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  • MICHAEL SHORT: So today is going to be the last day of neutron

  • physics.

  • As promised, we're going to talk about what

  • happens as a function of time when you perturb the reactor,

  • like you all did about a month ago.

  • Did any of you guys notice the old-fashioned analog panel

  • meter that said, reactor period, when you were doing your power

  • manipulations?

  • We're going to do that today.

  • And you're going to explore that on the homework.

  • So I'm arranging for all of your actual power manipulation

  • traces to be sent to you.

  • So each one, you'll have your own reactor data.

  • You'll be able to describe the reactor period

  • and see how well it fits our infinite medium single group

  • equations, which it turns out is not very well.

  • But that's OK, because you'll get to explain the differences.

  • First, before we get into transients

  • I wanted to talk a bit about criticality and perturbing it.

  • So let's say we had our old single group kit criticality

  • relation.

  • And I'd like to analyze, just intuitively or mentally

  • with you guys, a few different situations.

  • Let's say we're talking about a light water

  • reactor or a thermal reactor, like the MIT reactor, or pretty

  • much all the reactors we have in this country.

  • What sort of things could you do to perturb it?

  • And how would that affect criticality?

  • For example, let's say you shoved in a control rod.

  • Let's take the simplest scenario.

  • Control rods in.

  • What would happen to each of the terms

  • in the criticality condition?

  • And then, what would happen to k effective?

  • So let's just go one by one.

  • Does nu ever change, ever?

  • Actually, yeah, it does.

  • Over time, you'll start--

  • that nu right there, remember, that's

  • a nu bar, number of neutrons produced per fission.

  • As you start to consume U238 add neutrons.

  • And as you guys saw through a complicated chain of events

  • on the exam, eventually make plutonium 239,

  • which is a fissile fuel.

  • The nu for 238 is actually different than the nu for 239.

  • So I don't want to say that nu never changes.

  • It's just that shoving the control rods into the reactor

  • is not going to change nu.

  • But it does change slowly over time as you build up plutonium.

  • What about sigma fission?

  • If this were a blended homogeneous reactor

  • or a reactor in a blender, what would happen to sigma fission

  • as you then shove in an absorbing material?

  • Does it change?

  • AUDIENCE: No.

  • MICHAEL SHORT: You say, no.

  • And I'm going to add here homogeneous.

  • So in this case, remember if we define the average sigma

  • fission as a sum--

  • I'll add bits to it--

  • of each material's volume fraction, or let's say

  • atomic fraction, times each material's sigma fission,

  • if we throw nu materials into the reactor,

  • then this homogeneous sigma fission

  • does change when we put materials in

  • or take materials out.

  • So you guys want to revise your idea?

  • AUDIENCE: Yes.

  • MICHAEL SHORT: Yes, thank you.

  • There's only one other choice.

  • Now the question is, by how much?

  • If you put in a control rod where let's say the control

  • rod's sigma fission would be equal to zero,

  • but volume would be equal to small.

  • Can't be any more specific than that.

  • How much of an effect do you think

  • you'll have on sigma fission?

  • AUDIENCE: Small.

  • MICHAEL SHORT: Very small.

  • So let's say a little down arrow like that.

  • What about sigma absorption?

  • The volume is still small, but a control rod by definition sigma

  • absorption equals huge.

  • So what do you think?

  • AUDIENCE: It's going to increase.

  • MICHAEL SHORT: It's going to increase a little or a lot?

  • AUDIENCE: A lot.

  • MICHAEL SHORT: Quite a bit.

  • Now let's look at the diffusion constant.

  • And remember that the diffusion constant is 1 over 3 sigma

  • total, minus the average cosine scattering angle

  • sigma scattering.

  • What do you think is going to happen to the neutron diffusion

  • coefficient as you throw in an absorbing material?

  • Something that's got an enormous absorption cross-section

  • is also going to have an enormous total cross-section,

  • because sigma total is sigma absorption plus sigma

  • scattering.

  • And sigma scattering doesn't change that much.

  • But if sigma absorption goes up, sigma total goes up.

  • If sigma total goes up, then what happens

  • o the diffusion coefficient?

  • AUDIENCE: Decrease.

  • MICHAEL SHORT: Yep, it's got a decrease.

  • And how does inserting a control rod change the geometry?

  • AUDIENCE: It doesn't.

  • MICHAEL SHORT: Very, very close.

  • Yeah, you're right.

  • The control rod better not change the geometry,

  • but what I do want to remind you of is

  • that this buckling term includes-- let's say,

  • this was a one dimensional infinite slab

  • Cartesian reactor.

