 Okay, so we're talking about fins and what we're going to be doing in this segment is we're going to be coming up with the fin equation which enables us to start to calculate or start the process of calculating the amount of heat removed from a surface. So I'll begin by drawing out a schematic of a fin which will be the basis for our derivation. Okay, so there is the schematic for the representation of the fin that we are going to be taking a look at and this fin theoretically could have capers so the area that could get smaller or larger as we move in the x direction and what I've done is I've sketched out a little differential element of the fin and that is written over on the side here and a couple of things we're going to assume that conduction is in the x direction only and the other thing that we're going to assume is that we have convective cooling from the external surface of the fin. So when we look at fins we refer to a fin as being a conductive convective system and the reason is because we have conduction going through our little differential element or through the fin and we also have convective heat transfer that is taking place to the surrounding environment and consequently there is going to be a difference between q of x in and q of x leaving and the difference is going to be related to the amount of convective heat transfer coming off of the fin and that will be the basis by which we will come up with our relationship. One other thing that I should point out here is that we have two different areas that we have shown in this schematic. We have the cross-sectional area of our fin as we move out and then we have the wetted surface area so that is the area around the perimeter of the fin where convective heat transfer is taking place and so don't get those two areas confused they are different and just be aware of that. So what we're going to do we're going to use this schematic to derive what we call the fin equation and so we're going to consider the little finite or differential element and we saw that qx was coming in and what was leaving was qx plus dx differential amount due to convective heat transfer so that kind of gives us the basis of an energy balance for that differential element and now for conduction we're going to use Fourier's law and we're also going to use the Taylor series expansion which we saw earlier when we came up with the heat diffusion equation and so what I'm going to do I'm going to take what we have here from Fourier's law and I'm going to make substitutions into our Taylor series expansion and then with this term here I'm going to use the chain rule so that's what we'll be doing with the conduction term and for the convection term so we have that and looking back at our schematic one other thing is I didn't mention this but we've assigned a temperature t that pertains to the temperature of that slice at that location we're going to assume that the entire slice is at one or the entire differential element is at one temperature t so that is what we've used in Newton's law of cooling there and with that we can put the terms from the Taylor series expansion and Newton's law of cooling into the energy balance when we get the following and so now I'm going to expand out the first term so this here is the most general form of the fin equation and you could solve this numerically for fins of complex shape we want to be able to come up with analytic representations so what we're now going to do is we're going to make some simplifying assumptions that enables us to clean this equation up a little bit and the first one we are going to assume is that we're dealing with a fin of uniform cross-sectional area and for a fin of uniform cross-sectional area what that means is that we're dealing with a fin looking at it from the side it would be a fin that the cross-sectional area ac is not changing and so it's a fin that we would say without taper and with that we can then say that dac by dx is equal to zero another thing that we can say is that the wetted surface area is going to be equal to the perimeter so let's say this was evaluated at x it's going to be the perimeter times x where p equals perimeter and with that das by dx is then simply equal to the perimeter of the fin whatever the perimeter of our particular cross-sectional area might be for the fin and so those are some simplifications that we can make we can come back to the very general form of the equation making those simplifications and we get something that looks like this okay so this is something that is looking a little better it's looking like something that you may have seen in your ordinary differential equation course and what we're going to do we're going to make a substitution to simplify and for our substitution we're going to introduce this new variable called theta which is a function of x and all it is it's the temperature of the fin at that x location minus the freestream convective cooling environment and so it's basically just a temperature difference if we look at our fin remember we were solving so this is x what we're after is t of x and then out here we have this fluid environment at t infinity so that's just taking t at x minus t infinity and that goes into this new variable called theta so when we make that substitution now another thing that we're going to do when you're dealing with fins quite often this h p over k ac comes up so we're going to make another substitution i'm going to call that little m squared and it'll make sense in in a few minutes why we did the little m squared but anyways that is what our equation is like so what type of equation is this well if you remember back to your odc course we could say that this is a linear equation nothing is squared in there we have no nonlinear terms it is second order and it is homogeneous due to the fact that the right hand side is zero so that's a differential equation and if you look back in your odc course you'll find solution techniques for that type of equation and if you recall for that type of equation solutions of the form the exponential of 8x those should be solutions to that equation so let's evaluate d theta by dx and the second derivative and making substitutions now canceling out this and that what we're left with is a squared and then we can solve for that and you can see the h p over k ac continues to appear here and what we're going to do we're going to bring back that m squared thing that i was talking about that term and with that a is equal to plus or minus little m okay so we have this for our solution given that we had a is equal to plus or minus m we can have the exponential of e to the mx or e to the minus x and linear combinations so with that so that's what we get out of the fin equation for a fin without taper we find that theta x can be expressed as two exponential functions where m is this h p over ak i believe that's what we said was yeah h p over k ac so what we're going to do in the next segment is we're going to try to determine what the boundary conditions are and with those boundary conditions we're then going to solve for c 1 and c 2 so that is we're part way through the derivation for the temperature profile within a fin and and the other thing that we'll be working with is trying to figure out the amount of heat flux coming off the fin because if you recall that's one of the main things that we're interested in is how much heat a fin removes from a surface so that is the fin equation for a fin without taper