 okay so what we're doing we're looking at heat source systems and what we are going to do we are going to work an example problem involving heat source systems so I'll begin by writing out the example problem and what we'll be doing is this will be an application involving rectangular coordinates we'll be looking at a plane wall or a slab so let me write out the problem statement okay so there is our problem statement what we have is a plane wall and we're told that it has a uniform heat generation and the heat generation rate in watts per meter cubed is q dot we're told that it is insulated on one side so that is a piece of information and then the other side is exposed to a convective environment and the conditions that convective environment are T infinity and H and then what we are asked to do we're asked to solve for an analytic expression describing the walls temperature profile so that's a bit of a hint there that we're going to be using the heat diffusion equation in this problem and the last thing it says is knowing the walls thermal conductivity case so that will be one of the parameters that we're going to use when we are performing the solution for this so what we'll do we'll begin by writing out what we know we'll come up with the schematic and then we'll work through this problem okay so that has what we know we know the internal generation rate we know the convective condition we know the thermal conductivity we're looking for the temperature distribution as a function of position we will assume this to be a steady-state problem 1d and the thermal conductivity is a constant we will also assume the width of the wall and so the width of the wall that we will assume will be with L and so what I'm going to do now I'm going to write out a schematic okay so there is a schematic of our wall one thing I forgot to add here was our convective environment okay so there we have our conditions now the boundaries on the walls what we are going to do let's check the problem statement it said nothing about the temperature on the right-hand surface but what I'm going to do another assumption I'm going to make is I'm going to say that the temperature at that point is T1 another thing that we can say if you're called the boundary conditions that we had for the heat diffusion equation whenever you had an insulated surface that meant that there is zero heat flux at that point so with that we can make a comment here we don't know anything about the temperature at the left-hand surface but we do know something about the change in temperature with position the slope of it from Fourier's law and we know that q is 0 if q is 0 dt by dx is 0 and consequently we can write out for the other boundary condition dt by dx and I'm going to say this is x equals 0 I didn't put the coordinate system but I will that is equal to 0 so the coordinate system in this problem we will assume that our wall begins here and then moves in that direction and consequently this would be x equals L this is x equals 0 so that is the schematic for this problem in terms of analyzing this we want to come up with an analytical solution and consequently we'll be using the heat diffusion equation in order to do that so let's begin the analysis okay so that is the heat diffusion equation looking at this we can reduce the complexity of it right off the bat so first of all partial with respect to y well we said that this is a 1d problem so consequently partial with respect to y and partial with respect to z those two are going to go to zero and they're going to disappear and then on the right hand side of the equation we said that this was steady state and consequently the temperature would not be changing as a function of time so that term disappears as well and with that we quickly are able to reduce the equation into something that looks a little more friendly and i'm going to bring the heat generation term over to the right hand side of the equation now looking at this we can do a further simplification and that is to looking at this the fact that it's only a function of x we can change that partial differential into an ordinary differential so we can write so given it's 1d we can assume that t oops i apologize that should not be an x and a y that should be a y and a z so t is not a function of y and z and consequently the partial with respect to x can be expressed as an ordinary derivative with respect to x and with that the mathematical physics equation governing this problem the heat diffusion equation reduces to the following okay so that's something that looks a little bit easier for doing analysis and so how do we solve for the temperature profile if we have an ordinary differential equation we're looking for this the way that we solve for that we integrate and we integrate twice because it's a second derivative so we will integrate this twice and so on the first integration we get a constant of integration we don't know what that is yet the way that we're going to solve for that will apply boundary conditions we integrated a second time and we get a second constant of integration so in solving for c1 and c2 we will apply the boundary conditions and if you recall back let's look at our schematic which was back here where the heck was it there it was okay so one boundary condition is what's going on on the left hand surface over there another boundary condition was at that location and then there actually is another one and it relates to the heat generation basically doing an energy surface balance right here and so we could do another one there but let's take a look at the boundary conditions now so those are three boundary conditions that we have for this problem we only have two constants of integration and consequently we only need to use two of these what we're going to do in the solution here we're going to use the first two and so let's apply those and use them to determine and you'll notice the third one here this is the one that i said was related to an energy balance on the right hand surface and that's q dot times the volume and that has to be equal to the energy that is flowing out and going into convective heat transfer so that's the relationship that we've made with that third boundary condition so let's go ahead and proceed to use the first two boundary conditions so what we find by applying the first boundary condition which is the one of the insulated wall on the left hand side of our slab when we apply the first one we get c one is equal to zero and then applying the second boundary condition and so this is what the second boundary condition reduces to it's not as clean as the first one but what we're going to do we're going to take this and this and we're going to plug it back into our equation here so let's plug those back in and see what we obtain so this is what the temperature profile in the wall turns out to be with internal generations so recall we had a wall like this we had insulation over here over here we had a free convective environment and what this is telling us is it's telling us temperature as a function of x and x is going starting at this location here so it gives us the temperature distribution and so what we will do and it assumes that we know the temperature on that wall and we know q dot we know k and the width of the wall l so what we're going to do in the next segment we're going to plug in some values and we're going to see what this temperature distribution looks like