  • That little hot over there means we have some extrapolation

  • distance.

  • Remember, if we were to draw our infinite reactor

  • with the thickness A and we wanted

  • to draw a flux profile on top of that,

  • it would have to be symmetric about the middle.

  • And let's say we had our axis of this is x and this is flux.

  • Flux can't go to zero right at the edge of the reactor,

  • because that would mean that no neutrons were literally

  • leaking out.

  • So there's going to be some small extrapolation

  • distance equal to about two times

  • the diffusion coefficient.

  • So the geometric buckling is actually

  • pi over the reactor geometry, plus 2 times

  • the diffusion coefficient.

  • And if the diffusion coefficient goes down,

  • but it's also very, very small compared

  • to the geometric buckling, how much does the buckling change

  • and by how much?

  • And in what direction?

  • AUDIENCE: It increases very slightly.

  • MICHAEL SHORT: Increases very slightly.

  • So the buckling might increase very slightly.

  • What's the overall net effect on k effective?

  • AUDIENCE: It goes down.

  • MICHAEL SHORT: Should go down, you would hope.

  • If you put a control rod in, it should

  • make k effective go down, because there's

  • a little decrease here.

  • Things kind of cancel out there.

  • But the big one is putting an absorption material,

  • like a control rod in, should make k effective go down.

  • And that's the most intuitive one,

  • but you can work out one term at a time what's

  • generally going to happen.

  • So let's now look at some other scenarios

  • for the same criticality condition.

  • I'll just rewrite it so that we can mess it all up again.

  • Now we want to go for the case of boil or void your coolant.

  • And now we're getting into the concept of different feedback

  • mechanisms.

  • We've already talked once about how

  • raising the temperature of something

  • tends to increase cross-sections in certain ways.

  • But now let's say, what would happen

  • if you boil your coolant?

  • If things got really hot and the water started to boil.

  • What do you want to happen to k effective?

  • You want it to increase?

  • AUDIENCE: Decrease.

  • MICHAEL SHORT: Decrease, thank you.

  • You want it to decrease, or else you'd get a Chernobyl.

  • And we'll talk about how that happened in a week or two.

  • So now let's reason through each one of these.

  • Let's assume that nu doesn't change

  • when you boil the coolant.

  • What about sigma fission of the whole reactor?

  • You're taking a little bit of material out of the reactor

  • by taking liquid water, which is fairly dense,

  • and making it gaseous water, which is less dense.

  • So overall, there are more fissile atoms in the reactor

  • proportionately when the coolant is boiled away

  • than when it's not.

  • So what happens to sigma fission?

  • The average sigma fission for the reactor

  • will go up ever so slightly.

  • Probably not enough to matter.

  • What about sigma absorption?

  • If the coolant disappears.

  • AUDIENCE: It goes down.

  • MICHAEL SHORT: Yeah, water is an absorber.

  • Hydrogen and oxygen-- really just hydrogen--

  • have some pretty non-negligible absorption coefficients.

  • And if those go away, then you're

  • losing a bit of absorber, aren't you?

  • Actually it's interesting.

  • Oxygen is the lowest thermal cross-section of any element.

  • So we can treat it as pretty much transparent.

  • Now how about the diffusion coefficient?

  • We've got the formula for it up there.

  • If all of a sudden your neutrons don't

  • have much to moderate from--

  • there's not much to moderate your neutrons.

  • Yeah.

  • AUDIENCE: Your scattering just disappears.

  • MICHAEL SHORT: Your scattering just disappears, right?

  • But so does some of your total cross-section.

  • So chances are, those neutrons are

  • going to go farther before they undergo any given

  • collision because there's no water in the way.

  • So you'd expect neutron diffusion to go up.

  • And what about geometric buckling?

  • Diffusion goes up, then the geometric buckling--

  • I'm just going to make it really small.

  • But the net effect here, once again, k effective goes down.

  • We didn't talk about anything to do with the actual temperature

  • effects on the cross-sections.

  • This is just a density thing on the coolant itself.

  • So let's now look at that.

  • What about if you have some power spike,

  • raised fuel temperature?

  • I'll write it again, so we can mess it up again.

  • So let's say you raise the fuel temperature.

  • And that's going to cause every cross-section effectively

  • to increase if you're doing this average scenario.

  • Let's talk a little bit about why.

  • It's not as simple as just saying,

  • the cross-sections go up.

  • So let's say we had two different temperatures,

  • cold and hot.

  • So this would be your sigma fission cold.

  • And this would be your sigma fission hot.

  • For cold, sigma fission looks something like that.

