 We're now going to solve the second part of the example problem. And that is to evaluate the amount of extra g destroyed. And remember I said that there were two ways that we could go about doing this. When we're doing extra g destroyed we have two ways. We can look at entropy generation as the first method. So there we have extra g destroyed is equal to the surrounding or ambient temperature multiplied by the generation. Or we can do the extra g balance approach. We'll start with the one of calculating the entropy generated and then computing extra g destroyed. So we'll begin here with little extra g destroyed is equal to the surrounding or dead state temperature multiplied by the entropy generation term. So we need to evaluate or calculate entropy generated so the expression for that is the following. Now we're dealing with a nozzle that we said was adiabatic and consequently this heat transfer term drops out which makes our lives a little simpler. And the other thing that I'm going to do I'm going to divide by the mass flow rate from both sides of the equation and when we do that what we get is little s jan is then equal to s exit minus s inlet. So that would be the entropy of the steam coming in or sorry exiting minus the steam coming in. We can then evaluate that using the values that we have in the steam table that we pulled out earlier in the problem. Now for extra g destroyed we take the entropy generated and we multiply it by the surrounding temperature. Surrounding temperature was 25 degrees C which is 298 Kelvin. And when we do that we get extra g destroyed is equal to 6.29 kilojoules per kilogram. So that answers the question using a extra g or sorry an entropy generation approach. The second approach that we're going to look at is that of the extra g balance. And for the extra g balance the equation that is the equation for a single inlet single exit as well as for steady flow. With this we know that we're dealing with an adiabatic nozzle so that disappears. Nozzles don't do work so that disappears and the final relationship then turns out to be extra g destructed or destruction equals mass flow rate times the change in extra g of our flow stream. Now what I'm going to do I'm going to divide by mass flow rate on both the left and the right hand side and that will convert this into a smaller x for extra g. So we have that expression there psi one minus psi two. What we're now going to do we're going to expand this by putting all the different terms that that consists of. And so we get the following. Now here we can cancel out our potential energy term because the nozzle is not moving in the vertical in any way shape or form. We have to keep the velocities because they are there. So what I'm going to do now I'm going to plug in the values that we extracted when we looked at the steam tables at the very beginning of the problem and I'll plug them into this equation. And remember you have to be careful with these whenever you're mixing enthalpy entropy and velocity we need to account for the fact that velocity is in joules whereas entropy and enthalpy will be in kilojoules. And so in order to correct for that what I do is I multiply by a thousand and again multiplying by a thousand to correct for that difference. So plugging in all of the values what we end up with is x or g destroyed is equal to 6.30 kilojoules per kilogram. So that's the answer to the third part of the problem using the second approach. Let's take a look back at what we got when we used the previous approach. We obtained 6.29 and with this new approach using the x or g balance we get 6.30. So you can see the two approaches agree more or less off by 0.01 but that's not that big of a deal it's probably a round off error somewhere. Consequently we see both approaches work. So now what can we say about this? Why are we getting x or g destruction with this nozzle? Well most of this most of the x or g destruction is due to entropy generation. So where is the entropy coming from? Where could we have entropy generation within a nozzle? Well for that what you're going to have to do is think back to your fluid mechanics course. Either you're taking one now or you've taken one before but when you look at fluid mechanics what this means is that this nozzle is probably not that well designed. We'll always have some entropy generation no matter what we do but this could be there because we have a poorly designed nozzle from a fluid mechanic perspective. So from a fluid mechanic perspective what are we looking for? We're looking for a number of things. So you could see these are the design requirements. We want the exit flow from the nozzle to be uniform. We want to avoid separation inside of the nozzle. That means we do not want the boundary layer to separate off the wall. Usually we want to have it a minimum length because you don't want to have a long nozzle and another thing we want to try to minimize the boundary layer thickness at the exit because that will help give us a uniform flow. So let's take a look at two different types of nozzles. Let's say we have a nozzle that's like this. Now what might be happening here? Well if we were to draw our streamlines sometimes what you can get in here you can get flow separation along the walls depending upon the pressure gradient that you have along the wall of the nozzle. If you have a pressure gradient that is so strong that the flow in the boundary layer right along the wall of the nozzle comes to a stagnation point that is the velocity of zero you will have that boundary layer lift off and you have separated flow. So that might be a bad nozzle but let's say we were to draw or design a nozzle that looked like this. Well here it's nice and smooth we're probably not going to get any separation. However here what's going to happen is our boundary layer is going to get very very thick. If we were to look at the velocity profile along the wall we'd have a very very thick boundary layer and consequently again we don't have a uniform exit. So those were places where you can have inefficiencies and separated flow leads to turbulence and that leads to viscous dissipation which is basically conversion of pressure into thermal energy which we said is a disorganized state of energy. So this here would be the process of entropy generation. Now there could be other ones that are at play as well but this would probably be one of the larger more significant ones. So that gives you some background into terms of why this nozzle may be behaving the way it is that we see where there's actually X or G destruction taking place. So with that I'd like to conclude the lecture. Thank you for your attention. Bye-bye.