 I have discussed let me finish this vapor compression refrigeration on the pH diagram I already basically the basic refrigeration system looks like this there is a throttling valve there is an evaporator after the cooling occurs you get vapor which is compressed is compressed vapor is again condensed there is a fluid flowing at some rate m dot looking at a steady state operation the throttling operation starts at C so this is CD then BA is the evaporation AB is the compression and BC is the condensation you can hardly see the refrigerator in the back nowadays everything is closed these days but this let us take this operation of compression the compressor you can also see the back usually at the bottom this compressor is usually it is treated as an adiabatic compressor because there is no time for heat exchange I told you BC is the set of coils nowadays your refrigerator gets hot on the side in the back that is because you got all these coils and it exchanges heat with the surroundings at the room temperature so Q condenser it exchanges heat till the material completely condenses the classical refrigerators at tree on you have a throttling expansion the throttling on each of these is very easily analyzed the basic equation all of these are treated as it is just a flow system and at steady state you have DH I am sorry have DU I will write the open system equations if for each of these for each of these it is an open system so is equal to TDS I am considering reversible operation right now minus delta WS plus HN DMN so if you are looking at the evaporator looking at either steady state if you are looking at steady state let us look at we will start with AB for AB I do not even have to say the steady state or negligible hold up my system is the compressor the hold up in the system is negligible compared to the mass flowing through the system it means DU and DS can be neglected because you refers to the total internal energy of the open system in the open system consistent consists of mass inside the compressor so this part this becomes 0 this becomes 0 and you have the compressor work delta WS and negligible hold up first of all DU is 0 is also equal to DS is also 0 and DMN is equal to DM out I divide through by DT I get WS dot is equal to H out minus HN so minus delta H into M dot delta H of course is B-A this is HB-HA then you can do the analysis for the condenser here no shaft work is done again you are looking at steady state or you are looking at negligible hold up so you are really looking at DU DS being 0 and WS being I think I should write this in terms of not TDS is not 0 I will write it in terms of first law itself much easier I will show you exactly where the second law kicks in right now in terms of the first law I have this is adiabatic operation so I have to add delta Q is equal to 0 this part delta WS is 0 AB then BC delta WS is equal to 0 and again DU is equal to DS is equal to 0 actually the other way around DU is equal to 0 and DS is also equal to 0 so you get delta H into M dot is simply Q Q dot is equal to delta H into M dot and delta H in this case is C-D it is not C-D it is C-B it is C-HB is negative so Q is actually lost to S around X process CD I have W delta WS is 0 and so is delta Q then you are looking at negligible hold up so DMN is equal to DM out the same thing holds so you have HN is equal to H out or I will write it in terms of in HC is equal to HD then the last FDA if the evaporator is exactly like the condenser so you get Q dot this is Q condenser this is Q evaporator this delta H is HA-HD this enthalpy is greater than this you are simply evaporating the whole thing it is completely vapor and then you are compressing the vapor this here if the operation is actually adiabatic notice that this work done is equal to delta H into M dot is always valid so first law derivation if the process is isentropic you go along the isentropic line if the process is not isentropic the actual work done is more than this is work done on the system you have to do more work so you will end up at some point B prime I do not know the path you just draw a line arbitrarily you do not actually the irreversible parts are all variable but depending on the efficiency because work is a function of the path I do not know the path beforehand depending on the condition of your compressor you will do more work and this the actual work done will be always delta H this value here is delta H at constant entropy so this is reversible work is this this is the actual work so the thermodynamic state of the system will be different that is all instead of B it will go to B prime it will get it will go to a higher temperature and therefore to a higher enthalpy the condenser and the evaporator effectively operated constant pressure thermodynamically but if you are doing design for industry you will worry about pressure drop here so basically what you have to do is design this condenser first for the heat load that you expect and then the rest of it will design itself because you know the total heat load here divided by delta H will give you M dot that determines how much of free on or refrigerant should be circulating inside that is a sealed system so there is a whole you lose it but otherwise it is a sealed system and normally this is operated just a little above atmospheric pressure because you do not yet leak in you would rather the fluid leaked out but this pressure therefore is determined below is determined really close to atmospheric pressure it used to be just above now it may be just below this pressure is anything convenient for safety in case it blows up to make sure nobody gets injured too badly so this pressure so these two pressures then you look at thermodynamic charts to see which fluid will work with between these two pressures and which has see this is the cooling you get which has the maximum da that means pictorially you look at a graph and say which has the widest on the same scale the one that has the widest liquid vapor envelope will be the one that has the largest enthalpy chain this divided by this is what gives you the efficiency or the coefficient of performance that maximum the cooling you get for unit of work done okay this is about vapor compression refrigeration one other thing I have to mention in the work cycles is the Rankine cycle which is still used a lot in boilers have you done the Rankine cycle in the other course I think I will just give you an assignment very similar so you must have done the TS diagram