foreign [Music] of thermal engineering basic and applied today we shall start our discussion on this new topic that is vapor compression refrigeration cycle so if we recall our previous classes then we can understand that we have talked about Vapor power Cycles and we have discussed about those Cycles essentially in the context of the development or generation of power but today we shall discuss about this particular cycle that is vapor compression refrigeration cycle essentially a vapor power cycle but the sole purpose is not to produce power but to produce Refrigeration or refrigerating effect so as I said the sole purpose is to produce refrigerating effect so purpose of this particular cycle is to produce refurting effect now let us look into this particular cycle and whistle try to analyze the processes those constitute this particular cycle so as we can see the sole purpose is to produce refrigerating effect and we had seen from our from our daily life experience from household refrigerator that a compartment needs to be at a temperature always which is say for example at minus 10 degree Celsius or minus five degree Celsius temperature so you know if we schematically try to you know draw this particular cycle then essentially this is the space say for example and this space should be always maintained at a temperature which is Sub Zero temp at a sub zero temperature so this is SEC coal space so if you need to maintain the temperature of this particular space at a temperature which is for example if it is minus 10 degree Celsius then certainly you know that to run this we need to have a few mechanical components electrical components so as those components will be running heat will be generated and so we need to have always transfer of it from this particular space to somewhere so as to the temperature of this particular phase will be maintained at a sub zero temperature now this heat will be taken by a device that is called evaporator so this is Q in and if we recall the steam power cycle we had seen that the walking fluid that we considered for that particular cycle is steam water or water steam water mixture or water now in this particular cycle as well there is a working fluid but that working fluid is not watered that walking fluid is a special type of walking fluid and that working fluid is known as refrigerant so I am writing the working fluid is a refrigerant this is a special type of working fluid we shall be discussing later what kind of property this particular fluid should have and this particular type of you know working fluid is you know given a name start with you know R and there are different you know types of reference we shall be discussing this particular you know this particular type of fluid later now you know that working fluid will be on receiving Heat from this cold space that working fluid will be evaporated and that is why this device is known as evaporated so this particular working fluid will be having a few you know distinctive properties and those properties we shall be discussing later in today's class so now on receiving heat that walking fluid will be converted into vapor and essentially this evaporator is designed in such a way that at the exit of the evaporator will be getting saturated Vapor of that working fluid so if we give name say this is one so at one if we write at one the refrigerant would be saturated refrigerant saturated state and then as I said you that we need to complete the processes and those processes with those Pro those processes will eventually form a cycle so now that saturated liquid is taken into a compressor so this device is basically a compressor so this is a compressor on so this is a device which is work absorbing device this is not like a turbine we had seen that turbine is a work producing device but to run this compressor we need to supply certain energy always in the form of work so this is W in so we are supplying work to this compressor and then this saturated refrigerant saturated Vapor to be precise this is not the saturated liquid so this is saturated Vapor of this referral and because you know evaporator refrigerant will be evaporated so essentially you will be getting saturated vapor at the exit of the refrigerator and that is at the inlet of this compressor now that the sole purpose of providing with this compressor is to develop or build up a pressure and we will be getting you know if this is State point two so high pressure refrigerant Vapor that would be available at the exit of the compressor so essentially compressor is a device which can handle two phase mixture certainly but we had seen in a steam power cycle a pump is there to raise the pressure of the working fluid from a condenser pressure to the boiler pressure so now that high pressure saturated high pressure you know refrigerant Vapor will be now taken into a into another device that is called condenser and while passing through the condenser it will release Heat so Q out because we need to bring back the original state of the evaporator original state of the refrigerant at the inlet of the evaporator so evaporator on receiving heat from this cold space got evaporated that refrigerant Vapor that is saturated vapor is now taken into this compressor wherein you know we are supplying energy to run the compressor and the pressure of that you know ah refrigerant will increase and that refrigerant which is available at the exit of the compressor is now taken into this condenser and by releasing Heat while passing through the condenser the refrigerant will be taken to another device so if we give name this is point three now question is in this case we could have you know install another one device to produce work and that is what we had seen in a steam power cycle so you could have you know in a steam power cycle we had seen that ah turbine basically steam is allowed to expand so why steam is expanding it does work on the rotating part of the wheel and we are getting walk output rather we get work output so here what is done instead of a turbine a special type of device that is considered because here the purpose is is not to produce over but to Forget You Know Refrigeration or refrigerating effect so the special device which is considered Here is known as throttle valve so this is known as throttle valve and the sole purpose of providing this throttle valve