 The last cycle that we are going to look at is a reverse cycle. So, this is a power absorbing cycle as we had mentioned earlier. So, the refrigerant executes cyclic process in this and interestingly enough even in the real life refrigerator or air conditioner the refrigerant executes cyclic process. It may be treated as a system and it executes a cyclic process unlike the Brayton cycle. Now in the case of the Rankine cycle again even in real life installations the water executes a cyclic process. It is in a separate circuit and it executes a cyclic process. There is some makeup water that is provided at periodic intervals but apart from that we may consider that the water in the case of a Rankine cycle executes a cyclic process. The Brayton cycle is the one which is usually not run as a cyclic process air is drawn in and then combustion gases are exhausted. In the case of the vapor compression also refrigerant executes a cycle. If we represent this cycle in TS coordinates this is what the cycle looks like. So, again this also operates between two pressures. So, this is the condenser pressure denoted Pc and this is the evaporator pressure denoted Pe. So, we take this to be the starting point of the cycle state 1. So, here the refrigerant as you can see is a saturated vapor and it is compressed to from the evaporator pressure to the condenser pressure. In the case of an ideal process we end up at state 2 S. In an actual process with which is adiabatic but with internal irreversibilities we will end up at state 2. The high pressure high temperature refrigerant is then cooled at constant pressure in the condenser until it becomes a saturated liquid and this temperature is taken to be the highest temperature I am sorry this is taken to be T H for purposes of calculating entropy generation. So, in the case of the vapor compression cycle this is taken to be the saturation temperature corresponding to the condenser pressure is taken to be T H. So, the refrigerant leaves the condenser as a saturated liquid it is then throttled in a throttling valve as we have shown here. So, it is then throttled in a throttling valve to low pressure and low temperature the low pressure being the evaporator pressure and it becomes a saturated mixture at the end of the throttling process. So, the evaporator temperature or the saturation temperature corresponding to the evaporator pressure is taken as T C and the refrigerant now picks up heat from the refrigerated compartment. So, as it picks up heat it also evaporates until it becomes a saturated vapor and the cycle is repeated. So, let us look at an example of a vapor compression cycle. So, an ideal vapor compression cycle using R 134A operates with the refrigerated space at 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 this is equal to T H. Refrigerant enters the compressor as saturated vapor at the rate of 0.0041 meter cube per second determine the power input to the compressor rate of heat removal from the refrigerator space in tons as COP and the rate of entropy generation. So, we assume the cycle to be ideal and take the isentropic efficiency of the compressor to be 100 percent which means the process the compression process is isentropic and goes from 1 to 2 S. So, from the temperature table corresponding to saturated vapor we can pick up specific volume at state 1, specific enthalpy at state 1 and specific entropy at state 1. Now, since the volume flow rate is given at inlet to the compressor we can evaluate mass flow rate M dot. So, M dot is equal to V 1 dot divided by V 1. So, we may evaluate the mass flow rate as 0.0495 kilogram per second. Now, it is given that the refrigerant leaves the condenser as a saturated liquid which means H 3 is equal to H F at 30 degree Celsius and S 3 is equal to S F at 30 degree Celsius. Now, the throttling process undergone by the refrigerant is actually we can take it to be an isentropic process. So, we take H 4 to be equal to H 3. Now, we apply steady flow energy equation to the individual components and evaluate the quantities of interest. Now, before we do that we need to actually evaluate the entropy I am sorry we need to evaluate H 2 S. So, we know H 1, H 3 and H 4 we need to know H 2 S. So, you can see from here that S 2 S is equal to S 1 and P 2 S is also given. So, that is P sat of 30 degree Celsius. So, P 2 S is equal to P sat of 30 degree Celsius which is 776 point kilopascal. S 2 S specific entropy is equal to S 1 equal to 0.9345. So, you can see that and the state 2 S is superheated and we can actually calculate H 2 S to be 271.46 kilo joule per kilogram by interpolation. So, the compressor power required from SFEE comes out to be 1.184 kilowatts. Now, of course the most important performance parameter in the case of a refrigerator is the amount of heat that is removed from the refrigerator space. So, we may evaluate that as m dot times H 1 minus H 4 and that comes out to be 7.622 kilowatts. Now, in the refrigeration community it is customary to express heat removed from the refrigerator space in units of tons rather than kilowatts. So, generally we say air conditioner is 1 ton or 1.5 ton and so on. But the definition of a ton of refrigeration is given here. One ton of refrigeration is defined as the heat removal rate that is required to convert 1 ton of water. 1 ton here is 910 kilogram. It is not 1 metric ton, it is 1 ton which is equal to 910 kilograms of liquid water at 0 degrees Celsius into ice at 0 degrees Celsius in 24 hours. So, it is the rate at which heat must be removed from 1 ton of water at 0 degrees Celsius to convert it into ice at 0 degrees Celsius in a span of 24 hours and is equal to 211 kilojoule per minute. So, if I convert this to tons I get this to be 2.167 tons of refrigeration effect. And the COP for the cycle you may recall is defined as QC dot divided by WX dot compressor and that comes out to be 6.44 for the ideal refrigeration cycle. This is somewhat high because the compression process is taken to be ideal. Practical cycles will have COP values much less than this. Now rate of entropy generation in the universe again is a sum of delta S system plus delta S surroundings. And we have taken the R134A to be our system and since it executes a cyclic process delta S system is 0. Delta S surroundings comes I mean delta S surroundings is the sum of two things. One, the heat that is rejected in the condenser. So, the entropy of the surroundings increases as a result of heat that is rejected in the condenser. And the entropy of the surroundings decreases as a result of heat that is absorbed by the system in the evaporator. Accordingly this has a negative sign and this has a positive sign. TH and TC are known. So, if you substitute the values sigma dot comes out to be 0.6193 watt per Kelvin. So, in the previous two examples we did all the calculations on a per unit mass flow rate basis. Here mass flow rate is given. So, this comes out when units of watt per Kelvin. So, this is the rate at which entropy is generated in the universe as a result of operation of this cycle. So, as I said this is an ideal cycle. So, we have taken the isentropic efficiency to be 100 percent. In reality we have to account for internal irreversibilities in the compressor. Compressor efficiency here will typically be of the order of about 70 percent or so, which is quite far from the 100 percent that we have assumed. And other additional complexities are also there in the cycle. We can never guarantee that the refrigerant will leave the evaporator as a saturated liquid. Usually it leaves the evaporator as a slightly super heated vapor that needs to be taken into account. And once again it can never be guaranteed that the refrigerant will leave the condenser as a saturated liquid. Typically it leaves as a compressed liquid with the state being over here. So, here the state is usually over here, here the state is over here. So, these sorts of departures from the ideal cycle in real life have to be accounted for when we do calculations for real life situations. So, that concludes the lectures for this course. I hope that you found these lectures to be useful and that you are able to understand and learn concepts in an unambiguous manner. Hopefully the lectures would have helped you learn the concepts in a much better manner than what you may have understood earlier. If you had whatever you had seen before, hopefully you would have learned better now and whatever new concepts you have learned, hopefully you have understood the fundamentals and you would remember them as you go through the higher level thermodynamics courses. I hope you enjoy the lectures. I wish you success in your pursuits.