  • And as the temperature goes up, these

  • resonances, which I'll just label right here--

  • resonances being specific energy is

  • where the absorption suddenly goes up, suddenly goes

  • down will actually decrease in height.

  • But they'll start to spread out more.

  • That's about as well as I can draw it very crudely.

  • And same thing goes not just for sigma fission,

  • but for sigma anything, including absorption, including

  • total whatever you want.

  • And so if your goal is to get your neutrons

  • from the fast region where they're

  • born into the thermal region where you get fission,

  • broadening these cross-sections makes it more likely

  • that if the neutron loses any amount of energy,

  • it's going to hit one of these big resonance regions

  • and get absorbed or taken away before it gets a chance

  • to go to the fission region.

  • So what this is really going to do--

  • it's kind of funny to say it in terms of a one group

  • criticality relation, but your fission cross-section

  • is actually going to go down.

  • One reason is that the fuel physically spreads out.

  • And so just from the density modification,

  • you're not going to get as much.

  • But then you've also got that effect

  • of increasing fission from these resonance regions spreading

  • out.

  • The question is, which one is a bigger effect?

  • Can't answer that with a simple statement.

  • You'll go over a lot more of that in 22.05

  • when you talk about what actually defines a resonance

  • region, how do you calculate them,

  • and how do they Doppler broaden or broaden with temperature.

  • How about sigma absorption?

  • AUDIENCE: It goes down.

  • MICHAEL SHORT: Yeah, sigma absorption, it's

  • going to go down because things spread out.

  • But it might also go up because the cross-sections spread out,

  • or the resonances spread out.

  • What's really going to happen though is the reactor

  • atoms are effectively spreading themselves apart.

  • The coolant's less dense.

  • The structural materials in the fuel and everything

  • are still there.

  • They're less dense, but there's not

  • fewer of them in the reactor.

  • But there is going to be less coolant in the reactor,

  • because it has the ability to sparsify or get

  • less dense, and kind of squeeze out the inlet

  • and outlet of the reactor.

  • So what's really going to happen here

  • is, we know diffusion is going to go up,

  • which might cause a corresponding change

  • in buckling.

  • And the net effect, as we would hope,

  • k effective would go down.

  • And so what we've talked about now here

  • is directly controlling reactivity

  • with control rods, what's called a void coefficient,

  • where you actually want to have a negative void coefficient.

  • So if you boil your coolant too much,

  • k effective should go down.

  • And that's one of the mechanisms that a light water

  • or a thermal reactor can help stabilize itself.

  • And you can see that now from just a really simplified one

  • group criticality relation.

  • And if you raise the fuel temperature,

  • let's say the fuel gets really hot because there's

  • been some power spike, you also want

  • the reactor to shut itself down, which you can see that it does.

  • Let's make things a little trickier.

  • Let's now talk about a sodium reactor.

  • Fast reactor.

  • This one relies a lot more on fast fission of U238.

  • So if we were to draw the two cross-sections of 235 sigma

  • fission and 238 sigma fission--

  • remember, uranium 235 looked like the one

  • that we drew before, whereas U238 goes something like that,

  • with no actual scale given.

  • I'm not going to even go there.

  • But uranium 238 does not need moderation for the neutrons

  • to induce more fission.

  • So let's now write the same criticality reaction, which,

  • again, is a super simplified view of things, but that's OK.

  • What would happen to each of these terms in a sodium

  • fast reactor if you void the coolant?

  • So nu won't change.

  • What about sigma fission?

  • Well, if the coolant goes away, then on average

  • there is fissile materials contributing

  • more to that cross-section, but not that much.

  • So if you want to get technical, might

  • be the slightest of increases, but doesn't matter that much.

  • What really matters, though, is the stuff on the bottom.

  • Sodium does have a low, but non-negligible absorption

  • cross-section.

  • So if the sodium were to boil away,

  • then the absorption would go down

  • by a non-negligible amount.

  • And then what about diffusion?

  • Well, we've got the formula for it up there.

  • If there's not as much coolant in the way,

  • then the neutrons are going to be

  • able to get further on average.

  • Let's say, they're not going to be scattering around

  • with as much of the sodium.

  • So there might be a small increase in diffusion

  • and corresponding small increase in buckling.

  • But this is where the one group kind of fails.

  • What the sodium is actually doing

  • is providing a little bit of moderation,

  • so that some of those neutrons when they bounce off of sodium

  • leave the fast fission region and get absorbed.

  • And that's part of the balance of the reactor.

  • If all of the neutrons are then born fast

  • and don't really slow down and just get absorbed,

  • then you might have an overall positive void coefficient.