is the sole purpose of providing with this throttle valve in the circuit is to you know expand that high pressure refrigerant that is coming that comes out from the condenser to the evaporating to the evaporator pressure so this is the you know total circuit so let me tell you once again we are getting high pressure refrigerant at the inlet of the condenser by releasing certain amount of feed from the condenser that refrigerant which is available at the exit of the condenser is taken into another device that is a special type of device perhaps you have studied in thermodynamics that is called throttle valve so why will stream is you know passing through this throttle valve it will expand and this is as you know very fast process pressure will decrease and the purpose of providing with this throttle valve in this circuit is to reduce the pressure of the refrigerant from condenser pressure to the evaporator evaporator pressure so this is the circuit now as I said you that ah sole purpose is to get the refrigerating effect and we have also discussed many times that whenever we consider any mechanical component or mechanical device and those mechanical components when when those components are there in a circuit or in a cycle essentially if you need to measure the performance of the cycle itself we need to map all those processes in several thermodynamic planes one of the most important thermodynamic planes is the TS diagram wherein we can represent all these processes we have discussed that if we can represent any process in t s plane and area under the process line in t s plane will give us at the direct measurement of energy that is either added to the you know device or energy getting extracted from this device so following you know this discussion which we had earlier and also with this understanding let us now move to draw the TS diagram corresponding to this cycle rather corresponding to the processes those are there to constitute this cycle so if I try to draw the t s plane so now try to understand this point is four and this is this cycle so cyclic process what we have discussed that this evaporator will be designed in such a way that at this evaporator pressure and of course when the circuit or cycle is designed certainly evaporator is designed to you know produce saturated refrigerant vapor so now State point one is saturated state that is saturated vapor of the refrigerant right so that means let us consider point one is here so this is point one this is point one then again evaporator pressure is less than the pressure at which condenser is operating so we need to raise the pressure of the working fluid from evaporator pressure to the condenser pressure and that is why this compressor is there so the compressor will now compress this saturated vapor from State point one to State point two following if we assume that it is a reversible adiabatic compression so you know basically I should say that with this assumption that the process that will occur in this compressor is reversible adiabatic process so with this idealized assumption I should say that this is ideal Vapor compression refrigeration cycle so you know this is I am writing ideal Vapor compression refrigeration cycle why because you know in Practical case it is very difficult to have a process which should be reversible adiabatic but here we are considering this particular assumption and that is why we are writing this is ideal Vapor compression refrigeration cycle because this is an idealized assumption not only here you also will consider a few more assumptions to analyze this cycle and that is why I said considering a few idealized assumptions so now so you know this is so this is the evaporator pressure and say this is State point two that is the compression process and if we consider this is the you know this pressure that is p condenser and this is P evaporator right so what we can understand that now one process one to two is reversible adiabatic process the compression process now straight point two is the exit state of compressor or Inlet of the condenser now that is there is basically you know condensation but we have also discussed that even in the context of steam power cycle we have discussed about this particular aspect that designing a designing a condenser which will allow us to have partial condensation is very difficult so here also the condenser is provided with this cycle or circuit to allow not the partial condensation but to have you know the condensation up to this saturated liquid so this this is State point three that is on the saturated liquid line and this is the process now consider one important case that what about the process three two four process three to four that we can see from the schematic that is the throttling process the throttling process is a very fast process so it is basically you know kind of the expansion of the refrigerant vapor and the sole purpose is to reduce the pressure of the working fluid from condenser pressure to the evaporator pressure now perhaps you have studied this particular you know topic in your basic thermodynamics course the throttling is a very fast process since it is very City since it is a very fast process so the intermediate States may not attain equilibrium and hence it is very difficult to you know math this process by a solid line because it is very difficult to understand by which rather it is a difficult to you know get the paths by which the state point three will now change from this to State point four let me tell you once again circling is a very fast process and the intermediate States may not attain equilibrium hence if we need to map the process three to four in the TS plane here then it is very difficult to get the path by which the state point three has now changed to State point four and hence this particular process is shown by this dotted line so this is four so this is basically one two three four that is the you know four different state points we could map in this t s plane let me tell you one thing because ah whenever we discuss whenever we have discussed any cycle in the context of steam power cycle also in the context of you know gas power cycle we have assumed that all the processes are internally reversible those processes may not be externally reversible but we could map all those processes assuming that the processes are internally