 So, there is two problems or first one is Rankine cycle, second one is Bledam cycle. These two are proper producing cycles where heat input is given, a heat rejection happens, but the net heat given to the cycle is positive. Similarly, the net work done by the cycle is also positive. But this third cycle is vapor compression refrigeration cycle. Here the main objective is to transfer heat from low temperature regime to the high temperature regime. So, that is the thing here. So, here you can see that R134A is the working fluid and the problem is in a vapor compression refrigeration cycle operating at steady state with R134A saturated vapor enters the compressor at 1.6 bar and saturated liquid leaves the condenser at 9 bar, ok. Mass flow rate of the refrigeration is 5 kilogram per minute. So, these are the given information. Determine the compressor power in kilowatts if the isentropic efficiency is 67 percent and second coefficient of performance, COP, third is the refrigeration capacity we can say tons of refrigeration, refrigeration capacity in tons of refrigeration. So, here first we have to understand the basic control volumes, basic control volumes in the vapor compression refrigeration. So, basically there is a space to be cooled. So, let us say there is a space to be cooled. So, in order to keep this space at a given temperature Tc, I have to neglect or reject heat say q dot c to a device in the system which is the which we call evaporator, ok. Now, then this evaporator the refrigerant R134A comes in some state, but exits as at least a saturated vapor, ok. So, here the condition is at least a saturated vapor, then this vapor is compressed. So, that means I need to supply some compressor work here, this is vapor compressor, so that is what the name is vapor compression refrigeration cycle. So, I have to compress this vapor from this operates at a say a pressure, so this is state 1 let us say, so this operates at a pressure of P1. So, evaporator is a constant pressure operation it makes, where the pressure is P1 and the vapor compressor will compress the vapor which is a saturated vapor at least I told you know that to state 2, high pressure. And now this, so it is given clearly is saturated vapor enters, saturated vapor enters the compressor. So here I can say x1 equal to 1, it can be superheated also, ok. But please understand this name vapor compressor, I cannot have a saturated mixture entering here that will not work properly. So, at least a saturated vapor or a superheated vapor should go into it, the problem clearly it is given that saturated vapor enters. So, now it goes to higher pressure, so P2, P2 is the higher pressure, so again the pressure ratio of this system is P2 by P1. Now, pressure is also given, P1 is given as 1.6 bar and P2 is given as 9 bar. So, this is the pressure ratio, P2 is given as 9 bar, ok, 9 bar. Now, a device called condenser, condenser removes, see for example, here the pressure ratio is 9 bar, it is a higher saturation temperature. Now, this can reject the heat at rate of q dot h to an ambient which is at T h, ok, because at P2 the saturation temperature is higher, so it can reject heat to higher temperature. So, why it choose a lower pressure here? Because Tc is low, it has to take heat from a lower temperature regime, so the saturation temperature of the working substance in the evaporator should be lower than this. So, lower pressure, lower saturation temperature, higher pressure, higher saturation temperature, so that is why the P2 and P1 ratio will be fixed appropriately. Now, exiting this basically, so this is 2 and this is 3, exit of this is it is given as saturated liquid, so here x3 equal to 0, it is given in the problem. Now, I have a high pressure 9 bar saturated liquid, but evaporator will operate at 1.6 bar, constant pressure, this is also constant pressure device, constant pressure operation like evaporator, ok. Now, I have to reduce the pressure of saturated liquid at 9 bar to 1.6 bar, so what I do is I just put a adiabatic throttle valve, I just put that device which will it is actually irreversible, irreversible operation, irreversibly it will reduce the pressure of the saturated liquid at 9 bar to 1.6 bar, in that way you will get a saturated mixture. So, this state this is state 4, state 4 will be saturated mixture at 1.6 bar, that is all. So, from the saturated