  • So this would tell you that in a fast reactor where you're

  • depending on your coolant not just to cool

  • the reactor, but to absorb somewhat

  • and to moderate somewhat, you don't

  • want to boil the coolant in a fast reactor.

  • And is a lot of the reason why most fast reactor

  • coolants tend to have extremely high boiling points.

  • Sodium is approximately 893 Celsius.

  • Lead bismuth is approximately 1,670 Celsius.

  • Molten salt, about 1,400 Celsius.

  • So all those coolant, except for the sodium one,

  • you'll melt the steel that the reactor is made out

  • of before your boil the coolant.

  • So boiling the coolant is a bad day in a fast reactor,

  • because then things will go from bad to worse,

  • because in this case, the feedback coefficient can

  • be positive for voiding the coolant.

  • That's no good.

  • So you want to keep the reactor submerged.

  • And that's another reason why a lot of these fast reactors

  • are what's called, pool-type reactors.

  • The reactor is not a vessel with a bunch

  • of piping under it that can break and fail,

  • but instead it's designed as a huge pool of liquid sodium.

  • And then the core is somewhere in here with a bunch of pumps

  • sending the coolant in and back out, or through some heat

  • exchange or something.

  • So there's not really any penetrations

  • on the bottom up this pool.

  • And you make sure that you maintain,

  • either when you have sodium or lead bismuth

  • eutectic, or liquid lead, or some other fast reactor

  • coolant.

  • So these are some kind of interesting scenarios

  • to think about.

  • I think one of them that I put in the homework was

  • imagine you have the MIT reactor and replace

  • the coolant with molten sodium.

  • What's going to happen?

  • Well, let's say you got all the water out first

  • and it wouldn't just blow up.

  • What would actually happen to the criticality relation?

  • That's something I want you to think about,

  • because one of the big problems on the homework

  • is doing exactly this for scenarios

  • that have happened to the MIT reactor,

  • except for the sodium one.

  • That's never happened and hopefully never will.

  • I can't even imagine.

  • But now let's talk a little bit about when

  • you perturb a reactor by doing something to it,

  • putting the control rods in, or pulling them out,

  • or doing whatever you want.

  • You're by definition going to take

  • one of our first assumptions about how the neutron diffusion

  • equation works and throw it out the window.

  • So we're now moving into the transient regime.

  • So to study what happens in a reactor transient

  • or when something changes as a function of time,

  • let's first go from k effective to what we call k infinity.

  • The multiplication factor for an infinite medium.

  • We're only doing this because it's

  • analytically easier to understand

  • and still gets the point across.

  • So we'll say that our k infinity is still

  • a balance between production and destruction.

  • The difference is if we have an infinite medium,

  • there's no leakage.

  • You can't leak out of an infinitely sized reactor,

  • should one ever exist.

  • And so it just comes out as nu sigma fission

  • over sigma absorption.

  • A much simpler form.

  • And so now we can write what would

  • happen to the flux in the reactor as a function of time.

  • In this case, it's going to be one over velocity.

  • I'm going to make this a very obvious wide v. That change

  • in the reactor flux is going to just be proportional

  • to the imbalance now in the number of neutrons produced

  • and destroyed.

  • So the number of neutrons produced

  • will be proportional to very sharp nu sigma

  • fission minus the number of neutrons destroyed,

  • sigma absorption, times phi is a function of t.

  • Y'all with me so far?

  • So this right here is a change, which

  • is proportional to an imbalance between production

  • and destruction, times the actual flux that you

  • have in some given time.

  • So to make this simpler, let's multiply everything

  • by v. Where's my green substitute color?

  • Multiply everything by v. And the only unfortunate

  • situation is we have a v and a nu next to each other.

  • I'm going to try to keep them looking really different.

  • Those go away.

  • And then we end up with, if we divide by phi,

  • then those phi's go away.

  • And we have phi prime over phi, equals v nu sigma fission,

  • minus v sigma absorption.

  • And now we can start to define things

  • in terms of our k infinity factor

  • and a new quantity I'd like to introduce

  • called the prompt lifetime.

  • It's a measure of how long a given neutron tends to live

  • before something happens to it.

  • Before it's either absorbed or leaks out, well, not

  • from our infinite reactor.

  • And so we can define this as 1 over their neutron velocity,

  • times sigma absorption.

  • And just to check the units here--

  • velocities in meters per second.

  • Macroscopic cross-sections are in 1 over meters.

  • So those cancel out, and we're left

  • with a total units of seconds.

  • That's nice.