reversible and hence we could use this solid line but in this particular case since throttling is a process which is not an internally reversible process as well so that is why this process is shown by this dotted line what we can understand from this test plane also from this you know schematic defection of this ideal Vapor compression refrigeration cycle that the process that is there in this evaporator essentially you know conversion of State point four into State point one so that process occurs at a constant pressure so we can see from this ah t s plane on the other hand you know refereal and Vapors changes its state from State 2 to State 3 in this condenser and the process occurs at a constant pressure process at constant pressure so this is this is also a constant pressure process so these two processes are constant pressure processes while the process that is you know there in this throttle valve that is not we can say that the enthalpy is remaining constant all throughout the process but enthalpy at State point three is equal to enthalpy at State point four so enthalpy before throttling is equal to the enthalpy after throttling so basically you know if we write here that enthalpy before throttling is equal to the enthalpy of the working fluid after throttling so that means H three equal to h four Maybe the enthalpy at intermediate States at intermediate state or enthalpy at many intermediate States made not be equal but enthalpy before throttling is equal to enthalpy after throttling so what we can understand out of these four processes two processes occur at a constant pressure while another one process that is very important process an interesting process as well which is occurring at constant enthalpy so next objective should be to plot all these processes or map all these processes in another plane that is called pH plane so try to understand for the steam power Cycles we had also tried to map all the processes in PV diagram but here since two processes are at constant pressure and one process is at not constant enthalpy but enthalpy before startling is equal to enthalpy after throttling so we can map the processes now all these processes in this pH plane pressure enthalpy you know plane so if we now map these processes so again starting point should be one because that is the saturated vapor at the exit of the evaporator so if we consider this point is one right certainly pressure to be you know ah constant pressure so let us assume this is the pressure line we really do not know where the fire point four would be but we can understand from this t s plane that you know 0.4 is basically you know ah it is ah so at the inlet of the evaporator the straight point is not the saturated liquid so it is two phase mixture so now we have identified State point one here then one to two that is again reversible adiabatic process so question is what would be the enthalpy of State point two so now we also can take point two ah towards right or where the where point two should be there in PH plane so if we apply this tedious relation that is t d s equal to d h minus v d p so process one to two is foreign so d s equal to zero isentropic process so if d s equal to zero then d h equal to v d p compression process this fellow is always positive and what about this quantity because d p is also positive for a compression process because the sole purpose of having this process is to raise the pressure in the direction of working fluid flow so deep is positive so essentially d h is positive that means enthalpy will increase so from this you know discussion now we can map we can identify State point two say this is State point two so this is State point two what about process two two three that is at a constant pressure process but Point c as I said you that partial condensation is not possible so we can have the process two two three that is a constant pressure process and this point is three right now three to four is basically you know the throttling process so we need to map this three to four process so what I said you that enthalpy at State point three is equal to Intel enthalpy at State point four so that means this process should be like this so this is point four right so this is the pH ah representation so representation of all these processes in PH diagram or pH plane so we have discussed now I forgot to discuss one important point question is comfortably I could look at point four here now where is the guarantee that point force should not be closer to the saturated liquid line because then if 0.4 is closer to the saturated liquid line certainly entropy at State point four will be the deciding factor to look at point four in this plane so if we can somehow get a clue about the change in entropy of this particular process from there we can comfortably locate point four in this plane now let us again do this for this particular case so again for process three two four we can write this t d s equal to d h minus v d p right now State point three and state point four as I said you that if we if you you know start your process at State point three and if you go to state point four as I said you that enthalpy at State point four is equal to enthalpy of State point four so that is equal to zero what about other this you know these two V specific volume is always positive what about DP because if we go back to this this is an X you know this is a throttle valve acting like an expander so basically you know pressure drops in the direction of the flow of the refrigerant then d p is negative so this is negative so essentially you try to understand this is positive this is negative so total is negative which is again multiplied with negative s of positive so that means from this we get d s equal to positive so that means for a pro for the process three to four that is throttling process As you move from State point three to State point four entropy will increase though I could look at this point four comfortably by this but this is the correct so that means entropy increases as you know the process three to four you know takes place in this expander so this understanding now let me discuss one important thing that ah so even for the you know steam powered cycle we had tried to compare the performance of rankine cycle and its you know modified versions with the Carnot cycle so again it would be wise two