 Alright, what we're going to do now is we are going to solve a heat exchanger problem using the effectiveness NTU method. Let's all begin by writing out what we know and what we're looking for. Okay, so we know that we're dealing with an air-to-air cross flow heat exchanger. Both fluids are mixed. Mass flow rates for both of the streams is identical of 0.5 kilograms per second. The hot air flow coming in is at 400 degrees C, whereas the cool air flow coming in is at 20. Overall, a heat transfer coefficient is low because we have gas to gas. So it's around 40 area is 20 square meters. And what are we looking for? We're looking for the exit temperatures of both fluid streams. So we're looking for hot exit and we're looking for cold exit. So, how do we go about doing this? Well, we know we're probably not going to be able to do LNTD. Could we do LNTD via m dot C sub p? You'd have to guess C sub p and you don't know your exit temperatures. That would be hard. We are going to use epsilon or effectiveness NTU to solve this. And if you recall back from the last segment, we looked at the two different procedures. We had a sizing problem where you're looking for area or a rating problem. And so this particular problem turns out to be a rating problem. We're looking for a Q as well as the exit temperatures. So what we're going to begin with, we're going to begin by getting the properties of air. And we have a little bit of a problem here because if we look back at what is known, we know the inlet temperatures of both of the gas streams. But we don't know the exit temperatures. And usually we're supposed to be evaluating the specific heats at the average temperature. And we don't know what the average temperature is. So we're going to have to do a bit of a guess here. And we will begin by evaluating at the inlet temperatures. We know that it's going to be lower, but this is kind of a first step. We have to iterate this problem. And when you look up the values for the hot and then for the cold fluid stream, we'll take properties at 20 degrees C. So when we're dealing with the effectiveness NTU method, we have to determine which is the minimum fluid. And the mass flow rates are identical here. So the minimum one is going to be the one with the lower specific heat. And that is going to be our cool fluid. So that is C man. And then that will be C max. So we're looking at the step by step procedure. I said, don't use it all the time that we will. CR is C man over C max. Mass flow rates are the same. So with that, we get the ratios to be 0.9408. That's the first step. The second step is to estimate the number of transfer units. And remember that is defined as being UA divided by C man. So when you compute that, you get 1.59. Now that we have CR and we have NTU, we can then go ahead and evaluate the effectiveness. We can do this in one of two ways. We can get the effectiveness off of a figure. And typically I'm not going to pull out a figure. I'm just going to sketch it so that you get an idea. But it is probably in any textbook that you're using. What we have is effectiveness on the vertical. And then on the horizontal, we have the number of transfer units. And that is UA divided by C man. And then what we have are different curves for different values of CR. And so you'll find these different curves that are on this plot. And so that would be for CR equals 0. And then asymptotically, we would get CR equals 1 at this point here. But what you would do is you would take the value. So we said CR was 0.9408. And we said NTU was 1.59. So you go on to the plot. You would let me do reds so that you can see it a little better. You find the location on here. You work your way up and 0.9408. It's going to be pretty close there. You then go across and you read the value of epsilon or the effectiveness. And from that, you get 0.59. So we get effectiveness is 0.59 by reading a chart. You can also obtain it by using an equation. And this is the more accurate, but it's a little bit more laborious. Depending on how your eyesight is, it's not always easy to read these charts. Sometimes it's a lot easier just to plug and chug things into your calculator. Okay, so you get that big, long equation there. But you know everything in it. You know CR, NTU, CR, NTU. So you can directly evaluate. When you plug the values in, we get the effectiveness is 0.58196. Which is in pretty good agreement with what we get off of the chart. So I'm just going to take the 0.59 and continue on calculating. I should use the 0.58, but anyways, I'll use 0.59 not a big deal. Because they're so close. So with that, we can then evaluate the heat transfer. And remember, heat transfer was determined this way. And with that get 112.73 kilowatts fourth step. And this is the last step on this pass through the problem. Remember, we have to iterate because we guessed the properties at the beginning. And so we get T hot out 189.1 degrees C. So it didn't drop much, but we went over going from 400, 420 down. Yes, it is dropping a fair amount. And TC out. So the cool fluid out 244.2 degrees C. So it's going up quite a bit from 20 degrees C up to 244. For the way that we have this particular problem. So that would be iteration number one. And technically what you should do is you should iterate this problem again. And to do that, you compute a new average temperature for the hot fluid. And a new average for the cold fluid. And when you do that again, you find that this is going to be the minimum fluid because the mass flow rates are the same. And from this step on, you go through the exact same procedure again. And hopefully it converges after doing the calculation one more time. But that is an example of doing a rating problem using the effectiveness NTU method. And with that, that concludes heat exchangers. And it concludes this course in heat transfers. So I hope you enjoyed it and learned something about heat transfer from this course. Anyways, that is all that I have for you. Thank you for watching.