  • We would want a mean neutron lifetime,

  • or a prompt lifetime to have of seconds or time, at least.

  • Yep.

  • AUDIENCE: Can you say again why the [INAUDIBLE] squared

  • went away?

  • MICHAEL SHORT: Why the d phi dt squared went away?

  • AUDIENCE: No, the [INAUDIBLE] square.

  • MICHAEL SHORT: Oh, OK.

  • So that's because we assume we're

  • going to be analyzing an infinite medium.

  • So right here, this to relabel these terms,

  • this would be the total production term.

  • That right there represents absorption.

  • And that right there represents leakage.

  • But if we're analyzing an infinite medium,

  • you can't leak out, because it takes up the entire universe

  • and beyond, depending on what you believe metaphysically.

  • That's different costs.

  • So this right here, we can rewrite as 1 over lifetime.

  • That makes it easier.

  • And this right here, if we note that nu--

  • that's a nu.

  • I'm going to be really explicit about that.

  • Nu sigma fission over sigma absorption.

  • This kind of looks like this is looking to be like k--

  • wrong color.

  • --like our k infinity over lp.

  • So all of a sudden we of a much simpler relation.

  • We have 5 prime over 5 equals k infinity minus 1

  • over the prompt neutron lifetime.

  • So if we solve this, this is just an exponential.

  • So we end up with our phi as a function of t is--

  • whatever flux we started at, like for your power

  • in manipulations, it would be whatever the neutron flux was

  • before you touch the control rod, times e to the t,

  • or e to the that stuff, k infinity minus 1 over lp,

  • times t, which we can rewrite as t over capital T.

  • We're going to define this symbol as what's

  • called the reactor period.

  • What the reactor period actually says

  • is how long before the flux increases by a factor of e.

  • And so this is actually what that meter

  • was measuring on the reactor.

  • It's the reactor period or the time

  • it would then take for the reactor's power

  • to increase by a factor of e because it's an exponential.

  • To tell you what these typical reactor periods tend

  • to be for a thermal reactor, t is

  • about 0.1 seconds corresponding to an average prompt neutron

  • lifetime of 10 to the minus 4 seconds.

  • Seems fast, doesn't it?

  • Like, really fast.

  • So the question I asked you guys is, why don't reactors

  • just blow up?

  • AUDIENCE: [INAUDIBLE].

  • MICHAEL SHORT: Yes, there is something

  • we've neglected from here.

  • It's like what Sarah said.

  • And it deserves its own board.

  • There is a fraction of delayed neutrons.

  • We'll give that fraction the symbol, beta.

  • And for a uranium 235, it equals about 0.0064.

  • So there's less than a percent of all the neutrons coming out

  • of a reactor have some delay to them, because they're not

  • made directly from fission in the 10 to the minus 14 seconds

  • that we talked about in the timeline.

  • But they come out of radioactive decay processes

  • with delayed lifetimes ranging from about 0.2 seconds

  • to about 54 seconds.

  • This is the whole reason why reactors don't just blow up.

  • So you can actually make a reactor go super critical.

  • But if the k effective is less than 1 plus beta,

  • then the reactor is not what we call prompt super critical.

  • And so the reason for that is, let's say you raise the reactor

  • power by some amount and the k effective goes up to 1.005,

  • there's still this fraction 0.0064 of the neutrons

  • are not going to be released immediately.

  • They're going to be released not in 10 to the minus 14 seconds,

  • but in 10 to the 2 seconds.

  • So a measly 15 orders of magnitude

  • slower, meaning that there's actually

  • some ability for this reactor to raise its power level.

  • And these delayed neutrons, even though that's

  • such a small fraction, takes the reactor period

  • from its t infinity value of about 0.1 seconds

  • to about 100 seconds.

  • So the same reactor when you account

  • for the delayed neutrons increases in power

  • by a factor of e.

  • And it takes it about 100 seconds, which means

  • this is totally controllable.

  • Now I have a question for you guys.

  • Would you guys like me to derive this formula,

  • or do you want to go into more of the intuitive

  • implications of it?

  • Because we can go either way.

  • There is a formula that will tell you

  • what the reactor period and time dependence will be.

  • And you will hit it in 22.05 probably.

  • I can't guarantee it because I'm not teaching it.

  • Or we can talk a little bit more about some of the intuition

  • behind delayed neutrons.

  • So a bit of choose your own adventure.

  • Math or intuition?

  • AUDIENCE: Intuition.

  • MICHAEL SHORT: Intuition.

  • OK, that's fine.

  • Good.

  • So that was the derivation.