but you can trace the same thing on a pH diagram so I will skip the Rankine cycle stuff I will just summarize for heat to work devices what you need to know is the Otto cycle the diesel cycle then the Rankine cycle and the gas turbines and then I would expect you to then vapor compression refrigeration I think going beyond that I have to go to mixture thermodynamics because chemical engineering thermodynamics is all about mixture thermodynamics really about multi component systems so let me start discussing multi component systems if I write internal energy I know it is a property of state so it has to be a property it has to be a function of measurable variables for example it you is a function of SNV for pure substances I told you the number of variables is determined experimentally you will come up with a phase rule but that is there is empirical input in applying the phase rule also equal to function of SV let us say it is an R component system then I have N1 moles of 1 N2 moles of 2 and so on NR moles of R components N1 is number of moles of 1 and so on since it is a property of state I can use calculus partial of U is partial of U with respect to S holding V N1 N2 etc constant since all the moles are held constant and the volume it is the same partial as in a pure substance so you will get TDS partial of U with respect to S is T and the partial of U with respect to V is – P plus sum over I partial of U with respect to Ni Dni holding S V Nj not equal to I is Nj not equal to I means all mole numbers other than I other than Ni the reason I can use calculus is because I have asserted from the two laws that U is a function of state and S is a function of state this incidentally this is what I define as Mu I have four equations of this kind because I have simply introduced redundancy and write them all down dA is simply A is still U – TS so we will get – SdT – PdV plus sum over I Mu I Dni here Mu I is partial of A with respect to Ni the only difference is that the variables that are held constant are T V Nj not equal to I if you differentiate holding S V and Nj not equal to I if you differentiate U you get the chemical potential if you differentiate the Helmholtz free energy with respect to Ni holding T V and Nj not equal to I you get the same chemical potential because A is still U – TS so DH is U plus PV so this is TDS plus VDP plus Mu I Dni this is H with respect to Ni again S P Nj not equal to I then Dg in this last partial g with respect to any extensive variable with respect to the number of moles holding T and P and other mole numbers constant keeps coming again and again it is called a partial molar property it is written as gi bar turns out to be very important because ideally what would you like you have a mixture property what you would like to do is simply multiply the number of moles of I by the property and add you should get the answer it happens exactly that way provided you use partial molar properties not any other property so for example I from this I am going to integrate this I will integrate this equation for convenience exactly as I did in the pure state pure component case sort of central to the entire thermodynamic argument for mixtures if I integrate this I will show you g is simply Ni Mu I sum over Ni Mu I that means if I know the chemical potential in the mixture of component I I can simply add it up and get the answer but Mu I in the mixture is not the same as Mu I pure in the pure state chemical potential is different from the chemical potential in the mixture because the environment is different you are talking of molecules you are talking of certain energy and entropic properties if you have a mixture I have more rearrangement possibilities so the entropy is higher the energy is different because the intermolecular forces are different between like and unlike molecules so I am going to do this integration what we will show is integrate and show that g is equal to so I forgot this board exists so integrate I will call this equation one these the results I derive can be derived from any one of these but g is the most convenient for chemical engineering purposes so I use this arguments are identical integrate one we will show that g is simply a Ni Mu I to show that it is actually a trivial thing but it is conceptually very important it was done of course like all the most fundamental things in thermodynamics was done by Josiah will ya gives during his 10 years of meditation in Yale apparently nobody ever met him he did not bother he just walked in did thermodynamics went back for lunch then came back and did more thermodynamics and then went back and then wrote a 150 page book he wrote three such books and all of which are absolute classics is not a single conceptual error in any of the books this is after 200 what each 7 after 100 years little over a 100 years okay what you have is a system with number of moles Ni when I put a squiggly bracket I mean I runs over all the indices I have a system in the final state that is k times larger I have a reservoir that is connected to the word reservoir means the properties here do not change at all it has mean 10 to the power 6 Ni or whatever it is same composition that means I have x i the mole fractions x i this is automatically i is equal to 1 to r minus 1 the last one is determined same composition but infinite amount of fluid I just open the valve at T is equal to 0 this is in state TP this is also in state TP it is K times larger all extensive properties are therefore K times larger here so if I have the volume is V this volume will be K times V V K times V any property if you take internal energies will be times K times U just a larger system and so on now I make an assertion that G with respect to Ni doing the two extensive variables and I make the assumptions at G is essentially a homogeneous function of the number of moles so partial of G with respect to Ni is an intensive variable it is actually an assertion in thermodynamics you do not need a separate law if you there is a whole theory of homogeneous functions of which this can be shown as a correct result in this is done in I do not know if you have seen this book there is a book called rational thermodynamics this is typical of the author trues del passed away recently Cliff Clifford trues del was a great expositor of thermodynamics he said the thermodynamics guys are loose in their nomenclature they have to do things correctly so uses fairly rigorous mathematical language to describe thermodynamics