 So, we begin a new module in this lecture, one on exergy. The first thing that we are going to do is to motivate why we need to introduce a new concept which we are calling exergy. There are some textbooks which also refer to exergy as availability, both mean the same thing. Exergy is I think modern usage, so we will adopt the term exergy. Why do we need a new concept? There are few reasons. The most important one is given here, so let us say that you know we have a piston cylinder mechanism, say something like this and there is a gas or some other working substance inside and it undergoes an expansion process and produces some work. Now one of the things that we would like to be able to do is to actually come up with an estimate of let us say an efficiency for this process. And as I mentioned in the previous module, we would actually like this metric or this efficiency to be in some manner connected to the amount of entropy that is generated in the universe as a result of this process. So that is the total of entropy generated, that is the sum of entropy generated due to internal irreversibility plus external irreversibility. So we want a performance metric for non-flow processes like that. You may recall from the previous course that we have not actually defined efficiencies for individual processes. We have defined efficiency for a cycle, COP for a reverse cycle, but not efficiencies for individual processes except isentropic efficiency, I will come to that in a minute. So we would like to be able to do this. So basically what we would like to do is, so a system starts from a given state 1 and goes to a final state let us say 2. Now we would like to find out what would be the ideal process between these two states. So the work developed during that process would then form the basis of defining say an efficiency. So that is something that we would like to be able to do for non-cyclic processes whether it is a flow process or a non-flow process, it does not matter. For instance, we looked at the example of a mixing chamber in the previous module. So we had flow two streams coming in, one stream going out. Now how do we define an efficiency for the mixing chamber that actually is quite interesting and it is not possible to do this from an energy perspective. So what we need to do is we need to come up with a way by which we can define an efficiency for this device based on the amount of entropy that the process generates in the universe. That is what we would like to do. We already pointed out to the fact that this is a highly irreversible process due to the, I mean mixing is a highly irreversible process and so there is generation of entropy because of this internal irreversibility. If we allow heat transfer to take place then there could be external irreversibility also. We saw another example involving external irreversibility. So we would like to be able to define an ideal process for this case. So we know all the initial states. So this is an initial state, this is an initial state, this is a final state. So what would be the ideal process in going from these two initial states to this final state and based on that we can then define an efficiency. Now as I have written here energy based performance matrix do not account for entropy generated in the universe. So let us see what we mean, what I mean by this. So let us see, we just erase these things here. So you may recall that we wrote down a simple block diagram of an engine. So this is an engine which receives a certain amount of heat, let us say QH from high temperature reservoir rejects an amount of heat QC to the low temperature reservoir produces an amount of work equal to W. Let us say it executes a cyclic process and we wrote down the expression for efficiency as 1 minus QC over QH, again when it executes a cyclic process. If it does not execute a cyclic process then the efficiency would simply be W over QH. And in that case there would be no upper limit on the efficiency mandated by the second law. If it does not execute a cyclic process, if it executes a cyclic process then this is the expression for efficiency and there is of course an upper limit which is a corner limit on this, on the value for this efficiency. Now two engines, let us say we have two engines, both of which are supplied with the same amount of heat QH and they produce the same amount of work rejecting the same amount of heat. However, one engine is given QH from a reservoir at let us say 1400 Kelvin. Another one is given the same amount of QH from a reservoir at let us say 800 Kelvin or 1000 Kelvin. The efficiency for both these engines would be the same. However, for the engine that receives heat from the higher temperature reservoir, the amount of external irreversibility is higher. So although its efficiency based on energy is the same, we suddenly do not like the fact that there is more external irreversibility associated with the engine which is receiving heat from a higher temperature reservoir. So which then points to the requirement that we need to develop a metric which will take into account the amount of entropy that is generated in the universe as a result of operation on this engine. So energy based performance metrics do not necessarily account for entropy generated in the universe. Now exergy, expression for exergy that we are going to develop and the notion of exergy meets these requirements, it is quite general and allows us to