liquid at 9 bar, I get a saturated mixture of liquid vapor at 1.6 bar and then now this liquid will evaporate and get a saturated vapor at the end of the evaporator and this saturated vapor goes and the cycle continues. This is the schematic of the vapor compression, if you do not think you know this, correct. So, now the properties are also given, ok. For example, saturated R134A, the pressure 1.6 and 9 bar, the quantities like Hf, Hg, Hf, Hg are given. Similarly, at 9 bar, superheated the for different temperatures H and S values are given. Now, I draw the T as diagram for this. So, now let us draw, this is 1.6 bar, this is 9 bar, ok. Now, state 1 is the, this is saturated vapor at 1.6 bar that enters the compressor. So, now, compressors can operate in an isentropic manner, correct, straight S1 equal to S2, S. Obviously, in the problem, the isentropic efficiency of the compressor is given. So, it is a, maybe the access state 2 will be here and this is, this process is not known to me. So, I can draw this. Now, at 9 bar, we get a superheated vapor at the end of the compressor, then the condenser, this is the condenser where heat is lost, q dot h, to the temperature of T h. So, now, the condensation occurs, it becomes saturated vapor, then the light energy is removed. So, finally, I get a state 3 which is saturated liquid at 9 bar. So, this is the process and this is compression process isentropic and this, ok. Now, this state is fixed, this state is fixed straight away. To fix this state 2, I will first assume a isentropic compression process and get the state 2 S, apply the isentropic efficiency to get the state 2. So, these states are fixed, ok. What about final state? Final state 4, I told you it is a saturated mixture. So, it should come somewhere in this, pressure has decreased from 9 bar to 1.6 bar. So, how to get this we will see, ok. So, now, this is the cycle. Now, let us fix the states. State 1, 1.6 bar x equal to 1. So, now, go to the tables. 1.6 bar saturation x equal to 1 that is hg, sg I have to take, ok. So, 237.97, 0.9295. So, h1 equal to 237.97 kilojoule per kg and s1 equal to 0.99295 kilojoule per kg Kelvin. State 1 is very straightforward. Now, state, I told you no, state 2 can be fixed only after fixing state 2 S. So, state 2 S I know p equal to 9 bar, 9 bar under s2 S isentropic process equal to s1 equal to 0.9295 kilojoule per kg Kelvin. Now, go to the saturation table for 9 bar. I find that sg is 0.9054, correct. So, sg at 9 bar equal to 0.9054 is what? Basically, is less than s2 S is 0.9295, correct. So, that means, state is, state 2 is superheated, ok. So, how will you find this? Go to the superheated tables. So, superheated table for 9 bar S, s2 S equal to 0.9295. So, you can see that this is between these two, correct. Interpolating between 1450 degrees from the superheated table of 9 bar, ok. Interpolating for s equal to 0.9295 which occurs in the temperature range of 40 degrees centigrade and 50 degrees centigrade. I get h2S equal to 273.73 kg per, sorry kilojoule per kg, kilojoule per kg. That is it. Now, isentropic efficiency I have to apply. Isentropic efficiency of the compressor equal to what? The isentropic work given to the compressor, because work is now given, correct. So, this actual work. Obviously, for work absorbing device like compressor, as I saw in the previous problem, isentropic work will be lesser than the actual work. So, for compressor, which is adiabatic, q dot minus w dot c equal to m dot into h2 minus h1, correct. Or I will say wc equal to minus w dot c by m dot equal to h2 minus h1. So, I can apply this. For isentropic, it is h2S minus h1 divided by this is h2 minus h1. Okay, h2S I know, h1 I know. So, I can find h2S, h2S 0.67 is the isentropic efficiency into sorry h2S 273.73 minus h1 237.97 divided by 0.96 plus 237.97. So, you can get this. Okay. Now, I want what is the actual work, actual specific work input to the compressor equal to what? Wc equal to h1 minus h2, sorry h2 minus h1. I am talking about the input, work input, which is absolute value. So, that will be equal to 53.69 kilojoule per kg, for absolute value is this. Do you understand? So, that is what we have got. Substituting this h2 value here and h1 is no. So, that I know. Now, what is or I can do it in power, compressor power Wc equal to m dot into Wc, which is equal to m dot equal to 5 kg per