he called this thermodynamics rational implying everybody else is a rational that was typical of trues del and in rational thermodynamics he will show you what happens if it is not a homogeneous function like when G is not a homogeneous function of Ni and so on so fundamentally delta G by delta Ni because this is a property per mole this is assumed to be an intensive variable intensive variables it does not depend on the extent of the system so the assert here that all the intensive properties let us say let E be any extensive property extensive property simply increases with size then E I bar is intensive I bar is E per mole specific properties are also intensive if I if small e would be intensive because it is per mole but E I bar is the question thing in question so GI bar or Mu I Mu I is intensive if it is intensive then for this process you can take this this this equation applies for the open system it has change in number of moles so I want to integrate it at constant T and P so this is 0 this is 0 Mu I is constant because it is composition remains constant if Mu I is constant all you have to do is integrate this you will get G is equal to Mu I into Ni actually you will get delta G so integrating 1 you get delta G is equal to K minus 1 times G because this system is simply K times larger than that this is equal to on the right hand side I get some over I Mu I times delta Ni which is equal to some over I K minus 1 times Mu I Ni so if since K is not identically 1 you can cancel it off you get this result this crucially depends on the fact on the on this fact that Mu I is an intensive variable fact one of the things about this nano technology that they are talking about no big deal just suddenly nano has become very fashionable molecular scales were measured in angstrunks and 10 angstrunks is 1 nanometer I do not know if you know this started with Feynman's lecture he gave a lecture typically saying there is lots plenty of room at the bottom what he meant was there are lots of space and you can rearrange molecules as you please and so on but he was actually it is quite an interesting lecture you should read that independently but nano has just become a buzzword but the factors in small systems if you are talking of systems nano systems 10 angstrunks that kind of dimensions you are not talking of a system with very large number of molecules when you are talking of very small number of molecules all of these you must remember classical thermodynamics does not depend on molecular structure but it assumes that average properties very continuously that means you must have very large number of molecules so that if you have density variations you are actually adding or subtracting molecules in a change the density but if the number of molecules added or subtracted a small compared to the number of molecules in the system you can treat the whole thing in calculus if you have 10 and you take away 1 then the changes too much you cannot use calculus so in nano systems that is the problem the total number of molecules is small so the property fluctuations are larger than the property itself and in such systems for example the chemical potential need not be an intensive variable at all it will depend on the number of molecules in the system in that sense it is not an intensive variable but for our purposes we will resolve in this thing never to treat less than 10 to the power 10 molecules at a time it is already 10 to the power minus 13 of 1 mole so you do not have to worry it is very hard for you to deal with less now if you take a drop then you are already talking of 10 to the power at least 15 so if k will cancel if k cancels you get g is equal to n i mu i you might say I did the integration for one process but notice that the final result has only the variables that belong to the initial state and the initial state is entirely arbitrary the process is described by one parameter k it is I have chosen a particular process but the result is independent of the process once k cancelled out the result is independent of the process the result is applicable to the initial state which could be arbitrary the one is full of bugging arguments of this kind so you get a result and then you use it everywhere then you get all your answers nicely but you have this nagging doubt did I do that right and I did it for a special process but actually in this case this is what gives us a book is full of such thought experiments he never got up from a stable to do a single experiment we did everything sitting at a stable so this result is very important it is central to make sure thermodynamics and typical of a thermodynamics you would not stop there having done that this is I have integrated now I like to differentiate you know you would think it was a kid playing but differentiate this you get dg is equal to sum over n i d mu i or sum over mu i d n i right n i mu i you differentiate you get n i d mu i then I insist that if I differentiate I could get I should get the same answer as before so I have to compare three with one and declare the two are one and the same if I declare they are both one and the same I get n i d mu i is equal to minus stt plus v dp this is another central equation in thermodynamics it is called the Gibbs-Duhem equation and stop there this is again very central to mixture thermodynamics I will put it in another form that we actually use the when I did the integration I used a physical input that mu i is an intensive variable therefore I was able to write this form now I do calculus and I come up with this you might say where does this come from it comes from the fact that if mu i is an intensive variable the state of a system should be described by temperature pressure and r minus 1 intensive variables right if you if I use composition I would have used temperature pressure x1 x2 xr minus 1 so mole fraction satisfy the condition that sum over dxi is equal to 0 sum over xi is 1 sum over dxi is 0 looks a little more ugly but it is the same thing sum over n i d mu i is equal to something it is entirely equivalent to saying sum over dxi is equal to 0 so any intense set of intensive variables we use to describe the system the independent variables will be only r minus 1 the r intensive variable that describes composition it is differential will always be determined by the others and that is the constraint that is the Gibbs-Duhem equation so we will continue from there I have to do some work on this to produce a final form that we actually use