 Alright, we're going to close out this lecture with an example problem looking at water flow through a pipe with a non-circular cross section. So I'll begin by writing out the problem statement. Alright, so there is the problem statement. We're dealing with water flowing in a duct with a cross section 5 by 10 millimeter, and we're told that the mean bulk temperature is 20 degrees C, and we have a constant wall temperature of 60 degrees C, and we're dealing with fully developed laminar flow, and then what we're looking for, we want to calculate the heat transfer per unit length, and so in this problem we're told the bulk temperature 20 degrees C and the wall temperature 60, and we want to evaluate the amount of heat transfer per unit length in that section of pipe, and consequently this is not something that we would be applying the log mean temperature difference to, given that we're interested specifically what is going on at this particular point within the pipe. So what we'll do is we'll begin by writing out what we know for this problem, and then we'll start the solution procedure, and so that is the information that we know, and what we're trying to find, we're trying to find the heat transfer per unit length, and so I'm going to assume a length of 1 meter, and going through the solution for this. So what we'll do, we'll begin by drawing a schematic. So there we have our duct, and then looking at a cross section of the duct. So that would be a cross section of the duct, a rectangular cross section, and now in terms of the analysis what we will do, we know that this is non-circular geometry, we're dealing with laminar flow. We have constant axial wall temperature, and if you recall from the last segment I mentioned that any heat transfer book would probably have tables showing these values. Sure enough you look up the value in a book and you will find that would be a rectangular channel with an aspect ratio of 2, so we would have A and B as shown there. And you get a value of the new salt number for that condition of 3.391, and remember this is new salt number evaluated using the hydraulic diameter, and the properties, where will we evaluate the properties? Well the property that we have to worry about is going to be K, and given that we're told that we're dealing with a section of pipe where the mean or bulk temperature is 20 degrees C, and the wall temperature is 60 degrees C, what we'll do is we will evaluate our properties at the 20 degrees C value. So the property that we have to look up in a table is going to be the thermal conductivity of water, doing an interpolation, and the hydraulic diameter is 4 times the area divided by the perimeter, well the area is dimension A times dimension B, and the perimeter is 2A plus 2B, and so when you put in the numbers, or the values for A and B, we get this. So what we can do, we can take our new salt number, which we have right here, we've solved for K, well we looked that up in the tables, DH we just computed, so the last thing left here is going to be H, the convective heat transfer coefficient, and that's what we want to solve, or solve for, so with that we can isolate H, and we get this value of 306.6 watts per meter squared Kelvin for the convective heat transfer coefficient, now that's kind of a high value, well although we're dealing with water, consequently we would expect it to be quite high, but that is the value that we get for H, and now looking at the problem statement, what we were looking for, we want to find the heat transfer per unit length, so heat transfer per unit length, that is going to be watts per meter, so looking at it in terms of things it's going to be Q prime, so coming back here, first of all, let me express this in terms of H, A, T wall minus T mean, or bulk, and that would be H length times 2A, the perimeter of our duct, now I'm going to bring the length over to the left-hand side, and that's going to give us Q prime, which will be watts per meter, and we have the following, so we have all of the values in this equation, we can plug them in, and what we find is the following, so we get that for the heat transfer per unit length, and that is for our rectangular duct, main thing is just looking up the value in the table for the new salt number, and computing the length scale being the hydraulic diameter, so that is an example of performing a heat transfer calculation using non-circular geometry.