 Alright, so in this lecture what we're going to do, we're going to solve two different, well they're the same example problems, but we're going to work them in two different techniques. One, using the approximate solution techniques we saw in the last lecture for transient conduction when you change the convective boundary condition on an object. We'll use the approximate solution in this segment and then in the next segment we'll use the Heisler charts to solve the problem. So let me begin by writing out the problem. I'm not going to give a problem statement, I'll just basically write out what we know. So this is an example of the approximate solution. Okay, so there is our problem statement written out. What we have is a fused quartz sphere. So we're dealing with a sphere diameter 2.5 centimeters. We're told the thermal diffusivity as well as the thermal conductivity. So we know the diameter. Originally it's at 25 degrees C and then it's suddenly placed in a convective environment of 200 degrees C and the convective heat transfer coefficient 110 watts per meter squared Kelvin. So that would be a pretty good flow coming across it in order to get that high of a convective heat transfer coefficient. And then what we are told to do is to evaluate the temperature at the center line r equals 0 at 4 minutes and then also at some point off of the center line 6.4 millimeters off the center line at that radial location also at 4 minutes. So we're told to use the approximate solution technique. So that will be the one for the sphere. So let's begin by drawing a schematic of what this looks like and it won't be that complex because all we have is our sphere and r or not. Radial location r, we're interested in what's going on at 6.4 millimeters and then this is exposed to some convective environment. Okay so the analysis, let's begin the analysis and we'll start with part one and that's where they ask us to determine what is going on at the center line. So t at r equals 0 at 4 minutes which is 240 seconds. And so we'll use the equations that we presented in the last lecture and so we have to evaluate the Fourier number as well as the Bio number because that is we need the Bio number in order to get c1 and zeta1 and those will come out of a table. So let's begin with the Fourier number. We get the Fourier number to be 1.459. Let's take a look at the Bio number and from that we get a Bio number of 0.9046. Okay so with the Bio number, this is where we go and we find the table in our books that enable us to get c1 and zeta. So what I'm going to do, I'm going to write out the values. It turns out that the table is listed. There's value of Bio number 0.9 and 101. So we're going to have to do linear interpolation. So let me write out the table values. Okay so those are the values that are in the table and we're interested in what is going on at 0.9046. So we can see that it's going to be closer to 0.9 than it is to 1 and with that what we need to do we need to do linear interpolation. And so when you go through and do the linear interpolation you find the following values. And whenever you do interpolation do a sanity check. 1.5075 that's close to 1.5044. So that makes sense. 1.2499 that's close to the 1.249 or 2488. So that probably means that we did the linear interpolation correctly. And then the next thing we do we go back let's go back here. So we've done the linear interpolation and that has enabled us to get c1 and zeta. So we can now use this equation here to evaluate theta naught star. So let's do that. So we get that and what I'm now going to do let's see did I did. So what I'm going to do I'm going to take that value of theta naught star and we're going to use this equation and what we're looking for is this here t naught because that will be the centerline temperature at 240 seconds. So let's try to evaluate that. So we plug in the values and we get t naught 192.1 degrees C. So that is our centerline temperature after 240 seconds and that is answered a part one. What we now need to do is go on and evaluate what is going on at radial location what do we say 6.4 millimeters. So for that we have to use the spatial distribution solution. Okay and we'll notice in this equation we have r star. r star is the radial location we're interested at 6.4 millimeters divided by the radius of our sphere. So r star is 0.512 and with that we can plug in the values because we have theta naught star. We solved that in the first part of the problem and we know zeta one because we looked that up knowing the bio number from the table. So now let's write out what theta star would be and it's at this point that if you're sleepwalking while you're doing this solution you're going to make a mistake and the reason is this needs to be in radians. So remember when you do solutions to engineering problems be awake. Don't sleepwalk your way through them. Okay so you got to convert your calculator to radians you plug in those values and what do we get. You get 0.0410 and we can then evaluate what are we looking for we're looking for this here because that's going to be the temperature at the radial location that we're interested in so let's isolate for it. Okay so there we go we got 192.8 degrees C. Now let's see if that makes sense what did we get we got 192.1 so we had 192.1 degrees C and that was at the center line so that was at r equals zero so does this make sense. We have at a radial location 6.8 millimeters at the same time these are both at T is 240 seconds and what did we say we said the convective environment was T infinity 200 degrees C and H was equal to what was it 110 and T initial was 25 degrees C. So what's happening here is our sphere begins at a low temperature it's exposed to a very hot convective environment so the temperature is going to go up. Consequently what we should expect we should expect at a radial location r equals what was it it was 6.4 millimeters we should expect that a radial location further out so something out here should be at a higher temperature before the center line the center line is the thing that will take the longest to warm up because we have a heat transfer moving in and it takes time maybe because we have the time constant there's a transient term in the heat diffusion equation and consequently it'll take the longest for the center line to warm up so based on my hand waving arguments here it means that we we should be able to say that this does make sense because the center line temperature is lower which it would be because it's going to take longer for the center line to warm up to whatever temperature it's going to well it's going to eventually go to 200 degrees C if we let this go to T infinity so anyways that is an example using the approximate solution in the next one what we're going to do we're going to use the Heisebler charts to solve the exact same problem