or the cold reservoir from which heat is drawn by the refrigerant. Remember, T C is the temperature of the refrigerated compartment. So, this is where the heat is drawn by the refrigerant. So, this corresponds to the saturation temperature of the evaporator pressure. That is number one. Number two, state one is a saturated vapor in the case of the ideal vapor compression refrigeration cycle and state three is a saturated liquid. So, this is a saturated liquid and this is a saturated vapor. So, T H is equal to T sat of P C, T C equal to T sat of P E and state one, exit of evaporator is a saturated vapor state, exit of condenser is a saturated liquid state. So, this cycle is usually referred to as the ideal vapor compression refrigeration cycle in the refrigeration community. Now, you may wonder or you may wonder why the diagram has to be illustrated in different in a different set of coordinates. Why not just stay with T S? Why do we need this P H coordinates? It turns out to be an interesting question. So, let us look at the processes that the refrigerant undergoes in the cycle So, four one is a phase change process and this as you can imagine is an isothermal process or isobaric process isobaric slash isothermal because it is a phase change process that is the process that the refrigerant undergoes in the evaporator. Now, two three is isobaric part of it is isothermal, but two three is isobaric pressure remains constant as the refrigerant loses heat to the ambient. Three four assuming that the change in specific volume as a result of throttling is not very large that the specific volume more or less remains the same you know that throttling is an isenthalpic process. So, this is an isenthalpic process. So, we can see that two out of the four processes in the cycle or constant pressure processes and one process is a constant enthalpy process. So, which means if I choose P and H as coordinates then the process will be straight line in the in a P H coordinates phase and the third process which is a throttling process will also be a straight line aligned with the axis. So, if I use P H coordinates then these two processes will be aligned with the coordinate axis and this will be aligned with the other axis which is why we would like to illustrate the vapor compression refrigeration cycle in a P H diagram also and this is extensively used in the refrigeration community. So, as you can see here four one evaporator constant pressure process at P E and 2 S or 2 to 3 is a constant pressure process at the condenser pressure three four throttling process constant enthalpy which is now aligned with the H equal to constant line. So, 1 to 2 S looks like this in case the process actually has internal irreversibilities still adiabatic then the state point shifts from 2 S to 2. So, the P H coordinates actually are extremely just as useful as the T S coordinates in depicting the process and it is very widely used in the refrigeration community. But in this course though we will stay with the T S diagram. So, this is an ideal vapor compression refrigeration cycle. So, these points are very important let me emphasize this again T H equal to T sat of P C, T C equal to T sat of P A. So, this is the T H is the temperature of the high temperature reservoir to which heat is rejected by the refrigerant, T C is the temperature of the low temperature reservoir to which from which heat is absorbed by the refrigerant. And saturated vapor at the exit of the evaporator saturated liquid at the exit of the condenser. Let us go through a work example now. So, an ideal vapor compression cycle using refrigerant R 134A we no longer use R 134A because it is not environmentally friendly we have much better refrigerants now. But we use R 134A in the course for purposes of illustration only. This can be replaced with any other refrigerant once the property values in the property tables are available. So, here it is given that this is an ideal vapor compression refrigeration cycle. Temperature of the refrigerated space is minus 5 degree Celsius. So, that means T C is equal to minus 5 degree Celsius. The refrigerant leaves the condenser as a saturated liquid at 30 degree Celsius. So, that means T H is equal to 30 degree Celsius. Refrigerant enters the compressor as saturated vapor at the rate of 0.0041 meter cube per second. We are asked to determine the power input rate of heat removal and CO P as well as the second low efficiency and rate of exergy destruction in the individual components. The isentropic efficiency of the compressor is given to be 73 percent. So, this is an ideal vapor compression cycle. Since it is ideal, state point 1 is a saturated vapor at entry to the compressor. So, V 1 equal to V G and H 1 equal to H G and S 1 equal to S G from the saturated temperature table. So, the mass flow rate of the refrigerant since volume flow rate is given, M dot is equal to V 1 dot the volume flow rate. Since volume flow rate is given, we may evaluate M dot equal to V 1 dot over V 1 and that is what we have done here. So, the mass flow rate comes out to be 0.0495 kg per second. H 3 is equal to H f at 30 degree Celsius because it is an ideal vapor compression cycle. S 3 may also be retrieved from the table and H 4 equal to H 3 because throttling is an isentropic process. So, H 4 equal to H 3. State 2 S is superheated at a pressure which is P sat of 30 degree Celsius. So, that pressure corresponds to 770.6 kilo Pascal. Let us just take a quick look here. So, state 2 S is a superheated state at a pressure P C which corresponds to the saturation pressure of T H or 30 degree Celsius. So, we can get H 2 S from the superheated table through interpolation and isentropic efficiency of the compressor is given to be 73 percent. So, from the definition of isentropic efficiency, notice that this is known, this is known, this is known. So, H 2 may be evaluated to be 280.3 kilo joule per kg and using this value of the specific enthalpy and the pressure we may interpolate and get S 2 to be 0.96285. What is that? This is more than the value for S 2 S. S 2 S is 0.9345. So, S 2 is equal to 0.96. So, the state, the exit of the compressor lies to