  • I'll post that anyway, if you guys want to see.

  • I think in the Yip reading it says, let's account

  • for the delayed neutrons.

  • Intuitively we find that the answer ends up being--

  • so I'll skip the derivation.

  • And it comes out to phi naught e to the beta minus 1,

  • times k minus 1 over lt, plus beta phi naught over beta minus

  • 1k, minus 1, times 1 minus e to the beta minus 1k, minus 1

  • over l.

  • OK, so left as an exercise to the reader--

  • AUDIENCE: That's intuitive.

  • MICHAEL SHORT: Yeah, that's intuitive.

  • But let's actually talk about how intuitive it is.

  • I do want to give you the starting and the ending

  • equation.

  • And we will not go through the rest.

  • Yeah, Charlie?

  • AUDIENCE: Should we copy that down?

  • MICHAEL SHORT: No, you shouldn't.

  • I'm going to scan it for you guys.

  • So don't bother copying it down.

  • Let's talk about where it comes from.

  • And the answer may astound you because we're

  • going to bring right back the idea of series radioactive

  • decay.

  • So let's say you want to relate the change

  • in the number in the neutron flux to a 1 minus--

  • I'm going to take a quick look at the original equation

  • because I don't want to screw that up.

  • That's the first page, and that's the one we want.

  • Let's say we had some equations that looked something

  • like this.

  • Phi plus phi naught times beta.

  • This is the original differential equation

  • from whence it came.

  • And the intuitive part that I want you to note

  • is that the jump from changing k effective

  • is moderated by this term right here, 1 minus beta.

  • So that's the fraction of prompt neutrons,

  • that as soon as you pull the control rod out,

  • that's your instantaneous feedback.

  • By instantaneous, I mean on the order of, like,

  • 10 to the minus 4 seconds, or something

  • that you can't really control.

  • This right here represents the delayed fraction.

  • This is as mathy as it's going to get because you've

  • chosen intuition.

  • I think you have chosen wisely.

  • It's going to be a more fun.

  • So what this represents right here

  • is your kind of instant change, because whatever you change

  • k effective to, it's going to be moderated

  • by the prompt fraction, how long the neutrons tend to take

  • to undergo that feedback.

  • Yes, Sara?

  • AUDIENCE: Was that the average?

  • MICHAEL SHORT: The average what?

  • AUDIENCE: Average neutron lifetime.

  • MICHAEL SHORT: Yes, this is the average neutron lifetime.

  • So let's define the average neutron lifetime

  • as simply 1 minus beta times the prompt neutron lifetime,

  • plus the beta times some delayed neutron lifetime.

  • So what no book I've ever seen actually says,

  • this is what's referred to as a Maxwell mixing model.

  • It's just the simplest thing to say, oh,

  • if you want to get the average of some variable,

  • take the fraction of one species times its variable,

  • plus the fraction of the other species, times its variable.

  • Folks do the same thing with electrical resistivity,

  • thermal conductivity, or any sort of other material

  • property.

  • And it is or isn't good in some situations.

  • Like, if you had a piece of material made out

  • of two different things--

  • let's say this had thermal conductivity k1,

  • and it had thermal conductivity K2.

  • Would a Maxwell mixing model be appropriate to describe

  • the flow of heat across this thing?

  • Probably not.

  • But in the case of neutrons where

  • they're flying about like crazy and their mean free path is

  • much larger than the distance between atoms,

  • this works great.

  • So we can define this mean neutron lifetime

  • and use that in this equation right here.

  • So this term right here describes

  • the instantaneous change.

  • You pull the control rods out, and fraction 1

  • minus beta neutrons respond immediately.

  • What about that fraction of neutrons?

  • Those are being produced with a fraction beta

  • depending on what the flux was before,

  • because they're still waiting to decay from the old power level.

  • Does anyone notice anything suspiciously familiar

  • about the final form of this equation for flux?

  • You've seen it before with a couple

  • of constants changed around.

  • What about the form of this differential equation?

  • [INTERPOSING VOICES]

  • It is exactly the same as series radioactive decay.

  • So the horrible derivation I was going to do for you guys

  • and we're not anymore is, use an integrating factor.

  • You solve it in exactly the same way.

  • You bring everything to one side of the equation.

  • Find some factor mu, that makes this a product rule.

  • Do a lot of algebra.

  • And you end up with a very suspiciously similar looking

  • equation.

  • So it's exactly the same posing and solution

  • as series radioactive decay, with the difference being,

  • that's the constant in front of everything,

  • instead of a bunch of lambdas and fluxes.