again C if we try to compare the performance or if we compare the process or all these processes with the Carnot cycle so let me tell you once again if it is a corner cycle so if we go to the next slide so if we try to draw um we had seen so this is what you know just I am redrawing the TS diagram here so we can see four two one constant pressure evaporation one to two reversible adiabatic compression two to three constant pressure you know process that is occurring inside the arena condenser and three to four that is the throttling process so this is the throttling process now had it been a Carnot cycle then certainly the processes would have been different and we now discuss the if the cycle you know had we tried to measure the performance using the corner cycle what would have been the differences so now let us consider this aspect as we had seen the corner cycle again this reversible adiabatic process would will remain reversible adiabatic but instead of a dry compression that we can see because you know this ideal vapor compression refrigeration cycle this one two two is basically reversible adiabatic compression so this is dry compression why it is dry because try to understand the refrigerant or walking fluid available at the inlet of this compressor is saturated it is not a two phase mixture so basically it is called dry you know compression dry compression or dry compressions are preferred to the weight compression because of several ah advantages that I will be discussing today so basically as I said you we are now trying to compare all these processes with those had we considered this cycle to be a Carnot cycle then the this compression process will be the reversible adiabatic process but the actual process between these two temperature limits will be like this so this is one prime say this is two prime similarly you know instead of a throttle valve we need to use or reversible adiabatic expander to have this reduction in pressure from condenser pressure to this evaporator pressure so that means this process three to four will be again this so this is 4 Prime so now if I write 1 Prime 2 Prime 3 4 Prime so this is the Carnot cycle right whereas you know ideal Vapor compression refrigeration cycle that is one two three four so what are the differences first of all the compression process is now at compression so this compression One Prime two two prime that is again reversible adiabatic compression but that is weight compression why it is wet because you can understand point or state point one state of the working fluid corresponding to this point one prime is not a saturated Vapor so this is a two phase mixture so that time compressor you know needs to handle this two phase mixture now handling two phase mixture is indeed a challenging task why it is so I will be discussing soon and C two four Prime now that is the process that will occur in a reversible adiabatic expander so this process is also reversible adiabatic process so three to four Prime will occur in a reversible adiabatic expander because the sole purpose is to reduce the pressure so why Carnot cycle is not preferred because the most important part is you know this compression process that is weight compression what will happen because we know compressibility of liquid and Vapor you know is not same so liquid and Vapor you know since the compressibility of these two is different then liquid and Vapor being compressed differently will create an very you know detrimental phenomenon that is known as you know lubricant toss out phenomenon what is this first of all since now had a considered Carnot cycle in this case compressor needs to handle two phase mixture and liquid and Vapor these two are having different compressibility so liquid will now will rise with the Piston that is the working fluid will now when it would be compressed the liquid refrigerant that would be there in the cylinder head and when the Piston is rising and that liquid you know refrigerant will create or will damage the you know valves and etcetera not only that as I said you that liquid Vapor being compressed differently the liquid of phenomenon will be there wherein liquid molecule will try to penetrate into the gas between piston and cylinder wall so typically the gap between piston and cylinder wall is filled with are lubricant now when that liquid molecule will be there or the where liquid molecule will you know penetrate into this you know passage that will you know wash away the lubricants and this phenomenon is known as lubricant was a phenomena so considering these two compression is not preferred and rather dry compression that is you know one to two is preferred and that is why a variant of the cycle ideal Vapor compression refrigeration cycle is preferred over the Carnot cycle so now with this let us try to discuss that essentially what we are looking for by studying this particular cycle we need to obtain the refrigerating effect or Refrigeration effect what is that refrigerating effect or Refrigeration effect is essentially if we look at the schematic what would be the extraction of energy from this core space per unit mass flow rate of the refrigerant so that would be the refrigerating effect typically that particular you know effect is represented or expressed in terms of tons of refrigeration I will be discussing this so let me tell you once again we are studying this particular cycle essentially to understand the refrigerating effect what does it mean it means the amount of energy that would be extracted from the core stress per unit mass flow rate of this working substance that is refrigerant so now if we consider this particular part because any cycle as I said you finally we need to measure the performance and the sole performance sole purpose of this cycle is to get the refrigerating effect at the cost of some input energy that is the work input or work added to the compressor so we are supplying this much amount of energy to run this cycle and at the cost of that input energy we are getting some desired output or desert effect that is the refrigerating effect so now let us consider steady state ah steady flow equation if we assume that the steady state has reached and if we consider that one kg of the working fluid that is refrigerant so then if we just apply the steady state steady flow equation across all the components if we go