calculate a limiting value meaning what is the maximum work possible for this process. So it allows us to calculate a limiting value against which the actual performance of any device executing a cyclic or non-cyclic process may be compared. Remember here this engine is executing a cyclic process. So even for this we can come up with definition of efficiency which takes into account the amount of entropy that is generated in the universe. So exergy in that sense is quite general. It can be applied to any process. Number one, number two, it allows us to calculate the maximum work that can be obtained as a result of executing the process between these and the two given end states whether it is cyclic or non-cyclic it does not matter. So it is very, very general. Now let us turn to the notion of isentropic efficiency. You may recall that we defined isentropic efficiency as let us say in the case of a power producing device as actual work divided by isentropic work. So these were situations. So we defined isentropic efficiency for a turbine, compressor, nozzle and diffuser. So these are situations for which the ideal process is actually an isentropic process. So we compare the actual performance against the performance that would have been realized had the working substance executed an isentropic process. So consequently this applicability of isentropic efficiency is restricted to adiabatic devices. So if it is non-adiabatic. So for example for the same turbine or compressor if the turbine or compressor actually loses heat to the surrounding which is very realistic then how do we define an isentropic efficiency? We cannot because the basic process is not an adiabatic process. So we cannot define an isentropic process as the ideal process for this case because heat loss is taking place. And we cannot extend the notion of isentropic efficiency for devices such as mixing chambers, heat exchangers and as I said even turbines or compressors when they actually have exchange of heat with the surroundings meaning they are losing heat to the surroundings. So isentropic efficiency is very very limited in its utility and so it cannot be used to meet our requirements. So we want a very general definition of an efficiency which can be used for calculating limiting value and is applicable to cyclic as well as non-cyclic processes and flow as well as non-flow processes. To develop an expression for such an efficiency we start with the notion of exergy. First let us define a reference state or what is called a dead state. So dead state we denote the pressure at the dead state as P0, temperature as T0, velocity is 0, elevation above the datum as 0. Now when a system exists in this state it is not possible to develop any work from this system. Normally ambient state at 25 degree Celsius and 100 kPa is taken to be the dead state. So when we have a system which is at 25 degree Celsius and 100 kPa it is not possible to develop any work from the system that is called the dead state. So exergy of a system at a given state so we have a system which is at a certain state maybe some other some pressure, temperature and say elevation, velocity all these things are known. So it is at a given state and the exergy of the system at this state is defined as the maximum theoretical work which is very important, maximum theoretical work that can be developed as the system goes from the given state to the dead state. So let us if you label this state as 1 then as the system goes from state 1 to the dead state which is denoted as state 0 the maximum work that can be developed during this process is actually called the exergy of the system at that state. So exergy of the system at this state is defined as the maximum work that can be developed as the system goes from state 1 to the dead state. Of course it follows very clearly that the exergy of the system at dead state is 0 because we define that no work can be developed from a system which is at the dead state. So exergy of a system is at the dead state is 0. So the exergy of a system that is not at the dead state or any other state for instance is non-zero but is it positive or negative or can it be both that is the next question that we have to ask. It is non-zero but is it positive or negative that is what we will discuss next. So here we have four examples of systems which are at state different from the dead state and they are executing a process to go towards the dead state. So here we have a system which is at the same temperature as the ambient temperature but at a pressure higher than the ambient pressure. So in this case we can actually place a weight here and as the system expands to the ambient pressure the weight is lifted which means positive work is being done by the system. So the exergy of the system is positive. Now in this case we have a system which is at the same pressure as the ambient pressure. Now by looking at these four cases we have not really lost any generality. So this is without any loss of generality it is sufficient if you look at this and we can actually then make a statement about the sign for exergy. So here we have a system which is at the same pressure as the ambient pressure but at a temperature greater than the ambient temperature. So maybe this is gas contained in a piston cylinder assembly at let us say 500 Kelvin. Now I can use this as a high temperature reservoir and operate an engine between this high temperature reservoir and