 So this is part two of the solar house, the passive solar house, and if you recall the problem statement that we looked at earlier, what we're dealing with is a house that we're trying to maintain at 22 degrees C throughout the evening. We just calculated the number of hours we need to have the electric resistance heater on. The second part they want us to calculate is right here, exergy destruction. So let's take a look at exergy destruction. What we're going to do, we're going to approach this using two different techniques, one using entropy generation and the other one using the exergy balance. So we will begin by looking at the exergy balance and then we will look at the entropy generation approach. So exergy destruction, we're dealing with the fixed mass or closed system version of the exergy equation. Okay, so that's what we get for the equation. Now is there anything we can knock out of this equation? Can we knock out the heat transfer? Well it depends where we define our boundary but with it currently defined around the house as you can see here, we can't because there's heat transfer taking place. But we're going to play a trick with that and you'll see it in a second. What else can we do? The volume term, the volume of the house is not changing with time and consequently that one we can neglect. There is no moving boundary going on here. Okay, but what we're going to do, we're going to play a little trick here in order to be able to figure out how to quantify this heat transfer term. And that trick is, we're going to apply the balance on the extended system. So not the immediate system of our house, we're going to go further out away from the house, into the air surrounding the house on the cold winter night. And why are we doing that? We're doing that in order to go to a point where we've accounted for all of the exergy destruction because outside of the walls of the house they're still heated because of the heat loss and consequently you still have exergy destruction. You've got to go all the way out until you get to the surroundings and at that point you will get the total exergy destroyed. So with that, what we can say is that Tk in the first term of the equation is equal to T0 which is equal to 276 and the beauty of that is that that term disappears. So with that what are we left with for the exergy balance equation? And remember we're after exergy destruction so let's isolate exergy destruction and I'm going to express the exergy change for our closed mass system on the per unit mass basis. Now with this what we're going to do we're going to spend a little bit of time looking carefully at this term here. Now what does that consist of? Well that's going to consist of a number of different components. That is the exergy change of our system and it's now an extended system because we're outside of the house. So that's going to include the water in our containers. It will include the room air and it will include the surrounding air. So that's what we need to include when we're evaluating that term. So let's take a look at that. So if you look at the water the water is going from 80 degrees C down to 22 degrees C so that means that the exergy at state 2 for the water does not equal to the exergy at state 1. For the room air for our 10 hour process or time that we're looking at it stays at 82 degrees C for the whole time. Consequently what we can do is we can write that air at 2 for the exergy is equal to air at 1 and that makes it a little simpler. Now what about the outside air? So let's look at the surrounding air and if you were to go and walk around the house and with a thermocouple and measure the temperature you'll find on the inside on average it would be like this. It's going to start dropping down when you get near the wall so you would have some sort of thermal transition in there and then when you get outside eventually as you go further and further away you'll eventually get to 3 degrees Celsius. And so with that we're going from 22 degrees C all the way down to 3 degrees C but we're looking at an extended system so we're way out here where we're doing our analysis and consequently we can say that the surrounding at 2 is going to be equal to the surrounding at 1. It's not going to change as the night goes on. The temperature in the surrounding is going to remain at 3 degrees Celsius. And so with that what we can do is we can rewrite phi 2 minus phi 1 or we can express it in the following manner. And some of these terms we can neglect. First of all the specific volume is not going to change. There's no kinetic and there's no potential energy. So only water contributes to the change. And so with that what we can do we can plug this value into our exergy destruction equation that we have here and we'll go forward and evaluate the exergy destruction. See here is the specific heat for water 4180. T2 is 295 Kelvin. T1 is 353 Kelvin. So that's the temperature change for the water. And T0 is 3 degrees Celsius or 276 Kelvin. So we can plug in the values and here we get exergy destruction is the following. So that's the answer to the question using the first approach. We'll quickly take a look at the second approach using the entropy generation so we can write OR. Okay so in this equation the error does not change in entropy because it is maintained with the temperature. The first term enables us to evaluate the heat transfer terms. That's the heat loss. And with that we get the entropy generation as the following. So we get that for the entropy generation. We want exergy destroyed so that will be T0 times the entropy generation. With that we get exergy destroyed 292 922 kilojoules. Let's compare and see how it is with the other one where we found 292 913 and here we get 292 922. So not too bad they're quite close. When you look at the analysis however the entropy generation approach was a little easier and entropy generation is usually the easier approach here. So we'll now move on to the third part which is to look at the minimum work input which we'll do in the next segment.