  • So what this says here is that the flux as a function of time,

  • this is the prompt feedback right here,

  • which says that if-- let's graph it,

  • since we're going intuitive.

  • There's no room.

  • Even those all boards are full.

  • OK, here we go.

  • If we graft time and flux right here,

  • what that part right there says is

  • that you're going to get some sort

  • of instantaneous exponential feedback.

  • But it's going to be moderated by this one minus exponential

  • on top.

  • So you're going to end up with a little bit of prompt feedback,

  • this stuff right here.

  • And then-- have to draw longer because it's going to take

  • forever--

  • you'll have some delayed feedback,

  • because you've got to wait 100 or so seconds,

  • or whatever that new reactor period is,

  • for the delayed neutrons to take effect.

  • And that's the whole reason you could pull the control rods out

  • at almost any speed you wanted and the reactor doesn't just

  • explode.

  • If you pull the control rods out fast enough, such

  • that the change in k effective is greater than beta,

  • then the reactor goes prompt super critical,

  • which means you don't have any delayed neutrons slowing down

  • the feedback.

  • And you've kind of turned your reactor into a weapon.

  • A very poor, terrible weapon, but

  • a prompt super critical nuclear device, nonetheless.

  • Did anybody pull out the control rods too fast

  • and the controls took over for you?

  • What about you guys in training?

  • Did you ever do things when you watched the automatic control

  • take over?

  • No?

  • AUDIENCE: It'll just take over.

  • AUDIENCE: Yeah, it'll kick you off if you don't pay attention.

  • MICHAEL SHORT: That's what I mean.

  • The machine takes over and it will kick you off

  • and stop responding to you.

  • AUDIENCE: [INAUDIBLE] horrible noise

  • and so we don't want that.

  • It's more to avoid an annoying alarm.

  • MICHAEL SHORT: I see.

  • But the annoying alarm is to stop you

  • from doing something like that, like, making the reactor go

  • prompt super critical.

  • AUDIENCE: [INAUDIBLE]

  • MICHAEL SHORT: OK, so that's what I would

  • call the machine taking over.

  • AUDIENCE: Oh, I see.

  • AUDIENCE: It'll kick you on to manual, and then [INAUDIBLE]

  • still don't do anything [INAUDIBLE]..

  • MICHAEL SHORT: Yeah.

  • So if your blood alcohol level is above beta

  • and you try and, let's say, increase the reactor

  • reactivity too much, it will then take over, insert

  • a control rod, make a horrible noise, and say, go home,

  • you're drunk.

  • Something like that.

  • OK, that makes sense to me.

  • So what did your guys' reactor power traces look like?

  • Did they look something like this,

  • where there was an initial rise as you pulled the control

  • right out?

  • And then after you pulled the control rod out the power kept

  • rising just a smidge, right?

  • And what happened when you put the control rod back in?

  • Let's say you put the control rod back in.

  • You're going to get another prompt drop, not equal

  • to the same prompt gain that you got,

  • because now the reactor's at a different flux,

  • and then some asymptotic feedback like that.

  • And so this is why to those who don't understand

  • neutron physics, reactor feedback is very non-intuitive.

  • It's not a linear system.

  • You can't just pull the control right out and change the power

  • accordingly.

  • This is why there's automated controls in systems to stop you

  • in case, like I said, if your blood alcohol

  • content's above beta, which is very low, by the way.

  • Though you shouldn't be drinking on the job,

  • especially at a nuclear reactor.

  • Plus, you're all under 21, so what am I even saying?

  • AUDIENCE: What is alcohol?

  • MICHAEL SHORT: That's right.

  • Good answer.

  • What is alcohol?

  • AUDIENCE: Is that going to be covered in [INAUDIBLE]

  • MICHAEL SHORT: That'll be on the exam, yeah.

  • AUDIENCE: What is alcohol?

  • MICHAEL SHORT: Yeah, cool.

  • So that's all I want to go into for the intuitive stuff.

  • And it's about 5 of 5 of.

  • So I'd like to stop here and see if you guys have any questions

  • on neutron physics at a whole.

  • Noting that we're going to take Thursday's class

  • and turn into a recitation.

  • So I would like all of you guys to look at the problem set,

  • because it is posted.

  • It is hard.

  • Trust me.

  • This one's a doozy.

  • So I want to warn you guys because you've

  • got seven days to work on it.

  • But I want you to look at it so that we can start formulating

  • strategies for the problems together on Thursday,

  • because there are some tricks to it.

  • You guys know me by now, right?

  • There's always some sort of a trick.

  • Like, do you have to integrate every energy

  • to get the stopping power?

