 We're now going to work an example problem involving exergy analysis and an exergy balance. It's a rather long problem and consequently what I'll do is I'll break it into two sections. The first section will be dealing with things that pertain to the first law and the adiabatic efficiency. In the second part we will look at exergy destruction and we'll use two approaches. To begin with I will write out the example problem or the problem statement. That is the problem statement that we have before us. Just like any of the other problems the first place we want to start is we want to look at the terms and try to pull out the keywords that will help us understand how to solve this. To begin with we're dealing with steam as the working fluid. It is accelerated in a nozzle, so we're talking about a nozzle of a turbine, but we're not dealing with the turbine. We're dealing with the nozzle before the turbine. That information don't let it confuse you. It tells us that it is an adiabatic nozzle which is important because that pertains to heat transfer that we'll have to think about. We're told that it enters, so this is the initial state, 7 MPa 500 degrees C 70 meters per second and it exits, I'll change pen color here, exits at 5 MPa and 450 degrees Celsius. The other thing it tells us is the surroundings are at 25 degrees C. Whenever you see that that usually hints towards exergy analysis because you need to know the dead state or the surroundings, but anyways that's what we have for this particular problem. It's telling us to solve for three things, exit velocity of the steam and the nozzle, so that would be velocity 2, adiabatic efficiency of the nozzle, and the final thing, exergy destroyed. Okay, so let's begin the problem by writing out what we know and what we're looking for and then we'll work our way through. So the known information, we're dealing with steam as the working fluid and it is an adiabatic nozzle. State information, 7 MPa T1 500 degrees C and velocity 1 70 meters per second. Now we're going to have to use the steam tables, right now we do not know if that is compressed water, if it's saturated or if it's super heated, but we'll extract that information when we go into the steam tables which we'll do in a moment. At state 2 we have 5 MPa, temperature 2 is 450 degrees C, we are not given the velocity, we're told to find that. Another piece of information we're given is that the surroundings is at 25 degrees Celsius and we're told to find the following. The first thing is the velocity 2 or the exit velocity, second thing is adiabatic efficiency of a nozzle or of our nozzle and the third thing we're after is exergy destroyed. Okay and the next slide what I'll do is I'll write out a schematic, it's pretty clear but I'll do this anyways. So we have something that looks like this, a nozzle, remember it's taking a high pressure fluid and it's accelerating it, this is a very crude nozzle, but anyways that's what we can write it to look like. We have our inlet state there, exit state and we have steam coming through. Exit condition 7 MPa, 500 degrees C and 70 meters per second and exit condition 5 MPa and 450 degrees C. Okay so it's at this point that we're going to want to jump into the steam tables and pull out our property information. So I'll write out the properties P1, 7 MPa and T1, 500 degrees C. So what you have to do grab your steam tables and go into the back and take a look and what you'll find we're dealing with super heat or with water sorry and so what I would do go directly to the 7 MPa table and you'll see it's 7 MPa, the saturation temperature is in brackets at 285.83 degrees C, we're at 500 so that means we're well above it or up in the superheated region and the exit condition we may as well look here as well, 5 MPa, saturation temperature there is 263 degrees Celsius, the exit temperature is 450, again we're well above it. That means that we're in the superheated steam region for both the inlet and the exit which is good because that's what you'd want for a steam turbine. You don't want to have saturated steam because then you get droplets hitting the turbine blades and that becomes a bit of a mess, you lose efficiency through that and we'll look at that later in the course but what I'm going to do now is I'm going to write out the property data and when you're in the back of the book you might as well pull everything out because you never know if you're going to need it and as you do more and more of these problems you'll become more knowledgeable about which information you should be pulling out of the tables when you go back there. Now the final thing I'm going to do here, I'm going to, that's P2 actual that they've given us but I'm going to assume that we have P2S which would be for an isentropic process again it'll be at the same pressure and the thing that we can write is that S2S needs to equal S1 if we're looking at an isentropic process. So with that what we can do, we can go into our steam tables at 5 MPA and with an entropy equal to the entropy at state 1 and we can then pull out the enthalpy for that state and you're going to have to do interpolation in order to get this but it comes out to be 3301.5 kilojoules per kilogram. So that's the property information. We can now go ahead and begin our analysis and the first thing we want to determine is the exit velocity and for that we're going to use the first law. Okay, so there we have the expression for the first law. We were told it's adiabatic, that disappears. It's a nozzle. Does a nozzle do work? No it doesn't. There are no moving boundaries or anything like that so we can get rid of that term. Potential energy, there's no change in the height of the elevation so that disappears. We cannot get rid of kinetic energy, that's very important in a nozzle. So we're left with kinetic energy and enthalpy as being the first law and I'll cancel the mass flow rate with the zero on the left. So with this what we're going to do, we want to isolate for V2. That's what we're looking for so let's isolate for V2 and we get this term here. We can then enter in numbers. Another thing you have to be careful with here is you have mixed units. When you pull your enthalpy values out of the tables that's kilojoules per kilogram whereas the velocity term will be expressed in joules per kilogram while joules would be the mass if it's per mass. So consequently what we need to do, we need to multiply this by a thousand in order to have the enthalpy match the units of velocity so be careful with that as well. And what we get is the exit velocity 439.4 meters per second. Let's move on and take a look at the isentropic efficiency. We also call this adiabatic efficiency and we saw the expression for the isentropic efficiency of a nozzle was like that and what we can do now, we can determine a velocity using the equation that we just derived for the first law but we will use it for a process that is isentropic. And so we'll use the enthalpy for the isentropic state, H2S, again being careful of the units here and we get that an isentropic velocity would be 471.7 meters per second, consequently we can evaluate the adiabatic efficiency as 0.868. So that answers the first and second part of the problem. I'm now going to go on and in the next segment what I'll do is I'll cover the third part which is the extra G destroyed.