 okay what we're now going to do we're going to solve the exact same example problem that we just solved however now we're going to do we're going to use the Heisler charts in order to come up with the solution so I'll begin by just like we did last time writing out what we know and what we're looking for we're dealing with that fuse-court sphere that is going through a change in convective environment and we're looking at transient conduction analysis okay so there's our problem statement fuse-court sphere we're given the thermal diffusivity thermal conductivity we're told the diameter is 2.5 centimeters we can get the radius from that initially it's at 25 degrees C and then it's placed in a convective environment hot convective environment so it's going to be increasing in temperature the convective environment 200 degrees C and H of 110 and we're told to calculate two things one is the temperature at the center line after four minutes and then the temperature at a radial location of 6.4 millimeters after four minutes so let's begin by drawing out a schematic for what this looks like okay so our radial location are and this is being exposed to a convective environment okay so analysis for this we're going to use the Heisler charts and we'll start with part one which we're looking for the temperature at radial location r equals zero and 240 seconds and so in order to do this in using the Heisler charts we're going to begin with the one that gives us the center line temperature and for that we need to evaluate the x-axis value which is just the 4a number and for that we get 1.459 and the curve the curve that we read is I believe I said it was 1 over the bio number so K over HR naught okay so those are the two values once we have them then what we can do is we can go to our Heisler chart so let's take a look at the Heisler chart and here we are very very busy chart a lot of stuff going on but it's not as bad as it might immediately look when you first look at it what we have here are our curves K over HR naught so we have to find the appropriate curve and we're dealing with 1.106 and so looking there is one and there is 1.2 so we're halfway between those two and then what about the 4a number where are we with the 4a number we had 1.459 so there is 1.5 so there's one 1.1.2 what are the increments of this one so I would say 1.459 it's probably somewhere right in about here so what I'm going to do I'm going to sketch up and that's up here somewhere and we're between those two curves so coming across I would estimate that we're probably somewhere right around .05 so let's take that as being our value and we'll work with it let's go back so that gave us the Y axis value and that is theta naught over theta I and from that we can go ahead and calculate T naught because that's what we're looking for we want to know the centerline temperature and what do we get we get 191.3 degrees C and when we did this using the approximate solution from the last segment there we calculated 192.1 degrees C so you can see we're pretty close we're a little bit lower with a Heisler chart than we were using the approximate analysis but that gives us the result for the centerline temperature after four minutes let's move on now so that was our Heisler chart what we're now going to do we want to find the temperature at a location off the centerline and so that was part two and so for this what we're going to be doing is we're going to be reading off of a curve r divided by r naught which that is 0.512 and the x on this curve is 1 over the bio number 1.106 and those were the that was the curve that we were reading in the previous graph and we're looking at the Heisler chart for the centerline temperature okay so with that now we go to our graph and we take those two so this is the graph for what happened spatially you can see on the left we have theta over theta naught on the bottom we have one over the bio so we were looking at one over a bio number of 1.1 so there's one there's two 1.1 I know it's probably someplace right in there you can see when you're using the charts you're doing a little bit of guessing a bit of a guesstimate for this and then 0.5 well we don't have a curve the radial location these are the radial location curves we don't have any 1.512 now we have a 0.4 and a 0.6 what I'm going to do I'm going to go in the midpoint between those two so we're probably someplace right in there that's probably closer to the bottom which it should be because we're at 0.512 so with that we then say that that's probably the location we come over we're a little bit above 0.9 so let me guess this to be 0.92 and with that we'll go back and assuming that the y-axis which is theta over theta naught assuming that that is equal to 0.92 we can then plug in all the values and again here what we're after we're after that temperature because that will be at the radial location at four minutes and notice what we need here we need t naught and that's pulled from the solution from the previous part and so that's why I'll probably recognize that number there and then we multiply that by 0.92 and when we do that we get t 192 degrees C and so that is the result that we get and let's compare when we did the previous example using the approximate technique we got 192.8 degrees C so slightly higher than what we're getting now but that's consistent with what we just saw let's see where was it here the approximate technique was slightly higher than what we got from Heisler charts for the center line temperature and so here we're finding the exact same trends so anyways those results are pretty close you can see the Heisler chart is relatively quick I didn't use a ruler where did my Heisler chart go there we go I didn't use a ruler here and if you had used a ruler it would be a little bit more accurate for either the center line temperature or the spatial temperature but when when we compare the results really not too bad so anyways that gives you an example of transient conduction analysis with convective environments for a sphere one of the three different shapes that we can look at and that is covering everything we're going to look at with transient analysis and heat transfer