 We're now going to solve an example problem using the log mean temperature difference approach. And for this problem we're looking at a feed water heater. And what I'm going to do I'm going to begin just by writing out what is known. I'm not going to write out the problem statement. So we know that we have a feed water heater. We're told that it's a shell and tube heat exchanger and we're given information for both the shell and the tube. So for the shell side we're told that it is a one pass and the fluid going through on the shell side is condensing steam with a saturation temperature of 120 degrees C and then for the tube okay. The other thing that we're given we're given the overall heat transfer coefficient. And we're told to find we're looking for the area of the heat exchanger that would correspond to that overall heat transfer coefficient. So when you look at this we know a number of things. We know the inlet temperature on the tube side so this is the cold fluid. We know the exit temperature. We know the inlet and exit on the shell side so that's our hot fluid because it's going through a condensation process so the temperature will remain at 120 degrees C. So those are all the elements that we said that we need in order to apply the log mean temperature difference. So let's go ahead and use the analysis for LMTD. So we're going to use LMTD. Now note that this is not a double pipe heat exchanger. So what we're going to do remember I told you that there are correction factors that exist when you're dealing with heat exchangers that are not double pipes. What we're going to do we're going to model it as a counter flow double pipe unit with the correction factor F that we get from a figure. I won't give you the figures but I'll show you essentially what would be in the figure and you can find these in textbooks. The equation that we're going to use here it's going to be the following where delta TM is our log mean temperature difference. So what we're going to do we're going to begin by sketching out the temperature distribution remember I said that that can quite often help you when you're solving these problems in order to understand what is going on. Okay so we have our two fluid streams in the heat exchanger as shown. Now what I'm going to do I'm going to label these in order to apply the LMTD. First of all this here is temperature hot at one and this is temperature hot at two. This is temperature cold at one this is temperature cold at two. So what I'm doing is assuming that that's location one and that's location two. With that we can calculate the log mean temperature difference and the other thing that I am going to do is when we're trying to determine the F factor that we have for LMTD given that this is a shell and tube exchanger and not a double pipe. There are other symbols and I'll show you a schematic in a moment that will help make sense in terms of where I'm getting these symbols from. But essentially these are the fluid streams that will then enable us to determine that F factor or these are the temperatures for the fluid streams I should say. Okay so notice we have this this this and this those all lead to the F factor. Now the equations that we have here we're going to begin by determining the amount of heat transfer by looking at the cold fluid so this here is the cold fluid and this up here would be the hot fluid and it is condensing and consequently it is not changing temperature as it goes through but what we can do we can write out Q and this is going to be the heat exchange is the mass flow rate of water times the specific heat of water and the water was the cold fluid but I'll just put water and then Tc1 minus Tc2 so that's taking the hot temperature minus the cooler temperature. Now in this what we need to do we need to be careful about this because the specific heat of water or of any fluid is a function of temperature and consequently what we need to do is we need to be able to obtain that at the average temperature so average water temperature with that we get Cp4182.6 joules per kilogram kelvin. Now I did say Cp is a function of temperature for all fluids I should make a disclaimer there because if you look in thermodynamics helium the Cp value is a constant it is not a function of temperature and I think it's for all of the other gases that are in that column in the periodic table it turns out that the Cp is not it does not change with temperature but anyways that's just an aside. So we're dealing with this we have the value of Cp for water and that's at the average so with that we can calculate the heat transfer so we know the mass flow rate of water was 2.5 kilograms per second be careful when you're pulling the values out of the back of the book or wherever you're getting them from sometimes they're in kilojoules per kilogram kelvin and so you have to be careful to put that in joules then we have 100 minus 30 we end up with then 731.96 kilowatts is the heat exchange now that is giving us the heat transfer we want to determine the area so what we need to do next is determine the log mean temperature difference so let's do that next so if we go back and look at our temperature diagram what I'm doing here is doing a basically the first one is T hot 2 minus T cold 2 so it's this differential and then the other differential that we're going to be evaluating is that one there so that's what we're doing here and dividing by the logarithm and remember I said that the numerator was just going to be the first term that we have here so it's this one and then the denominator is this one and with that we get delta Tm or the log mean temperature difference is 46.54 degrees C okay so we're almost there we have Q is Uaf delta Tm and what we have thus far we've determined that U was known we're looking for area we just solve for that we still need to get this F factor and that is the correction factor for the fact that this is not a double pipe heat exchanger it's a shell and tube heat exchanger and this is where we're going to pull those different temperature values that I was showing you when we looked at the temperature diagram and for that I'm going to draw a little schematic and this is going to vary from book to book but the one that I have as a reference they label the temperature of the shell or of the tube fluid stream so this is going through and there's a loop and then they show that there are baffles here and then they show the tube inlet coming here with capital T or sorry that's the shell inlet and the shell exit capital T2 so with those you can then just by looking at the schematic and reading the problem statement you can figure out which is the big T and which is the little T in the temperatures and then what you would do there would be another chart and in this other chart what we would have is the correction factor and that would be plotted against some variable P that I'll define in a moment and then there's going to be a number of different curves and these curves are going to be for different values of R yet another factor that will define a moment so R is defined in the following manner and P is defined in the following manner and this would depend upon the figure that you're using for the particular heat exchanger so you have to be a little careful with that but what you would do is you go back to your temperature diagram so let's do that and let's begin and what we're going to find out T1 minus T2 so going back to our temperature diagram T1 is here is 120 and T2 is there it's also 120 so let's look at our equation here this here is going to be 120 minus 120 which is 0 so we're going to get an R value of 0 and there is no curve for 0 so you're going to wonder what's going on well it turns out that this is a unit with a phase change and whenever you have a phase change so either condensation or boiling with one of the streams then we can just make the assumption that F is equal to 1.0 okay and so with that now that we know F we know Q we know delta Tm it's pretty straightforward then to evaluate the area so there we go we get 7.864 square meters this would be known as a sizing problem you know your fluid streams and you're determining the size of the exchanger that is required so that's a case of LMTD it seemed to work quite well what we'll do in the next segment is we're going to solve the problem again using LMTD where things don't work quite as well and that will lead us to a new approach for being able to solve exchanger problems