Will it Blow? Joule Heating in a Fuse on a Circuit Board, Chapter 2
Some examples of resistive heating can involve the device reach high temperatures and so they either structurally degenerate or even melt. The design challenge is to remove this heat as effectively as possible. In this second part of the three-chapter series, you will see how to set up the physics to describe the joule heating and convection cooling of this fuse using specialized multiphysics interfaces.
Transcript
00:00:01 [Music] now first just please note that we have dual heating per default taking place everywhere in the model it says all domains right now i want dual heating to take place almost everywhere in the model with one exception and that's the tube as i mentioned there is vacuum in the tube and since not much of interest is going to happen in the vacuum i'm
00:00:40 going to remove that domain from the interface just like that now with that done let's send in some current and i'll find all the conditions that i need by right clicking on dual heating i just did that um i'll select for this condition electric currents and terminal let's do that and this lets us send in a current i'll apply that to the patch right here and
00:01:06 i'll just type in 15 amps um for the current to have anywhere to go we'll also need a ground condition and i'll find that too under electric currents it's ground i'll apply that to the edge of the ground plane now these two conditions the terminal and the ground will together see to that we get a proper current distribution all over the
00:01:33 model from the dual heating interface we automatically get an electromagnetic heat source applying everywhere except for in the vacuum which is not part of the model right and this heat source contains the heating that results from the electric current so we actually already know what will heat up the fuse let's now define how the heat will be lost to the outside
00:01:58 world so um well the fuse and the top surfaces of the board will be in contact with air and this means that some heat will be transported away through convection this can be modeled with a convective cooling condition so right clicking on dual heating i can find that under heat transfer it's right here convective cooling the convective cooling condition requires a heat
00:02:24 transfer coefficient and we'll set that to an estimated 5 watts per meter squared in kelvin to make it easier to select the boundaries where convective cooling should apply i'll hit the yz button switching to a yz view like that and i'll use the select box tool it's this one so with this tool i can select exactly those boundaries where i want to apply this boundary
00:02:51 condition remember though we have vacuum inside the tube and uh convection doesn't take place in vacuum so let's remove those particular boundaries that are in contact with vacuum from this selection they're now removed the only heat loss mechanism that does take place inside the tube is radiation and that's another boundary condition we can find it under heat
00:03:17 transfer and there's one called surface to ambient radiation i'll select the exact same boundaries as i just removed from the convective cooling condition the um radiation actually gets increasingly important to take into account when we get to high temperatures uh because it's proportional to the temperature squared uh not squared it's the fourth power actually which you can
00:03:44 read from the equation right here you'll always have access access to equations in console and if you don't want to see them if you find them scary just click equation again and hide them uh what you do need to fill in is the surface emissivity this is number between 0 and one i'll uh go for a number8 so one would be a perfect black body and zero is no radiation at all
00:04:10 finally for the very last boundary condition let's just assume that we know the temperature on the bottom boundary of the circuit board and i'll set that to the default which is room temperature 293.15 kelvin now that actually concludes the physics for this model uh as the next step we'll need a mesh
00:04:37 you can spend uh any amount of time on just perfecting the mesh if you want to but the defaults work fine in most situations and i'll pick a fairly coarse mesh i'll actually go for a coarser one i'll click the build all button in the study selecting compute now will solve the model and yes here we go that's a temperature distribution look at that we
00:05:03 have a solution so it ranges from room temperature that's again 293.15 kelvin up to a maximum right here in the fuse wire which is approximately 937 kelvin this is exactly 4 kelvin above the melting point of aluminum so will it melt well yes barely but the fuse will melt and just a couple of other results i wanted to show you in the model in order
00:05:32 to get the temperature distribution console has had to also compute the electric potential distribution i can look at that in a surface plot let's just add a 3d plot group and to that i'll add a surface plot i can type in capital v for voltage and just hit the plot button and plot that as you can see from the range right here the maximum electric potential
00:06:01 right now is 17 volts and most of the voltage drop as you may have may have expected already is across the fuse wire and finally you can also plot the current distribution in the wire let's select it from this quite extensive list of things you can plot so current charge and we'll find current density norm that's the magnitude of the local current
00:06:26 density so you get the currents computed everywhere in the model of course but i guess the wire is probably the most interesting place to look at the orange parts in the wire are mainly the straight sections they have fairly constant current densities then you have the inner turns with a greater current density as you can see in red and the outer turns with a decreased current
00:06:50 density in light green so very much like a car in a racetrack if you will the current simply tries to take the shortest path