back to this slide if we assume that processes all the processes are now at a steady state I mean all processes got steady state ah you know ah situation scenario and then if we consider the working fluid is one kg so per kg of working fluid if we apply that steady state steady flow equation across this evaporator what we can write we can write that so evaporator we can write h four U in equal to H one so therefore q n equal to H one minus h four so that is what is very important H one minus h four right so ah this is of course unit is kilo Joule per kg so this is kilo Joule per kg now this much amount of energy this much amount of heat this evaporator is taking per kg of refrigerant flow at the cost of some input energy and that input energy is the work input or work addition to the compressor so if we apply steady state steady flow equation across the compressor then we can write if we go back to this schematic defection then if we apply this steady flow energy equation steady flow equation across this computer across this compressor we can write I mean just we are writing this and we can write H one plus W in equal to h two that is w n equal to h two minus H one so now try to understand so this is this is basically input energy in the form of work so this is input energy and this is the desired effect right so we can Define now coefficient of performance for any refrigeration cycle we cannot Define efficiency that we have discussed now you also have studied ah when you have studied second law of Thermodynamics so coefficient of performance cop that is desired effect to the input power input energy I should say so that means this is H one minus h four divided by H2 minus H one this is also the unit is kilo Joule per kg so try to understand this is the mathematical expression of the coefficient of performance and essentially per unit mass flow rate of working fluid the change in enthalpy is giving us an estimate about the desired effect and input energy so if cop is higher then the you know ah performance of the refrigeration cycle will be better so higher is the cop better better the performance of the refrigeration cycle is so that is what we can understand so we have to calculate cop to get an estimate about the performance of the cycle higher will be the value of cop better the performance of the cycle will be so now what is refrigerating effect that is also very important so though I have already discussed about this particular term representing effect or Refrigeration effect what is this and this is typically denoted by this symbol r Now what is refuration effect so this is essentially you know had we applied steady flow steady state steady flow equation across this evaporator so this is H one minus h four kilo Joule per kg so this is H one minus H4 so as I said few minutes back that the amount of energy that would be extracted by the evaporator by this evaporator let me go back to the schematic so the amount of energy that would be extracted by this evaporator for getting the state of the walking fluid changed from State point four to State point one that is the refrigerating effect and that refrigerating effect mathematically is H one minus H4 so this is the refrigerating effect now let me tell you one thing we shall be solving one numerical problem from this topic as well but if we try to go back to this t s diagram we can see that state point four that is at the exit of the throttle valve or at the inlet to the X you know evaporator it is the two phase mixture so we need to know what would be the liquid Vapor liquid you know or quality of the you know refrigerant at the inlet of this ah evaporator so if we try to ah if we look at this stress plane carefully then we can write if we go to this next slide then we can write you know that h four that is the you know enthalpy at State point four should be equal to so this is at constant pressure and this is P evaporator right so we can write this equal to H you know ah so basically what we can write so this is H one Plus export so this is h f P condenser equal to x four h f g at P condenser so this is ah P of operated sorry this is p evaporator so this is p evaporator right so essentially this is ah foreign h four that is h f so this is h f plus h f g now what we can write x four therefore we can write x four equal to h four minus h f at p evaporator divided by h f g at p f operator now what about ah if we go to the next slide if we go to the next slide then we can write that x four that is h four minus h f at p evaporator divided by h f g at p f operator what about H4 because we have discussed that you know we have discussed that h four equal to H three that is enthalpy of the working fluid after throttling is equal to enthalpy of the walking fluid before throttling and what about S three that is h f at the condenser pressure that we can see from this test diagram so this is H three or minus h f at p evaporator divided by h f g at P evaporator so this is h f at P condenser minus h f at P evaporator divided by h f at p foreign so this x four this X for this quantity is very important and it is known as now flash gas fraction so this is known as flash gas you know fraction what does it mean so basically you know so ah Mass fraction of liquid in this liquid Vapor mixture so basically this is if I write so this is Mass fraction of liquid in liquid vapor mixture so this is important to know to the designer who will be designing this evaporator because you can see that the inlet condition of the working fluid is two phase mixture before it enters to this evaporator so now the designers should know the quality and that is what we could establish here and that is known as mass fraction of liquid in the liquid Vapor mixture and known as fly gas fraction so to summarize today's discussion we have discussed about the ideal Vapor compression refrigeration cycle we have tried to understand all the processes constitute this cycle we have mapped all those processes in two different thermodynamic planes from there we have tried to quantify what is the coefficient of performance and the refrigerating effect in this context we have also discussed about several critical issues which are very important to know for about about this particular cycle with this I stop here today and we shall continue our discussion in the next class thank you [Music] good morning