ambient and develop some work from this. That means the exergy of this system is positive because work can be developed as it executes as it supplies heat to this engine and the engine will stop working when this system reaches the ambient temperature. So we start at a temperature greater than T0 and we supply heat to this engine. It develops work and then the engine stops developing work when the system reaches the ambient temperature. This was discussed in the previous course in connection with second law under the topic of finite reservoirs. So reservoirs in the context of heat engines may be infinite or finite. So this was discussed under the context of finite reservoirs. So the exergy for these two cases is positive. Notice that here the pressure of the system was greater than the ambient pressure to begin with and here the temperature of the system was greater than the ambient temperature to begin with. For both these cases exergy of the system turns out to be greater than zero. So here we discuss a case when the pressure same example as this but the pressure is less than ambient pressure. Temperature is same as ambient temperature. In this case by placing weights like this as the system goes towards ambient pressure. So the piston moves up in this case and the pressure of the system increases. So as it goes towards the ambient pressure these weights are lifted in the gravitational field. So work is done by the system which means exergy is greater than zero in this case. So even when the system is at a pressure less than the ambient pressure the exergy is still positive because we can still develop positive work as the system goes towards the dead state. So here we have the counterpart of this example where the temperature of the system is initially less than T naught pressure is same as P naught. In this case we can actually use this as low temperature reservoir and operate a direct engine between the ambient and this low temperature reservoir. So this engine by virtue of the fact that this is a low temperature reservoir can draw heat from the ambient reject heat to the reservoir at lower temperature than the ambient and produce an amount of work which is positive and so exergy of this is also positive. So as this device as the system goes towards the dead state it develops a positive amount of work which means that the exergy of the system is positive. So you can see that whether the systems are at temperatures and pressures above or below the ambient in all these cases a positive amount of work is developed as the system goes towards the dead state which means that exergy of the system we can generalize and conclude from this that the exergy of the system exergy of a system is always positive and non-zero so long as it is not at the dead state. It is non-zero so long as it is not at the dead state and it is always positive. Now let us develop an expression for exergy. So if you take any of these systems that we have considered throughout any loss of generality we may write first law for the system like this delta E equal to Q minus W for the system. So the system is initially at state 1 and it goes towards the dead state which is the final state denoted to 0. So let us take each one of these terms in this in this expression one by one and develop the expression further. So delta E for the system may be written as final minus initial so E0 minus E1 remember the system goes from an initial state 1 to a final state 0. So delta E is final minus initial so E0 minus E1 and by definition the total energy is the sum of the internal energy plus kinetic energy plus potential energy. So internal energy plus kinetic energy plus potential energy and by definition the dead state is at 0 velocity and 0 elevation above the datum so these two terms go to 0 and this is the total energy at the initial state. So we may write delta E for the system like this. Let us take W system next. We decompose W system into two parts one is called the useful part the other one is the work done again in displacing the atmosphere. So if you look at say a system like this piston cylinder mechanism either this or this as work is being developed the atmosphere is also being pushed aside. So there are two things that are happening one is the weight is being lifted and atmosphere is also being pushed aside. So we must do some work in pushing aside the atmosphere. So the system must do some work in pushing aside the atmosphere and it also does work while lifting the weight and the same is true in this case also. So the network developed by the system is the sum of these two work spent in pushing aside the atmosphere displacing the atmosphere and the work done in raising your weight. So and that is what we have done here. So we have written the W system as the sum of two terms this is the work done in displacing the atmosphere this is the useful work this is called useful work because this is what can be used to lift the weight. So that is denoted with a subscript U. So this is W U plus P0 times V0 minus V1 because this is just displacement work. So the exergy of a system if you recall we defined exergy of a system as the maximum theoretical work that can be developed as the system goes from the given state to the dead state which means that exergy of the system is the maximum value for this quantity here W max for this I am sorry W max is the exergy of the system. So we need to develop an expression for W U max.