minute, it is given 5 kg per minute divided by 60 into 53.69 kilojoule per kg equal to 4.474 kilowatts. Okay, done. So, compressor part is over. Now, what is Qc? For that I need the state 4. Correct. So, state 4 for adiabatic throttling device. Okay. This is 3 to 4. Correct. Q dot minus Wx dot equal to m dot into h4 minus h3. But this adiabatic, so this is 0. Throttling device, no work is there. So, that is also 0. So, which implies h4 equal to h3. So, now what is h4? h4 equal to h3 equal to hf at 9 bar. So, go to the tables. So, 9 bar, this table saturation hf 99.56. So, 99.56 kilojoule per kg and p4 equal to p1 equal to 1.6 bar. So, that fixes the state 4. So, straight away I get this value. So, now what is Q dot c equal to m dot into here? Q dot c is, so now state 4 is fixed here. This is state 4. So, this is, as I told you, irreversible process. Okay. State 4, I have to find, straight away I got the enthalpy value that is enough for me. So, I do not need to fix this. Somewhere it will come here. So, now, 4 to 1 is the process where it takes heat from the cold space, correct, Q dot c. So, what is Q dot c? Q dot c will become m dot into h1 minus h4, okay, which is equal to h1 is through 37.97 minus h4 99.56, okay. So, this is equal to 138.41, heat added, heat added to the cycle, okay, in the cold space. So, now, sorry, this is into mass I have to put, into 5 by 60. So, which is equal to 5 by 60 into 138.41, which is equal to 11.534 kilowatts. You understand? So, that is the heat removed. Heat removed from the cold space or heat input to the cycle. So, now, what is COP? COP of the refrigerator will be equal to objective energy transfer, okay. This is the objective energy transfer for the cycle divided by W c dot, which is the energy transfer that costs. That is, we have to pay for this, you know, we have to pay for this. So, this is equal to 11.534 divided by 4.7474. So, we calculated this, this is 4.474. So, that will be equal to 32.578, that is COP. So, please see this, this problem. First state is fixed as saturated vapor. Second state is fixed using the first isentropic process and using the isentropic efficiency. Third state is given as a saturated liquid at 9 bar. Fourth state, we have to apply the first law for the throttling device where throttling device does not have any heat loss or heat gain or does not do any work. So, it is an isoenthalpic device, isenthalpic device. So, H3 equal to H4. So, from that I can fix the state 4. It will look like that. And now, what is Qc? Qc equal to Q1 minus Q4, is it added? And W is Q H2 minus H1, that is a work input. So, from that Q dot c by W dot c will give you the coefficient of performance. Now, finally, what is tons of refrigeration? Tons of refrigeration for this, that is a simple thing, ok. One ton of refrigeration is equivalent to 3.52 kilowatts of heat removed from the cold space, ok. So, that means, if from the cold space 3.52 kilowatts is removed, that will correspond to one ton of refrigeration. Now, from the cold space, 11.534 kilowatts is removed. So, for Q dot c equal to 11.534 kilowatts, ok, the refrigeration current, the refrigeration capacity in tons is equal to 11.534 divided by this 3.52, which is equal to 3.276 tons of refrigeration. So, this is the value. So, in this problem, we can see that there are four components, three. Basically, the third wall is irreversible, vapor compressor also acts irreversibly. So, because I think the efficiency is given. And evaporator condensers are constant pressure processes. And the states are given for state 1 as a saturated vapor at 1.6 bar. And for the condenser exit, it is given as saturated liquidate 9 bar. To fix the state 2, we have first found ideal state 2 as due to isentropic compression. Then isentropic efficiency is used to find state 2. State 4 is got from applying the first law to the throttling process. So, once the states are fixed, then first law will lead to the calculation of the W and Qc. W was found as 4.474 kilowatts and Q dot c is found as 11.534. CUP is Q dot c by Wc for refrigerator. Using that, I got about 2.6. Similarly, it is about 3.276 tons of refrigeration, keeping in the mind the definition that 1 ton of refrigeration is equal to 3.52 kilowatts of heat removed from the cold space. So, this is about this problem. So, these are the three problems in the thermodynamic cycles. Thank you.