## 字幕列表 影片播放

• The following content is provided under a Creative

• Your support will help MIT OpenCourseWare

• To make a donation or to view additional materials

• from hundreds of MIT courses, visit MIT OpenCourseWare

• at ocw.mit.edu.

• MICHAEL SHORT: I think I might actually

• use all 16 colors today.

• Oh no, this is the most satisfying day.

• Whereas Tuesday was probably the most mathematically intense,

• because we developed this equation right here,

• today is going to be the most satisfying,

• because we are going to cancel out just about every term,

• leaving a homogeneous, infinite reactor criticality condition.

• So we will go over today, how do you

• go from this, to what is criticality in a reactor?

• So I want to get a couple of variables up over here

• to remind you guys.

• We had this variable flux of r, e,

• omega, t in the number of neutrons

• per centimeter squared per second traveling

• through something.

• And we also had its corresponding non-angular

• dependent term, on just r, e, t, if we don't care

• what angle things go through.

• We've got a corresponding variable called current.

• So I'll put this as flux.

• We have current j, r, e, omega, t, and its corresponding,

• we don't care about angle form.

• And today, what we're going to do

• is first go over this equation again so that we

• understand all of its parts.

• And there are more parts here than are in the reading.

• If you remember, that's because I

• wanted to show you how all of these terms are created.

• Just about every one of these terms,

• except the external source and the flow through some surface,

• has the form of some multiplier, times the integral

• over all possible variables that we care about,

• times a reaction rate d stuff, where this reaction

• rate is always going to be some cross-section, times some flux.

• So when you look at this equation using that template,

• it's actually not so bad.

• So let's go through each of these pieces right here.

• And then we're going to start simplifying things.

• And this board's going to look like some sort of rainbow

• explosion.

• But all that's going to be left is a much simpler form

• of the neutron diffusion equation.

• So we've got our time-dependent term right here,

• where I've stuck in this variable flux,

• instead of the number of neutrons n,

• because we know that flux is the number of neutrons,

• times the speed at which they're moving.

• And just to check our units, flux

• should be in neutrons per centimeters squared per second.

• And n is in neutrons per cubic centimeter.

• And velocity is in centimeters per second.

• So the units check out.

• That's why I made that substitution right there.

• And this way, everything is in terms of little fi, the flux.

• We have our first term here.

• I think I'll have a labeling color.

• That'll make things a little easier to understand.

• Which is due to regular old fission.

• In this case, we have new, the number of neutrons created

• per fission, times chi, the sort of fission birth spectrum,

• or at what energy the neutrons are born.

• Over 4pi would account for all different angles in which they

• could go out, times the integral over our whole control volume,

• and all other energies in angles.

• If you remember now, we're trying

• to track the number of neutrons in some small energy group, e,

• traveling in some small direction, omega.

• And those have little vector things on it

• at some specific position as a function of time.

• So in order to figure out how many neutrons are entering

• our group from fission, we need to know,

• what are all the fissions happening

• in all the other groups?

• I've also escalated this problem a little bit

• to not assume that the reactor's homogeneous.

• So I've added an r, or a spatial dependence

• for every cross-section here, which

• means that as you move through the reactor,

• you might encounter different materials.

• You almost certainly will, unless your reactor

• has been in a blender.

• So except for that case, you would actually

• have different cross sections in different parts of the reactor.

• So all of a sudden, this is starting

• to get awfully interesting, or messy depending on what

• you want to think about it.

• There is the external source, which is actually

• a real phenomenon, because reactors

• do stick in those californium kickstarter sources.

• So for some amount of time, there

• is an external source of neutrons,

• giving them out with some positional energy angle

• and time dependence.

• So let's call this the kickstarter source.

• There's this term right here, the nin reactions.

• So these are other reactions where it's absorb a neutron,

• and give off anywhere between 2 and 4 neutrons.

• Beyond that, it's just not energetically

• possible in a fission reactor.

• But don't undergo fission.

• They have their own cross sections, their own birth

• spectrum.

• And I've stuck in something right here,

• if we have summing over all possible i, where you have

• this reaction be n in reaction, where 1 neutron goes in,

• and i neutrons come out.

• You've got to multiply by the number of neutrons

• per reaction.

• For fission, that was new.

• For an nin reaction, that's just i.

• But otherwise, the term looks the same.

• and the flux.

• And these two together give you a reaction rate.

• I've just written all of the differentials as d stuff,

• because it takes a lot of time to write those

• over and over again.

• And then we have our photo fission term,

• where gamma rays of sufficiently high energy

• can also induce fission external to the neutrons.

• The term looks exactly the same.

• There's going to be some new for photofission,

• some birth spectrum for photofission,

• some cross section for photofission, and the same flux

• that we're using everywhere else.

• Then we had what's called the in scattering term, where neutrons

• can undergo scattering, lose some energy,

• and enter our group from somewhere else.

• That's why we have those en omega primes,

• because it's some other energy group.

• And we have to account for all of those energy groups.

• That's why we have this integral there.

• And it looks very much the same.

• There's a scattering cross-section.

• And that should actually be an e prime right there.

• Make sure I'm not missing any more

• of those inside the integral.

• That's all good.

• That's a prime, good.

• There's also a flux.

• that a given neutron starting off an energy

• e prime omega prime, ends up scattering into r

• energy, e and omega.

• So this would be the other one.

• And this would be r group.

• But otherwise, the term looks very much the same.

• And that takes care of all the possible gains of neutrons

• into r group.

• The losses are a fair bit simpler.

• There is reaction of absolutely any kind.

• Let's say this would be the total cross-section, which

• says that if a neutron undergoes any reaction at all,

• it's going to lose energy and go out of our energy group d, e.

• Notice here that these are all--

• these energies and omega is our all r group,

• because we only care about how many neutrons in r group

• undergo a reaction and leave.

• And the form is very simple--

• integrate over volume, energy, and direction,

• times a cross-section, times a flux,

• just like all the other ones.