 We're now going to take a look at ways of performing exergy balances for both fixed mass systems as well as control volume systems and then what we'll do is we'll work an example problem applying the exergy balance to a system and we'll be looking at an open system so we'll be using the control volume approach. So what we're looking at today is exergy balance and we'll start with a fixed mass system. Now if you recall from last lecture what we did is we said that exergy itself was representation of useful work potential of a system and we talked about exergy destruction and we said that it was related to irreversibilities that exist within the system and so once you have exergy destruction that that's essentially a bad thing for a mechanical engineer because you want to minimize any kind of irreversibilities and and consequently you do not want to have them so you want to try to minimize the amount of exergy destruction and we also saw that the exergy destroyed could be related to the amount of entropy generated and I said that sometimes it's easier to just calculate the entropy generation than it is to do the exergy destruction which we will see in the example problem that I show you at the end of today's lecture. So we're looking at exergy balance for fixed mass systems so what I'll do is I'll write out the exergy balance so we have exergy n minus exergy out minus the exergy destroyed is equal to the change in exergy in our system. Now if you look in any thermodynamics book you should find a relation where they then show you what the exergy balance is for fixed mass system and I'm just going to write that out now but what we have is a term for the heat transfer that might be taking place across our control boundary and again we're multiplying it by the reversible thermal efficiency or the Carnot efficiency. Next we have a work term that we correct for work done on the surroundings so that there is an equation that we could use for a fixed mass or closed system if we want it to perform an exergy balance. The next thing I want to do is take a look at the exergy balance equation for an open system or the control volume approach and what I'm going to do I'm going to run out the equation for the most general form and then what we'll do is we'll reduce terms, reduce terms by making a number of different assumptions and so we'll have different forms of the exergy balance equation for an open system. So in the most general form it looks something like this now this first term was the same as what we had for the fixed mass system however now we have it in rate form as you'll notice with everything with the over dot so that is the most general form of the exergy balance equation for a control volume. One thing I should say about this first term here sometimes this is a difficult one to estimate especially if you do not know the exact temperature at which heat transfer is taking place at and you'll see that when we work a couple of example problems of ways that you can get around that sometimes you can move your control surface further out to the external surroundings and with that you can actually get rid of the heat transfer term well what you're doing is you have the temperatures being the same as the dead state and that removes that term from the equation but that is sometimes a tricky term to handle the first term there. So let's take a look at this equation now and we're going to make a number of approximations we're first of all going to assume steady flow so if we have a steady flow what can we get rid of? Well steady flow means things do not change with respect to time and consequently any of the rate terms where we have a derivative with respect to time we can remove and looking at the equation this term has it and this term has it and consequently those two will drop out and what we end up with then for steady flow is the following relationship okay so that's it for the steady flow the next thing we're going to do we're going to approximate or reduce it further by making another assumption and that is we're going to assume that we have a single inlet and single exit stream so we will call this single stream and with that the term that will be modified is going to be this term and this term because there we will only have a single input and a single exit okay further reducing it what we're next going to do is we're going to assume that we have a system with reversible work now if we're assuming reversible work what that means is that there is no entropy generation and consequently x or g destroyed is equal to zero and so we can also write this in terms of our entropy generation equation that we've seen earlier in the course so that would be the way that you could get your entropy generation and then t0 your dead state temperature times the entropy generation would be equal to the x or g destroyed however coming back to our equation for the x or g balance what we then have is we still have the heat transfer term which i said is sometimes a tricky one to deal with our work term is now reversible work so notice that i've changed that i've added a reversible here because we're dealing with zero x or g destroyed and the x or g destroyed term is removed from the equation oops sorry and then all we're left with is that so that becomes the equation for x or g destroyed is equal to zero now making another approximation we're going to say let's say we have reversible which we just said we have and if i say it's adiabatic if you recall that was the term that we used for an isentropic process so this is what would result if we have an isentropic process and if it is adiabatic that means the heat transfer is zero so the first term is gone and then all we're left with is the following and similarly for the entropy generation for this type of process we would write m dot times s e minus s i because the adiabatic assumption would remove the heat transfer term if we further go and we neglect kinetic energy and potential energy what we end up with is the following relation and what i'm going to do here is i'm going to expand out the x or g change term that we have for an open system oops sorry the first one that should be s 1 minus s 2 so this term in here in brackets represents what we had in the original the previous equation was psi 1 minus psi 2 now if we've neglected potential energy and kinetic energy that goes away and that goes away and so then what we're left with for reversible work but remember we did say that this was reversible and adiabatic so if we said reversible and adiabatic that is isentropic which we know means that the entropy does not change and therefore s 2 is equal to s 1 which further reduces the equation then to reversible work is m dot times h 1 minus h 2 so that becomes an expression for an isentropic process and if you look even at the first law and compare it so let's take a quick look at the first law for an open system we would have written it as follows neglecting the kinetic energy and the potential energy and we said adiabatic consequently the heat transfer term disappears what we're left with is work is equal to m dot times h 1 minus h 2 and you can see these two equations are then consistent provided that we would be looking at an isentropic process and and that would be one where there is no entropy change going from state one to state two but in any event this is the final one that we get from our exergy analysis and you can see it is consistent consistent if s 1 equals s 2 because then we would have the same sort of process as we're looking at with the reversible work one