the right of state 2 S. So, from steady flow energy equation, we may get the compressor power to be 1.6211 kilowatts and again from steady flow equation applied to the condenser, we can get the heat removed in the condenser to be 9.4264 kilowatts. And by applying steady flow energy equation to the evaporator, we may get the heat removed in the evaporator to be 7.62 kilowatts, but we have been asked to evaluate this in tons. So, 1 ton of refrigeration is equal to 211 kilo joule per minute, a heat removal rate of 211 kilo joule per minute. So, 1 ton of refrigeration is defined as the heat removal rate that is required to convert 1 ton of liquid water at 0 degree Celsius to ice at 0 degree Celsius in 24 hours. That is the definition for 1 ton of refrigeration and that corresponds to a heat removal rate of 211 kilo joule per minute. So, we may then convert 7.62 kilowatts to 2.167 tons. So, you may recall that when you go to the shop to buy an air conditioner for instance, you are normally asked I know what do you want a 1 ton AC, 2 ton AC and so on. So, that is what the ton that terminology that is this is what it means. So, C O P for the refrigerator is nothing but Q C dot divided by power input to the compressor and that works out to 4.702. Now, if you go back to the cycle, notice that and in the condenser, exergy is recovered and in the evaporator also exergy is recovered. Please go back and check the module on exergy, this is the low temperature reservoir which is actually from which heat is extracted. So, that means exergy is recovered in this case the sign for X and Q are opposite in this case. So, here X is supplied and X is recovered in these two devices. So, X supplied is equal to the power. So, X dot recovered is what we get in the condenser where heat is rejected. So, here the direction of heat flow and the direction of exergy flow are the same whereas, here the direction of heat flow and direction of exergy flow are opposite. So, this is X recovered that comes out to be 1.0088 kilowatts. Notice that T H in this case is 30 degree Celsius that is a temperature of the reservoir to which heat is rejected by the refrigerant. This is the temperature of the reservoir from which heat is extracted by the refrigerant. So, the second law efficiency comes out to be 62.23 percent. So, this aspect is very important. So, you must bear this in mind that this is exergy that is recovered. Now, if you apply entropy balance equation to the individual components, we can get the following things. So, rate of exergy destruction in the compressor works out to 0.4182. There is no heat loss from the compressor because it is still adiabatic although internally irreversible. So, this loss of exergy or entropy generation is due to internal irreversibility in the compressor. No external irreversibility because it is insulated. Rate of exergy destruction in the condenser again this is heat transfer across a finite temperature difference. So, this is due to external irreversibility. So, you may see that when the refrigerant enters the condenser, it is at a much higher temperature compared to T H. So, there is heat loss across a finite temperature difference which is why we have entropy generation or exergy destruction in the condenser. Throttling as we already saw is an internally irreversible process. So, the exergy destruction is due to internal irreversibility, but if you look at the exergy destruction in the components, you can see that the maximum contribution comes from the compressor, others contribute quite less than what we get from the internal due to internal irreversibility in the compressor. Now, we have looked at the ideal vapor compression refrigeration cycle. In the case of an actual cycle, it can never be guaranteed that the state at the exit of the evaporator will always be a saturated vapor. In general, the state at the exit of the evaporator is slightly super heated. Similarly, we can never always guarantee that the state will at the exit of the condenser will be a saturated liquid. It is almost always slightly compressed or we may say this is a compressed liquid state. And notice in this case that T H is not the same as T sat of P c, T c is not the same as T sat of P e. So, T H in this case is less than T sat of P c and T c is I am sorry greater than T sat of P e. Now, T c has to be greater than T sat of P e because that is the only way heat can be extracted by the refrigerant from the refrigerated space. Similarly, T H has to be less than T sat of P c because that is the only way heat can be transferred from the refrigerant to the ambient or high temperature reservoir. So, we see here the actual cycle illustrated on a P H diagram. The open circles indicate the ideal vapor compression cycle. And the actual compression cycle as you can see here it is slightly super heated and here it is slightly sub cooled and that may be a better analogy. So, compressed liquid or slightly sub cooled. Now, let us see what changes in the performance matrix this makes in going from ideal cycle to the actual cycle. So, we will redo the previous example, but now we take the evaporator and condenser pressure to be 200 and 900 kilo Pascal respectively. So, this means that T H remains at 30 degree Celsius and T c remains at minus 5 degree Celsius, but we have changed the evaporator and the condenser pressure. So, interestingly you can see that T sat corresponding to the evaporator pressure is minus 10 degree Celsius. So, you can see that T c is greater than T sat corresponding to T e. So, that heat can be abstracted by the refrigerant from the refrigerator compartment. So, refrigerator compartment is maintained at minus 5 degree Celsius, but the refrigerant is at a temperature of minus 10 degree Celsius when it enters the evaporator. So, that means heat flows from the refrigerator compartment to the refrigerant. Similarly, T sat corresponding to the condenser pressure is 35.5 degree Celsius whereas, the ambient temperature is 30 degree Celsius. So, heat transfer can comfortably take place between the refrigerant and the ambient. So, state 1 now is slightly superheated. So, from the superheated table we can retrieve