 Alright, we're going to finish this lecture by solving an example problem of the condensing unit. And the condensing unit we will have what we call shell side condensation. And in the condensing unit we have steam at one atmospheric pressure that is being condensed. So shell side condensation, recall from the last segment that means that the condensation is on the outside of the tubes. And we have horizontal tubes in the tube bundle and they are aligned in a square array. And they're 20 in the horizontal and 20 in the vertical. The outside diameter of the tube is 6.35 millimeters. And the length of our tube bundle we're told is 1 meter. The surface temperature of the tubes is maintained at 88 degrees C. And given that we're dealing with an atmospheric pressure, that means that the saturation temperature of the steam is 100 degrees C. And what we're told to find is the condensation rate. So that is m dot and we're looking for this in kilograms per hour. So let's begin with a schematic of this problem. So there we have 20 tubes in the horizontal and we have 20 in the vertical. And outside of the tube bundle we have steam at atmospheric pressure and saturation temperature 100 degrees Celsius. So that is roughly a schematic of what's going on. And just like we talked about for two bundles we'll get a film forming that's going to drip down and the film is going to get thicker and thicker. And eventually we have a very large film coming out of the bottom. If you sum up all of these that would then give you the mass flow rate for the entire tube bundle. What we'll do we'll begin with a single tube looking at the relations to give us the convective heat transfer coefficient. And then we will extend that with the relationship that we had for two bundles by looking at a vertical column. And then we'll multiply that by 20 and that will give us information about what is happening for the entire tube bundle in terms of the condensation. But to begin with what we need to do we need to evaluate the properties. So let's begin with that. So if you recall the heat of vaporization we take that at the saturation temperature. And remember to bring 10 to the 3 here because this is joules per kilogram. And the density of the vapor we also get at the saturation temperature. Now for the other properties we evaluate those at the film temperature which is going to be our saturation temperature plus the surface or wall temperature. And so in this particular application or problem that turns out to be 94 degrees C which is 367 Kelvin. And so chances are we're going to have to do some interpolation when we go into our tables because typically the tables are either in units of 10 or 5. But at 367 we have to do interpolation. So let's go into the tables pull out the values and I'll just write them down. So those are the properties at the film temperature. Now what we want to do in our correlation we're going to have the modified latent heat of vaporization. We won't be using that one exactly. We have to use the modified or corrected one. So let's evaluate that to begin with. So we get that for the modified latent heat of vaporization and we will be using that in our correlations. Now the correlation that we use here we're going to begin with that for a single horizontal tube. And this is assuming laminar flow. So what we can do we can plug in all the different values that we obtained out of the tables as well as our modified heat of vaporization. And with this we obtain a new salt number of about 157. And so we can take that and then evaluate the convective heat transfer coefficient for a single cylinder. So that's what we're interested in obtaining. So doing that. So we get about 16 or 17,000 watts per meter squared kelvin. And that is for a single tube. Now what we need to be able to do we need to be able to extrapolate this for a bundle or a vertical column because what we just arrived is for a single tube. But recall we're interested in being able to obtain what's happening when we have a two bundle. So we're going to look beginning with the vertical column and we'll understand what is going on there. And that will be determined by this value here. So let's look at the relationship that we use for a tube bundle. And if you recall from the last segment we said that what this would do it would actually reduce the convective heat transfer coefficient. And the reason is that as the film grows on each successive tube lower on in the bundle the film itself acts as a bit of an insulating layer. And so that has the net effect of reducing the amount of the convective heat transfer coefficient. And that's reflected in the negative that we have in the power here. So when you do that you obtain the average convective heat transfer coefficient for the vertical column is actually a little smaller. And it's 10,216 watts per meter squared Kelvin. So when we obtain that then what we need to do remember we're after the condensate and we want to know m dot coming off of this tube bundle. So now that we know the convective heat transfer coefficient the next step that we're going to want to follow is to determine q for the cube bundle. And once we get that we can then solve for the mass flow rate by taking q divided by the modified heat of vaporization. So that's where we're going. First step is going to be to get the amount of heat transfer going on within this two bundles. Let's begin with that. And what I'm going to do is I'm going to determine the total heat transfer for the condensing unit. So remember there are n by n or 400 tubes in this bundle. And this is Newton's law of cooling here. So we have to multiply by the area, the wetted area, which is going to be pi dL. And L is one meter. And then we multiply that by T sat minus the surface temperature plugging in values. So we get about 978 kilowatts per meter. I'm assuming the given that we have the length of being one meter that that's per meter. So now that we've obtained that we can then go ahead and determine the mass flow rate. So let's do that next. So we then determine the mass flow rate. And that's kilograms per second. And if we're calling the problem statement, they wanted it in kilograms per hour. So we multiply that by 3600 or 60 times 60. And we then end up with about 1500 kilograms per hour is the amount of water that this condensing unit would be able to produce from the steam coming in. Now one thing that we should do, we could declare victory and say that we've solved the problem. But really what we need to do, we need to check the Reynolds number on each of these horizontal cylinders. And if you recall, we were using a relationship that was very much like new salts relationship, which was derived for a laminar flow. So this is what we were using. And it was modified a little bit to take into account the horizontal cylinder. But what we need to do, we need to check the Reynolds number. And so that's what we're going to do in the next step here. So which Reynolds number do we use? Well in convective or condensation, I should say heat transfer with condensation, you recall there are many, many different Reynolds numbers that we've been talking about. I'm going to pull out the one that has the mass flux in it because that is what we know in this problem. And we divide by the perimeter times the dynamic viscosity of the liquid. Now what is the perimeter? Well the perimeter, if we have a tube that is like this, and it's going into the plane of the page, and so the length of that tube is one meter, the perimeter is essentially just the length that would be in this direction. So remember for the vertical plate, the perimeter was B. So when we were looking at the vertical flat plate, the perimeter was B. Well in this case, it is just going to be the length of our pipe, which turns out to be our tube bundle has a length of one meter. So the perimeter is one meter for this particular problem. And so with that, what we're going to do, we're going to begin with the top tube. So the top tube, the mass flow rate, and now I'm going to use it in kilograms per second. And I'll divide by all 400 tubes because that will tell me the mass flow rate for an individual tube. And so with this, plugging the values into the Reynolds number, we get 14.27, which is less than 30, and therefore it is a laminar, so that's good. Because new salts relation was derived assuming that we had laminar film flow coming over. Now let's take a look at what's going on on the bottom tube. And so in order to get that mass flow rate, it's essentially the one for a single tube multiplied by 20, because that would be the number of tubes in the vertical direction. And we get our e bottom of 285.4, which is wavy flow, or wavy laminar. So we are actually pushing it a little bit. The correlation would be starting to break down here. But I did not give you a correlation for wavy laminar on a horizontal tube. And consequently, we're going to have to take it as it is and assume that this is the correct answer. But realizing that we are pushing things a little bit because by the time we get down to the bottom tube, we have the case, even by the second tube, we multiply that by two, we're pretty close to 30. So by the second tube in this vertical tube bundle, we're starting to transition and move into the wavy laminar regime. Now remember, turbulent would be at 1800, so we're a ways away from turbulent, but we are in the wavy laminar regime. But anyways, what that does, that demonstrates a problem for a horizontal cylinder and then how to determine the mass flow rate for a condensing unit when you have a tube bundle. And again, that was for a shell side condensation because the condensation was on the outside of the tubes.