  • No, you actually don't have to do any integrals at all.

  • But you can if you want, and your answer

  • will be more accurate and correct.

  • It'll just take longer to get to.

  • So there's a lot of diminishing returns on these problems sets.

  • If you're willing to take an hour

  • and think about how can I do this

  • simpler and with fewer decimal points,

  • you're probably onto something.

  • And we'll work on those strategies together.

  • AUDIENCE: Is this due next Monday as well?

  • MICHAEL SHORT: Yes.

  • So I posted it yesterday at around noon or whatever

  • the Stellar site says.

  • I'll also teach you guys explicitly how to use Janus.

  • So we got a comment in from the anonymous feedback saying,

  • we have to use a lot of software.

  • Can we have some sort of tutorial for dummies?

  • Well, you guys aren't dummies, but you still

  • deserve a tutorial.

  • So I will show you how to export the data you'll

  • need from this problem set for Janus.

  • So you can focus on the intuition and the physics

  • and not get frustrated with getting data out of a computer.

  • So any questions on anything from the neutron diffusion

  • equation?

  • Yeah, Luke.

  • AUDIENCE: I'm not real clear on [INAUDIBLE] neutrons are

  • and how those are different from the prompt neutrons?

  • MICHAEL SHORT: The prompt neutrons

  • come right out of fission.

  • If we looked at that timeline of,

  • let's say the fission event happens here.

  • Two fission products are released in about 10

  • to the minus 14 seconds.

  • They move a little further apart.

  • And then some of them just boil off neutrons,

  • because they're so neutron heavy, after around 10

  • to the minus 13 seconds or so.

  • These right here are prop neutrons,

  • coming directly from the immediate decay of neutron

  • rich fission products.

  • Some of the delayed neutrons come from radioactive decay,

  • but of the much later fission products

  • with much less likely occurrences,

  • which is why the fraction is very low.

  • But also, because it's much longer half life,

  • those delayed neutrons take seconds, instead

  • of pico seconds, to show up.

  • And that's the whole basis behind easier control

  • and feedback a reactor.

  • Good question.

  • So anything starting from neutron transport

  • to simplifying to neutron diffusion,

  • to getting to this criticality condition,

  • making the two group criticality condition if you

  • want to have fast and thermal, or any of the time dependent

  • stuff that we intuited today.

  • Yeah?

  • AUDIENCE: So for that cross-section [INAUDIBLE]

  • you have there, so you have one for 235 and one for 238.

  • 235, it has to be thermal neutrons [INAUDIBLE] fast?

  • MICHAEL SHORT: Yep.

  • AUDIENCE: And you said that [INAUDIBLE] different

  • [INAUDIBLE] as well it had--

  • if you have [INAUDIBLE] into the--

  • MICHAEL SHORT: Was it on the other board

  • or from a different day?

  • AUDIENCE: It was a different board.

  • MICHAEL SHORT: OK.

  • AUDIENCE: Yeah, so could you explain that graph?

  • MICHAEL SHORT: Yes.

  • So in this case--

  • let me get a finer chalk.

  • This blue one would be for low temperature,

  • and this red one would be for high temperature.

  • So this blue graph, there are resonances,

  • which have very high values, but they're very narrow.

  • And because the width of a resonance doesn't matter,

  • it doesn't affect the probability

  • that a neutron scatters up here and moves some distance

  • down the energy spectrum.

  • Thinner resonances tend to get passed over,

  • especially if your reactor's full of hydrogen. Some

  • of those neutrons will be born and immediately

  • jump into the thermal region, where

  • it's easy to tell how much fission they'll undergo.

  • As you go up in temperature, you undergo

  • what's called Doppler broadening, which

  • causes these resonances to spread out and also go down

  • in value.

  • So the actual value of the cross-section

  • at these residences is lower, but the widths are larger.

  • So there's a higher probability that a neutron scattering

  • around and losing energy will hit

  • one of these higher cross-section regions, called

  • a resonance, at a higher temperature.

  • That's the difference there, is these two plots show

  • the same cross-section at low and high temperature.

  • These plots show the difference between uranium 235

  • and uranium 238.

  • Good question.

  • Anyone else?

  • Cool.

  • OK, for the first time in history,

  • I'll let you out a minute early.

  • Bring all your questions on Thursday.

  • So we'll start off with a Janus tutorial.

  • And then we'll start attacking this problem set together.

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24.過渡現象、フィードバック、時間依存性中性子 (24. Transients, Feedback, and Time-Dependent Neutronics)

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    林宜悉 に公開 2021 年 01 月 14 日
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