appropriate property values H 1 and S 1. H 2 S may be evaluated in the same manner as before by using the fact that S 2 S equal to S 1 and P 2 S equal to P c the condenser pressure. So, once we do that we can evaluate H 2 or we can evaluate H 2 by making use of the definition of the isentropic efficiency of the compressor and H 2 comes out to be 292.82 kilo joule per kilogram. And the corresponding value for specific entropy may be evaluated by interpolation from the superheated table. So, the power required now comes out to be 2.177 kilowatts. Now, state 3 is a subcooled liquid. Notice that state 3 at state 3 the pressure of the refrigerant is 900 kilo Pascal temperature is 30 degree Celsius. So, it is a compressed liquid or subcooled liquid and we evaluate the enthalpy using this approximation. So, this is actually an approximation and the specific enthalpy comes out to be 93.688 and S 3 comes out to be equal to S f approximately at 30 degree Celsius. Now, H 4 equal to H 3 still because it is an isenthalpic process and dryness fraction at state 4 may be evaluated from here and using this dryness fraction S 4 may be evaluated just like what we did before as 0.3646. So, notice that from state 3 to state 4 the specific entropy increases as it should because throttling is an isenthalpic process. So, these are consistency checks that you can use when you are doing the calculation just make sure that the numbers that you are getting or what they ought to be at least qualitatively. So, rate of heat removal from the refrigerator space is 2.184 tons compared to 2.167 tons. So, more heat is removed now mass flow rate of the refrigerant remains the same. So, the reason for this is that state point 3 has now moved to the left of where it was before which means that state point 4 is going to be further to the left of where it was before. This is very evident here. So, which means if it is over here then the amount of heat that can be removed from the new state point 4 to the new state 1 is going to be more than what it was before. So, earlier state point 4 was over here and state point 1 was over here. So, the amount of heat that could be removed earlier was less than what it is now. So, the COP for the cycle comes out to be 3.53 which is less than what we had before. The heat removal has increased. However, the power has also if you look at the power this is 2.177 and earlier we had 1.6211 kilowatts. This is 45 percent less than the COP for the ideal cycle. So, heat removal in the condenser again from steady flow energy equation comes out to be 9.857 and once again exergy supplied in the cycle is equal to the power that is given to the compressor. So, that is 2.177 and exergy recovered from the condenser and from the evaporator. So, together it comes out to be like this. Notice that T H is still 30 degree Celsius, T C is still minus 5 degree Celsius as we wrote down here. So, T H is still 30, this is still minus 5, I am sorry. So, the second law efficiency comes out to be 47 percent only which is actually 45 percent less than the second law efficiency for the ideal cycle. Again, this is due to the fact that this is due to the fact that you know the if you look at the cycle you can see that you know now the temperature differences in this case or external irreversibility increases in the condenser. So, the loss of or the destruction of exergy is going to be higher in the condenser. So, earlier all the exergy destruction was in the compressor. Now, the exergy destruction in the compressor more or less remains the same, but the exergy destruction in the condenser has increased considerably and the exergy destruction in the throttling valve also increases somewhat because of the fact that it is now entering as a slightly sub cool liquid. Because as I said before the vapor compression refrigeration cycle is used mostly in domestic applications refrigerated air conditioners. It can also be used in for cooling of buildings and so on, but a single refrigerated single cycle or a cycle of the single refrigerant will not be adequate for such purposes. So, normally you either use a cascade refrigeration system which which cascades one refrigerant cycle on top of on top of the other. So, that is that is one way of doing this because there are also there is also something called a multi stage refrigeration cycle which uses multiple stages to to do this both are acceptable basically the idea is. So, if you go back to the block diagram the basic idea is this heat rejection in the case of a multi stage or cascade sort of cycle this heat rejection does not take place to the ambient, but it actually this condenser for one cycle becomes the evaporator for the next cycle and that the condenser for the second cycle which is the upper cycle will then reject heat to the to the ambient. So, that is how these are coupled and together this can give very good performance and can be used for even larger applications than the refrigerator or air conditioner. Of course, one important aspect about the vapor compression refrigeration cycle and indeed refrigeration is the use of environmentally friendly refrigerants. We have not paid any attention to that when you do a course on refrigeration and air conditioning the details and the impact of refrigerants on the environment will be discussed in great length. It is very important because the refrigerants have the potential to cause damage to the ozone layer and also cause global warming. So, we need environmentally friendly refrigerants which is what we are using nowadays plus other refrigerants which are non flammable which have much better environment friendly properties and so on are also being developed. These are usually discussed at the next level course on refrigeration and air conditioning. So, this concludes our lectures on thermodynamic cycles where we looked at Rankine cycle, air standard cycles where within which we saw we discussed air standard Brayton cycle, air standard Otto and air standard Diesel cycle and then vapor compression refrigeration cycle which is the only power absorbing cycle among all these cycles. The next module that we will take up is psychrometry.